From e084960b431da54e703b5a7b5eee5eab761364db Mon Sep 17 00:00:00 2001 From: Alex Jordan Date: Sun, 19 May 2024 01:34:26 -0700 Subject: [PATCH] Latest build deployed. --- .mapping.json | 2 +- appendix-basic-math-review.html | 16 +- appendix-ccogs.html | 16 +- appendix-unit-conversions.html | 16 +- backmatter-4.html | 120 +- backmatter-5.html | 383 +- backmatter-6.html | 22 +- backmatter.html | 16 +- ccog-mth60.html | 16 +- chapter-graphing-lines.html | 17 +- ...ter-linear-equations-and-inequalities.html | 16 +- chapter-systems-of-two-linear-equations.html | 20 +- ...xpressions-equations-and-inequalities.html | 16 +- frontmatter-2.html | 16 +- frontmatter.html | 18 +- generated/.html_assets.pkl | Bin 8720 -> 10386 bytes ...oint-slope-yequals-130groupx-2021+2975.svg | 241 + .../review-graphing-lines-10-1-1-2-1.svg | 127 + .../review-graphing-lines-10-8-13-2-1.svg | 126 + .../review-graphing-lines-10-8-14-2-1.svg | 132 + .../review-graphing-lines-10-8-15-2-1.svg | 121 + .../review-graphing-lines-10-8-16-2-1.svg | 126 + .../review-graphing-lines-2-3-2-1-2.svg | 189 + .../review-graphing-lines-3-3-2-3-2-1-2.svg | 197 + .../review-graphing-lines-3-3-2-3-2-2-2.svg | 217 + .../review-graphing-lines-5-4-1-2-2.svg | 132 + .../review-graphing-lines-5-4-2-1-1-2.svg | 192 + .../review-graphing-lines-6-3-2-1-2-1-2.svg | 139 + .../review-graphing-lines-6-3-2-1-2-2-2.svg | 191 + .../review-graphing-lines-6-3-2-1-2-3-2.svg | 185 + .../review-graphing-lines-6-5-1-1-2-2.svg | 126 + .../review-graphing-lines-6-5-2-1-1-2.svg | 154 + .../review-graphing-lines-6-5-2-1-2-2.svg | 205 + .../review-graphing-lines-8-5-2-4-2.svg | 151 + .../webwork/images/webwork-1-image-1.pdf | Bin 4861 -> 4448 bytes .../webwork/images/webwork-1-image-1.png | Bin 10728 -> 11077 bytes .../webwork/images/webwork-1-image-1.svg | 93 +- .../webwork/images/webwork-1-image-1.tex | 22 +- .../webwork/images/webwork-100-image-1.pdf | Bin 4816 -> 4808 bytes .../webwork/images/webwork-100-image-1.png | Bin 7291 -> 7291 bytes .../webwork/images/webwork-1012-image-1.pdf | Bin 7002 -> 7004 bytes .../webwork/images/webwork-1012-image-1.png | Bin 10192 -> 10192 bytes .../webwork/images/webwork-1013-image-1.pdf | Bin 8774 -> 8770 bytes .../webwork/images/webwork-1013-image-1.png | Bin 14470 -> 14470 bytes .../webwork/images/webwork-1022-image-1.pdf | Bin 5385 -> 5385 bytes .../webwork/images/webwork-1022-image-1.png | Bin 28973 -> 28973 bytes .../webwork/images/webwork-1069-image-1.pdf | Bin 4524 -> 4524 bytes .../webwork/images/webwork-1069-image-1.png | Bin 7528 -> 7528 bytes .../webwork/images/webwork-1071-image-1.pdf | Bin 5028 -> 5034 bytes .../webwork/images/webwork-1071-image-1.png | Bin 7602 -> 7602 bytes .../webwork/images/webwork-1073-image-1.pdf | Bin 6802 -> 6807 bytes .../webwork/images/webwork-1073-image-1.png | Bin 10105 -> 10105 bytes .../webwork/images/webwork-1090-image-1.pdf | Bin 7455 -> 7454 bytes .../webwork/images/webwork-1090-image-1.png | Bin 17307 -> 17307 bytes .../webwork/images/webwork-1090-image-2.pdf | Bin 7479 -> 7479 bytes .../webwork/images/webwork-1090-image-2.png | Bin 19044 -> 19044 bytes .../webwork/images/webwork-1091-image-1.pdf | Bin 8642 -> 8639 bytes .../webwork/images/webwork-1091-image-1.png | Bin 18551 -> 18551 bytes .../webwork/images/webwork-1092-image-1.pdf | Bin 4864 -> 4861 bytes .../webwork/images/webwork-1092-image-1.png | Bin 10728 -> 10728 bytes .../webwork/images/webwork-1092-image-2.pdf | Bin 6857 -> 6853 bytes .../webwork/images/webwork-1093-image-1.pdf | Bin 8056 -> 8055 bytes .../webwork/images/webwork-1094-image-1.pdf | Bin 6815 -> 6813 bytes .../webwork/images/webwork-1094-image-1.png | Bin 20817 -> 20817 bytes .../webwork/images/webwork-1115-image-1.pdf | Bin 4726 -> 4725 bytes .../webwork/images/webwork-1115-image-1.png | Bin 12486 -> 12486 bytes .../webwork/images/webwork-1117-image-1.pdf | Bin 4668 -> 4668 bytes .../webwork/images/webwork-1119-image-1.pdf | Bin 4974 -> 4976 bytes .../webwork/images/webwork-1121-image-1.pdf | Bin 5227 -> 5224 bytes .../webwork/images/webwork-1121-image-1.png | Bin 11196 -> 11196 bytes .../webwork/images/webwork-1123-image-1.pdf | Bin 6765 -> 6764 bytes .../webwork/images/webwork-1123-image-1.png | Bin 13690 -> 13690 bytes .../webwork/images/webwork-1125-image-1.pdf | Bin 6649 -> 6648 bytes .../webwork/images/webwork-1125-image-1.png | Bin 13274 -> 13274 bytes .../webwork/images/webwork-1138-image-1.pdf | Bin 8309 -> 8303 bytes .../webwork/images/webwork-1139-image-1.pdf | Bin 6427 -> 6425 bytes .../webwork/images/webwork-1141-image-1.pdf | Bin 10456 -> 10456 bytes .../webwork/images/webwork-1146-image-1.pdf | Bin 10114 -> 10118 bytes .../webwork/images/webwork-1147-image-1.pdf | Bin 5548 -> 5552 bytes .../webwork/images/webwork-1147-image-2.pdf | Bin 10149 -> 10152 bytes .../webwork/images/webwork-1148-image-1.pdf | Bin 7031 -> 7031 bytes .../webwork/images/webwork-1148-image-2.pdf | Bin 7051 -> 7052 bytes .../webwork/images/webwork-1169-image-1.pdf | Bin 5184 -> 5191 bytes .../webwork/images/webwork-1169-image-1.png | Bin 14796 -> 14796 bytes .../webwork/images/webwork-1171-image-1.pdf | Bin 6100 -> 6104 bytes .../webwork/images/webwork-1173-image-1.pdf | Bin 5571 -> 5573 bytes .../webwork/images/webwork-1173-image-1.png | Bin 15635 -> 15635 bytes .../webwork/images/webwork-1175-image-1.pdf | Bin 5573 -> 5572 bytes .../webwork/images/webwork-1175-image-1.png | Bin 15235 -> 15235 bytes .../webwork/images/webwork-1177-image-1.pdf | Bin 5562 -> 5561 bytes .../webwork/images/webwork-1177-image-1.png | Bin 15040 -> 15040 bytes .../webwork/images/webwork-1179-image-1.pdf | Bin 5570 -> 5573 bytes .../webwork/images/webwork-1179-image-1.png | Bin 15007 -> 15007 bytes .../webwork/images/webwork-1229-image-1.pdf | Bin 0 -> 4450 bytes .../webwork/images/webwork-1229-image-1.png | Bin 1547 -> 11077 bytes .../webwork/images/webwork-1229-image-1.svg | 200 + .../webwork/images/webwork-1229-image-1.tex | 148 + .../webwork/images/webwork-1230-image-1.pdf | Bin 0 -> 5741 bytes .../webwork/images/webwork-1230-image-1.png | Bin 2187 -> 27080 bytes .../webwork/images/webwork-1230-image-1.svg | 272 + .../webwork/images/webwork-1230-image-1.tex | 140 + .../webwork/images/webwork-1231-image-1.pdf | Bin 0 -> 12204 bytes .../webwork/images/webwork-1231-image-1.png | Bin 0 -> 58204 bytes .../webwork/images/webwork-1231-image-1.svg | 408 + .../webwork/images/webwork-1231-image-1.tex | 150 + .../webwork/images/webwork-1264-image-1.pdf | Bin 0 -> 5614 bytes .../webwork/images/webwork-1264-image-1.png | Bin 0 -> 16704 bytes .../webwork/images/webwork-1264-image-1.svg | 261 + .../webwork/images/webwork-1264-image-1.tex | 138 + .../webwork/images/webwork-1266-image-1.pdf | Bin 0 -> 5597 bytes .../webwork/images/webwork-1266-image-1.png | Bin 1181 -> 13716 bytes .../webwork/images/webwork-1266-image-1.svg | 261 + .../webwork/images/webwork-1266-image-1.tex | 138 + .../webwork/images/webwork-1268-image-1.pdf | Bin 0 -> 5316 bytes .../webwork/images/webwork-1268-image-1.png | Bin 1237 -> 14955 bytes .../webwork/images/webwork-1268-image-1.svg | 263 + .../webwork/images/webwork-1268-image-1.tex | 147 + .../webwork/images/webwork-1270-image-1.pdf | Bin 0 -> 7419 bytes .../webwork/images/webwork-1270-image-1.png | Bin 1684 -> 40673 bytes .../webwork/images/webwork-1270-image-1.svg | 324 + .../webwork/images/webwork-1270-image-1.tex | 149 + .../webwork/images/webwork-1324-image-1.png | Bin 1194 -> 1684 bytes .../webwork/images/webwork-1325-image-1.png | Bin 1684 -> 1843 bytes .../webwork/images/webwork-1343-image-1.png | Bin 43477 -> 1182 bytes .../webwork/images/webwork-1345-image-1.png | Bin 1174 -> 1220 bytes .../webwork/images/webwork-1376-image-1.pdf | Bin 0 -> 7530 bytes .../webwork/images/webwork-1376-image-1.png | Bin 0 -> 14489 bytes .../webwork/images/webwork-1376-image-1.svg | 372 + .../webwork/images/webwork-1376-image-1.tex | 141 + .../webwork/images/webwork-1381-image-1.pdf | Bin 7986 -> 6111 bytes .../webwork/images/webwork-1381-image-1.png | Bin 51758 -> 29510 bytes .../webwork/images/webwork-1381-image-1.svg | 229 +- .../webwork/images/webwork-1381-image-1.tex | 25 +- .../webwork/images/webwork-1381-image-2.pdf | Bin 0 -> 8839 bytes .../webwork/images/webwork-1381-image-2.png | Bin 0 -> 35076 bytes .../webwork/images/webwork-1381-image-2.svg | 303 + .../webwork/images/webwork-1381-image-2.tex | 154 + .../webwork/images/webwork-1382-image-1.pdf | Bin 9644 -> 5734 bytes .../webwork/images/webwork-1382-image-1.png | Bin 35012 -> 22158 bytes .../webwork/images/webwork-1382-image-1.svg | 253 +- .../webwork/images/webwork-1382-image-1.tex | 23 +- .../webwork/images/webwork-1382-image-2.pdf | Bin 0 -> 7475 bytes .../webwork/images/webwork-1382-image-2.png | Bin 0 -> 27419 bytes .../webwork/images/webwork-1382-image-2.svg | 289 + .../webwork/images/webwork-1382-image-2.tex | 154 + .../webwork/images/webwork-1383-image-1.pdf | Bin 8215 -> 5791 bytes .../webwork/images/webwork-1383-image-1.png | Bin 40875 -> 21575 bytes .../webwork/images/webwork-1383-image-1.svg | 239 +- .../webwork/images/webwork-1383-image-1.tex | 23 +- .../webwork/images/webwork-1383-image-2.pdf | Bin 0 -> 8492 bytes .../webwork/images/webwork-1383-image-2.png | Bin 0 -> 26650 bytes .../webwork/images/webwork-1383-image-2.svg | 303 + .../webwork/images/webwork-1383-image-2.tex | 154 + .../webwork/images/webwork-1384-image-1.pdf | Bin 5721 -> 6383 bytes .../webwork/images/webwork-1384-image-1.png | Bin 12546 -> 34047 bytes .../webwork/images/webwork-1384-image-1.svg | 222 +- .../webwork/images/webwork-1384-image-1.tex | 22 +- .../webwork/images/webwork-1384-image-2.pdf | Bin 0 -> 8795 bytes .../webwork/images/webwork-1384-image-2.png | Bin 0 -> 38015 bytes .../webwork/images/webwork-1384-image-2.svg | 376 + .../webwork/images/webwork-1384-image-2.tex | 150 + .../webwork/images/webwork-1385-image-1.pdf | Bin 0 -> 6167 bytes .../webwork/images/webwork-1385-image-1.png | Bin 0 -> 39006 bytes .../webwork/images/webwork-1385-image-1.svg | 286 + .../webwork/images/webwork-1385-image-1.tex | 149 + .../webwork/images/webwork-1385-image-2.pdf | Bin 0 -> 8703 bytes .../webwork/images/webwork-1385-image-2.png | Bin 0 -> 42654 bytes .../webwork/images/webwork-1385-image-2.svg | 337 + .../webwork/images/webwork-1385-image-2.tex | 150 + .../webwork/images/webwork-1386-image-1.pdf | Bin 5753 -> 6098 bytes .../webwork/images/webwork-1386-image-1.png | Bin 39322 -> 42146 bytes .../webwork/images/webwork-1386-image-1.svg | 173 +- .../webwork/images/webwork-1386-image-1.tex | 22 +- .../webwork/images/webwork-1386-image-2.pdf | Bin 0 -> 8370 bytes .../webwork/images/webwork-1386-image-2.png | Bin 0 -> 44876 bytes .../webwork/images/webwork-1386-image-2.svg | 326 + .../webwork/images/webwork-1386-image-2.tex | 150 + .../webwork/images/webwork-1393-image-1.pdf | Bin 5917 -> 6174 bytes .../webwork/images/webwork-1393-image-1.png | Bin 44522 -> 38060 bytes .../webwork/images/webwork-1393-image-1.svg | 208 +- .../webwork/images/webwork-1393-image-1.tex | 22 +- .../webwork/images/webwork-1393-image-2.pdf | Bin 0 -> 8606 bytes .../webwork/images/webwork-1393-image-2.png | Bin 0 -> 42205 bytes .../webwork/images/webwork-1393-image-2.svg | 328 + .../webwork/images/webwork-1393-image-2.tex | 151 + .../webwork/images/webwork-1394-image-1.pdf | Bin 5726 -> 6192 bytes .../webwork/images/webwork-1394-image-1.png | Bin 14817 -> 35285 bytes .../webwork/images/webwork-1394-image-1.svg | 202 +- .../webwork/images/webwork-1394-image-1.tex | 22 +- .../webwork/images/webwork-1394-image-2.pdf | Bin 0 -> 7782 bytes .../webwork/images/webwork-1394-image-2.png | Bin 0 -> 39391 bytes .../webwork/images/webwork-1394-image-2.svg | 326 + .../webwork/images/webwork-1394-image-2.tex | 151 + .../webwork/images/webwork-1409-image-1.png | Bin 43919 -> 1180 bytes .../webwork/images/webwork-1410-image-1.png | Bin 0 -> 1181 bytes .../webwork/images/webwork-1415-image-1.pdf | Bin 5709 -> 5950 bytes .../webwork/images/webwork-1415-image-1.png | Bin 44050 -> 44418 bytes .../webwork/images/webwork-1415-image-1.svg | 170 +- .../webwork/images/webwork-1415-image-1.tex | 4 +- .../webwork/images/webwork-1416-image-1.pdf | Bin 0 -> 9927 bytes .../webwork/images/webwork-1416-image-1.png | Bin 0 -> 48691 bytes .../webwork/images/webwork-1416-image-1.svg | 387 + .../webwork/images/webwork-1416-image-1.tex | 145 + .../webwork/images/webwork-1417-image-1.pdf | Bin 5678 -> 9143 bytes .../webwork/images/webwork-1417-image-1.png | Bin 28406 -> 61609 bytes .../webwork/images/webwork-1417-image-1.svg | 253 +- .../webwork/images/webwork-1417-image-1.tex | 17 +- .../webwork/images/webwork-1418-image-1.pdf | Bin 0 -> 9969 bytes .../webwork/images/webwork-1418-image-1.png | Bin 684 -> 53691 bytes .../webwork/images/webwork-1418-image-1.svg | 393 + .../webwork/images/webwork-1418-image-1.tex | 145 + .../webwork/images/webwork-1419-image-1.pdf | Bin 5567 -> 7986 bytes .../webwork/images/webwork-1419-image-1.png | Bin 16482 -> 51758 bytes .../webwork/images/webwork-1419-image-1.svg | 248 +- .../webwork/images/webwork-1419-image-1.tex | 12 +- .../webwork/images/webwork-1420-image-1.pdf | Bin 0 -> 9647 bytes .../webwork/images/webwork-1420-image-1.png | Bin 708 -> 35012 bytes .../webwork/images/webwork-1420-image-1.svg | 337 + .../webwork/images/webwork-1420-image-1.tex | 139 + .../webwork/images/webwork-1421-image-1.pdf | Bin 5745 -> 8211 bytes .../webwork/images/webwork-1421-image-1.png | Bin 39257 -> 40875 bytes .../webwork/images/webwork-1421-image-1.svg | 218 +- .../webwork/images/webwork-1421-image-1.tex | 4 +- .../webwork/images/webwork-1431-image-1.pdf | Bin 0 -> 5936 bytes .../webwork/images/webwork-1431-image-1.png | Bin 0 -> 41756 bytes .../webwork/images/webwork-1431-image-1.svg | 282 + .../webwork/images/webwork-1431-image-1.tex | 139 + .../webwork/images/webwork-1433-image-1.pdf | Bin 0 -> 5967 bytes .../webwork/images/webwork-1433-image-1.png | Bin 0 -> 45914 bytes .../webwork/images/webwork-1433-image-1.svg | 282 + .../webwork/images/webwork-1433-image-1.tex | 139 + .../webwork/images/webwork-1435-image-1.pdf | Bin 0 -> 5806 bytes .../webwork/images/webwork-1435-image-1.png | Bin 0 -> 26151 bytes .../webwork/images/webwork-1435-image-1.svg | 282 + .../webwork/images/webwork-1435-image-1.tex | 139 + .../webwork/images/webwork-1443-image-1.pdf | Bin 0 -> 5706 bytes .../webwork/images/webwork-1443-image-1.png | Bin 0 -> 29446 bytes .../webwork/images/webwork-1443-image-1.svg | 254 + .../webwork/images/webwork-1443-image-1.tex | 139 + .../webwork/images/webwork-1445-image-1.pdf | Bin 0 -> 5731 bytes .../webwork/images/webwork-1445-image-1.png | Bin 0 -> 36987 bytes .../webwork/images/webwork-1445-image-1.svg | 254 + .../webwork/images/webwork-1445-image-1.tex | 139 + .../webwork/images/webwork-1447-image-1.pdf | Bin 0 -> 5728 bytes .../webwork/images/webwork-1447-image-1.png | Bin 0 -> 33546 bytes .../webwork/images/webwork-1447-image-1.svg | 254 + .../webwork/images/webwork-1447-image-1.tex | 139 + .../webwork/images/webwork-1449-image-1.pdf | Bin 0 -> 5755 bytes .../webwork/images/webwork-1449-image-1.png | Bin 0 -> 40936 bytes .../webwork/images/webwork-1449-image-1.svg | 254 + .../webwork/images/webwork-1449-image-1.tex | 139 + .../webwork/images/webwork-1451-image-1.pdf | Bin 0 -> 5734 bytes .../webwork/images/webwork-1451-image-1.png | Bin 0 -> 37308 bytes .../webwork/images/webwork-1451-image-1.svg | 254 + .../webwork/images/webwork-1451-image-1.tex | 139 + .../webwork/images/webwork-1453-image-1.pdf | Bin 0 -> 5742 bytes .../webwork/images/webwork-1453-image-1.png | Bin 0 -> 41128 bytes .../webwork/images/webwork-1453-image-1.svg | 254 + .../webwork/images/webwork-1453-image-1.tex | 139 + .../webwork/images/webwork-1455-image-1.pdf | Bin 0 -> 5712 bytes .../webwork/images/webwork-1455-image-1.png | Bin 0 -> 46186 bytes .../webwork/images/webwork-1455-image-1.svg | 254 + .../webwork/images/webwork-1455-image-1.tex | 139 + .../webwork/images/webwork-1457-image-1.pdf | Bin 0 -> 5548 bytes .../webwork/images/webwork-1457-image-1.png | Bin 0 -> 13590 bytes .../webwork/images/webwork-1457-image-1.svg | 254 + .../webwork/images/webwork-1457-image-1.tex | 139 + .../webwork/images/webwork-1459-image-1.pdf | Bin 0 -> 5759 bytes .../webwork/images/webwork-1459-image-1.png | Bin 0 -> 14842 bytes .../webwork/images/webwork-1459-image-1.svg | 254 + .../webwork/images/webwork-1459-image-1.tex | 139 + .../webwork/images/webwork-1461-image-1.pdf | Bin 0 -> 5580 bytes .../webwork/images/webwork-1461-image-1.png | Bin 0 -> 16293 bytes .../webwork/images/webwork-1461-image-1.svg | 254 + .../webwork/images/webwork-1461-image-1.tex | 139 + .../webwork/images/webwork-1569-image-1.pdf | Bin 0 -> 5885 bytes .../webwork/images/webwork-1569-image-1.png | Bin 0 -> 44339 bytes .../webwork/images/webwork-1569-image-1.svg | 282 + .../webwork/images/webwork-1569-image-1.tex | 139 + .../webwork/images/webwork-1570-image-1.pdf | Bin 0 -> 5678 bytes .../webwork/images/webwork-1570-image-1.png | Bin 679 -> 33496 bytes .../webwork/images/webwork-1570-image-1.svg | 254 + .../webwork/images/webwork-1570-image-1.tex | 139 + .../webwork/images/webwork-1571-image-1.pdf | Bin 0 -> 5708 bytes .../webwork/images/webwork-1571-image-1.png | Bin 637 -> 45539 bytes .../webwork/images/webwork-1571-image-1.svg | 254 + .../webwork/images/webwork-1571-image-1.tex | 139 + .../webwork/images/webwork-1579-image-1.pdf | Bin 0 -> 5929 bytes .../webwork/images/webwork-1579-image-1.png | Bin 698 -> 32125 bytes .../webwork/images/webwork-1579-image-1.svg | 282 + .../webwork/images/webwork-1579-image-1.tex | 139 + .../webwork/images/webwork-1581-image-1.pdf | Bin 0 -> 5727 bytes .../webwork/images/webwork-1581-image-1.png | Bin 707 -> 35579 bytes .../webwork/images/webwork-1581-image-1.svg | 254 + .../webwork/images/webwork-1581-image-1.tex | 139 + .../webwork/images/webwork-1582-image-1.pdf | Bin 0 -> 5723 bytes .../webwork/images/webwork-1582-image-1.png | Bin 700 -> 40338 bytes .../webwork/images/webwork-1582-image-1.svg | 254 + .../webwork/images/webwork-1582-image-1.tex | 139 + .../webwork/images/webwork-1583-image-1.pdf | Bin 0 -> 5548 bytes .../webwork/images/webwork-1583-image-1.png | Bin 753 -> 14648 bytes .../webwork/images/webwork-1583-image-1.svg | 254 + .../webwork/images/webwork-1583-image-1.tex | 139 + .../webwork/images/webwork-1584-image-1.pdf | Bin 0 -> 4965 bytes .../webwork/images/webwork-1584-image-1.png | Bin 0 -> 26613 bytes .../webwork/images/webwork-1584-image-1.svg | 205 + .../webwork/images/webwork-1584-image-1.tex | 139 + .../webwork/images/webwork-1608-image-1.png | Bin 0 -> 684 bytes .../webwork/images/webwork-1609-image-1.png | Bin 0 -> 659 bytes .../webwork/images/webwork-1610-image-1.png | Bin 0 -> 710 bytes .../webwork/images/webwork-1611-image-1.png | Bin 0 -> 693 bytes .../webwork/images/webwork-1612-image-1.png | Bin 4180 -> 682 bytes .../webwork/images/webwork-1613-image-1.png | Bin 4555 -> 677 bytes .../webwork/images/webwork-1616-image-1.png | Bin 355 -> 752 bytes .../webwork/images/webwork-1617-image-1.png | Bin 379 -> 734 bytes .../webwork/images/webwork-1618-image-1.png | Bin 394 -> 707 bytes .../webwork/images/webwork-1619-image-1.png | Bin 0 -> 738 bytes .../webwork/images/webwork-1620-image-1.png | Bin 4550 -> 753 bytes .../webwork/images/webwork-1621-image-1.png | Bin 4555 -> 742 bytes .../webwork/images/webwork-1666-image-1.pdf | Bin 0 -> 5475 bytes .../webwork/images/webwork-1666-image-1.png | Bin 0 -> 4180 bytes .../webwork/images/webwork-1666-image-1.svg | 90 + .../webwork/images/webwork-1666-image-1.tex | 146 + .../webwork/images/webwork-1674-image-1.pdf | Bin 0 -> 6455 bytes .../webwork/images/webwork-1674-image-1.png | Bin 0 -> 4550 bytes .../webwork/images/webwork-1674-image-1.svg | 194 + .../webwork/images/webwork-1674-image-1.tex | 147 + .../webwork/images/webwork-1675-image-1.pdf | Bin 0 -> 6451 bytes .../webwork/images/webwork-1675-image-1.png | Bin 0 -> 4555 bytes .../webwork/images/webwork-1675-image-1.svg | 194 + .../webwork/images/webwork-1675-image-1.tex | 147 + .../webwork/images/webwork-1676-image-1.png | Bin 0 -> 341 bytes .../webwork/images/webwork-1677-image-1.png | Bin 4983 -> 362 bytes .../webwork/images/webwork-1678-image-1.png | Bin 4918 -> 361 bytes .../webwork/images/webwork-1679-image-1.png | Bin 0 -> 372 bytes .../webwork/images/webwork-170-image-1.pdf | Bin 3559 -> 3563 bytes .../webwork/images/webwork-170-image-1.png | Bin 6151 -> 6151 bytes .../webwork/images/webwork-172-image-1.pdf | Bin 3200 -> 3198 bytes .../webwork/images/webwork-172-image-1.png | Bin 5185 -> 5185 bytes .../webwork/images/webwork-174-image-1.pdf | Bin 4205 -> 4203 bytes .../webwork/images/webwork-174-image-1.png | Bin 6812 -> 6812 bytes .../webwork/images/webwork-1756-image-1.png | Bin 0 -> 3691 bytes .../webwork/images/webwork-1763-image-1.png | Bin 0 -> 3597 bytes .../webwork/images/webwork-180-image-1.pdf | Bin 8034 -> 8032 bytes .../webwork/images/webwork-180-image-1.png | Bin 5685 -> 5685 bytes .../webwork/images/webwork-180-image-2.pdf | Bin 9783 -> 9781 bytes .../webwork/images/webwork-180-image-2.png | Bin 9560 -> 9560 bytes .../webwork/images/webwork-181-image-1.pdf | Bin 5521 -> 5519 bytes .../webwork/images/webwork-181-image-1.png | Bin 4413 -> 4413 bytes .../webwork/images/webwork-181-image-2.pdf | Bin 5552 -> 5552 bytes .../webwork/images/webwork-181-image-2.png | Bin 4567 -> 4567 bytes .../webwork/images/webwork-181-image-3.pdf | Bin 5523 -> 5525 bytes .../webwork/images/webwork-181-image-3.png | Bin 4416 -> 4416 bytes .../webwork/images/webwork-182-image-1.pdf | Bin 5519 -> 5520 bytes .../webwork/images/webwork-182-image-1.png | Bin 4413 -> 4413 bytes .../webwork/images/webwork-182-image-2.pdf | Bin 5552 -> 5551 bytes .../webwork/images/webwork-182-image-2.png | Bin 4567 -> 4567 bytes .../webwork/images/webwork-182-image-3.pdf | Bin 5520 -> 5521 bytes .../webwork/images/webwork-182-image-3.png | Bin 4416 -> 4416 bytes .../webwork/images/webwork-1868-image-1.png | Bin 0 -> 5031 bytes .../webwork/images/webwork-1869-image-1.png | Bin 0 -> 4846 bytes .../webwork/images/webwork-1878-image-1.png | Bin 0 -> 5064 bytes .../webwork/images/webwork-1879-image-1.png | Bin 0 -> 5047 bytes .../webwork/images/webwork-1880-image-1.png | Bin 0 -> 2230 bytes .../webwork/images/webwork-1881-image-1.png | Bin 0 -> 2261 bytes .../webwork/images/webwork-1896-image-1.png | Bin 0 -> 2597 bytes .../webwork/images/webwork-1897-image-1.png | Bin 0 -> 2415 bytes .../webwork/images/webwork-2-image-1.pdf | Bin 8059 -> 5740 bytes .../webwork/images/webwork-2-image-1.png | Bin 29830 -> 27080 bytes .../webwork/images/webwork-2-image-1.svg | 244 +- .../webwork/images/webwork-2-image-1.tex | 17 +- .../webwork/images/webwork-228-image-1.pdf | Bin 5898 -> 5900 bytes .../webwork/images/webwork-228-image-1.png | Bin 4771 -> 4771 bytes .../webwork/images/webwork-230-image-1.pdf | Bin 5877 -> 5874 bytes .../webwork/images/webwork-230-image-1.png | Bin 4622 -> 4622 bytes .../webwork/images/webwork-232-image-1.pdf | Bin 5875 -> 5880 bytes .../webwork/images/webwork-232-image-1.png | Bin 4593 -> 4593 bytes .../webwork/images/webwork-234-image-1.pdf | Bin 5899 -> 5899 bytes .../webwork/images/webwork-234-image-1.png | Bin 4790 -> 4790 bytes .../webwork/images/webwork-3-image-1.pdf | Bin 10117 -> 12200 bytes .../webwork/images/webwork-3-image-1.png | Bin 61392 -> 58204 bytes .../webwork/images/webwork-3-image-1.svg | 575 +- .../webwork/images/webwork-3-image-1.tex | 34 +- .../webwork/images/webwork-35-image-1.pdf | Bin 6505 -> 5612 bytes .../webwork/images/webwork-35-image-1.png | Bin 44083 -> 16261 bytes .../webwork/images/webwork-35-image-1.svg | 194 +- .../webwork/images/webwork-35-image-1.tex | 21 +- .../webwork/images/webwork-36-image-1.pdf | Bin 5580 -> 5617 bytes .../webwork/images/webwork-36-image-1.png | Bin 16238 -> 15872 bytes .../webwork/images/webwork-36-image-1.svg | 24 +- .../webwork/images/webwork-36-image-1.tex | 14 +- .../webwork/images/webwork-37-image-1.pdf | Bin 5486 -> 5601 bytes .../webwork/images/webwork-37-image-1.png | Bin 31253 -> 14221 bytes .../webwork/images/webwork-37-image-1.svg | 510 +- .../webwork/images/webwork-37-image-1.tex | 161 +- .../webwork/images/webwork-38-image-1.pdf | Bin 5577 -> 5575 bytes .../webwork/images/webwork-38-image-1.png | Bin 14397 -> 32930 bytes .../webwork/images/webwork-38-image-1.svg | 144 +- .../webwork/images/webwork-38-image-1.tex | 19 +- .../webwork/images/webwork-39-image-1.pdf | Bin 5893 -> 5314 bytes .../webwork/images/webwork-39-image-1.png | Bin 34531 -> 15620 bytes .../webwork/images/webwork-39-image-1.svg | 180 +- .../webwork/images/webwork-39-image-1.tex | 30 +- .../webwork/images/webwork-40-image-1.pdf | Bin 6168 -> 5321 bytes .../webwork/images/webwork-40-image-1.png | Bin 39006 -> 15862 bytes .../webwork/images/webwork-40-image-1.svg | 197 +- .../webwork/images/webwork-40-image-1.tex | 30 +- .../webwork/images/webwork-41-image-1.pdf | Bin 5082 -> 7309 bytes .../webwork/images/webwork-41-image-1.png | Bin 10787 -> 15883 bytes .../webwork/images/webwork-41-image-1.svg | 237 +- .../webwork/images/webwork-41-image-1.tex | 31 +- .../webwork/images/webwork-415-image-1.pdf | Bin 5901 -> 5905 bytes .../webwork/images/webwork-415-image-1.png | Bin 4966 -> 4966 bytes .../webwork/images/webwork-416-image-1.pdf | Bin 875 -> 875 bytes .../webwork/images/webwork-417-image-1.pdf | Bin 4922 -> 4923 bytes .../webwork/images/webwork-417-image-1.png | Bin 4744 -> 4744 bytes .../webwork/images/webwork-42-image-1.pdf | Bin 6205 -> 7451 bytes .../webwork/images/webwork-42-image-1.png | Bin 10878 -> 40563 bytes .../webwork/images/webwork-42-image-1.svg | 251 +- .../webwork/images/webwork-42-image-1.tex | 31 +- .../webwork/images/webwork-43-image-1.png | Bin 927 -> 1993 bytes .../webwork/images/webwork-44-image-1.png | Bin 909 -> 2165 bytes .../webwork/images/webwork-45-image-1.png | Bin 15434 -> 2106 bytes .../webwork/images/webwork-47-image-1.png | Bin 53543 -> 1997 bytes .../webwork/images/webwork-49-image-1.png | Bin 2768 -> 1461 bytes .../webwork/images/webwork-5-image-1.pdf | Bin 2305 -> 2303 bytes .../webwork/images/webwork-5-image-1.png | Bin 9918 -> 9918 bytes .../webwork/images/webwork-51-image-1.png | Bin 1464 -> 1954 bytes .../webwork/images/webwork-52-image-1.png | Bin 7582 -> 1875 bytes .../webwork/images/webwork-53-image-1.png | Bin 694 -> 2207 bytes .../webwork/images/webwork-588-image-1.pdf | Bin 3661 -> 3659 bytes .../webwork/images/webwork-588-image-1.png | Bin 6326 -> 6326 bytes .../webwork/images/webwork-589-image-1.pdf | Bin 5905 -> 5902 bytes .../webwork/images/webwork-589-image-1.png | Bin 4771 -> 4771 bytes .../webwork/images/webwork-590-image-1.pdf | Bin 5873 -> 5867 bytes .../webwork/images/webwork-590-image-1.png | Bin 4613 -> 4613 bytes .../webwork/images/webwork-636-image-1.pdf | Bin 3915 -> 3915 bytes .../webwork/images/webwork-636-image-1.png | Bin 6612 -> 6612 bytes .../webwork/images/webwork-642-image-1.pdf | Bin 5875 -> 5880 bytes .../webwork/images/webwork-642-image-1.png | Bin 4628 -> 4628 bytes .../webwork/images/webwork-768-image-1.pdf | Bin 5904 -> 5905 bytes .../webwork/images/webwork-768-image-1.png | Bin 4959 -> 4959 bytes .../webwork/images/webwork-999-image-1.pdf | Bin 7528 -> 7528 bytes .../webwork/images/webwork-999-image-1.png | Bin 14517 -> 14517 bytes ...-point-slope-form-from-graph-2-image-1.png | Bin 1789 -> 1466 bytes ...-point-slope-form-from-graph-3-image-1.png | Bin 2088 -> 2049 bytes ...-point-slope-form-from-graph-4-image-1.png | Bin 1649 -> 1823 bytes ...-point-slope-form-from-graph-5-image-1.png | Bin 2350 -> 2259 bytes ...ne-point-slope-form-from-graph-image-1.pdf | Bin 0 -> 5605 bytes ...ne-point-slope-form-from-graph-image-1.png | Bin 1501 -> 34347 bytes ...ne-point-slope-form-from-graph-image-1.svg | 261 + ...ne-point-slope-form-from-graph-image-1.tex | 138 + ...oint-slope-form-from-graph-neg-image-1.pdf | Bin 0 -> 5585 bytes ...oint-slope-form-from-graph-neg-image-1.png | Bin 0 -> 15870 bytes ...oint-slope-form-from-graph-neg-image-1.svg | 261 + ...oint-slope-form-from-graph-neg-image-1.tex | 138 + ...ope-form-from-graph-q1-large-y-image-1.pdf | Bin 0 -> 5262 bytes ...ope-form-from-graph-q1-large-y-image-1.png | Bin 0 -> 16186 bytes ...ope-form-from-graph-q1-large-y-image-1.svg | 263 + ...ope-form-from-graph-q1-large-y-image-1.tex | 147 + ...-from-graph-q1-large-y-precise-image-1.pdf | Bin 0 -> 7458 bytes ...-from-graph-q1-large-y-precise-image-1.png | Bin 0 -> 17231 bytes ...-from-graph-q1-large-y-precise-image-1.svg | 323 + ...-from-graph-q1-large-y-precise-image-1.tex | 149 + ...er-combine-like-terms-decimals-image-1.pdf | Bin 3945 -> 3948 bytes ...er-combine-like-terms-decimals-image-1.png | Bin 7175 -> 7175 bytes ...r-combine-like-terms-fractions-image-1.pdf | Bin 4057 -> 4056 bytes ...r-combine-like-terms-fractions-image-1.png | Bin 6850 -> 6850 bytes ...identify-cartesian-coordinates-image-1.pdf | Bin 7532 -> 7527 bytes ...erval-notation-closed-infinite-image-1.pdf | Bin 5868 -> 5873 bytes ...erval-notation-closed-infinite-image-1.png | Bin 4636 -> 4636 bytes ...erval-notation-infinite-closed-image-1.pdf | Bin 5878 -> 5872 bytes ...erval-notation-infinite-closed-image-1.png | Bin 4617 -> 4617 bytes ...nterval-notation-infinite-open-image-1.pdf | Bin 5904 -> 5903 bytes ...nterval-notation-infinite-open-image-1.png | Bin 4788 -> 4788 bytes ...nterval-notation-open-infinite-image-1.pdf | Bin 5898 -> 5898 bytes ...nterval-notation-open-infinite-image-1.png | Bin 4785 -> 4785 bytes ...l-perimeter-combine-like-terms-image-1.pdf | Bin 3727 -> 3731 bytes ...l-perimeter-combine-like-terms-image-1.png | Bin 5983 -> 5983 bytes .../webwork-rate-of-change-graph-image-1.pdf | Bin 4533 -> 4527 bytes .../webwork-rate-of-change-graph-image-1.png | Bin 7553 -> 7553 bytes ...ork-rate-of-change-graph-large-image-1.pdf | Bin 6752 -> 6756 bytes ...ork-rate-of-change-graph-large-image-1.png | Bin 10436 -> 10436 bytes ...bwork-rate-of-change-graph-neg-image-1.pdf | Bin 5031 -> 5028 bytes ...bwork-rate-of-change-graph-neg-image-1.png | Bin 7617 -> 7617 bytes ...olution-from-graph-point-slope-image-1.pdf | Bin 5939 -> 5939 bytes ...olution-from-graph-point-slope-image-1.png | Bin 37353 -> 42256 bytes ...olution-from-graph-point-slope-image-1.svg | 12 +- ...olution-from-graph-point-slope-image-1.tex | 4 +- ...rom-graph-point-slope-parallel-image-1.pdf | Bin 5922 -> 5896 bytes ...rom-graph-point-slope-parallel-image-1.png | Bin 13759 -> 40576 bytes ...rom-graph-point-slope-parallel-image-1.svg | 142 +- ...rom-graph-point-slope-parallel-image-1.tex | 4 +- ...ion-from-graph-slope-intercept-image-1.pdf | Bin 5955 -> 5935 bytes ...ion-from-graph-slope-intercept-image-1.png | Bin 42328 -> 35095 bytes ...ion-from-graph-slope-intercept-image-1.svg | 12 +- ...ion-from-graph-slope-intercept-image-1.tex | 4 +- ...work-slope-from-graph-fraction-image-1.pdf | Bin 4666 -> 4665 bytes ...-from-graph-fraction-neg-input-image-1.pdf | Bin 875 -> 875 bytes ...bwork-slope-from-graph-integer-image-1.pdf | Bin 4675 -> 4676 bytes ...bwork-slope-from-graph-integer-image-1.png | Bin 12459 -> 12459 bytes ...m-graph-labeled-points-decimal-image-1.pdf | Bin 6585 -> 6578 bytes ...m-graph-labeled-points-decimal-image-1.png | Bin 16021 -> 16021 bytes ...lope-from-graph-labeled-points-image-1.pdf | Bin 6351 -> 6347 bytes ...lope-from-graph-labeled-points-image-1.png | Bin 15581 -> 15581 bytes .../webwork-slope-from-graph-zero-image-1.pdf | Bin 5226 -> 5224 bytes .../webwork-slope-from-graph-zero-image-1.png | Bin 11140 -> 11140 bytes ...-intercept-from-graph-neg-frac-image-1.pdf | Bin 5565 -> 5566 bytes ...-intercept-from-graph-neg-frac-image-1.png | Bin 13491 -> 13491 bytes ...e-intercept-from-graph-neg-one-image-1.pdf | Bin 5573 -> 5566 bytes ...e-intercept-from-graph-neg-one-image-1.png | Bin 15143 -> 15143 bytes ...pe-intercept-from-graph-origin-image-1.pdf | Bin 5569 -> 5565 bytes ...pe-intercept-from-graph-origin-image-1.png | Bin 14910 -> 14910 bytes ...e-intercept-from-graph-pos-int-image-1.pdf | Bin 5184 -> 5186 bytes ...e-intercept-from-graph-pos-int-image-1.png | Bin 15134 -> 15134 bytes ...e-intercept-from-graph-pos-one-image-1.pdf | Bin 5493 -> 5496 bytes ...e-intercept-from-graph-pos-one-image-1.png | Bin 16036 -> 16036 bytes ...cept-from-graph-pos-reciprocal-image-1.pdf | Bin 5560 -> 5561 bytes ...cept-from-graph-pos-reciprocal-image-1.png | Bin 12900 -> 12900 bytes ...system-application-dog-cyclist-image-1.pdf | Bin 5470 -> 5484 bytes ...system-application-dog-cyclist-image-1.png | Bin 48687 -> 49715 bytes ...system-application-dog-cyclist-image-1.svg | 4 +- ...system-application-dog-cyclist-image-1.tex | 2 +- ...e-system-application-greet-dog-image-1.pdf | Bin 4977 -> 4972 bytes ...e-system-application-greet-dog-image-1.png | Bin 26680 -> 25780 bytes ...e-system-application-greet-dog-image-1.svg | 92 +- ...e-system-application-greet-dog-image-1.tex | 2 +- ...tem-application-hardware-store-image-1.pdf | Bin 6890 -> 8089 bytes ...tem-application-hardware-store-image-1.png | Bin 62939 -> 86283 bytes ...tem-application-hardware-store-image-1.svg | 285 +- ...tem-application-hardware-store-image-1.tex | 6 +- ...tem-application-party-planning-image-1.pdf | Bin 9515 -> 6518 bytes ...tem-application-party-planning-image-1.png | Bin 122191 -> 66764 bytes ...tem-application-party-planning-image-1.svg | 395 +- ...tem-application-party-planning-image-1.tex | 6 +- ...system-graphically-point-slope-image-1.pdf | Bin 5732 -> 5723 bytes ...system-graphically-point-slope-image-1.png | Bin 33925 -> 15975 bytes ...system-graphically-point-slope-image-1.svg | 134 +- ...system-graphically-point-slope-image-1.tex | 4 +- ...ically-point-slope-no-solution-image-1.pdf | Bin 5737 -> 5745 bytes ...ically-point-slope-no-solution-image-1.png | Bin 29971 -> 48301 bytes ...ically-point-slope-no-solution-image-1.svg | 134 +- ...ically-point-slope-no-solution-image-1.tex | 4 +- ...em-graphically-slope-intercept-image-1.pdf | Bin 5763 -> 5736 bytes ...em-graphically-slope-intercept-image-1.png | Bin 28393 -> 40351 bytes ...em-graphically-slope-intercept-image-1.svg | 142 +- ...em-graphically-slope-intercept-image-1.tex | 4 +- ...ly-slope-intercept-point-slope-image-1.pdf | Bin 5720 -> 5752 bytes ...ly-slope-intercept-point-slope-image-1.png | Bin 29140 -> 36089 bytes ...ly-slope-intercept-point-slope-image-1.svg | 134 +- ...ly-slope-intercept-point-slope-image-1.tex | 4 +- ...point-slope-infinite-solutions-image-1.pdf | Bin 5564 -> 5626 bytes ...point-slope-infinite-solutions-image-1.png | Bin 16249 -> 26553 bytes ...point-slope-infinite-solutions-image-1.svg | 142 +- ...point-slope-infinite-solutions-image-1.tex | 4 +- ...ve-system-graphically-standard-image-1.pdf | Bin 5716 -> 5718 bytes ...ve-system-graphically-standard-image-1.png | Bin 40115 -> 40040 bytes ...ve-system-graphically-standard-image-1.svg | 134 +- ...ve-system-graphically-standard-image-1.tex | 4 +- ...aphically-standard-point-slope-image-1.pdf | Bin 5708 -> 5738 bytes ...aphically-standard-point-slope-image-1.png | Bin 39145 -> 29596 bytes ...aphically-standard-point-slope-image-1.svg | 134 +- ...aphically-standard-point-slope-image-1.tex | 4 +- ...point-slope-infinite-solutions-image-1.pdf | Bin 5628 -> 5643 bytes ...point-slope-infinite-solutions-image-1.png | Bin 30121 -> 16456 bytes ...point-slope-infinite-solutions-image-1.svg | 142 +- ...point-slope-infinite-solutions-image-1.tex | 4 +- ...cally-standard-slope-intercept-image-1.pdf | Bin 5742 -> 5731 bytes ...cally-standard-slope-intercept-image-1.png | Bin 39792 -> 33567 bytes ...cally-standard-slope-intercept-image-1.svg | 134 +- ...cally-standard-slope-intercept-image-1.tex | 4 +- ...rd-slope-intercept-no-solution-image-1.pdf | Bin 5725 -> 5727 bytes ...rd-slope-intercept-no-solution-image-1.png | Bin 43427 -> 39788 bytes ...rd-slope-intercept-no-solution-image-1.svg | 130 +- ...rd-slope-intercept-no-solution-image-1.tex | 4 +- generated/webwork/webwork-representations.xml | 32506 +++++++++------- index.html | 2 +- knowl/index/definition-point-slope-form.html | 2 +- knowl/index/definition-standard-form.html | 4 +- ...stems-of-equations-substitution-intro.html | 4 +- knowl/index/ordering.html | 4 +- ...fractions-and-fraction-arithmetic-3-2.html | 10 +- knowl/index/section-standard-form-5-7.html | 2 +- knowl/index/solve-system-application.html | 6 +- ...efinition-cartesian-coordinate-system.html | 20 + knowl/xref/definition-point-slope-form.html | 27 + .../xref/definition-slope-intercept-form.html | 26 + knowl/xref/definition-slope.html | 18 + knowl/xref/definition-standard-form.html | 4 +- ...stems-of-equations-substitution-intro.html | 4 +- knowl/xref/exercise-negative-to-power.html | 2 +- knowl/xref/exercise-point-slope.html | 6 +- .../exercise-solve-system-application.html | 6 +- knowl/xref/fact-parallel-lines.html | 19 + knowl/xref/fact-perpendicular-lines.html | 21 + ...figure-point-slope-multiple-equations.html | 2 +- knowl/xref/figure-spa-collapse.html | 2 +- lunr-pretext-search-index.js | 1486 +- orcca.html | 16 +- part-linear-equations-and-lines.html | 16 +- review-graphing-lines.html | 1837 +- review-linear-equations-and-inequalities.html | 16 +- review-systems-of-linear-equations.html | 490 +- ...w-variables-expressions-and-equations.html | 16 +- section-absolute-value-and-square-root.html | 996 +- ...roperties-and-simplifying-expressions.html | 16 +- section-arithmetic-with-negative-numbers.html | 1590 +- section-cartesian-coordinates.html | 16 +- section-combining-like-terms.html | 16 +- ...on-symbols-and-notation-for-intervals.html | 20 +- section-elimination.html | 378 +- ...tions-and-inequalities-with-fractions.html | 16 +- ...-equations-inequalities-and-solutions.html | 16 +- ...-two-variable-data-and-rate-of-change.html | 16 +- ...ion-fractions-and-fraction-arithmetic.html | 2247 +- section-graphing-equations.html | 16 +- ...ical-parallel-and-perpendicular-lines.html | 940 +- section-isolating-a-linear-variable.html | 16 +- ...eling-with-equations-and-inequalities.html | 16 +- section-order-of-operations.html | 574 +- section-percentages.html | 3362 +- section-point-slope-form.html | 1645 +- ...ion-set-notation-and-types-of-numbers.html | 698 +- section-slope-intercept-form.html | 16 +- section-slope.html | 16 +- section-solving-a-system-by-graphing.html | 606 +- ...on-solving-multistep-linear-equations.html | 16 +- ...solving-multistep-linear-inequalities.html | 16 +- section-solving-one-step-equations.html | 16 +- section-solving-one-step-inequalities.html | 16 +- section-special-solution-sets.html | 16 +- section-standard-form.html | 926 +- section-substitution.html | 728 +- section-summary-of-graphing-lines.html | 20 +- ...-variables-and-evaluating-expressions.html | 16 +- 635 files changed, 56085 insertions(+), 27429 deletions(-) create mode 100644 generated/latex-image/point-slope-yequals-130groupx-2021+2975.svg create mode 100644 generated/latex-image/review-graphing-lines-10-1-1-2-1.svg create mode 100644 generated/latex-image/review-graphing-lines-10-8-13-2-1.svg create mode 100644 generated/latex-image/review-graphing-lines-10-8-14-2-1.svg create mode 100644 generated/latex-image/review-graphing-lines-10-8-15-2-1.svg create mode 100644 generated/latex-image/review-graphing-lines-10-8-16-2-1.svg create mode 100644 generated/latex-image/review-graphing-lines-2-3-2-1-2.svg create mode 100644 generated/latex-image/review-graphing-lines-3-3-2-3-2-1-2.svg create mode 100644 generated/latex-image/review-graphing-lines-3-3-2-3-2-2-2.svg create mode 100644 generated/latex-image/review-graphing-lines-5-4-1-2-2.svg create mode 100644 generated/latex-image/review-graphing-lines-5-4-2-1-1-2.svg create mode 100644 generated/latex-image/review-graphing-lines-6-3-2-1-2-1-2.svg create mode 100644 generated/latex-image/review-graphing-lines-6-3-2-1-2-2-2.svg create mode 100644 generated/latex-image/review-graphing-lines-6-3-2-1-2-3-2.svg create mode 100644 generated/latex-image/review-graphing-lines-6-5-1-1-2-2.svg create mode 100644 generated/latex-image/review-graphing-lines-6-5-2-1-1-2.svg create mode 100644 generated/latex-image/review-graphing-lines-6-5-2-1-2-2.svg create mode 100644 generated/latex-image/review-graphing-lines-8-5-2-4-2.svg create mode 100755 generated/webwork/images/webwork-1229-image-1.pdf create mode 100644 generated/webwork/images/webwork-1229-image-1.svg create mode 100755 generated/webwork/images/webwork-1229-image-1.tex create mode 100755 generated/webwork/images/webwork-1230-image-1.pdf create mode 100644 generated/webwork/images/webwork-1230-image-1.svg create mode 100755 generated/webwork/images/webwork-1230-image-1.tex create mode 100755 generated/webwork/images/webwork-1231-image-1.pdf create mode 100644 generated/webwork/images/webwork-1231-image-1.png create mode 100644 generated/webwork/images/webwork-1231-image-1.svg create mode 100755 generated/webwork/images/webwork-1231-image-1.tex create mode 100755 generated/webwork/images/webwork-1264-image-1.pdf create mode 100644 generated/webwork/images/webwork-1264-image-1.png create mode 100644 generated/webwork/images/webwork-1264-image-1.svg create mode 100755 generated/webwork/images/webwork-1264-image-1.tex create mode 100755 generated/webwork/images/webwork-1266-image-1.pdf create mode 100644 generated/webwork/images/webwork-1266-image-1.svg create mode 100755 generated/webwork/images/webwork-1266-image-1.tex create mode 100755 generated/webwork/images/webwork-1268-image-1.pdf create mode 100644 generated/webwork/images/webwork-1268-image-1.svg create mode 100755 generated/webwork/images/webwork-1268-image-1.tex create mode 100755 generated/webwork/images/webwork-1270-image-1.pdf create mode 100644 generated/webwork/images/webwork-1270-image-1.svg create mode 100755 generated/webwork/images/webwork-1270-image-1.tex create mode 100755 generated/webwork/images/webwork-1376-image-1.pdf create mode 100644 generated/webwork/images/webwork-1376-image-1.png create mode 100644 generated/webwork/images/webwork-1376-image-1.svg create mode 100755 generated/webwork/images/webwork-1376-image-1.tex create mode 100755 generated/webwork/images/webwork-1381-image-2.pdf create mode 100644 generated/webwork/images/webwork-1381-image-2.png create mode 100644 generated/webwork/images/webwork-1381-image-2.svg create mode 100755 generated/webwork/images/webwork-1381-image-2.tex create mode 100755 generated/webwork/images/webwork-1382-image-2.pdf create mode 100644 generated/webwork/images/webwork-1382-image-2.png create mode 100644 generated/webwork/images/webwork-1382-image-2.svg create mode 100755 generated/webwork/images/webwork-1382-image-2.tex create mode 100755 generated/webwork/images/webwork-1383-image-2.pdf create mode 100644 generated/webwork/images/webwork-1383-image-2.png create mode 100644 generated/webwork/images/webwork-1383-image-2.svg create mode 100755 generated/webwork/images/webwork-1383-image-2.tex create mode 100755 generated/webwork/images/webwork-1384-image-2.pdf create mode 100644 generated/webwork/images/webwork-1384-image-2.png create mode 100644 generated/webwork/images/webwork-1384-image-2.svg create mode 100755 generated/webwork/images/webwork-1384-image-2.tex create mode 100755 generated/webwork/images/webwork-1385-image-1.pdf create mode 100644 generated/webwork/images/webwork-1385-image-1.png create mode 100644 generated/webwork/images/webwork-1385-image-1.svg create mode 100755 generated/webwork/images/webwork-1385-image-1.tex create mode 100755 generated/webwork/images/webwork-1385-image-2.pdf create mode 100644 generated/webwork/images/webwork-1385-image-2.png create mode 100644 generated/webwork/images/webwork-1385-image-2.svg create mode 100755 generated/webwork/images/webwork-1385-image-2.tex create mode 100755 generated/webwork/images/webwork-1386-image-2.pdf create mode 100644 generated/webwork/images/webwork-1386-image-2.png create mode 100644 generated/webwork/images/webwork-1386-image-2.svg create mode 100755 generated/webwork/images/webwork-1386-image-2.tex create mode 100755 generated/webwork/images/webwork-1393-image-2.pdf create mode 100644 generated/webwork/images/webwork-1393-image-2.png create mode 100644 generated/webwork/images/webwork-1393-image-2.svg create mode 100755 generated/webwork/images/webwork-1393-image-2.tex create mode 100755 generated/webwork/images/webwork-1394-image-2.pdf create mode 100644 generated/webwork/images/webwork-1394-image-2.png create mode 100644 generated/webwork/images/webwork-1394-image-2.svg create mode 100755 generated/webwork/images/webwork-1394-image-2.tex create mode 100644 generated/webwork/images/webwork-1410-image-1.png create mode 100755 generated/webwork/images/webwork-1416-image-1.pdf create mode 100644 generated/webwork/images/webwork-1416-image-1.png create mode 100644 generated/webwork/images/webwork-1416-image-1.svg create mode 100755 generated/webwork/images/webwork-1416-image-1.tex create mode 100755 generated/webwork/images/webwork-1418-image-1.pdf create mode 100644 generated/webwork/images/webwork-1418-image-1.svg create mode 100755 generated/webwork/images/webwork-1418-image-1.tex create mode 100755 generated/webwork/images/webwork-1420-image-1.pdf create mode 100644 generated/webwork/images/webwork-1420-image-1.svg create mode 100755 generated/webwork/images/webwork-1420-image-1.tex create mode 100755 generated/webwork/images/webwork-1431-image-1.pdf create mode 100644 generated/webwork/images/webwork-1431-image-1.png create mode 100644 generated/webwork/images/webwork-1431-image-1.svg create mode 100755 generated/webwork/images/webwork-1431-image-1.tex create mode 100755 generated/webwork/images/webwork-1433-image-1.pdf create mode 100644 generated/webwork/images/webwork-1433-image-1.png create mode 100644 generated/webwork/images/webwork-1433-image-1.svg create mode 100755 generated/webwork/images/webwork-1433-image-1.tex create mode 100755 generated/webwork/images/webwork-1435-image-1.pdf create mode 100644 generated/webwork/images/webwork-1435-image-1.png create mode 100644 generated/webwork/images/webwork-1435-image-1.svg create mode 100755 generated/webwork/images/webwork-1435-image-1.tex create mode 100755 generated/webwork/images/webwork-1443-image-1.pdf create mode 100644 generated/webwork/images/webwork-1443-image-1.png create mode 100644 generated/webwork/images/webwork-1443-image-1.svg create mode 100755 generated/webwork/images/webwork-1443-image-1.tex create mode 100755 generated/webwork/images/webwork-1445-image-1.pdf create mode 100644 generated/webwork/images/webwork-1445-image-1.png create mode 100644 generated/webwork/images/webwork-1445-image-1.svg create mode 100755 generated/webwork/images/webwork-1445-image-1.tex create mode 100755 generated/webwork/images/webwork-1447-image-1.pdf create mode 100644 generated/webwork/images/webwork-1447-image-1.png create mode 100644 generated/webwork/images/webwork-1447-image-1.svg create mode 100755 generated/webwork/images/webwork-1447-image-1.tex create mode 100755 generated/webwork/images/webwork-1449-image-1.pdf create mode 100644 generated/webwork/images/webwork-1449-image-1.png create mode 100644 generated/webwork/images/webwork-1449-image-1.svg create mode 100755 generated/webwork/images/webwork-1449-image-1.tex create mode 100755 generated/webwork/images/webwork-1451-image-1.pdf create mode 100644 generated/webwork/images/webwork-1451-image-1.png create mode 100644 generated/webwork/images/webwork-1451-image-1.svg create mode 100755 generated/webwork/images/webwork-1451-image-1.tex create mode 100755 generated/webwork/images/webwork-1453-image-1.pdf create mode 100644 generated/webwork/images/webwork-1453-image-1.png create mode 100644 generated/webwork/images/webwork-1453-image-1.svg create mode 100755 generated/webwork/images/webwork-1453-image-1.tex create mode 100755 generated/webwork/images/webwork-1455-image-1.pdf create mode 100644 generated/webwork/images/webwork-1455-image-1.png create mode 100644 generated/webwork/images/webwork-1455-image-1.svg create mode 100755 generated/webwork/images/webwork-1455-image-1.tex create mode 100755 generated/webwork/images/webwork-1457-image-1.pdf create mode 100644 generated/webwork/images/webwork-1457-image-1.png create mode 100644 generated/webwork/images/webwork-1457-image-1.svg create mode 100755 generated/webwork/images/webwork-1457-image-1.tex create mode 100755 generated/webwork/images/webwork-1459-image-1.pdf create mode 100644 generated/webwork/images/webwork-1459-image-1.png create mode 100644 generated/webwork/images/webwork-1459-image-1.svg create mode 100755 generated/webwork/images/webwork-1459-image-1.tex create mode 100755 generated/webwork/images/webwork-1461-image-1.pdf create mode 100644 generated/webwork/images/webwork-1461-image-1.png create mode 100644 generated/webwork/images/webwork-1461-image-1.svg create mode 100755 generated/webwork/images/webwork-1461-image-1.tex create mode 100755 generated/webwork/images/webwork-1569-image-1.pdf create mode 100644 generated/webwork/images/webwork-1569-image-1.png create mode 100644 generated/webwork/images/webwork-1569-image-1.svg create mode 100755 generated/webwork/images/webwork-1569-image-1.tex create mode 100755 generated/webwork/images/webwork-1570-image-1.pdf create mode 100644 generated/webwork/images/webwork-1570-image-1.svg create mode 100755 generated/webwork/images/webwork-1570-image-1.tex create mode 100755 generated/webwork/images/webwork-1571-image-1.pdf create mode 100644 generated/webwork/images/webwork-1571-image-1.svg create mode 100755 generated/webwork/images/webwork-1571-image-1.tex create mode 100755 generated/webwork/images/webwork-1579-image-1.pdf create mode 100644 generated/webwork/images/webwork-1579-image-1.svg create mode 100755 generated/webwork/images/webwork-1579-image-1.tex create mode 100755 generated/webwork/images/webwork-1581-image-1.pdf create mode 100644 generated/webwork/images/webwork-1581-image-1.svg create mode 100755 generated/webwork/images/webwork-1581-image-1.tex create mode 100755 generated/webwork/images/webwork-1582-image-1.pdf create mode 100644 generated/webwork/images/webwork-1582-image-1.svg create mode 100755 generated/webwork/images/webwork-1582-image-1.tex create mode 100755 generated/webwork/images/webwork-1583-image-1.pdf create mode 100644 generated/webwork/images/webwork-1583-image-1.svg create mode 100755 generated/webwork/images/webwork-1583-image-1.tex create mode 100755 generated/webwork/images/webwork-1584-image-1.pdf create mode 100644 generated/webwork/images/webwork-1584-image-1.png create mode 100644 generated/webwork/images/webwork-1584-image-1.svg create mode 100755 generated/webwork/images/webwork-1584-image-1.tex create mode 100644 generated/webwork/images/webwork-1608-image-1.png create mode 100644 generated/webwork/images/webwork-1609-image-1.png create mode 100644 generated/webwork/images/webwork-1610-image-1.png create mode 100644 generated/webwork/images/webwork-1611-image-1.png create mode 100644 generated/webwork/images/webwork-1619-image-1.png create mode 100755 generated/webwork/images/webwork-1666-image-1.pdf create mode 100644 generated/webwork/images/webwork-1666-image-1.png create mode 100644 generated/webwork/images/webwork-1666-image-1.svg create mode 100755 generated/webwork/images/webwork-1666-image-1.tex create mode 100755 generated/webwork/images/webwork-1674-image-1.pdf create mode 100644 generated/webwork/images/webwork-1674-image-1.png create mode 100644 generated/webwork/images/webwork-1674-image-1.svg create mode 100755 generated/webwork/images/webwork-1674-image-1.tex create mode 100755 generated/webwork/images/webwork-1675-image-1.pdf create mode 100644 generated/webwork/images/webwork-1675-image-1.png create mode 100644 generated/webwork/images/webwork-1675-image-1.svg create mode 100755 generated/webwork/images/webwork-1675-image-1.tex create mode 100644 generated/webwork/images/webwork-1676-image-1.png create mode 100644 generated/webwork/images/webwork-1679-image-1.png create mode 100644 generated/webwork/images/webwork-1756-image-1.png create mode 100644 generated/webwork/images/webwork-1763-image-1.png create mode 100644 generated/webwork/images/webwork-1868-image-1.png create mode 100644 generated/webwork/images/webwork-1869-image-1.png create mode 100644 generated/webwork/images/webwork-1878-image-1.png create mode 100644 generated/webwork/images/webwork-1879-image-1.png create mode 100644 generated/webwork/images/webwork-1880-image-1.png create mode 100644 generated/webwork/images/webwork-1881-image-1.png create mode 100644 generated/webwork/images/webwork-1896-image-1.png create mode 100644 generated/webwork/images/webwork-1897-image-1.png create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-image-1.pdf create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-image-1.svg create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-image-1.tex create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-neg-image-1.pdf create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-neg-image-1.png create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-neg-image-1.svg create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-neg-image-1.tex create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-image-1.pdf create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-image-1.png create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-image-1.svg create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-image-1.tex create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-precise-image-1.pdf create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-precise-image-1.png create mode 100644 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-precise-image-1.svg create mode 100755 generated/webwork/images/webwork-determine-point-slope-form-from-graph-q1-large-y-precise-image-1.tex create mode 100644 knowl/xref/definition-cartesian-coordinate-system.html create mode 100644 knowl/xref/definition-point-slope-form.html create mode 100644 knowl/xref/definition-slope-intercept-form.html create mode 100644 knowl/xref/definition-slope.html create mode 100644 knowl/xref/fact-parallel-lines.html create mode 100644 knowl/xref/fact-perpendicular-lines.html diff --git a/.mapping.json b/.mapping.json index 5923d83c7..773e7ede9 100644 --- a/.mapping.json +++ b/.mapping.json @@ -1 +1 @@ -{"src/orcca.ptx": ["orcca", "part-linear-equations-and-lines"], "src/frontmatter.ptx": ["frontmatter"], "src/chapter-variables-expressions-equations-and-inequalities.ptx": ["chapter-variables-expressions-equations-and-inequalities"], "src/section-variables-and-evaluating-expressions.ptx": ["section-variables-and-evaluating-expressions", "video-variables-and-evaluating-expressions"], "src/section-combining-like-terms.ptx": ["section-combining-like-terms"], "src/section-comparison-symbols-and-notation-for-intervals.ptx": ["section-comparison-symbols-and-notation-for-intervals"], "src/section-equations-inequalities-and-solutions.ptx": ["section-equations-inequalities-and-solutions"], "src/section-solving-one-step-equations.ptx": ["section-solving-one-step-equations"], "src/section-solving-one-step-inequalities.ptx": ["section-solving-one-step-inequalities"], "src/section-algebraic-properties-and-simplifying-expressions.ptx": ["section-algebraic-properties-and-simplifying-expressions"], "src/section-modeling-with-equations-and-inequalities.ptx": ["section-modeling-with-equations-and-inequalities"], "src/review-variables-expressions-and-equations.ptx": ["review-variables-expressions-and-equations"], "src/chapter-linear-equations-and-inequalities.ptx": ["chapter-linear-equations-and-inequalities"], "src/section-solving-multistep-linear-equations.ptx": ["section-solving-multistep-linear-equations"], "src/section-solving-multistep-linear-inequalities.ptx": ["section-solving-multistep-linear-inequalities"], "src/section-equations-and-inequalities-with-fractions.ptx": ["section-equations-and-inequalities-with-fractions"], "src/section-special-solution-sets.ptx": ["section-special-solution-sets"], "src/section-isolating-a-linear-variable.ptx": ["section-isolating-a-linear-variable"], "src/review-linear-equations-and-inequalities.ptx": ["review-linear-equations-and-inequalities"], "src/chapter-graphing-lines.ptx": ["chapter-graphing-lines"], "src/section-cartesian-coordinates.ptx": ["section-cartesian-coordinates"], "src/section-graphing-equations.ptx": ["section-graphing-equations"], "src/section-exploring-two-variable-data-and-rate-of-change.ptx": ["section-exploring-two-variable-data-and-rate-of-change"], "src/section-slope.ptx": ["section-slope"], "src/section-slope-intercept-form.ptx": ["section-slope-intercept-form"], "src/section-point-slope-form.ptx": ["section-point-slope-form"], "src/section-standard-form.ptx": ["section-standard-form"], "src/section-horizontal-vertical-parallel-and-perpendicular-lines.ptx": ["section-horizontal-vertical-parallel-and-perpendicular-lines"], "src/section-summary-of-graphing-lines.ptx": ["section-summary-of-graphing-lines"], "src/chapter-systems-of-two-linear-equations.ptx": ["chapter-systems-of-two-linear-equations"], "src/section-solving-a-system-by-graphing.ptx": ["section-solving-a-system-by-graphing"], "src/section-substitution.ptx": ["section-substitution"], "src/section-elimination.ptx": ["section-elimination"], "src/review-systems-of-linear-equations.ptx": ["review-systems-of-linear-equations"], "src/backmatter.ptx": ["backmatter"], "src/appendix-basic-math-review.ptx": ["appendix-basic-math-review"], "src/section-arithmetic-with-negative-numbers.ptx": ["section-arithmetic-with-negative-numbers"], "src/section-fractions-and-fraction-arithmetic.ptx": ["section-fractions-and-fraction-arithmetic"], "src/section-absolute-value-and-square-root.ptx": ["section-absolute-value-and-square-root"], "src/section-percentages.ptx": ["section-percentages"], "src/section-order-of-operations.ptx": ["section-order-of-operations"], "src/section-set-notation-and-types-of-numbers.ptx": ["section-set-notation-and-types-of-numbers"], "src/appendix-unit-conversions.ptx": ["appendix-unit-conversions"], "src/appendix-ccogs.ptx": ["appendix-ccogs", "ccog-mth60"]} \ No newline at end of file +{"src/orcca.ptx": ["orcca", "part-linear-equations-and-lines"], "src/frontmatter.ptx": ["frontmatter"], "src/chapter-variables-expressions-equations-and-inequalities.ptx": ["chapter-variables-expressions-equations-and-inequalities"], "src/section-variables-and-evaluating-expressions.ptx": ["section-variables-and-evaluating-expressions", "video-variables-and-evaluating-expressions"], "src/section-combining-like-terms.ptx": ["section-combining-like-terms"], "src/section-comparison-symbols-and-notation-for-intervals.ptx": ["section-comparison-symbols-and-notation-for-intervals"], "src/section-equations-inequalities-and-solutions.ptx": ["section-equations-inequalities-and-solutions"], "src/section-solving-one-step-equations.ptx": ["section-solving-one-step-equations"], "src/section-solving-one-step-inequalities.ptx": ["section-solving-one-step-inequalities"], "src/section-algebraic-properties-and-simplifying-expressions.ptx": ["section-algebraic-properties-and-simplifying-expressions"], "src/section-modeling-with-equations-and-inequalities.ptx": ["section-modeling-with-equations-and-inequalities"], "src/review-variables-expressions-and-equations.ptx": ["review-variables-expressions-and-equations"], "src/chapter-linear-equations-and-inequalities.ptx": ["chapter-linear-equations-and-inequalities"], "src/section-solving-multistep-linear-equations.ptx": ["section-solving-multistep-linear-equations"], "src/section-solving-multistep-linear-inequalities.ptx": ["section-solving-multistep-linear-inequalities"], "src/section-equations-and-inequalities-with-fractions.ptx": ["section-equations-and-inequalities-with-fractions"], "src/section-special-solution-sets.ptx": ["section-special-solution-sets"], "src/section-isolating-a-linear-variable.ptx": ["section-isolating-a-linear-variable"], "src/review-linear-equations-and-inequalities.ptx": ["review-linear-equations-and-inequalities"], "src/chapter-graphing-lines.ptx": ["chapter-graphing-lines"], "src/section-cartesian-coordinates.ptx": ["section-cartesian-coordinates"], "src/section-graphing-equations.ptx": ["section-graphing-equations"], "src/section-exploring-two-variable-data-and-rate-of-change.ptx": ["section-exploring-two-variable-data-and-rate-of-change"], "src/section-slope.ptx": ["section-slope"], "src/section-slope-intercept-form.ptx": ["section-slope-intercept-form"], "src/section-point-slope-form.ptx": ["section-point-slope-form"], "src/section-standard-form.ptx": ["section-standard-form"], "src/section-horizontal-vertical-parallel-and-perpendicular-lines.ptx": ["section-horizontal-vertical-parallel-and-perpendicular-lines"], "src/section-summary-of-graphing-lines.ptx": ["section-summary-of-graphing-lines"], "src/review-graphing-lines.ptx": ["review-graphing-lines"], "src/chapter-systems-of-two-linear-equations.ptx": ["chapter-systems-of-two-linear-equations"], "src/section-solving-a-system-by-graphing.ptx": ["section-solving-a-system-by-graphing"], "src/section-substitution.ptx": ["section-substitution"], "src/section-elimination.ptx": ["section-elimination"], "src/review-systems-of-linear-equations.ptx": ["review-systems-of-linear-equations"], "src/backmatter.ptx": ["backmatter"], "src/appendix-basic-math-review.ptx": ["appendix-basic-math-review"], "src/section-arithmetic-with-negative-numbers.ptx": ["section-arithmetic-with-negative-numbers"], "src/section-fractions-and-fraction-arithmetic.ptx": ["section-fractions-and-fraction-arithmetic"], "src/section-absolute-value-and-square-root.ptx": ["section-absolute-value-and-square-root"], "src/section-percentages.ptx": ["section-percentages"], "src/section-order-of-operations.ptx": ["section-order-of-operations"], "src/section-set-notation-and-types-of-numbers.ptx": ["section-set-notation-and-types-of-numbers"], "src/appendix-unit-conversions.ptx": ["appendix-unit-conversions"], "src/appendix-ccogs.ptx": ["appendix-ccogs", "ccog-mth60"]} \ No newline at end of file diff --git a/appendix-basic-math-review.html b/appendix-basic-math-review.html index 7f106816d..e00d80fce 100644 --- a/appendix-basic-math-review.html +++ b/appendix-basic-math-review.html @@ -420,7 +420,7 @@

Search Results:

@@ -455,6 +455,20 @@

Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/appendix-ccogs.html b/appendix-ccogs.html index 856647d4d..53847be18 100644 --- a/appendix-ccogs.html +++ b/appendix-ccogs.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/appendix-unit-conversions.html b/appendix-unit-conversions.html index 6bca87a11..4179b3d0e 100644 --- a/appendix-unit-conversions.html +++ b/appendix-unit-conversions.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/backmatter-4.html b/backmatter-4.html index 6f5314c35..61de25e00 100644 --- a/backmatter-4.html +++ b/backmatter-4.html @@ -108,6 +108,37 @@ + + + + + + + + Skip to main content
    @@ -188,15 +219,13 @@

    Search Results:

    @@ -217,25 +246,72 @@

    Search Results:

    Appendix C Answers to Exercises

    -

    -I Linear Equations and Lines
    1 Graphing Lines
    1.1 Slope-Intercept Form
    1.1.7 Exercises +

    +I Linear Equations and Lines
    1 Graphing Lines
    1.1 Point-Slope Form
    1.1.5 Exercises

    -

    Applications

    -
    1.1.7.1.
    -
    Answer.
    \(14\)
    1.1.7.3.
    -
    Answer.
    \(370\)
    1.1.7.5.
    -
    Answer.
    \(45\)
    1.1.7.7.
    +

    Skills Practice

    +
    +
    1.1.5.1.
    -
    Answer 1.
    \(y = 0.06x+19\)
    Answer 2.
    \(\$29.80\)
    Answer 3.
    \(440\)
    -
    1.1.7.9.
    +
    Answer 1.
    \(6\)
    Answer 2.
    \(\left(4,9\right)\)
    +
    1.1.5.3.
    -
    Answer 1.
    \(y = 0.28x+12.6\)
    Answer 2.
    \(20.72\)
    Answer 3.
    \(53\)
    -
    1.1.7.11.
    +
    Answer 1.
    \(8\)
    Answer 2.
    \(\left(-6,7\right)\)
    +
    1.1.5.5.
    -
    Answer 1.
    \(y = -23000x+802000\)
    Answer 2.
    \(\$572{,}000\)
    Answer 3.
    \(2034\)
    -
    1.1.7.13.
    +
    Answer 1.
    \(9\)
    Answer 2.
    \(\left(-8,-5\right)\)
    +
    1.1.5.7.
    -
    Answer 1.
    \(y = -8.3x+356.9\)
    Answer 2.
    \(58.1\)
    Answer 3.
    \(43\)
    +
    Answer 1.
    \(3\)
    Answer 2.
    \(\left(5,-6\right)\)
    +
    + +
    +
    1.1.5.9.
    +
    Answer.
    \(y = 5\mathopen{}\left(x-7\right)+3\)
    1.1.5.11.
    +
    Answer.
    \(y = 7\mathopen{}\left(x+9\right)-2\)
    1.1.5.13.
    +
    Answer.
    \(y = 7.2\mathopen{}\left(x+8\right)+5\)
    +
    +
    +
    1.1.5.15.
    +
    Answer.
    \(y = 2\mathopen{}\left(x-6\right)+8\hbox{ or }y = 2\mathopen{}\left(x-9\right)+14\)
    1.1.5.17.
    +
    Answer.
    \(y = -4\mathopen{}\left(x-7\right)+2\hbox{ or }y = -4\mathopen{}\left(x+5\right)+50\)
    1.1.5.19.
    +
    Answer.
    \(y = \frac{-8}{9}\mathopen{}\left(x-1\right)-8\hbox{ or }y = \frac{-8}{9}\mathopen{}\left(x-55\right)-56\)
    1.1.5.21.
    +
    Answer.
    \(y = \frac{2}{3}\mathopen{}\left(x-5\right)-2\hbox{ or }y = \frac{2}{3}\mathopen{}\left(x+10\right)-12\)
    +
    +
    +
    1.1.5.23.
    +
    Answer.
    \(y = 9x-61\)
    1.1.5.25.
    +
    Answer.
    \(y = -3x+3\)
    1.1.5.27.
    +
    Answer.
    \(y = \frac{7}{3}x-\frac{23}{3}\)
    1.1.5.29.
    +
    Answer.
    \(y = -0.7x+2.62\)
    +
    +
    +
    1.1.5.31.
    +
    Answer.
    \(y = \frac{9}{2}\mathopen{}\left(x-5\right)-2\hbox{ or }y = \frac{9}{2}\mathopen{}\left(x-7\right)+7\)
    1.1.5.33.
    +
    Answer.
    \(y = \frac{-2}{5}\mathopen{}\left(x+6\right)-1\hbox{, }y = \frac{-2}{5}\mathopen{}\left(x+1\right)-3\hbox{, or }y = \frac{-2}{5}\mathopen{}\left(x-4\right)-5\)
    1.1.5.35.
    +
    Answer.
    \(y = \frac{-75}{2}\mathopen{}\left(x-1\right)+250\hbox{, }y = \frac{-75}{2}\mathopen{}\left(x-5\right)+100\hbox{, or }y = \frac{-75}{2}\mathopen{}\left(x-9\right)-50\)
    1.1.5.37.
    +
    Answer.
    \(y = \frac{-19}{3}\mathopen{}\left(x-5\right)+527\hbox{ or }y = \frac{-19}{3}\mathopen{}\left(x-8\right)+508\)
    +
    +
    +
    1.1.5.39.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(5,2\right), \left(6,6\right)\right\}\)
    1.1.5.41.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(5,-3\right), \left(6,1\right)\right\}\)
    1.1.5.43.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(8,4\right), \left(13,6\right)\right\}\)
    1.1.5.45.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(-2,4\right), \left(2,-3\right)\right\}\)
    1.1.5.47.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(5,2\right), \left(15,5\right)\right\}\)
    +

    Applications

    +
    1.1.5.49.
    +
    +
    Answer 1.
    \(y = 0.07\mathopen{}\left(x-290\right)+37.3\)
    Answer 2.
    \(\$26.10\)
    Answer 3.
    \(430\)
    +
    1.1.5.51.
    +
    +
    Answer 1.
    \(y = 0.11\mathopen{}\left(x-14\right)+20.54\)
    Answer 2.
    \(21.86\)
    Answer 3.
    \(52\)
    +
    1.1.5.53.
    +
    +
    Answer 1.
    \(y = -29000\mathopen{}\left(x-3\right)+630000\hbox{ or }y = -29000\mathopen{}\left(x-6\right)+543000\)
    Answer 2.
    \(\$456{,}000\)
    Answer 3.
    \(2024\)
    +
    1.1.5.55.
    +
    +
    Answer 1.
    \(y = -6.4\mathopen{}\left(x-7\right)+243.2\hbox{ or }y = -6.4\mathopen{}\left(x-18\right)+172.8\)
    Answer 2.
    \(89.6\)
    Answer 3.
    \(45\)
    PrevTopNext diff --git a/backmatter-5.html b/backmatter-5.html index 83adaab20..17e2ea46f 100644 --- a/backmatter-5.html +++ b/backmatter-5.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -1743,125 +1757,107 @@

    Exercise

    3.6 Point-Slope Form
    3.6.5 Exercises

    -

    Review and Warmup

    -
    3.6.5.1.
    -
    Answer.
    \(76\)
    3.6.5.3.
    -
    Answer.
    \(-{\frac{1}{23}}\)

    Skills Practice

    +

    Skills Practice

    -
    3.6.5.5.
    -
    -
    Answer 1.
    \(5\)
    Answer 2.
    \(\left(5,28\right)\)
    -
    3.6.5.7.
    -
    -
    Answer 1.
    \(-2\)
    Answer 2.
    \(\left(-2,5\right)\)
    -
    3.6.5.9.
    -
    -
    Answer 1.
    \({\frac{3}{2}}\)
    Answer 2.
    \(\left(-2,-2\right)\)
    -
    3.6.5.11.
    -
    -
    Answer 1.
    \({\frac{5}{4}}\)
    Answer 2.
    \(\left(12,10\right)\)
    -
    3.6.5.13.
    -
    -
    Answer 1.
    \(y = 4\mathopen{}\left(x-2\right)+15\)
    Answer 2.
    \(y = 4\mathopen{}\left(x-3\right)+19\)
    -
    3.6.5.15.
    -
    -
    Answer 1.
    \(y = -5\mathopen{}\left(x--3\right)+25\)
    Answer 2.
    \(y = -5\mathopen{}\left(x-5\right)+-15\)
    -
    3.6.5.17.
    -
    -
    Answer 1.
    \(y = \frac{2}{3}\mathopen{}\left(x+3\right)+-11\)
    Answer 2.
    \(y = \frac{2}{3}\mathopen{}\left(x-3\right)+-7\)
    -
    3.6.5.19.
    -
    -
    Answer 1.
    \(y = 3\mathopen{}\left(x-5\right)+18\)
    Answer 2.
    \(y = 3x+3\)
    -
    3.6.5.21.
    -
    -
    Answer 1.
    \(y = -3\mathopen{}\left(x-2\right)+-2\)
    Answer 2.
    \(y = -3x+4\)
    -
    3.6.5.23.
    +
    3.6.5.1.
    -
    Answer 1.
    \(y = 1\mathopen{}\left(x--4\right)\)
    Answer 2.
    \(y = 1x+4\)
    -
    3.6.5.25.
    +
    Answer 1.
    \(6\)
    Answer 2.
    \(\left(5,8\right)\)
    +
    3.6.5.3.
    -
    Answer 1.
    \(y = -1\mathopen{}\left(x--1\right)+-4\)
    Answer 2.
    \(y = -1x+-5\)
    -
    3.6.5.27.
    +
    Answer 1.
    \(-7\)
    Answer 2.
    \(\left(-8,5\right)\)
    +
    3.6.5.5.
    -
    Answer 1.
    \(y = {\frac{3}{8}}\mathopen{}\left(x-8\right)\)
    Answer 2.
    \(y = {\frac{3}{8}}x+-3\)
    -
    3.6.5.29.
    +
    Answer 1.
    \(9\)
    Answer 2.
    \(\left(-2,-5\right)\)
    +
    3.6.5.7.
    -
    Answer 1.
    \(y = -{\frac{5}{2}}\mathopen{}\left(x--6\right)+10\)
    Answer 2.
    \(y = -{\frac{5}{2}}x+-5\)
    +
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(5,-4\right)\)
    +
    3.6.5.9.
    +
    Answer.
    \(y = 5\mathopen{}\left(x-7\right)+2\)
    3.6.5.11.
    +
    Answer.
    \(y = 7\mathopen{}\left(x+9\right)-8\)
    3.6.5.13.
    +
    Answer.
    \(y = 7.3\mathopen{}\left(x+8\right)+3\)
    +
    +
    +
    3.6.5.15.
    +
    Answer.
    \(y = 2\mathopen{}\left(x-7\right)+6\hbox{ or }y = 2\mathopen{}\left(x-8\right)+8\)
    3.6.5.17.
    +
    Answer.
    \(y = -5\mathopen{}\left(x-8\right)-2\hbox{ or }y = -5\mathopen{}\left(x+8\right)+78\)
    3.6.5.19.
    +
    Answer.
    \(y = \frac{-8}{3}\mathopen{}\left(x-1\right)-6\hbox{ or }y = \frac{-8}{3}\mathopen{}\left(x-4\right)-14\)
    3.6.5.21.
    +
    Answer.
    \(y = \frac{1}{8}\mathopen{}\left(x-5\right)-1\hbox{ or }y = \frac{1}{8}\mathopen{}\left(x-77\right)+8\)
    +
    +
    +
    3.6.5.23.
    +
    Answer.
    \(y = 9x-55\)
    3.6.5.25.
    +
    Answer.
    \(y = -3x+2\)
    3.6.5.27.
    +
    Answer.
    \(y = \frac{6}{7}x-\frac{2}{7}\)
    3.6.5.29.
    +
    Answer.
    \(y = -0.2x+0.24\)
    +
    +
    3.6.5.31.
    -
    Answer.
    \(y = 4x-2\)
    3.6.5.33.
    -
    Answer.
    \(y = -2x+2\)
    3.6.5.35.
    -
    Answer.
    \(y = \frac{2}{9}x+1\)
    3.6.5.37.
    -
    Answer.
    \(y = -0.8x+3\)
    +
    Answer.
    \(y = \frac{8}{9}\mathopen{}\left(x+5\right)-2\hbox{ or }y = \frac{8}{9}\mathopen{}\left(x-4\right)+6\)
    3.6.5.33.
    +
    Answer.
    \(y = \frac{-9}{5}\mathopen{}\left(x+7\right)+5\hbox{ or }y = \frac{-9}{5}\mathopen{}\left(x+2\right)-4\)
    3.6.5.35.
    +
    Answer.
    \(y = \frac{-75}{2}\mathopen{}\left(x-2\right)+350\hbox{, }y = \frac{-75}{2}\mathopen{}\left(x-6\right)+200\hbox{, or }y = \frac{-75}{2}\mathopen{}\left(x-10\right)+50\)
    3.6.5.37.
    +
    Answer.
    \(y = \frac{-67}{2}\mathopen{}\left(x-5\right)+503\hbox{ or }y = \frac{-67}{2}\mathopen{}\left(x-9\right)+369\)
    -
    +
    3.6.5.39.
    -
    Answer.
    \(y = {\frac{1}{5}}\mathopen{}\left(x-3\right)+3\)
    3.6.5.41.
    -
    Answer.
    \(y = -{\frac{1}{2}}\mathopen{}\left(x-1\right)+2\)
    3.6.5.43.
    -
    Answer.
    \(y = {\frac{400}{3}}\mathopen{}\left(x-2\right)+500\)
    3.6.5.45.
    -
    Answer.
    \(y = {\frac{100}{3}}\mathopen{}\left(x-1\right)+100\)
    3.6.5.47.
    -
    Answer.
    \(y = {\frac{492}{403}}\mathopen{}\left(x-150\right)+364\hbox{ or }y = {\frac{492}{403}}\mathopen{}\left(x-553\right)+856\)
    3.6.5.49.
    -
    Answer.
    \(y = -{\frac{116}{491}}\mathopen{}\left(x-641\right)+336\hbox{ or }y = -{\frac{116}{491}}\mathopen{}\left(x-1132\right)+220\)
    -
    -
    -
    3.6.5.51.
    -
    Answer.
    3.6.5.53.
    -
    Answer.
    3.6.5.55.
    -
    Answer.
    3.6.5.57.
    -
    Answer.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(4,2\right), \left(5,7\right)\right\}\)
    3.6.5.41.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(5,-3\right), \left(6,-1\right)\right\}\)
    3.6.5.43.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(-8,-4\right), \left(-5,-2\right)\right\}\)
    3.6.5.45.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(8,-9\right), \left(15,-18\right)\right\}\)
    3.6.5.47.
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(6,-9\right), \left(16,-15\right)\right\}\)

    Applications

    -
    3.6.5.59.
    +
    3.6.5.49.
    -
    Answer 1.
    \(y = 0.07\mathopen{}\left(x-230\right)+30.1\)
    Answer 2.
    \(\$22.40\)
    Answer 3.
    \(430\)
    -
    3.6.5.61.
    +
    Answer 1.
    \(y = 0.07\mathopen{}\left(x-290\right)+37.3\)
    Answer 2.
    \(\$26.10\)
    Answer 3.
    \(480\)
    +
    3.6.5.51.
    -
    Answer 1.
    \(y = 0.13\mathopen{}\left(x-10\right)+18\)
    Answer 2.
    \(19.95\)
    Answer 3.
    \(52\)
    -
    3.6.5.63.
    +
    Answer 1.
    \(y = 0.11\mathopen{}\left(x-13\right)+12.93\)
    Answer 2.
    \(14.36\)
    Answer 3.
    \(58\)
    +
    3.6.5.53.
    -
    Answer 1.
    \(y = -32000\mathopen{}\left(x-4\right)+479000\hbox{ or }y = -32000\mathopen{}\left(x-7\right)+383000\)
    Answer 2.
    \(\$319{,}000\)
    Answer 3.
    \(2018\)
    -
    3.6.5.65.
    +
    Answer 1.
    \(y = -28000\mathopen{}\left(x-3\right)+663000\hbox{ or }y = -28000\mathopen{}\left(x-5\right)+607000\)
    Answer 2.
    \(\$495{,}000\)
    Answer 3.
    \(2026\)
    +
    3.6.5.55.
    -
    Answer 1.
    \(y = -3.9\mathopen{}\left(x-9\right)+144.3\hbox{ or }y = -3.9\mathopen{}\left(x-13\right)+128.7\)
    Answer 2.
    \(58.5\)
    Answer 3.
    \(46\)
    +
    Answer 1.
    \(y = -7\mathopen{}\left(x-6\right)+273\hbox{ or }y = -7\mathopen{}\left(x-15\right)+210\)
    Answer 2.
    \(98\)
    Answer 3.
    \(45\)

    3.7 Standard Form
    3.7.5 Exercises

    Review and Warmup

    3.7.5.1.
    -
    Answer.
    \(y = 2x-13\)
    3.7.5.3.
    -
    Answer.
    \(y = {\frac{3}{4}}x-18\)
    3.7.5.5.
    -
    Answer.
    \(y = {\frac{24}{55}}x-9\)

    Skills Practice

    +
    Answer.
    \(y = -9x-2\)
    3.7.5.3.
    +
    Answer.
    \(y = -{\frac{8}{3}}x-18\)
    3.7.5.5.
    +
    Answer.
    \(y = {\frac{65}{47}}x+-{\frac{481}{47}}\)

    Skills Practice

    3.7.5.7.
    -
    Answer 1.
    \(8\)
    Answer 2.
    \(\left(0,7\right)\)
    +
    Answer 1.
    \(7\)
    Answer 2.
    \(\left(0,-6\right)\)
    3.7.5.9.
    -
    Answer 1.
    \(-1\)
    Answer 2.
    \(\left(0,3\right)\)
    +
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(0,-2\right)\)
    3.7.5.11.
    -
    Answer 1.
    \(-{\frac{1}{2}}\)
    Answer 2.
    \(\left(0,1\right)\)
    +
    Answer 1.
    \(-{\frac{1}{2}}\)
    Answer 2.
    \(\left(0,3\right)\)
    3.7.5.13.
    -
    Answer 1.
    \({\frac{7}{4}}\)
    Answer 2.
    \(\left(0,1\right)\)
    +
    Answer 1.
    \({\frac{2}{3}}\)
    Answer 2.
    \(\left(0,-5\right)\)
    3.7.5.15.
    -
    Answer 1.
    \({\frac{4}{3}}\)
    Answer 2.
    \(\left(0,0\right)\)
    +
    Answer 1.
    \({\frac{3}{4}}\)
    Answer 2.
    \(\left(0,0\right)\)
    3.7.5.17.
    -
    Answer 1.
    \(-{\frac{5}{3}}\)
    Answer 2.
    \(\left(0,\frac{5}{6}\right)\)
    +
    Answer 1.
    \(-{\frac{5}{2}}\)
    Answer 2.
    \(\left(0,\frac{5}{8}\right)\)
    3.7.5.19.
    -
    Answer.
    \(2x-y=-3\)
    3.7.5.21.
    -
    Answer.
    \(4x-9y=-36\)
    +
    Answer.
    \(9x-y=9\)
    3.7.5.21.
    +
    Answer.
    \(3x-4y=12\)
    3.7.5.23.
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(10\)
    Answer 3.
    \(\left(0,10\right)\)
    Answer 4.
    \(8\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(8,0\right)\)
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(5\)
    Answer 3.
    \(\left(0,5\right)\)
    Answer 4.
    \(6\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(6,0\right)\)
    3.7.5.25.
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(12\)
    Answer 3.
    \(\left(0,12\right)\)
    Answer 4.
    \(-10\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-10,0\right)\)
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(18\)
    Answer 3.
    \(\left(0,18\right)\)
    Answer 4.
    \(-21\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-21,0\right)\)
    @@ -1896,12 +1892,12 @@

    Exercise
    3.7.5.43.
    Answer.
    \(\text{C}\)
    3.7.5.45.
    -
    Answer.
    \(\text{B}\)
    3.7.5.47.
    -
    Answer.
    \(\text{D}\)
    +
    Answer.
    \(\text{D}\)
    3.7.5.47.
    +
    Answer.
    \(\text{B}\)

    Challenge

    3.7.5.49.
    -
    Answer 1.
    \(14\)
    Answer 2.
    \(6\)
    Answer 3.
    \(7\)
    Answer 4.
    \(6\)
    Answer 5.
    \(14\)
    Answer 6.
    \(7\)
    +
    Answer 1.
    \(9\)
    Answer 2.
    \(6\)
    Answer 3.
    \(7\)
    Answer 4.
    \(6\)
    Answer 5.
    \(9\)
    Answer 6.
    \(7\)

    3.8 Horizontal, Vertical, Parallel, and Perpendicular Lines
    3.8.5 Exercises

    @@ -1909,7 +1905,7 @@

    Exercise
    3.8.5.1.
    -
    Answer 1.
    \(\text{DNE}\)
    Answer 2.
    \(0\)
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(\text{DNE}\)
    3.8.5.3.
    Answer.
    \(0\)
    3.8.5.5.
    Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    @@ -1917,29 +1913,29 @@

    Exercise
    3.8.5.11.
    -
    Answer 1.
    \(4\)
    Answer 2.
    \(\left(-2,4\right)\)
    Answer 3.
    \(4\)
    Answer 4.
    \(\left(-1,4\right)\)
    Answer 5.
    \(4\)
    Answer 6.
    \(\left(0,4\right)\)
    Answer 7.
    \(4\)
    Answer 8.
    \(\left(1,4\right)\)
    Answer 9.
    \(4\)
    Answer 10.
    \(\left(2,4\right)\)
    +
    Answer 1.
    \(3\)
    Answer 2.
    \(\left(-2,3\right)\)
    Answer 3.
    \(3\)
    Answer 4.
    \(\left(-1,3\right)\)
    Answer 5.
    \(3\)
    Answer 6.
    \(\left(0,3\right)\)
    Answer 7.
    \(3\)
    Answer 8.
    \(\left(1,3\right)\)
    Answer 9.
    \(3\)
    Answer 10.
    \(\left(2,3\right)\)
    3.8.5.13.
    -
    Answer 1.
    \(-5\)
    Answer 2.
    \(\left(-5,-2\right)\)
    Answer 3.
    \(-5\)
    Answer 4.
    \(\left(-5,-1\right)\)
    Answer 5.
    \(-5\)
    Answer 6.
    \(\left(-5,0\right)\)
    Answer 7.
    \(-5\)
    Answer 8.
    \(\left(-5,1\right)\)
    Answer 9.
    \(-5\)
    Answer 10.
    \(\left(-5,2\right)\)
    +
    Answer 1.
    \(-6\)
    Answer 2.
    \(\left(-6,-2\right)\)
    Answer 3.
    \(-6\)
    Answer 4.
    \(\left(-6,-1\right)\)
    Answer 5.
    \(-6\)
    Answer 6.
    \(\left(-6,0\right)\)
    Answer 7.
    \(-6\)
    Answer 8.
    \(\left(-6,1\right)\)
    Answer 9.
    \(-6\)
    Answer 10.
    \(\left(-6,2\right)\)
    3.8.5.15.
    -
    Answer.
    \(y = 3\)
    3.8.5.17.
    -
    Answer.
    \(x = -4\)
    +
    Answer.
    \(y = 2\)
    3.8.5.17.
    +
    Answer.
    \(x = 4\)
    3.8.5.19.
    -
    Answer.
    \(y = -5\)
    3.8.5.21.
    -
    Answer.
    \(x = 0\)
    +
    Answer.
    \(y = -7\)
    3.8.5.21.
    +
    Answer.
    \(x = -3\)
    3.8.5.23.
    -
    Answer 1.
    \(\text{NONE}\)
    Answer 2.
    \(\text{NONE}\)
    Answer 3.
    \(\text{NONE}\)
    Answer 4.
    \(4\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(4,0\right)\)
    +
    Answer 1.
    \(\text{NONE}\)
    Answer 2.
    \(\text{NONE}\)
    Answer 3.
    \(\text{NONE}\)
    Answer 4.
    \(2\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(2,0\right)\)
    3.8.5.25.
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(9\)
    Answer 3.
    \(\left(0,9\right)\)
    Answer 4.
    \(\text{NONE}\)
    Answer 5.
    \(\text{NONE}\)
    Answer 6.
    \(\text{NONE}\)
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(7\)
    Answer 3.
    \(\left(0,7\right)\)
    Answer 4.
    \(\text{NONE}\)
    Answer 5.
    \(\text{NONE}\)
    Answer 6.
    \(\text{NONE}\)
    @@ -1956,17 +1952,17 @@

    Exercise

    3.8.5.39.
    -
    Answer.
    \(y = 8\)
    3.8.5.41.
    -
    Answer.
    \(x = -6\)
    3.8.5.43.
    +
    Answer.
    \(y = -5\)
    3.8.5.41.
    +
    Answer.
    \(x = 1\)
    3.8.5.43.
    -
    Answer 1.
    \(y = -5\mathopen{}\left(x+5\right)+16\)
    Answer 2.
    \(y = -5x-9\)
    +
    Answer 1.
    \(y = -2\mathopen{}\left(x+1\right)+5\)
    Answer 2.
    \(y = -2x+3\)
    3.8.5.45.
    -
    Answer 1.
    \(y = -{\frac{7}{8}}\mathopen{}\left(x+3\right)+{\frac{61}{8}}\)
    Answer 2.
    \(y = -{\frac{7}{8}}x+5\)
    +
    Answer 1.
    \(y = -{\frac{6}{5}}\mathopen{}\left(x+3\right)+{\frac{38}{5}}\)
    Answer 2.
    \(y = -{\frac{6}{5}}x+4\)
    3.8.5.47.
    -
    Answer.
    \(y = -x+3\)
    3.8.5.49.
    +
    Answer.
    \(y = -x-2\)
    3.8.5.49.
    -
    Answer 1.
    \(y = -{\frac{5}{3}}\mathopen{}\left(x+3\right)+10\)
    Answer 2.
    \(y = -{\frac{5}{3}}x+5\)
    +
    Answer 1.
    \(y = -{\frac{3}{2}}\mathopen{}\left(x-1\right)+-{\frac{9}{2}}\)
    Answer 2.
    \(y = -{\frac{3}{2}}x-3\)

    3.9 Summary of Graphing Lines
    3.9.5 Exercises @@ -2095,12 +2091,64 @@

    Exercise
    Answer.

    3.9.5.33.

    Answer.

    +

    +3.10 Graphing Lines Chapter Review
    3.10.9 Exercises +

    +
    +

    3.10.9.1.

    +
    Answer.
    a Cartesian graph with the points plotted;(8,2) is 8 units to the right and up 2;(5,5) is 5 to the right and up 5;(-3,0) is down 3 units on the y-axis;(0,14/3)is 14/3 or 4 2/3 units up on the y-axis;(3,-2.5) is 3 units to the right and down 2.5;(-5,7) is left 5 units and up 7

    3.10.9.5.

    +
    Answer.
    \(y = 2x-4\)
    +
    +
    +

    3.10.9.7.

    +
    Answer.
    \({\frac{9}{4}}\)

    3.10.9.9.

    +
    Answer.
    \({\frac{3}{5}}\)

    3.10.9.11.

    +
    Answer.
    \(-{\frac{4}{3}}\)
    +
    +
    +

    3.10.9.13.

    +
    Answer.
    \(-{\frac{7}{8}}\)

    3.10.9.15.

    +
    Answer.
    \(0\)

    3.10.9.17.

    +
    Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)

    3.10.9.19.

    +
    Answer.
    \(y=-\frac{3}{8}x\)

    3.10.9.21.

    +
    +
    Answer 1.
    \({\frac{2}{5}}\)
    Answer 2.
    \(\left(0,2\right)\)
    +

    3.10.9.23.

    +
    +
    Answer 1.
    \(y = \frac{7}{5}\mathopen{}\left(x+15\right)+-28\)
    Answer 2.
    \(y = \frac{7}{5}\mathopen{}\left(x-5\right)\)
    +
    +
    +

    3.10.9.25.

    +
    +
    Answer 1.
    \(y = -7.6x+372.4\)
    Answer 2.
    \(91.2\)
    Answer 3.
    \(49\)
    +

    3.10.9.27.

    +
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(-6\)
    Answer 3.
    \(\left(0,-6\right)\)
    Answer 4.
    \(-9\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-9,0\right)\)
    +
    +

    3.10.9.29.

    +
    +
    Answer 1.
    \(5\)
    Answer 2.
    \(\left(0,-3\right)\)
    +

    3.10.9.31.

    +
    +
    Answer 1.
    \(-{\frac{4}{5}}\)
    Answer 2.
    \(\left(0,\frac{4}{15}\right)\)
    +

    3.10.9.33.

    +
    +
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(-3,-2\right)\)
    Answer 3.
    \(-3\)
    Answer 4.
    \(\left(-3,-1\right)\)
    Answer 5.
    \(-3\)
    Answer 6.
    \(\left(-3,0\right)\)
    Answer 7.
    \(-3\)
    Answer 8.
    \(\left(-3,1\right)\)
    Answer 9.
    \(-3\)
    Answer 10.
    \(\left(-3,2\right)\)
    +

    3.10.9.35.

    +
    Answer.
    \(y = -4\)

    3.10.9.37.

    +
    Answer.
    \(\text{parallel}\)

    3.10.9.39.

    +
    +
    Answer 1.
    \(y = -{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{19}{5}}\)
    Answer 2.
    \(y = -{\frac{8}{5}}x-1\)
    +

    3.10.9.41.

    +
    Answer.

    3.10.9.43.

    +
    Answer.
    +

    4 Systems of Two Linear Equations
    4.1 Solving a System by Graphing
    4.1.5 Exercises

    Review

    4.1.5.1.
    -
    Answer.
    \(\text{Choice 1, Choice 2, Choice 6}\)

    Skills Practice

    +
    Answer.
    \(\text{Choice 1, Choice 2, Choice 5}\)

    Skills Practice

    4.1.5.3.
    Answer.
    \(\text{Yes, it is a solution.}\)
    4.1.5.5.
    @@ -2109,8 +2157,8 @@

    4.1.5.9.
    -
    Answer.
    \(\left(-4,-3\right)\)
    4.1.5.11.
    -
    Answer.
    \(\left(-2,3\right)\)
    4.1.5.13.
    +
    Answer.
    \(\left(-2,5\right)\)
    4.1.5.11.
    +
    Answer.
    \(\left(1,-3\right)\)
    4.1.5.13.
    Answer.
    \(\text{no solutions}\)
    @@ -2128,45 +2176,45 @@

    4.1.5.21.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-8\right), \left(1,-24\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,-6\right), \left(1,-20\right)\right\}\)
    Answer 2.
    \(\left(-1,8\right)\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,8\right), \left(1,7\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,1\right), \left(3,5\right)\right\}\)
    Answer 2.
    \(\left(3,5\right)\)
    4.1.5.23.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(8,2\right), \left(13,8\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,8\right), \left(1,4\right)\right\}\)
    Answer 2.
    \(\left(3,-4\right)\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(4,-8\right), \left(7,-7\right)\right\}, \left\{\text{line}, \text{solid}, \left(-9,6\right), \left(7,-7\right)\right\}\)
    Answer 2.
    \(\left(7,-7\right)\)
    4.1.5.25.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(4,0\right), \left(7,1\right)\right\}, \left\{\text{line}, \text{solid}, \left(-9,8\right), \left(-6,9\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(2,-9\right), \left(11,-17\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,6\right), \left(9,-2\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    4.1.5.27.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-4,0\right), \left(0,5\right)\right\}, \left\{\text{line}, \text{solid}, \left(-3,0\right), \left(0,3\right)\right\}\)
    Answer 2.
    \(\left(-8,-5\right)\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-2,0\right), \left(0,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(8,0\right), \left(0,-2\right)\right\}\)
    Answer 2.
    \(\left(-4,-3\right)\)
    4.1.5.29.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-2,0\right), \left(0,-4\right)\right\}, \left\{\text{line}, \text{solid}, -44, \left(0,8\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(1,0\right), \left(0,-5\right)\right\}, \left\{\text{line}, \text{solid}, 25, \left(0,-2\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    4.1.5.31.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(1,0\right), \left(0,-1\right)\right\}, \left\{\text{line}, \text{solid}, \left(1,-2\right), \left(0,-3\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(3,0\right), \left(0,-3\right)\right\}, \left\{\text{line}, \text{solid}, \left(3,-2\right), \left(0,-5\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    4.1.5.33.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(3,0\right), \left(0,4\right)\right\}, \left\{\text{line}, \text{solid}, \left(6,-2\right), \left(-3,8\right)\right\}\)
    Answer 2.
    \(\left(-3,8\right)\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(5,0\right), \left(0,1\right)\right\}, \left\{\text{line}, \text{solid}, \left(-4,9\right), \left(-5,2\right)\right\}\)
    Answer 2.
    \(\left(-5,2\right)\)
    4.1.5.35.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-5,0\right), \left(0,-3\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-3,0\right), \left(0,5\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    4.1.5.37.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,5\right), \left(5,11\right)\right\}, \left\{\text{line}, \text{solid}, \left(4,-2\right), \left(13,-3\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,2\right), \left(2,9\right)\right\}, \left\{\text{line}, \text{solid}, \left(-6,8\right), \left(-2,-5\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    4.1.5.39.
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,1\right), \left(1,-3\right)\right\}\)
    Answer 2.
    \(\left(-1,5\right)\)
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-2\right), \left(3,-1\right)\right\}\)
    Answer 2.
    \(\left(3,-1\right)\)

    Applications

    4.1.5.41.
    4.1.5.41.a
    -
    Answer 1.
    \(2.25x+1.75y = 141.75\)
    Answer 2.
    \(x+y = 73\)
    -
    4.1.5.41.b
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,81\right), \left(63,0\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,73\right), \left(73,0\right)\right\}\)
    4.1.5.41.c
    -
    Answer 1.
    \(28\)
    Answer 2.
    \(45\)
    +
    Answer 1.
    \(3x+2y = 42\)
    Answer 2.
    \(x+y = 17\)
    +
    4.1.5.41.b
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,21\right), \left(14,0\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,17\right), \left(17,0\right)\right\}\)
    4.1.5.41.c
    +
    Answer 1.
    \(8\)
    Answer 2.
    \(9\)
    4.1.5.43.
    4.1.5.43.a
    -
    Answer 1.
    \(d = 3t\)
    Answer 2.
    \(d = 75+\left(-12\right)t\)
    -
    4.1.5.43.b
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,75\right), \left(1,63\right)\right\}\)
    4.1.5.43.c
    +
    Answer 1.
    \(d = 3t\)
    Answer 2.
    \(d = 55+\left(-8\right)t\)
    +
    4.1.5.43.b
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,55\right), \left(1,47\right)\right\}\)
    4.1.5.43.c
    Answer 1.
    \(5\ {\rm s}\)
    Answer 2.
    \(15\ {\rm m}\)

    4.2 Substitution
    4.2.5 Exercises @@ -2174,81 +2222,82 @@

    Skills Practice

    4.2.5.1.
    -
    Answer.
    \(r = 6, X = 7\)
    4.2.5.3.
    -
    Answer.
    \(C = 4, m = -3\)
    4.2.5.5.
    -
    Answer.
    \(N = 2, C = 5\)
    4.2.5.7.
    -
    Answer.
    \(a = -2, S = -4\)
    4.2.5.9.
    -
    Answer.
    \(k = -4, i = 4\)
    4.2.5.11.
    +
    Answer.
    \(B = 4, L = -2\)
    4.2.5.3.
    +
    Answer.
    \(M = 2, c = 5\)
    4.2.5.5.
    +
    Answer.
    \(Z = -1, r = -4\)
    4.2.5.7.
    +
    Answer.
    \(k = -4, H = 4\)
    4.2.5.9.
    +
    Answer.
    \(w = -6, Y = -4\)
    4.2.5.11.
    Answer.
    \(\text{no solutions}\)
    4.2.5.13.
    Answer.
    \(\text{infinitely many solutions}\)
    4.2.5.15.
    -
    Answer.
    \(T = 4, e = -7\)
    4.2.5.17.
    +
    Answer.
    \(e = 2, T = -1\)
    4.2.5.17.
    Answer.
    \(\text{no solutions}\)
    4.2.5.19.
    -
    Answer.
    \(r = -1, J = 7\)
    4.2.5.21.
    -
    Answer.
    \(B = {\frac{1}{2}}, a = {\frac{1}{2}}\)
    4.2.5.23.
    +
    Answer.
    \(B = -4, y = -1\)
    4.2.5.21.
    +
    Answer.
    \(M = {\frac{14}{9}}, N = -{\frac{49}{9}}\)
    4.2.5.23.
    Answer.
    \(\text{infinitely many solutions}\)
    4.2.5.25.
    -
    Answer.
    \(Z = {\frac{13}{7}}, E = -{\frac{116}{7}}\)
    4.2.5.27.
    -
    Answer.
    \(k = {\frac{7}{4}}, V = -{\frac{31}{4}}\)
    4.2.5.29.
    -
    Answer.
    \(w = -{\frac{15}{7}}, k = -{\frac{34}{7}}\)
    4.2.5.31.
    -
    Answer.
    \(H = -{\frac{17}{9}}, A = -{\frac{47}{9}}\)
    4.2.5.33.
    -
    Answer.
    \(T = -{\frac{17}{2}}, P = -{\frac{11}{2}}\)
    4.2.5.35.
    -
    Answer.
    \(e = -0.169231, h = 6.18308\)
    4.2.5.37.
    -
    Answer.
    \(q = -0.210677, x = 1.37779\)
    4.2.5.39.
    -
    Answer.
    \(B = {\frac{5}{6}}, L = {\frac{17}{6}}\)
    4.2.5.41.
    -
    Answer.
    \(M = -1, c = {\frac{7}{12}}\)
    4.2.5.43.
    -
    Answer.
    \(Z = -{\frac{15}{2}}, r = {\frac{77}{8}}\)
    4.2.5.45.
    -
    Answer.
    \(\text{no solutions}\)
    4.2.5.49.
    -
    -
    Answer 1.
    \(G = 0, m = {\frac{69}{41}}\)
    Answer 2.
    \(16\)
    +
    Answer.
    \(j = {\frac{9}{2}}, v = {\frac{5}{2}}\)
    4.2.5.27.
    +
    Answer.
    \(v = -{\frac{23}{3}}, J = -{\frac{5}{3}}\)
    4.2.5.29.
    +
    Answer.
    \(G = {\frac{1}{4}}, a = -{\frac{7}{4}}\)
    4.2.5.31.
    +
    Answer.
    \(S = -{\frac{17}{8}}, q = -{\frac{13}{2}}\)
    4.2.5.33.
    +
    Answer.
    \(e = {\frac{43}{8}}, F = -{\frac{5}{4}}\)
    4.2.5.35.
    +
    Answer.
    \(q = 8.66667, W = 58.6333\)
    4.2.5.37.
    +
    Answer.
    \(A = 0.229102, k = -0.0743034\)
    4.2.5.39.
    +
    Answer.
    \(L = -15, A = -{\frac{16}{3}}\)
    4.2.5.41.
    +
    Answer.
    \(Y = -{\frac{2}{5}}, Q = -{\frac{14}{15}}\)
    4.2.5.43.
    +
    Answer.
    \(j = -{\frac{115}{18}}, g = {\frac{115}{36}}\)
    4.2.5.45.
    +
    Answer.
    \(\text{no solutions}\)
    4.2.5.47.
    +
    Answer.
    \(G = -{\frac{15}{16}}, M = -{\frac{3}{2}}\)
    4.2.5.49.
    +
    +
    Answer 1.
    \(S = {\frac{3}{16}}, c = {\frac{7}{4}}\)
    Answer 2.
    \(16\)
    4.2.5.51.
    Answer.
    \(\text{infinitely many solutions}\)

    Applications

    4.2.5.53.
    -
    Answer 1.
    \(L = 2W-10\)
    Answer 2.
    \(2L+2W = 161\)
    Answer 3.
    \(50.3333\ {\rm ft}\)
    Answer 4.
    \(30.1667\ {\rm ft}\)
    +
    Answer 1.
    \(L = 4W-8\)
    Answer 2.
    \(2L+2W = 145\)
    Answer 3.
    \(56.4\ {\rm ft}\)
    Answer 4.
    \(16.1\ {\rm ft}\)
    4.2.5.55.
    -
    Answer 1.
    \(B = 32.95+5D\)
    Answer 2.
    \(B = 38.95+3.45D\)
    Answer 3.
    \(3.87097\ {\rm GB}\)
    Answer 4.
    \(\$52.30\)
    +
    Answer 1.
    \(B = 36.95+4.45D\)
    Answer 2.
    \(B = 46.95+3.25D\)
    Answer 3.
    \(8.33333\ {\rm GB}\)
    Answer 4.
    \(\$74.03\)
    4.2.5.57.
    -
    Answer 1.
    \(M+S = 30\)
    Answer 2.
    \(2M+7S = 100\)
    Answer 3.
    \(22\)
    Answer 4.
    \(8\)
    +
    Answer 1.
    \(M+S = 26\)
    Answer 2.
    \(2M+8S = 100\)
    Answer 3.
    \(18\)
    Answer 4.
    \(8\)
    4.2.5.59.
    -
    Answer 1.
    \(x+y = 5000\)
    Answer 2.
    \(0.025x-0.035y = -156\)
    Answer 3.
    \(\$1{,}900\)
    Answer 4.
    \(\$3{,}100\)
    +
    Answer 1.
    \(x+y = 3400\)
    Answer 2.
    \(0.02x-0.07y = -173\)
    Answer 3.
    \(\$1{,}300\)
    Answer 4.
    \(\$2{,}100\)
    4.2.5.61.
    -
    Answer 1.
    \(x+y = 500\)
    Answer 2.
    \(0.08x+0.2y = 64\)
    Answer 3.
    \(300\ {\rm ml}\)
    Answer 4.
    \(200\ {\rm ml}\)
    +
    Answer 1.
    \(x+y = 580\)
    Answer 2.
    \(0.11x+0.25y = 88.74\)
    Answer 3.
    \(400\ {\rm ml}\)
    Answer 4.
    \(180\ {\rm ml}\)

    4.3 Elimination
    4.3.5 Exercises

    Skills Practice

    4.3.5.1.
    -
    Answer.
    \(v = 2, J = -5\)
    4.3.5.3.
    -
    Answer.
    \(G = -1, a = 3\)
    4.3.5.5.
    -
    Answer.
    \(S = -4, q = -6\)
    4.3.5.7.
    -
    Answer.
    \(e = -6, F = 1\)
    4.3.5.9.
    +
    Answer.
    \(F = -1, y = 3\)
    4.3.5.3.
    +
    Answer.
    \(R = -4, N = -6\)
    4.3.5.5.
    +
    Answer.
    \(d = -6, f = 2\)
    4.3.5.7.
    +
    Answer.
    \(p = 7, v = -7\)
    4.3.5.9.
    Answer.
    \(\text{no solutions}\)
    4.3.5.11.
    -
    Answer.
    \(A = 4, k = -2\)
    4.3.5.13.
    -
    Answer.
    \(L = -{\frac{9}{2}}, A = -{\frac{11}{8}}\)
    4.3.5.15.
    -
    Answer.
    \(Y = {\frac{23}{4}}, Q = {\frac{5}{2}}\)
    4.3.5.17.
    -
    Answer.
    \(j = 180, g = 110\)
    4.3.5.19.
    -
    Answer.
    \(v = {\frac{115}{12}}, x = -{\frac{5}{2}}\)
    4.3.5.21.
    +
    Answer.
    \(L = 2, a = 6\)
    4.3.5.13.
    +
    Answer.
    \(X = -{\frac{1}{2}}, q = {\frac{5}{2}}\)
    4.3.5.15.
    +
    Answer.
    \(i = -2, G = {\frac{23}{9}}\)
    4.3.5.17.
    +
    Answer.
    \(u = 6.74775, W = 3.17568\)
    4.3.5.19.
    +
    Answer.
    \(F = -{\frac{16}{21}}, l = -{\frac{12}{7}}\)
    4.3.5.21.
    Answer.
    \(\text{infinitely many solutions}\)
    4.3.5.23.
    Answer.
    \(\text{no solutions}\)
    4.3.5.25.
    -
    Answer.
    \(d = 2, t = 1\)
    4.3.5.27.
    -
    Answer.
    \(p = -1, H = 7\)
    4.3.5.29.
    -
    Answer.
    \(A = -4, Y = -1\)
    4.3.5.31.
    +
    Answer.
    \(p = -1, h = -7\)
    4.3.5.27.
    +
    Answer.
    \(z = -3, w = -1\)
    4.3.5.29.
    +
    Answer.
    \(K = -6, M = 6\)
    4.3.5.31.
    Answer.
    \(\text{infinitely many solutions}\)

    Applications

    4.3.5.33.
    -
    Answer 1.
    \(M+S = 16\)
    Answer 2.
    \(4M+8S = 100\)
    Answer 3.
    \(7\)
    Answer 4.
    \(9\)
    +
    Answer 1.
    \(M+S = 21\)
    Answer 2.
    \(4M+6S = 100\)
    Answer 3.
    \(13\)
    Answer 4.
    \(8\)
    4.3.5.35.
    -
    Answer 1.
    \(x+y = 5800\)
    Answer 2.
    \(0.025x+0.065y = 305\)
    Answer 3.
    \(\$1{,}800\)
    Answer 4.
    \(\$4{,}000\)
    +
    Answer 1.
    \(x+y = 4200\)
    Answer 2.
    \(0.02x+0.06y = 204\)
    Answer 3.
    \(\$1{,}200\)
    Answer 4.
    \(\$3{,}000\)

    Challenge

    4.3.5.37.
    -
    Answer.
    \(7\)

    +
    Answer.
    \(3\)

    4.4 Systems of Two Linear Equations Chapter Review

    Review Exercises for [cross-reference to target(s) "chapter-systems-of-linear-equations" missing or not unique]

    @@ -2259,36 +2308,36 @@

    - + - - - + + + @@ -19,13 +19,13 @@ - + - + @@ -34,7 +34,7 @@ - + @@ -43,7 +43,7 @@ - + @@ -52,98 +52,114 @@ - + - + - + - + - + - - + + - + - - + + - + - + - + - + - - + - + - - + - + - + + + + + + + + + + + + + + + + + + + - + - - + + - + @@ -154,14 +170,7 @@ - - - - - - - - + @@ -170,7 +179,7 @@ - + diff --git a/generated/webwork/images/webwork-1-image-1.tex b/generated/webwork/images/webwork-1-image-1.tex index b0f23d5a3..912e69200 100755 --- a/generated/webwork/images/webwork-1-image-1.tex +++ b/generated/webwork/images/webwork-1-image-1.tex @@ -130,19 +130,17 @@ \begin{document} \begin{tikzpicture} \begin{axis}[ - width = 0.47\linewidth, - xmin = -4, - ymin = -3, - xmax = 16, - ymax = 9, - xtick = {}, - ytick = {}, - minor xtick = {-4,...,16}, - minor ytick = {-3,...,9}, - grid=both, + xmin=-1, + xmax=7, + width=0.47\linewidth, + ymin=-1, + ymax=7, + xticks={2,4,6,8}, + minor xticks={1,...,9}, + yticks={2,4,6,8}, + minor yticks={1,...,9}, ] - \addplot+[domain = -4:16] {3/8*x+2}; - \addplot[soliddot] coordinates {(0, 2) (8, 5)}; + \addplot+[domain = 2:4.666]{3*(x-3)+2}; \end{axis} \end{tikzpicture} diff --git a/generated/webwork/images/webwork-100-image-1.pdf b/generated/webwork/images/webwork-100-image-1.pdf index 92315491b3d760e625a3acfc23e832e6ba6063c8..bc0374d3b392c4b97815b0526ee347fb9180790c 100755 GIT binary patch delta 3451 zcmai$S2P<8ABSr+(O5BCEwy6QXhmzaR;k!4_Dl(l6`RybY(>ndJz7<2RZ&8WQdCj9 zuf3zFQMJ|l&i7t?=Y6lfyZ^=S>N)3m&VNYYtH1{UB~DsV5#;0R?P!MvJ)g2k)^&}P z07q~CVe^@R$HBBQjC`!uAFtK-k!ar%Gx5>}1TeWBH=jA7u%GOpn2(Wc*8Gd}mH-z7 zRUQz}5Lg=7r8$_+?<=dJ<^)jjCOR&z@ZX1&>V0djW#SDueQ4Idiu5CG7PyaRw}!SU zQjT}wcpR+eIHw)GWu?o;4Oppp8jk}AwX~bIQNMoMRfd>{nU+!b6U*pd`b)0>)r}e8 z@N_%Qgj-(Xo|5=CjxNgZBIj#rHW}0}kH`B83pf7A)E0d`%GWWvJU#}KRfo{j=Vaor z+=0ObrzR<-nux8>_CAUn96Rxg@M(YHhbpE2?f!oFnMu?hZD(>&Cn1cMg}_fGjg$B% z6|Q7g9V66@?)g!pK{+2aj*%CWvI@Tn8=%71&Fq}4Ny%>Th#P|Zq>A*~dJVZc*0R2a za?fDlXFN|EibBbf?opZVN_q9HJv)-C<adQZ z+sgy5y%#Pj79I#G)~nLj9NHixr0!Mpxz{yD$@bmmy(8Waz6aSjaQ>{$FziUojK4~l zM6+-`!w9opQx}N0@hxCml5}$?-dBJQlXoDQTmsip`vP%K^_v%}o6stW=6s~yPD)-Q z5r5j4eg;FDf?RO4p|`B|Y@xP<=GDUT8@yUCpWG6TEF6+%P53Ab*ixnpWTnLx&=%9gdN@6{t5FdZ+MY#N+6suU1AM)wE3wr*BF-zl+tWQ4HHvoW*@y10@? zrCyVX5=BxSQSOcUD0PU$@cR!CoS2<8biXxWU;kKtTJmXQW&(;`BO(GR+Z0GGN;bgK z;%`Q#-ud4~2cDGi*sQu_$)&7;LnAKRN6DPtdC{hpx0 z(oFY!+om5<&dzvhdZa)?LUzwfV!Wuw&TDsH#4Nn#F+5Pr$vmdQF0t>AMhfl&8@Z=~ z27;BAH@Ft;hDL3cxtgHk4;g0qvqp@$GW+y%?~1Sr{Mif45bKDF&$Sjcdx?l1$N-(m zPF3NSV;tU7JHKVz9zD5Mx%^>v$hg-`h08dDpd{0ZuU=BV=^E4PJLl{u*mBfzv2(iX z@U@tr8&hvM%iM&Rd^>B9q9=XkLX@nVVHc^Z80r16Yd$>!uw5f-JVJ(*0bI$|+TKPX zLMW75k>{u|8SRm^<>Y+f&omX*(e>?)Pd=_xhi|&4T`QGx(wZ%oW+ptx?ATRoVetKV z$T`CieC?rv#%yEED)trh%cXM!3xOh&Jmx=(&3KoJ-;w9suEqqAdb8Jy(dybu-ix}4 zLzz-`0FNNj3ns?etC%Fb>4WLuAN`!r3KxB%_PNAB|C!H9MYg&ekBg2t^tjzUF)B zZn*$dwl^qBdBV$ie54H@>tLuq+sWTrt*E$HXL`Hl9*SC>@y*D{qP?jQ7?7JLxJd)Q z_6gM!-$;ar=FH8Va_~0I24K*LzT#4jkd3LhkkHN@3TK&kPu$Gj$!B60*?MDGIiDqT z^O25q^Ynse`m?qS^HX2uyb%^oQfdhZQF-3ZQ3uqD<4Zs$kTerE%&htIZWENknkb6U%l@&oUikCrNVPn3i*q;v;ue1 zb_-xk>fp|8v>kSyd1Y!$;wa|T^VU3rREWtqO<)Kt&bA=mwY>FNl~~on! zNUZMo#{Ef7QB>7~oHFx>*s6sJLB%v{y98+;fo=qnTr2Xt^~A}gtYc9FYT*iZ!yWE! zMnc96d(1wd8gCBRc3VuyTd@tYXH{@EE#}uSd85bHE%R$@0Xb2%W=yXM-mKKsF%#OG z{Ar+YNLuHW0DSIC>dMV<9uayiA(*BwGkPRzPh8hgC4Ty5<}$eTPLBoLZm@Pua!(BvR%_cn_8W~!dXeR@U{dMJjzlfs6wEU zSjVW`M_yTD`^DOm0$mdJm1^^A7@k`!@;slX$&XJ;7h+`&YfY_5O?}tqQFazJ-+WbyVrcL4TWGk&$>p-km<#w8%O-=_LUhmKQ^dUV+q%`(cdD zv-}>z4s;@>A@s}Ear*k5E-+C<1$}SN8uL|1|=ODQxqHER&}d+~w*EH8*QYg3=~4cEXZmk8H4mK~D=x&pp@j%I!};;#e? z1mq{a?ob+9xxU>hZw**qb(1c>Awm0VC?gNMRtV&+fH}K%zqS5^16y>POf1hoDSxRB zlgbH_a!C>td6yrby*BJLWWSG#{;cf!KLgPJy*voXw4yl4f489BrIQETa{BkjBN&wr zF{q)2Pi)l)azFJ=1tr8mW6na}+&UbJ;NM$m$qG>2IGnL$37ZQyqU6VaSr3i3icq=X zDVfeH;PhrG<;S~cKd)8Z8$tZQD}U;&{N!JC@Wb{4vDGz>aY$D2#_S4HFF(#>(B6t~ zppuKrQ}4=Jh(|-`hxVUFO%w>`L#_t^q>$`E-Paf9Etg#!XlhuQ2vPL5=$#U;9k^Yx zzshHh@=zIFwV`!fohlT+t`7J--0XajGV`IN7sk>kKC5_icJc6uXM?Fm6iG( zSpMm4s%S(Pg%b7(oqTLPFx>ko1q~xND3YcPxcG*JN~s7OO2^^J0O}BsC<|W5O!C1| ztgb>t#?++5$xo01AQnk6?}$^?IOp7ci6|>ix^A^9^)DtVI(1ouou?Yl0$snZ-O=;= z7_^T`8ZI1v)UM-rVs84A>S6bnH4!1I^IUx3Jak7bXv7 z{N0bW*w9D@C`n$KJR&>rte=RtHk!HmA{n|DIo8>)4Mjb=^~^N^K6h>0 zKdI0yg>Ss`Ozee)i6ue(JA3It?Lh2C$rhi-lF%&Ywo|fF#6F`kK%Ix`lHP&m7LP|U zG2?!PuY&zPm)1pDz_hsETm}I&`{DbV-q@ zn2MvloD9+tDeDM@K0w;r%h@?NI3SS@vNCd#vXYKc(u(r`Z$SVs8|zmyh4QdQEK(bT z!GjzXvtM6Vh|hMISYPdFS!q(bex*_lg0~0j8GPf5k20rVGxreH$C|R7&n0N{`>Tt^E~HA5S|pSrJ+D8sH-ys1p2!=p%|Vo*r%F$N6E0n zoT?8gKau`ZHA+rP#2>IWqREH*wGR!=UdoVk~kG85p}wd=UslF${|BSqFS z4?=@&9kRw$mHD20@*v1ju-In>7dJq*Uu(DgZk&^PfXk?eTk1rHmApwh>m|>K%KrXA zK)+wlj**XP4f*C5#>FpB=Yb_mOZe&-i$khhjY$km|Ir>iAW0ku{!Lr6NO%6xrx?+u z`O_2hQ!kcyKL2F6pIzO$y=p7}P_IhGI0nERkgQlaT<91Z7K%+{r8Gp#{u>B?s;3FU zo$KSo&+?6q0A23s_<5+>F1}UvE~k|fa+asQ>}kA{a`DI$m}g=-!nmj{sh#0t++!Mw zSYrmL)Y^miS2;YZe^V;~xO1Hi>0MlQDV@MPK6_5m#&hTES(n|H*As`uqD?t?51R(} zWs$<&W%NblR62@oW9Ki6)4h?5VTIw`bkfcDe^+o6z!fW;MH50udf6#+aWDbyAEAOW zJMi?JQp|-{lS3gE_gREjcmtNT^dx}MFB7G0VUr{1U@*5e6RifEa&5_Gq)MAdvGcn< zyZfC{ADn0(SjfIA=jGi3QInloI{;_%$IxR(&~2W!9pCEOQQFN9@3tij*a;%Ud>P|L zfpEIQJeF9vKQ!?x-dM))1HLi>kF;cS#)*n0ENJQ?Y{X(0?)ix&l+LEVYBP18QWHzA zuq?NJXAXOI>>$UpT( z7xZhWh73gZ4@5$cuP!+DUYdc{tC>y18`gr#z`nf>oK8LY1_t)g>X_~!tI`H)GPfq#N7CEEr z#&xtCH4|0_gY`JDDL+U=MVn&3*O2@&6exT$A9c$&i#(HNWt;02eGFZZ8W~fBx zFx9(YvD|B_y25~$GrUQM+Yt%Ou*u|d)eOd?ylA8Yg{J##a2<0t&yMQt9A{#;tCGLJ z=YHHNnT&X!%(OhN#R#(NDNsZUKyDbj2+iB;S%4Tb`iFW@txq$$$H}j@&zF^U@mAVa z+3M%JFI4|fM#)~8@$&IA_Q=~1{Wo6VjpPOoeepBIT@UH%nu1z>(WH^c7FH)2_dW^Y zz84UoYO+pLQN6$L=rBVb^!VX)fd?k!5tZqcI!OllBFvGb;z;NK_#5t=jS#ZS2kN+& zmXly0&zW6*k!BG$lhH#N$kkfER^#{@B1y6pcaV@G{Nv7echJc(ktW%#*^<4t#*#NI z=X3hz<~2Elxm@rco7vN(XhqP9|2Ep*+|;vQ2GDt7wfRL>Z@YS%gTmG*fpLCt`c_b& zmoV4OMd-NTb$D~`acxwzIx+3&{_G_o8E76BBCZ0EK^Ciy2OY2#eT4J&o!}Y)wwaxI z(L%aQ45w0eqF0YjUMLp|Y&EYZsoP64g_6hgGx=Ndqx-@~&gyJ8BG{FpV(P(U16FxZ z+U~wx%lTUZ7CZgoM+-@{?Q{0kfr$Yc&?0(GL(6r)?Vz+q26qu~h?5>zxa3B4aj3din^D?CxMX>; zDcfw!tKhzvk2q}lxlPPR{U|QSA+*HYeiz-|)Jb9P6pF{=ndwXKF^k<^L;GCg6fa3C zdHt}sWf)isPGyKaCoSoZh>I)bNMmYliz8xW}RQY%RORN1^`k<~+dm$3-d#T96XVTfsZY`B_A`G{PtIdoBJhz`(l*fYRSUg`n zr796LczOGL$o{*I?sTP=0p8irb&u}V@DxQ9s<87cdX@)BF*{SlnBTcw;<8QpG(_)a zdP)Xf?KPkk(d+D#0sdJ!AojK((bkH;ruDXH4#emttA()LFtV%W@c4Qq??LY*4SqAz zlYRqLGRZt@oPL|X`PgWtKCY5=y!Y(2dBor`S{$%>Hz%hD@g_te&mu%Uwp!TGbr(Ui za6ZmbJF+vNsZ^?T+A9-xUs6n4`R&DEZ)F8-@PXs251;pA!tD1$4PHd=Ew0h6t)~WZ45RD_&$43KX1TQ*e*AHqBg?2?N#VXY&pHpr2 zeR_t$L|qgkFPIzm4@M?h+#|2fYN=-UzP$DQH4+I8tfNeQP<-4aJHVI14EmgX({_#V zA%Bx|wDp(%&D$mqEKbEdJ93>SkLs^QFq%gHE5Lt$n}1YD6QHC{!!U}VA^-N^KYNe| z2$GA$vJr|dkkfsyU%z*W9Z*aNAapx!$KDF6O)B`O2kDk$<#l&7PYnNR;(8?WfPLIs zFlKTMDKWx0sTZSW3^$!_s8Tk*mJ*5Hu59P`S)TZGhCR&kwA%2&`L#mYJ_(^YLWy|` zPk~T{AYTv_B|*xzA2!JlR~)3KuDcy2kF68ptG3Kd4LZ2u6lo-fijT}nb=SPj8=NZR*?8*v?j|1P-KOQw5yzXD0twzI z7VFe_?1%#XAB(gmfF&&xfIa|_INQL=6hKaeoJvXkpIYVuwf^%EHl;6mFKDvF?j3Rj zaGFzx);|0E`Zn0>p7wO3=%hUBN}nvRGriY{gT?ttz5zO(1 z_SJkNU;wW07r@`FY+O(V6s_HGq@gc=*za#`Y*eij`CN1v<3H}x}hQ7%P)!GwUSF&hj)5gu_1 zD}PmeIp!?06Zn@50;9_5N{SxG$hM6dn03FcyX8P-8&_0KvgF=kxID)n)Myt6Jia!{ z=U!}CVy=^uuYBb`hT6{!h}jU|GC^0*SS$EF5%4c|_u%-xMX#G1T=AZT9D(|{I2c#* z+QdQen?uU6+IMk86~K3RN%02i&P2BF{h1(D+krN(DtBoQ*-(e-mAXiWJTX0i+%eMv z`H2wBR5J06sU~%rVosZ+n3<>=%DcL%@&&4*S|iD%k-)B6K=-r%YGSU_-VAMa$H?5g zk8;4Sk*Z`yi6+R~r7=Wqu2v-}>2S}QX9JQvQ!Rag86WMa$CuOQWMvZ@t@1H5!8rd;0OfRmK}*68sAP|A z%lWfp>>%02Uu)i(67uG9av|to+F_ADy6N{pTX~WB4vFGOKd*u!Zwy)foco`^BKR=x z%EB*rIuq7CC%X_PduZWFN?N#`5kAFTzU488am1GpefG*Q8c`gypmkOQ(I8OCLSzTs zEvf42{nA|W;yrT`I=OzpymMFFtZTh|^Okv-tyhG_=TO@F{1iLb2({?YufVo5Pk$mg z>^eRV7s~cl#eaG5*-|^LN-=W}?AkbrB*fHqbc6di1>+YI4>)e2_3EaVI8`#aZZu6x zWtGr|L1m#XP*<3ni#*gB262PQ$T+*nD#&{%ueSuO=~OwfrV%J=2to7NJ4GnDm&*anjtayC!Hqx;<+ msa}j`jS)yT!ti_RCuABVO4_$+MQpd|<>VNIg>}vJ82$@I9faTj diff --git a/generated/webwork/images/webwork-100-image-1.png b/generated/webwork/images/webwork-100-image-1.png index ea98177c1647ff4edf7c095e4d62b24e9be22da3..bdfb69f6d08cc4d8e340ffab220a37d01d2489d0 100644 GIT binary patch delta 21 dcmexu@!Mj91Q&-e2Mf=Aw>dXAD{w^%0RUJO2T%Y2 delta 21 ccmexu@!Mj91Q&;Zm?}#ozrvi&3S7}b08VfP{r~^~ diff --git a/generated/webwork/images/webwork-1012-image-1.pdf b/generated/webwork/images/webwork-1012-image-1.pdf index 42dde01066b0f26ceedf68e5c2ac0ffe309ee55a..3712f5edd090cb40470d825951e96b80c145b4ff 100755 GIT binary patch delta 3674 zcmai%S3Dbz;>D{%BDJYKirR`IM(wEG#H#r-8bWIXv3IJ(Ry0(t7PUudNn+G2wfElC z>QH;ndmsJ}_y4&M_w78Kr}J>m_sq)9%U1UTFw!b2U~eBUM+6ESFl%F|?;at^IDf(D zkx^DAX{aeiYYdI-;LQW~khF#w>URfA$GY=3~`$ z+&#_W5g_H)&^PiV;_ns|F*T)Q_eH95ntd6dK980%y$@_{cV@Z{j5@WKIORPC%<6TpNt zHPqp_4@kLWE2K4|gnp#$@)0|1Ivjn{*q;&TG^o0HIVc@nbe}%1$oXt`xP&T?eUU02 zGhf>~1H7Noj%gI|w&^@eJr#H$?n$Shb(kuaE9BC?(4*?4s36nTQ%J4Dt0Y7LB)@isDLwIQ4ztq!^~1Q8hv6eLVAV|!s}+`O6{hIsDGR+2WaE^yI2@% zT@VsGO;&s$VRccng2KYHvcuA5EfZtRt`xTlO!UPm?DtcAPsL)7HVf5l?!wXXz~@Li z!S;mnROMH*9olUEWiax%eAdFXP?t2mT*YqgBC{Tq%NRd zMyk#JT5QCL0Be}}xnX1t>5Z|hOTwBf$oa|Fc4D zzX9`Q>}~>aht8kZuJUln>$rTh7v~7$%;L!^Z}cud=gsXxm79w^L%#r_`J`ikfeyBQ z)E2FgkFmAXRo<|-8^!CWLHNxh34jzy3aHe?`iirExu~UTeE3RY7xNm!a`_;vgPNc% z!Mn}P|6P~^_zxs~IODfSaD*XVwDVoc@&PZs{^zOH&zoQtTh;2r)7;8UStT`^%Le8H z^%0)fi42HfW!E417xShKn`-(~P3;aZUC8&}3f-H}?d=^Gyz{o?>K9ocSv8GgDDdul zt;Uu!^C~C(#kLf$oC4+pz8xV=qydAQ`*}l6C=Ps14@IKevY)|+b+y4XQ2!Hk7BmYZ zvh9pNF>WRO82bg1FRHk%N8MB8!O-N^@;awHZ%DOMbCq&q*F%hwyTN1NaESCabMU(mU(m+bB-iNkg(X)J1|BXI?{PwH!NNa~U?L0? zUmrF)Kv&9c=lpyZbe+^>9o)#d{ZVko=Uki;WHYG#!w7}XYqv($6L9^pFbXWoV{6F+ zilV)(=iek0Fm<0$Jr&(T67saevE6a#Ny8K?VDZS&qIb)D3mx;Iy3gg?L~=&@9wy6b znXkOXN<;@{H?=nP55V(iYtA3b+w1Qa$z}Sig`?6C4tpzl!bWM`TkfRNZcE+kr0e?` z*;&X!(`&Xc$O)!w9zX(ibVvSQ`TkpZ*t_7;z``UCaV_9`m=`FeI@Is&Lb=Eh{!%qs zn(I)q0j{8zh^Q0W#jFCiGA&V>o^m9qT0OQm$wT*2YGU*(vz4u@a2FCH{AB4?>)1j`#O z0Ky=(;9ef^h&hH)NVx=vPUmy^CRio6bAmgGhsq=h#YsVI+}zvsnVJS7ib zb$P&H{w-daq^i76*VFR3!lzGFRpqS^4)Y}Tg;!G(&L_{xbL7d0IcvSj^Spiy0Qy!( zDmXM~(A@OwDlC3s!IiGbhPhX?dPy_;G-_BhAoLZf9lbI@Zf@cBqAziaT%sWua~`wY zX*R+WpbRp4M8m7l=~{C4mv0PI#;S18`Hk)copG%m*4STVu;P!LDx-cx66kS8>sU9X zBGnK*&dbd$NI|74L^M_u?P%|(loLS5{-LKLWQly^?|6>8v;}G=Kl~WP${9uLeNpHg z=&GxnrlX~1|LvBs&+yXtf<37K=2TL_HR`OiUU>bim)#WT+P`@`rSuOKpu1z2mj4sn?+<4wb9u$Qe4T&cLcv@?l z>CG^e>v--~5PP_hFq?}ix@L+jLE;kGM(OQ0NIx#BkuTCt>od9b3+x&>5W*Ve*#_@^ z+0jbIzT>6%r@zQgr(5XZUJ>1o z#y==;cC9Vf8IkUTTlccZIwUjM#TM%tmT}5+?60-DshNtupalmCj_wAh@;z2Js`Kg~ zb}@FTn?yzRQb#Cdnv(=Vs(96UZWeIe#Wq&Q`iWG6>l|E~6v7oh5~t$L*YRFfsHVl7 z$&T`UE5*(#+gNkyjZ`GOwCc^N$gf%9qZH?j%qD-wIl@6$LKf_Mzv>QGXW|50O(*1Q zmf5UR{SJ+PO#D^NC`}*v|0kcy)H(M#dD^I*3r25qcC3amuW%`d^0eAL5%s2hP*&a4 zMu`3+s9IA!zvfuwidwi)#iwREJ`E8Yz2dpCI)q>p@{ED*hM!Ts1B&gpks1J6TWP9? z!cDTNUS5y{aCv6a-wOeA0fnK-w0e!V^J6qXm;^F1p;ttz~YO#)Ea#C(0UuIyU$Q!N@D5$KH={)o!0gdw`s;N>%2(o zH5x~ETdVTNf=OST2iEveFQeD=8$9mles?aZS%bP~w4cH(8r6S3;M8C5v1x zY6hi2z%r#vtV;kK&s#h{=oJ`0k5s{jDyk>3Y&R!yw_HpZ+L!5qi>>XSUHkuRB%foU&#X~*N_$8 z(u!GL5!O|<^v-$;V~>GGmsD*oZ1gw|HZiE9le9F#0U@s-CFdxqfRuEA$RQw3PO`EN zPEK-8_77E*{+9z6$EHalj~ixV_1v!~8c`Sx&~axmj7@EZ&CO?&PxM2}h0M4FcS)U=L0|uiR{nh)5net4 TUXD&+Svg4vn4e$MKnwgYGK>7m delta 3692 zcmai1Ralb^pr%F+HisC5SR$a5CKW0q)T8( zcSv`~`7h4J|D2n1_g=hL@AJGhB|RaH?HS8Bfrl-^VKFb?{L1ECqu z>>|3VaX(SMI2a5v>1HCM$gOZ62@3K113V|LkA%NH5tDerS3GH0QV_aiBegTwERc%- zzBcL=M$BKVnH8KkwT1eA)(b|iOGvQ0XlS0V{B8gg$5x&9yf(5ADM>mg*LKJ`CxMIO zk!hw#Oo;|98cCGAQB))QCbBhqkW}ex9LMmoFl)}Eq+ax|iEWUF#`4?b&(vOy-fdMs z+ePEkim8xDM-Y10h%EKAriaUx`Anv=pf>@Q{CjJBor{2&+lo4b{Lo&tdI-d6)^RWP z11oC&b}gpPGWDFb(6AF>2Bj7$stG?6*&kIRZ>q2(AT2=pziO~02R<=$ppV>HPCI5J zKr(zab85!QX=U147RXi}S*K-!Begv0qD+Lpg;$$>0a;UG@9V`c6p=2sGC&1h8tY{+ z+fi~E^0Y*BM`DXHYH_Xjyd8A8HX#eE&~>(&Gwbj5Qj06^Wma0pr+nk2?3uVIU;?_d zp3a}#0?6ZzYhy5idr!@CR{6YQ*Y0ON)*mUr#Q?qDd)vZHO!A!i{`$AGG*n-|H#mF$ z!+?d^T5H^v!0QB>kqM$=hu&>yNq`vL3q2bpk9)>cF1@esq*5aG3FLt7b(d`4kZgg^ z_!gQ#4OG+aYinyCn8FOMp02b3gS3bWf$_)|Pky<0|rJSc5S z4Wcri4|1H8bz&9>plpnyEJz;>{@g2`ISfk3V>_Q;UFeVgXFlIa7IeiTy`02s21?^d8A)P^3xGt!!BKv zbxP8u8D=8W(kKRbo-YA|-Ix0bk^+{t+Y1g@KyFPNYv$K7H|!^GVKT8PAXG~jou&e~ z6t=cF7m!OrslgWgLBMF+y;6b*l_O*FTsavyIieGlCx^S-~NMu_A! zavu?A6_z9stgvt&DCW^z;Fl2y&}m14E0$DmtNZc{v}_XM_}pP$-JI_ro7JX!=Cl z^kS8Kd(+EqPmiZ>FC}h{y5=Wbja6Wz4$p9>bGGD|QyE4L$g~2cs%fjE2##+Kzr5#~ zBc)3r*QuB4d1;37T-5bsB8>{^<;Q4>%f&Y;1$9@c0vAJR?EUW)N5K&2@i7}Wn2C6@ zhS~yZ-{mUJcu4#FPdKA9yJY@eyoH)2Y-H!$$|8x#b2qcHx{!B!;`4On)@Y5{I!+-c zsctt_ntZnPrXt<>V&Vm}qCc(`xZd9_yvuWU)EL%tpQ3EcC;6F};5?p=PNp{?#8Ckg za41IsDAuj??#F%&x$2RJK0D5kO2ZMF9Y=QwLo7H&2B${5+5K<${)^mguqo8{h()nKs5yY|ReLWcN2X`!G_(zZIpCKUBw31w zM6KYR#s-;Z(^lqX3U;~uQE{dl+Obfi!t9hYBrvr;GJp$*Xez=88eymbo3`8}{xw zY4UG9$~XQxJ9=C@kJQ66j^M_n{UC@Vryxo)UvkFRJ}$1*`iJt{T}&kyF2onE$gEDd@@CM(Q*V*~bJ*JNW<>p{)PbdSQg#rYkj z5AW0`1hs?+js6}8M+I=`#n!I<5w#OQ3??+qg!S;=e|49bQtg1 zl+P}1`ar7-N>a*7%gO|8OJu*KmHAS^9n$;y%w5|!zq{~QY*+T#q#JSsRTuQDXznaW zQ|AsiRMege=i@H499cFnw<#uh4yoYfFC1X%GS=>YEJ^Bj+{PhR-oVlEo#s|^ME#s6 zxBB0yC_4uCe8z$?PU@kT@9GMWd|Qw8hPoNFrY~e`if6C&1ZWlcpv~8~c{zLX_wBf) zwbNcF7yUC6dPAG;rAzgD*IW<9(ycCWK<+BNxF$L@t)8YKSIIMbu__qISrdy;BGd}V% zw4b8p1%2dg(S9Bd);Pgpk^Oq~PcJXoH*2XOtu+?aYn4 zf@)-n#v+qmbYv2g_+Plc^yvcAEE?VsbXhl*RLaaTMn!YlYQydJ*I38>UUp|NNv+@x znpQ-9#cI4NnNyAjZ~w+R-g*c~{I|bRL;B94qEMnaU!YylCpO=+>?a9%ENKks+)p$mI0Sk@Le@SG zRfm$Sa(HNkfCW5=wKP6(7`TDjD&A5u12haC(bqhnLoQpYni@MhlcM+TUkiWH3ifuk z88LS7(>I9rcJsA$VVS7D6KagGL zr@Gi{jxx0?ecn^}ewr3SE-{_q+BiZhy6$VfDZ^SubMpBW8g=a!5SC|MgR~h_1WK>1 zr{*pREZFBg4M-Aqwe43{_G*>dInIpxFkZ2fT@A9xqmSkegb}vWBBpa#M%lE>{eMF< zq8nDEBrfPTR+!_SUbnruKjwv1F8AGS3^mD#hz;8Du~057{*^iUZtwBxTdkrgr=r(J zjkv?Wn(5z<-#qE=Q0{iA76<9@1Gfh*&UJNn4pI4^dX`UVQecbvT!K##iIkiMbbJP% zC0&&14lqeiLIlV!ak1ypLALJA%nw%ev#yEOjUKH`@^3tJ3rh+=Gf36OI;SKiTvqgM zX)mU%E$b8F*c$DmqM`!&Z@CtCiaLMut;tY$vJF>`-F8SiQ6}_#eIP?Q3m^$X8#Xt? z3j5;$ztMX(DhOg0No^c81I>bqWqk);0ui55A@dWD@(aNU%Y&2~nE`m~1xLKz8L^Y*K=D4e zG2NBAR9%6262^_GwgKTL$yVQ#{ed66SbvtT%*d`ZJJ9>Egn6say8?x-lE~eGfql&t zVb9UgUn`E9Bs%-B+sU&VmSbpR?;<%q%ZWfH-w79!*IWA?0zrVy1S4*bdp9ciu_@*z zbuZXSeUx{UF)w=HR`WR6ibjMakxMD`9{$6LU#t-*I#di^WcpCK`;FU)&{l~ynl(^6 zA)$at_a-ru;)btrxXY99{_etv8B-{rd9FHjGw<^H@lvDiQaW`xF`QHm=Ipu567j;a zHetdb$j5O(jh-AJ<)LIH#9|If5iU8#P>uBDt9=f+2^lC$7&5%!}-qv)Cc7lvHARk+pIYqL6wuHVFrM85WL|j)pu{6DI>#^@R9FfbFc32tvylPOI&s@9p6XL5vRsMhH di_^bP(%Q@0&kJpj)q{~yic|CQB6KyV{|Een4gvrG diff --git a/generated/webwork/images/webwork-1012-image-1.png b/generated/webwork/images/webwork-1012-image-1.png index 844e3b6d8e74b5995bb20573e5d8427d9e4be771..34e52660b8bf964e1c33a938bd2b900fcc5a413f 100644 GIT binary patch delta 21 dcmccMf5CslI&KbO4o+2#sHz{Ew{c6!003J?2bTZ< delta 21 dcmccMf5CslI&KaDF)hK(UN?4Z-o`B@0{~&)2uJ_` diff --git a/generated/webwork/images/webwork-1013-image-1.pdf b/generated/webwork/images/webwork-1013-image-1.pdf index 7fa63810539e97926808ede17f39dab08c3e5886..cab1dcd7a4730d698129b396106c55ee0907ad50 100755 GIT binary patch delta 3392 zcmah~Sv=H_qa_TDeWx%8*WKSk!jGZVu zSwb;oY}x+z-iQ17-KYC@9?ru#pU+wP;dppmF#`b8&;X;*{%%g*;GkJMvmVbVC`;F- z)JAI=N6Ey*3%Wi;D&|+iF5ccnepb_1J2CeBG?z!6oEjCpzcEZStlIc0z9pz?JFJF= z>WNB^Jges``izu(=`JV|SPzE1&KA-SaDL;)u08QQ?EU5`Dxkfyd%h;%FX&2al3y54 z6d*l4Tpu#TRIbB#Eg_OZB+J4Vb4VG>Toe-e=W7e+GtqBASyrb0SHGV zxj=R_Nn4(`Sll3^5o0{kK8#XmoRLl3q2U_8Y(cf&*2+$dl1PVI2>&R>!@rqFjllw5 zox7}t0x;!~967zeih4T8>_CWn_K7|NE|-uI%%_aaC}$r={YWR16{+$Ep<86I+D=`L zaT)r3d`km{9l|b328}#%31-gEc3MAaKoAHrSyBX7S<_X%%rX>16S3cR>)?s%j$cD` zQ<~KJ79TFZ?0W`obztqMTSw5e)+1E(yApsy$N;a$)w^Ap21>fiZWc+6)QhP?!kVN0 zicckS36Bia$!j+=r=qCt6-9QA;mzDkf8)H*u_HdzNGh6rNfn}jgYc6`iwkRhtZ#MD z`Uhryl19)btxw1bBwRrHk8{d2s6Q1|_lGq!B80?H0K+I>KAZFtkm+e$Fg9|$0Z7M6 zKcJOZWc;c+`~bmco-L1Y&=8gAmB1 znZiac;@tV(%EmDBSWPZ#vm}F3gmT#Y2LFlb@g0K&h{&0hM4Qojue&`v<71V(x_0>_ zR$N-61J)Cn(NpTv_H5Z43m87gZ={&27BTGyLPI^v+{5q^|o=g^!X?HRMfmHMx}7*ytHDcs)iu z?(Zl}K6ANgsrcS&X4|7WW%b!_(ypMZGiFQD+{Q@>6%B!|r6y(DjFmcD=?zU{U+Cl& zFaBA`GdTsj`MCa1{c%@VHGv+VC_$F4eNNQ;8xJ`19KGDP059!(b8l)O;ZU2GVCvt? zOIBk)i)eY=gZ9?D2$Ijg>(CB-=*+kU;%Pz(Jn9IFANyXxNT4^fi77E`c@=fT`U~BW zkXY0^uj7OBzYmodN4E?EYxfnz`7i@v0d#%i`?eh;lsA1q^ogwDKSO}~uUB=&DoV*T;Pq`l1$e zk!PxAYkOe`pIzBMa7wlrE$X|-SHHvW1=|#vPkTvInY=A>FvWf8QYx(~(y}w0mdapU z3ubpyB5x0sTI%jd;F4w1)m&rCW;Dl58rqSjlT?qkw|`!D{NKaohDm9lb0Gc}l z%%F}nZ&gaCquk`0TYkK7;5?^VQGGTD8zx#TINu9uGDI_Ya(5`TVp$Jy^h zzJ_{Tx#yVmXu$~B@w3t57&EO$d03{-4>>zAK7QmRm4%oND?yU>B(l_^{6 z>dRZbW-F;qU3X`m=I8~_lypLeb!U((1Dgq7nopi+(Docjot$k1KIg>;0b0(HBq}vl zQcgQh3*!^$D{?6xuhlHwahnefdo~}I;KGs+)%wlKt4CZt${KJ-kK@8Wk1dr=)A^dZK$Ng7YKVh!@r=$ZlK zS{$kG;mZ#C77L_=YAWfor~N0*Yn4&a_9>ZEZQ4>6EPZf24fVC|P8Tt{7FzcY*0dt? z=K|e>=cgBb6a4pC^IsS8`woo&vYYY-@!_Dr*LxtnGe3)|qvzo@_mae{xmegtoEjiM z^M^E|KH&3g)ods&t1^dV=4NZBO5|WaG#R5n!MiC~Qt(6iAB^7}b3BCxekl6C%$OXY zGFx50OwVQRP(DksZV6*Gos%%e&3l~0OPbMmZf3wM*qX^UNb^A2yUVZ=ow0bU5|+d-2&BZP@KJXd{-3 zl@=gdW;2aL3>)SNQ1-%yZQ8jJBZ+BccGVc5 z!^VI`iw^;&i34#UV;*w|#$?m6BDCmG$tsi{aVqR_a=lt2sms)I$))Cc!lEKSDHOmG za1T1+D~Bq&j))36JNwmo>pXXh@|+kd;Hu4PWu^2x)2k2mQQ*uoU5dG77nNRE62Wq| z|JYr=ujOyw>Nl}l9gFEuJw!XK*ozEiU=&yx|q{G9w{XYz&CqoFYiMN6I z&ceMrW6mO_Ya!iB-z1Ec^s9G`HEvVE#|C~e6;r0|T@7B^OC~Q0MEsvuRJd{gYECGL8pCeF#LleJZoStA^@lvl} ztM{c|&qsCx_|K^2>ZRN>_|7(E?R(~F<9Tpj%Z9CmyV#7Gkw#31e2Z!ITYU#{x5?Sx z_h$ubNa_m!mLg@sG6XfVFFOM}g7jm{{v6(HAI@!vQEXj3P{Lp?kW8i9qd>cKf%Hqs zk5=9r+jn*N)cB9?6hd<-N8~%I5}I|WYu_!NMQg{G?=(G5BP;9Hh=U^FRAPnlbo0q( zslCKE>o%Vjwq?Ti1roYqfgk;j_dT9mDj(N}PKVWo=KGL6;I*MZUcR~u_XlTP2~~Va zBL7yH?U2qs>#OGMh!L;ntH9+3tk94Bm{NG|E(1XVfhipLBFphbwGVu$vV!3v+OG9C z#^4sh!%uZmL#%4dbr$MxA>jeJ3^ zU@39~(N{8PsVj6eGwwb$8^o_!*g*z^-2%OSof^2=gy}P-2tFG1gXGX&igmec@cfde zRZ&9sF2l55a~=T$!^`2JBwnCu`I_c662T@`%Cv{(`?ap^++#vn{Gt6B)nqgm>;HXR z9H)?gwvwWf9Mnm{$yq^8!Oc}!#R&>?lXr4)SAe;wDk?h5YpDN!3|xQ^+)YLG-%}_` z7-rPeW7g z7i|gJij~&*p6|Q(p7($CzdIM_<~+|izte{tM7|}k;gnQWS$zF`F!mlSK_r_8|9HmU z1a?z+e3!~xkcN3esVMT`jQk>uye|}@#ScNzM+g1R{;+}EmLzsfpx7BR>4ibyPfWM_ZreuCr28iP3F$X@KiNxwKG$@1`g4^*AZQ6^%q0v^j?vZ{n?!>))N z+JLs!VkK5_LpsXhbR+6?FHE?<+UN*4dvo@eIC+DMrXy>C3H={!zv?4MEs-xKE0fPv<%{Wr4i|D5KO6lFJt4(77eG z89uK`>dpFjc)vKUv^Wb-82;j38X7!j^wc-5rWIQA%o4V zNPLE62me6MmPuLDaTGYKj3fe3aXts=E5$tF+QxxwGH#z}WocP>vwgoBc!|2l`1iE` z^|4dVwsT*DJ&b8dXb^NaTx%jh+je*Q(LX@H)Q5=JdMsZ-e@fX(ww;&Y;@UmL>gTaj z-zPYQnkd7J-Gyk79AHa3QfwQ@D--7hGaaB9=@q0^D(tOvp!Z}&e{sNB zy-7poVDV?G;~d| zbA7}1G#0hnrAV-PSSoji`nk%u^)cJyZd@Lt&>q!doa9M+F@-Ovim}YCMwar6-%iAr zhe-Ca21tM3{5JAB$00b4@}OYI|8zofp8b*haPg20zrgKy;3V5=Ugae8S41|aRZxwx z|6#}$pclEU#{aD97&I9yW^Sg#!p-LsT`>62{#FR)cMv{r5lXj1lVCM&yDOo*<@cj< zDfcW}goSWH$sgc)^V|C2&cTyE5p&5yaI4rFn)T~~R-1$d6o>ioQ2dTr)kUGCQT5AG zSoL7$Be7zG@27WHcgzThitVhercsm!9& zSjDk#n#v^LMIAc&R_c1aVEXf)TB32`+tr(`A4IrA$M5dlqrcI<4T0XLY5ajR^Ja;N z>bG3D_T2k;izmc`L20QPW=L9SXH0)a_W+%Zl37KdsR>L7bw6jmn%FxVUS%8r|5wkYcaP9b;o1BE0ye_SK$>lPIJK=VV@+zTHf`mn!=> z=<46$++5Wrzp{YKUkX0{{N5OmG+>(Z2+Wbh`K#^ayldtK z9#sX@#HDjn|2)WCOh`ePMh=8?qh#5hb`_%YEvgfNn~u@syu9?2YSCN$4+{$=lPbuDx_HglyxW zQw#?Tm~wZvSsmL3t7iwVR}3;k_eZ-YtUg_dk;jPn_&YcSy-Wpow>f&7a&nfo%h zYaN9$vr^Q;yMoe(V)Fwl*}--z=5JSIuSd6$qtvtw`X9SXf>?~}v6dHZ;33~Id9K_v zNW5s}5~Afe}2SK{ySEXeJss z$6;P4|8vkT^vwZLMQ*=|`KSJ}-fbmiUW);d8vtKQJxo(Jh=~$4_PtS&`FYWMonrrx znb)Ib)$xMBd9Gv)r7^IazA` zttca37@m(u1tL9NUNLBb&wfZj{DOnVFp(}#WYu7BRkBd9O`wXfOYdtl8Fu(wfsT{c9Y&*v46xA zg}bzv-oKA%8CoX)`Beh16NRbS%dNEG^6iKT67)$5d#bAd%`>d!!;^^Kx`)m2(yTOn zcWU++LlYLvccgmSnvJCG{-O4o}-znhS5^VF9?AKcI?dZdiG zTPk1-Hy{4{@oaRH2n6N#NAgz^!gd$!(;d{JnL1|__X|J)2hO_isD`>Wd7DIB&}OJ) z!rvwGaxmhKYoQ>5TDDU=-u`y)Boq=TFs#ts5rKXCC5S=#ImD;v?+bGnw{BiusFK|M z?0{y{`JSr)Jqw*Efc9-9?M5$wMhPoG7jtu0uT0W+W zvr}z3O9o||a|rmo#gcY`!2SwB=Mn5F1H_D~W%li);*HAFofigUrVaN}*o(XS`s4$w z>kY1?wumq9OU97eFRKl=!_FeYQ|s27lEl=Oe_RwT6^4bI-`k_T^kup6gcvnss z6@rcbtWB_b6Q3GEK7qsIAmWvB-6Nr1k$Pokp1gpH5dH1Nh(IUuo=C;qrT5NM^Cy!V z`eP^SN2(h+T}eEpKgoYT;M{sKsF`@{WdrV7^er~WJ7xrdx@jbfw%Ww+keeU(x+z@m`-W@J-$|Rp``tE3=RPrHDOiQ>&EN zTaD^K%^IP;oA2WNy*KauxjpB1VjjdisIH;_q{Nk!XnnBWXnS|sfCZ$n!Sg;SLpw?E zTXS(r5XXW@go-Y;j&9Qwk`^Z^H$=-0ail*#mFC|mpz?JJ`hBp%2dOMG=Suc)rcejB z6Nz6@PA~>Jjud_?dIU|U|G{nECcd&Kn9j54q^Z(hFj8>ULYmY0$kq~y0IgYtZUg;y zH*rSK8}+%g%W>qIs{bVCiVjlMV~h_KctI&1omt;6*8Q2+vU`r;zA(O9*Zn0 z(C~WI+T?G%P(c*A(r8YmP&8bgrhCL!vB>P{9Q@%*a%FpK%-wXs8Yu5zW?83H7(3sx z#v{U4Vx1t6@H*GZ(ZXB-aN6|fpv9L{+5+?Dx!4LK>K#^IG?|!T!rhmoa$B{3s`gie zXA3^NHEwY}L2Ht(bhlGhX8WU%LiX1wDv#ncrh>RBJr;rJ!mc3NSPWl|$#wxe=sY-x zJ81tJ5a`+95s0NVocb^(Wh+;1{J<#H0)@|_Z@l){dF{;N!)j0axFotax+Ojd2CT@9 zk_NV!d8iLbDru{y=yNoXrY*5{cErFH>i}mN+)9;b`KV?uB|F+`E*z6ypR&F+J#B{g zs3t=A^!y~kYPJx*_DH(y-tjXDN4b41$GAjjZh&q=8lSr0tY>!%odjv??RVpbU_leOOAm23Nr(r*pUBPGnDOHQk9uCb+=MV_9Q8lGb56`Ct z&#Q#%=Y<5$8T7ACroHB6mTS=MT-lTWX+0TV*=P(QuZzcNJKw^WZIyU6G}B{GxTwy} z?|74$2ZchO;M6;!Q4C>U?Of8nb?EDp|ACe_>U1rrw`ODA-VB;~JOmhlAUn|FU=Yd^ z6e&I|y;F@{SeSMlhz+1~Vqwxdi1a0q*>!SF^lYklyw_g8(MkC7yj%gd`@rtWz-BT< zazVglKL=8P4F|^okT5aInCy&@AE@lI<0z?qW|L*pl?B+wWDV?m5u39G$)HX!q{O(; zwwjZ!$TR8T>IR1T4*?KTrXDlB1=)?%fJoZ`x8<*`+pV|eJj&RT|Ywp)g%^Nc(=FTs_a(0}mN)UtrI?0;G75ia&``?kf|oNE@f zn6DzVD2+Bo{q*sp-#p^~y7E1yruuuO)AM4EXRPk(t$X|X_hYFXLK=Bm2W-OFaApup znmuvcL0RV(3<|h5H?QUerdyDxc+FZ z=#F+T>NVdxsi+DepD@JbuVU2+zsu}9$;AXgJn_W(KxwVLH!1}D^3$ip>-xp4sY+dZ zR}s79@sh*#crR3mz3%Ov`!QolGl?n&T-rc1+*%bC-hU~BKd12p7-~bIwFY4&viLU}X&Iq_y1;2j}ADXORPPk^jQF-R3**Zbb|JmY5Fd@3@^HsQcj>PO${{>}Jq z4t>9KWSk?gcY`(J4XuTDdWo-cks7>cW@|DvVweE`htra(NotO?MM=xeiFceG=Kogs zt|4u@GxhpqXgg(Am%aa7H+?GCS>tXT=0$_GYxV5CS}mXV1mEWAZFax2yM@bI=)x-- zz_i-KrjW(6D4WfuD|OQQKI4G6bioNhYD!uLE{y_IgvX1->H#|^zXRv%vmNG3-Hqz) z59BZh!xFPNBc^A}D8pNDpwB5!k3(s-1()(-JF1grG{g0nkb z_0|vQzvDcH4+w&bVHeUucWk92GY(oT(PiyQ_)2)xpIkOdfjC&JUhs?oKy`{&95ncZ}j;xFRbsaK_*j!(jQSsy4V26nWc z#`_mGE5}7JL1Orvh~etCZKSnRT!5f?u~5IpmTJscvaTcKj^iMxi(uQn8+@!iHLx#j zGjyQ=*3M_85352rD;_4E{aN63TNbnj(%%bO)O-;fPz*@T{GNT>&Cy=(%=gMjv8&F` zRbBy#odVy%ebDRUe5HcI9(#|413)%zum`b{RF%#zYuPUqlz-G^Q4g$t+C4uAb32*T zG^ilw7$4XAg0K9n4GS88zBV)x)mg8@jMj<$zP4XbCp5_x2j|XV|9q58;lOZMwDxNC zuvUdLW|Z#D~22*}l$&j8sp zejKbf{u5{~WQCpZs@qFwS$(@LRhHaYs@U+#b7$0OFp=o(cyJE;9fa6ABQI}sm!m~v zQ2&2dXbfHfYy$HBVA&%6B(J$*o?LxK{E88gpxMBKRVgTsWeNSWn~W-EpLNd8vIP?B;`isG562 zCoO35Ut>eVH|{ThSQwkBL`tK`W+6~lIu*~P8(~x(+As==*l;ihIc3-(4Y1dHAb#^G z&^9qnV8)cjz7G`WS(=VXN2UZS%s35;T#&;j1bZXn1SY;-Ojs*N z`Y-QM4V&_TH8Yn?JW=r?k~6bLt`*vOFS0}4R2`!4{Gv5Qx*XcL&>HR_etw>7*?p;# z`^L(E{5gic6J)XZbivLq3m7XeGa0CsRMdNC+mUs5K2a1Le$pz-nIRYYrY4IJPk$@vBky#LfKM%eLlXQa`X$cU?c+z8 znQHkQgToP_grm%o>?%EEspOHRRrgeSW?Cu@_q*h!!43@q$hrMJ(YNRC`RmDplY#vm zw?R$hJHm%%{hN24rp=xgO!UqesYCL7vlQ6&$)a*iYua^;T0Vc8JEey?>flQcbkjad zyE92w?WpP#6|D63%_akmZB%Td!60Y%54-pq5G%{q1H2K5SY>V`}9wYYi| zp2#GuiWZZVMMvDIu!;El{%>CQQiKmKLReJ>InB5Fkk zc~Pb~9(crDGK}EP*eGtM)5-CKQR@hM#a^+pka>ihqkjjl8=5@7{8C%p`24B-_cYG{ zD$FHeHYc!<9Nxky#(Iaecp2ftfR6j}s!>VVy{X!KyM=Sbf6XJmImxeAzJ$6@*HWm% z{Tx&~Fi)L`%w^K+x1iQC&7w&Db)Iv8MW?9@;8r(eA8;la)l7s%z~7QLh`3uGAU0Mb zPSe;kCOx#9`>)0e0=PvBRdl|1mC!hO;_B{LV|=RN3T7^ih|WTwp37Jjz+RWXtkllc z1;HIVTDUl6vpDW4lkX|-ar>zQw&k=kPgkiNEZR~)WmAsqR8?cWvuX0wUtpy`0noBC zda!myjJdtlV3Gi;Au^XiUdq$no9)&m@>YmEL)XXiQN0Z+MM`6cSP}KSRUYj@6ooa0 z@DsMLf`m&(^Y;RrE`O_L+cG$Zh{f<=ln1mBW1D1oHTqzj=}59rrR8je;L+ra)Z5aE zia(;wM;v8?3yUCoe@s(amk71QA`P7l9)5FlpJpm&c~m={QNE)Y(y`>3^ZJavim({i zXb}B1pq#x|)q<)i`7^OHnUcBQe2*Fvejd0}_aQkHmjxPYANc$J7$xW=gIp_3^iSrM zb$;OiEhvHi%fxfupHGj@n>W^kn^y=(ScF^4`^=^B>MZ|BU$@Hb0Iq4UL zitP7kbnn+5ZSFi+xoGORa$}5j(On$8FaUm4$E%c!RK)I<9M7t7{*|I4GFP%D*7jNR zg}=Uc40!tw?n8jH!kn0_ENlnE;k)#?Gb_iu<<% z6$>m>@Moj=yDQC(Mc0>YDg&eRhsPNm6DJu=ys6)mulZ1Fd~3KyDi{U|`QO#XKY9z~(>T~&KnSY4H)s$_={w`rJbI!P`D|wR4;os+yzoqM z98%ZQ+qelub&o+kL!MM3cS(wGe$JIStI`b9ETHZm+&gQ#wRP;DxQz6i1&>EM@Nvvy zv;;s_NWNE2jq)>%=4o3L{GB!A58c@3?zQh8ew9U%?(3f`m9V4Jmo5>o- z2TZ`_uFCIZm$+<1f-0LM-@c}a&bZyfs}at=cgcc`Y|Vx6du^*?#Ico#zAzOBhhfZu zhvDu_xt`r*GH0k~gWooa@l~>=(K(bk)MW9u`szEE+Gdl>b6No-^dsCa>HrM*6J>N( zX^O$e&YxH*@wok`f9rAxgf|1eYN8<3OFWtK7ET(ud2Dg_E221+Hy@yKiKQ>HQBsvX zcWu#g$6NaQTf-5yA2sgBks(-HB&4s1QRV?gZoVfhCh7_64(q2FOVI$-Bow3s*ZbAWp75lUv46q9V|wMYw}vPHujF zW75}8Cso!JT5qyA=<(KdYBIF!F&uOU-l5eEnvC}r$G65xEwZ!Io?2Q~K`m0MYg<== zt3f_lf~9$*z!X|&m)&afjB)y?v;~FdC(CZ_yvsM9TATlw*H2;C@nsUkHY-_{ z-#vm*`a$M755-bKFn8@3wMP`I3NOX4?%!*?-)*Mzl^S0yMSjovZoSOh$wmkb-#whe z#E8&lbTvL+w!13K=)gT*+Z$#B_m1uc_IUg*O+x-K&M$2FSc6wwT$Wl~JqL7MBN~uN zQu86LDne6o5(HZLg(vh9c2~{T`a{%Mfw%kkUkerSL!tCj7!pI8fvdaD&Oi(Ub^_l2 zmrP2qv+LdT9~dy<)Y3uc3O_RuHu}|dj`D@KEKDhBO@p-f0A0D^z`tXsjfTQB!MrR- z%OkR3JbPI*F~54aeXsXMTGM)B1m9I>zwr%+A}znyjM`U^yt{%YM^DJnM5xV!owA#{ zRh%!QTey?-AO?lloaZ}_{B{;AD%tA60*G;l4Qz%E6W{2cN zgdSU4O>i!2qI$D6nb{JBHp;mZx_!594uX!%YV@sMCVh-~P$r%<1eO{~qmQ)h@f|%s zOwCPQaci&nJxHNb%!ni}Te9$3_c*F;(RmhBt7(lXm$2};TV>7w!X#n<=lC#Wu zeM;SHnl|eB_Mx-y&ROlL;`~Us{{V*Wczetsf~LM3HZc%ROxQt&MzmNh%tqDn75OFL z-g)rodkyNAcPW8g%cWiKahG^ol=1eLrQQIJFYpIA3I-TPTJnE>%k|wqzjY<*|D8ji z0q{7kFPoW4&??VlC?%!VF@6Y|;`d>>KF|>GShu+H9Hhi%81x;>@U);ii-+7`K{&N~ z>8&P*&rJMNO8wEa>dcJ?JR+-vuZWR?drQ4E?QKuI>qQV9HnW!KPmDuc3 z`hi;?UK$to`T@!NS&vn&YlZ|lC!O$@bCv&aBX7mqC0PApbCC@wL#l5)>%!B${0}## z|IH0fJ=1L2XC788GP|~qM2GZlX#UE>BC^{#lV5Kipo~1fE&r@|2-5NCfL>hSnj~Hu zUz7I5dl>zr49EWf3kL(3QUZ=inVdi~UCOR8l&$O;M$!T~C`>8kyA zI3vl&P<@A@FrrR-U*8ePGv%yHD5yJN(gH2eG!=iJiJm0%v*P$gf>?cmpA1=bAljup^&tQn$o z;1{8uuk0)GfJ%DfV+0Zu#KFT^_7Y1s<-k?_F$BX6LEw?V2C$l0oU>8K0C7+vZUkHS zbw6Q_Gk-(z&F0U;C*FxSE%ZXb9ZYG#V^ccLYAvp_GiNvHLd;!VoI(^z?-as@D__la*y+0Btzg~m=p%pqQ2p#GP;@}b}P)wd^ zbgc3wFX?i0o!WgI98H&lK^R9HRgXo4CEU9P__wPTnxr(cpSB*;f5EG(%^tTn6w6pS zY7{MGUHLT|f3jrgM87I{k#@rHp`)Ln;4pA;Tp5|#gS4-uKCwlu-KHAKydj87uDBbsB=F&e)SezrOr|)Vvlz9O~vRMS8yAjH4gh=T6>2yIbuEo7I+VNd{M-4E^4dqdkbYqd3 Uo>-~(Z6qI=ENFQ+$dX=TME2t$QD+m95 zEm_T>2TAN`W_FmQY7UK(4w4V6d5aEusf5Kz-z`VzaFB|m83@T;oWjjVW=K+{ehKr7PKQ^vyhq;dg77%%Q%}h7T?kJ{mUK zcvU#RI>~&^3T2bp1W122K0XR_baG{3Z3=kWCDD6KQ&$|o@q63So?%sByP|R4UiZWe zv9KAyIZ%kWv9bAN$_B`w&{o=FxodfdE%>7RpfH($Ee?E)f>s6s6N88-Iv+&Y@WnD6 zJ~pRuEL+fX**x-#@?e$vHG7kx(>E%U34GuS<-7Gd6!=RfKn5m__XWoSrE0 z5oS}ui?lFDB(P#3P%?+2sBwR9f1U2_EkAEx^+b{xKuKsC@p(Iy%C|z4%CJ~NPQFo- zrRKSy^m$xpWMp_S7p7E3a!Yk8qb5B~&n2YsYLzaH_YB$GYD2n4#piQDi`6`zvuMEr zv)LSyrppa67_)z(<^^+R4X@@_sd80DlPZIYGw6Blqcok0^O8fni8bhQzTj0xF2RtY zGU_Rc7Ar+>iy}Nge=2}_o?@sOR3J5rLL!Q$Bvc7?fqs{+5=DxJ#3@4QG{I$ZP58R# z&ii=(eX&r4hZ%Tk9n@Y*y|-xJ8Xy6XJio}06=Ms6PvC!j`)fV#9AA1Z?-caF9Y^mN zY8$rShe1SmGRSBk&S@IA4!}d`gCY2#=5*s#kr=n)4UE&#<#z75FmUcP{D$z9*KZPk zB44ZwOFFu3Pvap+d*6Dnu+3L5A3Oo*_nyV!3V1mtn)s5bB!JAAX|aM0j7U&OoAGj= zY-y=^>v~Jx_N~!S2#CZ30z`#HSjC$a_EUL7oO>z5)d#-GW;KNM%kXAj*eO5&>WI$%wT0?4{Eu6=tC zwdw23`5TKjgpMe&4EsNMjQubQ|C{(K*28CoDh+xD6=u19e$ep^ee5x=d?>~&{)K_t zyRLqFzNPnM=S66Mv^ zy1(lfQZ>x*pSR8AWBu1|vc1r<>1%akCZxbR z&q+0BZAw<>r^5fN311Is0H5s@<5B!CbEfG^FN~l^Z}f-ET^KW;h0Hq5L4Ccc(sq9# zgZQnxtw!ed zy)ngAC70iD%BQlvlJExkbNJ~C{{aO2P82zlo(WTbP1ArL+HPV@T8)iKZJX+T0K({! zR3L~4|GtQ34?O@g!+S694K3=j*2)9@c1E=5VNhB+oXz3{LXV@;BuPOk;g2rfV&az2 zqT%+FPYCp_eDBJShoWifggsunDtQ8l!w%_Ii;y0DUAF7&ajNxTAj=2<{RxV*cPpG; zrRnj1yv@FxM5pu!r)$y1NxNA4ixX-OjO2NgOLkAfNSrMf=(rw{WQ^sIRaH(<%YT3A z&gchmr3Pvh)rS3dXFj==Yog0keF^WqBP!jW{urg@%#ZV@NbI> zWo~41baG{3Z3<;>WN%_>3NbbyFd%PYY6?7)?+PseF_Rq&Cx10djTU8PDQO+s&CR)| z+v%K}4_otzpPNqIocMUPTxm#x$N@ZD|Jh>F<_wmWT~==yu!kQ93o#XFC@+x zC|ai?jf9e9w3LBl5~ZHFZc^ze#!h4vX&J_RCO6k+v&l$HgkbjhNFN@Jab=@LVa_oCQdk%TJkP*00WJoa6dZ* z5p23)(0^&5E&A%$wivz&b89~Tux3uZ4A}NNP7)5dpUw0=}8gko1j!7Y4BLmLVtZ0!N42+0%KyzRT{hc+4H< znt17ZrXO?qK;J4Day;h>k@ft+(2$R23E-Wvqr4sPQ!~j6w)Th66)^5KLe* z<~|ow=cuZTX;$!U*!~vbQZaD@%PWGx5NBX_#mV4qJWW_IDcUn2O6wJ+&!l}x5Bo(T zZU_Cqf6m1h%ag?oaDVqGLJGD>$)!qxrjZB_jy*Ulf!DH^l=SO6Yaj`U`ql@pcXrNP z&Y1~;g9~8<4hShha1fyb5~Le|0LP~%2~I>_4_p}loep{E;(Opj;D;0d5$AAr#$vtF z^|VSeq?{ZkY@|y=?;}oett}{qE`tHPt#vjaB?x+118w$O`Tx_VOwbSQXzp0d(>DOOU%B<)DW4?Jh0|I z_!Zoce`CSyO6ASmX)){ubCnMDc5YJ$PxeU>;fOvL`}nxV)abec_c+6Q&>K^23`j2B zKKJ?}I^}{#U4P3s-(*;Kw)wNF^5uF94;?~D9ti%>?e6j@T_&l)PI77!a8E=w+6#N! zKjHkOyDQmtdmz>JAEzc=>4MyI&k=C=IDE1ega`z<5nsHrw}WES&7-k~eGzP?f0ioK zIf4*vnO`b5PmSG&c9g1aO#GDA;D+ov-QPW*4bukObbmkYF7NLyzs5I{u+5)flhvvs zrR@H+TX0nW%}->iZ>WkV1AgeR^Cq(+^dpy9jJ@{+8DnCsvh%W7A#gS|xW1!>WTRDD zG~l;6hjwl~$A_NuXm~u}5O{pi`bKJnwAapVF{pc5#}500wPw0RW`1k7q@`j@^^B}h zTh?~eZBJvrr)zvq@B@{9OX(g->3xt=NdBg!km60uVRA7`uBq0|SnyABhW}g6o}k@z z51PuorU8F1syvA5hMz?Bk44?Gmv_1yaiOmN0st9MUXy?jEJ8FkF=RJ3VKOv1G+{R} zVK8MdGhtyeV>D%9GBYtUHDNwHG&V70H#T81G&wY3H!)!_WiT^gVKQShWnnTiF)}q_ blkN~>1~^j=4RE59R}m@%e(eXHlZ6o{<0ixM delta 3033 zcmV;~3nuieBdjB^+yj3>YQr!Pyz3QvOew{(9YPMq5F89{leEOi17vCK3TjEn@*%&k zC965~Ac-B#%np-O&9PC^LGooaZ_z<7m9RMJ*>Z#q2dPM!fsowADcpQyh9uRQnF|{c zlGEYz@qT%&&e#3h{Z^+t5ZJGT+ruDaac++7V{$k)s@G5y1R#I8E`F1<57-=p?(oH9 zpH`riYJknXY6H^F#{A%j$^j*18SK(E5!V7I!#G-|1|$WgB@}gC5>(>fUsQBG|7MmH z`F>$ix;&m6(d>^C?Q(CIXDgGI3$XRTZUwecx|01y-^{ZZK3ltF4)ujLd~m7o*|5>Z ztHSvO0sMTR3T2bp1W123K0XR_baG{3Z3=kWCDDIS(`6jM@$YvV+Y?R&>MMHYcVGL7 zg)k@&F$0AYOU29|6cixAV4G~i@ofVYH_(*vkwKDx8wR2=)JY(ac!9M;uyXWqG_xb1ZyzY5uN+O|XnwGCxx@zT$)o*{s#;;xIofl*g`#-&t zMLxo8N_deL=7+hG6N_HO(Q;Uw^I35h*B9IYskqr zYO>Tk7nDAa3yX}52E@-iu=W`Z? zhMLXhkThLxh{1oD6*VuIGi!J?w^o&_GMZEwT%1AAbF0&ID$Yv|@g~-w%lVvF8M$~v zhRUd?C|ax(y)9bk0s2z`)bkWW&7cCQSrigcG$o;ms0;MFbh#)}G$c+Dil+%KlWWq~ zMYr9@_4ma>0Ul=HiFHtWDfQl>gX@6=K=S+oLspC}2tI#;_wBEBzH@5nwY;;?33nV_ z9pPj4x06GYz&qIqYZY3Aw?5w6X^1fpZP$={{Lzu_c&QYo%@W zJJ8W_7utX1uejB+;Nq%gg?G&@{O80UZu;y5<{yc16W+);9qn%C{tNvBz3>~tQ(nJG z{E2+AGCb+T&i!>q9WC97U|}1tUOs#p(C?j#BNXs*Of>N&Q%L}sG1FoN8yJzGkT&Aw zKH1V@^NvJI-WNNfVJjdKCw)f8fo>tiV;{aQ7TABu*-R}|nAq)5Qc_qzhM%U-)Z(Uv z9=J7%<+ajLdnxqym%^wA&dZ{TE1MO!AG_zumIyw~Qy;Qd=KYH!6XQ6}g_+T+TKb6_ z+a8E9Ck!yvyITv}0b&0e_Z*cw?Yk5*oip-_R$=aM?pKX7Mi! z+}?Zj+w(`d&a_>GS~yi)uTHV;)_@i1>bHL!I?!0z@_owbFrpxH$xM<>4&i|J#@J5i zuI~7*^~CXZxQlRL&;uf7V_{=5pM)?&t}_XsZQkbMJ8V~=1&#MyyUBLJkoOq) zKK7hcbJpf$b$%-R&zki0koxhtE-@a(|1y0ISGr&XJ$j=jWbT5P`7C5sYYyvcOl5zz z3mL?3eNGIlLqvqdt1ykHkKcaG!bMH{)|xHOl06H05D9(A=xcHtp$T0aYW;EKQo=G8 zK1@kkqes#4OqmT!m(^IEMctd=0~EFBlgRMs^S4;I-ZwCG3{5lD39$}r*jG^}C25s1 zx9^Q9u5!8jhEqP3^_4`_%AdneUibwMF~}4Qlbs1uf2L_b4{bLwCL1><9BiBF9@-ni zXh|v%#DjldM6-t;fSKXFm-mKN{K}0EKtH^dEe060Q66Wzc!|*Cs13<-P+EqgOShba zBeZISdlM4^?fCasS3Oou!!!1D>+0+o6b?J2TOC4v^!(bcbHIt4!9r0G0Qwmf=ipa3 zU6k$de_@-wdxc(`5l*+NO_O$U?iVLC0T?OEB$w@;q*XXOA<+vnA}JUvp{VPUpi%$+ z@}1ES@<(`5yu3yvWATagd9TljcOW)}E$cn1T)dAS(n3wnATS_rVrmLJlkN&F0Wgys3nzauDh(ECWo@K&Y&SRO zo^GdeZa!?yCw^`^b#vn5)pDgRMUxd2K}2FfM%>k2)P`?9U%ua0-w%oi zg$RnG;u{re)eXgIg;H5jBu_78#$@2{(?8x!{wTs9L;2hc9vAS3h4{u5;XhGEE`lJx za?<>{Kg}Np(ma2aLWc7oB1BQF$qjP(S(#j3tgu*}w4qMV5P6z>qNudAq<|=v%S(w1 zCWYu3Iz)ig6NoWqO9Oo+~VLI2d{tc?ss6H1GgqRmK3sTqGo5EY_E1Q`g52$5H)o9_|x zpF7yd7DsMHSoHBW_i0nboEgA=G@|m|%3Jopgx+LLJ>dCv%2@Etvq66$S z#ITvhA-8{twi#<)I}-RB%&!N6Bf5F*3Sc|$I7u|*g&wIFdb@jl^~7bn5iqT5E-nQ& z$c7V`m~Tsx?DFO^%b#IU4*3{f5%4X1j--DS{4j`3v<|y*54bwL?%uBZ4c$(k&1dbj z)+Wo3bi40etN`qGM`A$=FUMEon`6)W!!U)tFfMwl4N)RTVY-q|^G zIcFvY4nD*^@IXijf`f=XkRbgC1b87mN$`5)4ZxQX(CLwfE@1#6L}5$;(BmA=&RC*X zx}H{PhLn@TgpG7*=zYW~&e?)u=+f`An_6d6G;?+{9I+X04SyT2(*l#-Z?OA~7ybSx zN>CNcsxIycoEn35$%go-GMs!F>;V&O^rp_Rf!GzAJy%HJ!X7o2#uBsd2{lA!Gmosf z4}JkRm>ONT;1*|i3wmR!jRDD} z-{nD9gr_3#Wq;c;&Nms>t!@6Os(iWL!d;6{l1GBywcEQqN|(dbU~7146YxNUHrk6j zJUrq2q`fQYx;>C`{U@nOSGpkgJn#e@J`NwO1tB5E_;8!@dYM)89*# z=^Q}_SLUb6%|m1N!HrVYjY%HT8vK}Dru%ykvO(IQOMiEHetGfj`g3wMiCz97Hd(DI zQp)a++k&I|uYMv^eMMC~7zksJoi~{sVc7GT#n^jCkTE94DmyQW6#{2dgX1;&_>RiIrSy-a4Bkm8B!AUXNbxG>Z2bGzETLLgW5GYj8UJrNJA!uC zKWHlVnnwJssPZ7HAAb-vJQj7!p5N$p#D%*43yV@vY?FWxEJ8S9Wi>cvF)?N}GdDM5 zF)%VPW@2SDI5IUgH)dsGWoAA+IAUcrIA$?1W;HW6H)Am{GB9RhWi&W4H8nS8WnyJ! blkN~>2DsA=6>y=GR}m@%e;Ws%lZ6o{Tidk~ diff --git a/generated/webwork/images/webwork-1069-image-1.png b/generated/webwork/images/webwork-1069-image-1.png index 1e39387b4b8e03a658b48968572d180bf23ec808..54e58a6f49e6e98aa011bd2c27ceec6071bf9bcf 100644 GIT binary patch delta 21 dcmaE1^}=eyPHqli4o<73LFK%g4{|RT1OQz#2hjik delta 21 dcmaE1^}=eyPHqkXF)cOj@3q@EALL#x2moR`2&4c2 diff --git a/generated/webwork/images/webwork-1071-image-1.pdf b/generated/webwork/images/webwork-1071-image-1.pdf index 6ccf21c03bb290fd24307882bfdbd975ee8842ea..3fa34d445569c58709df3d902ac7c9c3f4261ac9 100755 GIT binary patch delta 3079 zcmai#X&}=N1IHS}>%uO_N zjTyGT(cGNH-x))qRL}GtwoxASS<_jV zN&_x#;r62_ip(q1XUjHv5Y7 z#f58`YNe*+Kujmt9GtQ_$DGCy4M#4@3W=7dCjI7v3NM;PQpA?OUwp9~Ee-G$U^$kqQ`|Ya6V7rXP1^0hjB7651y7(@~q~Ae*mL zM+p0H4haO70QA>lpsg?U+LZN&mJW}Oi0!t0%$gAj!OztVMdy#`pNe`4vh(6}WtoP< z*HDgXU7M*Df-iF;RK!ysU)k8iFJyU*90`Fl1>LRq+nnkM-TYKb>%_!<2~!E`xkX!B zXy%EFl1;<;C@gvh?(|X9S%L}*_b_}H&~EU}O?p4}0xpygqvpOFd!rYwaJT7Xh4I30 z-mA=R!uD_%+(W(-Lh1yYoKZoLI9P>&w=sHd^9M(dL{&uLt1M)wc%V^HA*hH)_1v=^ zpA^NOyCvUO?SqTHyIf8_BS>i$G{1JWTYA%Msz?$S{A3vR?4jtGXWaA(PTyQdt4~?W zW$;kNq`-iq%O@L-vuiH&Nz`k1CA|ORS^5D8x<6T-DRW`%n(;F^((RVa!lw_RfT&sW zE^SrQB`qn`kAsL;wL);DFv7JN8)11$)<4R2G^@PW1xL(B6CQ1BRFVx#Epik;HKkZo z^x4ZVicv5PVj-3Ecv~)#ZMOtWks25BWvJe(1y>zbbPxTu%XctO&pOzFf{}xw9uE3Z z%s+8g)5pnBhqC)XCTRGH-CPOk<6_uwhe!bF?$a5uT0pjhX0^z zaBDDM-?iBuk~2**>v2oey`z+K5X9GvK-%s8bpcv-nx}V2Fsr)oYPo#io_dRJHI|d* zpMshi2@uk{&_3@U|Mii!Bi_8?-xw?6OjcNpwuS>k>m+Fr!;1PW?$u_T1Yj*#)-=Yg z#;qsW;il^VpWlSU)N9};b>qz@d@^9GIx!2&U#C8pFS;I7f-&aLnWQjx} zHU55E!_}dv=|*4C^QlG|h-8B8bj+qANP(R2*b?t;w1zPK?7Ns$ae~e_+i&wbl@*G# z1M6#Jm(~WLB4~Og;1l--dY^I0@!KCviKG<3`Agh;kUN9%16Yce1wyy{evFPr>eS$_ zL`ilA4@Y)eZPe5oa{BAcB{pP0d8MxI82puBl$UxBdcEh(BawsSxT_-C^fsOL{k`eoCofxunOB>d z>OL^#Y4k7n$SU(VZZyJ7A!Y-$k=xl1g5$FrgI~HEm8$W_2_(GK1&&IRM8e|NP8%ome$k9(MJFd6UY z_DesP^%Y37Fxl(eGg0j&ek{K%>oeBg9_G=I7tMOJZa@88AAJY@)3@SgM$<*R1oeN^ z%5&el=yoS`hgI+dZnda+Gy0QPxDgUTDgTZp!86F+4?;@duoBPU&umJ0m>FOZ1xsY{ zuos0BGfIoY15l`y7lWi-4h_IGtI(zZ`|?(~&CiaSxm-eJbIy;VP1#iMJbA&?a_Rg%X_8EUKv)d8de6yY2#CeKo~&Lb(F`V&-?C^e=B--DwEp-U^Bq|P;ya zXkWSSQPbtW>-erJJgJVawK}R~DTc9`tZX{Bw0JilYW~ob)vH#hw`5({X}(gnBo6G> zfkLY6-&-=Z!h7;QvdU*@1XV0IpgALJJGo2u4^+EliiXq52Ssm`&OB!{3018B_5uEMfne)5J}O}d=n89qm1 zgZyV1`MD1&DfPVc>{g4u_8*n7ZkZ|@+bl|?w|7KZ5J*yvJ{#rATrZ3`ITYXG;X>5n zG8Jp(!OOGr1%d&au}qUYF>25>CNOUD%-{!fsF1LLUhp0jhz5CkUoyGntJl|Q;(iI@ zp=!>Tx`Iw7TqYN!_;qb-6W%%zNw0>D)wG5d}oXvYCVJfXC!2@gmq?wSbB-ij?%CL1l#qlsi6>n1I3gc^Se*}I#u)EW|d zG$Ks8-U!Hf+2PUEONI%jVQNKkYIAuA=0}sVoihw>qR2Wj=tEv^zmCcEyev^ojM!z1 zr+}GAQK0p9;UmxK3HS@!H}=)2tJ6W^;Ovt%A%2fogZY39crE^9Puk7Oo zH%|?uhnE)0O&#y1rS7h$`~McCNnV_>5fsZC87fxeoZCTr-^@Ppejuk^K4$v7YTVO! zKK>LrcWKf5Gn2kcYx)&rSgObU0OQ`sPei(TWs|L-ZNQZBEco~s+(5d=C&7kPfk;bZ IZW=@W15R?z-v9sr delta 3075 zcmajgSvVAo0tR6FGWKDL&|vJ@XN*DF$&gTFUj`#$F^PG$GKj-4ye>dNE@qXnp#WFQaOc)g%9YAPU2pSa#h@5n>dV8-& zjC1A~fVH!=1tygn3cR++=I?f?2+^1}eUE&8NiTu(_>_fv=4&VZ>@0tCr3ftw)zaHp zs$3LwA2Eq?-dL0Bip9Z;i%9jHO=b>5UC)cIu}55h8tlZZkAK{0y0$YlHLy>!t0o1v zLX`-H7*dnB=o~&43MD5tOV{`x#+h#mi_GSTZp(QTxE79i$9Fd@vHz$D?Ld&(t+Gmy z3+1Osn!GbHb2ag0<_LO1Z|0CopM1Vw>mj&=-F}JmHRjhKTZ1;QO@I1+8WVJ)A0m+J z{{*X{xn3mS3%=_e2n3+-dHts``W}vv`4%fzO4v?V$XJ~-vot*uKMb9NOJdP9g(3bi zP3C}Kw+gTE;Yo@^Y4%S$HWxv|HFp)5Z!&)^zT_?BnUwed$DbHQhsfbX$>pJ!Pw&}` zQXBys!__!8*2@^>FuiA&v1_Nh`>y%b1>i%c9FbVhMutWQNU?a8<=|CE5x)t!zEB2U zm)yr>b?8QhbHmYcJ-ByT1%R?Cc`oWy1WE=u_r??ki0^i{QLaMj{`9CqT{HVYp>Uck4g^ zsA1SZtLiV%rDjEW;d)t38B`EB9yUSD8ruoe!qLfRpB|qY5{oL6D1l_H%Ad~B!(b6r z-;4KXqm5S1n@n}s2(Z?a^)|Sd$3HWl&UtrT_Z=yjW7sIH3-O7m&9WLCFo=Bo={RHv zs}ErI5HzXtFrF>y4V78n%l({oDCB+V?D5{zg%TJGvI9!9f7vroBG7A}@Kq0Ui@1eJ z-P_+r$n=QWNkf>sYsOtFzFC-${ZiY{FvEWHqq>U^!jOG^LqmBIUTa|!T?OI8zUo3( z5~3m0XW@_x5TRxSRhu#nDPFQysUoUBC?8ld)0iA9)g~@aBX2Y1f}9ehemV|2NZ|od z82|+Z_@uFT74h=ZSyZ+2Q;absb4wQV3dE*bdjsZ@^p&lr&{~#R4*g6YuxP>~s9>HQ z-CfMqD*cD+To_%&13beLi4s!I=4INGwOg)VHVNBtQpx;oUnR>w4WlpQPf)A(jmq{M z=_8M+6l{ho%fUhkQ9$B|--~n{_ z<<~z8th0bQDP5dVyUt2Gb030HK4m{O48^SKl68(zW{NTt@A84@?@d>g?Dm=0NfZkA z32cFE|7zWT<7@jdP?VAnk!u2*HD4Dcyzv8{-Pn~#Soa4&-N-$2$};g`1plhd@?R`? z$^CX7^?Mw@zn)+`!=1vj@4+H-JKd}VYt_GxoQl(Jc$Xdc{4#D}ZB>kBWQzE%0ZY`X zKIl|L+#6q?4XT3b$z?O+HB)B7ae*~T-tHzB3i*k!vwQty>EoMx0`HXI2A9)iKxr=|I@hCsPed_Vj+Qcbu zQfhlQUIJ4IDt9>jv45SrUOjwPu9>Lc5GG<#0)Lnw`2{?$+}W_qvC=$5+(?R0KqA*E zA+9`9EY5wv44oCcVbsSDZ0++H&Q}{N!e3ms?MN|$`sV&Z0-gQ#DK5+EEU6U1ag=Rk zmX47KMnkN=yzH&DsKu&9+K4}_O);-9I$q++8zF=Cj5BsSgl7oOkOPV7^y?AhZ94^p zPTxwAeoEyKukq5D7oVYR(CUVf@-X=iBD;s`(i?@W54@wdc4oh(6f_OfuTmOod+FNX z{R>`qmHD(p>ct(n%?4^Be&NZ%33t+i!5&5=P=|AS^%vW3+i7jnK@Sc9fK#pzClQ18 z!?S6n+wY$fW(}-)=}s^p5M}-%f|$Lz?fc55&7gXa?&+iNrATedCE2gt8w7|ij9)rZHBt8%lmkgT^yJtyzJf9#p!tLVzx=d&G}z0Tt+ zE;r+6Bpki**`cwPcFd^V9)$mgm|j61egIqw=Pi+8S=peAd~y@S0|QQW5Bp1RXr{Mm zT2uld)UpwG#0keF*S;w(UORCpF|);2n=1AuCH&d?rkf8tM|2hF9V^gdIgZH^(R<~q z1@@VVVtv)01u&Oy?0hT~_#!DDEMj6Lo1)&LSBOxylLcgZr#=}STNjjYdS$qH5d>NW zG6RhH?td~JeyHv%VIxTq4T|`V+FvFKN8FgVIUwOfJcJY_&uyXM2-Nk_&wd(4$uM>`3_r65Flx5itY$cnOI` z(Av*OQfDhS;jM$ur%)GSTDxN=`9j>98fovztlGjVY&@vandv}EXOgC&V%^5s;x0kh z^I@wE+B;84jzEXgZ6BT3xbE0s85T$9ZAth`@_32|z zY`6cuW3$48WSZx-sz*}j*qPsH%Kh_8J&pnXXUBSMrEf1L^{zPw{;ZcNiDxaG(?IB) z#vCJWp(7bM)krgUf>KY4$qc{UnZapWJ5tXqjL4g|qUvjA75Qr44TMLIopL{jb9$0H zpS$Y`meyzv#1ediM{Pbf_#Y{4JWdVqeNmjelv8N+0$O_t^zzZoSCS`Q01E;shpEtI zT8EbrgNJHk)S~V>|HV<9ACC=2^}k=Z5*HXWslprEacDQ)7$?1!+Y|L#zF0b&yK+>tvZ@EH*ecZmB3B z&O%d#^&(G$dnganIG&Xl7tDm1pS%#veL1h?ec5&#Sr@YmMd-?)sQoF(%e_E|wIB{f z9gfr~UMwBIH|n`g z;r-$U1-O@(n~lOpyUtm6g@HHn9uDztta^{7Rw#6Vd8~ubsAp9WaiSV1&-d);g2Ej(l#EuT|1vQ$(aNHx!oWo38t!h|s3fzXM zwvxt9h_L8If(DA}EL?5!6~>$*H%K6g5*a)5dkEVPe_}c?H2)bviq~AG2JV;bCDSZt zT>7-88jIG>x=pWFC>ZZgFAV+kQc@tR&;9L>8LGihEiXPTx1jVDKctzc$fVt<+Y$!d zP41EKKZNG2uF%quGX}QVTG1H*(xR4Dz>~7vbbws|gQvY~Z2GH~ERCtv>71Twm{cLF*ja*U z`1by! zH=NI>-Yx_Q1a7^MpaiBk?M}{*mNVC#Gp90FPT@j<3I=d5Rdp2<8V-ZPJm3g#6dLLQ z^FVoMdTXN8ywyArI#>T+0`~*x%*`ZV1(XRD#^w_{d#^U*%=twP`Z7Q8feFdKMC(N0 rcSon>Q{~~mFK#m)rL*S!uh*ersF1M85VSV{rU6w4NJ$x6ngIRJMReB_u%KRck~^huzWq2Zt}?T(G9CQIwv;xj5< zwh-GR&V>+R$gg8p;fZ`A1)1^jc%Jf1`@*8IUvsd{f-oySTlj3l4lxTQG)Wwfqc!Gr z%k?n1Mg@)7v73szRn|95TJ6vQVpHBy=xrNb=$&TEI=K=ZW_Nj$C(^z!fI?Lf2bp}j z4VDTFn^4@NWRtzHB`_-|T(+nLu^+jOKaMYB)`NcT6||&*3wVjf7COQdpn7IX`b*r3 z-1m2?s!OMyP8vza^gP95KU=9L?j>ggFMa)+MO|8P%pV^b*)SFJuXvtKtjBCMG`-r@ zvv%$$N+Ug{U~=!^%eKB|4ay*OOfeuZMk4j*9GQEt$M2ZdXmkh;_ycy|S^xVOE(xR* z!3(FEFSymvctlO!JOW}ygY@8d+Kg;0T-bgJyd;JLae%t4iptDw@AL2v2-~Ydt%SXU zY0RsHjiwd3cgO8h^0}C|Q_8QYNmS<=gRY}LiC`0F0Lz_9vxS^88+ z0$E&aEb(q%>ZrO$xsKy}QR#jwO5Cb?wiuhkk!9|xzEpMjUX%_U4WpA&A%>A$N zI?U7yZrIS*?u7p!g+)eYL1aR&U)lvRdpcyt`+oCz*x8K1`~ag$JKo1!(Ola(n#_Lu z^6z)Nq*)s{B;36XUcDurhnAs#3H-5>?!5h~BpFB4zGQ$DB+RME;ogUZtjyY-##4Om z5b{;&H&WozxrvR(+fa%gf(QfYmPb~!1X8n>kMy8W69vMY66DnU_D?c7FIU_4qm8zDzWjW(!l@r^Xo5WzzV{7Idh9S7%c&yRFtnT{qg+i><7U%2E z_NF=(zy(_yrfl5>U0mmnP;D1&<~Cmp%n=_+vNSYt3Zqt7>&>iXFLk)Kf!FWY2nq1l zdN*b>hM!KSBX13@v==uWhC{}a&m-n#EdzWxt5BD#R}crGYw8OeZx&gvB2#PkNwh(W z2*Tu}Sje?zeoi zTB9J4%~xYwy6o%P6OA}Gr(|)iaJz==Ab|RbtqwfM5uQk|p1&f^G;NhKU9NKQDwaQ- z%nWBCQ7b0P>EpZ{8u0aD>d$!{3;?DOTGZht+ zTSy5Kdxl+0Fl{MtL7O!UnAY=l5G4^jW(?N$pkaP#d}EGe?#gV54PSV!QF#WG4t?Ha zyI=pbvY{-HCMc?aE|{8;F4zSX-g`>_FQLGtM4DqUl;X+GTWGS;bDw{d+VD>!JIXrO z28+hj`us>rdH*eWY8_evQb!L5AB4A~tvhK7C#eup3<4qjpYFG6nWXO;CJY!t-_c$! zi$pINVH8ld7{H>(JYDmUxRhlc566LW_nlVFSw4}*mF4}r)^?O*XZ0s$j@#x;u%r_I zr#i{K5pR^3FO)!^TbFVbJy zNS_qf9LiQp$k7Gwbg$`;3a&Pz@MpzA`J^+e1%?k~O}`29?h@jk)?Db_HUMqTX$@M` zy_D$V1Cr(n7W{3YlVlj25nZrNtaajN|8H;&wznXKWj2DAGWGXR*FPN*Bauyl& z)l&UdJ7_usH0=tafcYixlYAoog&e zaI_YAM-_pS;!7rl6c_nFYw7jEzOw^5|2^?pheq>6Hj{Da)l=YSZp-V9!{9#d3jzApSkbh973H{GAth>*-0AVIOdEY|oOX3%$R**5T9Am4E8 zC_|U{?C;F`?p(IYW_QsGMEl~3m^w3P_${)D_<9pu>hVQRCytezmf%sIwIck&BfLLL zTn_SmWk3U=nc&r8XqkzKY3i2Dzvxjnqx)@63e48LOCt~d(k6w)!dhoP7mScbvvYo}OT5Q@ zog6Sj_UsQ#2-XkI6+$0Z9uujel<-ycu?zf~hAsmx;ZRd6c3lDP2-L&zoO}-A9@xjG ze0?a+N`Dd=el_U0C9iJ=`oP{@54~-bA7CSb%6>wsWd3C5RMD>*A@II7&9Ja46EF(1j)q8BSQX!5xjSsb}C99-Bv(iZ}v;Q%Ra>4>k{V^iPO zjznjjQ!@73PFwv^mQ#CCnxkqn+NlvO&L9yL0+*twjpO?3w|zw_!ffpI=nRFB3%9@Hud(o zcw?#y4eE9IvtuP*;c41<^13;|fYZ}R3pL~_tjym%;is!`_n*hH&KL>)kfV;DTuFIE z0Y$8PM}N*2Y0}&Xd&DPO(`lD+n4<0Nx&##K$Fjy=AR|17Aa zNFVlX{5U*o326TVM#yv+)fUe|vw$+QM{UMtl*>o)S08At;%f&%Lphg`Hl%^JCS}OU zIUbz=D4!qTCJ4`744sz4Wf(Eo%TBQBeh#Wkkm1{+G`i}&Vzovs`-!jmR$3O<)E_*5 zP0^@S-R%e`?-Q1$W~Pj=rOs+PYtINMt&py<9C&AgF;9Ip^gZrHEpd2XcG5-)yK*3w z+;%3_8J!0#)J>Jj!g7mI13tFIDHA?C;$NpfZpePDOo){D-%?f^xch8_0QZ~LU(uAi z5zgG^?^{QNX@01;M65rVmNgsLAYo#MdFh?0GIli;uXLTS1C%C1UK~AAV3d*+*ek? z7DR31q!qPwI0SdjE;MSyU+Q4b7el%xTQbf&=fT=7kD4%Dkv$I!aA>g%ufp&RY)}MF9 zoWIBK4J(A{Z~`|==hds;e0_4ciL4!z*at7B-=&wHkNbe!s%B;P_m9RSc17FRvx(=( zeg!_fFQzW6q?4@(JSPsYAC2Q!Mr0xHDumiWMo|PC5`ua+oBcwG z@_lUM>5dj^#^P^=O7>WffbFl(>f$lk{`%x+6VIvMi6g@&E;6;jx`U(_sD{8hpClnf z1u=ITS~({Zvr3uG^r53d`$4L_Ct=yW9z7Ovo^00Y1uhio{GAKA7)lu)U@HsI-{Dqt z3CL0Js0#sPrE9jdq6%WN+?Jrq$`l9ki>5WyJp&@Rt`5Ss-Jdg#q zDMe_^N(05Q-y;O75ucb!>7!36(=Na2cbAAYC*81hJ7k%rm?hBZ{gRu;&*9J4F?Mge z90d!TPdi^N9yIoVi#wE7*eI&pp7^Qk{=o)Z{XY458k@DD<)#+#~%g;%>JxJ7KaKaK*`Lx%8Mran4YpiaRP(r3;vU;dpay*)F_ z?JZvH;nIfjuFV6|{48GkNrwwxZ+TSTrZsIa=;$FI`~~mV7p99=b}K!m|` zt)!^Eq=yqUP1)yXe%lVk^TROls-lS!%nIYc80Jmd@Oxxbl>Qz{(AqU#(E#4 zMbAv)`jYIW6>c7FWQdX97XARxiR!x&mjc1=P(?nvO1k-tTuCRj$6CPP6rX6P5#Fh- c#K-UFxbL|XNZN)(nT`FRmWpi|XW;LP zjhmr^ZN2l|z0)OCb7+)wkbGLrTXfJ%B`i+*ZaG4SgH$BVKuGT56mC8;Lz3#q%!!Q% z$=mJy^wm69$MgQ>da2VL2<%tFWi|*|oSQ>CPY%aM^%{zT03?6s#cy(UhRv(c9X@&N z!wR%g4Y0XZZ9v+)F+Vt>azII02D@}k#I?Z5Fpid~0ZBn=2}NC(1eN&r7ZqL4znLXP zzF(M>E|2F%H2dR3yWHF5*~+Bl0&G37TY+tqu4KQ_H}fcl->qFThx$SrKDbo)XxM1u zRpI;s5@UR=3T2a21y_H2OjB1J!0~(A(w<>`z;;FByuIv+8)9K2h;yJ2QMciI;$#D4 zP-rV{vD~$wVhcW?{ID>YfGrMujG|Tt0uzIXJaj&YvY{Z$Wcb*e#<6Tc&t=CYYh6zA zO@9CU&L`*4lte<&G%b%=zih>-@vvqb_c76K%* zDT*5R^Y+)N-rk<(?dzUMGW;nCO(QwJJwtG^u~mxj2KK=T@ZZRGgO_;!UhUm;D8=GIH^Tbd^z0QM6bodRH{x1N5W( zsTV1Rnob2!GbtpZXi7pAQkUuX>2gt|s9&5c6i*RcCf9_oi*C7(58W3F`FMbVr`A60 z)s%Y+4z2~_4@q5cN~8mW2kY!ejoY};YlB(fjFmW z+}aBdp%ePyhsy4{zlQf6ffLZbrB+$B78b!$SP7w)HOTZzT)q~N=XP)l$K9uEX>9pP zjIT}jG4;06Z1(fCgq&bNT3J1G!G(JCY@aNn*b>aDwbB;*9cXL23oY{3+-g~HaaE(j zyXH3jbNqh~H+_B_^N+;%B;Lq49j$KX{>!}=yWuy4C%t)#_!0R+Wmw{=o%`#KIhr~* zfQ4XU!I%B{)VlLYCIQL@gElB-M0N`MdG!xp{|*-O5JQb4YQ2aYxBtIc_v3O0rgE3pjw zJ$a0$VHEK<@l|Yq&$g;G=s8rF>H7IW^EdSA$GGgF7_;~n21fSY`1aD#j&m(npcc*) zA5tgViZozFx_a7_+O@{{(1)tqepLbg#?Dj%w-{?T608SV=A*XUc3?x}?VH zEbQC_AET%RpGJm7U%Ji0V9&+=<7kShPKb42!@i0-DM_uAxqXKwxyt48A*XyY>nn+< zmA`ej3NL>$K0XR_baG{3Z3=kWU6D;|!!Qhn@A(xv zc8rpbHeCiG(4{K`)@>xs?yjw5hQ>Bda_GNbY?U4+u+1aCk|jATt3##a1mU;Uyuk@N zsf5K*&z2)xu$L-`6A+S5ck(+Qmd@TA35QB`8uB~<2$wK9PWA??z0enYa=tGs&`QW-W1&)Ssv`qCF$$(|Dsv-!Oq9>oF8%qr`^mW`rNYP8 z7;RjXUw`XNh4hnm3psx?IX*rLb98cLVQmU{+FejhYsDZCy!$Km_(~};IkbHsfs#ic zhx99X&>oF$QlU{X=8%71HQ^n4POwZAx{_CQHlCb@j~B#gwxa)FNP2??XL z1XkB2K`sCN6+2@X#SeFcJv|2`rQr$nd9TljcOard%zBF|R_B6;w2+i88c`Z=JF~_{ zRyu!ql5funWp0zO3`PMkllKfLe<}?YxyqcRb!;~`=bmn-b8bFt%_n|tI(2K}WWn^n{ACQ*<+>B`#Ua3~`k-QN-?d*={H- zCTkNz(S*{XrD!vfQffgFe?*PcAc718MTE#J)Xn#b`OlqfWQ!xWA}soNoBOma;J$Ki zCBa7zw%1`09>R3HrQ1P+9dm_SX4tujr8n0`4R|H!t2!8Z?Pkaf*a=lqLVH5smkvX~ z?;nZ(UW>*7n|$W2ltN=`xh_e#lzQ?GbO8g+k?0^h4KZw{Y1nO|e{IIv*Nz0f2J`EI z;Ca$4(xo3ahaI7 ziRBgHaD+3ldy-^$FPTy0$>3H#(~8FaNsfm20?_f06|f!Km-wR4njmA zK~e!+Kp1d{Km`&w1x9HMAqJ6u0u;tm3gZBisJgu+1G6#JtV~uKZ=^6KBU4yBA5jeF zWsc5%6QDb8x?S?6GOL-k3-V<+B6GdBWV}r>J!qDGvy7MB?oSx0%+_U7}u!ZIb-rM zPwPeCZOBBYdD1ugg$vE3xJKc&~yC&W@pQtMclZ!q4I`>AD9%p z?MhB}D_ynr>2qGD>+K$Xw9tyvh|wSIb`vKz#W1nDHT>WLP{~5)EmVhw?K7C0axbL`-of-ha(GF3kQP}i#}FA-ulnSOq*?mz1hh4N14*^`BTPz9t{uUzlLSP*C9 z_8o@87@^8ny&~Iapk!wC_B%u&8EulRhS10CyM}jy5IS5s0z?!dNta$X$~PK`(+s!E z9tXKa{?L^nrqIOxZpa$Pk*xD)P3nA4d`4~W`#IU_%K~4O@&e^xJan>C&@(dM`rmug z(fd!6so5q4qn)XL&UTxma^|x`jfIShwef_=^4eC1RxdS=x=ntqQeEXm)*7dXLT(Rj zys0cVdh0-;NXXYfH#{2N16P#XyZ7NAfj>g6Xm|~E18x{!bjI2Ao-;ocAVqwEw+JW# z9wXp_A;}nzmam$1>e0*|Hrf-RFtxs=r=O%(zwWUz5BuMLqIYg8XH8QPPK^`}{-Wrs zY5xzs>6{?%+>xIPv)=vh#wmp(M$I66Z?dE!@l^MeS31aAlOwcnDw6y z&=g_{Wo~4XT^T4sFf(H~GBILhIb~#FVKOu@Gh;DgF)}n{Gc`6cV>4wpK0GipV>mJ~ zVr4mHWMN@4G%zz`F=8Rk z2xedcb_=myb97|`D8(V56<#<&F^X9Pu@DJRf%&m&=r^m5?OBg$uY}vSp7FDaOt00z cd{|8)alL>aA#KB}Gq03gfQ;kpKVy diff --git a/generated/webwork/images/webwork-1073-image-1.png b/generated/webwork/images/webwork-1073-image-1.png index 9fd67248398f5dea1577a0b589570610a9c27d5b..b3d25af9b5fb670fdd90c272467891f65aafce09 100644 GIT binary patch delta 21 dcmezA_tS5~RBjGo4lb@0%S4hk&*i=&4FF)K2rK{q delta 21 dcmezA_tS5~RBjFdF)hOml^+G0=W^eX1^{Jk2+jZi diff --git a/generated/webwork/images/webwork-1090-image-1.pdf b/generated/webwork/images/webwork-1090-image-1.pdf index 095c3c59fb316f754da4b5c62291c11d94b6c042..5f6d97d4a89b65a5bdad4cf4736b9789b16bed4b 100755 GIT binary patch delta 2492 zcmV;t2}AauI-WYPCkuZvK0XR_baG{3Z3=kWT~NVlgfI}j=PTy8r4*yfwmpbYT!b99 zExOx7Z!sAqP?Hej!M|@3?H+o_ka_dw@!n)LJ2S03WS`E64m}J?ONX;roIvPtRGQ=k zNG1Hmv4@QkvLl}&~ZH?DJiX> zY?_Lomj7nOw$Tqe^%$wgA{0u)KLDo)yllWJN&JzcT1z0iCmQ3<{HS7!s2>3}FOQ?ojHJoeWBU^rV!O)THMUl2cv`?%TE4`S18zc?kCy z!97lLTX-DJ=fVVA2!i-a8syx=LC$Op^63EhbD@NQBq_GILaBTqN~w%VFcnzMdaaIS zBD1$L(Q$FH3MNLWjAM2gHD-ObI-f~avpS7Y%?1_&v(J>R*RWP5GG52B1v|EFvsf&D zQEFphl*z1pWvha*=vf_;qAApvi!?b*qA8zc-cTDgOfYg(5D6w@!TYSn%p{v~H0FGQ zAbCpOb{;`UiBRGZViOTYgcBkjK@tL@oVY^1Oy1*_@Mbyt2F_RHTNC@p{(00k&*w_e zH5sgyPUd}|Ik%&AKj2VMJzrwd?si0f!a+!KJ~RB%(5@TB7hxD?J)_H1&o$>fOj4M| zlQKOddW%*{FT)~?!4&-H8gKdgcEf2n15<~amG1rUD(rzbA=1U5O zwRC%W$I`$?cV54GdIYH7hT~)9@YL>|Sd7=>P`oMJRtgRgV+1nNy=II5mW)?YdzK7liz5uzE_qNeOrk{NP2$>ZA4Um>_1p6PlI=k zX!O*BpxkEP&kKE@lINCC%_1Ms=vNv1a`O5Y-*$~&=>HCyVaVR5%Wzcap_DSV9d19> z>FWI^V48*n=c-Md5&!&u|smwji?48UCqg+(r4KAiwrDmI-$GJI2(3VA(8 zee9s)2k51iFWtCBk3!dfp-*%zxsU;G2PV^^J*3iEGatOZsD+!D2~>yPTI6$1?!K(v z1=Y|>;gdo$7^=+G2Wu+p?Ow_t)MIT(x3Jpdnrfil^abkoBkKoRXoJTggb2sk8okg1`rMTpfdkzYK$&Q~juOnW5|>uIqdwpsEJT7C3? zniALTlW`amlY%(mAVSdzLc5LHGWM7zrxR)tqCWKB&*Y>J z3q<(h<#M@%c~w-Z8%g+OwWwhvO{b)VK~Giyxk1~hfJH=hVqE0EqURz*q{*Jo&_JGblWrDO?_&|viCVSq%7tD=4key(@LRRzY9UDJJTj7&^T+~F z^51{9Y`Zn$ms=Ok@;RK+gk?@|5o+TVFWRGas2o+G-n}*qRia|AE%iz$Dnjj1F$%M6 z@w{z<+WXR%AsXE!T=$>N7n8MfEBq7rCcA!g3ZAcMv~j!q`~ymcj((GI7!$MY5!eQQ zI6giKb98cLVQmU{+Kp7(ZreBzeeYNBtu_jWq$KK61O|MmTcgRgkkcf)svZ<+q7bk} zfugDq5^Dk$iv~xnKMiXBLRsM;uG$9gmc1UpLoOzLc)n3qU{qu!JU`{ z2oOXtK|Dr6-zO1?eFPwq;UW&6VC3?DlFrmx=~Y{lfW8??=|VLI*B1=WJKIAvPR3*U zU8`~hCZpfy3wi^4LzkPPhFL#M`(c@n$M4WVTR*nCen4L5Sm>UTl0AM8QkIi)`IC*S#_E3F!UNKSxnV`K4WML z&My2IbN4Rz@%T(bVU!+_d$@snqPC{p8l=h#E||FB$w$UH8`^}cZ3%j?uXeJn9yfdP zbATw1MV$Q{4tqaRHS@y2VE)lMV1g(1G!I8Ce9zeXVSl9y_LP$ge^!`ct!muP1lH#C zb0NRj`Mb=EGr?FcSQZ399)_8J7sYwx^+S{i5#}*>kT`Hw5@x7_oHLR6ae`ywmN`d9 zUYzI84BBGT)%L+y!>(|8qkc59R285^T&#DpGEbe=mXm|jotZrqnglVuDvom@=(Om- zx%+FGrnk!vv+Fc={^!t@TDC~H)uL0FV?cO-N^r?&PZKO`e7bIu}0BEfREw~%YV4(iozqen>( z=s%zpRSi({x1x`72egxW+QAN*p*!|Rv4Qj4-emF2;%qs$w%uSaq@;KAKPOiouC7zH zDH_g3Db(9_p)GuM=rD(WW%H<9SJbjI`Y@PO>H&uIo3231rXDOSBtV+O5N~SRHEb*& z47>6T$J})*o!=2JX7tWJ2EXL|$a2iW*DUwEhRIn`EM~H+ogjz^k~69UpCQ)?avksp zRqjRkn>wA!voeP@xPE9<2bb*h3HIt6A2V~b|1W!I^^E#@BAtLQuR-!3Nxu^MZ`5XV zqI~p)erm8KunhEUt*5IhFa%}ar>K8>&KgJ@qxcu3=&*{D?;9*aI5J{0VPi66F=jC~ zVPRutVlgu_H8D9dGh=0AWiw`BK0G)wVl!c5GGsAkF*adgV`gG8Gcz?YIWjY2Wn*PC zW?_?E97R_)K0XR_baG{3Z3=kWCCb|kfG`Y0QMRgAqe6xX^?Xd2RXAb$fpC~&xH~KlBC$;3Z?ReD5Wwc(NthH>$N(TiOk;0 zM90O&Dwr6hGLG40)R^_z>U<_y&FVBpH5*tA%sx}LUc*|M$ao#g7VOxz&0?{CM5&F1 zQ6{tYm8}ZKqGxqXil$IwF4E*MNv3?3c|&c~Fu}-CK_r@t1@E&OGm~t}(U|iIg5)WA z+j#^bB|?cuh)qNo5l)DB1W5>pa^ed4GI@_z!kgvn8#rH)Z%ynY`{z;HJfAB;*JQ9- zI+^!<=G>0f{eVM3^?Zp*o8X9lgoE&!^O@n7hIZX3z6irG>lt09dagO=VUofuo|Ne! z$y>BidKngB45r{m*Lcg{w;N8w8JIfUtaR^(S7A523DLHFRMC@3C3!#v+`-R%=0Dd= zBHK@VbahQC>u^*R(C=oa7Aka9@_~= zfWk?_U~ye5%P7f*WOv+u&E%>dITrYc;U;8_6@6s`Kq2Y$oX2(Cc|z_8NR^++IjXW9 z2;86UF}TZW9Z(6>2T*F2;~|k1Y+kw!oHey|Zcp!pJMa~BsOpP9)>f+^2XZS9m@d7*zU>;l(ElAY!;rmAm*J?;Ln&o!JKTP%)7AS; z#z-`l<0o)9HsE#?x@Ujc`$aGGp@MPdWbMpV#~T50K^GlLUAuFxnQv4lt^`bps{T5hAbIB@&TTOjfN^SJ%}dfgvR zrsbN^`3aig#@+mK-m#QOs4@C{w?LA~<)*rKe?7FNt_{huh>& zkCWh(ZpiFXRAHqWti3@Cx+gTs^ROMm@`RDlSuVhF-9VJlapWZ1jU$gh%cuWrO1swL zk2@dEiZPthl;vJ;32GU%n6yT{qYBgpwGCP_)DE=>TG61Cq7u{^wLsx6D?D$NqSm4G zA*n`x3E%xk^TpuY*oyE(p~=1Hg;o!WG*JlHqCnBE-LLPEvJ)q0 ziwdyGArFT$XU;GsjD#dfNkF&|gmc1^fCwVOh;R}_Xa^)naVH@m0wj@462V9m1SBR& zfB-}?T*N_0MlLUZ=~S<^S+zw8=$nxxo$JQp`kdi;Zwo}@WIU#KMwctF8NHj$={@W% zU2KXPru{JKhebXfe?SLq{nVQJ5y6GEU>Z7uoho7Zb#eqOq0?JaLPt`6(fy$X1lR7* zKx18kZheXYq1k1LGmnQ~z#Z;y9hiPsRhI=GLvNu{#YFFaGlsV0>>@}ocX7dw$EOAg ztIdG?fP45s^wzdpi&Tm5g2@XWedaP0Lzi&1Ex`=-)lRk5(`HY84iF_+%-PT3u=k^M zGb=0%W}lq{CVAve%W%Y^kBogB_E-8~PdWMUr-d!ny2kBHVr@=8AM&f6f5=3VNyc)? zvM>zuD9S{CoaC|Shd7fm$`kG(N$9OK%1{S6XEFNglIJfBx?=# z{~2_pQ7zK#wCI)e=Kj}xWTd?YRX}zaN+*MVC=IE@xcP{qu|V2XdDYZwAdIdo zYcQ-V_aIqb1babzvNQ zb?7jEheh+GeOJ_~v*s|ks`Mia>93{&qndiKsE`0<4@11EZP&1Id@$_FcN}xyt#tN4 zM8fEUdkp@Q^CQa%i{7%_^BSdRMRAy`UF`)yWSE{&9R>`!PLb=7$Eb1<=kMxtD$mLu z*5Lc0(H&f}(xpzQ0=@?6KP3G|=>MZOqZ8$$YxmN6H8nY7G+{SoHDNVkHZd_YV>dW6F)=wbI59SpTO37JH9kHH zb98cLVQmU{+9k=!4S+BZLs2_yVTc5| uJD>vF6E)C0YG%3q+ delta 23 fcmbQ;&N#cBal;924goPO^F;Sp*UcBWPn!b(VQvUB diff --git a/generated/webwork/images/webwork-1090-image-2.pdf b/generated/webwork/images/webwork-1090-image-2.pdf index e28642bda603a2521c44dfd5dc86c31c6a464962..cb1c4395e26e77ce45138cf8995aa3d17f0197b9 100755 GIT binary patch delta 2479 zcmV;g2~hU8I=4EoK?{E{K0XR_baG{3Z3=kWT~JMHgfI}j=U2>eODTyi+x8$raS<$R zKUVh##AK8}O+t(Z|9z8a_s~Oz%$qll_l8!@xzWl4{c=XM=wVP=I-Jeo1VWFa(j?14 zD&hAoRWWf(Xm#QCgHH(b{Z_T#-D7pRzCPW$I(r6*!$#;V`+roM+3i*vzGI!#I%O(~7{>se#t zR(ijX6LJ6wWp0x~3`YSqlZp%}f79Jz+0mu^Ft`RR7~5!LIGz|&X@I619~dOC!N~Y! z3}_YzBnXI-Q5I)({y@7Z5l*5b`fKb*d7r!2-RGb8`~7);NJ1bWNRpK7O;IK7 zOMXt3xc|lAzFmi1|BkPfhj5P(+~Xv-g~!o+E=;h6Ac()DLC!rK9t6*Z3$~b11Noz4=Yx0>y4Xf9hG;Clo zGD+ragO;^1k@0$#E!eSbo7HNK(wGXP%og1%TNRAe!0MU(+Cr_RNSnhXnDbfYO^r#* z1S3ZUp)#8a-e0-)hGEt_x=i(4bIrpfg;_i)GeAO9(MsuMScEZ{f*;-EEq~vxKMiMK>QJ-NlMJuI z9(W6)?MYP8lL;kxKn2{v&wb`U*GwY&PkeNBO)BefRu<6jW~UY^fXWkfz$N&)gZi&t zlsL>Y0QK8&e5@Rv+PxEt@p>GJH-+0v z!3h?MlM6f1o)DwZVLh5=FaGf8PB;t{P7r?vi|blhMhQM7yW?plSN+Jjz())>A!|6% zS2h3?l1|Tg-N#(V<<5Xq`SF}1YI{oH{&cU=Q&!`I3ZOoKQkxtPiELnRqHDobT~q7v z_FlLHUqOeuuJ~hJl^Sv&w<5)yS!~KSrovu$qo19mTooTdB~ZTBQ{6rJ4OZKGWm$jG zx0OhQr1zK5MnuK_gT?YRc;~RzKs^Y`ZT9`V(Dx~MZV6Q{@)3=GmBBA3uYd7v*XV`* z@1Plm9Bujx=LrLpQl_>;?Wa23z29VvL}NLA0*7NGZbzZ7Ez!d;)-d>G-`O(*aF;@1 zkqekls~}6wrc+3UZ^}|3ujh!59dv*G0KL@mr5m^CQRq7OiM}NlGT`mNWLk9x)p}dz zgVz_ea1%3u>d?MLKIh`@%j(=v1+5f5DYSsG(o%Jxx}wg}L>Yw!tPANDR(ai1jkWHY zo8S$!s~1LK8Vj+a5XV6L<}y&9KM#W*dRPP~q8*7xCaDbp2bC9@>uXztSmPWP`Nh-g zd^HlubdzK~Ef&N!OCCb2kNyv83fJM2iV#$PZj55$TDyV>n_WR@*HPPtJ*LU&0yT!H z5B>KuIqAa!5x#hrTrP1@ZcC#Fh`(JYD;%KKMmg-e(Mg0WHrhly0j1?ID|}@uY!nv< zVZTI+5FdYz`;X#fe>iomFq|z;*o-&f7MJsmr=YN===0qoB(uwHb?*mkOH+4{Wf4Gs zJf-P=aCdlmSKSq#{o2tfblNoVbSbK^QjON$uthTfi}E~d$EiHoDeSF~=!9u7%Gfk= z$ad4nBQ)yYe?F&OYxygj59h@c&REKGuh)dM3R=$EklsiIX-oPHT5-~jv%e+l3HNAu+v=-kwb@I;}>z8~7);}vaZeO8!%ID3wBv-A=A z27keB+c*%t_bYg-jlv-*i4s)=24Z==b(?Hj+ihB?Iw;aa5nzh~MZ5NXeTS5tIKl2> z1=!?}kHeWaZx|Cs0wQDL6YhD0bHb%hJmLi*;lvNo_K6?kjwAsB1QARSkCD*#NkpWN z07Noe#K9AcTwc;dt(9K3c>(B~kp-Qp#(&`YjNy4_dx*x#cuYTQRV=}z^z(E^SFks9 zzR4>{`eD)!^K3kRgAUs2q1Dwrf^%a)H*^ZSx`4%>lOtdOo!aUGIug5!R)-c4T-iSj zjj;*Z^#lV#vrDBlkB6VY9qw-}n0~6uF7+LT-auLB6SYqn+JdtSUt;dw1wS57G=Jnq z=>ho;EBH>-*0fuLRC&P#6Bj&w&p2m8n{c@;Ko9oiu5QbR&7S-mAj)GAXFrF--cPNX zX>MRJeQzBw!DD-x$0HWLW$f*+ztRPJ%E^VF%|Z9t{blPp`Vd${>sbYB6>`;%Ll~wz{}#s(&>QMwbRQ z-$)eF**dQ}5;~s~TleX(eM@bLp-=Gy$4N&t-(T92m zw5#{DgB>(OckHix1LwKD%HoN|WInaF-C)n8-2S+|o+R(CZx(8kH=KTO0%~&@;!SP4hK=QeVOPH5n7eMJ(>vlx zM(^xn@SmI?S(Yq($#TzY7@rlzVyA;F6s_!CrjhV`h%_e`W8io>5;B(hGh<{qGh<^nG-PIDWH~cq zV`O1sGG#etK0GjDIWaM1WH2*hWH~cqV>dKpW@BVIGh<_9VPZ06IcAf097R?&K0XR_ zbaG{3Z3=kWCCW<yPOK*A`Z2!dh}MI>Sa tRABvF1M}&cSe`!Yug}aQx@OkfwesrEKWo4Rev=R#9|$uFB_%~qMhZ5esEhyr delta 2498 zcmV;z2|f0=I=4EoK?{E~K0XR_baG{3Z3=kWT~NzzgD?=h`z!XCs;V$VX?s9O6%iGO z^bts;9F4soOU1T~Iq>(j4b7nkue`IfqnS<0`dBOJAo;SIH|U_3N?4rqY&k-QgH$BV zKuGSkE>=D=Lz42u%m*70lI!)oy?sBHr}OL6rLEFu5ZJGTb}@el+1A&`W|(QbIiRE{qFlN<;#%Nj7)Q(0fF!51 zfV`>-f=c|GW$Q-QZ`6ID?lWH~^?wf>0`Rf|hbZw!4r(FwJv5)K^?u@MV$xtrD!gCM zXcM;L`vs}(au<_43`zkslZOl`e^c(j>gZAtgKNNov5hu{3mG_c+OI;c+yd3lnT12;whkkaG_QIkPdyf2RZB&xH~K zlBC$;3Z?ReD5Wwc!Bk*1>$N(TiOk;0M90O&Dwr6hGLG40)R^_z>U<_y&FVBpH5*tA z%zjh0Uc*|M$ao#g7VOxz&0?`csf~qECbRaHtqR7XXLZazO`*nIq{(3tP5CVIrrM}s zf{~+wNH7@--e)ytCfSsu0x{>4uni=C**9>$BHxo= ztto&*LG^rzNxRz-2?yad=QG1E4eh#7d=Z9W)-$?H^;~n#!z6`SJSo#dqPJ+J^fD~M z7)-&BuJM+?Z#SHVGca|yS?NxJS0M@Bf@s@*s_4nYl02XS?%?M>^Pg)bk?kjcKDxRl zm326(3g~yUQwtS9<%v4r5`5i3{nsGMwc8Z5E>GC+oQ1)`yU;I*@asehySqmol$k+) zuKeLAFRmc=cRug^997v41ny7w7~EyG z4yXj`11Po1@sP+0HZNTV&YIdfx2N~Q9ry}5RQ1IlYpYd|1G$w4Oqs=h#%x0>ybf>l zvy+sw@*}7M%GY|TyC=WVVtubHEBdw)iIDXE655EU*nhBCo(At6(delMLAlMopBMT* zCC@FPnngaM(XTT2<>d7*zU>;l(ElAY!;rmAm*J?;Ln&o!JKTP%)7AS;#z-`l<0o)9 zHsE#?x@U;gdo$7^=+G z2Wu+p?Ow_t)MIT(x3JpdnrfuIqdwpsEJT7C3?#n#vOlZX&he|C&w z;##|c2%DWCtlOyV!yeP*bV5x+)QA52nVjrlfe2r`OD>nVs1B9xMu@+hku{FcbXqwa z%;Y4(728fnJOS0oaZ&inUf3uuj>3M39w9D|->-G`_rtS&_zR_e*x8}_J2 zU{Rij?KqVujl#hSiB9Miqm0cXhpe1O9-&tM{_`d6+Q=W_d^j)WaK=)edp#wjZO~@X zp7cyANC(ne(3X=*(k5uff>K6GNPE(TM7!+hyj@D#htiiC{Uv<&|C%o+80Y3zgeMA3 z_WkHI{=K3#)@OzJ3p;_1b+hsj`UZb(Z`(Ey{@!1~Z?#c4e32+oMPMMd=hkSmb+(tZ zQ1wBPE(!r#6eQZS_v?40?8FJSVFGOO$cM+f=bmFi7zv3=NIiTi`K^m5L~&x7HaDfbn8b71;Kmt7io484U? z=M!^C8QPMwt3YAy;)RE1KF#G5nFv%l#TE-(5y=UzGu)op=d(O#+Kh15v zF%@p75^Hn*`H)}i>|H8UDjCZpOT#eCq9_%y%3{$EaVlk$DefUE^j3corKp39Gnob| z!7=g6oTDS5vh0~bS8Tf4Jy>VB6;9XYk6IU|1eA!6^mG^meJd{~2^es~e=-Y0)d`)%`E~$Vl@UboFFOqyl==6;1|G z5>kh8^ASg5fwbx3rmlZBKp0(F*xr#iqVr8&btLjWC$9FRAR;3tbIxK$V#zYUH`g1$ z4(iqJqDM&&=wHwpYHFbNPeUK|9%!czw1Yj=Lx1eod<&Pky~*O4#pz<^Y`4K)NSWXM zn5-5bzTPg)Hm^AwEuq?N^47su#}0E?)Q`q@MWs929tSt2d4PW*{jDvb)pa#klt_TK z$06R-G#l7DJ{Wf88;-f}Ryw;ULNR*p9)thn{KT?i(QB4_UZdoqC=PS8ue>0L43i70 z!+;^z3346s7*#If>`k4{<#jZc|*+W(cki+aX=J&{hp*C6?a zq+bdBf7E7lrhFmvnf_8^OJEu3#ad5SnPUjXy-#uf_FOcOJVo(05eu-GlLs6uLSbQJ zVKy=`GBq}4W;Zi4GdE>sW-&B3Ib<<0WMw&FHaVgwwcz`*GIM z^ySOn6hLbM34`cG5EPRrA`uIq0_*GQn0HrXx&LE*JTtfG8d)#b%<}*~Z8imqlMo#r M2sH{NB}Gq03VempSpWb4 diff --git a/generated/webwork/images/webwork-1090-image-2.png b/generated/webwork/images/webwork-1090-image-2.png index c94ed04763dbe9d8f5e10dcd8cd0d6c3327e6a26..6f71ab939b350e902b75c4079d85ed947d249d04 100644 GIT binary patch delta 23 fcmaDdh4IN0#toi49Ksx2B0(!P88-*=2-^VwV95r* delta 23 fcmaDdh4IN0#toi490FomR`TBze{2rq5w-&WYE=js diff --git a/generated/webwork/images/webwork-1091-image-1.pdf b/generated/webwork/images/webwork-1091-image-1.pdf index 955ddbc63da3a231d8a80ec08930b2c81f726634..0cfacec6309214b1192dce14967449c53d80bee3 100755 GIT binary patch delta 2798 zcmajgX&@7h02y6JmFkkOr)~H$c8;x+33j4x7 zK)k+MfzENA5%rKkQuVgm<=d1@A(1L+yM)9pWy^Jr%PV?wJ2TVIhARTGwm`L%G?ZpG%sft}RZ5GSZ<9*R>vJ ziXo#N-pzejo~$%7DdJ05Zy75PBz)WeJ$!CkcjV3BS_Rcv7@2mSoq7o@ya6_}v^x$B zZA-bIi1!u)J4n4K9H;y9UWr*@uRfR2%wRXf&$6iVw-t;MGI*;yulHxdSp2}pjPvHf z{K;BNOm`WR`oa7ie_}7N)0Ob*xBEdw+bL&x5Be)q^v$0gLA*xX7Un+3L!|T%cPxv! zB~zPM1GwdT)a->)aC11hba^9jxRQtWaUnh3Q#48NrDjC{NMX2y{H~WZb=g z(jK$WaUPBIhz!YpVIh}av8MY%O>N04{+dj%bDFj};Ke2qopUt~#ev;r2M6)9)VMHb z29?PL4vmda-+9>7@Bd-oLiWX-VNn%4o`@U8Q8HG=DORfdxx$*sfkkXXbDz!n2B7G$ zELJnc+%gAY&1jf@ZpG@90`3I`1yWkbd}Fugd)wke{?BLRp59Tv@R#<^ynZSIvO%0{ zpG7=Ix9zhl(`0W#as!F`^SUrx>b?RxI_nM*fb4mjb~|6v(|6JV$q%+m@hJBwy`z^J zk2N@y?!R$y5&@Gfxd$`8X)kpkdcgBl*6z8Nn^c=sov)|hfj#KI({vY&dP0tv@Cl(wpHt}}BS9QU&u-q291 zU9atV;lSTB_93@kZlOi&P~eJf5ofc@M^30Ty`|gUqgSA+(GSSh@`Gozrwv@(me~-e z!HaOoK{7;DUf|mrM;c#7-nX&W)tmdQYK<=hEBm)q#Y6AYjodHCPoce|j#WM*`l4?o z)z(Sf`%$cmbZNP{>+KZ#$u2MQnU;6QFA}vbWIdeg!Dx~Y{#cvQFJtlV??AV6u7%QN z)P9R>LkD>mu%6=_oC9p3(D zZ#E}RAJ%&u)+IEya{lrku=a^J>jE-BMNvl~nvIR&ipJ;(qeFU}^sAZ?Ih!XBO-)_? zYE|sb_UEA$(*hAknpoXxPBRV=Z)f+qD9p9^a(YH2O)ml*F3PsE=CeIn+NV)5kWoG- zQj;+ZNqqj@9m`Ge4)R%<3J{Q0b{95CpMEKlj70=h>QBqb3}4TiqZ}!ac-sHAm~`eJ zW}E7+*)E&g(zqCkKd4}Gc&;N={!6vMzJFDt0YvJtD}5##+;sisFji8|b@Pq&Z?M37 zL1Z11vrhk4?ZuDj5eVCHzBO87t1KOj*@h)eMfmNdIyf7vci)VQy*d%vZ@*R}u)(r+ zWn*bI$*^A#=;8UBY8h*wUx==Yr%#{AR^1X$L}tG?NR3NxJ9rdE)Syp@_h773-Fgr; z4eukMI+)TweiP>OS5gCM?I1&qM2kVe1%p#040fUoQKe#`D-}>(om}zDHOcAx7ocCK z3MkBqY0US98(|b?2<;AT%()#@rWrd|xZ>jCD$cfA3>h^MEXC>z_xV4^ec_lfUVA1E z&PdL^FcMylPxd1Hl@;F#j6J^vGRyORuLOVp(!PA`mXHi`_M3pQR~d?7I#>fN8l#TD zu%xG2il1x?b`*0FUhhtXuOe(+90y@TS*TY|`y$YBn2aJjqDjo+63)@6kXw@4mbnx? zX2mUV#c%Nz7oLEiolz0-y10ljN3q&5qP4akiS&=K1L(Ec;Tk+anhohZcmhcyI#*9g zkI)(S%*dj@E)H#)tuq$Q&!7Qb<(r>2T_?}ViqhJRxB{i^GJl@63(b*TyS;>%MjV-p z1V*zlHHBBX44)z?dyK)>BgnjnKV^k$cBZZErPfJU&~~QWc*%UstkUqMaq+5M?0|)g z{ieK;CoGF-E@exgSK3HGbr7X^kDA=DlgRrUOL0V*edZ^&(_ilX*`r>VuPIA1?}3W` z{PT;8`3dF_lPJN5p8<$6nf$*_;;?gYvKIG~9g6%9c(9_FxT*a(NM+dsA4&+gH0Saj`zED-%icq5n3HY%CGONE(yG%HI zY3*Oog^f6CUhJW8x`v9Vopj7CoXT5keyRDLW}wUD*z~RT#|QH%bC|&+irb0T^J~;y z^Sn`~itY{5l~AB3f@*Cb&EtFwKzf79VKUYz?kcRM;mv9~gG$GqUOA5T5KxWzi9@@A|ZF0dM< zER;&xm>l#cKgsu9fNd1+$DVk@SD&pD_zJ3&P1O?-Nv4`zf5>|6%N!GUjILnNUoJXX7+$N%&1VP3q2IuNg00AE^h}$(i zNIvqUil;ywaU)@4H~OzIEeA0wCmY0Mf*{2TVou_vIt+nyA$~7Cm`^Kz<}K7V*UM)I z9BS%AUKHumO#F&Y2?vWA8F<=PNt^U3m1A_-mSQ)R!7^c=TnC&l= z6sifCCGB7cEWE&yV>XZ9!mzbCfhM$U#j+(EgrjM_>rr({rT0Y*|>N@J) y-a48Pz5lhq>`3^qmCG4kP9-MaP^R0)ct~EOck5;WU0}FKW!&LHWoDdTGCOCFvqv)iXPo$>%s50g zS$~df%GsQ(p6A8$=Ktz>|9$=Ww(E52)HpB!DlizMub&UX$&2yzKlY{{y(8q6nFX)x~t=)0pGK?w}#q0+N*#tU)g~UNW z-gY!hM=yKOVjF6P;uGJN(McEE7%=&8(2EIkymfz@Yc2`rjqK5Ttv>}iMQ<1edd;tI z`%^C0FM>#|9`Kt*?%5e-Zd;EAje}Li(Scay2i(coK#Vm$G}0(-m+7`A4@0F5eUfLX zBx+{3KkCR~$a8o9Dl5=+ahVob%I+9oTJ&1{QBmw8`Lr}as2(c+df&n~y{;bp7@uTD zJqYfD1wZsRTmx6VG9jIpLWy*iB_VSa5 zAlWJ%j|D%b#vw-z=8Q{>yMxk+Ee7%%aM;>oeOZ!ep~6oo0Fxo9OE4#x$qVagek%x0 zaW1Dq-4f?CQxY%2mz$k3pT@e!m0Q z`{$x=wk1W9Gr}}G9k$Rr$L(RTO{*EbjhJR;TElYRTBL9f`(mF;akx^9$BY_HoE$yp zuEj;oLBS@_#G^vX$!&yL&Z+?ge36ZX==;-u*9G|1(p375m z41AjPpboA>?SZJ0yMnVFMdekFAQ=!7Pd!Wq?Kgfk(`oFllTc5}$*jn)Uu~D=Ex5T@`KBRV;`9Rs*GgrHgs>{mA!ZZ$h1|35t443$e)ku!o;OLcTKpNQduIUqK^h}~udZMaI3nP` z0foXcUI*p`Q(uvjL{z1kav1=l!_On+xafA90lfrJ@kue%ODl44iEYyBdlS*yW8W`4 zJF-0HlA;(iMx178s*8lTKQ)S-P6KTG}nkKHl=Fx zweAk8ecysssU`PzLRva%pZ5b<{(B#Cm(6q=KaYd8O3Fc)_Zchm&zr?q%Qf)3a2LyW z@K%mWLn+2z?kn|%rm}Qg$J&XKTZl%nzceSu+V2(S47kXJ)&{hJHDQ`mf!5|7E}i;< z)~f}Y>&mH1bzR)DbUOcV5>u_@cq)4cX>S2}|=+Q+QTpw z$~8q&wtwBoZMmO;O*4}MRo5vVehB+Z@6gC1r$!^^DH0*vR7$&sjpPjOxOw;jSSf$f<<={HoIkDm?V@leolGj4V zZN#+(Ec7qf8a_Jq@R^7ujveyjEoia`J5|3vvFm#eQ1Et{cE#AxlR$*Gh;5nvo#a{% z@`n8L`?@QJ4S#SLraG&TKd-X>nI!`JE!W)6NK^3}sAN}Of)&uufx@!d$=t#sk6c&CJpQjTr zrx!Q6nEPOdYVp0d*}P%42HG$4W|B__A2*7!N?MI&2rflzo+2FU(qThnNZgbZo)fdk zuXjq5+$JZq-~c!RZQbZK;2iY+T!d{?Q9HPk@O76XRzvsI8@@3Y3|H{SG#mSy3{HFn zrdVUC>Tf8L@SfVxGW1#T=JXM%i#>dYxH&wVxYK4gI3QiYM;`S1=zxq@EO=py4ScG> z=lB#;TR+&K*gM#|o$?DjW%PtQ&m&ow8Qz^wCHq?uXl8s#X8@KzGvdZ z-MT%IaU0%}xO4}!#UGg(b_1dfMJOq_C@HGmRYoG6p)O9U&QL{X6&HjG6p3_og}T7h u|6h^1b;I>~z2tpC{79tb9U4-qt*Gi1`!NQw)J#zkdR0boaXk}##{U4TU|<&j diff --git a/generated/webwork/images/webwork-1091-image-1.png b/generated/webwork/images/webwork-1091-image-1.png index 06501f48e176a31dd7eb00837ba76ca3283a1b78..a85f2f1260eb5d5d686962617c308f48948961e6 100644 GIT binary patch delta 23 fcmex9f${qU#tpGN9Ksx2;-?o>?B1NpBV+>rZ+-}V delta 23 fcmex9f${qU#tpGN90Fq6jN706oVz)dN5}>MblnLl diff --git a/generated/webwork/images/webwork-1092-image-1.pdf b/generated/webwork/images/webwork-1092-image-1.pdf index 4442f3fd376f152c03db9ec051ecad4aab2ed800..14f2f5abf42edbc88b4129fa0f2c3b449a57ee4b 100755 GIT binary patch delta 2624 zcmV-G3cvM$CjBL_90z|gK0XR_baG{3Z3=kWT~M)Z!ypjd`HCA;MU^2@st%T=$dMHf zX{wNP$8dI$Vn7%V{`(qiW#|CBJKjCLcUf7VYNG>W-`+}&0a|U8$H9(XA~e`(L)Z?K zmSL^ZEqmceR$her5wc zs)C?V|Ni#NMAOP&;Yq%aOiJfx#%#eX;=_UuNn>S4wV3)9T8`e&e$r}E(r8K>d~MHJ z7pKzs16r7IX9|-W2v>hyNKpl5 z<6d$*W1jX>*;GQEk1fM=GK*=%Mzz3bks`g>3q&If2{8yXg(yNO%>S`<(RTBJ-_`Sd z2qY3Af*_>2y2@j^${m`j>fM=LVonABPOP-8;x{3FGr{Kxc!htE-zdsM2>p;{oL`sm zOHRgZX#|#BBq9is^LEhmE;UUTYgnhxWi?wkD&JT@6_u5hs;FX`E~6^!jLT{?IH)QE zXJPCHEZ?0Zo*o^JFL&Z9pOcB9I8Rp?8Ff@66_7?%0YX zMkE&S?FAv*4tub(d;u7Ol8=yx`=<^zFzI01)(+sF5eRebWkfN_`$KV2k8 zrh*HQBvV`}DYt@l(7n|18otAGn1ql06GOjd`tHC3n7n@wp+hyW9}dECDDvw__ZDrl z4aoF$i1YX3Pa*{Ne-`2oD>8XB(Ct*5HP$t$fV7FD@Dkoc$$x#~hEBgqVXP503`Y`Nk_>iThi)Ka!?*5@I0ihvbFF8JW@s$NvOhm@4kk-~rzFag zaQY%+C9~Z_ye5pV5#iXh|v%#Djld zL9>S*fSKXFm-i;A*p+GJA^CDfwCG__S~{G~;sipEqtYbJKq}#nF5Y6|mLyff?FXL_ zlJdRm*zLJ$8kW+RdsnAtkT~p+ZnX%1+0nCWyUrdbrUwIgP5_ecO>y>ag;Oa_k56It z=_ER(M>yT9Hcr~b+FzVdd!VE!qFlOr5=P=|xj@JDh$N@9fV{2?f?EFj%XUUTh#&4q z{`$5dDczo}zU=iS@(x5$2w5Ld#p2xNAuc4Q3r3X2$Ih&=ffdgmCNyt?3T1ASHVj68 zF+M&Db98cLVQmU{+9i+eOA`Sc$M3dTmyVUuC@60i5h>Z6_CgE^Y?K--%F0sGI<}je zb5FO^IX54+<`X|Row_;k@oKr!mZHfDiXbAfu#o-&Q<8eo3nT7oFKWX#pD*9#k{jivr~_1HG}`25y6${VE@9VyGdiMz zvQcysr6DevO$>2`G*ZOwclmB8%|=TjL(zoFtf6QVl2mF&5u^&KMuY+cMMTIe)W!Em z_|F||c#9*q!YulDoBOmaR2+xyw$24ReN?XV|%kr8iec_4qZ= zRkkzq+RdN|uoD#tDeVq|Pc{U9exGkR_IoW718nk{=b8)}TFSHu!l}}cccBv)XbMLL z*lCDjGmS$oBW*R*ytc>jHJD%b2S&8>niary-f_}M&;va(5A=5Tdh3Z&n*lJ5b1o(W zHo%7Bn1pXjknHm2GRvP~Q4aYSS`qRse5SO26nrp9o`)%8qoq?p-Vg>~=?dK?X0!S7V!F&-+6#g}pE=ROMtYD_+WgGFG6fQ8g0APUahXQ>M2tVv_h@FWpNC`Fgg+(4u)Y8 zo3-@2nR;h+Rb0D@@4(J4w3of`f<@NRVC#0#raQ2`Z+j51vp!Ur9y?zAqqvQ2CSq zQ4qq#1xxf=n|YmPNSqvIY@#bm{t2hN)^<5`7!25r(b*jBlHH6aY=L{rrhl8Xz~t~7 z9e&f*VDNf*6B#RGv;Ytd9}jJq&0Cc_4z_M*jejRY>-$6#e$W6qyY zL1Z?|(3!{NS8zk|tpl@fW!@}Ri(%K8m+46FmySXNq91!8RNN2MF+VYwT5URThYP#| zy|wk$g5=WcbH6X*b1nq$x_^%IZHA_^&v#{=KW_GL-yxLbq2Tx3?t!P&)i|}-8J{}> z?u*!EOTWj1GtSSthb!A^Uz%#`Pg0w%bwS~!uLQW4w-23GCOF<-s*>qf)o+Xw3m z=Mik7e^q6sa|9t;;@@R%pBlFT?eI#|*yJfS;6?0vdOZ7zjnW1Wbbq&3Kj&9plj~X3 z@{h5}Oj(mQZe!ZHpWOczr}FD7EQ==#L8REQ$=uL_*kcak&OJfKlzJ?)VO6XVIGbDC z+(H2wKjOn^aE)(33umi-#OSpRd~3q+BB{uR5#(RZk1pVSmtQrYmHRO#cvEgT8ZM@nbl3G%@K4%B|69ABpw;wV1e8Ys zL;hYt`J$i~eNr$uRdCOq-|2SJ3swCWUIs>9lhhF`LS|!PHaBKrHa9UgH)CaDHDx$9 zW->81F*Y+YWH4l7F+MzIV`4TpW??orF*Y}2WnwjDI5uW7F*h+bGcsf_WMeUtLK0F% zCCSMRfG`vT(J@O%2qK~m&NM*(UC;vefb}m=V;SgupfH=s3`QoKRWxv+c14X!Z&Xq~ i|D*0P7j1V`F;&L{A2-+qr<2(d9tk%&3MC~)Peux^VAh@h delta 2659 zcmai#XE+;-!iJ3qBKAs4&4?gYk(e=pQX_U-YLB*N?V_Y=^+oORqAx{^npLAkHAa*u zsZwbvMQBOU7^Ne|`L6HRIX}*I{@=fz>wd2LMD8B!sW>}M1C0hnMn(90hk))ZxZ8Dv zrl|2Pod~1qn-noMP8Se(k<=uQ;IgP#QFY2r>_#k>eqpM)W@du* zB-~tcv)a*BSA1$6V3IX316@jS8V+-QIXb9Z9XGWjD#>@#{q9{4T-Cv!A4d^7r8Fgr zZI|J=9u6F%wJkG1!homvJ6O$}Y4g@Qv4INjpDH$4*y^7;0!FeB7)4r=Hk=5=R?;ySJ1)t6xbXoa3az>+jz_TPoJ-^r-ORZUER413I zESOWBLo*F-0-rqVIIP*3Voz$Xq+_J=Irst6a}tf(PO&~2u5P`016ekt!R5b-dz_D9 z#4y+|aprbD+`gEZ4o9`*v#x`!nIzKXe&w{mM-NQ_#u1e~<)4eOvu1&m)o54ifSd`f zb^ElsIyuFr2lZA@BE;6ZAU|R*H*ZX4+q|tamvtUJj{tAfMf@z+A=|syt;F;n9QM@T zrQBL5ZKzokAzM3Nw8@?wg(z?bZ8ZAEi(%>t4`v&$;VRbj$FvocgZv)Fq;fFZ2sP~c z!No$1WSC|0IJ&E2Qw!;n>$Pp0nb6Bb(cK+3>E-N(^{0ykPhYiHsfx9w`yn@Dke&4) zoXY#5)xo1Z1K%%I`l?>yUrn&0r2Hp?EYNXisr9(><8x8U-7nMWyW+4_-!tNLd)X_! zT_%_+E|un`wVjc2TO5?=j1S7Hp9lGU(&mkWl|RR8Y}1!F#0a6FNz**9{VgJO`rl0=86%D%wUS6+7M zJUw(W-C!Wrl4Lu%$wbP1Uf_GrR_sf7S79lK^3g(`CTpGl$NoESjI|4jx^H{>7+I)@ zCY=OVzUd24NYp_&{{eyexo0MEJkp;MVyFj@+g|mZc^d$!-|j+VD>t<7!aq@Gbs0yX z0i=OASuR%)trI)OY{PUpQ)Z&BLjIO;j;+Oy;U5+em$}B9o1|{OClG>tyZ(8}iQRwx zlg*O0k2nl(&dq-=>yoCmCEN7qrKdhIjTqI;AF|t_absyBbA$IzS{NBrIKwYchyf#U zYX3fiTrS;8FkF0$uo2*{i>S4?`>Z14@JweX86rT0YhA@tCh{rjSk%VO*iICA%JWUt z-A4qJ{OL>AwywIH&+f}Sus|U2gux=b#E-!T{R=4zyQHlt#>wW%aGq198rbSxpj9** zirl1CUK8wmmxMB7;7-veYnDQ}vU>nnz1)LTQ+-7_Wc%ealE4g~3Bh6QI8Kc$nL*VT z#N0nnEAWAM(1FF@UOLoG*t@mZQm%%qj7KE`%l;`Xt+4#s=oGgummcsie!@BsPBzjI z5>@K-!a3zUDO$@PbH=JAtObn$!eUyz%Mo ziZd8f^#P|Ra5^9XWmJV9 zxMOW3dD1My?9m9ka3yyNBw=t1R8{=P?p#6 zKik^s@F{e`h%b?TWzx+^czX&}!=V7lf1Iw!QvO&#LOg+bm8IQZm7n$@Hb$Bg`pwNh zF80RP$(>2l!J(8!c2H`Ol=cIQWB@Hq%7_hL@HpP@a7YleSp-d8VX&}p6k~b#*rcyt ze|oh&+Mu~WNCB3ifD@z5*N^GKhs=>zv1Zpg4rUA-@Kh-Kpi(}30- z<=*bG4Y3MzKYPw;jeE4o_gnzpU~%(u0HgN zJZVNa{3Fe=iep(8EsG1?2EJGoj`pQ);DM?>DAMYV64K&))GBOb=`fH zrJ_iE9hzOiusr3}EB18v%{LU|kQr5}c+aD}88_*(-7jkaam+P~zxFi#t`I0*m6nc?j9_}eXJw{2cZIZa^RaqoTz`*B(D z?u!a+^jQUpo{BmX3^Up1rKSo=PA)A?C3#LMeDFj368&CtHsyY4Ixs#AFu|HEZ5Uwp zN`2GJ69Tq9x7hJP_Ooq#b<-J3ww3wH|1|<(g4DUFrl#SqiA1US2WX;vynO?Fb#$~Y z`uHG`C@miiwEq7h3JnDu%#mpV(TcR2uN^hUj)g?nPJUA{{k3! B^FROq diff --git a/generated/webwork/images/webwork-1092-image-1.png b/generated/webwork/images/webwork-1092-image-1.png index 02c50a324549ea7502adc59197f96813acaf7f2d..1ed44cc6439060a704f174149c0f805a96adb3ae 100644 GIT binary patch delta 21 ccmaD6{33WmFb{_?2ba`VK^C^n(LBAf08-xuVE_OC delta 21 ccmaD6{33WmFb{`-m^RDt9$ul%(LBAf09fz`82|tP diff --git a/generated/webwork/images/webwork-1092-image-2.pdf b/generated/webwork/images/webwork-1092-image-2.pdf index 2688a3ca08ed66d560e3e3f3b13b06f342105e24..a869f11c925dc580282bc072cf5c7cc12df85b84 100755 GIT binary patch delta 4351 zcmai1S2P@ox<-W238F_G-RPasYqZgVgfMzHj4p-*6Gn>>y(M~y-doh@QKGkr=mcTw z_3m@mIcuH$bk_PGzK8$ut3%@b;?;}zh;ElY7V?f8YM*68ZFs7He1sy5prCq5V!?h~ zovU=Bgw|&eBbiZ+d^Rn)o;eVVKHf8?V`8~EF_oar`=K7lc(4IfMs-L;94I-{HS(@_ zlgHcL@-M3q1L&$0K$r{1&`OMz$6Qj!+a3?#D@SBCW^ZqL)VAxBNOq(g7nV7qV<)bA zygK+rV}J4d%_(7rg#2%hcCBm&??W1VF>G$07>)r`jVWHn?8KeO-47edPZLw^l_p9M z2&C`lgN=kg915zOg(G#=jnJpZu>OI9nwoqIojZU4*Ab0^vpA2lSRo0xak~O#w0nzf zlmxROwD+*{fqUEAxc>`zqVw=^(aQh{Y%w$?0WYF@&|s8Lt#{UM8Q3Jc<(xWP^kgr@ zjjjl7uEQ=-_E;T7wd+9hUY+&{l-!9B5@C?`V7srjlu34LC67TWvQe-Bl*>v%RdSQn z8b*)~g$|U?P#31OA>Q{^qR*4-k`cR#{N$yYG~g3+Z9OV;Ds)J@jW4n*D^3+x$vX#> zfmkgr+Epgm|05&7DXOtae^T-`DG5q#qWJ@8oE`_WAP$l>`LsP6t2vIZB54r{=NvlL2#h1^ zeKjXq3#bfWhl*++XU9K8yef>2nXoXG`WwO#OY=0fy z#y4Wd(0_7eon{ti^B~DublG(BsZ!%0A-u14p?~7%N5$ePve2C!F{`PoJwt(F<+1K_ zu)--;Al+j7@J7TuQKyZX0&e9Q;tXGfEO)Gz@BNNPfJa8~!=`Hna z`-xrNw#+FG9aD#|=O}pNJjYeRNJzt0IOXmtx=zUaR+a~n6Y+zFjwpSXeodTpl=L*g zj^pgzuQWS9m3n>NY^D?sV@=h5o86%H)+>b@QHs=m@DZJZkB9zB#Dx$Ag9-o5(tt(F zL|xYiA@ZoBtAD992C~pd#75Qh_VH3}YveO=x-SN3^7U>6CS>qgkmgqUP{?Ln%-*Z$ zA;r^DpzB1J5J4sEL%9P$J|4{i!wG>pT-=Ur>3t|Ekv&@-15D0cxZapzXwHB|WOI$N z%;D?gB8$2RR*tsU*_rObQZJy1iT1cGt@BZ+Nd`u%=HJKP`@v!w^3fEebDxhtIGF6w zI8xzp5BXp~q%N;HL*x-+S{`4;Zuhl_J6-s|u?c!u3j-B`gi{YY^a*9gy< zTihmx*DooOEEzB!ScddX1x)nf(4!T-TlC~*L*orqut0~%Atkyib6K1B#pXD3Au4DU z{Q;j6O-%9{krzJ~rLP)3CS76h$q+6uKcJ1RQS(?^pZhHiNJ6;*9Hj@NG$$Ti!*xt1 zVOoNdPTTeNOGX7&-k%S|+}9w$L(*JVVA7HI{lp`+-OQE23a){R%gObz-2HuQoC&Qt zCh{vF3M+?*^a`O7{hw+*V#oYZHcs=nN0EAewvq`Dormu(t)_Bs9wEqXyaDbhr6z4UI*r$H}R<8|*8XY>ybB;ObYrA58@iInpQ6MltpM zocR_AL4CgA7yG`%z1nPMk15B#2_0QR^`JQELQ%FwB7(xAw7&+bW7CH0TNeV(BVHGA zD^9aXDM##1&z$queS9;))x$6idQ3C@vDKM2%ClQM%oCe4x66*Wzj{T$03S4EwqEv& zkJV+!W%!Ni9d8K$Y6BA`0Y3+BJMVni)K#QM>{-UuJ~Y&fa;-*#cwW^(y`E}M35#tR z0u#$JGI;5v^8BhcL;Iv#u}bKjdN{A2Wh?h8l!sY(C6--h;B^w*UzakscurP5R`Yrn z0U$+h@50G$W&nqb6a+0&9FhAobK+mQkG>s1lw>$OS7=*+4MD>E5q;MsXoBKj$k7$V z1ku+9-xXF3JpWhkFa!d5DpMh52Mx;&Imo36G8!9e$0Mt(+kKEAViZ%c&9IM& zfaeu_npf?_2Y6{-$Z{$0eTjU#BcqS{nU$X6VcBW!K#Aggh@f0n_&sFK8Ss%~EaCfa z?9>~xk)cN^kLXt=CWl&%3_UfN6DN%j$1oAIW)Fxd{n+Qs;+J1K%fr3amBSubby9Zj z${WtJhP~U|>C$qRGDt0z-c}v-c4^SGuVr(G1{2e)>IIBK@buT3;g{AJ``0yVk1^wK z*jeRm>WVkKg3GtP#)G)Gm)h|CnDM<&DtNTTk@_n{ZCLj2;mhzq$vkAWnG7Oo!rlH^|-&B)}nI@YENMHMf1Mb{a~4FwBrP$1m@ zoh{OG>vX2dFRcmTWlGI0&qct>P`IRl*>-v(7xb^s?dgs3h z3F+NP2)7>f*L+bz6a1N-e!xmk0>S>@k?twD5R(6}YlCi&h`&44Uw^xGk+Z6#a~pxM z^U8cc^-5)12YYNv0gjT~<%OD9@cg4c4gr6ita7ONR@9#)W@S;31nnfJkZOoZhfhGO z#I!^cwJ*QKG|@H#wIDt>L-@o$Fq;G8yEP^IyoLI6&(p#0`aB9>IiL`U?!LZmTa)Vs zyWg>FaapAC`zPDz{5+@XFRM>Gvs8=WWNogKeR`tiH^@gWWXT(W$Q-(CjoH zeh||3v9?dAoIW}9{Hj-Sn&!4| zv#$WfQ+V8{;QCvHdRzgQVcwet%|og6&)fngywo&;F%CR9=IoWYS~^h~J94oZie#R% zq^PpltMiH1CVL2uwd5mN zwxSmlh7PA&?K?+x3?~ceN=AoyE$; zK{01hz3nPS})y0C8%2#sBXSP!3nLK9U{euwm zYd7Ai>S`9`@SF=#DU`1_J_xi)hF&VF5bpP=cagk@3Liol zHDO^K=R#hoNZT_>Ks;dee1-QoC&WQju00~!UO6LxrzFR|nAm~F)t^|FIq6N}`##$> z10h;mHrn>p&*9K@p1PJVBsdegm-&{E`w>UsY-25E#8O0`|A+e=^6ECCU&u(YkeQ5=O9)_ z!v3V^b@^H>9^exbr@YE5voAVjcu+91k=% z#~6qM3<<`$RLiR^imt@%@T`O?biRHX1@MX-qAdOEmYCH1*@EmRWYm1)ie zRuCro&2(hKBE~3~^|Z^e;5yFs+_$t4J8P=|aqbo_4cav5JC)zPt*g}#{f3{}WHFM~Bl5Wum ziQMc&j8X37Kv2gix}aRIz~qaM1EWELR$T@@39lZH{Fl_1i%v~i3zxl`8Yp=x0u3C3aH#}uuSX}qGP`lsgA#*TOR470mW~JVbG+~~3`8x9 zi!P^tC*)-z{>{-Ah2Oaru&xD(11PUWcqjMGWz5PFZabStu!U%@8>K;PfGs0{EEZ1A^Y=H>tl^g2CsY8=FwZg zQbT+>gutRwl!qJIz{L1zc zQcxkNgT&t_35AMFNs5U{3rUEIKt-XpViI;XV7dSGfxf3xLx}%(QUZrY2*MPpR3`f{ zw~T7I-P*rLLSLC9Kbpm$U_DEW9)K`IdezggQc#EPjv2bcp&N8SN*E*@KuPHm5RgzpK^PDR8Ha8d z7&@dI`R@JJeYpRpd!F_>Yd`+hKKr0#2W1;(0C2QVFE?H15o4sDxp{w*$w{=Rxq;|K zy2>*D7Gv*>dNQG7)c)dRbQYMldq+wsce=Kgw#9<8Ny1$%mQ%63yceryz}-A0qJJo+ zC2$QvL8<26)GGU5^#I~_te)S>w5=Zoy@>o`S98tCAClykBlbO{ZW}C}#+e8SC1v3w zfKP{G*}L)_kHqy6hQ()_{EFp~Q#NS9kleu;C7P;7MkWl#;)bbl8S)VE-~RJ_j+so| zH8tE-tKYua=6Wa{pzk|&be9E%Gvy2A13C@ExMVFtj3}$Qic$EvWubaOLF$Uk{(;ll zMn~a7Rhv+MPYPEbm%yL^S4Y^tl`paga0fX@1|pP2-Xj-=H&+3FLYdiGTUW%t6i6C& zn1Qm4)}7Hnxg3;?Qz^uuSu4XQB*r6-hlez#f=-<*{zA3YuDR=?`_K5%b25H;t5HwE2MR zymz0Z=U2wUD?+P0+pT&PI}H$nL&g^gL^V~lwPxy!qd~0Zj?Vc^u?q&~TME+7qk)@P zE%Q&Ll)d5A*Yj?vftk!q?ejh@3ln8-1r?7-gnzNai_s;_M3eA{2rw=Y@87A&VMr*9eT5SLI7Xof*oFHJWsllwUejdyu2&cE&& z{t66Mt~_3is2;VXn;J+#!@cO5IuV2Lz zY`bs|mNv-m7d0zj9W;ol<>|&qqdaCA#)GCAd!iZa$eF{9rk3 zsE3#pFP0MI`fN;H2RZ{xSVU?kpDcDB*Wjpw#dIu0V==Lux<{>4Na298@|%f@_~1 z=V-5BgAh@Z+LIEC9#+Ajvb|GxYuHZgi@OU&G!H))e;oqz0Ru$pNunr$t9L3KBLSwrv3u%g)%!K0v`g3}VrK2AB$}A&*fyunl><_tKdsS-g zGcnTU5;C83V2|giO2MYE#+i}r&RQecQ zpwxlc_l9yf;Bu+$bWaKY<9Dj}zEW{~kQIc$J8Gh%^~jH%bcvU0Z)X-^3ZGw%Ed_ea zvybPKB&CyTH_Wvrn~DAygT-lweonr3FHMm*QV1A2w6t;uU$1i;ou}KM6i1W>tb`6=n7;pXg4Mj|K;!_40nfW$AsM zl2ItfJ){RaydKVgn&TQi-Z=_{9uO-fVvx%4nUJ(XkU z+l~aNY2)g`m*0)w=OCMN&*O-5Z=QfLN)xlRxANGaP}MC5MFPD0`a^#{9aA-kOS? zoqhdg4frNxcl?x@TyjaJBHX{2_YRF!{7MK9VFgj3%Nz6RM+#07Uq9X zHC2B4Ln}|LGH$S4`0IqEq0kTWn37earP1LC#8~!tE!36R=k4q`5x#1m4Gz!oMTgXZ z8Rd=zlU?WT@vnqtAiI8CuP7X!bIo{f^%Kum*zIxlQwmpaA!a0cz%M?J1-qU*g$UEa zb4V0eA?;wGUq?Z&cf43WT3mRcu8U(2!V zT5>I$^g5g@5F=QIA&~kgjW8 zWbJb~Ln$SU`Q`6+GKDGFedD#*oaQ&nwkDps|Ld@I9|SYI4gM;8>;&U&lmwyoU%nIhHPQK*zaKP6!Bw9 zZQ3K=Sm2i8)Fkv0y zGPekviM4Wcy}#*O?JGk(gzdMgy3b27l47{ja{`o+0{684dAbDBe6{ZRbM(XIJ}ScM zRAne?X4TbnS}VHfL`u?iO3m3kbvy?`0^xH$N`{p_x%D~qe?aT+eO4Y|qvoas^$Wl;{IwLHjiIBLtGEb!o~sHf`{GX0G_6lI04%{$v;87$ z(Lmbw&EzSV)@4*S1uK}@IXXTO&-#dEFd3+&av#Bhbaan2`gM*(%2O3FTY6NBdahic z!_J$fzc&VzruxwBaZ$@iexbMUfZQ-X^Zqeu8@|#?g9cEcA{ffLbsKte`@pI0l(Z1{ zKWfDQx;HZi2ET84{<^E@MV zvb5G2436U|+Ddfb2GZST<-a19yzDludD))r1Wpd{d>up`L`3@g>Ap|_DzX{Ma%$Dw*(#va4asyk3l2pVWc6zH^kgr zkSY)_B??C*eIQehqtd|W$ z1G^8JA$UQ<(&7>4fYIHIj2Vv+`?$ufOmGLsrD5F!dBo3U&+Gw=2ZvYPp}=NMFnx7T zVu&lyNWJ|hpZQ}4;@5Uk@%@eAt6tUmIJd8{ImM8a=cN{Ta8~BeU#0r-rcw|4Q(Jf} zz23y_21!>`!KDXtV}Rvyk1A&t2hFQSJ4tE1$~d`gH=mWqn2P6)!KA|bZE$1yD5QDW z;f!08LDOJRW6S^>#D7cj((2>e7jJ^5satA-F)74bPLJpG=Sj0FGYcos>7ZGdG3eHy zsyCc% zv`55mf#>tkHQAb1jqXD^>r9!G&?vVr!a4MLi~5|?Hxaiib?|7A{L^hxrRJ1R_h)_I z`~LYu%CcQ^OKP+3+Y19;t0%#{50e`-&mxh&cPVdKBr6kNl?FAE4JLH%V3n%K0&mH2 z^wYnQ2v*Uj3eIl-9%nZhSy^c(IXM}Svy8Kgvy_vp)W5dQk}Aso%YnRLRk#1gZzC|c zO&$$o0b|QlAkYu5xaASCU()t)yu-XLxU0a=JOn!auQt>~@{3s1=yTo?GJYd9*~m1Ig>0PUASns~0d+$SivI&>7(Gh> diff --git a/generated/webwork/images/webwork-1093-image-1.pdf b/generated/webwork/images/webwork-1093-image-1.pdf index 558e35a01bfb3140e9e2818a44bd8842468f5bd7..0842ca6046b7dee58a107288a73632c812a31460 100755 GIT binary patch delta 3884 zcmai$S5(sr*2Srkgbva}7f>$%+_f9AQgiDdo zm1ab`2vS86zdLKq!+i5HZ~Nh#m%a8mzx_k?Lbah03^ckANzknRX5p+)#SCfeGLY43 z@F6?XPgbW1jcIe!OCzY*s~7K8eKn}Sr>A)(<&kFyZ@=)*uW>CU#E~>St)2;qDZv&)bG!c!fHC%!szO6B0AMbfAu0Z7cG+v zIJy6#3M|X&%44(wwzXY}(^617H3Uy5D(|idLA*CQ^g}=t$qEQ5-RaGJX zUJCkXYozZn#K+M4>#wbj($ui8Od%<1V;Ia}8yt&xuO(7BUBT`<+UGsj@QPP8GG(p{M#|yS~1zZwHNC zHu?Epwr^SZkgub?WI({Bo!oZ7TW$k1r?e4_1hZczr_?=2)ZF3W4Lig`T1?Q$4l~3~ zOM+yde4$SYrisSK(r>7>|ARzQdL<)Rf0pUFZ3ac!kBh)W=HLB5rqF4o_k{Fn9&NsW zYOu{C8w&K>gZZ$%(C4_P-Iw5Muukut$|IagF*+ZzP4B^rB7KRqp>@8YZDDR|3Tqyr zMUqiO*V2e$6&VysC9wVsyntvtpIO_|8ax^RijFh#81 zb>qLhFLgLQ$d@M3F;PtfsU(@aktXNwTx~u&iF3o4x|WImV}JoKV$~+E`b4Hw{K6WG zi|FUS@0Fu9dfrppjcAl_b^0j;TQvkx9@&YN51&&!x}VUg)o7LeAkjP9Z{+m=wR}oj4Ze%ITA${`bmVqQgpawPd8(Ro!mN z!k+KW!Vk$@&K-qE0EaLO5$TBjHd)A-4R=Y6opxHjv5r{UpvhChL#KO$*NmKj7sUbg z*4Z)UXMcJdg;#obDOpzh94Brhj^aCnEcU`{y*+CzqNsXaLFuu$;)IC3h%)ce>K(o8 zBTccO0sMlTTpRM>=pq?kadkuJ?d=#P&zhY^$~=)9nf2aIz%rupY2>0AvYMDq+%BT} z7^OHGHSyLO)9xx)6Ao`^QYaPw?fKFF7YWTky~_5rgJI&gK5$=SORk+o+CW#qMuacL zw*iSO5{u_YpCaCjx;pnJ%dt}?heNtt=E|^t=t`Z8eNK8S_b;$zZXkOx55TLp zqy6ERA@W1-X*mpbBx+DKg6j$s^Yrxm-cG#+d#}3=ur9q&{z4DsU7-l=nC_pio9#ch zZ7QuOT<7r|io)em*8ItC?$>xne{juLS(fun$z`IPY&9Pm?^7=o zwcM%!tclo!UtcFov#E%;=)NHy$k$%`3Y@k*tvJ%qrEM3#0@qn79?xWUXhaO8#D{#@ zVH6ZBLl=MH-=aXCA4Csm$MCQG&N$Bc0cNd4#s)+()iJ=_|E3jvS#`9S@0U-)x>1J? zYz;c^p2r=Tb19#{$X|XCZF1Xvb#8;HhX}O>sOmUnxQ*od&x95s`s%*A?*^na`tXz> z)>Z{NFT~v_bj>j(JKy`F=2XP0B}G;W&=~Uc2#lk%Hw>UkhE~#fEqfr@#5i71!QD5d zob#2{N);Da8JfEqi;9K#%B>Yy3M&X45#$6y@%0Y-Jy_cuFLzv^>u)NBRL0eRegjsG z)f`YS*cS*wT33zdIo^gFp#J4gV> z*=6#8i&M$}f*4F!0Khv$(ZQGGqiF8;)Yuu(?daZK?H*-nRm6{*G6CmRZCfvg(yB)(>hx{!qu1jz}V|e)3(Sg60LM00` z_XOl)a8L;C4a0@?8LrdEBB2iK!b+hhS7|wy+D&tNU!~1F72JTtpVx_h0uW3W0w`YL zZyB$hb33?1GVT$9Qp|42fDM>2~A4E4f(Fc|`Y$60BlUGmoKu>+%8Y zqiJDF1H2sbotr@7}FupnWDlz6Hr?0kY!kJHBW8@SO1EI1ytqp`LO?$mxi zdw7zmK*`$=-;z!3)N#_#(oRFfqLR>AyR@nDwsn-yhpA81Ei$r0yh@VxaOZ@@bPKvD2jb z(X%#n^mEXj{k8Uc+eH>5yU<2yn8KW57e#h>RMm7rK5w{#V?`6*%@HwDw0I8cBw(v{3>M@LEvA80qiy3R^%2~kdMRVX=M$h;?M~u4PvA#WXq+W z!qp@L)jrZt1Wn*$o=>F%e1vq(2qzB(o{oNK+I<5jXZbiQ$8g@L92(0jF}`Na5K_(a9~t z3aEQ+6U%O8hP+Af_&0Nhl4Iw$ns?tVQ5z0;R2hYOsb*QZk&gfqzt{!oPU2i+m*+z_ zp_?AXdbp9?e=ZZ4alD;gnQ#N?vJf(HBwY1gt(Dm+|&O&%7OGW781w~sAU%G%C zMa0%yi*RmwlxVN(ASv`zV zuwA!iKg&nm%ccNhAMYOW&4jTssxwGAOP*C@_c}o;eLphTZ4T#>_en1MoI1 zBXuFG#L+)r=-;ic4u%AvTxkbeM>Tn}u8M}9r}mQO zJ%ZZSY*AB+#p>8&8_mb98z^_h2ZQM_tXkc|)mBg8V((Gml5Et*A;~c{yy7>vEG=Qe zVKwPDEJv#S*FkQ^X`Pkx*64sZblPry%#qy5`sptWM8!I_=t_B6|lfjfdm9!%Z?eZU&?iu@m<$zA{e delta 3885 zcmai0WmFRk!|u`fPpj;f^?~KY;=c+z(8VB5<>|I1wnc+N{~jRM7pGN z^e924-se5%JKuS}U*Es`j&+~vvUF-P#8QWq}Qgw72 zjx2piVY)?Bc}epttVIJiNdCpFi@DQ^5ay%9{GK}?=i8eG>_>~wiaaAUGCU=5`7xy) zZZCB%caYtX{`iB;KI=Xvg#y!Cd-L;s6X!b%Cy{Hyl2epKFW*URi#RnO7Y{H1>FMa_ z?~8GSWDJ&&gkE{W>aA9~B@uc)WJLLRl|iR&EHMd_K_XSz(kVsPj$x^P{H-`P7}^}(&M zke_law}PS^(YPJFgCGV`5lm;Jb0dGYR6*%W`?9z{#;f(rgTl6iQc)#N-v8%JhS2L` zut_T6p5y}9N5We=p#bTso@$z2F}{*+VqC%QZpV})$W z#GJi}kZmhxnBAAZH_LA7w}_M+)TtIE<^c#mSqUE_tOK1RrM8x_mZ5P?51ltDIc*@# zIaZkgN+N{~WVnN7xm|+s7${9|>2|D*VAo8*T%HYL<~4zR&6&@=P(k_=+3W%m{Etz~ z61Jd1Cdw(K3>e<7FPe(t=5Khx8U~XwO5Of`LiLU@wX$phcZmpS{diJB+&-Zof1NTi zFRM{c7%XY2U?u1z(y&mF=t+7mxen6Ndp%`4=$L{fs-yWP$T6+1nuYVVR~>9`DDL-5 zCm5{ihcJG2{Wb!T;}H>-_)xr)`kv`J__;YP-!&?RWu^@AaQkIf&#NSQg6xyq7YR8z zS1xCA&o}h`_&K7KWlk@i5{KuxTiI5Pf%24Q7ncv0=UrNyaRB0Nxw$ad|1}XdmC>2% zH@?rL-8~LV!la813!B_F;@ChU2}RC=%<=b&5AgflSJa0_yg#0PoncDB#>pUSZt{TC z`lx)c7l^<7i}ac(LvHC9xQE4~L=+sagUzxq-GXoo&|1Pu93}JeCnR%+KDo-;!n!^x za2WWnht&UY9e(FO>0N~#DifC4uAqmxlytaO7}`~}d!ieJb`0OH0sLlk{Rqc_1rzOr zZHeGx&rp#zeEwWNx>RQCPe~^BQ-%e+wASVq3ic)6+)Gj@Dn(F=JyleIKUFIwD4$>E zRa~=<^-cJz1Qu+VVB%KFn!d)UM81%v;YW?L0Ts35-*GI&hYbL7=ko3R!;C7jDY8~z z#!O@;2#q<+;T)zWogL*K_o++9yK-8D84Jjn1LEN}%88AJd@(L_ElIQC9-d6LNDe%kk_7FEJHv zuFe}jmxy;TCV%MeLrnnBW~@gLG|hUnkjIN1=t<>=)gZ6prNB#%uAyjP9mYSIEb~jf zAko^UjrMK(QBm5$iJyzY?j?k}EimsIqpIY)#Cq{m7cjq$E$O%87D@&azkSIl@t0=16nA6UjP1`o_zNj={h zN=sW($~H+13s?`Hp5E;ABD7$0v8`y;DuPz;9Yv`swo~5^S8h2@;d?Fa2Bw;c&n_!@ z%_2q+r;Q=3I1YPQ1JKoGUF=g0Mbqcwqmi1jjZqu1f%x(6s^$TRe-n%)oa@xEux$>H zc$~7)!R`nNbl(-Z&59YRYgpLMwKb$(q>gPm7`Z!_wn8yp3NuN2p2Ie`pMySnw9I?* z%i-n1rqYM`I9jiBwqJRxAD(rSM&HRQYrI^3;5{+CZ1wGH4uCese4I}=3B)Fa1Pdg2 zd)#nZ&<6$CGjyt=Me^b~s!XECYFE>|DNO7gP)`nJqTBY<&iSiA&qB*Fzt(mVnXfNnkmfmiIFY=>9J1G*yA9w{wY@dhIxyxdodzqpXOez24x}+{s!`kGco>C4On$nOZ3aCLEfM$ zv4AtOr_xOkSy3S1Rj6D)SFqj@>S*I|fxo#qO@~U))e9L(0`*Ai88LpuTRZKhIZB)M z7LHqypjy`J6kBfX4o|ix1`H12noIlBTIapA|I!1#B>KMN)Tp0YFU;%-*K~o>Brt}r zIhleZ^On6W>$}tIp8W!}ILfPn_nTvLP`>-Osboon`;LXvJ##-=C_?|4&M`iC+D3aE zuaNV2TidGqNZ=8V3cLpwt?%;9uM-C2dz4F&Cm#~WuKdEn&QA?7(0AVbz~0!fDQlE6 z@7!;)!tC&Zgg(QU2K)TBK3e+U+bED^r)TB^?zwj2+HrP6$^uGjm>ESfjbAVr@Ul51 zSh_lklyKG;eEB?kxH&#Jo7E6Do4vK6W$++#11XYa4P=bsXJs8V7{-d|t41>r#<}`O z1_%RZ=O>p>sCyV=;P@Si#>UcJtY{rn$cW7WpTLM-%qv-}v_52D0wPuzWA(nhqVo3L z16)~&D!pW2V#>d06zae8qClmZu5T0;g#ekZk{2HN$34uf#9QxJhep3HV3N}}w?(0J zq~BDUqaMx(#!R?5%M=zVaU${?_a5rcS2>^1TImX)wiJH^3(8G+HhdZ`DrKRu^N|$E z#P6V$INuDJOS#h(6iV8D2L~j|OO0O>MXoiGQZ{8uF@@u-5?xu0o%~Lw5;4P`=23Bs zlJ`KJBNnthtljjP5H1lA>yPWsC@@S%__)_<+NPr8hi$oLy$Bwt;SljCZ))}M#`58u zn~8e;epAQr9Bt0S4-(b8ekD-HF@;$2fBy_>ciZF`t(KHZ?gOKm8ZF|hb3oGA*PjUjxLukxfFbU0(|P}=Is>%2 zy9D4UC7cGzBDv{K^yerxu+ZVIece|@`r8xsS-tn`H|l@Zd7Er;oG%KZ zy(t)`8%4j*zj~O!<1!cO(yZI$#(OnZnc`VuZasSq{647?_P-AR+sbZ;R`_=YTHbg( zn1wKew;hY^b5s;f-u2lf|8QT{f#gtHYv74k2z7}aZ=F%v=knbtVs@>ph@r-qymsv4 z{g7v2Ap{73g&b1aP9i7_q0e5SvA_RhDTwydy(L+*=^$Z>WxXYO>ZF%6Ox<%70wBGE z%p{tTY?Jrs;kTL*h@H={pmzJ{ohpmXy@O=(KJh+E7m!Orc&FT!5KZK9s9QbA4y%nR zsLNk{&$;UlT@y9PkC5l{o!G&Cc*N!l1%84Gzw?~QhWSHZI7{akkGlVT)&D^XYL}cF z_L?s=bk0G{en-UymGVRCv~%o0V4_&HF3<380tEt1AWLAx{ON7z`n7Gf4qi7XGyJjQ z1M>xN`hMu~nKrWIbIa17JWZ)vZFp1j=#C>)R(~>FS#|tbvba%`>?v65U6Y3(FheZ6 zX#OaOZL7SRU9D#;NcwR{C|RANMZ$u@yTTkrn-h%!jutYv@TN@$D>Iz-I;`+~a1_b7 zhcZKJU)*!|H&Ad#yof3vrLLeN=$E24YDD3XA>mP4*>O3)dm`-Ri$V-5c)i zsFaSQk?DvWxMRkg_=+%_n@SvMr-zFCgT}#A-mw*Ng$}a}5&^=pTLZ^N6J%1Yl`}Ux z6d*~t*xVWNfMCk{DJY>H90#swTk)+1%qEV>kMH5NxWckaA*?{ybjMe=@zSD?Yei11MR?U z#_<#nFw}_uGXCfVOPYaOdovVr%*Vsr%%R7H6P*)%_Xi^17MD22>7E}qIP;W)J`zh* zm}D;S`dotUHu_8S?z-;E;`sHf(Z`bT_NSX)kxLx5v%WPg#t9YJjlNi!+0jOy>>+YNP%mF?^PRGbA+HhjL z3D2$ASLD&mbNtq6!{~`^oF}de(?fFY;jK@oPadrCNN=n@YHL O;CBk@39QK-q5l9(Bv)1d diff --git a/generated/webwork/images/webwork-1094-image-1.pdf b/generated/webwork/images/webwork-1094-image-1.pdf index 8186b49aced317d68e63027950a78e741432cce8..1dc9dbceceb831001d26b8b9598dd85d31dadbaa 100755 GIT binary patch delta 2739 zcmV;k3QYB%HJvrEvju;x8ygSXZFF~&9uY=MQh^{I{QC+L4?O^xH*elNUQ*VlT1f}V zm({#Q2fb9n;-qKG5jq^CBIyx?<|8vCDbLJ&vJoMV@P5~oQs+&#zlopw`>?3 zVgjh_FBq`rhTAW`}xTT<|_|0VAW)K4Zj4#zM<6kEDS4e(c z3a=9JFNoI>1b;E7NOI=+B=4_Ea#I2#ODZ8INyNDtY5I+Hn$9d>ZEm~CVC1MY{bnj7 zJ3C85WzuvuwcWzlO?sV`D$;RA#-igA!A$LA^(KaMQ)#(Ij%v3XW2UAXISu02F)>#-Tshplf0c)`x<$oAERm7Gvb}K=UDYR%?LQg@Y z603_g(gDY?Y zer_M{oPWRD)&rlzjbjlyxF6nyLihkOy!()2z1C9+D6t)4{Ds&z5fXc+Mfm!XLfPkQ zwyBQjOR6=1Dy4n!JzVWW|F%ghn!Fm7u~IS|cnB9SK7wKS#+Xs6X$lUilVk4Vrwf0^ z$gu^?{Vu{^;KRyrXe1WyxHd644!TN5E{Bt3SL$Jt?NDC z0(vr56HXK!9bPfhPYq^n;AgH;_zn@7g)1kMBMjW=vm!w##? zQ)4l6zTJ&3e}f;6BhMK)r}|%HVGe7fSot6rAs1uV_gI7}eo}e1@$0$~Kv)r<3ALXH zoK*V~o$61Pe_ZB0ka)zwkU3b_;sZaR<9|@=RtuM;Zt#Xxtq?fX(i#j6efI!B2d9+0?kB8+eR5Ec^ME)FJT=O3Th60xB zlX?nCf6HpbFbur=D|FmaN*;yMg9)@=vUnkVW%B`!t)zyHZ5-#2zi;dfd+0#|jb> zr|wc0FQ9O^F)ph^DEFEl+jR~&aWhz;lmKKuXWPIz_!SOs%J%pWu}`niYcs;3V{Jm( z#kng^XaeM>4-7p41e>*SZj(?A zI)5@gJ_>Vma%Ev{3V7OG(Lrm2FcbjbyT2mGj!`tW+6_e5w1Q*pMr{u}rpeO%V?NnTjgWv%2$!f&g210!jr5*7wMSq|h1om2shiI99;Bw0E+46Z#5IZtu$quu^qobtxLZ(@e(Ppa(UQD9CePj^g#W7BCpb!7|lBkVW&z zNmk7x4?u~3=X8T<*Ovb>>%wUDpgimFgAUfU(|74?EDQ17S`RXd{g zUI|52sKhJAnn{YnnsaHV-PUUtp;G^><^Jx0t}~Cu@X6Zwn|%9x1KeLp1&OzqiMSv{|5}l3r={uzCMou@2y;uPfNpU!w`S^1f1qwby5y$>&CAYh_~cc{V3k>cNo3HZ9b&YbIT@ zNXFwA=&3Gl>!!Ft^%S^2_#iZd3yri{^{O*!9Xsu?`w_hp^mZXe+10IfI;8T~mNc794-%~GQV zv)9%E6)du=J%2=lmz2I7c3Zk=k4d@sr>RLdy1;+wSm1g0UuwU}iwSyyE~;DYs!CC|rr8fJa(#nG_NmELQ&z=bkz>ZC*$;82y54`N ztpz}_sy&6;6{FcTa7oFvt+D?K`G8f@;1p}eFo=4mU~Mk$3nu{(e$=Dtdla!oh}B1F z;N=XHC)z$rdTsU_<*G@gTXjy4kDw>d_>h_7{w%%6qvNbQI>f244toEP^og|p!`Y0E z$VYFu}{MhM;YEmt73$?p>Ma5XCQE!WWp6U>PhzWnwpBVKZhiG-5Vm zVlienGGSt4F*0IdG&eChG-5V5K0IY&H(_BjW-&BkHe+HjW;ZfnVq-BfVqr8lF*!71 tHaL^c8DR~mMK_L43UO9~Ce)#xlRO$Y1XRojt&?~f9|$rEB_%~qMhc?%CA9zm delta 2741 zcmV;m3QF~zHJ>%Gvju;RTT|2Qwq197M;Kj_6$s+Nzpo(i&;yWp^XARtC1ri6m2{AN zTg@AE&`BjMj(V~jp~YS*lI}rB?qU^gJ}^U)^2p4I4G78U{pa)d37>tc6v(6ca z_`-sbXf2nv(DpXv+80?ZFHUF=>o5qFR#PBA8Fm|N5H-xn7?PL~=c1;HaZw@DEgJ^M zcrLs9uy$X5$^XkazyCQTAr=!PNy-a%6zwklU~_)K-nWu@t3e3-GqzZ_gnyObUnBW7 zDZEO=za(Bm5P!t1BFX6&lYFou$qfmJEUAQ;BoXIopy@Z$X*x5XwYlvkgOQ`s^qZ)R z?CdNJl}Xdt)HVxaH|cd&szApX8H#uzC%u=uzcfkKI(`r%A(}Z4DOl{@g96j2@5ro6+a%H>9_ zLMoKL;b9HieQEF!>R~4X(mX z__=MYV}I^mYd3rWH;+f?;C^@ycETRW@a{v7wOUUFpu~2F@fTtjBP8}tiSUg@g|gSz zWK$j07guQjRY-f`d$`t%{%w_(H+nTHV})cW@CYtldJIGI^)aJV(-<63C&xU%&*uM( zk)!jN`(1>;#7C6j&~Pl=etmpm41Ps;>Faj|g?}K=r85tHeX_lyKQwr?1iY%g+c&zu z1@v?zH%krcw&w{lVVRIBtbW;B13s`LUM=avJC~Rxjn4WKuj`ZgJg5dFXeG(ReSRB@ z@Zj@nJ>;D^eDp67K8+tMr>+#O6%_B67poJ6BgRqmRd-kH0@qb(V~^`Ka|Kv}UwQiE znSUC7*%jZ=ITfAw3GSo&8RP4yoT9w)?&_ZYNN1l41}fW+H4Ts54V(wm6R%&dh8;pfb6Msen`wy^(T^7B$7(RsEL);_^_)kL6VYaoULn7$fq%m21X|!1`kB8;UR5Ec^ME(L^SoI&-0Rooa zlX?nCf5A$`Fc7`xE9ST&lD2|)2nFp@2oe>$i+fFzX${yUr0Jo5Z<8&1@Gt}O=DnGD zLmPhN)B;$G#fCg}-ng!J!nLZIJ*i?_>tBLshb zIG@_`2{aB@#%c8k)n4;sx6TnKZU+a95diwNXAPWVSmE%ZU5|GO`v@9?u_GMLO_z~w z@%|Slv=ItY*LhrdJxMu@vzH2kv?CH$6g9B6tqEHFueU4uLH)=;*yHQQEQ+eh+#OA- zWbfuqEg@`s^AED4wA@&{y~#VDR5|_tc*AWwlc@_}2KV*!^^VZ94-7p41hVjPfRj)S zI)5`hJ_>Vma%Ev{3V7OG(O+wWFc1LncRxiQH%8IeYC8~N(+R@5Kec6SZ)tM6Kuto_ zhkpB+ob+LV1b)fgk>tXnE^8%65`I|C8yHC|m9Q}A$#NiP=%fm0OoZg)BERyUnIJ6o z%)GH4Nci(zoj$)FUiSyxGex7-0lMkJ?0@QXtY}IEv>1K78X)oFx@@k)2z{+uO|mQ? zB%IUaFxo4*dlLE#_ipdq3b0ajNM;?F@|IXnK8*N;c zpMNxClqj>q4i*Q0F+M&Db98cLVQmU{+O3pdZ`(Eyz~B2R?yWWohkqneQV|%4?Rqg% zcdeZ^*;G9!5~B#PMM0v|c%Obp%5LOz8TK#%5=rsyc=x;GF$xrXh(hpyc?=i_7J0yd zivSpS0j_=UDeeZ~^8h#p-}69#;1akGQs4`rAEE|CK>#6tFeWK*iuODo(258mI6o&- zz0u9OPBVoEQdp6>t_=R4Q?%yR3@y2dhiW((kICmoXKQ6*@_9BV*XqfT#WpR}v>PVf zut>(^H)yFY?(3$wL*djI)l_7r9?MLvew!QwR;tyzCQ~gi>Q1`e*9xTz`)O2VZ2b2B z6kWq*8%K73dC>kT+~WSu3X`icZ)4BNkfqAYbfTYPYnn58CIqlhp2_H-KW$WMbTfck zb**ke?@YZj2#Wh>RGcySo(bO@+K}^IrkcT1{#e%e{q_mI4^WEJfYI;6VP~h*)hsn? zFnezeP{E|tW^6>|TT0&!yCa>q$Do}5)6}FJUEp?qEHK~wm)dXgCgzc_-!K-GCW6Mk z?gEY!JQh4mlAkbb1(5ab z&e(ftE!XppZAeXeW}Xg!GH`sa?xhCB4p$fD`V#)hKmIPEVGNM z*eIZ6ZPfM~1d_}*Y0-joHYc_WMxKNb7kCe7Kp_w`aeMP}qwqF$C2zgQ@*a@CRHKQm z6fr+EF%OMWATe(pNXC8z8J>d7^md!}u=Rs^joWlFvunFSpE3OT$NcM;n|B|V zE4@uCMn@}E>^5m*#n<~5{asY|+I5wpY)!KtT;%!=Q}%n4tEQ}q!6L_qOS2#1Np-z{ zQCkauVpn?!wQEMR8{i=&H@3(AE93*#kjhi69YY!QvS4K{9t&3jApEFD)%Pf3jS#Dk zrGb|-NM6+Tv82~#zfrE6RJv8?^!NyR@{A9WIqt90drlo^UFs00!anHzL(nIs{U35O zIwBukk`EPL0!#zx{p)b$DY~F-dk?z}*4?`@(ZP$q0Ulu(Et6guEJ8RrGhUxa~b4qy)C! z{ue_&7Y`nwwz3~N4&7uUp<=_7>Bb;LWQ#e=2<%v(IhjJg;}a?%~@}1j6v&B4S=@=CRdmsLVi+L`1M)9Q-D@h4rLMIg8y&m&UPoK*q z%a);VJM*B}czXL=^nO}j7}4Xj_tDDZKj&zd(@R2@zV3{wqD$A>YtjaN17>8nf#q~6 zsX5`Y@DOJJ!3&v26!2_5r-CX_8E@Kf-PzpQYnbir1q?1^Yi79rvgOD%L_0UxoEtXU z@C*2ij)ARxokVwB~*pRjp;KJfN$le0&vR;zTXVGFh(#T+B69xa4)$>@H(b(h_= zf?f&dLm(P-XuPErO?Y-q2qO>0ro04N!UpxWW1pNA|3ojjU>~s#25zs|{atGsE!8Ll z)k#1ytMszj7g4Ww2`}Q6Ppe@9j!!tg{Ed5J-s@-2auFGY@MjaNs}<>9MQBDf+rM-` z&?7#F{SiujlHCUOmK=e8*N@DNw}4&O+9-> zu7O42Q~V?l6ZD$qp@G40d9mH_oxakbp2wB=%^B{!2~wu7!%ieY(kSd6yo=eC(vRAw z?`i(a4eStR3>w{A`=(@kCM1kcN&E3n`^Y;EJvi~}DrQ6ho5PT=bBPBcwZ7zw^xh*g z67^`P>(z|m<2EoxkNhiLN9rZK|HnuKjUgPv=`y5#=G$V5Ezo^7S6^t9+h1&{IgDf- z${FgveYLxs$kZ3Umf@3f3 zH9QuITe|mpHgO}4WDz`NjD-HgF#eZ2gdQHZ=+2?D&^a{JF;}m{DX`Zan70=#A=nQk zB6S(iYvAh5+LWShkURK+ z8&fN7D6S&$D@>}wd)HsXHV4YV7)#K*ayHG=jG*v{J9S9YFK{k@PCsKp>4EF4Un)0Ei-jYbC2Va}Hi zZF|fqjr^k+4Rk2B?d|fPx~v|Q`No5Gb_eB;=9vOI?fi0ETY~%5O6cDjJ4ko=9!l2X z$LJHCb0|#)?$aAMWmBd6I?YBU*Jo?tacgAd8~jO@wP}uz&EE(&$<;bESa0k;K1%1w zB-9zU4(v?jhsw*p*3!~M{iA1G5#s~Jc8BcL7c(+4C|-8ix%?AdCVE5;Oo3o*OHgmm z%L}&bm<5AG+c!D2K(D*+;AEMkF#0#B46IEey^cRMXSSHWsL^oAfIVlXzgnrpXn6e& zjMa+y>2OVi_otD;*(UdlR+AW`PP)r0**_XNYiWA21tz+yYRL&!Tt0y@x`EF1R$MZZ zJhXMF56b%O>hYtuIX+y{I1K6YY-74fCLYa%sbe??LTuI5XdryU z(hEjTor3vGBKDZ+-U{b<;6(A%Wob&yijPJ;M4w#z#FdSYQt`QK_tY&zk4=oE7Ka%id@Enf*zyt#|El@R=hHo? zCb9+_R(0B;n4-sLF#Mg?2n)}_Zz^E@D`p%VIf?b;p2iEs@p5>k$4mP-UHH`KceC{3 z%ZTo(02tc!em+EOD#z?9lbt?cXmK;+>X3tX4wuGai&6bIe|0t%sCjL+-c$Pt#Uh3< z;!5g`GFQ{51UvQ*(j4Sf61IgSLh(^?S$|&(BbD(-M^@+k-8bh^hc{MNb-1vM_(1Id zwhTrVHD`)zi`pLmoyw8ugM5nZkTK4uq7iJ`e6d{GKa?LhHR&tf)2t>i$=-~|C=1ru z%wDA1=C@G~eJCfov9^s%S9$KjZ@b$rdIMqQ)Y3gJm{Y;YN4>YQisX3iIlW<1`y{ZF z$s8a75L!7pjaCMFJ9(0{-5D)b- zV%(CO(rn-SlsN7zPEf!6nwb(6Dlg_R zG%nO959R*i7Su7qpM9KMygI#iqLgO0A@+$R@F16Jk+&)F0f| z&aTqzUizcWo@*?k@s^+S*@`J{Z62PICksgKmnA81XR5rK!tx2bRIyl$;-O6N%{qvo+wh;Jmc4>(vGw&;L{fW@r@9VzJFzCCij5O{6hs)B!xN3iE zo(6%Zqe!BK2axZb2!JnmpH9V4wP~XMJtFb#oc@<@yk9ezv%O}tU4u_1=hY=isb6Nl zM~sReoiF~13rduX3a@0ud@J}!LSP!9<}XH%P(wUrvhC@D?kS5Z3)Tz6e~j%T)K}xJ z|5%iV8{zu@6%u;}L2MPHZET^nN30@B6&L-Mi@GSS)T~*wR%^Pua}7Uz zn4~VonO%^{$2C8`-z#(p3=k-MvQo=EXDbGCdoMqNa>$d`knBfX1y7ISiC*#DGIrWJ z5nJ6sDh8=QP!sP+f=V)3SG~AV&$IylXs;kqDz#yUKi4eP1aT{*e-I%ddD`Ayw840^ zD3D&}-B&uLT_7~Jn#R3lh0hY#B&@Tm|FPUi+b6O(@VErOZprG9@*szFMnpyhLHEXM zv2#_#0~P~l)Fd-5&@<#=FwQODe-z%`k6B5XK%_?kNF5#K|24wr;NyVBBo5NCM-3C2 z=N`Rb;nJzxem-X|+oKe|oO4#s=MmUR|S7zM)pw`ZJ9TiXv^wZMRhv%PD75|O^M57U8%*Y?Q(+E=kdv9~L8d6rbeb_)qW(`Ex98xZMz7B05(u(3J+T;bSR zi?JQ5wLLxGlh&Ul)tzp(v)~doJid?*x<>qVHZAs+=3=Q?){#cR$e7awk$*z?5_3&v zwj+@)zLBP8e;_k6I}CPjzipP2-ht+Y{iZdFJ%~k0_X;QWfadIFzLOmtmE zoWQg#l7+Ium&qoXGq;SlB8y0294X<;f6LaF)|)E*)35D9bli0-ICb61K`$!g1Eu_F<1PXh!)$gMsaCUEEw7yf=F{bVYMbY-_P>{wCFx5Q zZM%-xJ!ikGW1k23)6WhK%2Yx|yp5K%rX+w2JHiJE0fb1y!&a!Oc~(LG6>3-HyC~iI ztjA0rX`0!E)ybtb$RBv=1%6a%4`SaArgyUAliT z|9Q>h{znF3q32<866TGa+Hth8y@}b+VxA2|adoja55S%feP1QxNlt%ysf>rLAm{0g z)qr>JhvIlzxX~GxpRv%xYv|z zr}4T*DdP(u@BRGlp2m1BJ16J0_B3Cu5cd;KmgrpKJ5`6gvX6YnBK1Cb<1?#BJu!-q z&*_7WTV1!V>D)5CS7L1W;7$(PZ9B4kK6>%h@=c!RLqcM0`Ry?KrikC)J07TX+eq}N z*Sfi&^Iw+Q4lcBj-ZT%c?K_yAJ|x?5Im-YHx{r5s>G;HWbO}DpC_w4oL*!)985k!5 zZav@)S$N}X$Pu6h6@T_o9O6$3#mtFyZKT_XBr;yMC3$gZUs3LkRG?z4cDdD1dck{G zGJnyeQ3Par6>gB+iiC~m`3fA#F9m~BVQBcmM}BZRb;$f6;<0RH(%Zv|=FClH$t`4m z=pz91W$D0LJB@}h5XF+h`DFFJCFIY-Z@X{&lR18Fb;}&i9R~R%q0A@)%`T8Jx+2); z633Q^bnLf{I75XyX>pdhv@%0^MV6Pp>z|E3+85VyQ>bA{pIEy~S;5{A!y8vK-zP`f zXhgJ!C$jvti4~h$j`kjH$>9Bb1AmI|WyTQ-S)Ak+s&DrV`tZCYB`3c0H;aTO&?(8X zUGwuan?Mm?y$UfWO=_s!Ag*)ks^zz65Tsb)K%4;`+VTM=o`S_~M#Nj)brvy~gD1pY zQYUo_9{z@P6&jpwsR;A!_Qju&L%-=--b2QUwAWkx6QTa;hOw1amSOISq^O}~^(Xq^ zQ*@4_bGf1eq_IuE53E6Pd|lMkB$+1pawz#P8KWd}l__^IOyD{WmgDIw8kjr%N74f@ zpaDlE)Hx1*Ce3dnCVU7vn+ZeB5phnRHxi`>jmG6(CN%jWm3r7FOgPWK;?9c|h2%z9 z(cHHh5?*6WtSWfXU|y?a&;*!=Q?$l{@2kYRZdl#7>!}M&4I3*?JCpC9oDa5U&OWDV<@*EzFZ5CtjD^D8F|vVmu`9&_js zLDxvCx+!cUMyG=DEb_edS&!-~8pUTJP;q)>N5c2)Y^urgEy|Jfkc+D&`zb$gk9#*D zD$_Fv5~Y}e`H`~7_ zCwSOZf%SH!R>1daiLzKmw+=NGrQ_%|Q!AV&7pqm#A{JkD)>y*4;~IY>g}R4J#2A%a zCu!cFp2lk9D*t>`eoT@x*W*X`_QjTTD&Fxp$kbPx+^Dww^x@};@K$tk@Md;X&KhsA zMYeK1BbVofNC8wfm-D*dH5_Xq;)imZr_ohWPEzQWuD26E10jvw8c_Xr|K3?)@N;_L z)?>S2xgzGH?xQ(FPAkPSeP3zjQn)At5*dWOO0Lg(c{meB*5l&(MgDU}Qxo5|^fyTT z1RUBy3x$K&Mfde$4|zemFN?f)QL9WHSOh^tq=7vk|<1E)X z=*+4=1$UpFlv+c8?dw-Os)vbp>SD{~RifpecQk9>TCi8kj!dh7BB~-bhHvh&G>Kl4 zW4?SCt!%dQm}T=qZRRX4lz}<%=Y~eJSl~NYI4j%U8tIx~ofq?E&i%fA@Fpo@>>9<9 zu5IQat@k|7P_KOC1e)kbU2fBt9Jtv@eKis-9Duy&aW4UOxOVj<`Q*^aRm!8EbI;R5 z8~So6M=L$zG6X`TaQM~K8&5gNvEbq>?cIC|-5<5<297#W-K`*`DoH3mmsLLzXc<<> z|I#-&P`p$sFC(NY4nN^M&xGv6xVM&ep%yp%P;IZGQ}Ml)`R!@8_b9R1!TS~HAgjx1 z^KKry@&jWQ>HS$;MhjbI>SsvrqC;#76jMDpglRKa_IH19K{)JZsDjn#O>0C32FP*w z-)3mk6b=T_wv|WIwi3T3vfVFJ%7~zB{v2ia2H7FWu@9YCM8EimdM? z(Y6V#eYH80uwkH<^Q5;2OO=MhLy#>WFZSa77Fp>p^}G+;H~pC=-UF*P!s=|N4pwU~ zY;@1`DpByzzJjm2j}R=O7kt?Si;n-}jS?I=1q|H0lo84rUP=h0qJ|>Q%|p@ML&FQ9 vqO6EOsCj94DCubZe+dGK(~bTytNxHW;X0Q*4VG|+3kp(UhQrb3M$G>KSxvzz diff --git a/generated/webwork/images/webwork-1115-image-1.png b/generated/webwork/images/webwork-1115-image-1.png index f3609634bef70a154d325126478323a00ba19428..c55dd2cf7a17eaecce883fce5d6af67dd2d67f94 100644 GIT binary patch delta 21 dcmX?>cr0;48xMyt2bbQIr5`qL?&Gmg0RUmW2yp-a delta 21 ccmX?>cr0;48xMzon6~Iz8{G$+`*|1Hf_8krfw07om&u;G8oro5+rM&b+cmgv_ii;f!N;9hNr}PR$IHRyCBth9%CPc92aJ9Cn8QO) zxU-5!jn>YYM^7=I`PGP;ta+dxpcZWJus1i6G{(dO-Z$_+iPJ4v{SMjSLzoX*`h3@Y zAZl>th)|0RGXW2(XzeVryDP!flQw$>KpzfBwj(>$E zbT%Mr4wpUMObEG<8!BGU{s=lbkrfzc7Q(t`%Tz?hs4`r~X!D!7@c@AKofF&BLCOQ~ zBJrAqFx`j~3{K`5$B;P8Ew~f!3@}Fr=^2~eZQT!=8p*Z^Y4y`sYHHqz2C3OdMtOf^ z9abb~@)^je&_>ja-0nv){#Eypt?vmVe!rqKv=8F!%LkK)>QP?}gMhx8eeUR9m5Oy> z41~%<&9}UgJPyv+{FM-qpTW0cKDz`QZRV4tx1PsGTYe$Qiw_C>LZ5w(2@XdGB=0GR z4RF$zn%zhkD1^Ki`8M_3DKN-8l_(x0(0fz}mcpGj*%ln&tP9|xT?u%PANIuoP*?~* z=N%xBmlPqK0-R>Z>MPQIGV1-xrOBc|cS%uQhSNi9qUbaj>iPb|ns+KDJbiF=UJB)2 z(U#890E?1zeJxEyru+l{V1nkhDpmdFeJ_8*Bs1;qwtlQ?+984f_j-k4N-=t3RAVL_ zXk0_0Im-J}>?Dl&pbs$g76yNE)=}dyTH7FQco~H|n`H_l53pJZP9zhs?k$llvuuoj z*WsbvFNb`wU_?xR{MaNFV6)T_wuAOosIVWhsC<~ zn=0hpGt}cTgL7!aV4esirBZCNYUnUuJ8pZk-3yd_jdF857=RG&r^DjQsDjJ`G-~sX zkGPK1?GDhU0k!5qN&GZ$-3jiehdvvV(R3(vF%raUt^f}HjpW_iV+OnDCIeibBvNiTt9D=1Qu;GKRzomm+iWvH1mdf zbKbsJ)i&7{O*Xhg0!q10a>XshM&);mP z4n2uZ$A#j0gKxDnHFSaB)j4EfcWY1>q||WGG3^xdU`7ja!by!GcpFi?S~}Rf`N!lm zM|x$8pLxD|##RAy0=f5VBp$1dew4SyPqm9mag z!Y6uj+TwH0S;+M45)2OR_WuzSFUKH`miS*$+g^mjSew^GR_hDc^CIgQRUy!-+`ReUu;vZ`0?5fIgegNJNt2i&oQ{Hy^TKlS^7=-JSjS4 zDnd2y^1?wR&t-+M7v9|bO-pCZsrQ>p5WwRO)`}=)XF1o|^01eheGczr8sIYePdIZqFma}#t<6GM9$=FGlfB^8_ zr=t9ls-l|Z)qnz>zcc^Qf<|EKuXab3v)M1f?Kx_{nW*E@jK7xzg10d1Ic5o=d_9CC zFq+rkpgR%#?E#T0D}hgt4^)q zKxr2N6L4N$EEo^Lk;JMko=Dsjs5b0|rRpLOeqT@qcV2o)qR4(l*)y>LMY+jWDCRV#hNczbE#B;zyk)~{pMSj-a??+ zg`O-JR%v!c=_f-2VcODVemHDKpDR2PHtN1pH)`FHnVX96v+-Kd-$>^br$K4Zse4S{ z{6Hmd%$qi{!*^=mF@N$~^HYngW8PbR#lz&@kLb!;Es#*=UpxVvjc-F&_qO@uX@G9! zE@X%NG7M$n7uQ#6Y{q44k)KgDEqPYn;34Qpg)l%9q8KtuyVy8LmEP6BnnDe-R+rFg zCw6Q5`39w|h*v-&gPF?ceRer#VfWi$dNzlt)Uu=V zrNwbB6Z?wf^Vy?Cwmssz25Jt3V$aniz!Kg4SsSK4i>)A&R~w}oF1>d z(z4+^!<*bRMoazor)5v4+!tZ2A#;Ru;ya|Z{D}$2Q+iu$An;mUS?WQ8p!y5?MTmsx z3}LP^w62Ci42fFW&uqv-06q8kV%h);a(r@8Jpj79;#Ab`eCyoP=?Y4(4XAt?m+9MB z6mIY_sWE&!<$^-ZJ|Ro8kuI2IdxX&eI^A;BJ|8R+V6#t`%C?p{l6e9WeNN7c|!IaFie1GOUwo)e{W#R_S?N zPczyEg4CE3CkNE+c;C&nGXIvP9VmC09YA4ii0S<`1V3-Q+M}i|Aor_@Lo~~RHm(@y z%gFl*6aP`Bk>c!GzvSC6$FYy#d`BE?{VmwhSldsm>Py?=QK-GKvS+wg=oggpxWp^r zg&Xsb{&fGA^4+ zoPWL^dnud6qe?lB%X~;@pnjQXvZhtS8Nx3u_rSONnev_=oYD7!Q$5iz49^M_RdckJ zkhHONl$5fwleLw0lz_|0!ENo}l8$n4gshyLl#;^#EAVfC_O##R^i(?b{_Vf;b+{$q K2A<=V(7yoB3do@V delta 3021 zcmV;;3o`V)B)lZBT?BtNK0XR_baG{3Z3=kWT~M)ZgD@1_{S|LaRaF>>q6`SBB}HLK zTZLrEXzT~IRBX$bA%9z(iJoi3@GL!+dF=PQjs(R zA-RiFxcSHoNvabwA8bTOj&H~JTl-v{&ij|^rA~JsuwMz6yFn?);@lkC`{ZzJRIi~Z z2taaP{3d7buz3}_!zYh@T7g!o0XEmF4M;m1lLiGz8joW5Z0(Xc)ECF@2YGrb-2^Y&#=Bsl?;gr*T+Sf*6I5vo*%#~JdAjGA0E&jn}A=fa|*B15=vr80_J zrc)U;S($n+F_TxTbeX(o$m5a>SsE2z#04)=^L+l|g$vDQb7-cnAk<*Yjh-LEnKitc zTcs*c8BMBxY%boQ=eZ9ubt=wF4)rF^pv(V)R~flPL$=DOrzl#i6ul{0jjem{qZid4!yrDr^yBe0l3RnqYmLz0)A--4($a6cmg+uOR4K%j=B*vGf z{FxTpj(qm>tmJ|aKw4P~^uyT}^klazr`!_4sdHDwx&UGS>vtSA zTkYEwHjl1!d-kR@%Noyf!>pygU|rS%bV2c zwlWP^k*;aO{=Ka=?cb-L2qOwIj|7o_JhC4LBs^q$ps%j`yN;uWyWlp$fx-8Pn2m$A zX?!Ze47tW6gSKsxi|@8whITa3fAt323kTMJt!~VLboj(`GR(Q_)6_*7@IPzH*Fze> zXM4qX9RJIlX}R1Bqv*kF{h{+BVi&NGQ>SUx*PAMB7qW@p>ik$pLPUhc?_(N&&z!jV zh=oB-*T(t{&Waro{fI2Q&*_|vHHnDb*STsw1a_z;?5Y82ue*s#00QA#pvWNzQ<(_B?@`E{p!I_oQmY>+>L zA3ygWQ5qBz3T19&b98cLVQmVNISMa-Fg`vCb98cLVQmU{+Fek=YQ!KAz55k&+)|1r zhqen6Sh55>q+6O@XpcrGS)oxe=8(T{V!|GJP-n*X9`6k;#aUPrfqn+38jR2xs{@X1 z3JRgc-dK|5ppA-qmu@vkPiR?7{~-qidJ(7ZAMsq)waD1Z?OJ88pt0K^*V!X~kAyQ;0I#Zopw<8W@{Q4U>MjG}Z|@6|vE^Cn<5nLNZ%3>OYt~0pHHENvNDE2n zRwKsZV`tv^m6gsPRJ3oc3T1ASXA4z--Ahw(9L9fVn|0E$G8zTt*@=jhY)-onLjsY? z1* zy!kvl&(-t6kVJwo43k|~7nGH&&lYMl>TFeF$v4IQf1m#GBniinZ~_z3L;@jy7LH1Y zjjNJ>Qc)U0=vQijKMyAO<4}TUb|G>eMiLCe8{D8$omZ+oS5xwhLKV4u?ZYnd_8Rrut$= zCh0J;2C{^%qnQR;M_#sCSn?WwWv0o!@5;R>w3==8EX|M_tBz(YXjiEbL#P~8Ah8Hx zNP=ErZlPNyd~V~yTRgQD=CH@x{HJX(|CN6$4?KFXy$=2G5T;tJoel=-a95~thMOB- zdUIpMNECyiw3TJnZUrrXpDKyVnf4&~6ocUR`G%su*QTR@PdxJ!E1{P{7Nv)#&Ev^S%&F;x9}K<(c~UX-`6h2tu!~ ze{|;i(vw}4!7(r?fN8gT8?`5$C2$AN`CxueEJAcF5^1^JTA?kq8)-`=G((lYX>k<7 zFftq*3Wi|=*nypI5k3i*)pJ5&C=}+++`c#;71~cEOL8Zr2L`0c zz0#D~2DnRt{5tHavec!LRZv{=@ zk2p@KDx?y|ineYQu&ov4Ax%tb3F0U*jPBR(+CW=c?QIWne0|UTI`_mtK|tJt02t@M zD6lvH4qQlJ5J-m45cJR!g5VPH4A2unCK?LPn;v z_&%Z-=WT{DcMHNvU74X9FdvYSu$Ryg-Z6n!TvX1^!pz$K~*fPx_CfvYOShE zGE`fgsmbTT0WeXG+0>b8Aohi3PnAM&;f`9B)(z&~6Kn|0W*#~75c~{o(7$nD@;du$XoQlF{pbitrGg)GD>6Zo#dZsar4`TWu_G%0-ukT_(?2D1x>d zINxNdZte1|uJYx22lp*RaT+mt-)`^oXj2YTt6IZzmw<*$x@aytRGv|K*4~%&%^qj+ z?I)>CSEfMsJPa5dKML=x1tKB_qDL3c!@E6^>E^*Y!@V#vGe1h5nOp&XW4_EUo!iI8 z$-!?@*Nsgc(;7rfzNUwJD9IpgluvgxxcYQA{g_-&VxNDgnyl6pGUepsMQ~F8#gD^O zS4tO;7GxZd%O-Ohls!Ql#@#y%jWJHFv&*trDWGI()%qvuk&ISp(SYdTtIb+&P9>6JjMtj+mn`v@kcxedJE$K2jl8 zB68MDNN7k_G(G=3ujlo8e*ZlGfB*Ts|M-4W+*NFG;UvNg47fu=Z{e{)+z|_y%R>Rf zFy6l7^P$@!awI-nGQWs&yu?CI7yjVOoMph3zzRh2Z&tvEgO)@|?Dl1Q$&8^1_$4magmd(pY^TAqS_QYEjpiv=`u@BW4T{Us zuoiRCx3RZbrB@-d!jaB-c#g0S8oZ4sAI4L$j#?H4W05ZfwJPj{cl4L*a=!Fyv81(A z$M3(~DC+%gmqX_xD+M9A@r0}YDaI2h&Ois?_cUI4KIR zEC<)_zNAeq@FU1oTcT7I_?HBDXXhg~C@KSnx~W0H|Kd>iz|J5r4k6jhf*Im4q(gscC9 zCg3iNzi;FGdS{hTMREs^tSS`g9b|cJYV3y zCiCkS8vdos#Z9MpVjF-=PMngb8mX)>p&xxd=gQ-$J0PjkS>tv7M8aDMXUszmpj~+SQ z4zf&@eO*;_4l8ey%Mq2B$2?k4y3_PQDhc=|6jU#(f z%*R3(IMn6&>)>vc<=Pfoa|FVh#RrR<#YF?ThQAPDU)Nteauf#hOYrtfX6h!7=6drN2Kn<~tcun`YI@4u$`2ZZ^S%&hxZ zJE6&!W~(wp=jUDPu2#D}v|hgezdmILu}1}?3(GqD>`@T+FvY@<%rR{=xea4!dt2wB zztF0A+Y_fe2YZVe(#0&(r-{B>e)^F!0V|_LL?IOA_|x>E+}qc40osKM8qWGux5*NF z{iF;?j$9o8Yo6qZJ=XfKCh9i+jWYlKQN4HX=7*Y@>dr&NZLRo zI(4rYQs3TJ>+90+k>wt`vM|xQrNGWZyOvsR*`dK^Fe80%bO)9BX={axCiu4Zo=UAh zaa7C1;@KYc3G zN8Y}6+g&4)iUoQwO44fMnBRwlJ!CXZm185X7o1Vs&5ttxx>Wfj=GShEvrF;PR_~+E zmJ2WKc-r0`VHMiA6!lSO3H9mgkG>obmwN2kB3#E`Cvba`8%;aVkC>cT0V39OvRbVC zTA#YGESrA+$-7O4mF=M;aAdIIRL9Im=VsjV=I!sa!Iq3qO0Q9&aM0KaJ)@X4{h1HbY&#Ud(js${WvTnlu^sAY&bc#05AJNUsl zX=X|D?lSjn$(;DR&tpHq8UpTUF}j%{FvRW>Aq|MgGNGB1M#sBwLu*6@@vGwvT>CJ4 zM=(iSZ?G1 z1oHFB6an{45Y+eNMUmA=MsUjUpXHo5-=m?nGaCvx?yA0 zocKnY9vSWOgm8$f?q(U+4&!q(qQ8N!1#9L(v~2SdcUI}J>Mo)f#8GWQ>DBXjc#l_& z9ppT%+~#NgtD{DEwB09Hx9q502qbP>M=yod;u!Yi-Oe=bCNt=!pk8FyQJG4GEEPhw9Pq*7StnsWy7cPt1T<)!=5-!t^nTq#yi)8S|l~yk}SZ` zS&QTK1RV%+gs7E9)dx3isReve!l*3FE)xn5{TXh7IZgFXFrcNR^E3IYm8*svl)t>+ zfMNbI0&>6NwpY~imksUe#$0^pQYqT~G2XK)l@hpNeRXDHehBYGJh{{hVRj^UQ$aJX z%i6nn6D_%?BSvoyj{M8x1Pq0eHPd-cz}rVEosSBX+_Z3HuMF2e;;y)9kG(}}K`B}w zL#TLAvGZTihVK-;xD^ykZPDES26_Wxe*gdg delta 2360 zcmajedpOez0|#(-voLmH47rpoj+y2%mqX@KiXx#%*u|wx7$(Xum(V09)8SZ2$izV| zxfP*ML@v2B!kHr#nQRS7y*=;q{&Al3&-*;zKfnKcKhN`B(_htp;3Ex~*xG{WG11f@ z8aSTo@7{IxDNeq9+c0JY`SdD;oS}pXliU}BqT{DV>TbRqp6ZQK+}a+m`4A*4ClnTy zp}8>`L+{0eV}378IxDi*Tc*a3q=}?+$R0d*H6xP%a?vb*y7a;Pw|3HD@)ol ztXmU+X?D_Y8a<2L`31M3wu-}_WkYmnP%t%u^1sPcmK#KtCV3MAA1WyLTkb}DQ)(oW@I$o9i*w)o68g9R z8fbDHaVc))eZ&6>8Sc^t@Rt83O4C;QNkpG14?6&%S(xZ&qfm-*7StWFw2|Bs+L^LH zI+($bU&ZiG6OH)_)tguRV9B#7o)UV(-jA1A>{R0WNZjuf%g|EZ$UVR8QJAmUDDXl# zIcc38TKmdtZ0%ZmT*&o(^G!eE!)QBwJ^4S zYH#3*7^evurW?!$q8w4lH|kxZ->kY=X0!6asg07kDRoe08&=MDhi^Qm_S1->-7~$- zq=a9Z`1*Kq>6x_>yZiSq%3gyye`<{qkGBnG)@@Gx!AK(C6Ud$voUi*S#qyB%0zMsq ziEH%nXZ}3W*B=VcYtG8d3qP8ulP+a)Si#%2_+D9TxP`@1Q3Qt`VH-Y(PKvOKrSlfX zqwg6xu@G>3M^cc^j&%8+FD0tZGvjAor=PbC=$P}FgD^o?K+F;zZ?wL@lD&jDObsit zX@!4S=y-DMM~mkc3j}}{@t*O(>CWA%Jio~+KOFb{0ZRGSe`tN_uRVZ<%lc`FeWKD# zC8s(S^z%hkYV?JUZc5Wwn~aeI<$nFu$KBm28oB{k<2^M~jroC}mv+_z{Kt20Dm+eFTcgyIJCgF^d|cP%;5jS+cI^$$-cq zXnz4)jj5lfZ{~!$k#^jWscm-1NcWJvyJO~K;_Zuf3VzJ=q{k7W(a%b97_G-1Nvs-f zc=G^ZAOn^@uk$jlAR*lJq{+btvDf_zd@p#K*E}7t_j~(d8096d7Z_r|gK`mKwJ}l` z&+tQq8_opqln~Eog#(4$!=_Q)cVFi+?xK*Pa#u2@PGttt2lb9h(ncoRn|jt1PRjp?of4_>7VUAUlDxvM|(D$NkxAwT49`dUBWZ#jaLcogeZzR1E=M(M%HbzgD zPQzZwrs7iDhmFVo4&{ZsbwSvONzUHA+R8UP1~t%Q73WKQs*m^e;(S6x$`1SpjL$K4 zh}jSQ&@z6is&|XWsAVI|0svs580@c|TkhTKngETkShU@&U*=A5a^D6t) z8D;r&f@Gq$_E8xKFRVL7ZRBo{2lq1OlI^D>2~)<mO7kz1~7LlE`WNln#HDsOEMsS|Nw_bMw7h-Tuj zA+i^X=P&b<#q^Q5Z(lXm^!DPEuugaad=MOfP3W@ElFHaMFJD$x+_|U~saWKwjO=lI zPj!_e*a2w!j+X;r%023a9tFSQ0#J<^@G(=Xr||5~&jfI)cq>*AUpaW|bPjR<7|Nly z3Nzj%GMnk>&FY$$!XrMAJ<1_S4Ykq%Gqn#(zf79R6WL${gdb5h3#t z`uzRS`D) zK5tCIdDp2p2X75pWeFT57u>pcpCT0*JW?IJO{G#(I2BTdJDd}vkX~&rqR;Pm0@|id zOW}86p-|7ihw7~jePlOCYwmxtRkzGz4UM8D`(b~VFh6;Dko1~Hk(_>BR&3krVKV5o zj%TS)^E?KMEp3#@H+RVpr6i;ffmh90gEn$C%f{*f1x98O%W)R8?G}b+eg8!4I8CC% z$?$FALdAleB+Dr9Bd7p={+au%9xA;Q*V1#Gwef7zPRX;QA!LGFp+PFOCaoHp#DZMy za~6zmd3@80+)ERuPj_lHgk0)e4$OSgyZA72>tGAgZVK1?2K8Xz^B3W}eby6g>iDmU ze{9Y@_hr9ZpNmijq?h9R<_Waa>EH4tBdSk5SPp0|E0<*vpuZ4VPgKor!bTYp7z?mV4q&z-Ua# zbCQlhvBWaXZ1|2hjKkUeSA!aI!&VSP557>HiyiQ9F8WcIUMV7oy$@ZP;n$DG0c E0ec}!UjP6A diff --git a/generated/webwork/images/webwork-1121-image-1.pdf b/generated/webwork/images/webwork-1121-image-1.pdf index 2aabd7583befa3be1a07e982b3ce271d816b2091..9b4ed5f95b600ebf38c54cdf33f9a0f7739baf02 100755 GIT binary patch delta 2280 zcmai#X*?4S1Audmp+6Rq`)oO~iy^I?FGA9&WHLuG$C#s>bI*~KHMx&JIxOdqE0ZyV zvN_6qR1&$bNc#8we(#s}{qlZ$K0KeEFV8Y~5nP?hL(J$j{wB=X>>*yL^Zbk$rpGv+UenBMc~deoNn9D2l|O z)+(WHn0%j7-yZMxZ-f*Gx9RPr3gjG9@ybC|Hs>o>bd2yPJ7-B+U^0ba0Ghzxd#Rj zJg~n1CcG>896y@(s-d}^tqG07@6XAt5iZC_<`>i_LQp8+f6RO17~40v&Of1kiUm&E zrKpk`1Vq7Oyru3pb1=I;K^efb6e#I15(L^EjMIm=?cKxQIb+k2!?&%jJK_@m?z)VB z&{bPL&0S`O2EZ3R*%CxG3^VF;JfKMnFE)w4+Dq@Qtu$p8R6bD)=S`d9ZR(d*AkKsb z;1;0s@;+aq1353!CAy%AlE?Am^*^fLq2ib>K7WzC$4V{Wp}$kP#`*4_FwN-Gx%}2k z`ZfAw>B=L2CUDl?UbyaH=2jbY+~!x$#UFjkvF<*hnFoy{-JKI6)Q+4TS$k^*4OzC4 zUGqp%&AplIyHOrxTL3F4zU)sMvXPM4e>YG`U%3S0knUbYghnXK;}1VZ*1V2*qML#0#oD_YpNd}abC>`V+km& z;n(1(J%IoRz!>F=JD9|P8zFqnt29gL^E!=-PL5`>7eZ6Ss<;K$W-HmCT*iY_n298E{z zl>)kMAG7+Dm);vKR9&pAI(JS~xuv$gWAp3v$Hu5chlB{J!ZR|`TylbLsw%hB<78Q@^sihRHx3gb59eunH_g(m&)<`6FgUEIcYkd|CCS>PL}nzveo2j502?tpH~e!V zlc^x+wij3O24aN`%rry}ExxQs_WrPr`S|Os_!ltKU`+QB+Lg3mGPi=FldSgcX%}bF`Q2>@7w^EVp7}FFneLZV&oK zpUEygQ<{sg2J!uj&}s`O4+tZOi|HF#-i&>{4Vt|}|0cxl9w_0Aq=!`6SEUpzE*qN^ z#$d@H<2af$YwtYWj-BLHf?d05&0L0DIr5WS&D4{~4_;1ZUydhQ zq4N?MJ}pJyK-Hg_fPPBkx7A<2j1u(K_1B9QmN;rHzfyQ@nx>xp{k|3c_T+L-$|p5D zQ%@=S)6m1#=?*tQg>LZ4$nPO{1K2*Gc*o#aI4BypAc@EAJvS)Zu4`$jeYgAixjym2 z@SNHu+t>b0P6ZnA9(6=gmEN1%6S5dtoKB=7ZOKCFO&$dHz75X?r;0(G0rbmmzwB}p zUPxt$m+ju_-~{iB%Qi5o(ZP`41cHPUapN>i&QAtq>wzxSn};&cuiAotk!00-s?a&D zcOF)no_*W1ohIs5Gm%!)foT4ytj=>j(cc#~go^|;D3{o05g*SE>MsrkQ9Z9Lh8kr- z#b$MBZ$GYok>$oBGkXT*!+Xw^}27yDcZnvQbn5#Qn#|>rh{|s^| z$dm(0UINK@qy_|R!3yNOrZ>OffkoORK`pa6Zc|a(^G@LN_LbUGi*nTV^&O?KT+q;% vr=E}5=m)h_Me$YSI#=Omp2+{*8t%`!u!Nv{1drQ5gdQ9QR8cX#YzF)XcGf=T delta 2280 zcmai#X*|;p1HdEa98(q%ikuDhlvfHj8q?ff=!9YYoFm z3k#!8HwnZq<~F|=_~PbLipcHTE(7u_KeBz~<6^`rMeOp^xI*5IqKQyG(V$w1@EB7~ z_DU5b+|pT`nvFlR9oNoios7`=zAFe;+uGV^a^Ab=;JU|j-o@6ocb3H-7QgnOB?F~teqz32 zQr`vSwf_2dV%kC?Qm=5Cv~_VO;AR3;u5Dg_uBjros>K1UVL+8zXuwV4{)8z7mT0T@ zP=M?5I51K($~bC0GKkQ9^GTlg)=CxxX8vGIGC@|n!tr682lJa}Yxi5jQf}ge z%PEbI;LWOu-hl9CM*C!t^FsbvU9YUWjX=a&lcwoE=>`0MHv)w%iSl zC^x^MZ|pdX@2-?77XE-Jv`;Tm9kO1+6JVB&TPZaujugF$@6E^nVm@^qd^IP848oDH zp)yKP`TD^ z{muNUr$lplI#wNJYfG;%$^k<+tql%Lz~?%q6s6gv;u9^)5BlE^w4Thf6EC^`iLJbs zKCYWdQEjs&yc|r@)YCQW8t(SlWbS+D&5TY~i+j|b`K>kx9OaoOd8hRnbw(1>pg#k* z020EMIqpaqq3r>>g1OEf0p3d`zSuXyqoH#bzAXT?nW=t#5v8iLvA{FhBnU%ua2JkW zLORw22V0jZKCxjcabguwb|0$%V3AF@93}r~b^gB6yV0HZ8|dfY>_&BkM6G4yQmbx!+m0^sZWd19YMORQR2uC|3WBbjIxCNo||BK{qkn&1ZeCe|o@vM!7MmPU9;dO%;A~o>;rRszsoJ^Prl>>N32suVfHU^ ze-ss8i3yE9ch5Vv;VGqbBuFeRw&-cvS%UYT)&VrI8r62Re`C5_*n(Hc=}Kp8pA-IuhaL@HvlqF7Bhaqo*j*AN;z?`Ri>QTm>HjS_45MR zavON_4Y4_jRLE@RTv0Sa4u5HlsFr>?QS=I>*IIZnm6U$^umsa&)4Kg)(c_fjTRNVr zdjxmHd-3f?ryZ-LU1G0+=p4TPg(A8)Q`O)%Wg5v4=}&Pe-}SOjf66Xr-Y(wHQrlTsfvq!bG8<2l>)p8Rkm7%t}eACn)hG zTmRiMK{MTG5#8><3RL?Eo;#>GM1#tC2q9>RavGXRx>v`;mU4(IxpR9(?`vqqc7!dq zPdzBtD=#o{mY3MRJAWsN#*s4tA`kDvMv~S;`S|5$`SVs)(EW)@O8DxYbr=fq`u#^# z+jOKVf+EyxSQq?4ki*O9!Im~WJZEP0Y-bYAGo*2SgX15q{m?Vw+ndPJy--eeOIniL zujD2~el@&?>YDeN8>dj)Z2e{QE-3G&d!xnQFk=X(B7n6!O9@HYVgm7v&##XO2tEC1j&E4V{?=ZXJ;;JI7N5q5etb zBwx4OaX=PSf>l;&=qmAeUH=_JHZ6=<3z|F48{c)vLubz<_e#4@Zh8HPKzd~XUXDA@ ztllUz>)H$=oYJ~Uh)xh~RU>@_bX$6Q9Q%ShXE~aaHev-_iMI);v>AN<2vs^SiSF@; z{SsMo8F79XWFLX3(vDVP%8qIJd)^A57e+KgEawZ;l^me-V@8f;{fNvF2FKmN2X*&ZSM?SCiv{N zL%hqceTUokSZ38cp~2-1tc_t9v<1S{0%L*2VXzoD9F0cc{48K-=ncdTvm3ZmC;y*8 zE0IU}K>zzSTJK=8u}2`3X$4UsD`ypCY5^rft!MpcpATclS%^NDHQlXSSmJUqK*EI>78F_Zg0{~&L2q6Ff diff --git a/generated/webwork/images/webwork-1123-image-1.pdf b/generated/webwork/images/webwork-1123-image-1.pdf index 56b0386eab5869574c02e28d7c07ea087b9a0985..315bf000f6a9df5f692325b8449410e9f2ab3b66 100755 GIT binary patch delta 4978 zcmai&Ra6uJ*R?@<=pMRj$YE#^1cV`!maZ8}Qc9$Ul+K}JL^`BPY5*w#krpJR8%c3! z{`cbl)_T`>^WB|`bG6sn&)NI@kvNj5|ALEc8_q5Vw zRUi_>7;KK$FVsn#^ws5jApew7PD;!ddO=_87fts; z0}v5!Mhe~Y%;%1!rl4*M+~N9|!>jiWep4_`ICXt-J{KZyTrP!Zw>*@=2~%OWIAKI^ z{JPj#SfP*i<|4pTRl+Lr&1*C9j_%LXeWuX9_=`{i$-M4SEoZ%NRDKWzg@TX4>O|pm= zm4akoAn`^kuD1he-H!h2K~S+q#xM`VUwI5lzkEY}l!KL<(MR>;DTph}^_v01C5@^U zO_6kkbR~aV5&5>(Gc9mBTMH^JmOnR5mv)x$UMl!yv*|3?!cRhNr1!5SjrSAlWwH&r zNqj&*ILplMZ`aRu4t?0sDJ?C_x+w@7@^B2ycup|*t>tn^`Ih+KzWa^`)mgBfB!O+r zsa?X<^R)hEJC+Z4z);VtO5cjJkCnL&G6b9Uq;bgodQx9yK&0t(%Kl)9moPi#y$6w_ zB|YOC*K5^;w<5PM-alWCG#Y zakkX)Bdps7X8v~76^*+$fnNBAwUADs(>3C}p@qVsVapS6+pJ~@ASB?jH)Rk^c z_W8+R#c*qUtg=zikeW4sis`(pmO_U#^Vvd3jOu<1$~-yv!B z;W)tjk$k0i5{usL)q(3RjMOBxw({tER!vKRqJuSn3!6LT3aRXce~Un0#Qcc}Gw5D@ zU!XXTLr(FJIF}DDMfuPEEQb(cmo|dsHe9Zao)00Go7z|_2y|6!{`080l1={WH0;*v zJKk;6Y^kgcRJ=tvuruSs*<1)1HTVcNHgmu1rwO<^uu->p@zsd#T*oQ*^0LhNR-M%} z%PJkqDJG`=%JSuAOFROSYzmzF%UmJqZO9r6e8KG#HZ@P*ru>wV<>bcyKy8SO?w_4w z*`8jIxPja7G|QzCbg*~0o@sC}2*|>U2+oZf5u6{ID3UBw<7gmba@Lm6jBPa9IY)Pn z`o9$EIZUZ@=;lc!J$F1s?miGKlXkrU*t^^Pe@^{pLf#VzVT&Qrlt4r&O|kNRIN{(R z9M)$4ZeNzSm~@tV!0+^giMlhIHO(J#@%#H}T9vG%rkuJ$&5p1fu!M@vvD+kRa+9A{ z145D3`HWccV>XAzJBOXGypdbg!GR|t$#d7a)8*#I3~N8eg`db*!ZRg|AsZXt12}*UNVkS8_fzx;@WO0bU7LI1 zoTigv883w)ZHce-mQ-H~HW~aDwZ`P%_bT4(%IBl5ww@%hBG2y=yl_1Aq_m5V zU=SEf+tBYcV__&Wf)3Ern2is)MyS%}w!frhW$A{ugOqEMj0SW5I@p#;S+`iZfcIXR z#b13hD5bGuyrioJ{q>I+)cJlT?EL~xX#*3s@G&=+(_6GuKl*J(MOA)0!=JyIr-ZI8e(Q&}LUUopMkFW9t>C&}Ve3_KSFYk*~IEpO#lYJ-cH3iB}u`TIu%;(pCC z7DTlF5}&r_=_g8g2fq$Z?Pjbxs6K>A>n&G*cE#sYvtcN*zgsef|xZbE@qI-~QVrt%q496wG z&8q_cMx|lO(*VUNrAHoA!MRLeV9)ujIjU5-s*5Ri^y(R$`!%i49SY2?ljKegbrWwU z@(yw&fPVZaHWfw87Xd|RU#`_Yf{qF63O}gy?3t&p8*p?BEg_zJG<9|EJiiQK2VPya z{NP<;ni~xOPnwyJplWzrIH)DYmjbGoeH+w#RbJ<$C4H;Wz2sdbAx#=8`&S#*5buGz zoJ;3oY!pFe#S_Bjc=rKUS7(_e3^9P%TADgj7ICf|Kak%`PJ+%KHVzA zWhJj$UQl{dZWIC<$X6H4jV~G1vH#8>E)W<_->x7lwSM=|AH?S8VxgvwB69K%-mb01 zm?<)^J|sKtKk10^@`={>%<|>(KaAO}ydpC(SBebZ^tcAEAXO<{Y$E!sAjrPR_!Z`EmqQ=APv?=XR+11wT#5u9>5I zb@Zeh11VOZNRu)O!ZkC(g)W6f}k^clFeku}oLVIQ{a*a`BVO zsRk3mgbd08H6LOOTo_kgmCNJ9ODMB}s3imaP(%yIcxO#Fy!S|iC(c@W2U_}^-+$}h z;_4)3Xp4pQMo@7o@cm_aOZQ(&)2eMTkGamy53lOi`@SrUPwAcU@%hW3A7?YxZv$@1 zMKJ&^2e^rJVuygIC%0eHSjNme+l=d$)_TT=uNcIL?wCS(`CiS2_zdgHVnc%9HyD=4 zkOdgeNAKF0@{V`G`IXL?dea+@%*nz>Ar*nTFBh^T0fEWXa*uemBoZl@sCQUuu$`Ug zK>Xhf&lax3fZwo-AvIiWU`mNC(@pBspyv%*G4J6afk-Vih};&c-SOU{=HbDy^M3ct zyOdP0<9>H-4)2kG$@o|R89^zBkpAzB@V{bx>Y`d>3Kgdz{$drk3e=WuK)+tt7XJY~ zR7k?&+Viw0FyJJuNY2>IUmaaQNgA-1o=FCU$c9bA>m)QFDVkR`ac#gaJQ{l5w<83R zp-JVd$-70h)#+#pQTrPJRcD4*jxa*Gb#AUb2b!!>`V=RDGIWcNKk4^){Zq}EjhX3N zFRFs+XGgWF)KACl{;&|tvht5y@MqPcB2#6|L?%5~xM1l!y`$oM727>M_i|Id8$&Fx zTLT;YTv}P!+Y`!YzP0{1ay?sHQD?1Vs4G~O{%&ANsnD$Xp=GR= zCd4G^BXtbO>fRmlgO*sPe6x9ytjCn*LHPa}RmEq|<@71ndjMJe~9Hz^O9wc1O z1%-wvbpsH@kj~{P8l91+-AKN ze_AcMU>g5!KuVSDdBxMolZya(R(5&R(?r0xDN8WW%&d@Dj;O7tpY85OHHOQ6W0+2% z88lhV!5Y}i8)Onsi6(x#^^E+aE*SAdQ9g8vSF5ui%K0nhXQo}glONq^xeC4uCF9xe z4P#N*5Q=C3r&p^amX#{oRoCx_)1h+{U`(Knip_IhVE)*6H ztlRg_HOXuEn-Pq%_^_{oMuo*W{o$lIBn}S}S_{{XBJPmuSjIn4ZlNTsu|0W9Z{6^sw*MxXh$Ay)?z9L5Bq8DrF^?p3=i5B{PU&P(C-Ix>l^@lkEbKEB zM2r4&O`bL^?~J^mHAIO0x7oYXSKcm)QS`3f8ZR$Br@n-us{xg`Q;mtWI6r-hgVp=7 zQ`wWK`u84g3aEi>Zl%v$k;nzd)q;}{1wJe?DO?2;b|LI=)bl5j!+6-4&_+C|EOv`} z;XI>0UOWPe{cxH!S*b4+GImOG%jFE4->@~v5N=CR7v6ML?V|*yG+b_jpC;r$a}ulT zTjl9^*Zc%&-<>918}lp%ab~%EXO8PRY9$XviUdBqp0DaSwr%oPMm6qpkCp3`Fgqps zk1l@EgdbVAN&$j}X2p{mLJODQT;+2J=#29X8|Dp+2dSr~nj{2qxhEh5>6_6Kqer&l z2oW`-%KiWu?V?PxWSJ38it+(bp-?b$8q#Hz=daAIN6tQ%I>od%mo!zw|f(*@3??SeE`L@eW;Mqy<0YWO;d5(-f?oVtE z$TngdDHWbW=~{s<)A&3@sGRxcIIRIiTQ)Ryu63>)-Na9@oq4i6?Kn1T*6;O+g9RjW?AZ}@*I;qT?JYrH2Xcr*)#@hR z4_JURIA;`T@+MVnR2v=a#V)6hjKRaPwQB9LMeu4zS-7fi+UNQK^EJQ>KwJw+gHb=( zeH^a1$C>S;bmm+5M5ljb65SpDvrTZZdOjGh^|B2nje9?PY03ET)WKKc5l0t z#>SaE2d8Y14E$aEK814dfjHvlUdsQpp8tHS{?h{A!kq!)66{D#2A~q$9wKdTFW~@p zuy+ubg2=$YG7>PjBwPjphe^xA6y(MJj}2rogP~3*X{Z>Dir!N;1tM&vumNR566mmv zgex_?duPuKU$lCLDbP1HAER9O7z}Gg6L5~nCJ`!8KM&?1cRy1FOqad5=+`dFUGjTOE5!K2S?3qt7Agfm2fc%5bmo+`tU92 zKxP?g=V`GTDZ_cz6P3oG)wTsvZMwX`&tCQ9{1X}yJ^NM-9aEM4f7imDu{2;k39-GZ z7Q|VkoA1*ZlD3j*6X$)DbYr0H;ykw@P%+K(OP_aoEl}Ap>@g+<{LfcnWWIlk<>w0B zv@&qH<gLHGBEvfa5mf4)J{Y>vKtT6loY+a_YQ z#o>}uCT3K{=BLHxWaNdJonLqbSENp3+CF#Vc;>$3%5&;VWisHgBGl*MNa@elLo_J% zdb2lYHb*Zk?CApa@}jpJ_Gh_WD~Gy%l=;V-Rc*-UC2ET8IM06OrK74^zsYYp^@v4M zL5rs7qUg1r5@sq($9|jlXq`biBTTb@{Q!{4T~}&U?X(TqTJgl+=Ii*^$7o9IlxpF@ z^0|m6%V%5+;=q#$=fydN0KP9~I)f_ITCYopoWp~|6gZYR4=4{^a8noS4>9?zyB^b7 zCOp5Mk;v7??^kaBWx4i|p@0r7>A4B0L%qVg?oNGYV}q)4(Z1zd9GBgwmIu`tk+%}~ z_R*&fag(oL(`^onkxv0%JpYvXmYn65W;lXyHX)?2&HZYchF-Gh>LT`wI^Jy=#hjBh zTJ6?xiHU{VSb(+lmAx5L6l@{EOAMxrweKJ(Opt0mp%E{Db|ZyI$T<1UTJ9Y|+woE7 zyC??Jl6!w}VxuF&WBX-lW=B)_&7Iu}{91Udg{ARgYv446{SVH%VdgQ#OO6C%6~`QT zkJ_{4ma!0&4aD5t%Dv;s-gw8KJvM(JZ_=NyvwgX8%{s z4AR{_vt?!b$`SMvj zZ<`QZIXY>0vm`jUnP2_pA>AClx08!~v_zc@eEiW}y;f2vD6o|~A<1|!Itx#d zs|B|65Kd80+Mda#^}EnP-?7vRmmMt{Tkrgoqn3zP|ATE`BIFj~D=xG8`(ma(&Vkm~ z@!hfT_dbK7rH~(kza`%xyMCRrXhh!$ONyQyh!Z1HE8}w2Tt?zUYsna0;1U`!Wkx$! z?d?N8D!|U8beatRlvXx}v+ z@zms1iCS_@@{=sOpLQsd!(&~p8Zor>v(F_-NDR{%XKfeDKG^{t^dXu zN>MBCj(KSL32Jnn(WS;^+h6H|)Ek53h~?L^tJ=*&jLHbfQp?Cwg2i_5-x=`h>V31d zV9?nxTeAazhiAeIzyA}S@1zl`@fWlGRwV9krm3ET4*Yc~F&XP~1(~erFP5ibzP`7` zsQInt=i6_$JskRTtPh!6==-ExQ2X8eR5q|qYrZyV*MH8+PJ^)oD1v4G128RODX7GM zuSQkh)xcnzRw0Vs(dwxZKu2#Wec#bopa2U<{|#$HugZ?oC<8RmMKbaP2Sl>*k?xOy!1!$^An(u+sT zV;7&EAM$=y!Cm}oqc&7qSiMkb1dc47r9As|V^k+HZSyvwNJuXN3>Gr|3DYevF!bbZ zUrehUHLRF6BMCsvaVU5hFrH7v2gDlE+{Sg=4O-v!JyTVMs@j_AE1Bd-Uo5J|8mg+r zDy0+(O=LD=e9BTjHPg;|n<16$r&H_%?_l+hN7>TzB8Et6G#7~#k`xeBhF9OH#0*lmk1v3s3D?VWieXwa#c>)ogaR%< z>fpNQw+#LeE#kz;0C2(i$*Kv5jie=Xtf;%{TL|-q!m^H{5ciN6A$Y+C#Vz!oJX~PXlWwEE+vk1A4k@_SCKWr) zxTicYT z#RN27?9s1>`A~Hq(RjK)n1-fK?j}F=JKo6r)j>@UR0}>|cxg{%Uuz+I6(D077}coA zdqOXRS|u)gRJU!S8al)@x=L^{y6!B&maqf43}%exdbIPK%&k@lw1X9WYusl!WT22^ zBM4ug4Cya)#Z}YVll{??5y63ANcrtjCo~E9rQF#4j6>Su4Ir?|MT6b{6f5C74T|-z zc4!!2^j$zf0DCjryUCiBW zp~?W=!?=VF%7`vcLX_dj)m4UkpTKawQ-@o6LBUPkRYQNI$a}hqH>oM9a>|Q6KRslG zV)@o7CtH%k@Yj239}=11Ah>w^j9;fOWUa5A89APiub&mLAo}HzzM6zTxSuODa=Oj( zf($fRT-^3YWD=<4Jbe>rAP3y;U z@D%WQ6c&iJ4oN+sx^1m;D%9oZ;;dl?Qzfnm$Gfj4_|dj;RXlNWq@cEU_^z31LUHWf zoB|-SG?Vr4Akh1d;$5p9+gL;*?nGj7T!Pv;Rs@iJHKw_O6iBb-f(UfXIrEmd zyUX3b%tekT14;g!A;vp@pJKl{=K7Bm(kNeGfe->j0O-GRq+0^w=3zJp>IHqStJ{$d z|Ab_O1IQnO?34esz>7RQ-9dFxZMrm;?sI+GyXUej@su!gq{B>maU zIT~%7{a*I0m)SN--$pNkM%s2- zlWC`=cjhxKdB_sd@O^DSG#f{0b+lDz&XD>6Z#by^Y6>W z>km-d&OVSNzSd+@(Y%-Xo$dU6g6b=veO`9^N8!#dUg@CchP~pVAH)iZ;Y+Se zKA{bwGmj?`(8q!mHy?3N63x=1fXI6G@YiJDX-V=V)WiNENM$2gFCpr!p|i`iXhKNK zb2HsW5T>*lG*Re|)J+G}#n#KJW$SM};p{y#;QYStJbJd;U3T zRefU`%pR2-dm?kQWgPO5u~oAi+q0}%daCMQ8NJ$41O3Co;qk0r=1?9;%RuOt4+-t` z3))#MReEMh+TSwFBny>s!?6Qvsdf7z*-H+ibcF$$_Zh4@SR!6%^NR4u1|sE1uk!fI zr+av08Hi2FU=r?_n4;i>6d$vL+?J^=IXQn;7#VgBEo)#{dFTcQ$lba~88p$Vo4H?$jC??~02w1iX*B$ZObV zSMOO~_JXhi1HMvQlm#Kk`AT`w>((?|@ zSc?a(&8$WVV2J4YML+b_gIw`{uA7Wj2`c*EFJF5yN@rP^q6>X(yuw&fw2`o&9ZEsv zIvqpSM7_k2-yiR44>J5O?Yv^wIZpm>1BPo>{xe&aBz~UUD#uh#D_gsRJU3S^UXG(fxf7n6v&)9SD#j@vhH|n4-`9xV)I3HQ7(v&O5O7dnQEZX zHMhB1`W0n~`6jt(<)V@%E9Nhn=({5qQ^jNIL_!m3GO7j_6p}0Z%g;4oQ@$+u*als| z{8$7>C6W@_HK#zTWN;j z68lJ31$2bvRO*J^_i@;p!bTnj7Q%tp{0Oe{P$whD+ohWa;|YMoA!``b$? z>haclOoEi#i|$<8wb8RpEM0P$L8;kVDRDdI5^BrASiTW}QIY0Fg zxyK}=%_W*<6gH^_$UeSQV?O*&cCCx?2s2Z!u|+)wpSgOx4z$c^f3keHTx}j2?CfmW zX}eh!W5~F32D*ov z{YCJr!o<<%TdqXkjiRF2+B`z-s9uk#U>U6CRLI;em%aP5MCs&h)3Rj2n)$qOo)0EK a%kttBXATRW0-?&lN+3eS#igpFM)W@#Mvu4v diff --git a/generated/webwork/images/webwork-1123-image-1.png b/generated/webwork/images/webwork-1123-image-1.png index 67554287883a979abf0fce16d31527947b914c90..86e3afd60fb352576dd0b81bc92d4766130aa4a0 100644 GIT binary patch delta 21 dcmeyB^($*b9}kBx2RGNoy{gKar|~?|003mF2x0&L delta 21 dcmeyB^($*b9}kCsn6_@K-PF3x(|Dd}003u~2+#lk diff --git a/generated/webwork/images/webwork-1125-image-1.pdf b/generated/webwork/images/webwork-1125-image-1.pdf index 6303c7bf77c099081795e7b0214a4381108699f9..af89e8650b734c08805652a8024284757e3adcbf 100755 GIT binary patch delta 4746 zcmai&RaDfEgT%FFylgVCqU3cFH3+9W5v|k%#VSbyYytQHP?eN1O^tFBaokL^!g7RvrAR7Uz>rpeoR+`$5F@JS$kxXr$Um zn)3u-)v|+CuWW0qEiFiFlJ;9nT>khjAdITzK-&oDEHaMfigo2_8)R1|a~a7a?WTUO zu4ok|{^pBAq`Oa11L*2x%PLkVY}SAZ^g7-eWnE?`n;4#@aLF+Q!I%x!;wp94QZmwK zzhJl^pXPd9XJf%GsXGCKa<7TTO>>{C|~6eDBA)Va@arO zTC1Q(!2RDA3lwb$^YJ{XK+u0CT0XEg(j4cXVt4;`BoK!vNt22nCn zf-l9X*LV@5lchf+MQ5_Y&=~{K9i=V#OL{@WeoV+NG;Ar=rJ0$}Q2{+BfR9t2-P})R z>8Z-!+=!XanPudLlA8-GsRK@!E{5L+euR~@{47-bk>~RFdIlC4-o_iv$1L{#|HkP()8HH3K2j-88xEsBQz2(8l*|Q@uirq69CPIUO_LLXPfU5 z?y0J(h)Te<4HjHNBY=vI#hgb&I0VE8Q^6UDi4wKhw}eQ{=Itm?2I zpyim$3FuyDPsyZQuNa{ndTX8TomBAe&gx%S3Kz*r>;5A} zs^f76V(M);LqkLG%GvyaxZ>HJhP)Z_bpiH%e0+Tncg5NAkVI~rK(cG_wM3Ro6^SY_ zD@AwjKMY#S#aM4s0v~1lLB2ACC%X>6X37Q)E54?nRv12|jP|8&JkR`&FOEHU-es3b zOdzcx{-2ELbNz3|@b9*vsLBLUgltr_ef=%4wV1}%cgfsxmanf z{F*Fc`v`~IiFWY|!P@44`CUC`jUm=itq}v?FLyL)N+lyiKSxpUEk7LOEs|UYbA71~ zF;{xMz^TajaCrqUJ-m@l~)NpmHdElw*jYt(+FR*j9lnBg@~+&=rhg9 zrl1!skfv)L`9n@__Y!YVeZ`kT`kQP1W#iJ+6GASU5lNXAgl2$;j)4=^w-c0?iFJ${ z_XJ`EB)YX*UmP7N9rMheuXKa$jVeDCfqt+4CmiuA2u}e-;DiL5PweX!7aujOc*}5i z@T~{BPq=a{g1AB_O5TJC`3c5+wSz49y)Cr|JaETxv{}C4&`dSm-z5e&oM^L`VscEV zKrqF}>8zjcGysJ7SD*~5?S2u)0Qo?2MY6{r8L7>k?Xvb)f&me09$ zXmdo)BbSV9}#h9~RS!(W-@GyKMcF98DFY z#(iZ`GwUp(j1*dMZGP}E$^l}fZ^}?(gUeKJ^%+sa`amTqFU1gp$c)&QhflVWZ1ob5 zGeczL*rHZKqwc`LH@lCm=yR`uPpI!s7PUH$2_eX zo3UrbgA|+)pHQvM5({qZUYwPsTvL(%GNecfFCo)`czCw!T8!SQAeeq^eqCGE_B=OQ zxLln{hFpjOfphsBV-#ZPP8Lb7^~_)IujdwZ1a-s$*XCuA;MI))vsSML{bh#-GrWQF79`_EC)BT1<+tH+u^6xZ_>pBk@yqHi>cId^+MYflYFRoXODsYZ)wF$D3PtIIGh@4wdBii8W z@JrUBI9fYad7e@yZ_3j7$4E&^0FVx!iOhUAFjXg^q@OIFqU8Cf+nc{KnE3snv@P8$ zh;ofVnU0(#hQ(k4M|`AdeAJrMNR@_Jx?_OU(zyFN%T2StUBQ53qfoTbM3hT~H<&p@ zNz)3>j@tdObuY`mZV0{fu=0pOJUw7E<~qG>uLkE9K8v8Oup<`;Fr@`{00PTe$6A7o z*nKsa-p$C*RGDMQW6eQleq|rnzqJCZ@L^ z7Sz`?JC9aq_P2@(nGM>|9m$vy25|jgXIFA!Hy*BTB&rJQ=}c6_;j=4~;=1fV28XqA zC-#N-kK*!ut6MG0bB1G-huy~a2MYNXTUgxJYSUQu3(0Xkn}3AOY9bshOkJ?F>cAF% zf`FRd-Q^RE8IkYzLrkzT%#Jkp+X z)iA&$3kwCtpK&q4g{*jW9!hm5%2IrHr6EN&HX+L4NK!w&sC zA3~@Y1D(W0^}7+~sO9HUPbg3@`K4zHQD_2U|2q%PRH%jloN)-~@tB{R|Kv<%7}(0j zCt`I)rlrD~t#+82gSt3rv|3>fY+bV|>~QIDbY1tD@bfK2wdU3;Xg3T7$Fr4I)~8v& zDVo(d;Kgoly@HTH<>P!O#-!+Ysh-+@;^PGg$b{ROM7}clte&rR1z2SC^`t%oeJ>~0 zSdK^OiX>!dA5v{!fQJ-N?vzUZOde1HP8EOLj1pI9S?}6$d4;*EnHR7q#4lv2gB1him;|%ot<3f2_Rf820;80MreNXz{uNx^T zK_q{Ei*%F6>Kv8~(8PmGkkB7W&fKh`$MPTidOnN4JPOfXX+oCwIyRe}5vlLHJ)TrS*lNZ?7$7v7R&6Yk+ZusS}P+1wU>Wf$hPVo9qdd%t~mA_^>`r5O9 zR^;C#Y>sAI?@rt=*ScNhQRzv#xUP)&>AUqos$G)3+cYGXtuKm7PQNApLW8|WFfuXOi?NeuIR=THtoHSf`lrL8iM{+fhqaFQ; z96R&$`(sl?fUKc9o^t;aB1&bWLMHybddtwwJQi+S2foE=E3jP<-r!$`*K%}MO4VNp zJwP7vqW-fk82uaJ@1`QASXH>5X3wUT$DA#}6_-Ia5H<$`&DSy%n}|Av9~S>^liAW2iTIuN5OPG!#?dF1}0C*z#2>FfiU<; zwK7h-ucN^_UP&*ojh`__W)LeRgwiVjiPSah;@-1;_*VZ;t+ERHBMU`BqOq-6Mv4Hp zDAnC9UG{!!7Jc|mA|5@|m?(fJd+TEPF-k@EWo8n!;eWznC~M7YS(vPM{mx|7LXq-P zeJj+ES|8u;mvI$=^Ea)#G%l{U*zhbmwd;iw_hOu7&5*?xoB^R%Yd6PxzDmU#N^}ue zXxrXXVyO*wS(}YNJTe>lBxtM!IGlQwEa`mDoS6Y zP5_(Wb z{3ch6wUfh7ZUcuK)xyJUvw)EW*&6PYIo7b;UATG;7Zr@Ak}Eb zy;O!hgwRUxh4<(hB_HH2OXO=F*H6Fn;i?8LBqI0K9)FN?G0g}0GSczgWxd)&&rh(j z5LZS>pOXE!N>cE*EGCHsi)H6l&80tOrr$)fH_ufz6ulVaj+j%qNC2jK7r#Anza9l= z=b`pcV9{rCG!Z+yzr}Wf7N@Z1bm*ADy=DHh)*Yltb4t-q z0=ELoq5OtZ4enZ$|3Xj4bKOCMP3A37UfRdo%17en?LAOCppIt+TU6N7GYBDUU0 z`z;NI?ZpU!r5#u8cmT^S-klLd`ZxzfWY#AfuUkF1NwT`HNnWO6-aG{lBv`*usdq** zqux0`<}_p3llvPI5vZ>r{xnSgSg*6J1(4A9`6RE4V~~N$|e3_ zRlNld-TkBO73vjWckx{9p-wn>6OYcSV@P8lknTP#8FgcX-gYpoQ_oJY8^6jvd;!%W zY|YU&>r(E@|HCXwazz)1mpt?eB{K1Yh4e(Q8*25_`hJX*M+V6EN!naTuNe!h7hmmr zj+}$sBqSZi_c7eecKA3G@+R#Y*@R@st|+lmX@iHqBd3d_sd{y&aP5WQj08gT>*o$@qIA2S5Z#U(G5d?%T~-j>uU z`D7+kh`W_HLZ6ZC&1pY=q9O;FoE{Rn;xRX+U*E*X19K!07ZnpB;N(=*QzQ5<@M-v4 delta 4736 zcmai$RalgbqJ=?vK$xLR7`mi8MCpcsp-YF7?yirN0US^o29R!~yZa|31ZhD+K)MyF zz0bZnXWyLrb@e{ade{0Rd?Q@@4I7Yo`Xus+vF0iv#5p5Q|fE$j?1 zUX;j8et!9Z{_=MPE90V%ikQTA_p1GYi%gRZ_qi@zWsV7`Bw@u*h8L7?_lij=Il7D! zrZ=Dzv1H(+sh}Me#!Mz%AFG{66#mS?)&0TIopBD|&Fqos2o3%r)zNL|tewaOW(raPYdXjuR zG3^zfBj*SFu{!c?_nSd)k{qXlPL%0fvd4N}PdA5RC2!z(&DEK&V@&W%LCoZW$E9TSIA9yopAba zPgc_JDAmgu-3itRc{;ocPSw@XZB%3mObtZEopJuveV9uoSHOWCaAhJ3xCji?P}6SH zDvD^ut_T;@4-v`qkR0zRde>YsLJI$gaFuz%qo&pMt?o&TO_C)uZlY*F?HcS)vQ`yZ z_jq-&;+kq1720a{vKA@DQL+1m2$K6=kXF?vBdyBSI82T@=F|Q=^h0KaOSrIr0(bq%0xFJ2>~hky8)KmS(QiI1dq zLTe4t03&|QvUkiMNI)mE(81v@2fDlZgmx8+viWG|DvdDg;V?0gIAS-6G?w2)>uz*5 zoRVj;4;ck@pj$hJvo>Xt9M+|TdO#u9Sj^>LNA@Lp*|z;xIRN&dCmo&P4pG+Z*YL86 zzf}>MRogg@9xCruR2Di2C7uD)7cjHr6TPZ+9Pbua2V}|V(3!?TI{Lj zrGcE2hj+5SF&3U91>*thH#>V@PChr2+)WFEQ0{B4_{`*C`6W?ljHQ#uMOj3Es(n8_(a5~9 zLjlpJt7c}*DUt6mBc9=S$Z}(2q3l6g#KPRa*&>Kq;1i#eu|eq1cJ$=0>iTdx;(t8F zQC^3A#eHV)-FdrP?v%;(yc3<=q>)HEn?Wm`Rw82jEH#p`PdoW5ff%+$UZ={x3q_X9 z2;xu?{V<4*Xj*w>;j5)GXS1!yqOBy(b9oXId*A);yVrI@A{Z`EReos+ecaFVu9V z&NV7_7CC=GZ|aj1Kn#G67&Gg_&V$G2PnEN5#r!Q!Io%PxvO=HZ<^E~zlx!Z~c6ZjV z$>+p><#`s7+(3A^7JYo`P@Pl#isu_fhuODx>e6%@#UU zkf)@<=6uv(u|QWy_Fmw_m*-sd;!{O)c!?wsMwv;6URnQarEDC5E*GKu*`?fx-?=F& zqKPcR*Yj+KEev*NOQw&gHx2^!_wo9F+`rp5Xh&ZVn)rtS|AF92s0=v4;6~t)6Sf*2 z9j3LornRCRDJv(zJff$L6z;(+pvD4f5qT&SlSqzenl$Rwo0vELW?{IvC8FM_UmA?3E=2liUTf1~s zR-Rxs(#PpW?`kP7-Ezq7ATPzOT5R0LExI7%yH_>^QjclRMhq zu;Jvn!mR$UOxkEu`ZUTKSScy?=oJ%N)XTag4X_vf+x}!QsmK|w;HOh|O7NsMwwinF zy&@HjVy<67cDuxiBvo5}RbS54@bFINKkVnfc!>j%gbo7oCcaqLik5gY+OHTfK$3~_ zAMnnrDx-x206mH!%$45F)-~1N8V&qN0mOxI0%?i6dqe;XSOrT2AfkH81EguGAN&LpGsO`9VOWJV z;u`j~d-hy6rAOe_x98!7Ivi)@ncgNo?W-Zz9>nsNIjptk@aYmDZrR{ltvqy40^oSB>ogPudY}^#&qfedCKe{HB zb1xc6s@?G9!GXLVei5%9eS7XjNGPVT1GrT0Y4Py1m@P~GBbTG9UPI2Bwil5(z;)-l zC0e`mS)t>gV}2mwM^cto!FoxR#v-*GS%zl@Q0V)K@ZA06fc1nNv)nP$H<(*^o)1-t>cBSTl$g|im07o{4_}|Z!1x@lWLRDe z=y1lzd{1%5O#$aJO(At8SR=v6*4&U4W5?&!nETF=#l>tZNq6P-Fv{sihRwJre#n(G zDS=W$){>ug!h4HG|0QMJyFY2V8&2COhH2g>Hu7Fm7Ph9nCE^^qZ609s*CeOUFX2A< ze(Oezn*shzC1*?@-Eqa49Pb&C!e^zxed#nn>P5YAfcgN=p-^C=3jVZ)TVZ*>yR1$9 zj%*HgFvkfeG3urDWyUN(kPP-7A1WHOR;dbiWKY;NyM4G@780C9%q~%Rr*6H;;xeBY zIfm&Ot*VyajQ+ux>KC6k3NbS>L(oP@b)CQevCeP$C-9wjhMyr z856<5d4_L%j{O9U+S;$h+u8*uS9{qPB=^ob0Zf(Lc(wm=X%d735dN<{(T)-Tcm3ws zX&t5p(fW|d*ZU5MKtu_H&J33^)rwryB~O2+)h|WHXs>JTc6D|~ov|!L#%zz%m=wmi zPLFNa*l|Ei5An3V63s#Hi{lLv`jesz`a_Wrrhv!Lt*0Hf21D>bKi9EK2)XZydV?_# zb@=tdx=WKm<+C(afHp=F9G2Z__QnrZGKBWI{zcv{_q$fQK}vgnY56t1ZcttkXMuTp zK$7x?O|&A1cOq+PV_4^N!U)s*prA!+7Y-nb<>sErSud-L60^rS1Ly9N-lO%&nwvl4 z72kwGz?UDi;n_KaYgVwcg~~p;El~}S)8Sa$F3E2?@4#dMQZ+#Af8$_FG*g9p(2qPm zy3DIBv`=$#PH8LY+cY0Cy+_)33z|}CX#DbBOgP${Y;Guoi@Q0MT(9hCigTC-X8mff zk|#T0ibD?eGY1j`3bB=!g157m6(j?1f^R~5_}BTBK{)-?l?mz3xMV6<@@EeS{hVe; zflw#z%iRsP*Y6q`)aYtcc#~Dz8V%*N8R2&ZWAzaE9;4zLC#9^HMkQ@Nkb1_-k58-% z#t2Q1a?Pv7bl&>Sx$ji6=PuhVY9CEbt)mrwIyBwrhod$oaI=apMck~MMSASH%Y=O+ zGdG;dTyC{wP8M67(pQcqi(NjOOZ#2*0@w6oE<0NKkS(iL?qStu%?&a>5zT^gk4J5f zN^e^vxdYx6Uiz{O)ZrtRWX(|!)9|!HJ^gn=Hux+!%Tdq$$sX97Dh7jU!%C^PH5$2urFdYvT<#=?1^xqtkQCDP>O1pd?@Cu5d z2z4*i#=qb$Tq)Y?>6*X?GpY? z&<~CY54`me$K&fKt7af&SfvdKs2I`A#f7V3O=(f67(>OVis#JX*b!OU_&AXkqtWGk ztl$>L8*Qh&f@}MLxbr=&OIv#Ym!ZNc5`n;Y7n?t(U4!y63S*)P5w3X+}NAJTYmtur!psK&B;qy zmqhuOXA^&mL`r3vhSIy`m7!pw-^BOST`m5_4J1{xKeDnP90!K8m|I_McQtgesX&*+ zWz2!k`4g~B5#d9nvPg~m<&Ty~IUTBtX&wBLOM3~S)=+16cA91l>w&exMi#;u_|9r1p3OV5a9MGYZ z*KW%K;NJCHuH))Ff;*17WPJ)ffY7PQla^qUQ}F%q>EsQ5Urq>+=dHdERuIGIb<{L43h%e#SMqVvB@@?2C8YI9WJem&;`> zAd@8l`1dx?vFJxFN{Qmvbf=s+mLD(7X zmzL*(Q>oQw?RWL{LK|NCR;c7kdtu5X??Ux&$Fhn1Cw#85(F+Pa*XwWtb32^e&s{`5 zKaYpI(lo7u`->j&K)ygPd~CBpCWkA}#%Au0SVrw{184QC zdD)KUdDDa_N-AUbxk7)!U)EjoQ+8vPoj!urm;TYT+ojy1Gi}T}G3@Y8{Gu_#BH|Jg zbV%%uqj*c?^H3e#L*0}Z@?Rt5Bat2B=+ZpcfF%Nb?duIC+MK5nmX@#-^mB4Eh&z9{M8y diff --git a/generated/webwork/images/webwork-1125-image-1.png b/generated/webwork/images/webwork-1125-image-1.png index bce11758ebd25759933882cf845e97fe98ebc98c..4a0ba73406cd99ab335f5d46017a068d1eb9a54a 100644 GIT binary patch delta 21 ccmcbWek*-L6c2|m2e*LF{X3DHlXxoB0bWxF_W%F@ delta 21 ccmcbWek*-L6c2}hn6?pz*^KVZNjw$m09h;tMF0Q* diff --git a/generated/webwork/images/webwork-1138-image-1.pdf b/generated/webwork/images/webwork-1138-image-1.pdf index cf1c560e85aabac76433bbfb88985ef4076113e3..7a47045030c98f2e0d201fba82812b7972076d87 100755 GIT binary patch delta 3608 zcmajicQo6NyTEZfq-N}*F=7)js%pubj>F{}2bR#j`Xv}kQb z?X8LuTiu@Dxqp1`J?EbL|MSoD{^NO_hoHKrTDMFAAe5EC-acMvlq)!3#>(J>a}R>L z^Fq|;CmY`jx~DPpLc+v!?df;#9&JwRI-5OPk&C&!a-KS>iy#XM(jR>^(!eM!uj+YV z;zf;cDpsuVJA|Nm%{hz=4J@K9OT<{0?vLa}T|jQAOV<7QJzza6-4*thCe*YuqI>JZ zrhkko;5~#4u_1*I>alFUBw|uI33u67;7t6ScnYa^ul?pC-Gg!rR6kvxlYl#tq`h() zZQb@vZ33GL!h%p3n6J64fxnf;w=$r!icK((QCV8J+Hn~pQ&H4es?Zh)EaZeGbM~#x zEFL}_ZZFe400l3&3V_jW_W!*x+6}i1G6K;MsHygUk2fk%3J{ZthNH`IXexQ&&i^hG ziJ$qX!b}Yb181?sB?Jxb%CIi26?C=_1#Hh zG&j4U2}w(&2omaVyBV>7dHv4-r|bXrKeG0BjG4j5D+EU_zUxc`4*+ z)y0L3tE$bkN}^ckHMogWZOsDXmQYcvJq1NrVh&aAGDEe3Wp3WsaZqtnS*%H1lJK?} zr&v&C4c^IaV>mZ^?`!9VLMH6zE>$WFuTDA@MWGb}VS=>J$x8gQL38b+zkb)loAMWF zYL3gN5t8AFFgTz_&p0froaeR@!7dAL=nk8UsuR@dea(6wmLH@Gn^)UjrB2|r9+vtw z+83W$Tbl%05l=als!WqgP`_Dhr*3D~rpMl65d*{EEGm-g+2b$3cfSZ+h+DnP$XI7D z);lnES-xToMmT{@h9f^4*}pV47h|k4sUJLND~Ycc4hClEUzrH(*`jX>XmW{;l~F!# z!ck^EA>>lAE{EC)s4+BBnsz5-*=Z+7Dy3*!@~a32@6P>fCtrGQShKG*oa>zDBZCvy z_ZVU%&P)Ia1>nXqJ(SMtZ3vh}D{e2HH}c8`8?+1E7CkY$`S>Tk3Rk>MJsM9xu{m=h z6vLH!8-WmL33f4V{6+4q01Es!x~>dA(*!!MiYVN-e5cej*XZ;*3R!HXAEPcFnu{mu zKbJRaWz1+4fVk{Y?I-9sR9Ks)>AB;gO9F^==_fQ;Eo-#h>0T^@n{PEKs!iJkP`bJ% z74$h3YQcXCiMcS%)&yd0ke6s438d|D8&Q5Y@T464w6bJ*eS?rTYRioP`WD`(p$%k!1}M29>6cFzRz%e3#ah3Ly=mqB*Vn? zzR*p*zRrRYnPX3F3&rQ^pO|5DtgUMJ+>=&ZfgAf_m$B_r^`6Rmms*xt)9rOSVGMYn zrn)}Y)&+u_$Tmtkpcm_P)7Q3%Lj~+E7HtQw&*ykRf1GTazdv=Qj;$Q3P;jeoD=3sa z*iS5){M8Z7hux+)WQ{{LiwGz@$nM>qWPP=BNT|DkhDPe1TWS1sUMi|$0R5tr3-Y2x zKTkCdV%;RPeZPGFWa}qch;v)BZK4V=BtWT69b8!IAQ5kNqk;Fjd-UGHzRMTI5#8bN zpq=+n`p#KbJ8DplK4;1<*v%x6Sn-#mvCYaF~wm3lfB#n4JdNCAzBuAJOc7wZMKTFQ#Y z(nnt*PQmZoP#4K+UC=n9?8+ogcjNR@G;H0Ca5L4G!7uEJzX4dy)|+8AhqJ@^Qj z<}H1#JY%Vy7aIOgJ{2P2y1@4Ief5}WqOYZ68-;+JsTp6=D66Vj_>pB1S zTULqnjf2KB*dHOkv)|-DXz|2rxGP!?K<1ZBqicK4x5$s}ZVM#}i_^#XtB*bD$rsG^ zeJn1I?+4zKxvagt<@k8}L#6^*0NM2etBS7g#2TAtpRDo@GsxkEJT;bvl@)s8lU509 z4$qLZkq={AdY=1c^^Sdtwi)wyIP&swwt1n1#bnfJ$ycGrXkF4=^{Mk5fTR7^o0kUB z35Ec zSU!CyUk^nhBbbq6j@rRc+!)g(W*z7)w(`PBGBM)L=>!3(|NG7tM~y~MR`UkbJsH2k zX2r`I;1C}p02^Go(h-96aJ@4-5SR9Ff^C=QOh7dF=r4ejrX4AoY@Upu(;fPPoTohc*Nop+^sMZjL`xxZL3PQ;fx1JZe;8;lKMy6e%3vCjH>kO{62mJ? zx!};h05a8y3d3A6t#hB1d9VHhprVHflc!~nijGB8B_l*9i`{%UkT}qal{(>kN`ZL= z29VWX5)N&>^3qd+i`1BZD$NZfY)*3k^juMY4P5zaU_smg$UD|m_Fo3#6qzV-!VC{7 zsHlFQ(2O>QFenj|Ratrl|0mElh#(;OKLU*yxEBMy8z==us0?-R&7O zf3PxltqKHqwl%J=^pNLusw6gE{;;;XvM&|`DUlA9fuwv+B@h-0C3D(UTD&oF!i(e( z%T?2PX~xLDlx*lEDV6(1&z4WvwX(I(SEhtX*3H+T#+_N9xk#Ado81w?3ETV6PD3Cq z-%+#k?STetvIi2^iY$C0_V0XnAcp81?Ce_YP~$X&6cQOC>1pWs_IX*N zV~bRcG|89Kcti@5V-0)BW88W*o|8dHs`$bDYCNeV05y}5lqN%wF%^?sW;`Fcn=Pbl zgg!)9lex9^ZLdb7)I@fBUxKr`8PVcgzCn{=TP$yR9oDXsMwb=v4lL-E8R0gz~rKYj{P95GL!9fbQmmCIzs zCU}K?pzDcAM^=~1Z&q4w*7iT!#kipEuyt`SKd(rNG@9{!F#P6$m9O9TNA&}@CKZmV zNloH4(?WLqI8>s9waX{)Bib8S%8I#7?9u&GATRmk2SykA)6-D-66HLWK2WHTA*MIq z$6K&jo!Kfejo_a%{rEgtGdasUkCFid)9dDDA&Em^QuKHynRqkQk~LHIaN|ihgsO-^ zuCCAlA<~HDh5&9gyM_^z=e}3%(yGEFF__t4_EL{juPd>Gw=t<1SIzpsaKi62%On<9 z%sFb4zwfe(4`r8l-Ibt1EqJI>;;%AFXb_RDV?<6_V+6;#j!3J91g%z+SXQBnl|frD z^}0q!=cib6vD(ieC5B1e%3g{sL06zv+IQ9^feF0Jc-mZeP_md3KUtYRm`O|IAatrv zSHnwIXR#af!GFJ9sKs}RBLq2hZVzxdFm%y~$rB5F#d@Bp>qScRVs`Can`8H?nH_{qAoKKFW4sEbd>DC58A#SIAtEeUM^_rM*?gEP%4JR zt!UAb+EyvQ&%RERJ55p5omnldG@tmU{J3_Fb7|e&OQCK8(v8RmxwUHc3@(`%a41K=ybNIDhs^_u{Zk zruiN0lXfz$6fiUQCVHz?+w41(vgzU6YF-2tJJ+01slU_7_j_LO@&T0&(Jc1K^-W82 zFb|tQs$w1ftVGWvc3GiqQ#AFB#bUcnZXduL>Xg*8APwqqL(mwi5^}2^%i|1x*$=4sNqv0m?-%Y zLG(U6Qn=TUd;YqDWoPyT+AD9o3YCv2h& z!JJJz{A8sLpN+owFs2MUiq0x%Qy)JI2)vY&$AnxT9X^$5UXu{y^hjFShM3m0-?QEb zf8<0-YotQ5ISKeJa$bS+OdLW&tgWl2Xnwlxr#%L_~6>|L8){PS?ItwQsF zD!|j77wq6+@B7rp0q_1FILn*rRXkgLG^l)~8C5ec)0mhD4Qpy@0nr@u}i>=DKpD+3ByJJ-o$z znid^&wEpd@)|U%o#1`zlIrp#4qAX3*{NpFmig8hfaM_5o*VUV#3n4UL; zL3uk$-`r3c(c91Lhooce(3%`X*ZVpR4*cGjKPu2-JUu1X$j~V+?ymZzBu@wa(L%(C z>MkbVJk?#7@$#kFdzO*s)c>7ya6F6~P=GKEV&n?(D^WQZk!E7pY)suWI zP58V*<%?TQIE6mcUcDB(sI@hezR!a9z`+>t#x>|kQU*-|-oZfSmK#?_&cjO6))(Jw zW=+1BVtPG)G`xS09%5IiEY~bp?CgosAT**JU!#pGC)8K6Z89?)6z!M|`pR|z7o{}K z^oFR+${dJMzTrDK^@t9yC4Ssjz%u@KycxO5fPf{9lx-rH0^_N&hp03w7CQw!~m0;lO9jy=Cf2L}KCH6%it!Rr(NJT)LC%)hy>Qr~= z-N%A3hUq1IP(QmXrUJp(<3kd;F=Ef6iz{Db95Xe(aPBFU`ZV zRL7hwmd-_^X(Jn%KxWodv;jn)31!=#vY?ORTiWA0w2wC3>+x^7!x>dhhAFakzRxI? zoRPkxx~Ik5%nntwnBjBwHn&etz=1V6dYj$27qgfEv(L|&vY&}`mG1+_l_ej>4ZP1> zbv>o7686XFe;p2f7}ZP&=*gLjt4hg|@{5gn>Cxi%`V?1G{uA-6r%p^PJYco_K;&vy z+MH`sA^CoN9jMk%ot5>?T8!H$FhCoQd=OeU%(^#q583swR%9~$ul|GPPKrSJFy*7h5Fce3Q8p%jQl>vUx!z&iD<63p$jx75E)h;Z*Kjd<(r7C z{@hvgMjv-Ijk129=vy=|vrnb`%Ri6(Q@5jb@{8QhQmoZb7Ms|e(l3X3zUCnOMtA=j ze|vIBPq16rtgJVdugRhs;LtC$%x%=Ar+m=xC;|1rLV+!+-r3sQC1^B;p@sY; zooW=pbgHVz*#mGiSCovh)c=j&=MNnxp7OlL%QKUrOrX;96-ceG>R*0rOTxQ(cdsre z3KP2j-flYLM49bK9AdD8~#hXgMEq<}f7a;NVJ@ zuQ+*;2XM4;H^=J5-qvT>$rOY|3y;_>fFHW|6YDUZeTTRGsdKsd^usR|~Vd^7M~OM2a(+*UJrW%V4aAbTT0etT7GB zS;SH48%Zlw-JJNVOjcQIXyc$J0vl!wt=P7=n2>+~R{`_ieNbh!86UfYYk^-^Wr+t= zCLi262uut%lmUk+UI*iiK@74W@&pZU8$uhS3?T8p(Dq&5?H{y}ECv87+)}FFr(CfR z{5TDXK#)VGb!E+RX^0@V8;pJ6OXcGqBGTU*(9lK|BB^{L zbSlI9l-1=#4iUd{Xo$;Q1K6F2=r!FxO*dh|SWdqo=44~F%`!w|6>#*nM2KP7_IS%N zyyy8&L+fq6<0elicTvSjc_H8I~Wry^%6{yBniwhk{csBGs%+}#nNlDuYcyqtep0emk3Qq zQs<}Vlr5bZ;5{zPbPH1chU9Yj1$85mnUxAa|IY^BWm=_9-)D~fb%_Wy(d#|q^ItaP zgUh|7n^8n$`?Hs}eufYd$i+1Wkh`@Dc2tg}qh>}`$uFi-Qd3i*o&9CZ3_T3|phL(S zxjwx(si^J8RKY{IiXlwrP6;D8fnCMYYg2Xht~!jT%beGmag|S~3xFvXAM1w;-ljOL z*6P!z_OM~B(1(UlO;Uj$G*FTi&%>;eRy=-lwahwJx#!4+d)p=n)8Y(u&kerGoupDq z7;Ra#8nX24Ikw3@0a#3(P9IegO>w6;B(~9xEc&)H{Yyg=EbP10lGpj=Myq{2fh8lj zt(|A~{azxqBphp3iO@KD%}jQy=3<N{SI|1UgBEr1n#arYcCw32r+*_3#4V+ zzmwQZfgh~|r0)v`@9>|}Vvx+I2tEaMq@4JZa$|DX?4Ma!Tu^gOET zL}R;8gB_-%XUYC(d0V^`$*BF$Fj*n!a0{~RGj)LA$MJ>dp>H2ih|S6D9gg9Qjhl6l zL1sDKb_TpaWG8%RJj_67V|OUe<;LRxFP5{QwGmcP>5$`zzgt~XFfdpKY*CrcaAZ2A zQm+15dffD$k!Q+$tVz6UAghDd0x?!xm&!SmJbe{D+x^-v!$=S|=(4a~LiYCvO$^D2 zMnohm$!bB+8lgTYL-XVecE&ot+FRp_jdE$wciI73;?LY|y6=bufs($m*%AQ-OhHQU z??>6Lr@5wQ(D-v6z;U8@*x8m(&9*xsdx3oYH62q=p*1;0WIl*?DB{mM&1Apgx^peN zBU9lRleC|KnesxGYazSKt`K+RQ9uN0g#61Ug&qSYJXZ99{E571EK{}%DokaJ??&po z1P?%YXrdTx*S4;?g{Vm?@U`0hCg5V@CbE9dBkEwd*M$MU0~#bW+?WcVyagnAb^i2DI>kZE zm@fEL-Y4_cqt@ZfHi2@rw8ruf%bb;KfsVe3akY}(w~LN6VfC95E;oJiPjxdGSPmL}50@=l{M332_)f8p@`K#!E_yiQT{Nh(Xz- zQMQs8v^YxgzO*A6FC}R&WhW(v`~Mt-3@F}+`b}b*kmf^!Vu2UMlnenI@P#WZ?cyb= zVY~{QR6##p6J5e~Z;6EEzKSsq;j1wI;*4*>y$q=%Kj$W4_t)8v1q_xh&uH+VPFUq0 Z=LX9eNEHP3B&frn)EF>7zlyFZ_}_CAts?*c diff --git a/generated/webwork/images/webwork-1139-image-1.pdf b/generated/webwork/images/webwork-1139-image-1.pdf index 6e4678bb9527a40a7b0f6d46781acd33511eaa30..656dbfc87d0e2272716aed2b2f4f6741d7691db6 100755 GIT binary patch delta 2665 zcmV-v3YPVoGMO^4)&&7Ilh*|ye+ojkExOB|WAe0tnuM4=`1dt2-9ryac$1lV!z5L6 zYLs-4d|AyqbTCLIEYA9FIYN)4R3yznNbcekZay(XlIqON2b&O*xAxY)*U#0tJ-%Eo zb$S4S!$!C)1|f@cbLy7a;nb)>Ls1Zbq+R`HXBXJL3f<$A$3Cq1FU6Z;`$sF)NW!C;$g!|`nc6*th7@sU9iKin`7iJ?vdfy6^Z z6ity+3V+e1;g2FbqN_unFFkhGX7}8^?>_&0p4UAOO-UpaP1Euf3F{M9#l0RIpRm|F zFUlhJf6h)7`3Q3;;dxq^D-u|-5Ga3{OHtIMzqh~6^!B!&w=a7l$qb+*G>!PY-Ad&f zp-N?VtRW}gsL4|ETyXk)E-W%KB7_T9DkHh&I+anAo~GyG(|EN?m&SXBY;KJqU8Caj zx!|R0p3hmbaG}|34o%bLh8m1nQS(DMvxZl5394L`(WJ`Y;tYD8`yfrH;=F(4P;X)l zx}48>m63}#WT=dKilW6z(VL>h9-u!JKs`$_)GR8HnoS`QMN<;0h`L0-O_z%zMT6oL zp?HSiGP$OFU3B{c-0(mw6yPBS9$N>rms9UAIj|N;03P&4a7N3lh!^MgB}=!AF9sQ|216M2uERXOP#WMEi8kT zuo}WFYmn)MxB@L8&+XtAj<`?M(b)2n7+;z4XPRt#a@fz(6LUiVX=P2&3+J2AlS)}; zu_c66Yo+b>yU^Kr58CA~xz)0e;_4QKcg=15=j0!5`phKeAByoYypex#Iy&6WeV6(! zoQ2;Ip7!c3;!osDmElRpckZh{>}c)T02a3S+LgvrfPU{<8liv}W1@*KnN9-8tRRaO zY+yu!LfVX1`eaLs%{w+&^1j#+4O;<`IO$V54s;7C9{bP@vA|BvVd|j5#BPU@lEMNq z^dvn;eWe~r$IWe#X5fK)z!Ze;aapw^W{hE%gwOeSMC3_b2BC>Fd(bwiS!!gu9 z*!JV_<-`>%e3X*3T92aRnKB!euBf#-i+VP}hbU^vCz0XN7jLt0jfyCw~S%e(pb%*c2j@;RX|vxeGRbUu%Of5CHIZ zKgB(6jH0ovZ9#-p*CDLksO<+xlhXxi5~4oz+fQ@ShXoS&C3pFEVO|!cRy~A2&WH+o zsFjuuJ2N#;3$Z6HkpfN)GxUGE<7@-EE{U#GcmC($Y0z{@qS{7f}i zd&d@a4^fb1emh3j6GmcZxj@HtgFzaNErawhwk$v`|E;qnrQK@r%iR>t@-aM$5|&L` z#-ufALrT0>^V*KICFP_&sUQ`k53e;N9Y|}hcuPu1v}ZymY?h=e9cksYiAafm)_niz zfm3Ckg%N{u<2U*8`2tLTr8GV(Y>l0=eE$R8Zj-X+^@f}16y|Q zdb&tpkNtQ&^UO1jlm8Sme>Gdvu9_^@=$pcl%uQ`^eNOSbw>hG5G9Ht!t;tu~CgkgE zPS*Nh$zqq4dfE?@epsaA@f&o|l$)k49}t{btJ|8)^i!4V<;Tehu+*K|x4G^h_7|;> zjYe?k{#spIm!MmpVnAqiiE`%Y@H4o>{k;Q|&sEVSfya9l&?J*5ASGZL_yX6%VhdxMJ~p$`~8Ego}Ny+rgoDs+wZ6JHU?tqBxBh z{V^Q&ekxPXGOGu(_s#)D(%tDG98vj}(znC@N+0YwCm;Sav)R^^xSfcA!TCqwA6EJ% z;VKc7rh+D67^Xrbe*u-LlnIYi8u5M*i3)-=iV`7Ys+d2Z5a}eN;xtxC5QizFeko;) zCxWXqedg1Zo~-r{)){Vvk+u1=u5wdo6pWAjzAEfv=alBfQMH|&K4vY1iri+WxsYU% zb=tf8u)JU1&fiV%mdg8|=`L@p2FZ8I^tyU+|F3=Ir}?C-fAZ0SQ~|l^awiHIhs0sr ze8kXLAc<9eQe}OSaRO3BF3@nab&6=4`h?e=#$;`?6kr ze0O`dG`p;3e{{6e<$jyB4!$~en8>2u7~d#m)!FtqxGBto9+E%WLbp|24i*KnP}$=U zZ)%#Y-Z@ezHtH*)nQve+y9cf)xp(iwe-eM9T2c9u>YiK~UsT3nZk|f73J_s@!FL!? zBs@mKLl&X%d6d2?*11M2d)#W@BDLxCH9dcVzWBzcJDs8L^3q*LPq5&uil zFVy}&YBM@hKKevH)YunT26D01(^X^`f^kn))IUcTT_jFX{0;v|wN{hs7c4?KIb}Ix zGdN^7Wic>iGc`3ZHZ@~4G&N#kF=aDlWi)0!JUKaKIb$<8WH)6oFl93}H83_cV>L82 zVq!66Gi7BoW|LVMRBk~MKrjG9(WJC!8JiAga6A{YbuU|slF8=H&jdhg0U1&BA_$66 z%pwyjpaSdZ>KHFqVtW3g+@6KY;Tl*s*Tnq=KAEluWC~?&WOH9 delta 2682 zcmV-=3WfEVGMh56)&&7Jlh*|ye-)%`TXdJD$Cx~Ape7;4gMVKW?H+na!kf&@8zw5M zQ>CPX=+kQ6po2~-VR6)x< zesDnLfRa29cJZ2sYk{L-94%8ll8n+Eva-wxD)H}cQ?wobW|m~>abZ%rJf6Fu$scF4 zOTAs1tW26Oz}^Fg71(dmmFzeAW*)`x$=W4zs4ukPgG+^vhK)8}70xds)_l8@;09%X z&L2=;(KEmM+D|ORL3xN7C>&TBWuh1qAi-dpY{T(w0~I&Wgz?cq5G>V1A)Xt zL=?>*oKpCUE(w()Jff>ZpD#Uj*Jk(Jz3)E%e4f`m4^2rV6iw6e)iE)NYd5?e9k+I= zcV3c3?EjpbD)JHLQNr`IFkd9FVj)0(GM}QTNk4CYo$2k}S>C?vi6qmXlF&5b^NN+q zH-eSQ&}c(WzEP8<=DEQ1g5 z^SQv~YM#$ows^7GYz|J-i-(7Y=C3XzokxDy$)8u zDp&&{mUv`(A+|sZ$a6cmg`@71bu_m8B*xdK{Fp}D?i}{B^n~0XKw4QN^uUEi^rTXj zSz-xd)mmwr{T_65+=n*#OK!C+sHD1C;azhV|2g@Gn?5^<`G;bB9B*WQoQ`(4bKm9O zi|61sgr~iFhxifsa%E`JiJklEk2qSo6T!kZ-MH3p8qn{Z%fl4#VpJsYCDVyNnLWp1 z1sfQVppZ7$&c7na+g z6wr3C<}1i4Sw2{tYz5fM$}6fJEoX+|JTxYMl~i!Rj`? z57|q@!lzqR8uSz@%ys=d+WHNB zavYb9i7|_RVPI(Qjc+dK<1Ea zatQmsGtPEGcTLBCcdaLmwZnac0|G~hn2m;w$$S#R47tT5fVO#?i|?>qhZZ!^bMp?{ z1&23(t*+066!_S4Qq5VLlhygD@IPzH*F);X7rMlF82`(hZ@k_GgXrOFJ;4i>L@i<= zvqsaPuQip~E@zNg>vEzX9uW~1zmI7=b7JTb3s*GlTWhy}I7@dg=|N=i7^APvZGz+I zN`LE*gI5z`SokOwmATS`4ybCgaPiw+36aeshKShp?A#JKwg;H>>LSb&hsyoNzwF}rJr0v0Pzb3DE zFp$76c`yH7n3qMVWKY5$t9b=IsihJYIz3np6R%P%}>w_cV_38^MNKrK#kGohXE2#E{p1a(e==ls@5b;146Z!5q`)sWdGul!6k7;}dfR8OKH%lvi}ugA53&M*#^sRn{H8e2xv!`QL_ zlz3UE3rxG!{Et}|PV+H5iV~VREn?IPwMHdg%V}+k+MsgO2dY37sJ+*Uq7JB)SG++b zDBLr_6V?kW)E2e$T1TkFKWnV!bTgH<@-PR&y)DG z7Y&>TOENw_3UhRFWnpa!c-oCr+iu%95Pk1g%v)^~4lfcZst64D61PU1-NMesEmS=y z7Oe=dMM0udd%u2%l$|=+-J$|)a>&Es%$YNcll~Mle=%FrEbDBo(Km$!nVHJs`i$av zZ*xTBWH=;W8na$%n~*Ql8M)I3OXj<*(35T$cf&j#4&R`Iy4ciB@qpmWTHRD+s-McW zUVIpz01MrkeY4gr#QvhYW33TfxW7hM)+OlHCm0Z#ZK9lcI{XZ7aewc?R7NI?e)q6~w?ob;VD|CymamX?pNg-|4Sl_O{+zq>6`ER9rFno{2E*yM*)o zS~tBz{#4faW_N(^Jw$ODG5WpV@BEaenr2q_rth5tDwuSqX*8hnEv0Y!-IYGrb51_| zNoKRHDR4Ux0fY08!oRHaO~O?oC`|=T!Z1vQe@FrX%r;VLM(x^(LdGF+7&jjE5uiyj5uBY*405L0Ple;BesnUYQUq)A&Ih>xi6&;F6^^ksst zQhBCwGM_rzZP1_0+~1EDtDAS9Zx?2le^rbQ7P{DPv&O-f#}*TrR~zFSr6^n59DAeO zJm^08v&nT+Rz+`~BMYTH_VK2=-s+trg<_+=BAWRICesygMajy&55FY-M75&wCDk3d zGQOyc!;GE^uL=-he8G1ZP$WD?!b29J@OhNJD%QD1OMBdE-y)T1^))?zg1-2MKBvq) z?f=W(MNy-!s7R;A2O|EDq+h80H)=CDQ$F}eZYt~xECab%>*&fe48gc3E9#!3i!KtU zDE=@%?QHe_UDFf%tYGcjahHDoqpH8(dnH8f#4HZnJ2I5%T6K0G#LWMeQh zH!?FZWMVaBHe)q6H#jvkVL3K3H)1$9V>6Ri7*uY-4S*m3L(vum6*OK**un7}%(HuW zHfYn!%Wn#x_kfHjMiB(XELM?;9Z-SwbPbG`D=|I)QEtz|?Ql)3&9$(-zz3RH2Wtvt oZe(+Ga%Ev{3T19&Z(?c+b97;Hba--QW(qbjG%*S#B}Gq03bfP-EdT%j diff --git a/generated/webwork/images/webwork-1141-image-1.pdf b/generated/webwork/images/webwork-1141-image-1.pdf index 892fb1b400b0210e111e64da9705439f4b81b03a..998d0ec84216795b991ea741b197a297f0366c42 100755 GIT binary patch delta 3296 zcmai#WmMA-!-j#;4NAiRX+&gz(kTOmBBgXoi7+}8;YSIhV{}SNx;bK!GHTK=@Sz2f zP^2bG$MZkW^W}ZtFYovJ%XQ9uu4~R@&IHZE2*|6e(*^|kyE=K(hA!Bpk9hXUFhh^e z60NGUR`m+yDV}-okyQur2NR$n82Mm|55g{C-#($hf=TrHeO$|23<{FN;SvMnQ@JJ$ zI2vWaHk`>=oM7`D9W3+aN5|^V#j-Zsw*VS&} zfHFAqj_*$N$w!aaKfN2>R)aM0roptBWD=cL!(^nDsS>Fb)$|P6YMD~484rTd88@;nHZLK8Eqvz{gSGs@=HAnyAmlvex!}Ley4IncCg*m1xPwSi~@O!2?pTc zF*5XBTen~D$QzR9NitH`R%ko3+E^?wQ71TM=pwWg4~~UsemTV}P8@}`{w>F-&l*@G zq}|r;SpTSErQCWB;VtXal4KPV5XOdW3i0QQn#{{w;&fwcBvw92`%Jh7jM?vDCDo&Q z)S2P+y81_*>eydiehp7Gv`TJ7{s7v?3LCAy`LdeD8~wvi&Rt8XxHR%Fiv{Q}eqIx+ zx0T`-9~Eu;#iE+rcp1z(hsiOk;E>b&nVv zPrYtFFPLeBA$uxS*)Wc1KvQ3y-V~WVh65?Y83IyKWOe9d0`LUEt1;t13 zpAqZ|kK4kQyRKgX_8~q6nfl@d)f*DS*pO<`u7*eRY~ROBC@QyC$~I{Eoj@Y059oNH zWVhMLgn1+qt$!c-M>OV6|LS~=13-SI8X{%S`*V4ph}+YciQmc|EH_2AYW4*PwiUJ^ z=H`AFOPiJ6`Nsp_qyg4y0!KWT zAq7q01Er*#S>KMC0Z*h=FuDNtMBrU?4(%8>Z|Cz>r-Q&PurI{kEh-JefjBE*;O-SQT< zhu0-E8Flv@awqixbVsj^YoIJL5W~(HJ95XPN;k&`8;9X8ZtmEY{&Vy-27Y=7Pm!95 z=oI>FBbRb*e>P(1VB32U4uRvtINj~i7GL_ioD4HxL><%_h9LMe0Q~J+auZ<1`4exu ztsXcU)2M(_a3J~P?(jO0Sv5CH|MigPno$X3^Fj})s4zLqhR9f|UAv3{ur2FJmkDSl zW>}88Ii3_QxcFTiBEdel!SD#gE7w%2<+`!8;o>Z!U>v{rOm(f5JZBc>UeDNfaNv~p z7jcLo5A)WW9DM+UpuOr>`x;KtD`fZCy?bBHq7Lm=Cbz(=+Y}Htq7VHjZ`bMam)0x( zcD*vv@x?;<`P3L|v|zZ|eC=JE7b?JLH?vl>z6_1o(boZm~nm>rz@|U6rYm%>0 zO4F0Y-Z_7j6-V5NfRqeeLvpR*fBJ5=UlSZHeV-86*?${(6f*qX^u+6ft#a+N$e$w+ z=o7w2-&>O8n5;Oz9PQNK4KN2Ma_b1cj)Kf^V6|1a`S#CVZd(rqvj@AmS92RX)0*w} z5O)9Gch?x`!5@GP5a%-YWr^GJbjaNI$)&n4MpL3Oj65$=?-?Z2nQtC-bqxpUNPpVR zLb-*DOfmcxJE(YfMa9C=DTmCusr-KoZ~OrsD?`70$OcB$)sa80Qx>QMiv@Cdn6&XI z$MsL@izH}so}6d$wwJ}+IXzWdoM>KihX;}@yt=JIFXK_6Ivz}9EyJ7eWNy4|rVq=; z6vYZPd&ZtK-ypk$FODQ=SQz%L*u!U)!zkYL75mxoX#?G}7fX!+Jqd+4Z4+javlfP; z7>92M@4eEDSkC{@cHI}^g+I+rhx_euh{=<3N{q-$ca#b@_RiBJ*ypjpaF#SsyIk^= zv)M(qO93EOY=GDd;Toy$U!JN-JOb~nuU)n(TnZ>9jYb*oBg zpRIdZXq=ID7${KwRwYA-=ml{<&hUMP?oZs>#qux?f2GlRUF2KYn5TL=Ai}}2=*v;i7ClRf4yd= zW@nvvO)FN9Vpea2RE|!ijahDxZZpkZiF&8er`6erfgelq-Y0L+dKwtQ$(i_z28%a` zRHG4y^`;5VAGCm7+JfTE@O{*QjTYPYw0@XvN_w1e@QEhj0k%3B8xS2k7qd_b%|fBU zj4=skviXt}-hQP=R-aubp}iv1IVK246CEy9RcabXx;4O;5j5A^;-uC?{)P=<690DO z;ZA+Fqf%Aqte4D{ zP8J+gB|dwS$nDS?{Hy3yElUHr^9lU?*!fVB0y5t^K7Y1w$fb#vE~R9D*x)^WzDw3w z>^We?)-nFHVb6u*(`{#~N-F{vUclRWw;gR#f?BtxGkQB0cv2=+acfLV)b6UF<=Y)+ zFb;c`m{+c9a1@ERaIF-2*^fJG+&iv86M6m55Vlk`6DZdL)tS6{{QCWrda8akH*JQ% z-9KJ?Z^1W=m0Kr6Qi&qq7?0`n|8isHl%iQxr|?)jvcd=eFxYR_ zB_c*3-Jb@%RVynbj%M0^$}5V8NKK7*k5Gc0At?^78HrltORp@DZ1L$)gpl!C)U^9xUH?H-LV3eQJenk|&!;!MR3(00?CRI?G>#6CJ$y_D;A|82 zse#Ea#D;2mUgN78`c&v^$wb|;2Nag=B#Kje46P$Wm@39bK2vw5-+>38FvgaytQHP8 zyQu>DT-ygk^Y^s={%by>06OKCk#U2)&EtbOqtumJu|3ludfMgUUdoQq^jyVikrBt} zV!+Q#_I%^_GYf@4zLM<{{ zOiEiuR@Ti8qNpV6ETf?G%vHfzN!CryS;iIO>LlZO-}Rol>i@Su9*F4zGXK-3BKQw1 z(VCn(-|RGX4iq)HsO4{=K{jQe_H^kVxYW->4Qrjn8mF6TB!a@eiV2T|qt_%2zQuAV zOtFHEqo$p+372bUSnu)88^>2IYE3KO8SA|Vh-k?ZYIDBQrY~>IT%RVzp%_j;v-I?o U|CK$mRh*kzj#fkjYNSj1Z^H#Z9smFU delta 3296 zcmajgRa_H}0tH||sZj$rLXd6*$LKC;CWs)2l)|Vy}fB*Y(@6&xhFXudb2m2KJw3?P02~|`CczSujUpWK(<}8w7h&~8q6z&Er>l2@n zRAE8NQC;XoQT|fu+Xgzgxr`yBZ7EpV z3eV*c5yzPI&K%7+H32!|7aH9o();5dn4Y_uH=L~?l%fh~Yom>yP39Tun_0OjJ96;) zv;Izj%N0o{et?P_ooLl!)v0HD)Snm}ww-RmOZTyC5}lWho2?dezjGwFRkd^XFI63L z(`O=Aa|4b4DEZmD3|5aYHrd~+ek8q;4yJRZMigEfN#C;k3OGe`(`X}MhzJO%hE>mL-noxz3O?tdbDo{{0- z&1~Aeq6%u4FlXT% za|o7x-c8;)bew*!1^@KVu`(PK(a+9@QeNfYh~XU;h|Kp=aWIdt_gc(bj_&lD*$6(c z^u-~UKFus<6>-zOEpj}+bacgG3xi|?;P&xSn(Jr`k~AF?9O0e%hf=tlfxO(5EV;ax z-)a2&*On>ECl&ZZVkAv}$4zQP5v^5_L7pFvUS5nIls+w^{``Ip#mMZidjFtCw8uMu z`okvmDoICABN7qy0mNhg7)MTQUMi|twxQdc>__#TVK6fK<-;_{`%eB*2aG~j=9LJ!%B;wLM*H>vT2N?1g4yI3pb5HP?7mVr=d9wtVi3KJIn@Nd;rjcO4R;LJ}jkqN@XAkrqcx!}s2!D*}hvx^kA zi9Knf;gD3cB?zF!hlnCap3^Wp6h&(3fI zT?@wp>Z8jHYm)8J?$bVP*?iH_m6W|~WK~$V=rdZRe>*31Z4ZT}TAubf#*qlkdmIH! z5Gx~31~o~0JTZ?~b)VBA5501yNT?wk99nZ>P3Iu3{j}@0iUF%V{fOXDt=i)nS!y?m z7p$jv@|tdu3Y(o@8JX9Vf`W>VxYoYPpkx>5*Jzt~s@tE)KJ#acilXj|*Zi2ygGJ1R z7qKB*!{J%PY<9EJEbslB9Kh49A=$fEQHrMoCg0*j*@#38cyJCrX$Jd9*1eO3$Pbg-)7NZEw_}$kAgR6#TMOYeuXB>aN41* z^?5&`AZolBUOyU;JF9^#Qn`18DHmG%PE?S)OI6}j{>j8`^6Aba%UNFpMndFMTD(VQ z3dAEGoyOuRZ<1Ojg9>PP$FI{#dcz7*vrLEO@|>ZQnPn_}Ew(VTJUT;K+E;kFjKMO7 zBDPvc{{E7Q$xgX6NFr40Ox~tM$O$DW4xK(f`rU9`@o;Y#gLP4k70`QsN4IHu_M_7J zHCtaGKgU5IcDTRn*+V_^R;bVY3aAvBb_3QlxLifYNCMh~+Xq*2d7L{J2l4C*#a5;( zq1Y7uDkWvN8==!cvPiN2@!g=u=^xgFHku1lOCo=&0UK#KND(i6w`F(>s3y7l{r=#} zUd*b|h(TPJvNEMo<8(~#p+fE!%f>?(%kGT1kOT6D#!XWgaueQazOVi8nJ-o9XV+J9 zRO6KW*Df~yiI|z<-NG>|TPC3{9S~m6MVAKWscsb^H@mL#<-bt59#+O4 z%@2;&hVdPs4oVl77PHM%u_r}r+?tmjHuxJ}+*PFxp436kSqK1kvHj1fp}&KTN&oti zANx|z)Ctim5Z=vUbaz9)VUR`2{LeD)o+82_z@g=9&?#2*?OZi2(LgeR)whR>vhttMQ{5;x*;VnglRyKC6_jZd2h;rQ8l3=nJ^YbA-OQJ zSY9AYc64uwtzL~RNohHHZT{$TyIfXKyI$T#ZmHx2(()0pcBsu-*(0<@49@(Z>I>PVh7tUikojlCg?KV15|P zzE0Ngv`wb>v?bKJ!3A#`d-sKWshQ&bR93*xky&^)HGvryJgIlDUka3!o&)Mf7I1si zCn)TT$X6)%FHZZ;^5*!AEiOj?b!520K4MPDEcWhNIdRaWr(1 zu@RvHq%D-K!@lHUa1f_7=qU*ejVciqC9R6=CTMh%DJRX)e?|P-h|Vh{@iHJUPMcK& zK*#A2A1zz9a-FPdEoTRu_hE@~lCx9rYPAS@x#f^wO0Y~F7MCpP|oXbQd_EbD42B%%mkDeG&p#y3><8#(?H?FP-i9DeBPp`Gz1Wq97SmajHmis>DEt79p`J;zBs!%o zObrz%e-yjzSaztVB16BVNO#1?z{#<5s9O+8ocfBNVtr?JAt((>6Sjwy7(nrrMPrI& z?`PCxvNtoNVa5{e(|vKS%R%j|yYLME>>8ctan+Emk)rZRk5mOGA)k~qV`Pn^MgN6L z10F_O#`!H}nxDjR!f7HZ;V>vKO~a39z^<3ZhzaaS4b421DyOoG@xS|h6kMb3pS;JK z^awNj_%Nn_qhi~)Vq`Q&E0AS}mebifm;yS774o-}Kq@J#N32=@+zO(*(t zV@yDaD1%LDL zJTt)?LH;M_;bL86$DSB3p9o(OeA~mHqA{V&IH%fGAoRz^*vN$}?5{-{kr}G*%8IT& zwsE&+y-{4mUYJK1Q+f-cw{|_VY2K%3(Z6xI_;nPFgWU$~JzxmyK+@fETv?iI%d|n0 zbFWmGSl*r)g5)Gl!7CX=!2V__Y$f|qlx)3^7FuC=qW88$)h1iOJb2=ac9bf8`~JM` zT;LS5{iP%QHE&!zOAY@ON!(4&KvvIY)(xdkD{HVm*X4w`55M&0SK^~a^VZKQys4fr z-m8EI=#eds%y*U|hJ7T?-}*kT?hmngd*7}vPOTsCh6>y&pgofGFt-oTDL2~a)sTtL zEU)!2S5>*wpBI!I4f$pfTALCU7>=$L!{YqgQrJ_Ey=QMsR&I%`=6)`q!}%Vn!fi!G zC2VXVQZ|xy5U8a1tJk7%2pnz$e+8G2hSHwG|KTz@*eJ>%-m*jzIL)z0oKO^M%Jfw zvOioM`>X?B%M=BFYqX!%ojY|1gx8sY-`qY2D_NCM$k}@{@LZB(34cye9RCA8x`CgW N3<}`kQP{h8#dTgn>ULjRGUxBB_*!G$`G|&>-C% zA}MnH*1GGid!O#xc{nfooOSm3?J(U)-D(~(pa#_iBVD+qWvqI)+NHb$#&cD2jC9t@ zDLfPJeP*DHWa?!&y^y^bB4gni9S=;-bsVp69mCy@s!v9SXfkbh)Mcd+bB@d5ufe-E|R6V7)}D@WwRqI(FCu@Q-t=(LpPe-uG$DK0z^zp`S28M zn5>Jm)*YgLeTz!6#l`hcapg3?s~lG@3<>#j zBC-p2&lu2Hts6NHF2TbEyAn(Wg%qw3#*&Evcvsr|NH$vUvxf5Q*H4E;j>3q19%xTE zu*5YddzJ|OwK=@jS`h!d;Th|8y~uk-if}A;#f9zxIDk2Y=M}C2NxNTJY*}r@XJ-fb zW|GjhUldbQKlTFxgBt67f(c_f)f|oE^!8=jMfdhSRs&#qsRSX5VT=0x*YuqRwTPYNR*iqpCLu)Y{B%N96>> zPar7~sjmSX9ZI;{H(t#@Nrh%lZixlxw!-qfGvA438Zfd=*Y@;OoCV{e6g_@oM#0WN zZ}6%ryp&w1gRMC$Q+=A(z@5wlxvw6-;y;%j%jEd^fD^+N4F3#e#LptcBZ-QpGNp5% z;hQTea8F_xd#l_=5p-2JeDkyQaPbZ;S9ODy9(S9L!~m9Kg1Q!Qg%rj59ACxf>=8H8MQpFD1=bdA{sof*1@ zn5r;E*3VOmZDLj8jPlx~LhG0$@tZA;%WLPn?|m9D7^-nY4M~Z=&!!K>+5nHQLdH#J zNwq>p%$|ThC_w#H6~MwAk^b5FF&iq6pp;-PCpP@Y8Np+C~^XglduNWPcoKMjjZ|Ysz!O`^=`?oTd0bIp4`Iv5s4T}t1$W%F%Vohnhbeq!;o6c zo9s2h3ZFv{EJ>!5=@cK#O=7k}uUq8>RLOt1n7u7LF<=f)>y@-z#^vuG>~?kZGEACr zUCKLnbmM*(FoRF?EQYPTa0Z;my3kQ9g;f^;2$`dc`rT^G4hElt{KZP z!j*e$)OqYK_9UWP63}dFZc`54QglqQ#b;9$R9oRp?h}~XvQOfR-j0dHkz`ZKb%NSl z4SFMuj8!T0G{oh0s^piYA6`Tp@KMn18%g^`8fUHvqDo6oLLj9<%btB6d0n_8JjHbA zC}$MYpBxC!GQXn{U~wMPW=3{TZ+pS1pI^i|?(;I)%+zoKpeJ{v?oM2~e1a^Iie)-@ zbHRU#dC8kd(9b%QZx@ZYf~XJHt}Mt(asJkO2MAvddQUs7H!^r%wtu^CC9QgW4EIg;6tuzsS*LGDM zxDCya5OK5O5#-zboN5!f-T828nY1@Nos(D>fE1O}Ug}4aSl(64!#RqRhQ*52br$USG?*Z}6;tsJntTji-&D@E7<452XIXNWX@xQe8SZTN!y zj0*p%hqNvHB~d-_;X{|UOSV!{SMlxyp=%rBhka>0W1WuOnE+xY8;h=(GZ|KU;2huc zdAOz~B_&0O&k}g`q^#h0m@I1sQK8>y|8a0K66Ay1Tx;lNdr|6*Iyc=)cX+}-nUWEY z!q3t-%NO6~jonlqM+9?d>*E@-)2eBfO}>i$0rhP`@M!^ z#d=XH*aGZJ$gu{tn4NQrGG)Y{`9CBy41B^qe_W^n6L!4{)ol(8V2nVb(N;*EoXRAdD)#-ej+Q z)ERgE!b$!dXopY3I`*#Jmv3kP!UXABPH@~Bb58XT57CTOeRS=a*9^hxaVHRow#?)V zgY|N$l|zpl`Xvf=`in;Xh0YK)9&LnV`hh9i#?HBuYh>|tv1cIugoK!=B1^Jdoypuc zpg%sAPPSP*u`lOZ<_bEqIq`&o#jAOli}CT^SQ-S#E@`M`8aMge@+aiPl05F{Q499w zr(kgZa2YzRcQYu=BYEG)qhKLcfNQ3$m%XF)OwJi zuD-nfE9WgutUE-y6CI7FfAD#9 zQ`=Mdbmb4O#DymeW5uiqi2iQ_=6-9b@zjGk^3EZ5_k**YC}^OTnv!ptiKI!Gv+N^yHG57~ zvgwu8;>B(dEMkqg6E?dO`>R*b@ZhojN3Dx8;Xx>Xlsa`dt#^Cfzz|~&6Vl~KHeFTA z&X$k4OF^81m6Tpa%1_&0A}u4OQ=u8AZ{0~bGk@kA92l1y-s7hGh?#so<}sl&{229I zU*B5JDBplqzndp=(%Dg@s92H(`Mz-jtB_Occt*PQ!Iu^5k^TGIVk>!+vS!vk;(p-9 zaU~iUDA@8gltenr@Th6A)TT@Bin`jA-Yuov6?WNt7WX98Dibh^6yuE0B!6#DS|C=+ zILU4i28$?sO8=&nO4_bt!MnTEeI3H{?$orSC=s2SgJop?)GcL`#dQN6quq;(P%Z$2 z3r2rGyGOG(LQl#{K~R7>R>vQ^+|unp8SntX$Hy^Mc(-ETk$!ZqFXBu)AS*>JH|N9M zlF=Ok^scoQUezxiSe$GM1+VrLLL7+BY#Fxs<|Tc4~Z7v8s=$V{p!sF77w=he37@s{kNB z2iImNg zgvWz*6DU6Z+8dt_VxSc!EO7UB-v@3An+N$?m0uX*?A2XmiJ9kYxF06N<6L4nuXLAfKevGDVGV3cx49)DY{{+`7R*1`Dyt@!xKJ+$PuJk3X^7CbXDP zy?sDH-7c)%Yn(LezY^{3^Hx-vYq&A(R_MX3Knmy=(_dT(9hpLq-AmhlREEiC6Tv{p zC;)BeP8ezHjq&!3P$ZJg7ZjRAm8o>GabCT^Y~D{A=gPd52(-_M$?mg<<}PB+ROW-1 z%&9}$h)M|5=o2P!A@zi>Vav9UteL;)*BDBEG!0Oi$DCDps^#~(dBB5ZzCJkVE-x8u zME=P4Xwe_ZQeQ4Qys$Cn7-?l;nGnOx2lA$4v`Crut-dfFL#^Eukucw|g(6x8%8wWA z@WKSWq~iXynFF2;(Z^PNq*xt|-Evg^VN+!ONSc!~B+7()(wU`_)Z(tcq)R3D~e`@BWa*GQF*oiVfc;2w4zakZTJ+5~U@a~gB& z^QBqt5bXxO^2!H;=lh5IL;JP0ND)FB0$4FS4iCo*T=yPa!HN(?u+R%Cf~{{#=+|(N z65lc&S9dBQH(Ud4!--d+AE7}A)_Wmpn;+`Iq7>pwFbT#3p2BJJWh?0OJBeEfw^LAL zRnIY2Z9{|lln)5k-n(lPzOt_Y6vpD*M9!A)z3FKM=bH)=i$-@`NnROl_V>+@Turbg z*36m+TXf$apKoiLpLP!j{)^`diF>90xUIm`rGi%kD?#C8Q@P~JMq8{?&%idVf|d8p^B&-=QVn>$h+Z<( z%%B4rS`YtRuDdqK$FGsoKzJ;Fn(J9BEE>jW$sn?Oo-wYH7xibk)QFCLCwLzg|H(9n zu%Pb+gCO7N1k*v2%igl&zKVg3P85~yOVQuRaBrt-I+q{m_u1a+$9=-OSk|?6METn> z?!jX(mqK?4tB;A(i7C21z}cJWtlq+IU(w9v$S&I{^k7S%T^)MYlO=@W8! z^%=N2nR;b&MxuJ&{IGnRKvmI?#>x8$WxEJj?CXz}S|au4W#0PxWpg*!bb~Q29P`knZkoq>)mI`@d(c zd)7Hm_kF+YZ?FCB{R`I))~*2(AiC{;@Zxv;<#a!xVvZuRi6CY>>T!J!-uJA@x8eL^ zRU5YrxVkGhoB2UVdUzzMs;#&}@9Tg@M!%c#9VNdu{euf9dnhW%#<)+GmfZDIU3%d= zMZc6MmUp4sr^nYEw+W}Z{#_Dg<7PMbGy}9d?LHV^kDnMCSvOZ#rew_of^4^ciG9GS z?{=zQ3nC)%1I5{%V6>@ok}?^dBV3edb8Iz{{RU0Ql$mDl(!_$S#5w4a%$f&*eEIWD zoi){5Lc!>6`*u{uR7v#h+1ad``&vd|BA{=tfBR8${dbAlC6g>`2=>DBY$_cw$q_L{Ku5@h~amAI`}Z>vzk%nHC$9KB1E5M zOh)Cbk=sy;Uq1P&XB3N;Bzd4gG62z$`g!HAiC^_JGxiCjCx8!uy)?gY0Q&-nZ}=&+ zD~hV(WbW+h`EgD4kkA6nvjMGAAN^#xE=om!_E!iq2UZH1Vv0IM*%U7mf9L%dn#A1{ zibRNktWFf-8+`5XPh2a}Y`{KV;nVp?d6an050RvZ-5iQH((=ES-kyWL0)YfY1w98h zvBFlIofcV(uDH0!Nn(41^53w99$Ak_lIT^dr70qE{3=w3*vCjdfHtk33%aJT6u;_r z1T$?OUyU6D^A@o#)5vzLoMM|G8FwA?AD6+qTpmQdPj=hF+TZMn1CQ*92o?pVU=~%o z@r*B};_HFjTsFt89|Bb)B%cYm^p!3+v7c~~E!a1y8R@y+suK4iHtKjm^fv*#X zW$*RK7@1U~Im_R0SVZb^yqq3o^^B1%K3FswqY;*3XV>ipvVBs8m#I4Ny`Sxaho{dt zXxej=MK3zS;$%mbmKJR#2R>C;w9Ui5^_6NCy1-(%p>e?t)FP5?1}0?Fa3fLfSix_I(p_{d2P=Wm7jM=1WE0T}Hx*H32P4_Gb%TMaFQ-BkWD-LoOIWy( z)rk$kEZuD9zG-(Rk9bLe{32m*i$ncIY4WYsMXevCoLwJlPlNJhZv0z}n}%NDiEj!` z2t4xK6`FIlp*(&V5a(6*i1#V74DA8X*@eVh zc+fT&@vys#lEtpG5&E9VIG>wgt7YRJvf@H)iX~R}DgHQ|O`!N#5xO9-r|eR+!G3;d zW6In_7Bs<#4YvB-C0e8K!lB{NOo6`%^_kb61h9}0yDs0oUeTPub8W4ZAIJ>hO_g>s z<*hwFLQr$J+@-6F_%cA3#dGNvDn2Z=+Jx$f9yZfQG^^_#G(pgkU`Ba0+Jn1{lCM+?}8gSjqz z6KN5-5_9z3vN;a}@n26~JJl_1R1YR}l`4?3v`L4gGICs60N5wOy}TReVmo1%HMfrLh|djQgpn{MQi z1Mts~T*z*I;?EF;Xm(h5W{b|Dd$E(f5|4Wura`k?-VDH`{cqX}+;UnZ)_ik#9fe|*9Vgx=~D9JIDnVDvmcZ?n} z+772PYV{#Mt-LrhbnIX@fgcq1f|B#FU0+wR)b|d3_?TKwr>bR#>SI8T@#$e(n-G2Lg;Q2O zE7b^6)jvUX?mw)(|E3l83(qz>DDhu>K2U)Fn~!3$g9Z{pHL%gNfWE_&0u#A8ZPUO~ zz~YKc;vlTWJOJ==M{$=J<>s5dHR4$XM+Z6j$Ps98uh8V;hadBQS5d?nlt~hGadT4t zfVZBU%eO)nEM83~RWV2O_ey1p{j_sV#6Kw8RM}i5+6eS5RFJ)^Iyyowlhwx+-*;sSzY^xea_FSf1eJaZ8 zD~5H9a)fG=d~_@1N;MXH`iJ+_gpG^&$o5Nb;AP7NbBD9Q^Ssb8jxo;Jw6euPg~}oQ zMj>EPULMoiN&zLrXByn~_~jl$vh;d*QGW?vio?}X%A1ZTO}%5mlV9Ug6Qvw;pT20) zpl7^IujV^skKR|F#tlT%)D29LCRbA};Z;)$({o^#rc=~I`^hV$_3Oxf-S=w}cw)h| zk%2o?#P=)IW9E7E;?i-N?FCl7zPKtjAy+?j&<0BTFTwRFM2L$D1XLe{-eCMnQ$v!? zvrC)hFHUv*1tV&sSK@a&rcBmuuMJ16S>M$u;9qIl1@9k{i;Q6z6o9n2up<$Xz$!W& z##~2qy3mK1Du=*Yq>8uMZBgaL0dJp=6`IdNB9K9UnI<1Cv6jvDufR4)f&It z{6Od#e+{{D_=0KCvcKL^xb9U_YMhR~)p?YF7KxY>Q2(_z)H!0?SJxz)b0 zYEU7FyoE!%rOm7uNGQX#L)SGggz?^Xxgt?EOL8aQ)J!XPY;tuHUJsU&ce4?gWgaT) zo5s|FA@u8c+r^_Rw5JY+34ah4-(GFz9p*QIEZIHS#{+ zcaY?H*oLfTE^~Ur*Ey*;3!S6!PEu`)+7$%1(=ZYa#HlXJUQ>cr#KUdx=vtVY0oAwj zH2WoYsT*MTgmo`gW+b-^vh0*xsXT$h2Zb& zP{}C#JSJC#MUbn%%XUQ7c;kG=Gh99P4Nuh}MUzxo$!s&V7`C@S_#H^+7`zRQ49-GVXEk;kYwljZ@l}TBw z&*}M?r$q|`!}OdG+xFoF&Khdqz0)EBBbY6p6(b6J)9?b@W z;rCbah)vL90NaZF=FD`PcbItwp(ogUKuc20Yk*D@MeRyrxUd{#M^31va^2&uLRBbQ zlQ)P%b93(8r6pMTK>oGn(FOJ<{W7zr=`Ql%U4H~QeO`aR`E2TH!lLVS;O)!LFOl3u z_1yzcnI0qh5{TdXF2Db2ToobkpBIWX!33`E0vLEM78fZRR*>(}N{D2D+0)s+wx)sS zalib`XU6}bkL3=1mvG#8bp7WlYx(4jHDW~JY4!1y5lglFm9NG)WY9Lg z9TOx?-~G%%{u`T-ooB>?v+kjVe`E|#N26L!B%H(}aMIq3+`d6kZ2oBP^&IT$vh||c z>1*zFGZ}KbhCcn$tgY9W%d$sy>!XWyfw{9f#YQ1vTOG5hqyiJ-cPEQmm)^K`ZEcBr zWp|ETIS2`gbqxbvr|@!{*jdK;ZoPr`;_{Er=J(R>6>G*L0)WDIarg%^*m@_5H&k*X zx~ye#&1hEG%F28IKX$LOf}znPKD%cL;HMbT>B2;@F-_`9x_FQH*0bbC;`8_$#hzRp z>`0wM`+Q8bv4P6@REoo3EhehdNZf2GHFecb4~QzoSOoIW*@dJ{h_fSWEU8n4XI2ti zJ<%D#wi;9mY+U>CG2;T~W@kHVD`G+$-)InNDK#_db~M9>MNG@dZ(k*ZX{E`d(6iOK z>o1msVZCd9TnwK1oI~JU5(CQHrqUqZY3K+PKd3YF-d9bsYkroRo^ft< zMr2}IpA{Bp^`cFd;>bXhA(LvaJXeBv|D3T1G2BMVMe7V{Qor?oxT$4xrsTuvASrU& z#T3BA0{vI;CXn65pb_%f4x>Gd;~P{-e!Qxub`L6?rQ%{Pj7vw$EMKP8nB!aiaw@xf9P-HkCKYD~RAom{8~KkvINmlL8bJY*&oKQWV{U%EX(UEmTd9u zI@i=ziQjg0y!Z`zXQp3xk3Kc#X!(B#{)ymFC!`Yq6S>NufraOKLqr%i`HhG1pSVs! z0P9B@Gr#8N$2;oVu_W4t?D zy$X!+Lkv4Pe)sb1{=%EoMn)FJ1{cke35^2f9pGXbmkNUf9Vq*O@O3XFGM}LUja%jo z+)gJrq!%R2G)bcvX^VpTLP;KzhH8NkNcG4lpymbEJNir%1 zsv9@LB1AtL8tMQSgM?j_9vGc z)bU>4&TVv76&)NnPkOebeV*IqbW|?jHh{oUxFjBrl?7`G%`y! zp@F7fdMOOB=*F^qLaiM~8WgQfGj)CXXT^bGhpj>r^QEO?8B1=4IQ-qV*pZ8wnSC<1(nyGzg5g^tW4($2+_x_Lj?c#gi2b04wB8@CVV?@-- z@pyvvdg|XFB4;u_c6EKo7vz|`d0J-My4!+{)8uD5sph*eg&*mF#N@W{v7Z9iis2)+ zx*C?(L=#p#$)s*#Dc-L6rS!Bg^^gWjzkbtu?^rS0PkYC$$K+1QpX@4k*tWZ{ zicG%uFk3Q75F@vPPVQFz#za_5%^?oOtx+E-d-YjvM}3U~-!2~F@pUj4rGF>8ZgvmT z)SFCh3{rHoM5LE7C^|(x<}Kq(iR8D$ii{=oM)t!8$Nb^ds_go{!`l#%9di2z=%#_Tp?6#C2Hqt0O0YreL1~J$7 z^DM@|?=Q)f{h{ugBO>A`X9J&!n)v>-c8bZSJuifs44C{v4rH8*>@@vO`v!N@ikahJ zq`j(?t!~JJ?C5d%|1nT>x!&^he?E*E!tv}TDv!P<6Rzm#P+=6l(+hnT;DT>SL;B|B zcz3%{uvw~E|Gl9|5g@aiwJ^+vA1cHz%*$^Dh6;%A3h?s_@I#@3U{QW+J`qtViT|$x znFUlr2>z$&xQ0haf@O$9x13}yYU|2#C@X@mEHMC41Dv~xQka%#p(ratrky{#U?>$1 zXb~ZU3QTO=+Z^r9QPCGc6!%ZdYDW|%q9Z`M2e>isgN>Bf2y?d0Xi<~IrBEOI?m(a| kn+v2y(rY&Rfd3tHKM{`!7mBn30dPb_a9LRuwUu!H2mbanjQ{`u diff --git a/generated/webwork/images/webwork-1147-image-1.pdf b/generated/webwork/images/webwork-1147-image-1.pdf index f6f68e7bd264fa2b62a08c9473f5a255428795bc..291fc272ff5b1efa04f9f5f61faa54bfa41140c1 100755 GIT binary patch delta 3078 zcmai#X&@601BVH@CAYbfFu7)KnKJh+A`+pEhGDZgA|pr6rlGmgG54I4quhk#roSr| zliZObXS`qD5AXYadB6X@JRhFtS1w;IUw!o=PDM|TEeL~#dm!0DXIyPNe7Zm!EhqAr z4I!z-OAnuM3(3i8V_cR6>PZjfvps53X`(v^WY_U7E}oTDh?$Mix2E`Xc=3OiE6nw^4Y;nbi~0q7N3Y!1UQ z=C4O8*GtdOejC+yRgI$U>IAaB@A&1=L@#mmk2RogiZz(%c+Ie!dQMBxxYd^9{YBWoBd4lCuJNoj)MXuDuy*{2KCOh%>9`a%134HE86YzgS(3WM5@(Cae`5 zp<@*}AsTozzZR_=`3J5AkQ8jR^w<=VS*+Ywt^_I;6d<177Z6kFMvZ5@eT&w&D@HJ2 zC)F9H&kr08dmeh-{F-(Q3CTsQDLqS0V8{0-Sm4PC^N}ITLBXm~jLM{cVEqMa}4jl7yiKhj4$H zeG?+2R;+2s(f?t?O)$V}rlhaX(}>WYX6T>Y(HJXS_*-9y#z0Lc3OD`CJ{r`0kiXn} z<%6gS2gzE!93^qERBc0kRXXZ}eH`xgwu^uiI4Vi$J54z}oI*5PlT(+Motd6Vcfl6z zE#Em;=NRJpmc}=P1@skukaz~(WogRl%q*c1mJ<=Fb1zuCGK0l zu1&FJDcN4sdva|+R+ucj(`+GwVt6>^0}MjncvApd4n97$F%IQIq}anhB=c&km`sOn z(_t{8QEvN;uS?7^#0Lp0N2|$q1~d6KB$y!OJ2pBZ1*6!2lN=kByRB(sHsH<|Aqsn< zk*G6$gY`k2#VcFR$=NzDlJ{$-q?)?m{=|p-OI8P|nqDI;0JAQH5=0*=_++8dz@_`2 z{z1Xvy7q9(&UeQpDdSzob)Gsh3Y(sHp+j*wqb z##U}v`3}ad)16NOSpLXk)0!&4V-{zW5bK+WIL$-cZbq;b%yI|AMY*mZi*y*W8}72y zi3oE87kB^+Vacop@%sm-Ki-UwQR;k>8(=ug8%3k9om?CuA!@Xt;=y66?mg|-G4P(f zJ0jNR9u^Qo?Y6S1CNfe=p*t~X;?p`o91}ExU8auLhL+SCkKN@tc|ie4t7noUI;t*C zit*c#AB6p5(C8lKSNE9vNPS15mBT?S-GoyI2maqa#)7z8mgRpo7V?pv)w5sa6P&i4 zeTgN;Q9Z#k?K!bRm(HG8WkB}QU|u`a#TGO3VE21rrme3ob3R_lc*$j=svSMzS*2#C znr41jk=!8EC~f8(6i#D+qEB?s@`i@tuC*Dh_mv%v*?5}x(b+gfX#M`i$N;G~dUfbl zK5x{IX5e$`VD(8_QFk%u1kLkm46;!P=an6VZHh5}6$i>C0g09Qu`jglDUG1!r26C( zuT+&iee?4CjFLGTOe5YPw6!RkniIWC4}4qj&77C{pc5u{RJ0pH6W9}1pI$EfU8Cyj z=ia@`?0qeul`Z`Vm+60hgfxE6!X72sgY(YCj@5txOiR;=2^oIYP22`g)6A32l*U96 zV)dllAZ2m<(1#<6`u9T}?Ap&yllJ|Y^?X*2wv3k(KKCB!b8YDKdmu10gSn@E4^QSZB1 z3gndBlUMI}&CGj>0(Pi2acgr)vBoc?D%q9}$vc$gT_6v32$4AzvMc+q)b8$}Rr^FU zuj89xMZ%=#lr*0t^%1Y6ImNmkR^%bmS}ri;dpo6`-Pt*~$<-`-J=3x1*wa`Bp^2kh zv@@6-x0I@ot#VRSaJI}wptqXP%53FcE({i`1v6#E!t65+KAqgxP3~P2kmN8Mkspx8 z^v4cw-hY%_Tn`of6PkPeUh!w!PQcQc$));94?6Kq47qN6=oaJ`hsI!F5U)jme+nRx z2b57?GcZ#T#k6go)0l^1ZZ8&ks)u_WpmiLsD9RMElU-~`UpE-%fEp{+unl?3;Pi+i zj=2$$Ft5sKprmT=T^5SOU-kvHQI{o3b0(|kT6dolpD~E9IEKC{Cil7+xK(J?xJnMqPU%bvO|OoG>+v~Re-zS&y%rq z*UYylkX#gMGAtJ(D+XwcS&jg;$qGp7h$p^(S%<(DVvT3)=oM|JC!4h{N(zawaxgRZitw(&J`eXc)InzAGzf>#op4tB^1Rx@UVTGQ+r;DAG95j0pZSI zuy156VOl_?U10>Mq+MrZYTb?wVu@hY1C=ob?)BoJOvS|o#Nl1^oll89rhHeypNSWT znfd0lvfr{dEQE?_zT)~Iw+ky6jGf(MhLumoa|oi!!gC9ooa4u>msueEt3kJmY#xb< zkou)p*C!Q5N5evVxmT}0$V{xff{E@4h%;XN8fo65R_`B9?HaHxmx<Dwsez?Uz3ZZsa?5n!u+u5oN=dUNyWsB7M0%V!S~`0M1O`IqurrojcX zbYDAhD>ct)Q18`bdj8NK<>r0|z#4(vIqvlVLM>cpa->tkR^QlIxi#7r2EvcCMc~%* zf|0}b#N!-+(iFD9;J<-um>9oGRFvkT&v@@`zci!z`wmv#fnv&-isvyY=Ny(3QKby- zXnfWkiO>G;nt&g>Y)8F4PbCscZ8x&&gxF4|R$gYETU%~2*pJW%vAkpZ*Aci$_HN! z<8FkUiji2jRM7W&8)X8*XMO72C)()3461PHL3|2OkcSrBL)}YN1MK0c0oR79gVn*> zAT2l;rVUf|1cUW-|KB2gfzN|ID>^&t+4CgiRPxJoD}FUbAx{?b^f-PA7ELy3X(JnB GwtoRMmew@@ delta 3077 zcmai$S3Dbz!iJ3=VsC2iQA&-7)!Jg0gqpFVMq_V6)NHLFs8V|rwX0T*QUrgL*qf4~ zQmaO(alVW1;``3cxqok-=i>RjjS{sIjRIsSX;oFqfIxp&Cojs-B|Fn5&jD%b)f*O$ zhOF-cT6^>)iD-4P`J5~0?(Twtr#oUxQR}wHt*E0U_Fj0n362SWQ{5&)aszluks{6h zfwbP6iqdId(F!Uz2_A&vYgzXw@NRqS+cD}6y=1CfyqDI<-p4d=>Z8a~r&IkFpOBy= zf>4ZqIsEp1Z5Mk0h;F)QVJ-`szBAqfiE-WJseKFVw4`k84EYzm?Zv648Li`6KbQ4K zJz(AD_sa`1fX`+%r-LLlKHQxpIlr4hQ^${3jmx$b>t_>ISF0uC_Sdds31X58tcN}) z6Qa-3)PmZ)_$Xby;r~nQ>W$$ihmxPsQU%cl&3p_e7a_hUBC7dt{tRisgR@vq()ph>Rf2EqSOZ*)jkeD1z)zBta&}_KDPGp<1*maqE z-B6&xfBd%GVyWr+l2mXh;4@~O!U`2>H0H;J?d?6-lh}C>HF~r^FCEr7kuhpBRo6J8 zWi1%+?rPSq^D|5>1YiAl+>RG6pDN`UpTwlk17-{xO$YFrXddJ)uLVKHs=uq9`nQc1E1Z1j{5rtM=-TH8^jWY|d=)=NVVTLf9b)`31f^GatzO4C@)R z$9BQc`GU?lnB)M&& zdzV{APzj`0R-w}@3U9VFAM6m9iaWnOO{iwoqdZ5)5KT z?>0A8x9g(`HFv~kR9H1t&1T+*Ik@kkf|g9{jCY;Nn&}jB-PZ*~Q{e=R`+-wtZ23bK z!o7sH?_TjghDmf`g%rDrMirkRtVAna`f=VM8hhj>RU`m=!NfhU&?i?tm+O&aOUG(fbhhegUTEe(6CoNSqB` zEJ$svT+k9PtS(+jtlgR=4IfZHG@_oj1JT|e>*rgZsfv_J8+gIqb9#=e){9NZ z6o#|Wv(2+RM+9b4rY6ZGUPm%9olnLp%k^A@`p8N>?au=&TKd-X0ScOwH+CzF>1_z9 z!ePWqRqY{r7O|&7Tc1vO{ndwS=G*pFdUt{({jYqLEH4G!383Mya{y{^E!j*C%|Py? zLY~49O^ASSx#VshmD(kT>=WeNsfW&e+mA$y^G5zUa7}`X%O7H3Jz>zNTe}tgd|v1{ zhz-eyn#|(ayW6DHsoCnu=ws=lbXVLo9Q7= z&kU#Bnx7wF(l)K6z}xzRNll@kQ~vnavjdk?IzpMmC(zWGeoJDO89qfNc;NZS^Q%J9 zqxHiNhvYii0{3(v`uZN$#_=az&k%j@s@a(<`b1e4xvaFr8L>=w% zj!K5Tkr>2`6+*hjz%AP>t#3(^{fDuWdFehoaur%z8?g+C8ke#7&wYzqel(}3wHF5g zp>_6#=n8&FaOt<83r{7-w$3d>*-)hpp5?+&RpuCWLcCKo$TDAXU%z@o06X?z~4zrvqq&>pvMa6&qt;B&CoIq&Hz>nRrE)`f8edkp1_b zgRB5?Q|_|zIqy~++cd_P!3?FK_*dK@w{K4VCac?h2C zOz0SN(7JFFa&EoZCBwU!PO}ETS@Wos z__?P{D8P#yFnZ`mR}ryIQaD+3AaN9n`&EtyFzuJkjer1ZSCx+V)S1z+*z$C5z^5LlGq`@1@Gnzs@I2GQG)4rt12tgVcr1&Orex=8+|* zh7oMdqvzdK;&*(&P4QI7lZ$qncViVUpL87}P-&8Es&-uBZL0#9lM$=FLTOp!Ly&%! zl)SWke7i}`kBY`J=qH&GDyFVDB~yQHp4OYj$?T6G7VUgPZsptGULT2HklJa7W%3KD zKxc`7{ixlhGy+ugIGGr)wB7XzaUs^k`gs+wZ&do*xC=Mg2Qt+V0XMkuF|*hAw-dcV zSv>vQ>c*X@iG9Op9j5rXzPBpyE4;!lI%Lap8Vf;W1T{ZM4t70y0^0|3>$2am=OT#A zbicq;bIo~&6Fx2uo&mH{)tI*K)Qv8`f2sXyTIEs};e4PX^NhT*_^a?>?@Gmn!4lCA z(95?)55k6c(aa+`=1McWvj$f$LhVTl6;Suk#~e?FtZCK<@i%QmE~_c`Z|+TAt`l(v z?3}DmaHFtdT`no_Kr-GVwgtA`tYcSff@}5Xm=NZNnoe*8mxuUF-3u&|B^ GnDXECVAr?+ diff --git a/generated/webwork/images/webwork-1147-image-2.pdf b/generated/webwork/images/webwork-1147-image-2.pdf index 143f8501508c0a1c23f0602d720a06c8da411675..f49b0de681c0c6fc349c718a33a02837a6a190bf 100755 GIT binary patch delta 5434 zcmai%Ra6@amqtr*4-^Z*i$ifq(2Kjf7S|%dJ-Ay5PK&z~OM#Z+6pCAMibHXC0-byB znptc9d71a~@}0HLclIvRsMKgE!-VlkNZ@#SdDvRI;P@?>Y4$kv@!)kla(Eq4^u-Zc z#{d?lqjo0iM%mse4C-skSc3*3+ouNYnk}^(t1QM9)hXc(7Jj~^cGQQ8E$q+16-IFK2qibIz@bYfNG)PKfNgpR@Kw=H#Q?> zxkn>Q+2bjzTk5bWuj{i&`&cJ>?U}8aR8|t@^$@XW( zaoI9if+vRpx~S;|hRr?uA8SzFt{ly{3rMTG2mJh4;8io%1uj7V;MEr%O78t6 zgK|32(ZkqEz(LYQNl$F&q^woE`t^@L8ne4a=JAxIkQx?2=80{JVEo z0_GL70~eQ^;lDoEAufjGrx*2@!Ppv+pMIj?~GULu0>DxU`ICK3z8 zl6-UJUUf-y>>{yXX26tt>dH*7 zv(kAlu->4fcU9JHhTcaGHwPnJbkpO6!3sB&XFq-}5xfBw=E=WAA~9yPiqNChF+?+a z(?8pUF_`RooWRUXQd?neFW+IR{^ocpFel#3z{|k%UuA-~h4H>)qVjZaFRXO9RewNg z7(0+vrD)l<;cP*8RTcy90A5~H;pW$9^_B6y8KyXg_xtz;V;LkzwCLj17iuC0rBi}# zkCYQU^PQ=i8hpL)eL1hFf}!V9W|zktA2qRsO=p=S9$}$CoW`<%<*HYonecD%TkR%! zDAi>gg~a8;$wM|=rB(K?Ym&0K1$qKir5PS76KDDcuC3UaIghOjq~^+~1vkBKUpaI? z)p5V|9Mf#wDXn5h#cLSI!XF=fbEtL?dhutT66=c1JGFg`Z(8NK>DazNpbv!fSFT|e zV~2`h0%2vlGjli!jgOWh_{KSKEVgV6z@k-es^ycoBxFxO*rk49mraCXHr{cYWzP7$ z3GQ{uCI|v_&gnEoMEEfE&cS=rVA~_*X---UwHMuS=)m;wdpA z)r`TAMruj@veE(K4B@lc_1xD^*BmJUSmdlJpKx^v8M@@3QH_wg$bTZzB~;x`#R3!^v(|#`!j`JiEW^Q92Eb!j%nS za9dpAASP9-mv{uyI}N9__`|qzh03{aO-L)cp?IXNX7r2-rdhu_DL*(vxKeG==$wwb zXZNVp4Lb4ewmshFmXLXbznkgU3o2hufA|h8Ls7n1NlEBadbiP5onLga6UkU25K~+2 z5uWzvQAK8ur>Vdk>Krk#1~%Pkd)}pApb9?m-NO1`!1Le85nD%C|0F-YfNI?Gvgy z9y6D$j}|pm3{tEm+xd!oL^X}_^ZYxs6|6AmbLac9`wP3NT01{9|}3R z*sy3)UAd7YmjNipn*Px+`l1mloHW+RY}ad@uiTY$#$1)3D?6fb-SnkvQvTzD8z|LAJpq@I;&UE{s@M$ay$@XZ>1>B05B zl60o};cJRSg61J7ugG|(q96LMyb#lHpZdj($}vSHsL015d1texG4tz( zWX2vxXm*se&c~rui)sMDf)iD59!$PQ$B>|UG;`in&M05)O&%Z zgm080{Nb``e#P`ui>m3wy#OFY-IW(O>W$6ug)YFlAMKzWVJbHjN#-WF{Utwt|HN}+)z9HX;Vst z*;64I7C$2m0|ntV+=R7XGGo-~S0yAgjl9LJt6iDv_O6`k_vlrARQBL(w*p}p%*#m9 zBJ$-A32%n`?Xa8R2Fx2@SXr|D>S#7}HJt5kW7=M0v1aCUq9=v3u9)8xor=c$O?$){ z+DT9A**fwkKwkU^7?rXf=B&7Ef|lcsbt9AiNVlA>>mZI{>6`t1cDC!~!E%jZiw*5; zY-P^Q^F7G>2WExE4n((fe1?L3LRpF9D}HBB%H!amskL3FAaW)UmUXHT>_NUzmIr*0 zUGTJVYYZcs@g-h4nM5Sd*2XJ;CmqdPCOw{OY04~bnG-I2mRx=zK?_JKQe?3&1IGop zW;IpMeEcq~%P|oAwzt3EfalW^nS-XC_l=nf(w6Bg&9Is(`eX)~qd8tTCeUW7RH?t| z5L#l69`GH@A3VYV+f2wJ&y=9n)8MrF4oAw3VKA?2ti%$x^PVhGhoCWwh2Sph5)4)= zgvssCtV!qPB5c`r1VR3qMw^_ryzvs-1Yv@E zU)FOY{o8(wo0%DHW3GE*QE=%?V}HXhb1!WiHg-q+A__pTta9(-S*g8keJiDL#ew>M zM(dd-!zsX!bWCA*I!2G^&Tpm*tl6miZ$UAWu>4<+>2dm(V>ou(%1BCZUC6^HzbcBi znN{D=z^CwQ|KXT5gOa177)Cc{H=0wmbkg6_9E4M|K&_$}b;o^o0r)GdH#f>~FiKVA zT8(=`mL9Vw#^s4V}DG^}a<`Y*i3f=Y4Cv~nq~RWa<>P=smW)>nmIR<*Dl ze`U=P7;Mm`__#kwv^Ti`7e39Xf*lx0=;u!sv~!3x(RRf6&VZ=hpJF;UckQVssG7o8oN8+IT`w`XBcZW26};*7J> zrK}sB*8WP}((D6m-I{_3yE&I1w&rDx#bxFjFS(HF8xj)9GLn%^t4zY#l*FUxe*dHy z-Wd5ka{GAoWPvqE6m5AC-)zg2!Q$(TSeg#I4_a#329#zQV z;$E@pzv{S(2~2=^z-0&)YKZjZ5u~_A~t{@7o)s zfb0eVp73&gyJ01h*m*S3^TwFH+wg%>^RdB|%DY5p_eeQus4C zLlp&j*m#N{cA`|la~v)aoX#qu23Kk@mLPc&iC=q*n{yl@OoNUoSt{0tyxM9e&SWKF zDE+;nQtuP*3w8D*{!-|6D^H#^4xJ37DNpbAJwJy^a|u;AK7T*LHDBqoe)&LAkQ$$4 z)xt1MNWaJIDs&w%By@&&{1$dN?w6&*0yo&hA z$wCFrRdaeGYKg}T0zGB!TMI833%j`IKq}pDHp5;|59El?>TN|Ud7zMmB;s&j`BQT{ zZJtM3>==~q-?8)Ia#N+%on+dg+d9+O1RKGR(uN{)lK!C3S&8NSkL|#L09<%J1~A9-bfkgX+ht zEBF|e`F&=Ra*tux9*M;33jx=*b6DrYwptT?vB7L6=#G>xB2{uI?VDQkh?T-v6P|p~6%HQnHKUX*bDY>_g)?{~G$!!O~gnxX3N2 z-;qErNIxXee1({9uIQbqDTxG&^yh!OXt>(dFq@KZbg z+I*;rJu|MJ+>7|r$9};(Y62Ep1u^-f5ss^+QFn@eL>J=QhprZy_P6LN&>&s8n74TYmTipzTDlJc0%k2e#)RhWpiX{ilzH#c?STdow6iQX(GMYKpvFv{FLmN zw#fIg;oT}|O6sYDMPWcjL79A6iCIO4I?@Z0;&g1ZBn8Mz^mI=`sn=04Vmy^W4H|%1 z9GE%!cM=juo{}CuIn_|6hVIh!y*rBAX}R2#W_P=8mvH$zF*W6ENfwJ=No6kRAhn5m25-eEq|3KdU(=8qU5SSJmYvf)VT6+o*;^#W9>ICsJTe z6*xw;C_?@@QGj0A2yDssBn{+SBeNd4kBEMw&X1vO%|^2QY5Eq zXqvv`%O}^V;OLg&Pw8|0nomr@zd?-xTK8sL+|(fqrv#Vzr>`GbKC7gn`li|{3vNG8sIao4 zxpZhMFZ%sR_oPw8Hop(k)}gm4-1#}fIVm9^Rw&!PxNfWGlD~O=(1}15l{shZP~S8X z5!S>bSaT8Mro}aJ=Th1F#u}Jn2}{O$rtI`;x3v7wH1x~p_Fz-T4gCDe_u#QSdk2bv zZQ(pw=G%unfz)dGSZ?k$%1+;&Ir27FDK%tN2A~@D`T69^;zo$1T19R_9W#YBk#|8v z^61+AY{yGMN8J*WQ__DK-D@5yW1cMff8%MW)19^d6(jAuVf&#rzU9>&8H`Kqb+UNM z8bJzS&_9eL;HBN+xP!oN5OTP_uwk$QX#q#TT zIwnPGiV`)8odULyP#1Lab4pQKrzvu+)mdwZQy$R}&T1!5TSc(=x zUOb)brf9s4De(d_r+djW8^7sYiBBoObq`1tD?Le1+AJO|FsZn_f~_LXZ-ZPfz;VVA zkVxvR^$EWWTij-^Jm80mN1Ga)y|rF@f}Yl!#~oTFw}l5;p~cS9Jy*8c(S4n5K9pgF z`R2KkG*wg+^emTs?cm{`%7cG$KuOZh?~pY9YfAp! z&|v=q_3U}Qfgx^D_MxJe5m+L|(!`qJ>kQ)SmzR4Q$-@D3>lbSbvi5XvF=kF_J}@6d zPyo!sD-5x+1q<=<30c}$3kX5@t@wE$!eBuOvHx!ZUd=53pC5{8Xc<4;0uVU`{N_GG ztE-v#@RwC(k0H9CW~AGk1-g)hl?sU^Y8yNla&Jf8Eg*+w82cp_?YlCXR(#z;f-+U? zOUe^c)gc|E+i3ivix f3;;F0!f&U!L$yA_+l7a)&|-mcn3&`=QDc}OItdZIL}w5!GlVH?ZM&|=`@h&di@UMtxYaak$3~UYo>wGj9ffP#e-<8pD2o@r1 z+cg|@Q{zT@kdw0V2?T|$KJ`?q6b3ovfLZ!KKDL7spy%;J8AGiDBT*$g%c|xqf5jp_ z?aYCRbZ=(uX}SZ@@$B$GLYug8LvdY=+ymG});K`>YW4D}a^YZp$0e@)rO3KuXb5r! zu#cLOSa3Gx8;_x<0^(qVFPpej+kBh}DaA1Te%RwGrufp;S1&8uqjbJQGOvydU32aZ zckKlZDC>rX4&%eg2RMM9CKO3`Yby|;(1d#}rN}Ffwdl?J^q5H1FI4dWZ7O zZmfbQuY{2#(WkZVQFYlf{s0;xMfcDyiz(T1uL@Y{+`bb|pBJJ6zSUD}^9NyJnaAdx zWk5F~SOf4vq)ZJL(kP?Ngd^J%CkmHpMuT3Bh7F`YuQpS#%AeVvrQr{u&45ZEVc5rF zQD9kpF5n>vONM6sYh`5m6zd7V+BtpInAV-4wlA<}>(JE@HpLtm5?mW=&^Tvq-1;dC7^^k=_v zpQa>s=0HT?2x8#(wWPq3ILS3ekTJUz<9hR=aZf8PF-;z8A3*zjAS@{z3C)RX&z;_XZTal>&vmQ!!Kc@cQzuXcjyv`#EXjfH)T7}3ZWMqd;)*XA^%*z9{ZTQYWtCP zmA_3WyZ(s!fK71zal3%YtQ<+&p-S*Vjq9M9 zy{NaNL&l4U?a@31xD-~JKMNTs9ON807OO(U)()ocT5jd-%D6LL9#Y=J1jlqi4^{K4 z^yh^3*F6Q@g}pGkHL4#>2qW*1fBl%wRxTFy1R`h-GEG$bmx~z3@l=VC6!ZR^oLDMP z0L{=WCYFHf^J>K}SxFY$V#*?sT3piA;iuJrpk#{L=F$D4k+kM&bwA;;Ml+KBAOwEn zka$YScI%&Ai}IGmF+s@gmdpf%RLM%DTS#3U-qn?hxmxPeyy2nISY4awJ8$ApxV`a| zwY@Xkm;=QsNF^~+PoX$Z(R~LM7|q`7@rPjVH%1Q9ggQAM&D+%%+KYjFQhdYa^epS} zhmu!J(#_%%15V{KC%qVI!7yQLm-VoAioD12;oS^@mg4(n0O?^r4qu z$&c^x_SN&PP5jDL8IG_oi<4$m;-He?`xL1RtsImDU9hc0p{(WSzBg-jqOWkW3Wm+f z*@6=|MQIZiyy6#>Y+>=LiSo0-Y#9+ciZrZj1B3kFs3rk(4(jLc%hFV#sq<>Ebh+af zC*;4^JF#p(g}uWpdAYt4P>HegcHm8PPcSL&Pa^L{gb5?{8t(p$vYrWyqg;<5y~;MA z-(Lf@tw4KqEiYHq2K|hHKRRp{$9q44(vL!fzXG>IXd2FtbX64<0)`ZgU;KJcR8A#4 zq_wan5AB=d`gQ00$|s4P=jUeTMy`4`-d|~bz~lJH6{6x;wa!Qq=uN9c`I;9=|m zk?ctC#;h^Do+sJ=Lguc?s0pp^K#(2=wLww)luHt|Hir#n1jxqoCEd$|^Cif2LcueS z(9$5*Y|AB3FsCbNdkEfcAx!ANJeyQD2ro8a>w#i0RH=&kGz)JjCCTDU8_?3jS5fh> zZMgcjFOs8FlRwut<#yL)mYFFtD0x>aK~~eUSca27Z%f=eRE{x|=6y8X&E+4>V^dVS z+{KWIS*o+mJG9xt&(#7cqZex$9#Tp(xCU)JJXD54hR*sPpQL$~jii>>g8CmF@2^`; zXTw;tGLtvh?VA{eKRMfytm&1e{qe^S>@*Z2zfbIdr_VrL`S(+)9(gUA+=0HDCxuFo zhKA|&Q0{7j2w6FmRRWJF@>ACo!(J4%pdjCe16`b_)dxd?$``oc4bAUq6n?@F%M3^& zOo7S>r{dteZPngmE(Z090Keb0AtCIquWeALuxF_@*seXe70;N!p|G*iSU}mzmGv1v> zjhO)FX)j8P8c)m3E@W8?^gL9^D3JjBWWA#@cS(t~Gvnvjh@aBgsUGeBke&n4q36Zi1?Z;*G<`uN>7h2@PS>=AbQV4H+v#UwJj|L?bVKq z9U;DQw-F-|aBAJW`F_mQ1jE;zMD^pH9a`?p4#<1O_@R*s326~qqjtO zOMD|8<(2Al9iF4kA;H~0D3Kw?;Q08@Tb2Cqhwq=Z0-vcqZD67A36~-*(o+%DFUiC# z9qXfY-h&+ET#_YcrKV;mY!~VI=(VSt6B{in7QcHrk@n8Hd{!&9+z6|uspuo;IlZlP zO7gacl<2!!wwd7{l0IA?-BGlXQ2lO6n)=u&ZQHV^vCZ-sAp3j29be$9v3jFebM9~Y z165Chg@)Yp;#lfH$!x^k8~`OrskonuL5l*^V2T2^?&7sHtzNYdo(_wMtSL}d*_^mz zHrS749)ypFh3sGgbUePyqd2{wj_KdA*$eRMrnVwvhskVE>-kJJsJn^XY733iSEZ*L4gaegbxbU04<2M&Ck z6^g)s8oZ!DdC3RAcVZ;HlS(et6!x?PrIbm=<%Oj=RFXweF`?JwG@SkRelj-m&L!SQ znVhNXQ(ZtX64RH?Bq@^EqB*yg6mwF8NIVp3Mc8QE_BnQ?!G8ITY|F&FeW^O|$norroC$XC z+|JFw3vcm1QtjU#^F;}IdUa-Uz?(9mW zMn(6*m76&hOXYT7kewJf?Is3asyRZv4-sa~y;`H!!RI7$VYzIVst)#FdSJp*l1cc# z5JxA`2>(a#i(ifWC`1-={2{XGum*@~)+b;EI=wbu<)4$4SBFVxp~)_~8%)uvf?mvl zF77s9#9b@ji!RPTE=JR|oO_LCv?>sX$VrQd5NcSbWR`f11Vi}gT`TKB1`*(4ciXGt z(^chx03K_%8EOEDq-aBlp%#R2L$h=xq7=i$1)_XGob)fI;HtW$ou(t#lKVJbwqRp_ z(I%!bEyPsMQNn__?!*hIo4l~Xx^=n}b@Q?IkgHyBE;v_3ka&KOxve6>hikFpYEl0i z!#&4x5$RM3Nzio~=UBcUCN4_vOsol5%e?Kw(FjPK^`uo{l7N@G zdMf5p+rV^^t%>y)=CEV_E{sIFCCYW%f-RFcoMCx9fBlxx8CO?r`}jlXTUsLQ>N+O{ zG$EZS9>VPu|Ach9){X9;R(o)1Q&^ipFM)$83_;Jbu%OrlG@>Y`fs5dlSyGbEk|~|_ zj%beCUst0~y1@B#<@He^yBnK`o1ILME-860xiy7PH~}%Lmqy9;hC;g+Us;dks0$n5 zD{H~HK>?${#}p)IhbSww9`7`FjHt#K%J43P1(Gv~BR zRzZ&!S8n!@8Q>}L)ojU= zhQ^b@VcZj&%ZuTizS8j_kty|(zeNeCoiTH>__d*TBa!9Inpm5LDAD-%kHBP^_KB!# zlHi^a5#>m{9pUthlieQ%W{$&-`lA~7Kc#eMP=((EJRDs@;Jk#3<5yiyG%2FPhu*h; zYd!EWd3j?56N@q-CK!zAT=;%VzA&@V;P;r00SggUD_6H6_@!lbUyMbcHENI_U+92n_R#vOy7$w|E`}ZGn#!f7eJZoc5%Wlr{os9e zs+JaD+4oCQ%}G3HS0A=*GKC z<4>ctgYv?X{t|FKXj1+9n-s9}>kFoC6ehVpNcO#-`Mb=TsFAL*y_K*ByVI8ASyFX~ zU?M$5=X{RZap=uG;#J;V6nrC$0KV3<>h7$IHsh_{CpITW`$)u4<9agp0S3^+d>u+B zn`g>2sjB3Kb+%{KShw;jgc#P_O69x^jkdc(-JSI{#j?L-fkuRaXb=7RtWJGYqgae2 zi42kq%M>Ii-7?qr3Fyw;%9yH&c6OJD9=GP&V^lgo-TLj4kX{)aL`?v*wuf z&|H8{&}^XaGPc2e`cJ<_1K0Gj>a#Df%BR!DKZWN5b{RT1J-u6Tj}JEf+FS8!p9Xz? zEZu{D2xg=Q82di&$Qj5ucW*gH35J0CF z-rsdumFxGqfc3CxU7(1+X%9s*lv~v+joee)=~MoeF&0p2?bm$f8x#JvXMu_2a8b6% z$<9>Pq&C#EyV#{#Qr;E{+1o4$QQmN^PDw9mk{MEHFN9cxatvX zE+&%+qZTz0Vn>K)wla^T>W#@)&62_pw=joK$va7h$RIJK?aCniwc-Bng^87<=Dv44 zTM=vu5+ZlR4l_$E$NpW}S~^8IuW;XFtgJ^Wy0Rd$rLE4mb$J;cv%lK`jVw2>1~Xm` zD;&H3^*nr;ma5-BBq+@@RE_x&AZ4UTp(S~dOmTsEwl@(s3}aqDU7LTjxkt-_`4q*> zpNWWwi-=oTiHJyASc{5@m|IF%iHKT3tR-wL#BF3{{=W%y85s8O*~Br_MI2;C7CZ&g zQkZ1Y(~5o*KJ}DHV=@@(U5g`~^Mum42Ef~yb-2BqVe(`K$YLg7!%6|f<{qgXJJ>)B{dKMd#cWQvfHcbu$1XRMYg>Fs h%avKgEjVf4;;oaKoe-^{138%h;shKV%9=2O{{px(S8)IU diff --git a/generated/webwork/images/webwork-1148-image-1.pdf b/generated/webwork/images/webwork-1148-image-1.pdf index 42244d278c252048a5c205e75e248417fc7de564..d2d4f471314d1c4707f03d901dc0ea556e01d311 100755 GIT binary patch delta 2572 zcmV+n3iI{%HupBL7YKhdK0XR_baG{3Z3=kWT~NzzgD?=h`z!XCs;V#~ih4ju6)6&G z(?E^-OU1T~Iq>(j4b7nkue`IfqnS<0`cy0FAo;SIH|U_3N?4rqY&k-QgH$BV zKuGSkE>=D=Lz42$%m*70lDGEt{r3G>o-fZ&?X^meAh2Hv*TsJzWLsaKnq_i0)~eT# z=L8_Rc)iKl1$Hk&clZ#nPb<($HNdVdn}{^CF+Vt>azIH@M7eZz#I?Z5Fpid~0ZC42 z0eMvw1eN$V%hrvq->CaQ-Dkc~>i-@%1mL&=hbZw!4r(FwJv5)K^?u@MV$xtrD!gCM zXcM;L`vs9&asQK14HJ_l2{!>UlU@lJe-qAf)FX7}v&VkoML4JqF~T2Qm>FHEP;d=c zW9&za;dt7ZiW~TYjgL%}_~DjqqA{>(Adql~D3N7xMxmlz8VVQL5&byi`_Pv^+Pu%* z>+b!}=kvPH=R*=A5kZoqbaUowZ|1!7d|Jkq)#3cI7Q6o&oUe@Fml6CEB)?q1e=|Zp zTC|)Xh`(iFPCOFkttDYT6N2!kC_+S%6jxTIP`ntgP$Z_AN^EAmR>!fi>XmFla&l4} zo2XDEv+Io-vtF$#W;0ZrPGeMYp-{wbHL3L)&c?>3=s2!q&8k&at2JI_ER8prwW%xP zSgW4Xv0F5y8nZ=Hz^0puIrc4;e^JARmE*%mGZ{-h;xuM9!&IO#7ZU_2PzYWU5QL0~ zA|54ZVkr?#NCX5)h=@w!GWjZ5D@YcM@Xm+%fF&>=86dkRQS+pbFGr&^%sGY)-{(!N zIkpXO6ewRPH)%JxV&PrbQlrSHtvr6NDGDuqBB7EIKE!0a2o%T3JXApO}??6w_UFecN9n?wUoQ>`BaGP7` zuh~C?j9lP`~%3B+22a4Qp{EUW}vg(ipn~TwtbnxwsW=j3^R2t$VWVW&8H5g`Gg*bnydq z6dmMqL++#3g*-FAjP^mDh1m_&)fMG?-#c>{8k@Lo3bk6g5`Vx}X}U-Qcjvaj(P%b}v&0?t`^;jh>G4-I(Scr81yR*w+~0tqdFL^I7$qS3Bth5?P_~VSIVIYmt(m`)VCxG0c zRXSi1kxC4U{9kn324QjJ_M3M=!rS$9x$WP-j%U+x&FK0J?fBqseZ3r5N+dKGeR*0S z(c-GCpZx$`r5i)CEFdI*oYGW3xF=cdrMbh~Ja=*uoYE~>U5nb!)T6Zz=s*vI26^tc zV|YAaB=nXGa9p<#WON!i$qv)V1JLr{I*T#w-iTlB=5SU_;gqH-1KO^nZ6zK%P9@7ep_SbEdTuhe?OL%lTi&5v)>Lb2!Aj+K0XR_baG{3Z3=kW zja1ui+b|4$-&g2u35HY`OTG+6fLxX>XuC_2u0uUPxHjV&NG!uvx@2EJN>019*oGNM zB#K8Ad3dCVFd|795sz>o22eG-sRqQ@tk z@qpkU5PyD7Cx3dQn`ND5fUy}_(3!3buFn|mceX$@PKHDJxzX7YOiVvdXLJR7L+9JH zfJrxuyJ4OThp#X|U99V-xI=Ji3^Wy;!b6$C;{EstSU{_HO$IHA+(D~D4G1pmUjvn~ z1={rqCWK}iht@o9ege1X-&rudDDyV<9EM&&UZ!KckAE53Sm~KWnuI*_yyN~!1F6wX zk9>m_d?R{i>YYKZ5X00NkKQuQdEZu??=on5`~0D-^YwO5zV{F%Sisr$e!q)T>S~%A z=uO|+08H`7?pA!j{MU@V?srGJY#)nq`A<@lZghd}Sb5m5GZyyS^a z+j+X&-5G1x6;4rU*W9I{o#vCau zvr$!SfH1l=u>DE`pUyUE(GuUq99tm_Jf94NVyORskwCG;`DWz?@LF`zx7nktAX>~h#b9uGU+qB|r zuz!GJw@DidUmjX4VP37ZyCFr{n&!|O<@yf#^iz{VQ&vT9o@4!`IrQ50XR9tx)i?09s-wDcHK9U;{cpBm0Z zki2NqV|SM3umN{9D&4|4d;Ao8_6?7fIXl|_%HBykgRY%OC*aEv{h{b*I{%N}3?3;T ze4y_tyb;(2da~Bhm8Y13wr^F?y+bGWM;){H1&dd+bdy9HEJ9&3Wn^J8Ic73tWHLB7 zFf?R1W;io4HDxzsF)=Y=H)B3LVKZf9VKO;pGG$~kI5;pgWH@FxGcq-0H)Js}F=01j zld>9AN=XrbFaSf*q%5UnyY+wp93cS(=tKh!kWk3v^5$m(pw@tlPP8Hjia|^w6Dyzs i%k63yUsunx|HpiKj$9sBWVu|4+W~y1b_ZpXP#Y()Dy=*uS7$_n9y+C^!_iDmAhg&kP0|dc z5`ODq6$7^eS4m$Dz`lfjlPw z$!*aay_;e8B~6QO0sD3mozgw*9;}H-GgrNNc{>eCg&DE@iZ}MFr_rUu4j!6 zTk-t@fPZoD3T2ZQ2}l7klVAxLf76`hs7L6`XOI2Fi*QgKVuU|-hgJvY`GbOMz#3ye zVhqRA##G!ulZ}r|lKA144bd3bG!RHQM3l&~IHOR}E)9i??1+9G@_p#bA8p>}?sfP6 z=ks~p=kp;6k%%BkQnn>imAN(RMODVOwUPX)9((^ATBwZTR}lPDB)?L?e=|ZpMzoS3 zh`;3#PCgdl?PU=@7l!bsXhKAi6jxTIRKA>`R3@n`B@U}mZ{XNC?P@kLB_%nYO;RdT z*o|hL)u`1Jvl$xBpfhW@a42H8TeL0AL_m;)h^Qp4kgt=qf)v3h?|zgI+JXzxAlW^ITBd}2IU1v3-Zf(SK5uf}@g0Dp zLH$y>MZd`t2XDh0-sk&Y>)Uv}>>~8TUEjbg)iLIsf)NUHc!V}Wf4bi`=Ng9lFbJdY zW5ZDMKey^n!f6=Y9Z)vzfDN!2wnCzFJ7s%1z1##;cshjm&qL<|By#>NL{}H&bgSo3 z3G<#dt27=clcW_c!Pl+SV!fo$?Tlv(CULj-F7)=^gKpWgA%i5|-Poasw7HG`n)@?E zUYtYRA3}5n<i7QCWCc94X+4(Wr8pWdk9E4h16GPxh}+QSs3Ni3zCYVp_Tm2Zum>odE`F$vVnTdw z*nRAVkY^TF&;h8kG5espx}qG9Eox&;=*-%%xGRm}OyUV|e+>)|*T6(roRg%v8#)ws z9)wm&HtBKGc80cf0=Ho>Z<4g3dzSOeGLb^2Nj;M zUFE@oy=v#q@Je%hMU54;9;gOtAGjO}+$V8>)6dj__i$}pqp$PA1bhXp>W|Al(buS< z01B&jTJp-wf7+rfcoW|0=0+%Q^+#|BsNnHqr#gz8?2h*<@)K_{AIf9GxFRFI(%P<81yR*caB{9;@kFt3*FyA0Q%f5hFnjT5nPnH zW%r3=Z4I5@JwLB5*Hp%9I4 z&{<&W*c;?}JwHGvHGAp$O=bYvcYSJTE`(foCtP`U{Vuh^kr)0ZTQjfnAxnK-vlwd{ zBq8bel3=Y&Htv@#Wu&6yfb0=8|HS_cIn1_`R1G--F))*l4HIA!s3b4uH91dsB z95N=1ND{>)AY2H-IpI-21Q9YMoCF$|0usdNjffPSD57vC7|~LbkVFzansCNLf`dT( z`F}Z0%*M3KCSL)@W@JHUrna~~W4Pbj0?{}bkLgWoR!gucy_wGFE$l6wZ}Sq8ewg&b zJR6VSVSuJwH*NWV;M`hhYdVFeY6XkWlOtdOo!PZ3=t%4jx;-?2;L`oIP+M1^TTd_{ zG`lo%=5g~2xI_QWf$5)B(WQaM&?_jae1Bs0DMK48J(Eb2kY_<~+@G|NThk86H@Jmw z#O!Rdv&a=lU zDX!ga5sa9A&)EB6f27a$xhS81lG}V^N_3|xzEfbAJg{`Dlrb0(#M{oF;TEsl&MWh@&w_3aiycU2cFd zy0oypBcZ0VOU%){f!0t{1GRq} zx<-}Kb@c?atM)^`>pI`UX=!hYcp)*FPo3=?>=!40+>CFT9|EKbAL8kK)KuG zt%ENQ9hNY!*T&zFvg&Mm7+e(Q0fzKzTR>aYG~;!8v>W6npiJk3E%n+W(WilXixEJCRPnmm&U_qF?F!H+nOArhIfsKh=06 zunqKNt*5KVF$Lq^s<3~DPVSF7X7Lku#j|3QL>eqYH#0RjV>UElH8VIhI5}i8F=J+A zW;A0mIWaV1Gd4FiK0G%wH8^87G+{L}I5jvqWHK>hW@KhGV=_50G-5M0H#U>98dOTT z4S+BJLs2^%A>ry$(6bT?umN*211m5>z;g5SEd!`EAfpql2!dh|lgPvhsK9c&8phYv cGwuH|pPnO^#}!#FSK@X6AEpZjT9Z&4Cv>LW761SM diff --git a/generated/webwork/images/webwork-1148-image-2.pdf b/generated/webwork/images/webwork-1148-image-2.pdf index 681009671e44fee6e7e0841e1b6400c744d11203..8af5cb82360a7b8864a5fadfe987b1f87da3b0e3 100755 GIT binary patch delta 2507 zcmV;+2{iVLH;gy1JqUk6Zi6roy!#b`1 zhUUr1=N4yRfT8nTQ4(7S)vo1IM>A{ zsV|KAzXwhMIB&oyO8k+7T1fp0%@=FEpLm*>G?|)GeSOEw2B~zzhz-gE(`Pa@-UwXLHJV?AtFhNE3Z~4 zUW`{L64OnkHnU!<H8_ zW9bK+#>{4#3N_{uf*=J7!Ak;ykP%VDV+2jCAfgG0fFKDGQAJ!KUnT1V$%0Yd`6wT- z1QsF#WcL(mnG*69XpDw=$B5zk{K*Z+cL0t8>O|S*tf&}|^ z%JNi3g#oD0bO`dF2haIPWdB)+t}e*vR@b3Y<~??em z;%@g{=kGsVlr zZD>nGvDj(dpJOloaQ{Zw0~F2>KU7E2K|U|!K6XRMe=`d!X&=;En0-)FQ(1w>7S%BZ zbY^`>+>y#~2JwWu7KVpwVIm~XNm8AS9r8O5f~zH)_y~G20w0ltZ_z$Ci!uU;o@nxs z^FidA79y6vOgC3|RrUadWaE>b#sltya#zT%>R{nsrF~~;rMaHsrph`O)Bv>)95y-b zli0xSe`V^yeYmc^$F2t_qJP5I?Ub#XSl0b9Gd5z1Zj5gY<4 zaQxV*j*@1p?fuGvgj)(EL9z!kXgQ*i{>72<9C&w+Mo&Ey<(~-rGTrqVd2R+BzAr=! z`jv(|N3MPGZTrB5?(e_{ea;qLo~v394$9cFfBVF-w#LqH^3EqGs ze93!xNnTh~Wu?_X!Y^k;4FhSEmJWI|IsxPctbd_!l$+Ccua7t7C;2z{< zC(R9B=ebWO!71I6mupe`nR>ML4jt%$&>+wKb_|avjD+5D0gmeyf{achC)s`)c>r4e zTW2w*-5T-B-5kz}DV)-j<+GwSDnjkN;7v?oaMhi za&(rDv*Hd>2!CyF+cpsX-d}OwYNK%YB2q6!U?8^V)o79ycDi&^^+B=DiU3;_Bs#PA z>vyD_)Csm>0&Mbl507`xJ!AqL6vPPxAfyB#K*Rw^kSYQJK}bsh1POU#P*Q*fO*6?M zR3StVEAoUOxQKuTN%+f4HZ@z*u9|#}6r016%}s4-eSgkrzqci!X)+$OudP|H(5CF` zY|ifRVA*1qmpJW*Nk1&I@%SADXv$5~mJbBZt;M!xGkmJnxcqf;0xYpJ`*w{TB>teg zV}k@P-Cv8fbp^WhDJ4W^m&VRKZGHiFC>h==bve?RPx^w~Za<@2BBHs6|(+^G(zU#B03Gb?+S z$~e`WXMs+&4l^awNXks}R4J8(VJf3Ii)26OG=HWNW95(#IH|-*mV{}L$SjFzP?_L5 zb*q9WK3?`hr|Udh?H{Z&+=^g#=38B@O@X8rpZ9%L*vHNZ&1<7-J3D>MTTn5(&QEi# z*d*`Jhnw7fncPl3OmCL4k3Yk1-Bt}z?}X`P_3HeeeqyKjgjM-yA&did)vcW-!bCBL zaewoXAY(xkR_m*}+#+zcvbg&VqL9tEdD%heV@{k9MnMQ8sX6IC;t*+`dEdO+B3+AK z`YwCY_kjI{tzo7{X8$y7Ln>vP>Iqp_9fp3_ZN9_v(%uyDLgI8WbGCEvpPYRA<97D@ z^@r=5rP<}R;G-p$`)%Gj`0ChE35$AT{C^E8tIoE^!Bt@%aLB&21-4aP4i*K~U)kf3 zPHLJh?i?MQZs;3|g}=FMb`LVL|4I3YWN%_>3UhQ} Va&&ldWo8OCH#Rj2B_%~qMhcYd!qNZ$ delta 2475 zcmV;c2~_rsH;Xs0JqUlnii0o^z2__D*kxIwwv;`HP;3REZWpy5pvfqKH3=~u{QFI! z+ru6*WZt}ayf-v2ucZ_g=(kapVP0f%yN_LI>T6O%m& zH%Ku)J_>Vma%Ev{3V7Nbl7CQ>bsWc^XB*oS&T`Zvbmp^1`-K(j-^3kG|1VQ{Q3v=?JFmEmk^T`l|KSdEDlBBruYK7vtc!eS{%~WbL>$N(L zja9E<6OxmY;@CumBAMM})R^^ZRSBD+;&d9LiVKBecB@IP*KjsAHbuvArR&$OwOW6z z@hW3kyveLhT@%Mz^_-5))Rbw=7EK|WZYtr}H&sRr8&-}FBh6$i{gBg`*$h*m##}-W zq(C8fUO*5sB8qsJpo!%~G$9cXBq1WIh)d*4WSt;cFv>e0ik$X1MrROryKh5p?;U^WmOc^G zN#dN|4tcoEP4w5?pF#5c9OC{EqEo1VZu0d6n+{(ZzBmNGQE182*Rc#sQxu80r}rLi zKH=*e%mF*ocIEQ%vq1gcpOPep$2V`lk$5SN!pmdq4sd~);^pEtv?Zcg?6mI5v6p|e zX9Mg63a5+jtE1>3pBHi;yC#3+nS~X!AL=d4Zm6lLtiWT7>KFq$wJs#?NM$&Kc*0!^ z!^5>O5fbMlsZLLa{MNnTD#=DZf?kZkha}-!w9n0=^Z=qKynb>%h+NY`#PS#E#_F!h z9-xqHeA4IH>)t1Kh3u;K74A~nw}n=k>nrwF*14btsNLYO$#I{=2KIj@rXJh}>gv6| z&T|v+6|^cpF8@SZtAs)*s@Z1BFE^@-v*9&(y_*}M+%*SbKTv_AM^1EVj+E!Xdpk9H>b@xdXyBLWuFuFbGw8rwA!5+4G~7CL<%@6I2hMeW z2Y%>tw&?O))p~GH#+H8_$BwjlI={&~n}Fqb6^_Bh_!x?MWrpd8!N%S%yH20%fjbn6 zj-AFrCJhRdTrP!Ze2vZqL&vTF*X#NLI;q(USFbYz(7ydsU2_rS!MmZ#w`#X5b+-J_ zH(8qbG#@fG)HjQ<$|DIzjxPz+Nu}dW(xptKD9JB<0L?%0KO+PB%)FD)788?v4K{yK zYlAQl0N{6jMIJXsF|lep5TR2#gtgsj`vcPCbb*?Ls1N=3)135S0fjHQ%O$z6sLD#K zfrQ`Ah#CgcC@mfIW^@9`4O*oG77?k$u*m;K$88W62X2qN0}}q6K2OK}+t=Z2I<6UA zouM7?+^w#cJxhs%2BR+z3nW@xmGyt4AE2vrV@Q?-goIO?>Ie5AFFR>&@H)?(oCK$I zOJ1%;?Pu!I+BDTL>~bjhtlrY2*QD`EQ-Yn09N#FL!e| zE2eNtQj>8puXxcrYKtmRAE?qRiM=)ym3T#4REok~Qi?N?SUtrG z8>~JR@`_OnKOrsfr5Z2h5#7nz!5zt}!-+C7GMr;`*F_;(l*)MB`*UCSO~Vue43b*V&w`^}&+G zE-Ur4A13{s|FqSA znKkVIe(1IS0cLNTy+y7NL*)fi@0k>0=qfJuxo!uC;<0Lq&F%m{2Z-V{Wc257*vF|% zJUto9a=Mt~w0;t}oe6pO^Nkh-VU~i=uiKBeU*_+wZp9)#YGOVEvUn4)LU>+3KC6 zgW?T+MKSX?m(1>fN0i*ThvB!BpGc0VdP#CmtBTKh;xJc_rPl*?JU(Mu1Qe-`k!pcY z4P!h^U$yC}J1cwKYJW9q)9Fk4^hx^S8=orkwEr)AXYCC8b|Re`UxxTCKZ<^#^FQd# z=!x>tC-R}j8-Z;gXKOuOMTRLD_g01dJ9Kt`*ND9|$xGB_%~qMhZuEvA6&L diff --git a/generated/webwork/images/webwork-1169-image-1.pdf b/generated/webwork/images/webwork-1169-image-1.pdf index 9e3e731a236833a8dd50d40f471f36feb1ff589d..5ed22345b2ee49b9fe2dd7b51639694146fbf486 100755 GIT binary patch delta 2341 zcmV+=3EK9+D90$U3I=~YZ^R%B-T4(Buc|5%hN?ppsh1N{hU*u(p(6n&MQQ*MGUVU) z;Lr^nY{}2>*?y*7f9kD?KtF<00wYYu>VT`;f^y_0faE)Py!>e{9J|yft zXbi^AaJY0LBVzOZ7ZYLO{!$??oMqXT=(W5WJhVav3P%zcRs0d`~hq~Z5NZl1rwA1222JGc5`zI46{)O zCjkT*0%e+$!37hOj|e7zG%!9s3UhRFWnpa!c-mc1!D@ss5WVLs=D4L4qg_f5A`};k zlx@qpTj((+qZ?S0Fvf#_-xy^NJtUc#y!Uu-lCnP3MtexU$;t*jblNDw(M}{0TI{u9 zX$DG5zjg7JgK#Vq9e-9*5d= z7V?|{B&VP^dYWPPrfiE3%j}C3k~Tf;E@cxZ&E(t-2Q(fyD~c$Wu8yQtI65Jbgz1sx zoEMN+Rl(4xe}CD^Xghfo#Pa7XAUO|DsP}8VkGu`BC@e|uLFGggJj8Xxbc+#f@V+r; z-NK6J51cJ;7Yb#Q!vztO!wF0Wi41dh^YqrU77EJ&1ddX-`IEr~6O;H0CIU1ulOPNn ze{DB0CXKr`9PFC9oAin>T9OqA;=#YKpxHwYz|8R8%X^cQ?8LP4kTlMS7CrPzONW!0 zoj~YtP@1F}NG1H<#aoQrlBBG;{o*4+a&A5^Kh1Ml*DR$kw`-NYg2aA*H1*BX36pg^=|TRm{$<9^yh` zx?n_UeC*5`8(8uD0aTrDt&#NUXj-f@y}zzuy;H}v=Rdm4!ns~#}5V<9R9 z*3SlGn3(T~lkDpDD$8GBQ4aYSTodqZe1>Fj9K0}$O|_3YaThqc-Om1=`%S%ekHur| zHrK~XPxLzP6_){4-G3EZlETZ$_2~A*^T8m@U@!DbH94C2S$9$e55RzTXkzaB%9Di3 z$Ry~cK)WoSR@G^HCEUUCKN#NQ%MhK6L^^JF)~c#3I?7lF?NIM)Tb_UrjE@CI10k5k z=FNjHrqNMb6Vt5Yd$9K{!lh#3W|mh3gCWkq9*C2{gLs;-U>HVpXjqgsAWEN4`;s2^ zi$vVkvHvE47wwb51rw7V5J7)4K0XR_baG{3Z3=kWt&`hsn?Mji-}4po7F(6Q7g#P) zWGTiawNkgP?WTzwKM>duBpVPcsq^=p1;;i{D^*nmVt00C=W=Fnfr1IH0|qdI0mIkj zfRC_%fnlR;fJr$p3l|5qU~&oqOoIYvAcPGT*oFYk!Lbamfl(9hqd$LUn{ataylkBn z^E?P8MxrnyewHdc_qCf^X6T~caJ-J2!GKJQESyWVB-4>k7ILe|coW3ZYnEZNj8}uf z8G6d&r@V+C(Kt{_7AYCYmn4+4>*3B|Cd;fULRo@a2U@)55{+a1DP*d2-TK_a%up*`$=7!rhUEtK}TWojVD`cRuq62qwA@3lo zR9-1G;bxzTK65W9W4x;?j;c@=oo)1zioSx1?wzE_0r+i>TD?tt;Mh2W1D|Xm#+5J>JE{4^{BL< zPZ(Kb-_s<_A_N)U8X3QhCoo0hD|W_S<1l%Ep=R55Fxwc$4|PQ@bC@ zGAxoD!RRrno8Ah4&|~iTCz-_03M|JUS7oR>Mv1m)c4Mq939r(4EuqH{t4ZccZf?P} zh`$cv5^P#K9qNBR^>A`~t`Nw_5n$Wys@D0*TH??)BdXKK@#~OHl2&E6Huw^3WUG#o zjQ{0*MzoWkR@4R1?ckGQpJ!K{A)nklIxlL-?nLN+osH)djFFf(OgH#9S1 zH)J+2W;SDCH#T80GhsP7GCn*uGB!76Vq`EgWnnioGh#PnHZW#3V_`QoVKFmdIXN0%YQ!)M-Tf6Buax4%L+KC#B_YMbwWUrd9UWWA85-L-&XB+F z*%xl;Ac3CVlk`Lz-g0Xq(9htMzzCDEI^gQIpb$o!jU^=mZB#s}oYf*dq0L$Pcex-? z{@o96?PGHmKQB*1-<8jxak?@3-6OE0=B?P*hzqxqgQ_9`^y_0faE)Py)2ntPJ|yfT zXbi^Aa2gts5wUsyiwoKa1*z*iuDqS3oW|8lg+baGNmUdzR9#mSwEAD~P>hqh$Z#+j zD_Q0qXi_D6cXw(F;kq~fAUjISjm7($yz@zw;}3e>Z6%Yz1rwA1222L^3=au)ce7Ck zCjkU1PGzW*!37hOj|e7zGdVs!3UhRFWnpa!c-mc1!D_@H5WV{qbKFvjCS6JoAyBde zLfLKEWT8D8ouopeV$30bUo~M5J*YF|dyn^qR?Vr=$^-p$MzrW*P+B^i&Ef3Att>C6^nDhLs~~l7mX;5_l;R& zBP*Rhz+rDW3T1ASv%T7>NA*|k|`j}y~_ zfjlPw$yZREy<6e*E=`Y5+w8+hUUW*2aJrXGoHUEIzc``xKuJ+Vxpem=jKtY;fsX4D zNls}2c~unzwfy&&?Tmg9KirZ0^(`PN4Ns^qdwq$#1F+|^#xhHpM!zTa2h4~mFH2#TVT za<%$~y6Uu2r7AB}q?b}-GVu55A8!VK6ycAde2##}h5TU=zHvo=^iM3vK@j9ucA7u; zr}^VRnx|3_I1eHs6vdj|pirEZD-=aai`7XR>h%nftH~n@OG=9Ki6Vufgt%a$XoH3{ z6DpF?Qznv0l}6&aMPr~CCy`sKXBg|bf&z!bAtz0CxrMGjpO;TK42+(rr0f*kOlgTr z7BfR!B~28u`(3_&8%m4G+Qd*ap|WTx+Ki->T2KV3MrshD06`HE@(OkHy%PR&CmY$~ z$gK#AKHla&Z40@t+*@hz(Sz-E7=(u~-EQf2&|t@0;g%V8Zer=pwNV3J0eW=@L$BQo znE^XdnUvC=5cp-o5b*m);=k9Talj^@c`Ibl*jlDb5-yd0p1cEHz(8{(I>=5#44Y{h zcAIFMvG%nifv>^*dLVdTH?Lg*Z08*(jfT9?EAv8MPoJ-WxNJ89rghE5Wxxj6Z~~L? z?Maed-dtw+Gc3v>AHyp`zLn3G4vc~yhOmjY5jXAuSC`k_*L|LAm z#y~|4@}ZWsHUZkSTMz&J%@)KK?~=>q(x$t!GdnNOOnsmr!1urb<}hIR`W*0)3K%#p z$_@z1fhk-()Pmp?BnXEBXCS2u3S375=in&^Twqk-ee}m%e*x#`WRR}Xe42%k#z+(< zWRz|Uo{y}XN@nPy+Hky%n_iFH=4mw5W=?L0BNFJXA>(xz>%ng6@0RhR*E>N^S^Sda z@iQ8S#^`)QhWagu^yIR?H<;)mE%Qhh;8%fy_e`U4Y(Kf)7+beJA7ExE7IV*v`^yL8 z0`;XeCRfR_f0#R!4Y}6KBEwtj)W%Q}t-uX$=%`h~&H9WHpsAS6Cc4UXD zQ%X34%ONYk!ksCtYA+V3DUbw z5~WKGj8|^&lcjkr?08hIOY*`DUc($H`4Jwr{jkuLv+;d!cRBnvxSl?GmHVMCqCCkE zj2)w@>5cH8ddxlhu9NuHfbtx2QAD<5lvs=1Zj7}h(ZwcSY3MM-D#+C6^$dg}qg5Ce z;L^%zf8X}0gOl6wr9?iS1lRQ!m2Q-*G!E@vJVK8B3Fg`qGWnwisWj8W1I5aabHZ?RiV_`8gVrDsHWM*M7VP!Crd=o`oGCn>E zb98cLVQmU{+9kW?C^6en<3hjImnu~R1B=E*x50IfM>c+m?dD3U0m z5;LFzm))wPzpVnp`ycVNIWS%SERW5|dRS#Kiizt6d>NAn29qEZ9|$oDB_%~qMhfa{ BSd{<( diff --git a/generated/webwork/images/webwork-1169-image-1.png b/generated/webwork/images/webwork-1169-image-1.png index ac1377250a4fe7810d711c09cc8c5138dc4ad2f3..44bf404d17a84d0b632890ee778c28cf1eb2434a 100644 GIT binary patch delta 21 ccmX?8e5QDV0uP5U2M=qW^SaW_8azvM0aX?U8vpy<5pHQNuR8^^if&-DxKGdPJ4yG->9Z=dICwDlR zK?&Y;T9h5?%ij_b%NYdXS_rrIsFMUVUQ_;!;@9dOi6IMZ&}7n0;3x!KrhQ&W-ES@^2zG~{zZO5LNh(yD+a zd%g4PTa&=jkDO7s1=Fp|D2`^buit+e&5)ci{W+j+gE8;kr!%2)DrotM5meXMR>X?# z3+cW04}h0~QE>A+T67k3Q)JjnVt=x8##g+%psRqv4~(`&{h7az)xKZEIK!*5W_ymv zR)P)K;@-`zoK_zMHPYAZKb~=cR1Yb!80z4(-{YSLsRT*LjCYXTxHF&~TGV>la%!F- ziSQU83p8W6i869I8$k&Ap%(Jk93RSVXcNdwr#u&yN#h>ke^y)dGlg?j@CA^Mu@HzNvY z2;7{E6>n1B42szbFiql$se%4ed#5_O}g{)@=J+4#A5QcL1YX{^zg0xbCAByJ{QPHP7cP)Col z`BANS<8tWitupcIpp~+vt(d+lrrUA(WTY<=<9Tv<-hh(F$V&p43P+zLMsr_|Ipm%F z6EWZRHw^9g)Gzmyu5fPUTF#aqs=WMbc7CI1cV1bt?%>xBQ43YPpnhmF3oM1qQjAO& z_5jqdyRnAGhfBD_LND?ue@3JA;p0gXO2wwFph{CsK@R;sY%ZVdaKftJ8lQ`68R8$^ z1isgEB@ftSFZ5NiDpw~(3SOPXyWH=1`O)lf&0e=Pcf;T(zGDy-lx=Rsj)JlXc+;&7 zusaU!nP=N!yPB^ymVULMi*S5&~2&YiG76Z z{#o-!>6mlGP@_J=?3D`HLY$mu;XeB0M`|7tT+IDThE}X(h>#n`suM4?f}t?KTZ!DE z62qf;48NJ2bQt|LBsP5W$gc7iNk!sArh!|G&II+!HxEP`o+&SW7taqYdNMYBl)~y{ zcj#)nM`b)Cei5MEUsVsWW^ zm$Es*KJCwHhV>&Wn5411r;A#`_V%y{VbwFbnB{3_XlD&j>|^H8Bxh`TIs&@Ew1Gwf z8!v~7XDEL?nhE)yXmi4=YU8nP1*A1};YpujQniu2=ORb79>) zoeocgu}cO37v;piuJ;=qxu1Ci?)HLh=I6>^_v_<4X2}|b-bS;Kq*Eh=R>>f z?wu?YcP+GXYo2JmT&2n?8?SrLSLP??6Gtxt8@)dMJy{K6Sr{BNCYE)|=u}RnR@}o{ z>P#-_9MJ10{o#R09rn@EJOE3W{1(ens}NB!7#UENEva2~u&5;jY2B=;y!YTK`CGtI`n7JOa`tgu0h3PRpa#PaFVmBPt9sFJDD zt*TP??;IB?%^bjVd=qVcAJU(B=zNP>ThJ#>~2R^7O$iF7vqH))Lik=i%aLb zXKO-wnae<^ZcK<>oA@>^RiruYPBgT=CyNKhgl$~hHhufPCcN(eE`H4v6>PyKNLFJOTqQ zf0yLwxg9DVp{IwAa?cts&wTd|Kdq$nIVe&(gbHL&1@M2SK+EdLOL+w(q2(<432O#5 zsYISvkrRO$&vu@FW20IGFhnksZZMK@j>J=dLCbRF0OPjGCY6PIZsitcadrynO~saqP9F73zPMyTn&Eq{ zt-jQSA6;xFrjD~@iqyo32c`&7<0x(J?GtR#L*%?}zwOx%`Qq4wb(u1{?Y6St4pGcw zf-Ua}m5VQe>rH|0-jJN3+#kdYE?jSI&2OTkD13F@BW_iCxV45)XhW4Sh7&|F?4-TR zd!MTPEqk>rl#93AcUbC6S6JAy5CVZTlXy2T8ZnRR8%Pw!$~a^g2gO zi8AwmQe_H<&hR*@{!YUgXwzN!B7-$$@Lk$T zl64A77rYWr9<3TyjaG7EQ(W?fp|f>lUfK%^O8qn{E{WC|P@-7Zh@-huG`-S1kF|Vu zVev=}S(4NT8C3&?;lb${Te;J2sZb5B$x|XIj=Jk0o^M4bF42z=vvPg1HI9jll!~gsU0m2^e85D#V?ff`p)eyU68GF1b zZ_Gn)&9P9_K}TY$4?svLj#4F?K_r>kOk?I&CZ#wX0V@^^(nT>_)zXWq{6u1ZCxqn} Z_W@vbmlA=i0doN*C8-4jv`nM(HDj-k#-33nB~?|kTC1ftv1-?zF>AMI zmD18!{myfq@8bK;&AI>Gych5LJikTBImx;(O0<-^Ixr+O*u&Ky7`g0h(d*L(rRlsB z3;oX8mj-f6p=T9gQ*RUtQE4u76U%q4&)nubIj?q}?W1Gd+gCF*Z`_VXnErzVno(h( z?A2rv2wI?P&xpOH%#_}MucutN(m-c6XI9Ssa9ynkdjm zKY0Bmx?Y^&5kjBRRGVaPC~SkVkD|lwkRrxlF?xy>Q}`faUq`S1;97FRi{HB3|FrcY zJSUi)Me;UE1;tF&#yn7EgvV01);5A9~V!e1U0qtVQP z%L4qVVeU34LBwAi=z(zmUt|w#ArmDw8el?6n*j(sKb>os00@!(Bg^Mehs|bU!cqeb z=ordtoEO$WQ|^s%FA`a?u0SRH2mZ|HP$-zOoPs; z)6C17X3TciC0F$y)BGC~PHm6=z)0aZxf3wyzeP=l-C>GUZ z^k!$R=%?B}ZK9>QCC#pW?HhN#^a=rM zayeMH88?GJ^KU@2T>R{g(^>e;T1cz6+1+Q`R1|j{h8hXMkF$u%T&a@o{^AN?fyIOk zwMCoXuyxiS>9dE1xQxo9xpasJKIlD=7u~0q5f?E}yGMRTXR# zJ&_IxTan6s358FAu6eTVNcl=9>fviHr;px*HVSsf#|+!wM%7I+t}PfYT@iXgP@dWv za*EAh>U)*hyDN8lYwn31zH6uM>^v`aym@IXPTkCP1UQ~E+3v>O1W&Wzpy>OUm7gwy zaKd)*OHQ^f>3`5@9=y*f*OvB9efsfOLs606Rr9-$2O#^!g#to>B*{gCq`WkVykI|5 zz=s!qa^75v+twv8P?65q;%BQZd}O#6-9(66nyTL^b#ltH=Ie`$hr>I}P14E3qsV4S zhD^S{v2Y+dDCOIa{3dLjL0Venp6oWwELHVERbyB^*S7HJyA!r64A9f5+mIEZNGUOw zows8Rx&?KgiQi%f3cuPYTaj8Ahbs;;8U29A5lyv!Q+~j`Vo#fe!kVv(DQXM|*5L=E z3(FJS4U&o3B#Z`EhP+BisP)L6v)-Px&qo~-ulD1j0hi?COi~4a=!QQgiEPp%?y=3= zOv3BF-K2$Rub|@vu4eA#Gt_IuSc+fdI!ftFJj&r<=H(E7{&HUDno-}M;BMB*t-sHW z#T71Gfotd>r~s7DRg9V_EIWRIv}!^LWYri;8pX=$=-6K|SNvQa$WOV8@j;#ua7v;e zT^c^d_MW|O8IC1fnNK3FRl~&HD5;O!A5|Q{W2E)1{D)?^e%tB~^ynYUr)Lm(lvQsI zWT~;ZcbFnQUz1GOLoWF_PTyyY9<^h;sOa3&RHw%)y!XeE#(W$*<{SFv*PvK1-s0ar z$AE;<(EkWKB9Vu_^YnJeYK6BZ$X`K9$WdON)32aCdWqwsm!T+qe**fxZbQnQ*!;~- z+<9_v5c7gg@%`RP3jWZ+gV?fyc$incLY}*lSJ~F**N(I^OrL!BHFjb(iTj6x>)cV! z85lKZNt-ho_72+Md^H%l>wIf<%&lK*aY74DPjb--c_K9ysOU}D>Qla?VNa-6<9vrV z#Co3GuuwKKu!@Wd;QU`bn^i#4nXDP-g;8JaY?aZFM*)fdDm=EFnI8Lr-sK;8BU)xG zm;r_^{gq<`54SC`wgnGc8K?1X{=NK_C0wZSZ_SN zemr}{l^KQaLoG7|-fwHk=_#XB;eY}_t`%#m{%?Adm9)hhzF+N}6nt|-;u%BdxzIG7 zk7c}WeYduO!8#H)uyhIQUpdmQU9$Q~3XoqapTXRG+vK$rbCpjdLP^QSH-~vt$l!j> z-isdPwaLqbGsP}CYL1(px8Yh1z{MXbW{gzGV!MO5u2<>-WN=Rzl8T^-q4FMzMGH=* z(?hBp=cRkD#E`q6c)7K!Z*}EP=AdaXaiHY7k!A*#9~6&pL-Pbwcm2?26H?qcuU&^& ze}~XHdOLRQ2S(CpbGQGVpYIh>6Hrmll05K2)qkr|_;oyG$-07~`X!2JXz3Aoi^j5e z%{Cz8qPz_F<3SQP=IEzjs;N%dVT-c$flu`->3JB(9MKUs$V=l^)|>r+RimO#X*9OL zJFv=UYgsOaCA?d=E!1SS|0LZ4GRC%qqLOapo}_0s&7h-6lfKZ+4`I76%}<)0x}ti* zjmNX}|4S@uwBrAjs`i} zfLoDlU({m5i1cN`*C0)%1Ng>XmJFH>hZ7iPc95FrGjJ0C^3P0KCS0P@5u*7e89uIA z+oiL-5xUHoP5C(;E=4B4mY?=YW5Vj9!W7>3*32qtn@3&cu>6!^PZ`|f6I8p-(?N!= zt+8W5^J~)$a2s}ULPu?S+l%VJLUzZNH4sz=JX-l^;Bd1uDD5IwPDo({kAA6~OWQVo zD)8O&n}}|{af>)xVAZluv??Uj>q9AN7P6F^*D*iFV=6vFKG(lO;yJj~oBGvzK4;m5 z(-wiE?z@EJIZZJ$XT72qsD>xT^y8pw-ABpMfYyzd9c26{uh5a54oLX8QNhE<~-|%6K1HI9^jlPbAU`S((%E=gZ zqy464<~KHJQ0cQtttq$g2dzBAcF}^|UZ~T-JuYFkziiKINUel>5+ito=1G^vj}eg? z{T5|e{eLv*t-2Ove5Jro>Ziq}BJ@qT$gHNe7y~HN?GnYjX`ee%o>{T5jaZtFWnu6e z+}Y1YURaNCx~ghc9ljSq%B9~>=?-R5fC9?MhBm4$)mjJbTgdb@o`&8ij4YG1U$u}8 zN2n%6e8SEuPs&AUY}r-7JggW$+Fs#qwi_H?=Wg{2q(AvjUCq8Nv;=jhT%;kZ?7 zo06#Y>myY<ksxo)!Caq`swm&r%Rq~{uJ-t+c0A7ThTqBr3gwzY(2{z ODySk*P*B%g5BNV(#{h-^ diff --git a/generated/webwork/images/webwork-1173-image-1.pdf b/generated/webwork/images/webwork-1173-image-1.pdf index f4814c0f629a9b6da0be51ee3ce8ec02b8be066d..2e95230129731aecd5e5282dbd4be74d570dc75b 100755 GIT binary patch delta 2247 zcmV;&2srn{E5$3Y*#>`V#4rrq{S_Lol;Xrg=@0@XM@kCml{yFQ=-5im(AdUthWvfc zK666{3H0=yq$k>R$Ij>g^wV3(F~Fpa^0?Z~ON0?;ZAi&LX&G{rvRs5CwCRQW5DNkw z-uu(r_hZxZpO>fe(3a1jaJn&u-661CcSpXj0hi897OIK>(63|bz%}?C4zJ3N_z zWOHln1x`^?M^3Zt3UC4hTM!HzlVJ@ye=t5i3UhRFWnpa! zc-mc1!D_@H5WV{qbKFu&Ob%@qLSV@@ASG?7yX@KMBo!JJV-ETI8WZ-=gE}+5_jqrz znw^=^0kSV|CC320Hp=5M0ZPlTcj=a+a3rf+;XcHOklpyHd4GGX+m_|@ z>3V7MXHeL0kjvr_e~P_lXTD4UM`n5pWk~?C@2EHhzrf+7Y=;l)?9(gs+6-{G);vx4 z?A$MoXaZ1DRY@-2JV~o?azdgPWT0z+~6+xr^{S_Od>*bH|q*H1* z6K_YX3TxJTR5^PeJ*0)CbgL0<@V+x=-O5Vm4?&J^n+j!albQ}j0Wp)|4kv#yG%Xfs zq$z0~+s)0nr`zeAn-5#_iJzNJ-JJM%wOnaS(PRZh5Rq6|NPmGTNxkTW5qGs0wc(r3 zm+$x0_k$urA%dc)xcX*s>9vyc3Z=3ffV~uM2MnTixcGXi!!-9Utu!aXV3nEZhRfMG zgiXh2iAu^!(JhpkC^Z=w;yP)di2d)f{ZN<;=4OVX38hI*(MBYp)QW#1NDWep2vQIf z5hAZpC*LRLKXSXARTLB|rXDZ_o+7$q=bOd}}?`ZV*MmP%C^fPyb6q?$~G;zYA)RK3h z8yIK_g@@Q#h+uQgBTjz8S3kRaIGZl(7!lq2AZFJOLpX z9}A2ILNJZZn+IJ?qocMardh}LVDDRmOU1;^EUyR#L!5y<5GR8N@ibw=f0EcqstT!u(XkHN0b4s!9@5~Z zksyf@r=$OV$AQv>ZqhVOpxEc)bNPG{4=C8+xnKda7%=>M9PlRuFt8koYOpD)B`ywH z!R7)4*p>j!Ku8V}9Ge2?;7S+Jg$f27<1yEQ^K;^->olKbL8vejg((@PiAMd9Vwg&1 z7^2#6e>{$xUXR@5X*g4QL2d>^GFMwoMw=i~{%+~-meI1;JHbd<^qA$*13CvD%aitPI6s;Tmy&_~2Ziy)@3`Dqa-} ztMVaZwTgp&x?NC|x#e|j4>*-wxj*jbD$r@(e}X$TS9g$>Ix97LaJx%Imw9KDG2S)} zugXy6t?lYL&Q_0`E!?-zi&KZu`*wR5Cr*<=pjB&dW&$8+&zxqmL**%@r|n(Js@T__ zs`C9n2kSIK^TOqFd9o0eu^ygA<1ny2x;(UOu8GXb2W_kB*$=l0KTt)O#~I==vsB%DBmJ8l zJIFq%IC|0`U5i{4p}96njK^*_!Vcr`B8k=tIt=03WTw^TJJ^y8*FjW(L#v>Db6FjW zLdP>h&RhYG<1H)OFkUO1+})g-@^J=Re`FIUWznu2Cc#EE`Z%ii&%i_jm^=a|YQ%*7 zhU|Lsb$U6$RMYpw_&*Sno_%1}VJOtG(LFKem6)splhEM;zXudQ3XY%+U``X5{+W_T zeB($FZ)@CY&HObsw}H(&!p3-w8@>o{;BIFZ!>f;&Ydm7~9XjiOqial@!+$$*82lBW z?H&W`Z!ypjd`HCA;RTVZv)xomVkXrJPrYgyh5yRO@HknM6BNb;*2&zMcOuxtM4Z%r*ZL8VUTu0(o|IoP1m&qt^U_5ieXegG886b zHOt(BCRMWcaHrM~ZhP|&vZJ)zSbV(6JD*fJ{s3LcZ3&ZM4HJ{j2TTWba`ShNbk(yV z2qpmp2i|3jlVJ@LlU@lnf5B>mFc7`xE9SVR6r;OH4MB^^ZHR`UiO^im0nlb$U{=x~sV#0dz=-Cl*8kIay$sG0d< zBSLijsXyDxYf<0c-`aZ_ry#K32=~PxBzs+*n`Lr1R;t&KWdtC)f313xvkM$PgzoUg z_pVl;m1=-PTQs55%*H%$MCE{zJP&g5=7?*7lVKb!Qv;HW(j2m~%n2&-@9(%xbUpuN zmSo3dWm396D?V-bujW%fL-X0%)lXPWNa{^Vh0pC7ZTwU?e_hLQatdW` zWOHln1y@Z=RY$Y!3UC4hXI%^+lVJ@ye=$Bj3UhRFWnpa! zc-mc1O>cxC5WVv&=D1DMfF9azVoVx0HXLl5y4#N9e}=hD_acEYok0)cJ>mX!$BL8<)E|-dzWrG3P)(w2=^jJ1p4{){QXhiR!ze* zc6T|~**z%iH^_N$e+c>B^HaM_0Y`3n3q?Ty==-WT1;4=IN!bo>;_THc^x6z?xKwSL zw6k-+IHC!_NLePiZ1W_o!pRAVUYG$%!B`1JU6%xn`uCS_jINhI!jt0RaYZt=K5Kp4 z>SN;Vh^P>=UZcv{`_)5QNJ5>{+ej@^tK~{tiY6;4f{4VzLi!6#N$N!}jJT`4s14tI zzI?y0z8@433K0}V#Wl5Ms_N473Z=55K%QDk3<>|=r++-@{7HmAh4R^HJTBmm3-PUM z!hfQ)Yy?4mWu^G@aEd>UqXnmuOA#&8YL_u+J zQ65n!mlqRPj1;X`lO{q*GCImgGRe|R+%&886yqRrN^}fkxs;!8x7%f;(Izv~jhA!t z2)mxq5ml6pqFX5qQEE0Z#C6h05&Pd|`=Kx!EiDX16H2p&qD@FrsSSTc5EW8~2+|M~ z5hAZp7vC@DKXKJq3ZgBQ^Tm!xL zn)@7HtJl(FX-Jfx=yTn@S^-#XcYH|-FDKVy+Y`@+LNJ59Ff4zm^0bMwo|Fn6ghAi% z#N79lCrOpjNiax(c3Zt|>eG%YxQ!KlFuuo^AvPI}cGh&&sjIDe%2W>>(BN-ho`5in z-w%!j!!V7_TZY_Ble110*RJDxu=g#>rDNh2mRE#AVa~`NNRXj}c!scOMs#RIlrbpE zoX_}@83~9)+|~xM|75)w1(RV76O)_~K>;3(=>r%pNr4s^GST5Ai(#)0p>7Z`1d*BPfB3mxD?eOD5@1ME?Plw z2@-@OfisYbg92Aj;2b>V0lHAhfMY!7IdFbXf^?DQ<1CCcW}+}6{d8qe-=`R+k{O1m zHXM(CvT>qQWn2tdHjsdp)opNk)D1_ zB0ak7ZXHItNXtCZ1^88>$$O^JIkt~nuZ->6>I19{#bWANaeH|0T%f(Q&g3eY7gMM5 zAwxY+!fv{rQk1#lw}lU!sAbO?za`oHWx0l@26}PoGWyhPZsH{AsuvpF=$+XB2wK_GLba$mrS!D9Nm&)U>!~U~ z2u-+1V>C}au4X4w=~(OjX*Dhb;nUf^WqVCzTs|8sSPdhS^mdg*>0ATjRq*>{Ze9z2 zJ0sQoC3#_j*Dwc4zK8p5H!O4&tlPg19>0DGhT})C3g6R3lqVVDv9na&d@cQl9XrUr z=_GzNpgf0M6p_6)O036bH^2^)=wcNwG_)AP6=ZDm@(zR|{Y4lT;L<8+*Iri3q0sW} zkTXw$>-w|G)=w51CwDWawtSqy23aP5tFmZT4nM-BuJv(L@t=Xo8ZdPPOdb%E;se?B z{qXAc5>t&n6XX9tO!;EZti@0`#74Ko+#NA>2uwkXOZ*;C{3y7R)_}QnVCH8^Uhs`0 zMY2BNcF@dUWAkd*{3C3PA8=!weSo`vxO}|pW3J(V(NE}{|BbFTaW4Pu#PJ;#K;az& z#@TNH}9 zYI0SMO$a$c{E{5y?2vxX^Ljmh{9do;kLSP7AD@3d@4r4h>fP#(UxENM9uEnR2=ny{ zf<(W&bgna?OY3mQPxXjZL{}=*n+jXDC zG-o~bs^yWEloq{cfYzwVu-oSkG7UmNe1m=dyP4J_=m^@qvIzl?HY$Ksc?G2PeT!&T zLI%7#v|r4W_Xuqm^cLA2{bhrEZ?&9?3{ed^vSI7s@X?AOINUte zZY-rT z+7LH=9x=lkZAAO5cBC6Jc+7Sf`nWdsyy>#2N8Cm~0C{3363UXaUB)zOLuJq7%8;By zw|3lu%bX9wvVSxlS%ING)Cff~RnEbRO;_a|)2*mE2VnbqrtIc&khs zG1WEiuIK;8X7=|&ydIyjdCp?PAT3PV^0)-ZKNd-^=Zkif5U4%a%MZzDvOkH29c;EL zNV%3&;X`(CU!%GikNvKhIogBMsZX-=f^~@+_^7{%J)A%7^O}{1l>9FcT8qXM)#cTq z&5KG%O+U2mdrgApHK(WK1)hmjOcu~aAJusC(t!qr;((hb0_Nu_Ee$TtF4x`5?fw*A ztf}hG{_^|Uk3QN^so6JswbJ6=4}1qK*n5=Zfq0d#GIy{D9FzfVE-gcB-pSxP4G z9v3%-e2=_WOG%K%9^=n45r=ashEyPF>FJv(CN1p+EMg_8<1R)Fl{TrKxmLAm38M;6Qxrsu#S}LokzcP755gs7X6IiPYo+67m+)&4)}v`x z!rXA;j4dL8sx{@cUO(d5MK6Dgi}DIs8qA~sM((_=j5EozEC~#6R5e<$e8%_-!IHhP zmDzhe#EckkdveR+ba_p!O_2f}B^M*vdgI2OP@k0IBsINp!v*vkC({{ntnB;v9$UcE zrtN_I43K|IvwL209dx<5Z|r^e^P2Jbr&o{7-1%WPnd&~?!L(vC+7_(sqH!%p zrb?pmhycT8Bp_+I_i$7y~kS;ay}LboZHgON7nz zz-HPxwE?L~ej#)Oj05}EV2Y?nIvQkFxFzL*R_4t~nKIaw=Q$)fOG^`@0J#$c2Bbz{ z*m%5_|83+w!^+emiv*+C>D`5n4gIJo*S=u5PmU$+AOgJ<^WPjP9uuEZV zc9x%snh=s6Gac8mkVe*sDjw`>V*^zvdIe zQ+HouX-p(4-)YUX&M{A*E3mCnExiBp)R9os&oE&6xI4}?Gs4|}doHRp z4Zk&{=G{(~&SIRCC=S}DFAjh36W){CUKsV-el|7JLuiPAlg6Zq3wglrbnfFe&c|4N z^1qVbpt<|>erw!GZl^)|aC=0HTd_ux??dYSVeH|+TXmwxipWY^E5*mC{$K4vp*0~V>A)5J4@621s(JT7*tK;aMqQE55-!Y0z<#bv+{efkC;GrMqwbN?wg>Ef~_&9$PiiK7e*tUY0M|K|pK+BhC)QsHhgPk=rcx25--;1Z9$N z!3d}t@LLpnkF`^s0ySIc0GWe9URzm@wP)vgv9bL6#nZOE_5+bdUfXg8pMeUpExijk zLD`p-C|f(izxYHX)Y@Ho+NHyD>^V=xIx~`%3D+z`ubiN97Z3bPxSNE zLu0gjbhW(DL>;{0|7*~G!p%J}|2!iquR4X|wcv^Kl>uaCEx7I@d}zTOX4smcGnRbi z`!`db(Uoki6nCt#SPv3i&w~o>cxdpW@3l_cT1wj>JEEx&TBO`hzTK?>X_ooMD?S4@ Sry0P(~Ok*5h z(Z9&I(c#6=Z!05xrH?c1h*H5}whW69h3H(GH}O!tiP)B|oZ!y;b1f+|=W+8>dJgER zJXUdJll~3|MyT<=tkFURBc~{EZP(jOMI>@N1wE5++#w!F) z!Sy>zU323ff7?^84K>jZv`-?<`Wb=YsUzDAu&E;%mD|k4tWa4&=eBPwL!MKI}$xPwB1Ty^E zB8*YMDCoIHc45{!4O)H;K}N7G^>S-apT?qz5aB1bl*Bt9!dzeRdc`qSGKOoHtZbJi zCPtRG&c~RZIN_rgLrH48uI4RANV*digS(W^;Eq=MqMamf%zoZf50)Qo7U6KwiwW+D zXR83QBUPlL*rW6Er(7U#&A0IU=UH?)ltMS#0I8x0*D0x7zFRE4_v`zz)yTJlL6};W z8QFO+F5Z0|_A$Qe%FyVjrA}E(>U>?pq=UCTc0U%AytSeb*6w0P@G0PrivEKl59B0d zxirTHtYz0aA?ib%b|}_Ux=wfb z3s=J*zHJK9OlYik-2qAe01xWQ!6vT7hUPJ5dhrn|EK!TEico^anj9+L_8s?~>5o|H zqCo*A-9Q-)=B1Sbr-qc_n(6TT@CU?fl!xv>)NvJtrj81=>Sl4?-Y*)7y*b@|fH5^G zaCh(vafd$K(>o#cw4Jgk<=WQ|l!$`qb3&~`;y?H8e@UC>3HTL6H2%9d;QjC&uvL*Ecl9 zirgcvAuXy6P(|rqtqyx}gWHcp2;y^ff@Zq25*c4eTBQVkOSzjWtp;ay8&7l{6m~Yt zCALLLN|nIh&+;cvKl3huJ5qEYp64pg-?|h7$v)@ea?cJ5hgx?64_Pk{m!&yoXJ#4P z#nOx+FMZB>gY{r=^9+^jg$q zSEAy<^CLeKrB+@A0c(mb7aB=#SlsRtXy!CeUY({f`R;X_$ddtzf1{(Y?fN}D!L(!{2&LUOf;Wr8|~XF*TCeKL?L=_#CkO|mu9j+ z^d#AML`#o6?5Hy71t*kx*3EM>=ZTPK$@?i9O32S<-vLpGa0eod>AAzfHrF*BdG6D7 z`R@g<9J`LgByFmAa(iJmhpd9WwvKn==EO}EqnaHG^$UKf9g>#JXe6Qc0M-O~-k)DA zkn$x+PhhG*Pl&a(dDXzHV{nO($al6DrJeYVsWp@%Wx$^DSQPrhJ6`vp!?XYp&>{>$ zqyjoxSpb4Fzd%oUrI4jlbz~G9bkpvRijrtyqKgugev^OUqo%|_@@XNZ8GTmbyED>d z8Ab<#{mS%ZlBGUXXeI5KcOlCA5LS)cID(RK5u~y2-Gyh&9PjTt0j0(%;(310l2!&e zx=9`tIAS(!TmS8hFdjl;^U^mqaMGXH6|4PzWf}OQQ%!uWSfeWpBh#Mm*~_`L30?^| za!cH?$F)`=Yei!9)M`T$hxx7{&4gY@7P+P%{4q3ebvukJP*GJK+mqy-s?Jm-L{@0* zKn5Ea?}d)jFPs?*S|Va#lowo{>EM;~hJ?RpyqRc$E%3(lP~b+(*oSV+psSuy6Z=># zK(d}zp%&p3gZ&hRE1+qAZ>XT{su*!qNo=!CI^}?U`T6P}6~lCKqC@bHycJ~j`p#?j z*P;9FmyJ9>UQiE&#xhNNC=O@Af~^CnB?8}f&v;q~#yX8Xt=n|?j3-y|n@GdTTEbPfjBSQ7FWp*}`-sZ_4> z$PTJX-}NmBPDt$roHexItP68F!B<(!aePX3h!Xz`~g= z=!M+~d_x+J7om|Q-^vf2gjxKx%x64X-d_D)?Cejm1jf@Pat2iUBoe7_h(ZP!9z`Gn zjDrmQ0ue|*q`$v`AL_TF^i9mo{@(?e2etM!{P*&zF44*;JzCM!RS_Ts=Dt=$N=1ILtnA_5t_esSja7kL#H6qe>pWPY(w< a(K&4r&~TB{30a8NA( diff --git a/generated/webwork/images/webwork-1175-image-1.png b/generated/webwork/images/webwork-1175-image-1.png index 5ed5737273b6654e8b2915cbbb03533655d12480..697c4a918b1a72f094b1ea95277f6fc73b219518 100644 GIT binary patch delta 21 ccmZoJZ!X{9%flhe!6QER*L~Z~p*(u}095)1XaE2J delta 21 ccmZoJZ!X{9%flfcrek_jc=r3vp*(u}0988&bN~PV diff --git a/generated/webwork/images/webwork-1177-image-1.pdf b/generated/webwork/images/webwork-1177-image-1.pdf index 0ae8ed64bb21ad8a378d69db43af214ed08b6759..79be33c71033471b65fd21b57aaa0297fb29cd9e 100755 GIT binary patch delta 2290 zcmajfXFL>)0|)TSI*vFwoxL*8$hbjq_TDP1lTC@koij3yGP1&*71=8*Ny^D~kG<)! z$|gi9%1qDG%jffd@qhPw@q6|C{64>CX^QmAQ3hhY7ge6+<1bF0y7U#Q_7<2%U&{zY z$vnYsZq68ujLr1;$DW0cqVd{5rh|k0Qi-_(AKw7L$$IS?m(zq^2xfOftm|`T)A7qIP%)6B2;JIYYxm?{y# z0r3>+cQGQa+j5q@K;-8Z=K~)PWeiU~xJ~6SkvRuIU-=pSqNPBzw(C7p{GDuW!&?x< zO$Qqe__pI!=^n#7F{36uMww0A$yIm1E}LN&7Jrrtit77QkVnDf2DvJ$vkTTeJ~SYA ztQ$T6=kDzDC;7T~ffV3E@;IJOP?_YTlJ4L`B!$%w67DR91vFQP+VRpl} z1LO_FWJVs;^A+`4!bs~vqCugYShRvRX}fC(mQHc)(X9gJd$F7ej;HqH_u^tPb46Pt?lq0y?v#ga=KTeC zI1$LX`ZqSB%S+43*9g|)J9>AuqU)H}VbUM>UnX)q$eQ8SEdyg~{f2dr9}S(SJBtw6 z?xG>F6C3?6%swanL-{=&HXoFNI4??^&pWqS;DHX80|`X;ZD|oY_AiW+NATD6)_$)* zcl(gOwrA}854bWiS)mQrqU4uK;r-01Mn@&aZ?h?4b{}!4ou1J!%n8xdad8&@MPDxZqZq39;r*SdSYNbb z+oH`PTLNPYV?wbDfnpSLBgDGMN-A+}t86Hn+AUb8XFQD-YZUbV>aBwF?P3Ft*jCLrEv%Bf!^4<>S6t|lX=K68thlg%L+nVAsq zU$_t{OKkDoN3n1Y7-{^5ILRvK$(R$RC671O|{F)!M)9W|6$ zlJcJy??679+>0N)tfY;(VhX80ba(>=CHBuzE#O73?n=!im8RQU5a}$hARY_RYO)J| z2NCvLUC!|myt5m9yNrqGg=O?EV^iIS+lG7NkMb^ZMaRicVz&avoO&_KnRbB@xZ~j; z3bc2aPPSQN@3eC((gs$+q>9vp_l)C&mkv{?#5(=eNK>_aYont2My)JS3`{Kq+8i0R zVGxj-7sZ8~hwmVX<4D)t`}#DFaWTmnemCD1I?#I`@=L_Wwv0*t&~h{{sEcZHLD7^G zhI`987-s!wIQDl9Ne;92&c}F}0b#-JYMQo;(KME%D8bm5{cqS&UoO~WO@EvYZ?Es& zYB;WO_`BKduX=&QR%4?MY>@cnvOS!OY#ZV{Oo(N2hIzG^*sPJIx@%g!B)oaZ_GC1| zF@j{poewC1+)WvtE>_v9d3fFS8*nl-jA@0V-uATi36E*xu*TE-MHjbeYK>RHQT5|v z29h*4W}Xo`y5UV;Q%oOu+L+%hx?se`u`VIH%6!WvJ8h!e3n`vV#^t)eSRK1x#XFRm ztFz;U22>vbhGRDF3Fsy{lHHYP=R2nXj$GWA-jFQ>{d_sC4-t-HXEKc!c>-r&?yN_R z7)8p4=8=etcG{0kZ&y~qQiDFtsI>9V13}0m0ox842A1Dw;2F~e7MZMK9`6J%AQW2b zgug)JZ4B@f;$vlPLeh_P;#D=EFfPBOUR7qftA?pBCc`pg(x@hoa-v&07O|}l>AtDh z4SpZH>i^;3!J3?T$h7VNDq~uq9 ze`FE&8LqiMu{yHtEhXcshEDErcRCFqc+X-uo@R{BG3s^q5mg<$Z>9HK;;MT{w5gAf z9ZI@+Sn*v8Sm(@|9>$z9uW8!{o0*kl2Ue73*%6hW zEPilwuG*@j9;t)zVW)jH-4@SXOMot4fC?3?Ls6{IBelXKOFw!a06w?lASJzJiB})g zSnXep*q@$hVL<+-<6T8ylN!wZQhxOZEA(K<3MHcLg$?JZ<{m2L>@7#HI$&-=O2aG{ z5VY3XQ#;OV{fomwqWT{BAw)(?#oYs`jB-P$xTrWI;V2Iz+|5nNLjj3WMk18pN-Aj8 t|JNYzLELEtlAvT;PGwjz@b~|IG(Nx?7Z8kd_W&UjktmRun69}V=pXYaIrIPk delta 2276 zcmV=?s|4`YWAC<%d-b{%tj*wL|-)UepbaSr+W z)voDb2MP4_o}?$*^lfi+0Q%*vLtQ}lQyJeptKCRN?Fdr5!#%D`xG+* z<>&peJ-jw2{`2v6y>#U}C>$S*%jOVRuKSj6OTf7|qlKy>0Q7(RK5z+sgWX=)0be3^ z_6of=6YQ=HPY7R~`^6befP&O@8dp9}l2+m3ghVgQgrq8p8mg|V2^#gUmleY(Z^CCL zVWN%_> z3NbVwFd%PYY6_Fh2Sf)A_ILDg@zS#x2qpmp4mf3zlR^nef5B>mFc7`xE9SUMDMojZ z9z@u>2p(+9>Qd+}CZijuNr>^_-`7OhLk~&bOx`@+o2aO+m68skFROWj4tlAC#YxYW zBXl@OMdAd6{kfFiWyivN9=MpB0}seD*lqa2hmrwpa71pP~6|?dm72 zCM5Nyq{8R+j5dBMoIhAgacv4^Ze(+Ga%Ev{3T19&lZgse22@u~LRvwypO7;sXW?nwFc9FUZTC)DS? zJ}2IRhzc?5Evi_Z3m(!!Qo3kFX}s;s8XH;Z`~gfgZ>5ug6cdx64mJTXlh+O>f3fl* zSmYv2N$c2dZq7a3PUqZw*qTrL+;r;J#K)`UN?VF1D=319#KJ=Q3rtDsMK6rFtG%cV z-+aD&zpuU@6cGv$6h+0VvdYrx%V!lzWkrEJwUihV{=ZNEc+&af2!8_Qv(tE7z#kRj zn^%SZL}}Rwg8a%#@#n!5e;i8je@qf-j>Cu$MX?qa$mQo{a(SV`Y;n-~1|37>sB?*e z;^Lw_qEId`CN3E%TCXNegpy=*l#yhTrHQy>XbDG#*cpgov&|zeeVJsc6(g? zJ@=Y=9bT)~(ru|vlppJL-Mvx)*sZSkq7+_Eti`s*pAUv$8hc?_f2_&VCQf@&DtHhE zd_&{2-fnia` zfGBe=<4a~FAQEw#2S@$`5*ZielYbFoe_LmJ%SO?psw_S$wPUj3GW$|N{#}DWn z8l&?y>Feht(&MY%&S9*Jw9F%2fL|q=zGNDmWBbVU+StCWKETRQEM}e+cZUzo1=>sN zOg<)yV&+snWTY2K*h@DviZXZnf41;}Q`J_x<3X-NljaS$(^Gv1X=$?3pa&OiD%(t5 zFeyaSHoPb!oi{d%=OkM^t~PMrKrc>RM(>-=ZJZ=s_d}x_{RR0NfBm{nqI98w z@hbRjvM^7DossJPlDsg%Q(49Oe;`?EoZRi4+VXJ*8)TKN%c5C1d`q6Ksqhabujlf*W6b8Qk_U*J#M-BpY?#?|pDBF0M z@4x5+4&P3zD&<*a+%BJ7hV4c`HQ2>@QxDg&=Ka2q{~&M35`D5=~CO?9~m1U`VVD_ z?y<^rdEgg?kG9(W^j>%5>K;hh)7Q=4h}1^VK@HUyth%b(`3Waxh{I?FxxMC)HA%({ zeOTLbF;Ks%q`lmq zu!z&y05;8Tp9vXlW(Eq62n)c5fTE{;oL&%;l@7EEWW(P-2|5*ZIR>eo;RTgsg}587 z$zmH-WyKa>bm|tpJ1NQx!0`IQk`-(!ZSxZDzW?mj z_Olm!rb-;`VOZ=D#ed7s;(iK?jcbXzk(bnkJsMGM*-!q_YTU$IzmA_MDu#<$gbCss z8noqSlfP_eZkp9vXitiz#!kL;WE~RRk&^K#DSR)W5Z=I6W}?2n{SL5efzX%V;M zWY5z?0xcQj4|H-p?Rdea;S1>c9uc1;usjX7--L|%9~gWx@<2ORd!_j?q%`%pxC*3C zjqyC-L;I#2Z4CHEwKw`JKe%MQI8_x*+fv{=IM=5IeN_G#Y2CP5>~YbgV0c}`Dm7_b zm1Ci`uyt$}x16gQ(DPYCQHuJB8)ow8(a;mDW?YL8Z&OzF2Xwjh0y+5_-*fU9jh($V zYf&h3bGcs2aaQO0_T48v<;a~d>4)P0c|rhG9-ThmiG4_3tZc6QJe`b63?UmXruekI!m=wh7jg2eHv?=DcIAtwS>hVS$ zI{2DjpFq?pF0>L9Kzkrz^WO$)S$sPgqCC5^iPIWUlWuKEV%@eKh?oYTRByZCo8d=? zvfrRW2_rLEf`}20qSxQ$rQhww8{B&>z0hJ+^Feh&_=hlRCL3SYgn0wU#)V-auQs9p-%wD?2)kDK1Y=C_e{uka23WCnSe>r|`ykYRosFJZ8~-=MyPTn&ZlqB3RoLUo4@{ zMLC|ZkmR+8DZ5<9p6M2|UA<1;^oxc4npd_r@SK4mCgz`7`~jOAb!@Io-=;K6 zyrc$Z`molTgyKsuRsJX9zg7Z^*<>yhwCkHM0WjMbj`i6)xqoj8d~N(lC2JHYKe4xb zXIoQj0nJPzq%iY`d`(q)j|@NC+LVN@XuAu|ku>1te|%S6NyXKr)y%ZvXDER0LY{c_ zO+FVO<*jUZS~H#${@~!NulH}70o}`RN%w2kVZw^w`K64Lgl@o7!)L5&@}Kw{sST5} z_~3?nHsYBP&3qp%k>8Pa{+4@n+3r30rLSTENJ=MCH|GhoM5CQ3+b#K_qhss=-|iev zE9tVIyD+b==?yQ-^v;4r;hzN3IfPX-0)lUyta(kJa%g-EK2qFPjwM18k$UTj-$b!{ z9x=biG;7hH>f^6e(A-;eUQxo8uKajfC8s+tD8JvXt4TV(3*?TG)uIja5p#R%eCr zUI!gvigbDDZah0-Jee@A?@%6I}Oj!tb%!a zHzsoSiZOJA3lgqlF!R$8jOeQkA)gArE|h|RjKG*{SUw+2BPsMrqtD;ZCO#jlP;>#i z41rHhKr`7y^r5@?DtEZfy8WE1_Es&E{`Hy8hu3o#2oDgVNiyj<<>yG*W+mZeOoXAo z>Lpji6?#v5kA1GETW`Hcrz50eMf#zO0KROU-;rU~HRyymSSYoXDmkJyt-WO5QB_9v zzQB}hp;FRP-bGrTU&)ITR6hDp7S>EFZcC~eCp{iZnu*i;HEdp*zVS2& zQ;GA8cs_7vkXHU(PQhfguX1ShWE1O7X*~x%UXM0G3m<3&v{K$2s zT@|-qH7_<0Vdr12Ov6zMtPdilt{Z*oK4(@@$-%UiT%H#LrssP)Xy38gBndJn;5!0z zVh(I95NUqeSrc-ZNyW6Vj14|6fg}4bhE&yinR?p(@^LJ`&%<}GiblNIwSJq2q3=d6ZbL0*Oo9DD z(D~dp!)9j8iz>2kBcXbkmIgss85$z>12HJHA?EUB17ob7QJ_B#8;HW_q4d!H2Dr;+ z$NyhJ+lJsQ9TyO@_USgMMD;QfU6<@4#57R_mmG9-I)$RxBW{Acb*nS~)_ntCGF>E~ mkq<*wdRF?sT~tHe>HZ9ihYX3?(eh;##f(8}YNyXxgZ>4hCty7Q delta 2340 zcmai!X*?5*1IO87Hl8^~gj^wVA7ev~Im(eMlrtL}D)-StW*941WR7x0M2N_JOtUZ) zP|1m9ZqlXaG5Pksla)=M8B;~6f3udzvP?ESqsC zW&jA+`MDd14~HJDf8PGkvu3I$E1#PZE1tx1?3 z3mu9As=UW6k74`&B>~{m1VeLkNNijT799@xec9jjWdu=IWbjbq+r6eHU1ii+g*wvh zZ0jGX$~S^^M%>%9f}ltEb}J+|@MgkU)3T++$hdSu-oHjn393 zqN$GqxOO2Si+F*Lt2rFJZ<<$=T*AUba3xNkl?FTmDqd=g|9`cY}*q z1;tor*DM0zF#h58ex-NeHaXLt{Y7)XZIwu8XU|o)w^PboehN*Uz9wmbhATE2t4Nsl zgmms+Tsh*kRf^{I``1JX#Esw0{TvGA)v*nU0vyxB2|@k=CIsF8>KS$(Q4pc*NyPdE zIfV&ts$PI$FRr=56O5NMGQ(8bE!EC)Ia-}dL{K2^cg`ARLdqh#78rcx)8ug=cUdi; zTuHInh2P`a=9&R@g|=)M7P6iaZ39a!Y;uS>VA=a* zl3`*LU0`A1EM&63RY3YH=jo!9zYZbe^xE8Bc3m9)-H+b4^pW?XZi+aHZE?r<=3AWw zMcUFc{E?=y?`*wH2!;`n`4DUbxeRLeUk57r+m0|^SvsfNyYx=+jG9`cq^SB*#W$1? z_-c`~=Ipq~tVwryd8|v%9MV|6n+3BluIxwd?q2PiEqOQr$yQKmZ!7o(xLBApBY5hs zt2$R-7(dWqaGE(FTN|Zgwr&$^P3~Cl3GYV^o+VGuYmJgJ2!C_tEUFBDA!rXQ4>Wo6 z1rKVKlK09s9$YTSR2YByN}M3)v`eoMf32(-C2MD+mJJ`UtU>5pQiBxXa)|F1{*+Sk zt+ZxJN=UH8_#n2T(G%xqlHg%V9?G?~xyoeCypanXJ}!DtrGd+I-4emX#DoEVXX)@5 z1Al_v>C(182`5gvSuLsc-=7t^G^I-Fb(U7ga0Zffig2}$vT8z6*mtg-VEZ#&)dKE$ z^9m^5bR{LG#OCTvm7{fMqrs`W@6pHcgD>#QqA^$cdS#&6^Gmr*P|~1mT=7XEN{U7(V9Uc6+^BEH%YWicM01ZVe%6imMq9e@*!O+05!Ht?-`<~uZ#pzr>AywE%pdrc zjwuQS)ehecl&Wt-zDz9-i+bll@4B|Hy_KS!6BBl?E^Dp0#`T`TvqQ1qP>U+ zrNkQkX(em+d9V!tqt}zjv{y6^6RQiKadf zp;BYinuhBim-He+`N{=F0(}7ppd#{o)OAP&QN!O-@{CkeF5QnHI6Zy7;OSmWK*n8~ zlQ=H8Fhu$*EE_F-AAKA5Viyr%cPhr8HRH!t>63FAM}@=hN6>8e21XfLb8kcA((h(7 z+ttXRIB1s_)^2PQnK!>hW0=13C|#O#S;(M^_Y(RB>bUXZ`t2b(yxYFt3(+@#gs*)T`ZlgQwb{f#B!i8c=ICyt1b$T$ zDpOQ|Z^M8c35DrwZ7Ut0d$Fj36p1wk~#%{{ft{P8k3I diff --git a/generated/webwork/images/webwork-1179-image-1.png b/generated/webwork/images/webwork-1179-image-1.png index 290ef3063c946ca2deaece7bea48321f263cf93e..76c84829617557c4ca37df2ffa9747e86cdf2657 100644 GIT binary patch delta 21 ccmbPVI=^&-1rLWX2aoa|Kkbgq4m|OC08r2dxBvhE delta 21 dcmbPVI=^&-1rLXSm@do2YpbSgcHoKE0{~bA2dw}A diff --git a/generated/webwork/images/webwork-1229-image-1.pdf b/generated/webwork/images/webwork-1229-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..cdaab2ce486d8405eb00f1c6d773055325262e24 GIT binary patch literal 4450 zcma)92T)Vn+D+)hAYIU42-15(4P81aHGmPR0RjYwfh3gB6%a6J5EKPOiXdGjAVoxw zBBFrO6%_#$1ZheM5YUJJ{4@8xd+&e$H)m$exA&Ymd!Mz}S?fE}_GT8!2o+6`^yk&3 zEs!=04kLI6g7o!OEpWkPEK$`W7(>RIVSNa`SXFE6Ie&5h456k2H#7v1$V4m#4~m>v zavm>pHF@>~YGP3>XU!pL>hlB}n~`IMkj{=Ahx6dVrYNk10s zoV|9Jy5|xnls{D8*jbEw2YY3R8;K4JyZn*|`lQ|UtYJVI{uQ{%)3e7mCN3qaESKnE z^lgn;@6~k0PrJ_nufhIoj!^nxgwcu;vWo6$nEaR)O+OqPy~(b8S#Z^%;vK&fqHu6f z#PY?L#7C&{jkkG2849;ARr%~@-G1Bdd8LAXXRnJWilB%?sy^CPd)3o=cb+XU1QnmY zu^MsNi}SvU(B;g-)7&`ypjk5F4HmEQhn0tMh!=zs$>u%7V)|%42AU}WC z5v?Adtb$jRkc-aYk6(PNq)s?ICpn>mOV5;LJ;W7j1eZpRE%L~qiMmahCSwv^@T~m; zH6o*9l%@M7UCTDwbYQ?QqJHqftlHCooUy?y4N@QAH>SsqHN11BPx$eqa!uSj|%m@9k_8?KtFwu zVVxGhp(eJI)1pO9ACW&MshRuyWWEd?b*DFWj6y8XEbg<-m#cG>d|~r8jp!I|>6?Az z-gBkV3L-CW<@?Tjg}Tu~sDG96OHTJAO(Ju!+J=mtF+G)0ekcxNL(aV3hht1pBZq$)nD*5i|pwo8Z&=c|fnrL_z;L24Si z>6#hDF5RiLu(_ke&w~#;T|g!b-7DcqW@g^=ze*D8Qc9A&BXoTmR~8e0zAOG*ZoRc$ z+$FojQc=h%uyhuu;@s1cxbxHUPVoT0pZGbuXGZVcVqboA7oV&WKbTk5Ajugux|y6L z`DvB-eRMj6pd=;@sr!QNk$n6xa!L+yTIbf?-nb7c%24i;2_55mLo>#&vkPL|Zhb)) z6XX-lw$ zKuju+hLj1{@RmCao*)Ohx@LeMk~%Q zbRKRf$|NaB#yqGFl)L54AERie=3+x4u*G3PJK#p;;z%2Fo*CpnC*>N7Fae$92` z8?RNnQ`4TuxLx4g%R`-+O|u@eH{*gMxZWxmw#V2=B>!;CN&aLW;B@nKqYb-CWoo|2 zSy3z*E~l_6?fUFmeGow&C~az`2dKuQ!jx3+#C3PjbKcbG0V)kRA}?ZvMCLD3GzPlE zEbBKvz+#+MWI~(LmD6!!$~@ijmvlPIhCbg|`c;s-!Cq>x$bszZUfZ>rNW5o4xw5dF zR?irR=y7Bu!QEf}*qj#zZZp$?Z^wl^+R~40#QgH6lrV4HV-kJ>*;zdffUxI$|JM}v ztKT+?K>Rm;s-~g-?{?~L|MXHKoUe6GarO0p@&#)zZK3CPz&*^97e6LTb?YI~Xx#M}y4tAu&3=k6YpU3Zc!m{^&1k zwlC8cgL3v{1D*#{v0=dYMDR8*D z3LLIxN(iA4asB~hn4FJ1456c=tprnp!*yWBcq|d;gEQ!Yl~q$S_+B9t-=uWtHDx zO2CJN?ME`qW)U0r59QD$r^Rdj~T_4dv#yupbUuuFOlpM%0b>%X( ztLrj#3M^Gk<}^&rejq-%Q7yrFe2ZOpmz|l(x&O%Uimk(C@8{}DfsKm{+B_=*%J?Wu zoy;7$|Gc90N&O3I0kt``IKU$<1xHQb!WT!~$uG~Y?eTbwV2U#(n4-*J<{gHB)grWo z+IF+4p^jejkUGP~5Rsc>vO-O%Uhd?xTnu66nDA+8;DeYla#PwXYAx6D)6qB3R%%7) z?SMi*st5JV{^c3rAL<@J@uYGc3@_X=&1Dd-XBBcVG1;8epz#A`8DQ{u9+(FTV$&`F zb2B!Dw4}h`zfrfwAL{{&4 zc}?eEc24J7FWvdMtNY=GjidkKbW{c1(*6x0mhfGk9MpLyS-{sVu!vJ^|WdGyX&5|K-o{K97EA{md=X`r23v!8nVhS2@BTN zp2M%{=abgPbKAC>=12O|jPCbcZxFIMZO_J^mYlpUax-%z5h@yrjGgW@$#exUvpjxmKrc_Y@hND!rgm-*c(cRv<{{1bov$uv>*Q6_0)MaIP3Dg*gX=}1 zaQTF#a*{UA?%k=WgSR)%Q|y$CBlU#>m4m8YXOA7mk)-R7UvH$~z!%`UGvb?^7=ePi z&4>`qD?dKqoM22Z={6ipWB6a@VL26Jogd+&P zRtCDSrVzCFJXLMWbQIxAC`4D7$S;Nl@F#q zfpuIes<`{p>1?wGd-B}*l+W=7BA?Uvc9T>+TEY@B!~Kb27uR5UIi5lJIc`92)95M^ z)2U@!`_8a4epuGUxc#jiC6XBo3mSdxz9H)w|32n~s%&VdkyF^|`OIsf%pvQ&&%YXx z!k%&HN0D^CXV7osIS`kdTYZ@KEs4JA6{h(c~a`r(-UMtz*J-!IWktCr5dR=jqTVl%3z|FI(TBYj(z&c5Pn z8d?GY#Rx8YuQNCA*~PnGxf-9L^7G?RSjOg%(O22c9rN8Y$3yvE(~P)8Z-bh!s+N!P z*G-&%u`9hc|Mgd?BL1w^GbpT|>IpMfeP1n2HMp;i4#L;RR~@OLiP6Std8_NFVSP1` zT5vx#!~g#dcU2UDKnDG8rctUV&iN5wYX4I@;-awoEte|V6|p~~=Kc@g!k==WcaZ-$ zT*z#&MM45`k?vFkdbl;5Ak#U()FDM2`fA|N>!;ECyzg!rk6S=&F~=A6%JTQW9PZ!D nnt(X>7w;wTI(TvL{FMX}8ABvT60v?D4TKgFBrR=jX94;bT0s5o literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1229-image-1.png b/generated/webwork/images/webwork-1229-image-1.png index f7614cc4bdf187883b2936daea0cb0bb037113d0..a10c99e56627005a3feb738176516b194f72e3d4 100644 GIT binary patch literal 11077 zcmZ{Kc_39^_qXaBO-NGrlA*pOL*jB##!N@abh!v2^F>j)$lOHYUMgdRIEIUmIoFu6 zj2&}?RCEoAj2Yi^PT%Kwe(&@C-aj~J@4eRAd$0XjYkkhpt#%OiWCymoBOs zFfr`};U8enF1Y8)J3kjDroDnrYHIpUwzf=67avDQJL*T89&P@)pfDM9;gISheYK|| zr|w1_G!yE|pT6sNd9zbAGz72lL}SO%tI;Q1G!E|%yT9`Y`N5f;E%aMl1QpYp zqu9h1UggT|=f0h8TkaDh?w!AH?hpuhVE?w&*T#TeJM=yri#go(*QgLBIi_xYk-TfP zw}OKyx51Y_+W9+g564rdjqx{cDsTT5zE5U)Ib+|~rln^(2DWOmM{L(Qnp}B(^|R#Y zl`%%3_sr#vnlo$9zD~DW1aR$RIwriv!?qo~?fx^Ac<%n<0PWrN487Y#ALBTIXU|K! z_VUHHE*?Iq*(j(UU^urOy}_HV>5!SP?s~gLP~|#VTW%qcO*f%R&iv1lgp?9K~okcr-n^86@2Cb`);slT3Y`_Q6US@+!C$+*BQ zwA1M2Uh+K-iCrU;E2 zg2gk)e{AU{{>8rUwZG>q;praf)$z`8t;#4I*gqCKcWmdu6LlA&I4mAgZ*s7o#5`90 z>z#4hL1E3Qr^abKlWMcaRqc-azFYT_BqXSH)c)~p;CIR@jo4#M#AWaE*x7q;N@G}! z3aZmONs?xKxerP&EPS}fAd!#InJmIx)yu&Alt)k`!wYjD_@1?On?y&Pfc7y5sSJ!e zraQQtf^E>gHu3t(s{_U?EaHZtRqk$&_qWh9kb;tB&^luA|M@dD=N15KL zd=q@mL_ymx=T6D$`FX?9JCt>W<#&|k=ErkP}M;vZ&c zeeGH;PAp~~lpZ|EtL6xKFP`(^*mPWDvN(Au8XvP5+ZOXGrcyxIPT&A5bHvkHA?p_o z6bU6Q;fuG@PNXkiEYV`U)PA}3(vX&>)={mpOYD~eqpVxby=!QliJgq{J`wh;H)iQs z`!mH8zB-=9;b$F%n`1?zBgJ4WyYAHqHOLsJ zTcACk`h{{x=e?P*ahCSvv#$LaCnM{%uNh1DK0jgaZSOdKU|eo|$M|lpn9yXcB&~!N zPNVxG=F(av+9jGsX&Gq~c`-SCsmm96^E-?q2+CT8zDzaRtJ_3a0a^-tmna-1Wbt9D*w8-_>Wa-28KJGpLO6twNsun+^uIx1< zkbV-zQ-*8U>>EC{P91T*e8Zp7)$swZw*oemHanRGnfEfozGBS^J{#;EJRBUzUFoL*Wh|i$c>L2|iP;uP4K8PAcwdU!3n5wo*H9_rYy+Yko{Y0JjnDp2Q zrP#obZOJ;B)H61MT_T0}+afI@*?4EUL6Nl!i`o<&ZSS`0XJku7mOWm$U%m0kuTpbk zel`8!-G`;>S8YgN%1=5Q_!+nr+T8a1buCm^OC?mI_r%h*?2Dg8NZJp&`Z7JzyJM9b zw+^&ZEU;gn_dV}@{ns|@9k)JuW#XQW7o5W z_T69|9W2uEFyAVD|3i03Z$@AyWOUvWGHcme8mFSud?rYGq(qO4Frq@gfPQ5 z$}5&xde{q;56qZWW!~=mUKdhluEDG^t`XL_@x@%3d+W;l*RBy-)vi++r!<$XmrwoR z`RG=(t)8Xsq&wNLV`UeweOqB&VHt}oU!|$*RZHKHn?hfoyspbl%`wQmZ8dIVq}_Y7 z#KOaz>erqW|GS>!(WCGIKBrtGnppPc9sDxmfm|j%+=sgM%QnvTd2^ENs_f!s!&X)n zp|pTD-#wr9wT^S-v`<1?e0=D`3;i)+69_*j7^lI&6yr-G3HA_!HmRb+CTX<&Q6t}|n+j*Hog%z77RaOznCQ4TR zD!)|UYt%Jvgk6rkeYV{~w_L&Z)LM&rXH!zM$*%m7pG#A`ZzgUUx*Z6j-=du_BNz>PD0pEzYG zVnunud-=|N<@sL`#mYxcFC<4M$La~|NbB(EBq!f`QNC_79QHm|Sp2BN{;jChqKW8< zoT9*wTfO4@4(n7OskB_HSf#cle08IaF1LAiDEh~)x>{GyRkf{i@(=K@DKY!KFned# zaMswbYprMVYZEmfsJ*SN?ay!cB``41)z!7Bu&BGM_1jqU`{v?H zmo9a*w|yLF>F;Y9{`4M#THE78OANIK2i|Y5)eV27!Bc$bZ2dmoJU`nw(~^{SjHlJt z*1=Bq_f%=lnbS!J*b5Y-8eL2V5`4xho{i=Nb8xU$l!d>!tgNnvpPel6@-Y83nBDd+ z;ly!nDaq4DhAP$Om!HRwLxLQ$Gw(HCH+e_4Na5n@3Uh65Z?m?xGBhL{JH}J@w#v=* zI-OpYolWcLXzT1~Ei^N0qNQ6|S%rjz+_-V0pr9ZsDk>-_C{LnNg@Pp9lQ`Q6L75(~3bEKgr=%gE5X z4^e)ZlG5*MYU*lBy1H6B^3zH!O*2jlRo-#$eqWudty@K*mODFiRusg<#PIO&R99CI z^wIhGd9t!nJ3Ct^KIArf*roCC&;#!dHRjd0*=Gs~b#}Jd*;(GZ>kWIqr@J*^q^ zDmFIO+MvK1hby1orqHcTK@hBc9$)0(*x2wkH#fJVt){RrKP)V)zo)9KtgNQG^l5~r zfB<(m#l0Z^iLZ~BzrU}IjZGRDnWie2_dM`Pc#wsKMHcN@LPE5qh0XCn(n3Plv& zM)LCVj*gZ#HdeYi>Tz*#@0#eP#X0$TuO2^s{Gq*ENl{u|UA?&IwU3X_*|S3Q^8EAX zWiD!}y~;@y6%`T@7Jk?CN?2I9)yL+Zsb;v0n5?YK6$#O2oQH|bdmZ3_cU950p z@6C&*?r`53^v6A968@NAVj*AB(O_Zh=G=2YD&@EQ3=80lz4f;oTpiAS1$ z9lj6fD5UPxs@*krflxW8H0M^3vlcg-q%^gj+gUaxzc{^6zkGGwzimcg#e7t6siJ1o zFjE1di)3diiUrKkUQAC~u_`z8zxb7OtHjQNvsWq^s~p0%ap0%(yid%^^?970Zzl); zLivb0b!%jyI$-U4($~Zz+e*nl6fbYZe)G<(l&U6+374EyVqo?BE(U&-7gd_*9Vz)@ zo3yN(uu&4-TQZqk*;iqIW2Wu@@D?VtCe zQdyjwJ9l*QMO?N?qYkAb2^c?ZKymKXRVTY21*I|w^N$aMK4SjY$t?eB$%$?D2?CBf zV0xj!__IazPy$u1>@j58O*%Cf{p&&KoO&qqt)vv~8Tegf+q0tf@sE+KMijpM*E1{B zwef?XH^u^caCi|XmrDnlGY4+7f>=y$o=wjO(iQP2Zk3;jFFziTaBd{kq&L$_A=Tq% zDKhKV9q-GVDXx4jE>_XV^R+s7XmnY_<$e^x0og_{-tLVtAyly!0y?qv)9;PBixb0!E zgImm*tc__T@d01k6Z&zX1V^f|hnyhA2G0VzEEu%ILmdEk?tO3sfg)ECw;>G_>F0#6#?wUuHiO^!H+z{A3`Fp9R|;1hM}1_iBRYj)z!rlnW70; z`f?I}E(2k7UQPfPUjUuETN94jl|t+~xDx+)l8waCjvM&dI5r8ntd@2I(DGw5yp)w< zxA3!}P`XwYFPV?~IN&aw&3oDu1C|&F2MFKJ3D^+ep*T3fYMwZsSN9Fv)+@Rm`p-$k zr$ej{!kvy&WY#M}g0mYG;xh@G=icAPi18;{8M0l8m~bKwrP-SO1qMAofQwHgByJu% z+RB(&aFKu;56-=}!H8)BD?Cd;mVP}c4`BUrGR(R-mpI^jM#D?<^bOu{sEv|`X}F}D z1TD0#E%~Ka|K8oN;byV2b4lz0yVaYUV{5diP=2|eXTy5E-ApSLelpMF+~!;l^Dk^H zT6OLJ8x)tsEC8a)7135*!kuZZ23l2b)+gGv{9U%|A3b_hrk+mR_~6|S&h+K-rk43f zZ&NZW!yK2UJ=13yRlg6VTgq)YrFVq(Yv={E$yZf0G*=E)X!^DJR{3W7P1r;DwHt!A zBVUKbw){uNhlRHaKKdOnWmU4<1sXc%EIZT9{gncDBXygTaW%l*wABI9IqH z438$7(+fKSkuqi`tN6N6xvZp0Kcl>LZJiH z+4TLN;n*srW<(&7;7QFGK&NeB1goRAq(ly$ek%mza&w{UlA&e{q5`#ydl!(nDsdv7 zK=5ydr=Q?JB*nOO%8=QeM`;(B_)!5tL|{1_ewu#2TZ@uLS2x5!XN1i6t3 z8k;wC=aG|(&NT_~f-?k-lp7d({l&L$*eY`F9pO{T6Q^J>^HljZH2bA{8T-65(dcI6 zJY-Rvt}@tgtZA+wY{eeNM(5Tek)@!OW;uM^XWBn(eIfZ*Oy|YEO@c5NTFpXcl!t1yIra0QOJ^_YV};axTT_$-)97s6Y;|jtb;)|M>Ym1Wu05 zb~V%Z&;6)#6t5Q0_GpcDj>grih8 zek6$KB=YJVeRzQ?;NPeXHg4YYsErSdzB$y0FQ6I0(kKKBOUa>lpGtxt)W&zp?DEc4 zc>Ww<1z@j$&(A{2j8`7jZfCq5K($=}*`{zromYx36@acG4*D^V8YrVQ?bvww&ucgm znFJQ9=L))DX}aVg1QNqibeY4o&9H5}9Ey=dDYIi#1cV*fVIDRFKC@U}G^+Bj7Cb0C z^RUC`MsTqI&tixF425GAQD_fK4l45Xz}6C_eQu(lr`wZV&fG;xnFCoB!LmmSV|gBl zkEtlAcb-WJI7MN$r)Xgl51)PulLVb0f-v}95MFHjAnCF-v4!NJ)%*}50cSf?NnT zhy*YyE6x~yud<*2RmlVA`L`9&xBMyz$+grh<#q4*pZ}>MuSOOp%GB9qg{2qY*4NjY zr=rT!gvV3W=JM)R?KSlca^n^XF8ZS@o8pi8-AtV898U*G*jp@qYp(3Bxa`+fQRRDJ ztFJZHY0!mDgG&db}YW&0VWA8dc!KUBoWWg5XZ+WB;e)RN>>GiFQt zz*vNLqeG2}w=NUW^4L0Ki%&#>Bw~%9G2sOd0VbJEuvg`j&Bn~cuaUjXkhJE}z@VUq zaUYsOUJSWem*uA+PgdA}y6m5A@wtTF{DYtqr6V-*T2z0({57(pF?x)(jO7{B|z5df9*SUl%NFy<~xm$63*MGqI~juwg$Anac+nX^dD(XGj@ zh*{X$CuBYGMG*BbwCI4)a5<7!@mAmhe8`P372rgR%vCC18;Z+v?&0{O2!t1;Ve#-^`rlO=DbIz;RC!lSzxb0ev~P6TgNEI9;lDlfXE++YbG3tqKI(^H{u_P=0-Lz zLyQ+m$CJ_WD-On^#@+vUx`3j8I$S^(vmJ0^#ruMERKgb27)IS&0DW6O4=2TPOl~~N zMuUMRbUmejtTswU<-v+IRL2lxMxU2K!GE4CNCyDePfLiTe_@H>&KiUt_BOOM zKF9QqqAEQM-FOtTwG2=6rNj-dfhsqUa7HLkkab22Yo44bV3i~2w-A}9?tu~;1* z1cWzu{yfEaEH0rPVm#7+R@%gHUPI}U4mgP%3(W!)Mhi`mgc;W)V%!(ljj|3Skmm$S zpV}FD|HWiP&;&2Wt@s4xJV&S*C>JYnC2M3O_z}ZKKSo^IN6A24T5_XUs1Jz=Ps%<7 z`2=$y9i2lA%tDKX7*8}BL@|`vmf>S?&4E8tBhemMV+O&hnY51pT1{?XqyJ$tvmwZ* z2>c0p144YyB`bnO_l}8%&l69DJT-)koLDftvM9ANv>~^c z*I>HYO8e{XbJpZ{44%Jstzx8Q&EzG`niv;1UtfpSnx>)Ght?i=7e9u;HI9$)T)St- zWKAQ&dm=*NBhzuh>n|?VpMtX($!_sSW_Qm6t<8Y~b`cM=%8}Xk5x?1f7A`)?51~{O z?tI{ZOGDr?|4I^kVPXH1_g78YJ5sh!q=77Nm6uU|)Kty5*p=a<9RBaXlcK{4YaPk@ z?7FLSUpm122+GzsrwPJGZ+xQ@*Wh#<&xrY;U?UNj*Ob*uuE|p+B8cE4C{$R`xW0*Oe2P|-Aq-D2Z z**i&DCi4N|E7m<6;BKQyBeiH8w@Ib^`_4NUk9v{{D_3Wbr|3xMyf@PHjw3y;t~Q6| zO>BEossnt)GXBG+Pz?VMHsyt~j{k*Cp$zDMvMH2q{eRh10${>YfdEpMoA(dT>SBCA zNlqB>2&G6cNAW`^ONGYJw9Y~eM{y@dB7Z|^mLdqEGz;b|iYQU)WH3+@?tlq^+DMe5 z_n{vPdE6GfiZje`4QwafnnAUtL0?-`Jsg`G{%7?LGx70BA-#+<`JpG>+W?O?1E)Xhz#EKqnQAwPuR(X(SQT zE&~z&bnu}OKPRM9!UmM6zCF~la7Pak>~={ zfG|TOt-}@A{7;WOaJm@@axV%WN{i`EWDkTlVLX}`vuQ9d!%aOw=@O#Ti|{7og@qFi zq7p}9X&&-*{MkRc?*{yB>_apTBK+_%fq6cU9F}|BEZ=lUN=`6u>6Iq01UJiE*&lM> zjc%FeAHEA8grp);*yOCqH~yQ9eEAnEsmekMSkGQi3RT_(^gE#LVWGITA&C%rL7mkd zN<`M(z?ea6TLrWiJt$qQV%)#@hx8x*!2vEH?jVf-W#7?s{(`EDW?YZ@fAbFaKO)sY z&Ly-EYJyUKIKG6OGFs%F@qfrgFJoi_oiGzxDo4w#6?h3Xp9IJr`curQvnSCjnaY3o z2h1T*q73s7?f;v9ltB;Cit``-VG@b#6y_hZf-wK+86cHc_5}W2OM42X!IJS9);gae z(`@q1QaQJtgdM;V)i@HaH_q4aq=&rG(!F~IskQ9!*>?&4-~yb!Y2^@*v;J~ldU$WB z0T*S8N9P!vNUNdOAN^-OAFw^&1~LPYFe0*V;Dwgx>A?n&d9K#IC0OD{BAmE=dY#$c z;WPZ;+pqwgXK`JfgproahCi8U^NJg#2J0AUX|z=UWf<%LSgCm=*b&ax zi}4=kL!ok@{}4=eoL!4${Lv=Dbfl%Z*BeZlc1v1)zmbCMDe$=wSF2C<*?otvNF#F~`OuGR;}qc>f>(q@yoFPF4BN#(i$ z#C}GwD_qobMQA>`3y#NWIgGuT##Mk-z7yF5VNw>L3dj+>)Uj06Clq827JjoOcGDc- z5Lol~&mEqvJM-wrU_*W_7QPFEgR^m3kv%!F zAD4U1da-3MZf(w2LgeaEHb+eFKqo|pa&$wW`u$i8tHhxACJdP%3bQT1zx5IHGUxK5 zj%>f)i2Lb3dH`Iz_3X_^;P^vK)vo|=@TYJ7yK^O4zZr?_exG~BHg}~ApH4d_z1FAp z8<$^WPlvJz&EMAhuX?CbEIp`|ka=D#mX;eHWmzsMG02fUs06I1BZV-C3go*@iSlBo z53@5OAQdcpA|ej>5CzZZWlEE(kzy_5r8ayJTvFUsby=Bs$u)Q7?fvuXQ%Cuqilp;# zysORY~KD;#FwZ+?H`#`?y9rm~>Y1jxS`vKq&~^?{|aX|4Bci*o_&#Sve>RH=xqJ~yw?t;A7(gO>r~FnN4*`qcEw_UB9UYpbV~ z&W}q4Vz8TkN8*&1dNLbpIy)7Xu=67<`lyErU5FNi=^=J=xvmPHC_kF+3mFqt#~vS-GdIJsfZKInRE>|!*fUB=r@5O zsFl7|$$M-%j_n10kT}cZzbMY<*7LQS3n!vq(othS7aO>=+sK*KzvrfM^8rd})B+GI z;`PbT;(H2*4Kek8oWpms(dXI`PmEzITT0pTc<&8g08gHRN~(=ab^yDx^%=RE54t|& zSi_w64EJUNeHxJ8elmVX(d?No>xL>1E77*5Uy3WZ)MEJJFwRucXW^(?KUKykWk@Q> zM*N=Db#Pp_f;i318X?0msDnLS;IQa6x}n=o_1w}}@%8sNjbhYs=kj93_!069kN&ju z8}tB!80%k0BEONWPc5=6klePXdTt#u2kOOrI(VqtnYP4$ws$1=*L;c>VI8|-x~gCA8OL`{trLiI8S{NOq92#h$nad{AwfB?mN4(8LwD7@v9X zoqN0+TxZ=@$J{{fhNDjm-r3MnR;F)7Nb*J&Lk3baJr86RuS}kKQU7`-8Cg!#$lVyENre8p=9(hY#&B>0=K-h-? z(o5nlP*OcXWkbPm{?o+E4r_d4V?dS*+3Nj>fW=jqCT0<@uO<9p_fqz>PAF)I&OvT%gurT`1UoF9%;AuA_8iZ;4Y1z1dr{N6oI;I7wy`>bcYYp+(f%F ztr8cqM`JT1^_8ZYQN9MydTOjH>lMbo{9Dcs2_IDo^QrVyot~(<@NQ2^b%o|)Z8Kx% z(o#xLzx03%{f902DFxtkBGz@&bULN;1Gb#FnEqNX`l(mX?Ze^54KpJ>%g{?a`RViCP zrFuAHrdI+{*Hj5n*-U)`nJ7*qy!GPhu&BWE8_m8XlL(kd)w(I*~!h7>Hkr1H_%iS zxTK~gpnu!V!Rfj^=Ybu;urep2eACWB&e;C8JEEws4v68-N*l_Uy@iLFE@|kh=c!)1 G|9=3JE;_sb literal 1547 zcmV+m2K4!fP)zDP|zLe&w{5nkwzm!-!$%}EfxH#S8EFPxjv%GlgcGP^c?!=mJ*PT~$ zV%=#qr`Dad^6#J1Es8YAau@SH^5;?nwfsixOiAQo5bYH_94 z&*@j6Sh?v;&%ZNsn$OGv6B!XQx4=YZrY151}+*)4^PqsSMoMiw$6A`_kZ(Kvtj}ya&aKTOWLw*ZjM=xX=M?#5t0=O*4NH zY5HfjpT7}g+{)GbVj-96B&vw5un^=+MRtE?NVI#&rKU)*Qb}KwecrL4qO>EB@Bm|Ia$o?p18Z2mBoz1 z%&b28cKeUMe*QbNceZhunU`)Bjxjo$1?grSW)}3(w@=;n+tSfs-~BcYGfUFU;?Z+g zv#73SbR_rNZ_DR7!)6Dwo6zpb6Y51MTpMvj=(;gT_{ zjn3vi`)z$}N~m_otV=NCFtaY+Y#0daWHyL5!)B(BdQIjYecLkX`Z*p(RoiNA7Hh^~ zW{WtpX^6Xv*(A=49nATC+jc#HsdmV06Jy3y@&> zvW<3g{iVsB_fC;=v^;MO^~I_BJ8;>Q3G(SX$Xb+5nJ}|AcWy*|=8ze7 z^K122o!r%5J{w|muwTq2L|R|q^0(PS;5iBGP4(q`Zbl?V2vZuT}!78rS8pWrnl zf&C8;C8Zkf&2K=kM-7i@<$#e1_M{HZ_$=^jh_QnmLW5^OuswCNGd>NBT(DheF(rW8 z9`Z~`HJtIof~{(Noc1rcoSMOifUhGfEvbefhA4-Rt7k)u4{Q|tmI{2;5MW#CWTadN#z!@}C=UoGl=*gX7Z|&Mi4=jE-%|E~!{Z*UZ47a#Qx0(Mi|ZY! zj*ojZwlQ`BKPvfO1Plm%Oc9v*jb$^q^?r|V8)yF)2s0SFf$vk4GW_6f^&ipRt|K#!oNW{8CJ9 x`~jxs_ur0P4~^Mm1fv7Eeog6fRsPwz{0Ar_*Nsb%7+?SZ002ovPDHLkV1g9d2-N@p diff --git a/generated/webwork/images/webwork-1229-image-1.svg b/generated/webwork/images/webwork-1229-image-1.svg new file mode 100644 index 000000000..4577334ea --- /dev/null +++ b/generated/webwork/images/webwork-1229-image-1.svg @@ -0,0 +1,200 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1229-image-1.tex b/generated/webwork/images/webwork-1229-image-1.tex new file mode 100755 index 000000000..912e69200 --- /dev/null +++ b/generated/webwork/images/webwork-1229-image-1.tex @@ -0,0 +1,148 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[ + xmin=-1, + xmax=7, + width=0.47\linewidth, + ymin=-1, + ymax=7, + xticks={2,4,6,8}, + minor xticks={1,...,9}, + yticks={2,4,6,8}, + minor yticks={1,...,9}, + ] + \addplot+[domain = 2:4.666]{3*(x-3)+2}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1230-image-1.pdf b/generated/webwork/images/webwork-1230-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..e952b4d8ed49cf11901fbccc6098084fb3bc2134 GIT binary patch literal 5741 zcmbVv2UHX7wss;tGzFvxhA15cl0ZV0qJSvUQK>=#0fIm%p@V>QB1H+I3sM9G=~cR5 zp(;|PNeAhmkYH3U{{LO;d}n>6%`q6tP39FDWmOz!ecZsb~t;C%q@(YBi;!DlUIPNs)D@mo*0xX$ZvAX zLJv2MJl#sZyEXytf6}#cKNP8sb7T3Qw6X+Ynru`Lz*2XpI-Mv<|)rv>U z6@?v?_JSRDzp}9!>-Cyc2Fs(h{D32@{}Cv4-0z2?>NHtHS&C{#NS*rffY+uoDd1)A zQcLjoYEx4###JBRc-NrLkcan6d#b0kntb0ycjE8|QIk**^Q_?1t>oZATKTWK7EJ62ReUgQT+^ zeFcqj`c*;LeLiE?j(%Mu5yAaAZ`*}fA4$v9A5ePeH%%gvgQ0>(76m{k*aVSw?5YJ<)s0bl#OsmMrpOE1X6I+jhF;7C&HkC=$Ws3WFMC3Q6X$=4E#|ux#=if3z86*Z>R=gYc(4pmZ2g>)N`DOT*(T9Qeq77bmhaeIY zEvI3YN)95?2i~Ua3`>FT(sn(NOv;^dobe|!rGg~OxyWadGNOuXEw|3`o(ZhQh=ysp z{(h29BZOvdxh24RrrGOcNF|p$tmq7DQi&9c!3Ypu7rhXxE1uk(nDahDALOLrf2UJD zeLDZOrYP~fkcw=&ktkUK{+I~^S5&>od2@-UuSc_e)HHe&n_^duMQX8AKMM?3ilEt& z((Es4iVz#i1h;{m{i5lznYQQ}f!G0Pm1FFT{G0e2An$@UE7(KPxko|FgW;WQULPk= zY|;$toy+Q18aL4lsn(lwo*%(hY?I-A`u^^pCD^4Ib~~2=`kr%8{f8%~3)@QfitWke zO!vy^&U4Q?1|^?$oVb^|S1PZ%u1|zp!yyBIotQH4-MjC6*AE@^K*vL*h82?ifaiemm1c{!<1+-;Zgu?6^#Lq<`Mvg^hj(8d_*RrZI5mD+Ufd?nsxZRhch~5= z{oI4JD}0nXodN$8TXq{$=k1{zQVuL~A?77L`NJsk$Z$&IwxAIVC}t;*Bnw+Fwz6w7UKnXt?1re-$;TI}m8gbJ`OS zvx~m(sJqVt;cgNMiFkRI%PKNCjQgQR>8XJVf2r&~$Q4_e;+t9L8EDJxYl98lMXMd= zVG0#}+zq!x1mjdU_<$?7b(RORY*WUnMM)nMYLrN(MjaJWh0zSI^fckO4W}$*9$4;t zG~j$$o%A+99l)jKqc`2uB2orbiDi-(#bS*6ve7{cFBYCiMRVQ{FovoR-HFb4Hl{Ik zpuk5X*O}?LP;ds7_xaOkQ^66-`rsW*v-i2)V!gcWO;h!X$S3-*in}KFyKWYX&2H4C z3NNFRq;f!WXtEtz^ z#Jcwy{VPRXGnEHWi_V?3$Tuu$Cq!Ca|sZek|U{Z;p?OeR($KsCU=q9jT%$ zt=FIH$NDF!dBZWt>1~$i5VbbM!;WmR2_5}68Y-IN$5|%gsffmL!Nr2PM-`g0#>c_q zeYoFrB9jm6Ko~ds|0ckbyFZi(ga0So373Wal{&2rJDt0rEUnZ__)T7+Sa!SU)4UR4 z9g}rlffGYjDgBib+8%62)Y0;(`8;$pO$HV8>FzX`Bq3wP#60}8^%x_Z%C*3 zrS6W`@826)jACtV%N%WOm$GaO&57wxjq-9ImzCH$VX@`;w)VERC39(+7R2*7R5lD#->A$JjX zO5tmVKb2E>W345b z^bXY6SB1glVDNUfm2#8s;PN~WJ-frM1rQs-JpUK*K>Q;f|8w65my`Qz08Vl@E*yHg zb@kH91janC{yq4i;bmiq8R@!Tp=poYZ1t`_VmN-FM+6@xEe$LUEVe$w1iz@;S8vZg z!y2$gB(YtWRSf%xu8_MSn{<=%GOmHQQKYuTllq2%*1~#lwQsAnWmZk?X`R1ei3ts< zf(Mb@$zFt<#*8$%G~;tNQ#Eg+O8kBUlr}s}{snOI{{ZekPGlqk`d1u*A*Vur0RgnM z03Z;^qOGf=qib;Knx3)3A0OdJP5ANp=ph}2A3zZRQqG^EfN3Z^wC4c;z~-4hcA5Q; zojLu-ikz5J790RtAdnI77Y&8VN<*RWYdCj*PpqR89wKfh0f8wh%3p@Sp-@GLhAYMs zYlm`!=%MgV7*`bj#BqU`;Owv%ygx)-#R-phSCWzO_4SoTxq3+A)RkqX_jJE~O;}+hGmA}7 z&rn%Zpm2-oi1geS8CC7(gb;A?$Y4PWfpM)_MP85~bXAQCqOwvssNB3Ep@dk1CnH6W zzL5J+I8YSaJuW*M2MjB+mrNFE-ep@5J0&*l)4eZUV0adBb6 z)Zw3Yyw2ou1qO$hP*pJAnSL%Ip3vPLPncs7F2ggZ#*iunXAjCQFsM`L&ip-|iFgrPHd>uWko&0SGjBGx+lxrDeTuO2- zJrzY_Cvidh^ zuSSP|5*|>OoF_}qqL)a*n1QYIM8=-Fm#QDyKVYnDh;3GrlEHMbzv!_%gru!2%80%t z5Bpd0NvpJwBc`}yUuFT-g0H*S>|n)M{jWf_)auw|hx)?=A!l|Xmljc1-r%)P)@7cx zMsU&+kvr+`Z4JY|kdV)B*8)5i6P9p0z#CYVk8|_A4WGOrMV&U zK65<7B^X#j=vOzsC@lt6rmQi{~4E)MALLS8xi>PPpJX z(xj>3;b8Msf4z6JeQ!ped zJM6vc^^LLl>Cl-JVtihvIIEE}5~fLqEr^-fsh%-RizjJEa**e}x6{ zq0);}7gwIJzIQ*j)Y);ZE>nZkR!owUWAZ$04xaI+>KFG-vV}1H#FFd>goIoz5z852 zQX;f-RSf!(D}kkQ`1BixbJ14Xlg`Edu8E8Hq3zr3kDOK!mqNE%7KI<@G6WQW*S59~ z#yC@5=sV;Fo6@)@0Q_-w=r$>h>b}gr%OO!(ic5A*S{)6$=1+bb?RG+tLrR;kXXZA) zHwH9s%j1<>?o<>G8)u3uf7*KGTP09Ab$*XdqD8}0>BJ{7(l^|lj2H|ctSEKO9FoM*d3>7;;dA}0LgCVfg6$ZLko>H&tMJwR0U*2h#9Wgtu!a} z#vzb-YIW`=&&Nm*y}%-D{Nb?4f@5D<+6t-w${ESTR9&YXtPZ;+CCg%jQvkmkuvK@9 zyPP3;MNB?Fw0}f9c~9~dvE4dq3e&>Nk^nV+sMUl$q;6EQBy$K@Fr3fmTQ}zlc*^Iw0&*9?8BIOJ zm$-X2cscjJQOBjD1U_tSZg z8PK0<$o&B_UMzG)lI{cB?Qv19G=429@pp9PAt`qUNL_gMb9s1`(HT|noH z5MY%}Y0BSA0R9(A^Z!b~%oyV!b3@ZoMFD|Uke5dxQ3ypvd$_$K%udnX0SbrP!C)AS z0vexnRe(8a+ zg+jQ(=ef_G=HlSI{$ar(uhe(%ht4;WOo$2KrK0dHG?%_%|3yMdaVTh7x-@LQYX6aq z+LH#pv=6F%DB0?^J2x)pxwnT27An!@{v8f4JjxUA=ZSFuA>`x~K_Vhr2HK$i0vE1o AHvj+t literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1230-image-1.png b/generated/webwork/images/webwork-1230-image-1.png index 70b2289ee8177a2e4efe3066187d83de6a38c7ab..7fd16cc31f819aa27095459f39c904bf14a2a03a 100644 GIT binary patch literal 27080 zcmcG$2UL_3P{eFkc^1rq@bW6Fo0y} zMnWT!1SE$BT5@cf9Gbs&qw~!<=f8K||E_!2tQp6C!`{1g?W(7qs(O8{T~(mp!MX#3 z!O$ySmc4<&&+pP zO2UsT1^p$z;n83!r(g2Z8PjfqIlPK>?hnP-PHjEE(x|EWpdjP5E9TtPjL;vE8_pj> z>FNLU+&Xfe_V`xyH`{F=GYZgzl^zhsF!yb7y|AYpvvipD$kr`iw-y-C9mmKSaEkw# za{lx^jLI9P7~0rA$^5P9PrLf~SZISD?@@jde*TZ%$J_pV^7o#D2^Y@1-1~0#=y~}) zY*}814vT$d+wtV$+k3~ifAf6wHuf5$+79&#cGvdJ?X=xpxV7gQ?b{dM#f53s>D`_& zM*Puixa}5A>x&KnI{W8t{qt7BF*}A!muK`~xEHEr>@!SDqLRCI_J_|L+Pa6U?EGs+ z-Dd$-j0{Iud@la+RpZ^B6AEL&8t>RfFHQY*@y>x|&$2i7jvu?ieAnlZ{PNqS4{AR& z)C5y@*llfjJSNw3DIihyL&Lo@S{!LlbI+5%KOVVfd!P`b8~9K*FEQioi^h-<2bNu4 zj|~lL1?nRXD<3iuN@B5LY4$4cKUkq`F!cAWPrEee%%9JfdKornZfn0Xp*V4LVvqga zRSt!U6aD^Hf%yhHd-)hN1I_We74-c{Ry!T7i_i6LoqiltYFQTE$2QZmB(bEpw)Mi3 zQm$K!uG_7ju|L~!@nk=%q5q86U{JnZVlKQZ6oa`(OSnWJ(fLc7LxyN&kb5L z{!#JKGoycv(nl({9`1Vew0iVs$a#azT?C^d;#gEqU9^b4m2>Mw`^)x|JK_?q z_Fogd_UQQFYoXT?uWwry=X;53N-DdiI_4~$C0!?--5&Mq?XwrM39^~8*%c~2X1rl3 z7wq21Un764`k?u7pYXG!E5xRy&#|9rJ4HH=uwOE5bea=)W?VO2!RQ`Q3>6NW3$G3P z6jprr#GS*t=xKw3OOG4InD`4wUpaBv`W@H%h08ft=oRafY81P!C|qH_lBdX^DJj#2U;VJ-ENi_QqR^SV-Lvtr7>mL-5O69{q!J#JyAE! zO@Xdeu*Ik4>?fi89VOpf)tgdQ)DsJjfIcNX^qb1(xYVG+ubOdlIusk_{x`*aU3$HN z_89GUT96g}M=w4vC$Ana4>mJ4Qb0<;RK~Fkd!3R8c5P#A_;dJRyn;Vx#CPRx75nJ9 z2-hUUnig97sAadLa2UCF@6yf1X8d zj9$6czJesfx@V!j>JjR(q2Hs-*zddj(7KmsvCrhDCAn6kMYJrp(cU+)(uWxHvp7rW z?|tn#U3E66^R@HCCmCIMRxhq@p-Xuhe<`&5rUFydy{ZEW!681)CdoDR_e?d1*Xt7{ zw{gU`RqQK>D6p^|wL4m{lCM!9ZPRSOFW;`He2Q<1uDhx=B|}B4YjlgGBL|&ioMV>c zO(~E0GDjz8br-2c(YgBhDe~K8i#f9er(yG9?Sa*H)n@+kjH>tn&H=(t8BPPvYpjQN z{b1nL$q^tkAUlPqi#oZT#vdG&@pU+hBuI`-GLzqrS5k7u_Lx(^rqWmx_^;B|l@ zTf?)2OEedwFEN)(mW`DUl{xi`^m9pvdw6f0DHBaNWpu2GH%s^t?>F8Q;rr)0c~{QQ zDf_D^JJnj863gXXu#2(LdC=`ztguE}djHJxS+1;((Y=9!BlmB(-muIvdSpLq;CuDT z1z&*{u6cu$%YD4}l%F=WCfmJl4wtE^7GsI%q9VU5GbS$ZB66fMyGV91o&HF3Kn`BB4gnRl7C9If1-oL}YI zfVK=PQneEXq} zo{P8DapKFr%hD23Z=^laA2d=|Zn4VIwbKr8t&59VE@yo4BCwsqEKU8x>6CRx;f0Z> z=aLHpodZ^8jU$XBtK!6##OBs3C?7vw&&~WmY9_rSmhJ21agMHyii#SHR?_MV^AC#* z`HN?oVzFVPnflONmn|`EWWRW*^;jkUTbzH4y;P5lms`^Or! zd}3n97^~#)qx9sitfFGEs%6My7Eo70UuYT zNt4k#JwL{tKjb`#x2qc0TorcoYkhe&Kr0<33vUH6a-Rt(F^+k(z;o|f#D z_To6{%0+p%_HnIwg^=TJxi4x5zkgTIq72z-!SRUIoGQ6*iUl zchKf%%qBZ2ua~liLWfebJvt~Y{5$rmU=I}QtrRT<)J7960|*PXPW4i5 z;Y$wLoVBMnw&P(_!Pw zH&ief=VKU*=Q9ju1%C1z!C>qJFqkm|4CY(}2E%L>Q+QnxF3=h&Dac|r&_8cVl0)E^ z?N*m>+h8zTPEr50@g4U$)BU(sU*% zp8u?$T%`CkcseaFCmMfHEHk=pQjg1CvD0=vhrvReV0YHEXD4&ILy@8z{~5w$ok>7; z+{=Xliypn5A(n+*`2~9`%x+$Iu(N3U8j(~i<0xpwZ%$$?gjINxq14B^q9Vs*49@0#^J)k!i21>M`c}YEiDIGF*<|aYLqxm zG7Vy^CRY~naO`xu4+;E<$(Nspr%s!hnH^N-8w--S@1k-irj?V=-`jhI!JZ8-p7FUM zs5|mmhjQN_=V@>9`q4}aMg1S$-KK|)4@q3-VCShx92&yj%B`@#kBQ&rsJi-EVX*A+ zEhm?t$Ns?*A>Nvu!!L5ij5;i&AC?e>uB)pv$(&vBe|#(TlX`Q$M$h#wwraU`+B5im z&65eKML)FerbXS*3_VIqyd>k$DsarK{zf9k)4X{h)`cs)rZUsGD#2_~nr4fq`D+0> zPxEV{(YV9-yyTRc0m+=k=6nvfGWy4~mUvxhc+6rflv-}@wCz8h0^2ovE% zYzPaoYCVK`b`~R0eoR7Jnsu^1ipEob4r5Zklsy<48Y<{PS!?vL|8>tLeEey`+{mTI zmKLUxl9FJo$D{;l>eE@wpEekSm4KMjs_#<)afw=VA7o_AU|*J( z-4*LZz!FE>|F$-jT&-T zh&gwoRAlQ@1$i0!Q3~E47ZlE?2~tD zo^oUE=Aa5g*xA0J^)dk`dd7vJqGwU^P;}meZ=JS%<%5&b?2L>76Ws2fqKg$4m0Edu0V7~qi|?3)7sagy+->~3_)%HLPVg+4p?AvK5Ah9Q zZEb25I~gbz?|Aoy^vZ*pVFY+Y*5BC3<2LSur*Ga`Jd~*0==#}Q9_x_8#fTO~@@Q?puX32SfNPYx7|j#F9;aS~w9?#@cfiq{3aL3P6Gw zgR6K)^0;oWS&c$&vPUP*176(KisfO|R979PfW5?cj$bubP!(qXR)5Ps=tCM`kg3cv z_CvWJA-oY$+{bA>t^Ml##PCEFNxa7PeRW{a12xrhQa63#_#$GBEiF6J%_> z0wfaW`-0o&zCJ}-!WEjry530>?{rFwf#PQ2;`-EZOQ~9GXFax&#d31w#FG2G2vcJ| zFVkpYN@I3+M;xV{Qj(W2dd>7gcyTiCUQ*dp85cRKP$Mp+(FIWt7b7FT6LRVpY#?LckP%YRr1=!zv=YAYid6AX*VA3VQ&}q%Tad4Ii-E~!4#_hs*6U+Tjh|ig zg(uzl;@nKkfAHzEGEPtEV~6V9GQrkN8162e$4T4I$tIm<-dG+Kvi?+**SGq9{8hQ2 zp2%d^E8P~uUOQ377y4;4Ujeu1d`iyyVR5XHpud-qk%z$RSKqvy<{(jr@ZF+!?3ABt zc!z(BgZn*EQOFUMZZzgh@W+(HM^u)Wbw#&sUy+07-zSg_gdnc_-OT9vC zo@h+w3T6?0DOkVoS}QfZKj(h#+VL)S4+^n<$&b?kA5^p>h@`(9IkxoGC3QfvH{ADU z@mpe(s>fbZ^Vu_bS#RlX$Z9%#jQ^k*5zwJ*W+gK zE5d521Euf{uAD(n+O;}^;4)p(%KJ%OZe;Yd&!5?u8oDy(Ez0Lwd)VrDvKn0_*&ZYl zZm)&O>I?7ETiQLY&tG6YM=Y$|NfdinaF#Xf9oKE?HKj5#qwer4PH2Ie2)3oyr*cMe&q+5B-Lv!vAa5li!+yDZ^&`LGpNSK&WP?bI((m4mOaE-Y<8hrTbq5cl`l{` zWr*Ix-eAeYChzQvvSQc6_aEloQxQFqU2scUpS7f9dv;{0D&dshV~V$&-N+Rg0}AjbYp-dv&KHMXiIATMy4 zCn#|zg1d#CgGx!ZxP`;ErpBNqTnFj^^Ed1^nq`7Z*gimD%%YvVq zj+8Wp>U&REBzxY3WKu*oc>00(j`JgL!Ou3#R{)q5GU8;2-JH!O1%Q zTz{YCAhtu8?DU*O9(XgDuG?1r3bMlKDF=vXT6*;3t+If>8jEp9qwHmv!Ep;%M*vFn zew#fQ&EfX74-?VC((rjOp(K08;gV|G(HF+Wyf>J(Kixlsv9jv|hn_i9-=KhC1$KkT zteetHwJyEP^QoE@I~!yS4KTGI7V5G0&b?+rBMt?Ez!C@FCmWUHW2ujV1x_&3bpwoi z!m_i3pS3m3$8^ymKrQ84DJCMAh4ZYCUacMxTuyMhk~}fxKC)(gok3*gI1eslbIE>b z$P2#ZkxbQ;tx-~SDi_0~7qY-q6LSl@4}+)F(U;e=*^Lo0yputy?bi|Joa9|{&uJM1 zA9fTE*D;uz0aSZ0SGn{xlC%=F1Fa5&*`GLiEt!QG{cmj&zzA_XdD}lmy6I#Hz>KCn z)aSFAh?lyekD_hBV6OSM2F85_k%342BFMqmn4hBMPX!U_NAN@g*QFwOvlY)haY3x$qfIv?`WA5pwpWd8m_CX@RfFBy+YzU%Z-du|#&N+vFIIvjAt1()F2G~ZX`Q?Ijte}>7t$OFA(r#{-I#&QI90-}Y;T%9|+hbzU= z-wZ^JT}Nsiv3(`@n@m2J+tfSE4}t1o1hfUQj(S|~tq7ILi~Rmqe7{Wvr_P z@`ap?U1tODbYJf>e@UHvWFW~__t8?VqxzBh20w^U6=ZRbxoWpyFnq}++5F(0GmfJ% z8XJb|BTk^P`?3V3ADI;ABqsN*wzR4FV-pKY@Qg_AM$|c2&3^-Qn{--cEpV)ITBH~4 z8};%pVf^2&NIFaE`2*E3s%yQ!#NHC(<&6ZhhX@fZ#xwO69O@B?6y~BfN9VhtEDC5l zYEAqZVh?8cs>zidrjGOc{`7=j(xuyif#}hQI;54oosBnV*2m0wK^Q%HL@E>%+>Mr# z?|j!~g}XV7(_LYHV3E~QCv-3MGTK*L-81mWuDwXK*Zg6pqIcUJaE;$Y7k<{y3Lw(J zXznFO=DOZ-S=ot1*)k)M(q(=K;%Usa*VF65oEuZ2H;t%&7tn^y7}!|q?;@?q*y1y% zdY?>-WwklNF}d^y8e%^ik-IB8$6L(v`#P{U1NYU;9hiB$!R9FUsA?`@63L5v)l5+`W!6JtqTO_Ab~+f92epbM4vq&t}+%g=5KJ>oRX zIho_qC04nb!(c}{5{!xso$kKVJXR^L zV{Yh<#T|KMy7v7l-h$5_i>t6WCMYgGXKI-dpGvBaJ$Z1w*@Ev!mxOS%dlk{oHvTj{ zc)D+hRGG}V|Fd1FOHam)k22TNOybG-IY4I_cY4X4)4k!zQh>xOl1~?NRhr*0{l*^I zpY_TFLK=>Mp6t~5q#3T+)YNq5-Yy^|Tt?rA^R09<j^%SroB>dfXlTX@n82u`Y?-!NefVc}2y>e5Vy{ca7v(pnxDrkY zvg~Bu6@g!R`g6w8=9S+Hy>ZM`_IqpU+FnjyspvDPSX+K7i0f`>xg|Axh1gy2DniUs zfOjae?pBH0`^GNqy*?ENYUeog`&>N~VaYifauswke;0qqsGB$b46C<0YTi<|jplk- zL;+bdxnaI4ej{IqlETf|u#unLx4K@N;AGv0x^`l12 zZCvyKowS60Qm+Y{Q*k*DM+}|QUUNy!nf3drHQc)4fU(gD^^%bgAD1Bl&h<{8+6&v9 zV&L-%yS?=cIr78Vl4Uk95^o^vi_)L=g;TMOU<$qkul`X`aJNndQt8@j{-&L?!ILK@ zYF97&Gd!>i6PBgwo*O9Lrw(p58S9oP_>V13uvQ;SV4*{o1fB{?XO-N3U{`QqD-zWX5io4c zZ$GinQRNFJ8iWdt6dZY?;KK~+YR?Bp?Y8^1aEou$Wx-;b?bN}-`GcYJ`V*Pyh%^?? z&fZgI-iVa+H1eMi>9zfk-NzE-O?9)>c~VWV&Sx;3t~l^!;`m9C-RRllS6oc>AzJsf zMyyt;1s)-WkMIWM z^z}t9A5aeMb|lH;^389r3hqY#8=(@J+im4Pyc4ukJ=oh;R~@WXqM_6KzqX?1eMvg2 zqxa0tO1=NJCjXvAYRdBBNbY(8i-wYlf=U;LMlUgZ`R&0<$6)ddd%c2q$(0$V* z{(8z`L5?b7v;kl>gJKT;6Y>9sUz?K(V9C%f7LYD5vR)&LgkIOyefX04ii?}BFMd{! z$eH%ZC(76NR_;6g`zfot_9JU9&@31SFwRXyPFA*fEpgczOHif?DLmAyMsCQ8;%j^e z3HCgrptP-tB$)lOs}F!v2u@GUYNO8nygv<+O8ORL?x#QKP3RFvL)DEAQQsboo+|#I z%Rv|a+j7vezh{jR+zo>996t}N1dwkDhCQt1`Eit@{{cGCR0Y@&eFyxh(M-1ky>Va_ zayQhCL+|Kl0{nYtfkVLi_M>#NvaS1uLZh1~<7Za_SdB9;$VSv(bIg;qGgfNUa;@ zb+4%qQyn-s$|XdT_TB#sL}vQ}9UE{?XZ#P&65JPMl$Ws{nf-*SjbWP&-m84rw{N<{ zPG$JdeS6132m4r>V*v63de!XRw7xlu&`(RsOmqk~sdx{$ejcBUT{R`)2&>v5y-C<4 zqdE4LU(zv`g$V{P4?Vu;Gu;sOYImdPWCp_sDxYb!#4KVxtm%4~<2hR-m5RDd)4 zXs6ogkx8E_3HLq2y{mB!(oS3h6F%M>a|s7&1pEx+AOQo(9N_>eWl)*9Z_s+;nb7Hd zqmZd#=czN{((QTgo_@?a?UTCx^0y5!>6OePZaZ>S!u$(kjz!BbmC^Q&)>hFKlD1`u z$mk?xrur$f+Z3M7^3ZyIdiqqMvt*2!l=uTe>gVJt1{P0u$HnqPnEXeC4a#C-LBUxp zWj=$A7IRG}f`@s&ianNvW7a%xy^4Dvf` z-|8_FCb(f+Bd5PnLSS1VE3s+n4@u~9R*nC1ik9&s0UqO3G<{KtoW~wsHH8qD)uKIL z2vN!q8nWPp&8g@O`9AKhW|_4PC(?=wfjN&d$D_Hn&bicvlO8r%>0~1(;$#Mr*j9;N zmBUhB@#6FLw~Kosy%$f9`aR!C3n+k-GoBgfC8G*&&mPpos8WlQ@viK{SY@G zfSYgIQ2fD%Gpw%@BTSB^SJ!4750Z6qTAOHCkbiU?2VzXeaTOLSB|hpu|AWuLuI*NV zkk5#ejIEvKh-dGY0oj_otiso^HMF7G9#_vTFS_>KV z(m+1~bmj92@|e@F^4O@WN}8LdEc3hZBEv!9rXDe^yh##0JdVyMEc#1c{7ytD%b6?m zn--0m^KxxMp8Kf^xjH)R1@JI3M7Mts`tb||4s(vx7&?Ar*(2Dpq}Cg4KiEVWtzCql zb-L&Jy9RLUQ<@3<9?K4^>oD$>*T^<|7BHgl1>t#s^VK0EzT)ecofQ zlE-p+W7fR6^_(h}ZTJy0?Mm^_vRj4rcCIr@i$$*N$OYK&Ud>(xZhP=aTZAQnWD5j| zjqbxV`}e;*Nabrv4;eB9$S74%|AS0kC&MD_zbr(f`_Ld9{!|-her3qbqO8HPH|wO) zQf!Bq)XH}|QWW{$0FDQID_*@tb3ZYpfK;AW)m6QG-T4@!l1%&Lmp_qxEw~5vK4W~t zgB!^Y=={Ie<2EB@x$?A7@Q%(eF>oKox3RNvAMnFqR*<&Z}92$Y&ALkN^>YCX-{LyePRtr{RJb$)p;r z<$`fR7OEZ>77uoqOO~8DNMCc3o{=hQZ9(bJC_1E0w34O6y!hV?3^nEWKNuJYtdUTz zz5Y$WpgsII0b@x8_ARiRa2xC~ ziO6k=hoiP@FLfw?2pE39j>`5mFc-HuLeLBOl*<{)^5wHvF8zU?uhsayZQ%FLQ@4#8 zw8n0$a+H-tRM>8$Liw+L5p=&$x1iqL3~LPa5r}g)V+}M}h&2@I{qyYS_PgR%qI?kv zX|0~KIo18dTMm}4#45|G$J50CCI11iXbXM=*oL(S(pWN4Xd7WuC?__@kGQhAgXaKf zS3~AX;PU3+u(q#<>iwuwK|uL?B+2y)oP*zJ4(+HxE;Z1Ya<=4kUp75ZIZc_}y1DrM zi8+a;wl3L5i>#^{pL{n+rZ-4eR@W{=8HKkVs0NbH9e)PhBt4PBHpeM{r>$s>e5soZ zQ}R*#7Ie#g79@LZ3FjSHeK(a#AbNgn_~Fg#ni-%dy z%JU0bk&Jzj(H*kV^G7%`VlnPyfH9Abk@!Jiqo)gQWJgw%o>!EW{Weu zB;={!4v7|X{>cinS(Rn>0^BdGj@g?L9bTyDBxFR<*<$={3(1QPByo>$`= z5ql{JEc?h9U%O^IC!5=69KkFW6%qT%ev z2bOXMTc%?U#>)Y#5Im{;Ea$$_bkQI;BO@t93~S1Q!gAqs^^aqe2z_aT!^Xuew-30v z9TmcA==yz951&qHdgZ;?{nJtMwoRCvY;C;Qo|-V0raPC;MElvrpAP7?LHU4>c{N6F za7SLLugC2Qw3(2FgpQe-`hv@Zj%~5e*Z30wPJ>UIrK1skoviDW*tyXQOqFoA79<`K zH3l*N>P8Q+4dtr#Fs6|xn$%pnx>%2u^eex$CZjd}(<>`iOzxx=w%$(N@hAC4P;`yg zSxj5_ebx6aN)nQhr;PssM8YK{QlX6z*eS*`7u~|NGYV$AH~~3D$;fDuXZh67Wu5+9#1s z{A(+cKG;jd9Hdw>+17oUL?09XsyK7q{?$v^z{LrV>6XomOuc zX#$=Ey_r#dfpsHs8WMJdb0@)c!rhxWkIi)~-}6Uy6=sI)DqPgeqvlKe2D*`i45)i$<>C zhZpM0jL&Uu6Mwt;TKjhLA85?l`=HUV>BrAg*YuBvyN8H(>f zZ=@@K+8`?Jo=f&7QsoRl3>*>^Kn}RrHx_1ALto`c0>|~6c0pH7P9oo~#L9(kdc#6Z z5RRP^$8)190q}XSuF}v7bofDQwh?~Asd5Vlu}1;bia_eh-#y*1C+c2)sx)gGff-71 z{@MboNN0j48f}lC#P_p)cb>T{ITKlxkZW%yoQ@NV>&X_$id{zKM=C|}eyCarO81Pq zh-?WEB+$~rT=JugU4PatsVzX>0dhu$VRy5a^H<|u@cT|dl@b~VDs23Urk%uL1{H`h z;Ho$a+tcsti;@#m8TB4T#XFF5xXIjJURsSl8QmpCXhE{FZ zSz;y0Mp!kIUckK!XfCO&T>6ge<}>r}P;R*2wrF92X#9k-PD_W`_1wu)-^?~0MOOLF|GBzw~dXYK6511Y1KHgAPY zy-b?hG>t@wL1K_#FZraNtU8R6y0WsT^mLfwmjYIkCpSW>0&-GVXlSMnxtuF=J*JhK z73(u1t$qKn@!Wjqt<<4XEs+qYATH2}SPZMk+~*~HFk3L%=X2~-;1IFno^)`4aZd?k z4Gk4}-AO}`KvtB?k`$DzDiA4h)#RGhUAP_pwLrRxs3bL7&}vE!%_t^!&hoqVI}9i< z(px93HZ3|q0s{YReN@qMk#nPgnSZ5wpmKVxX}Vrw`m=_0aPzWkOuYn$>dzlte5Z0W z>gsatju)|yvOy!W8Y zD6b@bS})31!ZO{rD7<7J@zPN18ZE7ok-_1F&%z0xD;ip4v#lku?R|aU1&6ib>-0C~ z7m4Qjt6H+MYYUbhd>ofI`rETfJ2pRYoEc69dfM(Cue{M(cZ_@YuL517l-NQgkSp#l@gtih zx>b-4LGF#_AiRwcN_=riZ7IMmnQzPwd3(1*t!w5aY2z%{4k>qytFxr%P4Ty!Bd3KH zTeAv2YcP%ft_1#B0)$OAktA!bKAASmMJ-DZU%M}7YZnd$F3majlZsMK3640r`V83y z>{d2L$EZT4pdX)~dW;F@sQr;g;yj#Mboimt36+$hQrLq#Fjl~DbF0Qtb03JP1-AkO zuC((3u8Xgq9_RF(1yYjud^p}4D*bkCg99%R`zz)KG#nl9FIUT)|c<#Dgs$J&k5;?fLHVUsNhub zy<_HLKGdx48PaL~Csdj)q!!iceNamodlgrIVp^*LihgF5R8}@}7cLM<-SLs+j z`wb#W$W_2QE@cMVEy|M@dwqfSfcN~DhzMSO?b#+)^WRvTja=%=D?;AUQ@~UdA*p|- zvLQiT7W1!EHhTHZ|C!2$+YIZeE|8kMoyvwPg4ULI1I_bf;AGLvPiT-E(;-A;VuGBI z?7Q)lg#OlVOEYWH=-ABL^?1Q@;4WFHk3^y}t6|FdS8jK#v7w~tVH|Zf2ZO+OAzm3t z03K0;ZN|aeHlW3{`U|4As)@2u6GaGPaX0t07^^9C7#18EU;z(LkL9rA^rZNIVDkv@6trFagulyx<1g`R{_MIeiY8$+$!<6jtDeXe{L+&dT+ePJ>X1 zJzDxv^J|**D@F+Kz1!%7p=tvJ2s3Lw5K%c@{~!jD3h#G?`h4_1|H5*ZA-#w^hlNTS z!bTx;^F7uy6Ce?GEC0=&!- zb+_Q5gjEYOolz}2sgV(JyT)PZ1(H4x0MqWnIK8=?sHo^q>L|wD->#E;?BEeB=%pD;4QeOJ{53$klpt`h4()!JUgX=Y|r}bFqm2LvXz*KXf zo^$1Afp>R72m1k&n`-yw!(=TpmX&`(L4mkKy~VL6nMxu_$EiuE&6He@lhXRpc7V@0 zgDwy%2t?$1OqJ-rplS&Dr%wZ`!f&8*h^wl23F77& znCsf;5dblznCYa3ogTFrvWem3jAoW9If^9rx{&w!IhLF}=WgZJn7fROcYVGj@R2*A z9@>cfcob*rR-KTwWGaIbhZ-)R##qRq4h~*ZNrx%9~~XkOqcW zw+OKYi(^{fxw-ZNo1q-pfr=7aQ*P^|g2D(yTkrMx;(a_5SfM1>Zoz0JNqgS{YtU{{ zNdjYJE)2HL9M0#Bqr^X)*9W_VF_aDl8{^ogEpuW_x7m-h#7~&O(W2w!$_0fs#^q!l zlL>w(Vmrecrn!;8kDQMPY5`iCjWQt%ndRCM z{WstMJ08K6XUn(XBv?3NxwE8%HqxEojD>8VL;NS15xuAjx7hk=t;RgH21VAtt^l2Y z<`qd&YO8vb3-%Xd$ie6_-|V`UD&yv+!KcgFZvmNejO^>dl@(&fx!GpcYH$%h+SGK; z)!rm4bD}5WPN`OKb84|7AcvM!n{|U5{s0dQ{{b!#*P>Q=_XUIlgbP&W`v9gx4MwP40qUwRk=Birpu(LILZ)-91Zc&9DrOgX z%VuXQQ}lCDoa-H}tx(90dJhuUPzwVV%;lsm|^J7$;lt{q7{uvMry2 zRc!c7Equ)PhtZ6i6HgvnUujorfqMPOP+SOP7BB|&`8pOsF{gvPKj^PhZZAtXK?4#7 z!$MESdGHKON5u*aQOEES!HL_RaDinq7qbx&^UXs?C*N$Y14k+!l!?3 zbi5|J!xa+v-gC*VSGx$+_GU<8+mXaj!-DtQ^>1_yCHvZpFO}rlkG>-quhvUJ#Xa_e zF4ne_q6ZC9XnXxj)ZAQeSrRgbvn-cdp}6)8L2?f?Zdqw<=%Mv^o;5szH;1MTXu>(S zvP{qBAP-fzq=~pRl-Y(ht%Ab?++KW-3BNkV%n0&z&{~9Mf=YB*yCX{o2UiGUFZB4a zunoYXLT8dOl+;rDBS0m%j7F55peKNTZFyg_^LDDl46Z7^PeJ;Y0NU0l=FG;~1*uq6 zGb;yp9gz|@TKtTl*#~K}(8A(gbbM-0z&$#mpJC4mCd})T{o2q9ON%a}KY$2X%-Z5xxx(2n}dTMXN53v5E_N4j2k&hGwp(`68)(7|_ zZ|^5GC7)j&MklC*3q?Aw;RStrfh(GS67HeT_gx9Q(vqCHJnrJUeqYphel83j%$^wN!>%OS`p#W!Uw6d3u8~%`nwn zAXFc7mdFUIiD~{RHSMHYygU>g;_kLQJD6$KoCFbMi{DA&^6*B6qjN+2RY&47d01}c zCXTiGXTn(#S<_mc;ZqeqW1dIE2n~hU;9VcpU(|^b8lolw9-+TRXs0IKPmdFnahgs0 z>Pl`nw$?dNsZRR-G~S!ReS}lt?%ly_5gs!%HN-@m8tTXap33D|@euT$`XWnuR^+RU z>#TvStg9LEzKjY{a+N2esyd`#=^~L1YBOu&If69xoK91gH*7UG`+(#Bj5{8bOd-Zs z4R0)GRAe1B5z?rto?ViOqDV@wy$vP#=xduCHvUnKE=gF+u+u= ziCkvZTXoht;x}YSGvSmMkV1nj&TgC*BVmbuLqvK#8pnFlGOr{#C&kL+Zz8|(^}({G zo`89qp$!|T0YN={(8P%ABST*-kdXYMQnLo&ib(MrHEfK6Q*{K?e5R`^y%$KygeDSHPaBcXh+k5>Eoc6y+gE~$f`-Fn(eBSci`up&KQ@8pWMIIlW zE)fhnb=zpqz7HQFILSkGLf(qq?yHktR@GHOV*Zgc5jRVjXAO^O~!ihk&urTwtr=z-IaYwWF|_{ncAf!1JXxC~zp*BL*+ zH}_N|62P-je}b&6EgSw1!$IbAA*rdUet%jD3rB9-ImA}>hwtswQb4qYd)3uEe7K4Q-Q?yKx2vK=zbT^U*+I?b0!aW)-o6Tjf%9 z^1r_iQLdiMBXERQs&HM^o|hSaI# zUt7{GJ3G7U1y+Z-Cg^wGO3im}OYz?F^v&_S?ng?mkJ54vd@@u{_yLuQ0PFrS;#5=t zX`%loKk7Ju#K1`j{)s6iGeyvR^EYr7nYQ#v>S>T8`<+HE42Y)>X}YYan-h3sdhaKw z_y={`_7zOC((dcrWWnd3@W@g-P;}>!4+UhkYkU5LHPM)J{rOTdF00d;ra?2JQM!|Y zTfgnPu}TP%c&MKcbgbWhS=da*S#kcCug8Td`F{hk>dcl3$0$sW7rIs&S5Hfk7 zMey!59gWSd*V4lg*MJDN?($VtKwTsGlwQZxBkr_6>Yl0Oq^<1z%uZfcsE%mnu-)wy7S_~mVub+PfNTyOwUWzc+X`^7T$U#*Q6 zqLv0avzQ@F*gkcVq6naOu7vB7Q1CRq8{JarEIfDW@=a^(>&>M?;O0P0y9 zdAk2tHWb!fRxf(}(&qtCb4`mG0pK`PoTH(&hJC%V52}z(+4tr5=YRR^GO@V9+@Seg zp=@nz=ST2DC_eze^jLA-SEGn4>MBd6&%EEZQQoJda;=}p$T1v{OGmI!%JJu`@ZCu_ z)Dtt>SPK}aFSA|lY{MS#3yI5k6ZC9Z{hgCX&B(>Jzti2Dx*oW=UP5#rf~AheOviqJ zl|WsHuv6ZK0;9%vem0;6`lB92u3wbQ^ejXqt-7tc4Ybxq6kZUu2`d^FJu7$1&vR}k zr=kvgc^PcC#17F<+QbW0?+}03MKO3<_2)JzGi(f0RnhN)XFH`9Rz3FM$P&1H)$fK+ zjwFtn6n#D^@aSXPo$!0-|3dAf&>Y>#_e3cLJSWxJ0(?erXErM%)3m6VcF!TPLt+u> z!zX3P*t?D2zwcqBp?xB~cS!R^ri*7FYT)m3m_T6wm`2oewdQql(RuQZefNQLM2HVQ zM{t)#r}aOngt!PCp|8}s-1@1R6Lz*%NMCGM(hmi88L54KA=ToIqShLG`k}8fxZp2P0Mx3B#8~4P(OT8=)w~FBh7xw4 zkSCG0Qc_>>SZYFQm-2CJP}Xu+A~Xc3@6>Nr2$^9ef8=;~u0hADv=eJgWoi?zb1J{IuGEoB7_Ix= z-D)4yt!i_T>u54ZgX!p>nz`u&oU%empWFO>S=iO^jUaTCkaEl$PEg|a9b9F@11x80 zKNJo}5L^Yqv9X~BTAZ2;l1uF)A?D#;Tf#$X1{18y4^+`h>R;g0-IiX|7D1Tl)rXE} znKS|!z73%_``E5PjE`k?-luo}B^c=gDmm`E4Ly}R-3l8Fh^DLuj zvFDS7wVnD)MlsTp`Clp|_4}`ka2s+^zpT(ErsESR_5$B*)mEAD@uM#G&J{xQd)BrM ziqNAwz)7JdGtWFfRnJzpDkwTI1PV}Poz(aa z0Zui&a((I(?X?kWB2U#-CUd=f)dh&7gA{d6EMcZXg!Tl zaIWjks4KmsfLOea7<_62e2@RbAkknr!4SBL`)yHn>_&AWU%q(zLRXAs?;9d5Qr4UY z5EWHKNyu=b4{WSZg7j@q$b+Lm;Q|Wnp|}0nW9#ad_fff`k_S3I(8p-)!@w+Az@hbh zN0(NGF&)*GTrJSyqs;Pebds6Ci{J`T12+0-3EEhMv`~VUethf&b7^ZM;hId5>G)B8 zcH@U9BNsEb%iti=@$MHJ>wpwey-}^>2RIOu|B%)f%*|yiJvJ}hO)9hJpS$yz&pKna|e_RJnFxWRl?AaG(u!CnB~8sk}{qEe<>)FZ1q-Z(S}xwRT6!qLS)Q>LI6olBEwYGCm!~n*Xj^0tV_= zP4H5SFVsfmAdc>b`fq9YD@HZLCt(21GH*wJ{C^>p|3g-b3iLKfF)GlbKK}pPkDIhW zFVkIl2*b=*1CtY%D_cSJ&a4z5#;ZSl<(`g*KEx1Se_;F{;o&m}ivJs2CZZ=X8-S|u zfr$FA=i9 zKGH-9eHh0kKls59l{eN0L$`nwkc(<4_Rit?41QYLhiyBs9pQ#HuH0vBtbM%8^q zMcD+~@51=A)NeA}-JX3g!{f~nTDrO+C|7H0`d{5$dpOkT9-mTedm^jO=2Bv9skU+n ziCk)pIw*E)QNqOJei@-$GR9`BDV3!}ZqwnEC)X~JiBL~XU}ua`EM}4-}4*Z_x*mqpBrk^3W2=W(L3vwEMfH?n8&lH?pKSCB9xBl zq9iDiXteKz1#_zU^zalQHt9-1CsWIL1JVW~uc*}E(KeF3klx>4ovy^}>je_wP}YSV zA5-l#M777eX}#H{FtO;0^s@V#Je`!Na!Uo8NiSqi8rwGOCY27AZ1yki$e%SV@M&Kq zB_(BQIw4y=QvK6QfBs8#ik{KLBdeN;*-80gwWd z{Yqp*OIp3vKv!QBVcZ*Z8_&ERTG8sE1)J>zK#JIyKfoVT%q)w?KFX{O=1H3P336*H1Hmr~TH^2yzc->u6?@ zhHcj??cS0oqGl~&VaaqC;{HvTZ~dmnYio%jf`KNZaDp{c__hdR5`uEft;hCahK9xv zeVM*VbC^>ZHtg#f>U0~2Ow8)YiU6UJDZa|<98+`6OIc4?c#vLnXSa9-w*xwj;$u*= zquRM52e$gBk#lQwPsqCht!ae{H0~%*2iC3-kAj#`eKr{BCSKvFL4FBrI0PtzA{Bi%1Xbn?;Usrg+peLav<0 zzRKTYWPZ4Faw-Ryr_DQu64*WFc6N44E`jAPT)j-#WJ;TQA&+|uh!d>42CON#RZ%+toFK2y^Yr3@xY3E0OlnbjfoD@5Ss1T6 z`Me`Fe$MxdWIwmH>c#1LRfxbmV7uPj&{yqgvZDb4A%|5yuu=bk!aHwy{{Hp#dloXP zNEho|1&||Bt&CA{!)JlisDWHH@6QlT5Yg5Gy)r`jmLk;YfwvdTP6(Y5|1{_!;<&6G z=DutzAlEj>WaeG~5=LPOqVSGxr^9THM2~x!aMDWYM~j47cMSdhKOsl4GuID18SlJ1fdtqwt^YVk=e!^J-B z1@7+dMLYpz6tgOjyJh6tkVBf?RN6ya?L>StS~dqE;X&k`2O=bpe$x{#VV;T<=yg{9>L#E06cBxKZu8T&Lz8pl!`Mcw~< z<~P9=*dhQM?E&NfYNfoJ-HnSVW_!UhI(wBh_3Q^4YWk{1cHSq23ctwH6Pu)ds0irN zSp>@i{nh6fCceJq@|isuaArT0 z+iOdck~II_V9JacHpfgK0D)wh11mM)ts(|xG<>s98m!GK z%+m%z?H!x_gXF@nbt)?1Z)(OQk=Z7(hW?dxr;=D#Q>b5;mX|kop2uCoGx}6(xsMX> zqpU5)_mIZ=Bi5W{gSbYRY5|X7S>~`xVrG3zcK_HhKj=C!5F_(m7qKsmi@9T1o0YVa zj{mB)1j#*ap9#5L%f6Y|yvz+&VYkCNq+h@qn*Xjo98FN5QNReVAUc9LI8SL^I+#XqizWD4ZgQMNQinO?@Vg6*W@eV<*G> z{6Iv)eoATv{(?${2T+wn-6@Lt+!qO-18E~vR3K2*Dzp?a$jI^wIkfS2k+2kZ%i5FcGUbkF&c%r_3Q{r5NiaTiV4(fZU{jb-jtVR3M*~RT?O?icVRUqNk;u zR9+62atZ^2?Xt)@3L)NmCCgeE9c2OzwVrM~YR|tFTICRYUTK+$3l>z9pZvMg1ji^UQgq7$-SprDOTr169GyWzw56gcBL@=0;c&&(qwzO`;BfQ z*DBmUJEHv~Is4^SM7KiL4!tlnswXWRV78I0&l{gQc;+Pb(mitntq(i@oj<;U{YpU* zmwNzt;-3WR zwGhqQbDT5nT$hj_Dyk5VrzS8VrpM>&z7F?&qtVr+0Ex?WZUZ_a*v8RwLTVS_rsp6Z zS#lYow&pA>urDef;3B9axq~;3^G8 zM6Cff15CsbHb=S+E9PwnhUrB?9PKhH5^(66am8T6pS|JcRJqXCO#OO zxbukEBER@bicS#7713w3sG+@o9X0QD9o&4w6*IV3w}QQBFI*<8#^8COlkTqz9$4ox7(I7I#w(eNuZDFt?1H7ZHPG}>Ap933eIm&(#NWfs zJrHN|?v?Ux|3D8{mtf_CB-g+oCw~{^Z5o=IOG~Qr;X=)K7ans7Ao_dw1$vNtaX-5t z;K1ImmCZ~{l)v#Oxq5iJtdv{;n;f?iUF>(vRo}+NKLB0$4Svs7O*Nf^+wCgg$2c=n LOMLFvM^FC~@5VHU literal 2187 zcmV;62z2*}P)yPlh|ckd_Yk}855KLH2rJ#_QBE=$)VM66rLN9^?CXdgtw+YXy z95knlCu200US6y@MnO2u~W%ZVQ|KVo+urp88J5lm99Vi`g3TI#^ zsK!EDVdqPOdkE9zOdXHlvUS*up%Ygg-$Ux8OqDSzTL-oN(g#>2tSW(cQ@(jcR^c;~ z@fb$RL>`R#;HFMyAy?B!H18lw`yI$)4>GZZb7??-xQ-EXk?A0dysNd>!D}{hPb%vy z%PgD|54~N?etZzm7Dzp%hv6XIZS?Fdrf=)F=g0@oK{hQ_*Ta0F_>fa)c~S3%9!1C^ zl#oQmh$d)Nv3o)!Uv+0;K&KT(7Ljr#>iiN-J=K%sBKJAS;Ey0gJ#ydqpz`V*Q*IhL zi+##iWrS0G`#9ts!26K1y{Bh;Vx9MmRbO4t44BT_%=sCXIV{+zTjW*X8Spb}WJO2R znSnPJ&O>HQ#qNxAx#L_Oa(P_V+IIbAppWxDzr*uB$yuSfeTy1PvR5Sxo?FM{z2qb; z%a2_F?^x6awae2t8jG;3KjT^@I0+_ylEv;(6gYrteQZId1SQLo!avD)VSK!(?Pm74 zfpr|*P+xTx-K#o7p{h(sCZ+}}5=QFiFX9T96?%uTSPSPDyAwhdg{6|lew})q=dt2M zeJ8Kv4ztsbZ?8V5zZgce&M%{$+;{8D^6XiJ#^;PG3vZpXguVA!w05}3*0WR-&NtN7 zxew_%M>s%Pbrylt4B=NAN_8TG0D0>>6+7X#h5O9O5w=DQj{#HCU+54icC^lYZYS{< znhc?|Ll#oi_SrRuz^lVrvuf$q6OI4=$h75fj=bME9q z%ISZlI`0!^JBl50`o8y!%p*|%oPT%SLyMa&SPG*=dfT{a{mH)RulWS{jiO2y4bSK z)kKt@PySXWEQ@eaMZm`vu5gU_Qd)h$b~o6EKB!e!ExQ2oY2^0f;S+#KVlB`~N}|1g4Cz6F4mII1S#pcl`o1Ie~&YFxgq(QsYY1gxBg0@erm z*#?L0vdKF3C)U_38MZ0ZSwpu_h183y`C~=(FjqxyTE`3<`A~BjsKW~K96LtbE%3oa zr@Qg9f#)0fQ=1wWX)kvC7|BI`!R#M;f4S)WsP_+I_~iWnP9SFR28d2a@P2CGLTpGN%HSpw|-!vNqUIO~vG|VGl7krLIEMQMSzl##9 zlqX=)+E=ST*2&5z@15|zs7*kx8dwzGACukGA#Ua}<+i}GPm9be8=KQ+T}FUy25g>c z{u=USHv#>Q@7wF+eaODPnU14F!j_up_^nX33Gk~K0e1n{5Pk87H!-^EKT7Ni=yRQh z3;ehfJD{()O%?Lny_U+VVR6qO(JMRdN@bOf*^cp8sNXsrE2;yo(QW{^=qBY?!No1@ z_^sC4_<-{XokqECC#nZlVRJ*^%;uVXBnFlz>e%?mXykl^j_a*xrp=5%4vGD^HF0%c z*S5Z5GaO_R<%Sm)Um7i}dn;P^|^XcC%3V>K2rY`_l zgtoq#b42C%<{E5jSx1F|zv@7y;lCFYdnM*=Gu5xC3A5h + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1230-image-1.tex b/generated/webwork/images/webwork-1230-image-1.tex new file mode 100755 index 000000000..060236b31 --- /dev/null +++ b/generated/webwork/images/webwork-1230-image-1.tex @@ -0,0 +1,140 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis} + \addplot[soliddot] coordinates {(1,4)}; + \addplot {2/5*(x-1)+4}; + \addplot[guideline] coordinates {(1,4) (6,4) (6,6)}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1231-image-1.pdf b/generated/webwork/images/webwork-1231-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..cc7190263366b89b6e8fea5d256a0c5246d31c36 GIT binary patch literal 12204 zcma)hWpG{1vYo}u%*=3fj+iZGMvIx5Sr%H%%*<>tGc#Lkv1Bo`;xF}5_r2uiC4Y8J z&2-JKT0PxsbvKows5ldlnH`>L|LXh+o{N-))ZWMno{tY8ZfWCU>I4wCF?2B%H8r+3 zF$Kt)+L^mpkaB=PAOQh*XBQ_^LtA){wR7Ex7{(FD%~u91OMvF|BLqvMMK5;VW)rLj znZ+J8ZQn9(9Os8`ZrG&9{+%Sg?gF$NDke8;se+h^sHrI*sM7WQIzppY@cQ&5qHT*Q z_0Kc(e(KYsd+vv&Q}@X=F_@q+v%6G$(fhR#Icu%M5%=l6K2G1$HqPU=;4UcTr~g%x zzCiiGj2?cCzHZMmlb_FzO8;>G_tof!mtVOHPo|%@TTvtdQuOyR&ceOfM&H0YyRp$v zCfwoQ0b8^mET~bRAH2TAPunejq&>F`n~CwBf?$*_-1~Er4PNfP`1$K@XO?D-J3LAc z&&F=+_O$s;toTWgAP&Rquw4SZaCs1)UuTuZ`JF7jX}&!6Roq^54Kg#nzZ%>-J+8K7 z+h1G;u^(<~$*TQx8j-&Pos{2?$mQv{)fBanRN6aM=*haLn(^k_Fy1LsQ%oxtkhn6Q z{hD^F10k=M>n>kj;f=)Q`E#SZDG!-rXQ!A2NKm@T{P&FBz3P`;8W_!GT86#9lhwz| zdsd?P$;8U!?ONXtPwsLLd0uhF>8Ds<8pwU}A5 z<()uK_?;Se>wo{|?>Fsh$KmUgD?nK&X^zBEdF^#y-qjU>IXj*<#EytHM6(T@jE&AK7J2e;AH` z!N2fTVQF-y9NMb_+E;=7?|F*I{`R&91ixBgOey|x`9&W~g*SZE%b!1{cux+2 zYlH*)c9u+4(967otUKO00prdKeP~2hvRxxNWqH?U&7KAM!;RX%(?jVO+E#!>6_X3G z{wI#H&oyu}Q)uh;p$FpDX(;F8ch=RQcjZc1lD*^;kxtP2c$I7q3&X;Qm+6aQt!z&W z!vbJ$oX@CRIX2E-r{{uT1@Jn2U%%Zda7VB*0=M_VVyKe~fxA-rW%%B4s8yf}cjX9t z_`dwRqc0(-21SIU#Yf&C@sF?cReGsliTB{vE;qs}(b`~y*QBr^v9nCt$Oif$&^Xw0 z64x;2=hw0$M#T$9^JMRhZwJKgGHoN9$`_6>Nv{HET+4wTi7FTf`P{83dw#}kg;cV$ zu5=guoQ|Bt$NC?8V*R(#Y2dZ}kds{2L%5ebGGVB+!XE}Ce#W_e3PJ1cTa`uIg1Fl# zl^9AE%A)E4_Q*1@kaQynQ*_ff-bt5GZ#x2L`aZV>^TUrp4DZ&wR+qdwpSLkueIg1B zP#6BG>L*rd{M}t2AGwOpn?H>=w*5YNzT5Y_PHhigx%a$3XsihOy`D7hc@s3AKN&kp zjTrIWtaMPm-|3)3(2G|9dzNXQX^s=4>eNP7UW?u!edps;lV0ux$_0gTNxR^dLv$HovhW06i3rZrAGul|3+D3@60r8`as#gnld`@DkI(33_pmnEtMdEUl%Ms6>Rp652NQM*4WOi+w z&XKRGWo@*t9k?D6yGm+Pz7H`!bL%F>$ksQREaMJKlN)YAH0v&NTx<9W$f-;3({i?|NL*4|EADzn~$i(-HRl?Op0(E+~gUecPj9>4$M zODpj*BB~a^xDV-LY;h4UXg*df)EkHDGrON)3gVup6L-0IJES&sMqdwUPu+bAjdwjGXp>~g$NxJBxgEU&!qHe$TzHv)>c z4==(gX62j*iwwlt6Bs&UYJqpn%ZK$y4&%+-SY@!v}AWJ75vIo@4))mfhF%{=y7@S8~C>d5(viJF;Y)x0x>p;!Np7dtX)i%sT)|kw1#|n zlpHBYZ0Op!3N2F4FP>q;tgo^gl~Whj>&n@j#0+@2U1VER0j9iBqjn%nPBMdG@cwqz zaf^+fYF3pf>7Lwp+4x9NIxepGOQ9Oa$oTn&EEq2C`DZZ) z;e(DR`h1fM3os}?Z2vD3+MN0}v^j_yFT0S`US*>J@=ZoQO)>LIQ}~HbMvrbuiPp$u z3f1pAT1d25!tgB*Vnl@TCE{KVanSSO1%uyA%em*#1J{>XsSG3LJXn=Ef@>TV>8N5# zMMFd%Z=Ec%8e@q{Qh)HZTxBZ_5&ULw4VG5n&8pK#Y)Q>xpT>*5ha1mR}cF5D4fI&dN>TG(+O;OHYc7=1A?k345n{;~sKhUhirJO10VW3`x zuh(JZ6zh={Q+~)5UYioxGD&}gaic!bibB|jhUz4iq&b3)IXtdUpG{?zmYTiCv#ZV= zlrq?S8ga~m4_JbR=95|)r%;288nCEa)*B`5g zGDYBpbCG0GlGdqFslS{3IAV;$7M~s2mPQBaBBGUUoRlKJc?*azXFAX%ru|uD#7HH* zfw?{bE-oZAt>h?pvY(n-7FV{#J|kQorvgFc>A6@zPexLKwLWw}TwE9dl;4e9Y9LJ) znMh~w#|M8>#b!MOjB-B?YT;^V!D=`wHtd$6Vy^aw!$e9R zg^irzv?Q%-DdMZgHf8aq4QHLymxhIRayda-R<6r|ao$cUMu|tx>acFYDimvkOA!5# z91!Kzw{s|h%me{w(CN&77j0SML;+g<7Y6qJ+KHyX}9#lt4072Ryhq=@0-$ z{h#EZ`&zfr&;*tlFvN@Hir0|Cp~EkZTzl_oBz8i-OCjdPupl1loU0@WXO-{$d;Ia3oCuAkw zy9m?JB1;EyWc2&lPx5xLRi;nGcw`&x?Gd7unnDh#s40Kg&YI=ZY1ot+duNHeK7l|Sq2f&t7)U+lkud`0f20C(Pzy^_!3qS=(l5hu;CX01_LtOJF18bXc7@WrF zkps*a7Cd_ru^c`55~6dDl`3T0^>MxYqK&Q?E*sRSqB&Pg(Hb?27(H;4+TJw?HblPt2N4^(iADybHF)A zi|AkRQDb=p>1>G515**bC69}ID2oion)BJGtw0sr8iU?zvsH^}v%=PNKZi_jn|TXi z0FH#d)KP|(wu_ss>VU32a8{P@eGqXjZHTl0oWf0fvT8|YCtva0uXh7UBYcaQvDe_n zj5M%i%VdalKW9_sO;{&LzA(C+GV~6Zif#cUOVxgHx9ceT8q|D9G>#pHi^`w?mMs2k zJW=h@6~PY>6qQg@$r{8dESBf+o+LP<_To~}*?evGh7ZlB%*YZ^vxB}9)ofVFS`t z9fvOti~9-2!h?pA%yRO=^%mA)(wwSoZ4Jm7laeWlwLBl?>U3ibTG@gKll8FWy%S|b z@tcz`?Q-%KX~wBM!{gg~fEu+r^_#QndoHR!M?||d?U49^<2u6K^dUSSY*(wOVndzzj=zr2tdsHQ6~IT8z|u+ z6&XHdTeG)3!)2nZL;GD?OnAxRt{)y%=rFNhYW0f}HS9H5<9&^~>U7A8Cm|hTOQ^P~ z_gzxhiuqzc=DvHzM(bUe z+qLW@2`4W8;6<6YI<6RzqbhYoCUDirAAjG+UlDg3fyKzSE1$nQTT*RLJz!CK?LD1=)L>VgjSh0*%y5m$4vhyEd}nLjbPAk8%UN0Lhx zR?tyj51GOFQ!jf4J-l1=Sl;|5qt7zaKF39&u}YkMRZm79Rs2GdF3mV8!Go?%O~Rra z)x1sKQdEwzU*SMDtC~rN#tYJ21)m@kr8A7(UX`0yk&faZ9Q-W+1m!51E zUsE&^^fI3U`x+Ho-@!(Xj=I*Qi#p#px+p(XrHL7kLO^KvMRp&2RE1kqk{l^0SD>xX zGy`jW=8CweFat<^wPfdsELV3sI*q4%5DMvu7Hg55C1fxe#3FO37`5g+Q?+C(cDVg? z9PxGLzDW|6yq=q!==ze|1u$(`;CX}qZeD>o!9_`3@ivj~CBmz{Dx#W$SFkHk_sadRU^Qb5Za)_QbYI28@aqPTgE zIkzbO|9r&eYa++jsOiU0V4z`39Gez)lRRlx=<-n%o0fqA@>Ne$dj~nuqyAD2Ygp%T z=+xfuL+B;iUR3*A_+cn+mxCRFKZFB@-a*!?4J*KtD$AMYIEyse?N@DZXX8Nq1k~_K zfKtRt_3G`-fQu^Y$lr>`Ze80b>2}Q6oe4ZS1vRKut-)tuK^#~OLqq9OWpe&#j2t@( za4G5-;ffq6iVgh6hfGfdu5W(zS2Y)S9Z3*`)=O4eTm!>xubBJtb?UqJ+DHCso#?+> zrz`5z=^F_L`LE~&`eyGQo zqFx4w*zb15VQP%Uk`t~SZ{~rX7ekDa4i#0bZW!cZ@M#^xe~5-?0`$vKDNCiy>#|tZ zh)XUaRv2*gZAI8Bw2aUdyxq+XOd<%LB#Cd-y{@C=dyC^AalXKhUdmr_bMh3(*&9zv z)i%i0P{zO2-X28F+9EavFL094 zlG_XWzar{%j!Tcoa-%9wY)4W8)!ZzexiH^#TcSj1Q!Mema zVG-@<06E?qQ(FIIYD{ul%oLTRqev{XqY8dRD$=dgRON3-Cp$NSJQM}Pdjhn@<}23A zFwDoXj1G8YN7xwTb}J8@PE`(gQqM5F%1{s6@58rsy?B4i*Rr|FFYi_vbLvVQGprya zo~z5suTFtbs4pybSVO62NtL<)*saP^bD21r$M;nn$V-9e=+fers;tA5-D|d`w|_Yw zR*uo?OU!8PK*8zgUH|AhoIB4)@DZ(&g5#CD?e0aRYW?@w^~L~wB=s=s7a=BOukC@f z#o4TlPPTAljDuIFfamBC7fD7J$%yVRb1n#?I2}s(3sqBn%^L`RLbx}jSYv-ydQb3o25%lkQTHNZR5p1*fEWn9b7C4c-hG~Vn z|6GxK>fi3{?OA;^{P_MUNJ%pKDI?hPHsueo{y}H~Z)#`qAA^OztA9ojK+ylsOajEp z_Mb-*x{7kv11yMLuNudzz|d^$W+GvOqKVPGb&Iu&vsGyp#s1 zC*i60-P9Q0{XHx35oht{i6wYFLuQTa!%Vej#Y5O-^pNI(dc;MqaS-7C{;z)3%#R1Z zpWYk<)P7M|oYj0^I74l=*1OE~MC4jr{5I+g-5Wd634Fri0IW58;8^GC6>=^P9O`hC z&&y7~?|jGnT2~;s*YHPd+=?(;^_c2@)QJp!w6elL@Ql1z6SMsNj(h}r zM9_T%gc6D@h8pf=FcJmjCAIiDXZ=h1ra!&J-JzXT;4UB%GO-8e_?V*vqe;V7cE6Bx zosiccKW!O{ZP5p*Gv<k`j-koT^Bpn~Xw6W`!75ea(!`y3brgpHCoCI`eut}X7wGlZBg|fK`C|d?Q!Z;<-VPf^rrAw(MB`9J`d*;S{b2Eq9iKTdY`VqWCEB~7{}L^LFZ5~#f` zvAd*_vzQ>H0XBg$4sGFHE&6GOKgTVq^=2=8V6(Gh^K>y!7T`^rE7UD_(qE4cMg;|K zc43V^5A*jd3X6s}e^-`w&#$jQ_$yhwNwn1CM(!Te6Ubjw)Yv+<=t_byZ2Ih34RGh> zxDAa+86GBta6u~c=6guDHldywlyH^j;V~g_z}lqjVp?B34k3_{&}PxXhVeQ(y&+3}))EpcfQ(uEYKj|ch` z1~k<&j=5s;8o;P2!oi#Xf6z{j6u(d*Z|pkW9jk`J&^sLy?Kog!?F6`2IPmibtk~B% zfK8o3iA;lz&zXYAdW6(ldk?LW1w8gLnRN!CWVWF8eEaKdC;*BDI8vb%eEZrY4Jfy$ z1NTzMNm=x#U*N-GNIG;ovO0?0dP=f87Frmp{=E5j^x2lt)3*dclo2VW(F~afdm2>d zGxWBVb{33py7|HS1yYNdUI@vhgU^!<)K4d+SgYbC7()4MPCux~kLb}tjcdD?6|>)d zr?PjR8&2bOq7>cboSCB(9qJk;o5y@&)~_xc#y%kL^heCKqErx+@XonLjs8gL4f5rb ze{(8&^?9w`31DsZvaRS6g$YTj_gnuhGF8)@t1>fWK$jCUP!C2Tp0Cee+Og9gFioP6 z&9@8|ON)uk9_8r*ZlhX9Kug)zLOGUazjD)ux_ux5H`q$ZSajFCgdrMTrrCuAzRHQ6;*NH5Ywq9H{&=rZb!W29jghv z$_37sENHLmfPBs%+5()YKlV0kIKL(F6Q)Q+%KYDCM)rSY#{c_po0XOQKW9SS1ck^T z7Sy3jzX(G2ZxKqB&7}P`)uc9$3bwzT@g+3MfJI)%E&5Z5#2k576O);plY>(cy7_AX zZTcmtR7h2JYNp4_vEt|*iVO53x8Ih=4ar_pvu9&6iGn{+iGc3@<2{dpfXg4o+i(Zh zOYs@k^pmGP<6U8DsuZs-D}!$#k-DBeNQ+n%7vB=NqS_Lxx{wa1bMz`dglBHIO3OzZ zD-0Emsv3C@%I=eSlp*%yGuE7l&&Ik7GO-h)6p^^$a@EHZe6-3%Dqe6NxlK~FKHynP zy?-T5d+ryrK7)U7R)GAVUyhAliv|CK=0(1duiPq*}YjMbVH@k{Wm`0{FhJu7h8!)iHh6Xx%@*~EdL#8f!MfM{%5P#{19y@X_C$FwV7rw~bxLY+-e zEig6Zp^Q7xqeB^iW)zMhi_ycP#y<4gw&R;E{Ap>OEmEIKo0ltUJ$sdxuWI42?(V{r z=$_C&f=?TPbln2Z!fIBD)Z&*7&{qpj=J|wI-*@oPX4>euVtJZQ#an@1uwFx z$3F37a;RXHE?8^o?a6QA1OYz$@8v%26SK_z@o{ZviiC*$9w;re z+DJ91p-SZ_sbNGx07}9Gp1#@ZbqwDJy%HAh; zR+o^5F@4&%9tnNxO?L9V?M=U(1B29SvmqLU4Jkte~@6ogrjrzoOHiGuQ}l*1A0zZ6~3!iFFyyBDak zeJjGzb<(~ZZqvoEM?nehy~r%^u02HOi`%A`YrK`>-U^$?WP)<0>i&KDDOxS_Bm_)5u+ep$Ak*p;pTP;m zt_frd6O+<1FQG8<3WVE-o}ibhQ*Y*z`gBW500G*o1vx#6DtoXMKO93~j9yM&e)_-D z1mTdVC@%kENF?VRYVPuHFo$GfgeHD-{JCd?@-iBL?!_ko6C@sI8+>|X9VZ@8mhN8X zB6XJ%MfAm2eD}mGJoK%K(i`LUxFpf$Q`5+&)_f0#4d4A!qR0SS6ASM7k1~0{A<+N767tx|zz1Gd&gWpDX(A{Elz?*}$Dhy`40lxQ`tEBlP>;~jS*`MG?w zy9ak8KW6+=+I0vf@7r6an*fdRj7_*ZZD;SfN{f#b{t=lOtg`hqO_>>BS=(B$L9FH(XJJO|qyN6}{d?7G~F+(;|0W zAlVUC@X)xuQg~RW3}S_Sj$l=hMX_#IkVJR&4<;x?c#2wK`32+()|RE}(j$p#l}iKo zSzS8U*T+2d#CS7YST@Rp)%Xsv{`dehk2Ima{da`Y>0E|bSSYSrcV|lCFbr^+;4(?J zQ_q9#NN75mK;hNA&I0*RoWw=T$@VvY<&2-6r zNBA6`y)VKAbv{c;6{q~6gYxKTE?Jw0m?(?{@N8A`qyVwI*B-l!jOcEPj-Oa}nE3vW z5~2Z6RPgLkEMftYDyrk+JUkU2lJtd_LFVhp*;k?>BwS8;J>8KPf7r!1LD6##UDA1F zVziY(yt&N6<@#aI&xvg&^DD|*eODw2NV8^;_Q8`OV~wFzudp0`y?kD=s@iU^fhN(5 zTW&~hbGj>e#ZyBVZ`eP~c%RC3N*0`5VAY}(G5q4!s(4p@rDT)nh!7Kc=6CzMvuF>M zb`!NW&n0gN$ZO+^_hSlj!T`Cn4j|$ltB(jSHC=alR2TDV6>oXlj zLq1}BIpop&J>)yEqhD?)dLfV0o{CKtZUXwNh{)wRAw~UB2Rlt`e>IUu;MaL~#R!sn(-He_h?WCZhpQ&} zH~{~HXmvcpkjR&)?dUfS)LqH@c=cT0TqWiN#+#^EWeNA1J`rv2vCGc+3Qd;+2ULyKuf!Z;C?^)f_&CHS)G2z#G zh+!}*Ft1>0x)^J1H;wPM)z?%&s)o9o-f@&<+vrV=cCU&mFLZkoTFmWp#FmR9LU=9D^5GF>s^*DUfYp|A+_~oCWsxzRweD4_c9K!rAbsDL z@s7NCWyRJXTTnk)arr_vkbh1@RxPh-8T=H|maAqgxpWZXj~=DS^yYjy0vuUfF1VxR9@W4%0J!F#`Qigs$f zFa4t?VE_wbg}xIi7x=XV-q+2MXM9)Dq5bHctP6uKdvWF%J|mB-8ihu~LJcn9N}T;G zp;y2Fpmm>50l=hLW@>&oCQCV_+c%~BHaWF;r>{WS-LthsmT&0o&e(i4Step;OL?W56dUfu=a=bJfz~5SnZ8wL`tz ztm;Sg_?ml?P*{=I^+ZoRzT2$Q0 zb*S_NyoN|Sl8g5v5@?Wbrkb8ku@^6zx;F$wW2tA;>>OV;ojmsZAy&aI@Ik$e%-VBR5~p} zK{lH#Gac{WX)q}M@n@PRp!5I$$EmQZw!(CmQ99OFTP03+FoaC!vBs}M;}1VR;;4WU z?W}35o;lS%WA-yQc_MrB!AQG)(+)t-oD};>(oXn<|8KM)_rJ8@|6QMh{!5?7+Q483 z6;;td)nd#*2Yr6$%NHDEh}eH~H_N~7{zq5>6pXA?Tx|bNN`b6@`yx)xE~H$b|E590 z^nD{kVngbj>n*ACDkIlcdgTt;O#u#e0K+|H|raEUU23{k(itIyX?v&eJVf~s}zJDG& zxRPsV+CsMWs}XuBHKw(PA+Wp5&RlzV=eHk#bpD@?cDL2cj=afWG(FFzwaW2D3H6L| z+%JWKFfhdnF9ZTnT6PeeWeAm}Bvo&%{25ma5uj)Q-AGN+Fjh-lK&jwOS zkas1z9Rd-ml0?JRFoel?^Lo|xr%njLb0Hy{IOq~C?Nbveq5_Ab>@#1!;lqwOm^s85 zje&YNtDC=zEJsIn@K1}zetan`uL)SY%vghg_{zDu?V>{Ty?H6V+2MQpC56wVyF>b6 zWN;2F(G0b`JHZ)_V9UfwPfH~y$;F{p>Pi;Uq2fhLo8#3``H+j*z4$8}L*;UY%4Nyy z<80|?8n~=lYS*^_9o2`$k;Du$1YQZG0XTml&c{Rr8U}K7ag7VZXQp@aehh4Wn$AKl zWt@A*v`mo`@1Rjd-8y1uL`w}x@6qeAUGizYq_cWu1n;##l4`9Mhbdygu^it4XS1A# zPt(hSMn?^;B8)3jrO)`gpE;jFYly^jB%P%#?D8-v0eY)E?Gc2vd7Iie?w6}WjAOS< ze&BrX0|WnOfP#SkTYzdRo0z_mX0iw$giEj_2_gjJb!3^6xIE zYd>&)Rdby0&8s^P-aR-*XPR)s>A{t=(P780d57OV-d*Q$_3=F;_gRMYT^5CgqfbFd zw?nU`8876%`?2eph5EW51*=r&afu+Y+t#h+`)N2 zRQJzO!U0Q;vTYvXzN=?9=RId{b!j~8yqJj4os$-Jo1Iv-IQEA}Fr?iBi^hdOig7QB z>!hZ@xdaOD_2YMmgN9lzwTSp&3hDJfq9;Ec1?DH5J9ql|@#VXe+{g7Z&v?JMD0Cw6 z`!z`lx?jhF?_C(D*t<@7>-e!B$7?OAxG5f3vPqsUyvuJ-p^OW5hpkRMIWZPBAw=c-GGuPY{Z1M+ z=KKC1bMU$-tyY#lSNFtaY0l0i{b4?S;YRD-_={!{Z{07_-lBi4c=Crq_5~j0)%OP3 zOe;$3?26B>?LTkLwCCnjrL%eM`EdX9UcTOlfu3+7wa@X<7po7ZmELCE&qvye7_*jy zSKY-AyjZsPzE($J7VCPyIt%eRYVhN-A3aUv3oEN$q5kCS>ddyHx%6K2!;!Ty4DITc zOJAQ9(-=@Wz1;a8X*K-q)VS)J${NSo1>Z{~R^@h{xfu7@8cWkl04ycO9Ztd-yEY=%%8|dw+b)x zy*6)u^4s_6sm&{!lpbk&FYa?Ve5}d7phM{(xFWeCvckS{?vwi1b@YeupDPQE3!Gc} zTe@38!YRoa^jfzSol~DwK8YHCH_k_$@kMh^TU^_ddof-#UMAkiqoF2J(oj}CxX8b9 zPiph=rqr2~h|dvG_rKi#cE7w`5my;Pc5H1>t?hy~Ga5cgOidnQjF(K2*;V7z3oK)CON?dB7y;%j5coJPYBMfJXE zdksd#TdM|r-(+>Xo=;n7C=>m9dAt(c=fz)RQj>(c7{o zru_cn>tWlxo_>2fLkzh;OZF5i9MyigHo4Z>W5tJ~mRMs@EAKVzd%4D^zHMD#oocFU ztPr7Lj9xh=>(5Fho8ez3t1AcFY4r~X)DM!|72odPS;v3gciwg&1Yn#nCcjDJTEl{E zh%WeVw%@Z03T&2a+Kks}Mrfxl`O7>K=N8ZBqi0%Uo1&Mk?zft5@%hNk$$pCcCe{dB zj+&jiUe9jT_VR7~TPr4H#22L#im6I;ZL+PaZA+~Ib7FHhwc}&$Eo|NL7oX-ehaZ~bIt{5YL|1fljRydCWlo&ahhnY zwd-rIW;abT?wLExpwo}Z)Z?#iRb#Z)JSRtWt{YtErk{4HttU{8(0)^(Su<+P^F(4= zU$vS%pnS0SAiCq=w}}Gd;gikzp_%-~6CAl5$^nKmG)!1Cs=HH z%Ua8-sU%n%j9g94gI>9F&%gWHTJohxr^M4@@u|N0sC%WEkICEMzKoRpwu@0wvE!@` zCHndNg=GIbgynGQyt>%Hx5O>$Wb02|84`OE+vN7cf`Z3Y-|{zyH?!wkFVAoVruL?! zq%5YY8UIR*N&NJYoonOJ`N+$#$koY=DXV1pie!SvYKP$G=9n~JxoNM+kldYO;zWwC zagY_&$hQKe5r$!&HR@GHS&V0C9=fcaY}jZK2RD?r2)VbV`13()$3e7OlBYnQnP#nQ z5C^gQKC&~T%kTtZW@BfSrDiD**Xh%RH6%%fB)LABmV^!eUaB=&GfA)3aP!+v-L5{e z-7+CPU%A`GQZ8fW*SFzITguQN`ApP%F@xNQ-3{1vqeZZ=GH)^k2r-KA24C2;BwDq$ zq2ff?_+@A0BU$-^ZyaVH6dVi_XDdV9E~iin#-QVN74QKB|1gEkt+)z>Yg~BvM|hGJ zkA|2AlLj2_kycB3IvxEpiARvm_UvK&UinhOQc*c<;&4=u`ie%=wFYxy{odQ&)OokJ zvxMG&ez}mOJy)xyKaIVl%WUIpM0v{KG~*w}?E(hDL&OpC#14L3=l9T%g5vFWO&`59 z^Z;?XZv99{gMuQElY-)T1O){VeDr*og2G3Lf@0N@f$ zBj|5tW8O#b$r<-YMqU&Y$8P`q=k;m@u(IbAD);Z{`A@D*hnU9izStlO-k?u-eBtBu zsx#NmtT+r547i97To_49#yw1VGcl4TAm+3^_3izK)wd=RY7=SMbM`jE!a;XjZ`Y45 z+;^dND5}l6!Bgiz?LCssbLO}P_L1Jn^vCnpJ&*3$GC#{;V8!4=8;ON;fyvkvDtu$8 zmepIfqji{f+F6p;+ou~%Bv&}xtik`WU4PmjfA!Rm-GrInGzSH= zMC4sft2-Zox`EK?5`7R`kCko|+AmR^-xrxE$#*UB|!T z`RFR4F6;6gtAo*ZG8;G9Vo(XCU%^k>g7b6Misdo82PJ3wMC5(CuXaq2GkDqrcQ5dz z$d`N+(PYSr(FG&Q%8rS9e45v&SqX=`*42CrS}uBCcDi0Pnt!saa%gjwL_j&t9vWwy zgubF^2^Nw>CCVJp{QWI#pQ660YAV(k*v8*4FVqzB`+O%Z=}))+JN+g%~H zyC_tm7Cbbo#|Qd=;(9XGCoh)(^_Y997REV((;h{)R&a7mS1u2WBBd$;-?GRj&l)PM zqKM!VlR&cp-cZTWR)@k4uGxP%Y{;Z)biWqp}xXGnmES0Zi{PZ3ACi_&Iu^xSo0@^n& zOw70rrc|bi&BrQN*kDuNypGkGTtjoH^uV7Wz-{)J^}mbR|$!TYf zZN{q@OaEu%t_!lVVg7w+)PeREA>-U?Rbv*)g8w}6>^V}C(`?8DhRztQRnF{=zm`$` z{KZBxLEiO}VyBv5;{vy6#mVCo&n;QY#)_wlNv!^=(Vpc#gkcNFw6?*+VkwfvWgcks z*(yKsJ71C!))_)8oAbd3&2nP+YH6rm^fg;7G)}^Fl9-!o9qXa*`$UCrGZ<^~pElZ^ z;VtUv-S4_}8!z%kMx57o_few~lE)9ra?W~#S2*C+inb5Rdsw=N#Mr0&u%{u~<`gv_ z;^`4KYHVve8NCsgU8t9&L&NQ}t*#Sqq*PT&J-7&EBeNsgEJqgJH-vZ*PkC zS*I+|x>_mr(r43z%!5!S74WtFV^?trHsBut%M)dlIwn;HgsHM z+osdy4ZymqE0r9|b)lUJ^dUXe*V9{mMAl3?CjAay-Tcgb_xQaJV+UXJknTbCHs2k4 zNldG0PH14z6}C8A);|}s?TVj!cacrErey`7x1V16oCzM*6s1UWlE5s!RS}+V-WIeBL(M+x1Pe{amAYn+~pC zkR;YEPI1x^Q_3rF9DI@ZJ{@H!K{bJ!bsBTqwano)yLiX z@Y80b5R%jAj1{gH_ImgZFk+;jdm2?&eUpl>SxET&;yoRm;Jbn^*93KgCz=W-*60W8 z6m}))yI6i;-lhtvUFOsz1jq!TE@2FAJCwTtgj`4bMA_UwIB=!y#}xAH+DNA(u`7UIlj2cfHh8W;_?sY)!{*O(0&7JZ$C5TEhhFUg!tP%#2&)V9>L$JrnQp<4!cY#cobe( zTr9N`AX{OPo3_r=h7XtWq?**v`}GgkbhOhSte$Q5@=F*ya4(;fxoun_WCBKSsEb#@ zO@(Mf5C~pSlcuWm0YZwNZ^xi`0wRYKfY4amS{0Q|M@NyOf{EhncK@rT=4+ch`&(RN z^=syjYlwsL7>omBvnyIAR3V%L>t_%7`}lE!U~dz1HEg$JXw$QIV(F+{#bLz)zw!hH z@C=x85T$)@-};Wak9T^HuW>W_gtW)+`*7>m9>OOgi{?)od|HnKsHtBT8Z|G9L#hKC05}N+ zLa>Z`!JKDid)^5mV+ApnitVPSH{?hgj6t8$B)A$eHYQOysEVw%#n5ArA3Y2_p6}K& zT+7J~peevB%wkxKUTmpf@WPsf_u4Vfi6JWe8pm&79Fw)AOn~XKYt0aJO%(Zh0=Ny9 zHwa1wF7>5)*L4|x8-zV04s2Bsh{7QH|9YB1y2HV-GZH2hRYo`;J+u*M3I-TLsCR>v zfYBb}k)gxFaPrQO?QCg7gY4VrwXwvcqqg|~N!Z|QZ4vLU-0uV-31I60tSjZk8iD)x zGVfv$*LF!c8d?Le5f3(Ufbimh$Wy>QAXbs|!nTiAf)7MeebRc50Gx$XD>)kumNmyN zNoM`1e_Ee4{3SUQ1eywDxuwEbRI^z&25vHgmBi^@bDc0Q0GO@cusNKW5Dj2*-X8(u zT+fiWpEVQ5|zZrBur43e@o`#agFGXjykY z8l{@`Wp=tAevDA9PSwr&&`gdC*kzIXP8M`Oh|_!e!#HgD{OUey{=XgCNy?sw1#6Jb zk4PUsfTChK{IC$&WI&g=Y^Ol#$ea~~@b#vimUg3o(uGI5D|4Q#C_bZVGZ2ee-26*W z$RvKV0j%22j>U3hTkW{hYIa7-LaTmsNRuEzaEK7FYWA`h12c=2hM=h7KLWI3M_^OtY|3ajC1u_>DZe zR%%?#BLTbl%l+Gaj=n&g3}@%B)07|#O1HYgTK54rY*bYJDLtDaOuE#K6TG)377grT zH-+&V@I~-CLl6MP#8S=V(fzweO;j0M{!RU;+u$LRNx)E`h;Ot1`n7OAa4ES1+zqf? z%Q-NTRtVqjx6DTG3Wdu(iSqe?{YWMT6na2oOyoDq1i$A4O8BN08yjpnyZH(t`24rA+dgLH+L;8&Mp=dI0BB1 z*x>e^I$_vXi?EnABnMdOfwv$uoZ-MIvk+@-vjU9UcH@V&z(Nqy*PvzH>@`>T`lT`Y zMdQOEGfU&D?R#AqX}?c@Fn-E&s~xT?FEFapV)tQKB9JCnDywZ&cwR_sEraz606uB2n6)R3gl)%y#tWU2wmvRHIZ zeG~LM6p^O~@xWmw?M?_8YcYyyB8W|sQLs_naB;4_!QwI2x`T6BPN6lm$o*PTxAn?n z?P+;!*o`DN%;ZxG;tASF5Xpk64+%|Egp|4#JqRs1lB*N>P^jL(vj;P+TG!1q)+&ij zl76ng%!B#68&N#@EZc#-AnB^(XX+P}4Q9DBY|cr^YkJQ66QTXEb$IkjN=8_*U~q`< zL4bp_Hj=nw*7o7jyidcNm=Ahp1%JH6c|Kan_iy-%yW3y+Nm!V&vC*zApNV}8P3MGJ z1Wt0^xaT<_9VrqLX1l@UB4SZ9k&nD%ag>mSSc38_0IbY{X8~MJVFYt!{4B6;SStOf@C-j*|3=VO7Kh_PEBzDF@*D+NR%Q z9i0Yb<BXB|$!4tsj6n6ilx#5>d&MgJ~#%p{WF2jrEO@4wtm6x&@v ztHKrcA9;}J94wFyfRl95^@dQbPMsIIGfr)&P{?mG>mAYdVMnGxGhvDOMrum1QwEGI z-r^|cK@Rm@mzTT&7sWWFQEq#kr*RfFAWiiUkPJM6s=>7H&N3rO{bu?CJ#QM*>YY^P zgK3FgNZhapC08=7%O_3M7}Qddft0afEn`XWV5>7oH^rAf2TRIh`{6rzTK`5#_E@*9 z?3@jiX$j6$stvh$abh0pPd)VwbT`58+~@UJEz@sn4&5l4zOAf(S)T&K($`__EVy+-V&chfrIUQux4 z`uf9hW%~r5Ait}Ax*cyhqp!_i^ZTZ2NtOtFC2HGT!RPlQk$9UxurB-{hXy0iGdy|^ zk{^&Y4PbXm@ZQg4u|vI|{Fe9a*f2>KDJ-8jiHa50{(fi7>-fyUA>r18hsrBw75r5% zX-rV9_4I(qx#;2g4C5Y1K3zdR|0#SxR?x*)_3jB0&5|%dIVh?FL6TeOk~!|w6V>Li zi(&8;z!Dr#57qpxx&aR|1Tn(;LUrzvQ(u_eIpVc7yYEQ_9aH3W9D{IZ%UAwk!t?cD zbs$-c05v0)bD`9jk7j$ea&E7fO^&F1j6z7KwWd5~=&ED%#=d;pFSmPRBhF(`GzTn1 zE^;eF@T*%Lvng;~>&H($Ma|s#<}*tt{w#JRX}i)^ zl(xRYBZJR1_KP)y5Bl|^3IfxlP@3m=1@&i$-(}#<&6TAf7d=&(JHC(qcxhf)eO3iy z7bFyi4v@E2ne@vXLs(bM%3xM1*M_%FJwgI6BI&py&*OE)Ty*6qo7PlzMT8%D2>oDY0crgFizq+3aX>1Qp+S?R z8i;O~d$VahTDs*E7-KKY;8ZwS=9y%?gT;v~az~e?&l$#8WFX)#Avh{$;;OOgevO=8<^s&JPH@s!-jKD7PWAY_nu(g zoJg8?b*q+KDr0@~qKev(e|?l|GRg7@xC5wCzNU1r<39k?RT2*)-1RK2tk?x3@Oy`s z+s!#JJ?iyOaUDdLHvvx4<3m}eP9hfj&zlCx(MMv-54RP_+8;SJ7>7uwD6Yql7&~z~ zc&{KaE3Ra!L69~%K?%|?$)*DYkP_l73aJeR4b)>=gjwEm5d?1mz@hyya&P^%N!|Cy zLBDZo??Y-sZp|Je9S@k!bZI*Odx^OwO$n3$fUZ%(3_>Q$Tubp6VV%{9*=_lU!S`yB0)XagD6i5*1*e-PL z?C?(GMu^*^4LS)h83Bp)WmTloZo!mju0QZyzXY395Y9-hC1J;t@!VHs`zE z)|@?nqi1z=%Z(HQ167)v@4Ezh7~VzO&d3eBlp{f|0W_s$h3e)UrKmV4Hg#yFu8K5( zi;|!(`Xn-T{RJV}hmFbDr(q@g{Mu}u#!;0TZ0~Eww*y+Xs{F?3FG%@vsg$J7c6!-$ ze2d!@fKu8*kce?d?8MU6r>|^K z4(J5w&3otv0&(?V@(o=X!!TD^&%6ha1ea-H_Jd6)Dxs8hK0{m#uBRUJ8PnVr*!{ZR zN9&;Dc3ZXzaqV}WK%7*lF_Ju8-K+^l^q~1*bPg1QgA}zBSj|nd+2k6tQTTU2tpp3S z`vpTorDC6b<&z*P(eP?-O*D0$ zSn2oVih@fa$Qz`+_mZJ}-l#bX)$^_=NzmzFuZW0;Yx(oHwfq4SPaM%byRNnGE1*F5 z`stJKFjjP)_m`1iWY5!csASb*D~G# zcEkznrV#9VUGOzrS2K|4K&Cq~Shfn*mlp?Ej$`$s`sj zH!+!ul&>!aCV*TpHM*+zCCmLU-S(rkJkns@P^BCqrpuFP$|; zgw=02;T`oUE`_(t`}ICvc8}6`^YU(bSIVe=q%piIChVQz)4lhHdpfP4e~@tMXeFb?dqlJk4z&H3LeMCaaqbrmOsjmGDCvTPU+aX!zsq6*RVe0qHIa>NBSAC>2ys zbO0t{cUU^i5png_=@Jr+NZDFnNSM%tw?@&?Ni7|k1C;L4Qw8DN`yd#gZe8ZnamROz_=Ch9v0{n|cfwIedgj;2VwEjUFpP4jo z4vr2`qbLrcaujpZ1S+#@Eeh`^&qh**jQ`qRnk@}pwx*v^jEZ5|Y);u9~I{uZ%*=>Pj|8>9%mBJP*%|hlVMnPIF*=C39)Yo1M$ujt;zf zSj1itoZMgnh$EI?Jpstn)Kci_y*9YFf;Ke@z)({>x^BLCEq+qguht7>;p;rjAOi%l z(sC`8yezf^Y%!>abw>I*jU@?6tvdno{pPi>{b1`_Mv!ObSYAO>n0_vKKXqirb)~kN z%CT}ODwPG*vhVgRs6GcJY9lzWCA%3h6SUEIEX0?RXcYHaczZt`x3Vv{)#7iqLOs0g zkj^MUSj(J2e1_fxYH!@@othf(Ytr{{L=jfi)mVZacXnR`y$%Lv+Bxq^lWvzTBEM}j zHD)zR$abR&0SRoas$jR5N#}|4OHGGX&E_CX*eQ?`(09u~fd-+F0ho+KA8n9gI~GCd z_9oHt6uzYl|x(@03}6OGyiCn_h{9h zet+-@CAa|6Img)-(#j=$KvISBX?VXi$Zn8MiN&g^Awh^2TL%GyzYboGgleRwXz}uR zq(IdL(`%))OZ{CF!y>=4pMJ>-6jQ5G@B*1)$Nk-tyV=?P&CMf|;rc;QqZ{+*$Jo4K zO+O@O>$BS(LeJ&eyVY8;aXW#OxLVL-&Flu{jaxL&82q{pzi`v<5{N;+0Bo$&5-V%R zg1TdiT=`BIfz5~k>1LxH&A0{DzGHsHn1+xA*SfAI#HP9)763cStfyZ3yhSB9()1%DkO8A6V==iu^uNky{N|xO_iRzy zsnoDKgC{@y-!q1Pn53h~6ctS5A_^rWnrvF=B=VV<3sc(VeO-<5hZfuGn% zP5t+#Q2VOKpgaEP3q7<*%Kov;ZR7l84o^Hrg?~^xVICd;;ya{CL;hsMwC;M#-ED3P z&HVHGqeZO%=*A1COOCqQ1W{4gXDy9!+;P|Lb(a)D6)24Hf{KmHbHDTBpnB0jh0pfy zI$i(W-U@cl1|B_K3Ti4)Ai3;9|Ml@(Xbjt@4O{|(6CVRh4|Ngs&D3-Raxxq#N6JV0 zq@-v`pYl?qQXB?#L$fAlfR2ae6t)$&ZArW`R)P+CZrIs8ad*H2@g`7wWxdmN~-Ui z%~Xh4QAy{OW6ha1*O5z)$lo?8<4BF}o5o3dS!||b?JDc2qwQhQSXgy699w>{y^Eye`GU+MY#`~A35*3T!030Wx@HiwrFF1P0V0cG{l z(u^QhM-0oVO+yji?OQ$m-11n0NNAl)v2_2-C{qSEEIl7+3IIj^w7(U0s-IV*^_5|J z!=bpFzUy+U{d>7xG;89?CV~;u9&=Bh3}`N66pm`;_kP7e#WytHqk7LVhs+|RK-^|s zs9{pDdGqQoH@Wqmr;XQDL|FE-x^`HD|Fk^b{{!=tQrAgV>>DlGBBs6VS8)9tFr%}a z;g_Gd>AG1Q#hA*kE^e|chfPnnT3yTjTj~eR2mV4&u&X2%aiK5_G*|3qR9YQJ}i&4=TnOH@aNY!EJ{ORI}AoAd{f-zN8Cmd=EAY~jTf-pZ-Y+WFEglc|Aa;WkM2}Fa9n+!Vv-^YK37X>L^6Ac}`&FP) zqXKy)0ucR#4{PMiT&>jQz1{eqhFAQuFgrJIB0_$qy%(Cn5}VeF6iD>1>q_tc@MJd? z9yCk2GI~`6GhKK(y*z^N#uu4|^?hi}bRJ?0{iI=CYoE$!5mrZ=;9mNd&lqSsn(YW8 ztXHoZH_xuw1pkg0kaqiT`)S2ZqV)CERHroKe3jO2g^Vqkz&9e&)ESz+l8zUQ_?*OB zW=w&MiepFJX$e6%f0S}g=noTQ@82z4av$o;8Qf4)bAsD|bQA;=T~%|f9!E8`qT4V> z-2PFbmJQGzO4?RWZ%0N*+ME=kChdeV_ZA&c_w{=vp+;fZP`)3vizH=twbT2jK#!#e zl`^POtL#`5Na|qxnDEW&<~m9!Q1uowV6;~d=5+;H~S>Hemz z>aqILCiYAmbpQ8sY^rFN9YZa5Rt&gS=~U{7&f|H{0M(h`9J4)<8w64s?ETbChQo@QwBiN&eE z>rIdW$btj}#!veeJ16mCuGG*hf@R^C)RtcOESujV`;O5|1t8oL5aNLBFBgdi6nV`E zwv(l*8iP3lJ@r>LP$%T0aGF?-(Q8Rq_ZCipsngbMKGSBfEwE|eQIkaq`#?;50g(+r z)owu@5F2GGBoLY)J_scyp#C1v4XxB(36MO%CN@_)anE9_d|BGGQVRyHqhD=*k+pH- zqcJ}EV|m(`4WGR1bTfy9rh;U-#>TEpz|&AMGOUm13Ok^x7N|u4N&xk(@S2kIL8`sP z1C8~9_7kjD`L@!&Ah&#vWN|Aq(Ghyf%W!sf%?ZSn^Qp#L$IdLhEDBQe0#Ygvm3D(j zpg9E6h(YPBAb05_)bJxbNC^ld7PMmAvG|M3U$boLNtKPbg9j}9=n^nefhD15*sq0P zD(4o1%mg6S=Hbd4E7DUA`xaRj-#6y4%9>e{QI+>Gt*c;I&oikX>TYK^jGkU<(?~6N z9F;M-P?f3}_&5Wk@liWcF^l4m3DEbI!t)vwjtz`ieoP> z`P=z(Arpj&t8kkIMH*o@4b*&jOq=U-i7+W}KZT{tssB!bC^a}>5BB|?$qx}&&05jNXV1s;a=3;5u#oVoREat(zCU{P zJMkwEE@T;QDz4<0m-FRef)}+ZWnj?o&U^8zDT4OhP&?K2>~=eboxWLTJ*9o~PqeHn zx+l`!eJm*GWZ338N(0!Cvf)DGjP|1Q&bRznR(&ilXmoc<1ZSgFg-u_!Z;z2Pmb2qu|iXLrNC-t$g6FKY`yEz>H zt{a8xp2+cfL1zr&9HSj@2uGFI=#VJIVV{5Y_?;_BM&P6mY{~IsVPbp@bAjBHWowPo5wGMA#z&mAmdIkG$DJg2i zL}69~y9{6w2XKYUAILxdySe$+!ZS*r$+Gx{{gflTU6{^gt-p&;5fK6O;x}os-z-R9 zCCr-ZyAyby+)~sqb!2ZfG^vN? zj7N89&O2kzf6H}!+?mAyz;F?;>z+lh?Q-EA!rz(xx)30-09aW99Y0R0zZFOt_D;9_ z`&M`AL9hY+$+10SRtdK{W@XeijXIyFBum6^=!0dqw&=J2-{^z;hlUCP@qbn-zGF0o zFZa{{p74puka#8g|7N(i3_OWt;;U|YVl*pCh7?*3!U{Klgp~&Yt|*_TgE8h0xl~xXIzf^*_g_U`-QdVfz{)y7~`>D1N;x z%Egc${GSEKU0_lF*7?k)S%+Imk(ti#oe(bsG!!wGx^G3pst~|0K?GFgW_xCBqNO}z z`z_+_>c<1}`$%~O()I|4$`IDAV%vz2V8oL}Fb=lG2GDtd%vt^%Z93)UB$=RoQhARR zvBtsl$JOoVq|S(`06#O_e&e-&R2u7OpSYNyV z*o@t4=0AOcQ?sT}=7Yl7f*^B0Z>W7?r7F7zT(2xokKi-2QcVqbFT$#56YT6?Z5y($ z^W*ayJCFCliI;3ElVUxc6kdCS8WVDEs_5gYqJKrwA#@AUqgbqcBW_FCkR%`%U!1R}|O zS+2Dz3^rOiy8k)&$!L&>7jkg?3(KP7rjgnH(_7zBcSa>47&aC9&rzQ^M)&vt&edZB znUC|6&rhsH=-&ME$kUwv2rzK7TGNaV5*EO9-o(WIetPjA-n9S()*H3v08x`*bPae7 z1P=>+7{&j650N?Wd^3Tp|H3DQ&|LrwEWG2{iEkK_dj8K@X)b7Kpg2|l9NsbG`sO%>P{F^A`}@`kEzgdFeMgV+5r`S$%pmRD+jUIhLNC zl)E)4UF=wz|2}6?S7oSX`L*hgpUgXgCmO`e{?Q=n|M^zLaX0os5HeZ#&^gvN_3`HM zc_Rl?p)gy(fFQ1&ccKW;)e(M)m7)%{_;(+rR!gg^bFnRTo&Vs2o}JxeL+R)iy7sJm ziG_)&0KVlWU@!FUF0O4v!;$SREGz0;sHZB(B;wzB+3Srm|A%Nm#FX!`uahUn%w2+f z^-nv{pI&Snj$^c4?MlU0>?hWr%j2~ah;)e0Q$}hJ-tw;Z`7u0G{vZ(j0Jb;ykVRm8 zVaxQAr%2_myF8!M{_ffJcxHYyKfsmalLUZ&Nqqxy``#ocX}_+nAPSLEp%-_1zFPGdh87EdHTR zNiCoi03^L@EI&#kcU|gO;Wi^*h)E80^4-FFfAgm5;BzB0$weCSFV$eFjrB6NCQ&}) z68o;AVygqXsZ-|jkSowFZeFZaxzuL0@q+-mgmIoX#fd5=*yIUhJ^ z27LZq9wYmlJuG((THVN(AERPVxmeh#cSYd>CR0o6sTUtRlqc+uX`NYQkNigvT|r(I zZV-*rKyQTaO3i=aI$!VP0Qe*@a#w`iZw1-gZGMS-Pj&LLHCg5VdC)eE{IwWMR@laE z1U7igS3bW>E)A5rm)`*r4+%VOe0sY$3w^9ZMy>XJU0-M@Wa^Qb`4i8teQPiSJ_kSebul^FQI6bAg5*yvI$Y$S@O<7MX>-f+3>C8BRq z!JmorIlrj%n8^iT5z8-r5ouLWEC=@z zcFGTxd9v12^!a(t*pp7TodlZI(*Xp725QiP=8zI>#`n)lKi`H%d8*WTu9St=`#Lo% zD6n(s>e2<@%ybczkeA9oHV*vKgjNpcGvcEt)$q9!%1NHzDgfNZn3mXX;u)%zp& zn*djiRQ;HXFfUp0gFT2=KUyzKoHlQhPlhI|KljIUY$E%Lit=~6@V zBOw!cw*l{i+}Hwb5(>KyBVulF92VNu@7<>s{@)sZ+!8B>M?hGf5BWU*tQD z2X*TQ1dE<`7q~!*1Kc-A%nlfyOiBgqNc2mfVK9?t5z50p|7Lws1BjaMKXU%?J$%|s zxGd!mmNo0wR%NkEs!_)tK^yiFo6 ztZ?}ZW}-a2iq@;LV3ytD1H^#^l=gSn1Ce2Dbu2v77G2%G1;G7!v^9r)Xiol1G#-7> zX$opa+oBKGls@!Q#nG&O^D`DK0IU50WJgQ`^8bJMxw#x>Y~2KEpit{^?$p6Ka3G?_ zHe2=!)Uf~x4EqgasA%5ZH>>?*L5@1QL3zbbt`z93;7GxoE0hnO0P+F7dwwO=@6W7q z}4xqc`Z{C;ztOoGd2v3?>%bEY0NI=z!$Ay5CKI}s2feu&cnLjDw%@wLZ z47UgRFYE3%Q1gQdWRBqI3OJAeo$dg2GQFC>Z&0BZdOxVZu7h?S3ZdtqSgi-b;(ntU z_}{_yP_YXZ7o8z_t7{fu$L_XsMoKBurQa35?m9G^KcP6m3aRSMpZ7rp8R{|w#aiZ` zT^vF`sJ_Fx#19mKAicgPQD6YnX8Cx8XTCZA)l~e#LErX&2R4>U|IYdr@O~+(9uSLH z@Zx~XbW1%MD-sXjcIxjN<%B!-xd4g*Vx#@>jn#&0%-rWzGd=GPga?2ZJ?Y@^rPe|_PJgQZRqY-?lLDOs z0w|!gynyO}py|QJy_yOKnhS$QD|V-m;n3WnLm=)uzs)OG!AmK^1OWHfHU|Od5rEDIc`1Jg8nN`;LfXT!$w^uud47=c@!MmRSdLzb3bwzJDddvGMEWN*J zR(82-t&pc4L~M{5RaY*=)Wv?jL#uqj2sf}H9yQs8RCJ~0iJO`tk<)0?VW%94vi6VKYKfY6_C`&+YkvY2nCY!(j3RBSEm;K*xyzU{qs7Z`3lSFEmqI-!1GH01hod z?lAz|B&c}W6?FAn>0J?pvdNe+_l$}^8TZJ&?P=;=`B z$$%xPaxMX_SZ;)r)nd?F47Chw*YoOw_WsP>Gl(KLcd6vRct2;?+9w=NFalcw_%6gp zOvgT`$pN*qFmY;Vh_?rBEavKYCA{}DS2Dc({!H*nxofjHF?#>Y{4AHI`G6Cmxv-kx zJRHFtZmb0}|MSIllmKtcB(XjD0$f^5{ApQJ^>3L*0zQfsF`5jQlP~~@h-_$S-?gQ^ z-OWU`l*5i!&$XtS1l{0LlcqJ}!;ii+-ZZ5Xe0$~-zREbLzp`BP_QHDYEwgpz%bu8! zokH$;_9vYPbNxsE8%z(D$vvNlV_%idf8`uskD(%auvrRd4=KlAf&gH{Z zr(GxF_DXCHJfqG)d$K=v^!3LKw^qBPwqWz4X}{SV!+lh~HxKH?d1bVVw0S@jak`@eye)Q z9TofrobRVk3CZGU$8e_=O#d+MXs;mSFpo_-F=R7(cxNU^?&v(*aLKp145#(Bv>7mi zBLW(6jxaDdD3Pp{EOhUKH`+y%csn{aW6k?To$_>B=#C?D=j_?DC5q5bqF3Z|WwNRF zp>v{g8-q_GdFC^aGFLVN@ah($`6>qM;2p zo__IZT6)*BD4{@nm3xHDfghoMnIb_#p>cE&bGvd!z61u^NQ;|`Mq^-Rhte7Yr z2Ccfqa`K0BU4-Nz+0dfVth_1sC$6r%DJ`=jI_Qv0BF++wB|x~HpH-yKk&HQ!l1g_;|fbub|d;hS|sSC@8nyX`NpLV35}k7+zOJ1c(_ z6vT6@XRjhv=Bl5y-H(LL)Tc6HA;w(LhHuFxWyy7g76FPDR)-hu>3?DhC8^`OLzpT;0mv!f- zK`}urY1%aNd| z2ard%31TkkZez2ijuy@z9AqWk7X1I}OI{?qE2W`#>_es?#z6x~!@m8YQ#RbyadrK7 z9ZvtW6i|UqJ%i3W5I?-PkJedo`6@w-gWBx?!!cdI364X7V;~=HfpesdJ)F3|eIAF# zfO`yrCR+@|WUgmFF*b%6!(fa|yrYg&bkEM1l2$#4Q~v~PrAc`3^}+wg+q?&ZM4 z;^q`;>>Q*}gkK`P0BGhIR(!yEe(kG^tM67$e=qdyP&5%T52Bz#5291Y|4%!Eo0=&S z>696fT)bD@l0BHdvJC+8L8)eUO;lb!#iMZ<>lbG+^G_#=C5Q*CvxDekk(3Yr)daX8LOj<}0;Ncg9LcES1#EEMQBFO{oMlHN`?yF>hSJ0kUpHq%&zb(4! zFd+rC<;o9i+u!c#?O}LQKh9UpJ3aKxsgF0$|4?I$rGcaryflWICd%@;FAqNlx@EhD*yw<=wSh^EeDP0j>E zb#7_9FYb%5qT&v0+)R3g{l}Z#`@*tf51>G2Wi|G^3J@68|HUoSmPP0AUVr>bo2RBi z`GzgkH&5U4P|bq}9wByM#jusf)(4Nb)o+>Xpz(K4w1#d+wEEna!fIH-zHX+lwsUp& z;I$o8ty5Fr)&<Bnb`8cu|N$#VEobFQ>OZQ)m$FNIeJGc6X%I zoGG~Dd0DmM*|9gkA;%b&9Wn?uhxT6SIM;z1EiHDX!io=4V-L5nSN?TGFMVf+yN|ce zX%E}MtsFFF>4&*HH@kqMT)+X+I&WI13rxGhfrE^NZ{|21&6#0kCy+WZ+iEi_o0M-d z#dGQsEv8C#<2==pHGs9p9-BY;&Xi||^p;k65P&Q>wV~~8)JdyK@qH2%Z9a2+g6zQ{ zr7xidytnr0csBLxzYzdTP@h*0M6WQ@AKZ>5C9a)v=5EpL^tM&;j2o^CcHeb=y-{gR z(oi{X4wyB*^v^0lMN@B%qZ8(%4}kVROdX={;%}c}odu^(x)=z(x1qc&D70G#` zE8TDe;z5}ArGz?)6c_Hi>-6NEOuDp31Su$Z3wXG1&fUxL#u2=w&LNcmLfC|*Sg6_m z{qSouUA#G@&*^{%zR2&5IIf2>jw-kBOCkFR6tL~rFjeX0M_N{C&ph9ZkC^^4 zT8+yl?@R-_+_PoEi)^brw}Qz7)ZQx|%%hJzQec)g@`}DmT{`SFKG8jlF7pZd$B}%UOH>O-A}62PB!lZSE^GKcj_m zGqkG_<$G%(H)M8?0L(dLx8p9N93%HK2giMy0xsX{&*Bhf0BG4yBW6L#zieKXx!E5m zU#?yfHP(D-=ZyiQ-)8nB8|OpN!nmC-U9aD}4xb9@?~0>k-QR^TRYMiHn26#=z2Nb# zeZ40smi{ka9y#>jwZo8dre)54-5TcX64YK;b`KvoQ8kypah=_P%Gy5hLP{Y6_~C)b zw&x`#IKi`KJXblRERkq2m-~@}wn#R3zV@wl^?2nc`HPNxIc;tqb}_sEp^-)H;X7B9 zUmnAKqh%P^3 zDkLk8ZkG0Cvkoh7xlcq(-oh{w`)NH`I_|#@U8^6O@q-aBIFWn)LlnVwH9e7<%>9z> zKC7*q;5r-X)s_~Xo{|A**ORw7DMm}3TYmj&u-EVditybi4u$+jtLh=v#fwOoe)NSb zs;1QcKa-!mPeaY01gOrU(F*fRni%+X9)S+NPhA?+lHC zy8B}8D%`ePc9)jmVqozs13&BT1T-0Bf76UM>YzGx&fc;8uHZ=Exy9>6AJN6QJ63Cr zW1xI&FD>!(KgmWal8s)~MM3dkSA&MUZtyBcjI2L%U{?s=bun&4W9SVDnN>iKOxJRbAifyeL zj$k#AYB=)Y*52bofrX!&E`R1m!Kab{aesce?BZ}jdPb(Nl+&uUb}f6s-&Q^K9kluv zjs-Wgw@m zWuo2uuuI08ZyqXR%6GlLRb@!`#o~obW9(|BaamPQtaBFb7<8Bw->rQ`!SNufg6{Ep zT~#Niy5_toEySR*%Ozme66yne()o7V_Mgk)^Jff?4j-f&Tq9UMpw?|p0dY(HdA)9O zmc!EBzueyfG=YjLd~-yLj|s?D+~=U^r9NqcXSZLEd$L|bW%9X2_Z&eBIcLIUt8@WR z&K~Dq>%@-C+PLCF=c~!L+lh9AdC0hO8p2(t?PFXnpiPcEOG%r~!+BJm-B;d%w z^R)bMWM-UX!2Os~p`+Nw{%L%R=3ZlH>3N!z6CQTrdV zrq?=E4#v}_Gkh5LZG@!lZ#3L8%3iW+kFw3Czh0ErLv;ASzWvv+aflpJ=ZKF+b+MSvLt@ca)9y8OVEcU*4#8-T1vH5Vx+uWHYrr#qp?I##|Ejh zG!@;Fks6{g6~5!(y}q`h`?1saAHoKyiq|)(UwIOvA>q${A}?qD>)AQBiF>rP`{4Z& z2P^&Yn^vgm>!kN7YjY_5UE5`SgTA>p z9q_Rqw>z2Xf2gS#ZF7{;U)(LcW4rx2dDhV0>J zQ&d3z##5ErhgG!Vy!SmLI?~i-T`{YC`evtXBD#5zRgMP>cHLVLK;tvfXzYosd261; zK`p7RMs>ugx*h`3JwR-wSzWm&iT>Ep+Gv?0I4XL5+V^xvYYVT^;yzQ9V zP%Wjm1={?p5+k7sarfspE_0%}UKu4$T$@b`yUg{3O;kS=j|6UfcDz8VzwdFDx6UT} zYt{Q^xe`I^>}XeY(c+P@`+1$pF3&n?(p12U_M+dW%(;9Ma$3AOS}Xp9%S${rAV~m- z$sZoayPV7H3mXH`-P&x5Q?i%ZB7&`8bD6u(DLT+~2HBdPt{4{La-^=>Av3r^BiF0+ z?mDlxl0k92+zg86CaNMi4TWz#5O4Zt#omkP&Rw(R(LMw=3crABlsXzV<9i}rtTNj_ zs!@DzUg3^uJC||T3eT+q9Ryc>Zk1LPIb@AjY#UQDvu!FhO+Zjg`ZUL7|>?!gR-J}=WlLfdcSu|}vL$WWukh^*3T_9)Lt}|h7V?v9Zm;nyR zN}$iupVMulU@`FILed6y5FV8N_Ngah1U?PXc8_VZb!D3GM1xa(t&W3Bp+!tX|B1d; zy_B7ccA5~ldB#n|qmXx!P9nw2kr{vJJdE0FRoF0*L9FdPF12spI-$kfGx>J}ZjK`= zn`lLzq%GGJF=uu8ND}Q`p8`i>D$TvrZ$4R_=@}EPsFNK^sFN}E5j}**wqiDhosrig zI;AwX`xIm2&zux;<^z;-F%s)+zBZ@=+a@PxX;Jn57LCDfJD*3rx{%`nLmFD=4WYn8 zR9@;bvp{8oOgx&S!8;WzxBPpcpzUCj{ObdF_$bw&ko*!)!+)=_OOpB)T|?5LV(M1V z@o5G)t2^^!bB?y(Sx1DKGkAG#a;R}l>b8LnBm&n4rj%t%szO3GiVGHZ1PMya8kzo{ zaG&!2Dj64lBofvxrDMiYMNS6gX|!yx^#0C0mDR_KmNl1tJzV9mhe(~xJ*>hkuPnRw z`DxF+0Bh5YF_UOKP2lvLkics54o$}HFDICEie{}XIRQCdlE*yeNf}=OXQu;2R zJ7e7!Rz?^y>BgpAVdZ1vJBTtE1@uDxrJsIsv~B^v7#GL>Y~S`$#uoJwl#&hC2ap_q zwI37?YC%@$e}+_cbk^nlwDw=ykEl)O<*W>2S#bQ*w&;;t+d9iL|KJp1AwA|e#d(EI zv`Tw9-#f8rq^h|d>V&<`jzpHxRX=t5Yk(I8e73Xri#-PGf)B@<+X*Hixrrd zu%q@$*Y5@SqBq6UO}GDhL1c3j?oL9Nvtv9zuN7wB+@)pbh+S+r^^Lm{b{feFo}T<}JZvGc9-lI$ zm77)PZ;I{KmidnKr&xK}ooy%so^qsIMQm)*?MPEvFSCD@y1n4FX~GnaW>I!86bHmz z0PDSQcOCfOHR`U(5u&DRppC8gqy2ovf$Wq>YSx{9G#S4RpG4;!q!!o-kosPyZB@=% zql29YZYKt^A_zjyqaF_l_6UHdLPq5|!(YtNmmoZ=Xw3&BS0=b4f!8yi*e0=%8+@#} zvZhwsILrFcJAi=>VkJ(bTKeKi{#qpO^_8j^$PVH=LG+`Y-?5wBXIuiD5(79X#62gC z#zGncBASFC6y{KXKBLKK#!0@(>iv23Tca;J1gzW+K|%CKoky*p#IVJROO$MmmN<9K z={R|_$bL!v>jT|Db6+B@KY_UE69y^EBBkibQ?gx;f_HAe`Hx!UKyOT^3sx&C_=_UN z@lKqeS0-4z3yg~V zJEG==jM-20Bjsogj2s;?r(IE*$V1uVS$ zcMvKKf7;51eb;L+y_JAg@vJqk}inol@rUi0+c*J(d0 z64G_x_k;s)=CWSaU%tUhufVEn4HBk;& zI?i^&m*wvGW+#!lp~9V?ZdUI0J+{+@3nTL-TBDpmC?%rBZsEt*`90V1DzA0rvj6(P z^xJH*KrkG6Yn4Gfp-B7qRogwpIL^AS@hKWQdLbVL(~)L@Crf|)I0qQ)szF!DU^nr3evWQ-FNF;OY$;vhbC)czgeJg@d@g!geCrk9 zVTZfhb<>F{0&IUoHMj4#=pc6BHf?S^)&+*^nGEgXl40Iw4#(FkA03{9Q#xSuTc3%F z)7A%Dqzj&wSf#5)w3g|)95g&|;txR1UDbV}gGKSpbI*?r{#coBi-`i^%K}$_WUgC7 zS{nBRDXeZhbJGZ2r{uB1H|a>Ki%t9q;_o_mwc9^)Iq6F=iV9ObD3WhKVUuCU_i8A> zLE|9(+8p;sX|vndmg&9abfZ#wdHfQ)%!R-Eno2vCP5)LI_)OiC*iP{@x6GkLF3pBJ zUjtE?3H72b#tDSz$yAb+X0dBe)u)iQ+df8EO_;}^jI>tTy7|{<1)ZGm<^A(2PNc-A zT}F!n?sqZdr2q&9S_x3D8_0r||80@z;HrrK@vR&WDfK=!#SuGGWE2oTJoL%&4uy5--1&>8O!iZx8pEN!dc1*0 zB^#I`)OIl_Gc_{D_pwckV%1gL%iGeCfN?0d{$}dS;~JMPY5LC5oXLkSrgdPlLe9)f z4c+>^Qgy#yZM9paRA7}UzS4N9_tndd`bx+Va*{Ppq!Td+@5G-!kZ{Rmr#DABsKf^7 za^xY+u`AeRqqHt1V=$we({lQ!Zp>+Hh{>b*4uyAj2E3sWu)BQ8ILp+ z&|?=}9el~KtCf>A4t7?fhBd=g1&AG@aKm}ZgBNeK!y07ir)SMM-*VWrAO<1{G$1~M zpXTld_N}~oMB_)P?{vS<&|QFX->{@yXV{t*+*9{6Qi^$h~8SFKF)(rnbU<@goZ3=1#OM zEAv~r$Bz9}{W;9YzGUVNvVY54{h6Tkvnv-3kM_N&P$p){dg<@Cz4gBFM>8726bHCVlWDe{gcvfN$61XlDv9qL7h@J8> zy@F8{4+Gi0uGzt-slVAnHSr$WRT0{Hk5m)*hY=4aB2i9T&2Ew)Y`(PElJ(tTDr`)n(`iNI@UeN{wunwN7=69 zMK`%jbjnAz+`ZvIqvv!Ta~$jGcj?Otr52*av$INB&7)~uwphHDxW?O?6{) z#5FwhechdWQj9vnyRL_6XzT=15v1I9yFkf+_PP9jZwZaL7X~{#o0fDrhrQ=Rm_@a3 z((Aw#dDv?g-F%GPAhjt{7d-xGkJ5!|prb44r~i(3iNI zdkz8-8HrlcGA=Io3*~;%jQU8wxxCD+O4PAMS*e{GEB6dJb3aKQS`g6L%oM`w?f^ifilt4oRgxg$Q)uf zciyp@(GiZu6Y7>6;pPu_^Gp+t37jDrc<2 zQhn&7z_~R6FOHoV@HE&35&G7zG|+7GmCV&Wzs*t%Y@V{aAcQeo#HJ?-cBGYbcWG3D zCYGRETldZC!vVyJabb}EDi2n#LL%Du`rvaC%Y@pd_iQE^$&r)D`{xQvNQi*;HZW6tR10TJBGLAKk0ZqXkx_N%1jiCqI z15e?blE37-`Zfj&CH}C~2Uj~F0@6FD!@(Dv?O(MwKXv3uSjd7!9n24gb2 zt=#;m)i1iKLQenw9A6FAzF9n$3dkWLMOAo8cZ{*wnZulqX1KHV3kBC|$_kdkM>^4t zxmD&_!>{vQhOj*?$^JeZhNJyIHLUuaAR6#gdUHf8?&7GWjk9L`39x|dR^1%h_g`;T zHL@8(Thza%%v%DbF!Y-!e8hiu2jhSP^ge7MnfHq(xOAG=C{zTn1lb`-&G=-p*?w%3 z>rEN-n3k}=STwCPLmFn#OZ=8d!UpdKVw=tpGfHkjOTsIuv=wp}#vF|E#&u%1D&5#$ zvco{h^vI!39eYN&?> zwl14qF=;G;mx?DrfBRCHl!jLq=X@lnW(z%g4>?FJtcr!}kU zp28jjcBgmSyYU_-W-!bTSNt(?UEYN+=efZ-7Q=BY>>Afnm|L|DRclTH{yRzD?<-3r7$mR6= zluQia=ng`rh>w)mPstM`Nx87$m5qYVPsc+-%ltmv=oQ&W{|5__DfWB$GD%OvmGyuN zHe99PG=?_(Mb58R-OQ)f2BERnXUxoM zF6U79V}VK}9O$)7v_pGNaGW3^g&Tt-?9iO2mcK1zSnp_sQ6dTUZH;B1ht>adR^0 zD?*qmBwYl4)Ah|H_!V;^!vqp z`AA{#SO~d&e_>_>mNf@neb%rGL?V&afWA56`|wj`Wu2)waAVF}H`Y}=qpZH(x0|Wa z^bCY%06LKbjJHg|A5ht&lVdFoD=wq8#J|~z12+#=n%Jq;seqC4F3|Kq3qXmqidt+? zF<;xYR$;5@xbO;SCTedg)#7eY{a-mCU|g*TCTP*|QNzer7jT($!HE{n@~2zxlG{8m zow2?pL1X8?6p!aYq2fk|uQqMP*J#LYwtO`{t# z1jVC2zCst$50c^Y@Ui3%FDES}##=~fHveD?Z<9xH$Ss#v!M5}^>)zR{HzInI;^g|3 z!MH_N*>3#SV{z1K1zF18tFS@KSb6z8)?sNjFTqQ0VVYVvVCjBz>qOs;-qlm@Z=N?7 zq)eROEa%d5xVP*pGk7dZkn_T1{UGo^;5R(@JkzkO^zcA?ckm6v>JKsvU;54jshi3x z%*c&e@LJPS{nAy!{>j#zMQ0FdQk}9o>p;hF4h7KOk**(_DEq3paPxzm-h31z4h-vY zrIE(z?vnkREYXgTyC7QQ5plZSxm5;OSKYlM`Ppc4yjfL#a~SSebUdk&kf>>hk6oCM3R4=?ub39(|%Mng)Bx_uZ2z4WIXBuFcY&gFHFp!Tqc*T1KCd3Eh1Ao>TQ~qm_%Q%l?gQP; zU4*7~GH^)Sj}B&{7e7YXW>|22QOuLW^G1eEbY`#vIlr)67bvORI)$!ZV`UQbJ0Wsf zQqD}fJc_1TQaeE$PYO9d2nthKy1_e88u@_nG(u(q9nQRNyT@F`Rv)Vwm2t^6UdkSV z-j~saRQzeBMv6-dwK&`SRNX{3GiJlE_3!66Tai3f$9p`)JAXe4WRz%oe7ko*B{jwS z%>6=nyoFE`;SWHU<3!w~7wmHFGmbd`g?o!KsN46?sWo}?GyjN&z@@REhhh&I`-UKzwjXKC^-5IvcIsV=W%SxC4QnBsRdfRl>BJ~ z1~dKu40BlJg9K~hj#gbx#i)MFfX8~kt-LIy2!}~MamkPau(wbz8|m>{swDUX+?g&| z5wL7voEa57oVfQb$aqUxa!AHfd<*)^m#GCnRV#35jg!E)cCQ@zxWYsLI3=R+C;T_P zO5YOiq*Yka>fFTpQ}r^El%1l4_wu|iSt=+Ya&&AK3>2C?gIoBDp<_h4Nc37IIL6b=Sb65hySv7+!lVQzVzYmHUD>KOT%}@IRkT6I_h&bi zaC&}yWk0q~^Q7DrS4pl{#waD-u-3Z@`@0zLK7D#)6@X-^|FNa_g=AT6Vd@dsJ8HFW z&1q2Di9T#8)y$I%3XSGQZ}Vspje{$C(7e+fmTB|idV4<`gi5uzms~m)csB6MA`CA| zoa`CUEhxO(U7LFI)$MFX^3!RI;7#6MjvL~iq{Q^6QEMK%_j!+Rx|Dd-Z$h37Ig(6( zRS&gr`SJ`0t+PL$D46JriaCY%w4{pdViq+$@~`MHm9Z9yKl4DsV@+O7_(t>ikMR02 zR}+7R(pIZW$gR6`CUI*QD{$yeM*lWPR8A!3TD_URZnB01H|WjO%JRP9JF9F{6C*`> zOUVdmVQ|rrt2`L{*)PW@hh1~b1gXdGf99obz;23hdA%KNmXRolNd4*g zu>ZiAin__bBH}aG0g{|36@;f_2eU)l*Td}OK*Z!!6xpc_!4TH1cZC3j&B+VGw*jHv zA-rl@Vh>kCr;zYHi6dCS&T~pr1s!A~61{wwWu#vC^A~qLEh9~kkHx|UT{FvF$@zJ6SVyluU;bd4nd&7iJ7%vE%pPN=)WzQyEzXCTOA?=S^>E!ThPRzs+IUOtA?q_Gz-3|43 z#%wY=7R^hYd_p>wqUg%pOt|A2FDC=?w0Kj{$|LxL)36L4H`yoK>ltVOeSaz7<0JKUASVj9N}}z89r`) zR`Jf^*b(F>*9Wqs1E4w?-g20EW^Xd7nOn;Oz=c2$v?<3)Kfv--`scvcS4)Yu)3rwi zU1K|Pnu)fjL10TqsbQ76wDMk+oVV5SnleN7Vi)xog50OfFA7a6q-(sPK6uf^8`luy zEwZpv#YlHUWc|=omOEF&eIH!m!Ww)GFrKpFT zf#giE8y?qwa)C|=Z6DxVD5UXOd{gCKckVjqy+Ce@K!}jP1I+CSo2eVp06&fRW%fnR zfNzRj%0ga%^QX=B>qWvyaF&<1^0Nwi@u};61jQ|UU%0XWjDV`LL!&BKO#`Eu6mR5GX%9F2fY^7NI~uqFmCp<)6z^ zH-$CV%VYxsy?lF{THVfqXd=7K*I|Bg?A;dcKu#)P`g@NLD4?jTXj{^PUxzT^->d$; zumC?(dn?9;v{#Hc`@Lx1?o)H5eo;z&UAYml2Qsfa5{n#bt$NANV>_z=x05@_y#sU^ zP0U;2SRx(Z3ituoMl9j;zShqMPdWI-UJdmLc8h*(0`F81Xu&Zh=ifKCOP$L2kYoOL z>lgL1rk9}+-6St53Z{MJK5EVL@UapbJRrSCPuk#-v_UU}!PO1?HcL<>V?nSTb#c6Q zIqBqVIv<91+No-9HmWG z#CEaWg@#|srcb@$CkE=K;)%q0U@sU>X=Uy!l+|r0Gr_AQcYXDRjUS)W*+C@wshDR- za%oQk9NSQ(eYl3VO<2h5#e=`TqW2CB`L=mH+%nOW@QSj33;WZEQOm!7OSrd&kycOY zel(>@q4fIaO5%u+7AY|=OT=h{;rnYdxF6k+mSKMl21f$B-wmVA1Mz2jpNA5z5}1;5 z5{(TT^h7Y0qx7N$p^U5ocI_X~NZf_Y4U))Kp8Y-5K1VFj(*0Yr0D{4O|(wk zXA%mGbo7#jPuGlV8Ph46H%Tf?^k-;b9%BEw`P7wj?7kY8#I{KyT7!wr=ScSmlHwab zM)8rFOY!_HUW!px)v)7Mk0*Y-|D)dAWzP}q5QRzGH@ZUAPa3guR2GJ9);mON=uqmr zpo`F6lGiXrn@%O6)8+iRc^!NkK56q(rk(wz`704%KYW?TGClrNe=abzp%cmI0Y$G1 z89L|C%&QZPW?_tUqJ{8MjQvRw+4bhnl&uPYQRb&j9>yuV!eRIX;`nMQ9V9>HOVX0v z6C$?*Wvd>zqs1!8Rz|aTI(?4k$gi(^$Is1R(|IWBV)sNDq=WYF#Ju=(-ze~yb32HTH^~RuaMSjQsF}lF+98N!C4jRK|~>t z%g-SOm9`k;p$PvrHU!sq`uW@w{=q#hEVZq!aN_;=Aytq8tIsvjk+<9qzk7^6)!a2e zHhbkuAy^?who?6#P)Tchs_9GA*aM?0^O2~%yrn`hzUbwQX-epqey3w+uk(vt;rBpQ zLC47|`yU!?=PZOsLY~jfe~d)ilu=Vifm($$QoMOw<7#X3Ag z{1wYwWhXO_hNOMQeUBov8eM#f?>>MiND}(2q?2PwVL)Gy{I^B$QuJpRNBx4EXI zy-9u9zvzw_jTmx8X?ZL!G^N@8O1K%~XiFJal8c=@C|BJIb417kPf3I^?bz>UR}d6++eF@=~nQ;F3z+Hh6v zm<%tL*F`=a$}%kD8mI-Vg7ui=nfiEQh`JX)jhWU1G6a`RWN!!4pBp8$qL#4gSV+8? zhQ{WoUAISy4-L^5)5IZTfF0#tb0`j%tMg^G*h~G#8$fHe(4b|6_6+V>CO{z5V=|9GM>R z@9CG!{{HkNrKZj<4qly55U%XB?FfW~@Ri;U5y8Ao=-E8qz(~%|(PI}+zFhZcjv-wQ z#UzzA+2zpjZh8=Uajp-hiq+;kj|ebwO1m%+VD0zj*Nyb=cdA_6!+swvh^^gde&lMo z`|meRMkiw!lJd^{jW}=~NKlXMDUmq(;l-LTUr2%?Wd_Ga69f9^2x)aQF&ExOZ`)8` zDiYheW(Y=S&R`riecz{KO;|i;ldF#P_z%wshQ{RK)%%CDJr{Q;$y}w&yn`21Qc9bp zX<@nntrwP0>hY89!J)gXCQEKXT{`*oExpA>5N@YYnf>6^<`(fa)rzzaDTzJI8@D|l zniAQp5pDnn(^rk*lgCpY>vV-LqW$dpct_pZZX$mbd$@1%pi?Gm^EEMkFOO1##djtK z!>RX^;+on9#Dp}So0UlDWxN>no9G#PvQocHW%#FzhSZRAFrEwDsOfdBrHr+szMzEx z+w=N>@uc*>ZM`|B;Y~{WQp=BT>pLFleZ!atQ`v=a(vudYw6K()g&H15O(t1qwC~(` zV4p-ExOvlJ?1|#{m=gCDEB>at_6?1XUANI|j58JL9lYRjQe&(sUTLDS$s>+7V3fJ> z;DfVI9Cn3a(yd>X#Mo|u2)yL!czR`ZkMWc&~#lL?sKhm9JRnep2;c_uYH%BktccNd! z;LV_Ax_a%rmnXARbQLrP@2WR1ciYg?Xj-Z~N|hfoI(kCGLEm@X;#Z;*%ygcd#oxcP zCqvH4w9KF1c-#A__l@G_{?8*J)pX9d+M92lFm59phPcb{BtM^Iu_dk~2%@=FB?d0pK#Y3LnYRRxaS=n|^ z`QM?o8O?Br&Ck!+HKY~~PsD4!fAmE*Bu|Pr{?5zZD%P2aNKL)CSpn@%$D#IjQ&(&> zn9MdBQygRqEmCs+@w7pMy7M|8a zo9Ml|dP1||XtF)t2U{wP0UElQ>O`b|-j$0~Kfv5!27^p(JABg$X0G}gAY3s%&-QEXZ*?tPN{; zh(0%{)4k^m?DInQ3ZWbU!cfs&y2L~Ak|ZOK)xwiihBm_AJBx07u^fiqWUAjDHLv0{ z^Gaet0oL*eMgfEcedYxmhu+8OFWcq%pSiNK1jnMfds%+m27*c+$zSW&x`RO6?7n}J z$skBQ>A6Z9lCeOhMm(r0{-C(Y%Ll5=xns{R^nebkvSax>+)<)Vsv$=#wDCOALSgEz zE{H&!Wl@+e;6B!!h^zRIHeAnp4+k2@#?(wsyjR%FN^%D*38+5x^~!)3WP!AI2ow*= zzIfR==tE}-^;)Oxc{kaq!eUG^RUA&pOi%PBjz&Y{w&sO+#LEMn>YHX=s{ij=Z?9%@ z65S=XDxZQBaGbJZ9GF|W(x<#wD zFdn;wsOhgpk;!A(Y?jq}>GMX%_lgE-m%hgL04^6iui-+`JkO~wj9YJFAzg`~9zl&y ztB3Dp*X81lKUkTZ_TFwYc)LQaiK^(vh3X*)08p*)c4mDK$ zP`>wQnV(pV%gMZFTMX1nq@}gZPEBT2fwqt6 zVwt~CX}|_mw)F38g?(n5b42~!i6r!Aw6@Q^zAls8i=nein>sQ;)`4E<>5Hs#A%?G= zpnbmc@N%k|+;6|D?bVR{;?a0*3Sb$ZkHh!5Lx~O>z#B(O@DzT2_-2iwOOcDybP|yB znJTLL9YZp{fd{MvYx2!BxK)?uJ}4M`!~fCbir6F-^alMH>-O$c6r4pd&i(9SPUFIZ_oC z=$8kemjD2s-_A!z9R!LMz>e>My$VD3yvKBgt0`RNptO^SHkhmporC%@AJW= z0qjcDZazT_?b0IRN;G64DYb9nIv*cJb`6nw=guV7WDR-Eoj-O-cE&EC`maFXSO`G_ zE^?8G_)0EnVWYB_v?)YIuXG}Yh7gJpp(0Py>!+Tyo~(Skj`$U z_oa`MQx6^y=+98-iidtLv|xu$_5HBSIM`XocNdzxAU15vY%IA;pHXZu!N@D9WX8CpV@=KwCfv!?(R8f4WuV|!IH>YT6Li3-goeD<=jD}PXA zRW}-lQYgzcOJs=aAIN6AWjx%ssJVrxCYYbkncFLEgBK|W+p)gPVwrvZ!@hm@a-oIX ztNUHpW>;-H1!riTYiM0VMj>-Hf-WE$w4TiOch?U2m}rcC7KOh9W>9iv&v9Mm96-D7 z&Gq$;7#5%UtUcWL393`npA8$xiwC(^Lor$ccSlI&pZ3V0XBa_G{}^|*<}QFSGH{oG zlEUrYH8thyJbE_2m)-hU=5^~2+~40EXTtpL_}qV+1K2$$ySKEJZx3UN04ob(NKSG* zSA{8_r*4zB!0VR*3KGU7HoWf%u`!39HtN_J+RZLfdqX8k>cvzVVn$i1(I-!AM*?zq zsvrQNcH=lm5px2t**_P=p}Wxkfm2RX;{@OFt}eMZV^3B{{Y%T*Pje=-tic>f1Kc8! z0#pwuMBt@1x$=0&5vqYN2|31G=QM-v(cx0q zli#}sUCid84WI5JxBx-S{iFxrmel`ae!vkq#XNI$fXj>kJ3W=YJFtpjC8^=gO25=r z-A_5b{uDagPM)LLNgoEA<<0LtOMY3?0wbhZ?Qb^fsfV6T-4vrUko+QkaP(_jh+lQZQg_pSJ$?3U z;UxO=*bVxow7WI1#|@I5F#n|b6UDQ#o#SgOVR|e|iEFycCR5FxtnwxU-9g=C2A}w= z#zF4yl$8bo(F}p1O%m!6C-=8l84I{tX- zhR(M>9)kxD3@d&2YhJsNae4m1f=DPG1AoTdTi0@UmC?Zq7lceyw@>)gWBSL0cqHi( zeCh2gPSpx@BCpYwjsKx%QkQBSjkR_K1pI=@1`w78EbO zbpBy~LPB0v2e$372XYdx*rCwnZpOar0y`R=2U-)T9+)T5kXpfFpOb01f4|Ycf8K%^ zY9VG|^xqrJ#!>POmhDA76l-p~03hzt`Oa1Z;`HJMEm=1RlQYD#h({uP-As1h6dP# z9Zgp|YJED}{#BK+LmZAT^U;sX@DH44hOSb;SK%K|vlMO4g$MI1riO45YT0H z&<|Eg`2=g4ykjayLU770^m!v_r6f%tKpE;7{6M4HJD|-3^g}dtz6oF5`Gh^z?*Jd0 zl4q+ZXW&$b7{*bHU$G{rFu#I~Fqr}1W}2k_m2omUqrT5V5l0{j4CklZlk1z9by?Dl zGgS^c#L)l&8PAc}=F4@sAj?C~xBcjqyn2V}t zDSpTY5Km(GWsGz_1?v1>%t|NljY|^T#s0Kv=b1*y4y$3uw8yjP&bT)tXpC`=YBj73yce4n8Q{0shQYQo9_QMOE_4JGyE z(-L-++?(`trUGjuOg^UM^!L+)WX6XPJ7A^S27&R>O&dL)^txv42Dz(`+xkNjqA26- z@SCQiA!4OJ-yXDr5@W<$Rn~)MY$OHjMm~*|U|cE$tMTr45^n`?Z3ly+ZF1GmJdKzq z){C6!KZ_5c@3&1WhjX|$AO-UN=G`Vs9a9}jI!lT3!ol=y$f8+nCEktKzU(IAA5wT? zRuk9LP`m^-B@u!+Vv&+*Mj@@?PP5S7on}B~beSWG#&&qo%$rBjt-P8lCOQ@iLEIR7 zl0Gcizu<5~umQ_8IX3;v)6aD8^PxO8^ry$qF95lzl}zSgkkqh(_?4rs!I=!Fm(F!2 z78rHX;_eZe)?ehQZViqaYPy#9NF>N{kAl>t`&p#H@M?HMk=*}d*<^Z&C0-Uuxt-R~ z8d4w35k#x+T;i^t|9~5Vklu0zPB>JX8!7g9Xb_65wCF%t@m*!$=7ltSbX;+t(OYb9 z;zyy_m99}`PloLPXaLLb6Ce~G-&wOx5ff51kwgefji;RZd&KRDFq3ai?a1|di+Y5t zL-3!Ro&3!Oh*PEXUSgm_jtqScrAVR&UIM68N^a6dcH>rvg(1J~%H|6_ z1V34yNgXt9#U9T%u3ZLC6Jn@_`R;~?ttPI&|EtgDS0sj;CZ(KAJxP(>hBB1<4$){< z1$+lh;>e_W;i#>=Y zvx8R2#IY8*QPv}wlj)k!wclGox|h(j)!18-?FLb@3%llr;n=@DTRx#QM&t|$imW}i z8{MWpY${MzhGCn|E9Gsx2&pu_K^II=y=cG5KFQ3*+4R1PY5V_+Ijl0i_e8RNUPoAp zSH=uVK5EH|nxiI_L?fngNDZ-~AqD~6x>&h+xjkCq!=v%;jvn3^?lb06` zbyD5GIa2jr2k4H4hs6x<(l~-58EUTTCs^?O_OgU;9TTXI#zI}WvY>mk<59+if+iyn zR#s!lyp%skcR+pn&*5*xq=vtz2_+Jya_Cm|I{ByBtaLBKZU4#}Z)xwx*xL83B)ZeR znR2QtXeSSS3Uyj)(r`jAKH{wbug3+_4FJ@`{X5_Epn-4AGaQ;hn_jBbDh1q5tVUYP zGvJVULgz-4GwD$v%bGG}K{*7yl zj_DL(7eclS_|5fVJ#MGg9@A-jx5Wp>CCa}c{d(ly&nSs z_KfP34%^BL?W&neEH--Tr^a8Nl^Cq8QnNm}`k$#9+x}mtYH$Y0U4GP`KYh})LTjMr zj*x=~!-Y&d+PbcVOcm}f3D_3r%^Yvc5Po3d#!B4(vGI$Dt%>T?d{3ggLP9Uq5}xJl zEAmND3LOz=4iF#azRf}Dk@Fj~I%&y}nj)OSvE6C6dOcE;x3n2pr_u%dgM90Yv*dr* z!wopyAU-Sq2i-$tWG;n8<5sI$F8r1eEe$hCI2YW`p?G6;T{Ai~#TST!NCD4~^+<_k zU?ZwZ{j(wW344ZTNeiyG44{xkyFUQ`c9{PKN5%`$e4u zD6G87O66AVFIdtQzQ(-`s_vKp?b|IcXC&6n%+G&1 zE5o3*^%gY3xFIGwDCVF;<^ZfP(Z0cCG?7u->|)niqXyFuct$S5zr$eVGm0sKOFC35 zRe=4&f`X)W@`ZjQW$|{0xfhCNA~JlI%qb%gIW7f@36XWG`6i9Odf4O;&EGtZIB=eb zb|14K9Zqnleu8kyLx3@WRi_*<%toXEHskXu3udH$$EEq>r$_@M-i; zV9c*?)d`VZ&$L(T?YZOl2S=`jknAt_X5tG+QSOY131hdnjAlSS2HogBk{XB3p zHW9DOA3k${`53ul!f#o*$#Df9OX$j1Dqy+IEA@KghbX@__&phh3iryHZ&3*V8Jc}U z*cW31VBtzoN#J}P!TVX_EZ{&G9Mz8G7-v034(-g*(iyTgPwdW5eTQ<3B#FS=!`@GR z=u}wC`1p{BV$R@)k@?HArq^913NF~urSiALy%1`X+PGQ^U>zFBWsnnRJwnm~fl)Qm zMD-q|O#~$o?kAvMPy)$k0{w^p3#s1tL?(zAlqZ;yrcNEzr)k<@Ldzd@2qq-Als!#` zf9ZeAn&nE(PmN^bpn*RB_+(EbRkwlZ-Zv_gv!a{t zK|%6`St}icKkiIE9BHyUCRj3Ve!k;zy*l|#G2TzwBlz;O?Ctv077A2fhi;)l8gI07F*YNL4a=ozrOK-R2af`6hvfw;No&6EV<7g|*-e2hn4P7i}l_we& z`e>y^!|gMTQT@RRu}L1A&D}1Pp6H1vh_*3*BC6wHHaRfiYe6&mG@@}R?S7%JWEjI* zCTIL@reVCHbK65DU(3eyBb-OmF!%A}$5ndt*P(xR?o=O&Zcv(hl;yJHL*XH3=LX?@ z^e@MPj>ZKZI#-ZrGJ{ptZ{L{tIKtc8EcV1kljy&4Y~{2bi&UL3d*jF#XL8;}IjQ@x zXxPaXGqXd(F?&jCF2zutcD(9uD~U`}wJ5fg%f9e6vG=6>j=P6jsukt--;LaT*VO8m z#&F5p?!y^xeCytIjMt|-Jbv&(#zR`ApUnQP*A8rOX)X^M`HrB#lpaHWV7w z41U?D?JIuR!18gO3QIR(XovIl%t1011KVACj|Wxrf# z*P=F+R`rRPCp)GOjPc@nZi$u$2dUs9qi+w0D-W{NzuMWA7UJj4zm3HX7|otgYyLem z3#SDM7j%WhXnWIVYm1$>yOlqSbAS8wZCS?+l_iY|s?S;1<-cC|YTjTZGVJc1(8uJf zu5Q>7&k07a#Y@=}o^>iNZ6X6P86CSnbl@k~HXc`Ok$WeywLUF2m%snJuD^A_iX^AO zO%|L`mEb6=P(C!w@0`>Ezg(M#C^;z<^1q6E z@2IBMc3;$`qRWCTWhnvzDpmwR5s(s4QBhHPuL{xygisSgte`?bQ0YZLq&F#{M?`}{ zAoLCqfrK7pQJfy~UAIp61bf9;))>Ad{vyUx<{>)wW(bBy!|R^QsQ z)`WAbTYx7ktemQLos5gQokSEf7<}`_E-cgKOptpLQQU;onq0>lmN63$etv&(*yVq> zjaw1~l$c{(;kaD@i%2h@3zCy#X@$N*lOiL0KSeW>MrS>>NviDN!gh+slX>R62NXTEa(bQiFJtOb z#AKqSrNchXyV-riXorc+I&`$W3)!2=d#1z}VcMcaTDVo~_}QSG*m5k(#IAxboTH{x zbFeP#&Y3gFZda+ecT!nah8v9)`m0xZs~45%@3iVOs2so`LhXo`Yh<0#-)t|hfgCTN z<-Zpnz4HCm93z+(XjgaA`6`CqxW8Fj{cP~^s#m|m5qd~qLy+}o?{iU%PfJ95W?aGT zvVZz-R=lsbD6P;AptHb8<($XY8tVMD?jX*wk7?ZG)U;8cmp&O3Y+0gl z*DBLZr+DQAIw-H7=zIOstpyXIlfX?PTdQ*OyRt9ip*7j88YkxW6UCY%#5Uga7fD%1 zuMou!c1_>7BIN0WZ2S!qo|NpZU1MEp)i}#xLvkF~ zBQG@be3C{!jZBY}QiN1f?Xg+T6;_8>n#OT}+ zGyRlR7uFY-B-#J;5a4Qf9gd)ILMt8F&;H?&)rWjwWbpdm$3y?^GtvL|^!I;^+x`!Z zGeUT-_qkQ@w#{~@`@YAsNb)pF!}^%;&E^D|{^+?*XJ0vStMG5Qyq}d3`ALRy@ga7R zkA0B}om0?t=2a=Kb^DnM!OdbFQ$NyvqU!A@i%>XqHGnNy7u#`s>p!b+b8H@k+Tjwn z8Dpx9{zqwVoqPGRpiMv9`h%OWxJmJO8ihzaaa1vEtgiY`d%X*?k=8F+v5sF7CyQ<& zv8|=>Y?FAHgZ>?C`@z6#aV0Q!U4@B9G8#uz^ z{}=xm)&JkTw*S+C|KH6W{wrhfv+18y=yp@sB|`ug4daFHbj3>72W;LIP1q%ip^Z3< z&2_YqDXXka6`&NdE7*S!uuC8EDxfEkvuR-zq1Vn{7ZMn$&bE@~;{csL)&$w1CL+e{ z>stGpQl4aq*Gdj0*RZz*GNTOetT_yUaquf=0}byAb@=e8D0yAYN@agD%1N|S(6Yg3 z!xc^+f-Xz-kNq%&WE8IahDm=?LtBdp%`ts~BhK%T(ZU0aY0h9a{*yK?;$!IhVS@g_ zZ~r5zb6+jORgl zq0uTrh>eziI70lb&G+cWwszdZhHF26e~YT;-IqZ{7fh$I{`(?lJAU3AKMv2I|6L?N$eI`b0qR05{T3ro&*nEkXJFcmEqnbRMCq0U`FjLQLekQu zUsM>=y^27X){-KEn5&n5)&_PEQNBh|zkFGi!H&{cf}5R_gn7yN&VgXXQtt-K@%AiE z!%Hov+Xl2-?4qJ}o+T7q7~k!d)MsZWS(B!#_fXH#Q>0x&O0c3&ZvvSC$g6L3R14Kz zFib4f?YZ+C^2B|Ta#}Cse;PW*Gk8Zlrp%7TYJAGgoPQ&94Z}*oVxIc04*sMssyGpS zd>6+~&oMg5=KBX2?-z@PnI#{ZDFNiMhR0EMP(l^R#_~so5N=Up4{a+V; zFN*F;ZXE3#=o8vaCxwC3z}6N54mTM(4<4}lk6-*vnk=ab$l=W&2zjPB)k}R(`)5df z(_oq;9Qz?P62w&KxW|a;*!cY=3G??b{Q@1BcNt+bFRM9%7&e{Bi(}yF@ctw;r zG-gf3B~b|g&)HKwzu-%zQM8z`<97RFH=&yEaoGVAc<0b41~4D$q}hS}(U0DSl)UC) z6PC0R$qCE2BRK$g>ety5{Ck4b1X2|k&U+f!U9vj#y2X3^wsf#R?uH<=sr zd3%>Qax&>|Sq-s8w1XDvh7|1s_7jyz%rY3%)^7iY`{Zmre?j(M1s|75U1OJ~=5M(~ zdoXPuDaTB;$BT8v`j*V6oE+z=TF0TpW20^B;4}w2QYsO{^8MJ;%36NI+g6vnq8dzd zEhO;P|LQVncIVo~PCS${eVoxQziGOA!OgTR?mtM)iwo=9Ys`y19GOxXQme~|^~>wi z!Aebp%5_Rsp=()K0`L(lW0|yJL+y6M63p9#&uwm2R(Hl9yK{$)%^`U{75H=&JaEWr zn=;3|PEND4t7f{`_g;~Ti#vgs{|Uuv@~PiI#$Exv@K!G?qQC@~ViX`An7DAOZ>SchKov^zrUtxqUDG73kbN66BFGFT35X(b_4Kq@-CL;7_o zIw)%0*%)uskod&g`Oxy(A0>OrY#uz6#mf$fiWs~lkF%$==H%b4K=or^UV-O46fW{I zU9&V_qv5DO2oh&gC-rnFjVC(Zqo=&mPq%ywrj#&(Yz}oJkt6CF!7lB)ss265e{9OV z3XP&W25$^g`vn{BGJ@yT1x`niPG+JsZL-~yu+sNzGZLd5E$!TuwHTz>)t(h5yngPc zmX&S*yg|>C9KwO6!nB^AR-zqugf8H~h+kj!6hP4iN?%$nNAAM z%{0aXG2k0<)`g$oi#@G#t()TT_sQTLab}A&0w@e6(f{sXMPH3M4qX#O5z+wp=c%Pr z-4}-~H;5=VA?&-P$85ZApby6deM%R4rJ}-m*fyrIR3pXS?-F!2pK4*$@^JXVusGJx z7kk+H`>C^OdH^n<<3(Nm1gZ@DrxvrSD;u!003;KTc5aU3qx{P4=_9%TO5jtWbCwKF zvcxHR5)CP{Mo=~>Dl1$!zenWOPP=`LrW_N&@ z;Mf`>5pDX!;0s&W@4Yw|?&?hdFfrEo)iS-EUxO_!Vlu%{@{6`h=`%qWB-m80DVw zoXZ=JbGdPzc-z7Bj$!M2hkUE?#_<)>I%BtYTPxhNvE)vDEX*ZV`BoWciOY}O3#)3P zjY<<6_Q=l8S(B{ep~(-jYRmekz>mH5fKjpAEg^pr=8!S|2**3pRpT6gM_}w*e=py? z;qsdx>X}Z)MUV?=)bFEiQR|fcCVr2_;)*2lZoWM0KX7yDAb~|il_mPA1GNM)9N@~j zmpFO$Y*616+HP^68OZ4FsbiLh)_%hIYKW0LxOjuxK)E>J{l-&@v?5#rqTpQAxc@nC zqqJf;=Q(IUo&a?fAxT;6#qD@rTjrgYakZyn0iTFpF^fiwg96uCtgaQbs<6iakCITD zxOar@0z68V-lY5`8t69i5Zy8-cvjWxOZ2of#ifC$~NN*1U$u z{=RWzp|DS#_~c0S+&!@Z<;IfN(mdqj9{RUa(pGY|6eGv@yG6aFJbE#aqTscJ1(kN6 zpYyN+@GHw`R?`{THtdt_hM=HBDB~U!hkUNAoKuu@In@cEc|x7enwN^)EQKdj+SPH# zxA-k^eJezM9wBu3TcZ8SsSR7o5H@D=*8^h}XUh9!6fDe|xy+)bYVo!8466B3Q!^oH~xdQs<&;|jxZ`Pq{keP@4_4Ty7fI|Ab+i1de z9_e;vCG8Q~%XzwLGMQbnqDR<@s;J0ovdq2?zv>@vPTH&zgxB+A`3I~e;%VC&v>q_Z z=+~X#u_`7H)prU$em$~(ybK+b%1V}A7Uy6WR?unvL`%=%r*d}pOy%f|J?2h$=!wgp~P8y77( zDA)+_3;KrUwdJ5x%P%9?GK5OB;^GHE?5fMMILn7PK5k!U(eCy^QL_6Nzo4czGTh{; z)7*MZl}%&juLrZgIh39b!I$D!%&_1w2-9`Dz;7qsBy#UL~2nQ!~zgjHCU zv%6#Upy2}0+`IA_)MIGG?pXTwn8;bSj{K#U<7ltVi->85 zIcrdl*!oXgv@rQKqrLCgubhN$eDcIk%bhjSr#o2B;=FtU6pN?Vk1MnbXRQP-;56Sr z@2ZM*4Bss2L@4>283=Xod7mX2hdm&1A)@C&m#dR;Z)A3vlx3W483o}z7F2{zs5G0CrPpv52$KjrIoL%B4A7@ceo(@Uy^K4fgphI%_;NnA2dO2nt$TF##;* zSnwa9Pi65zI@S*t4thIoma-0O8Q_RJHojQP`0btMW z_U4WnHOtK7mTQ;6tZK{2McwQk|}#?-8&t(EkQrWc(JlL7+n_6 zk!(rhW-Z`~@H#8jx(`>l8qtGmw`hfje?#-ls>q?zs87!nqw|h9_U0f=zOQz7RGy$j zdMbmOx=e0>(PuD%x?zd}r#X^l!V@^Q)2I z*mxia`T#Cs$iX@e=^EFT_kbZJkh%V+iugel`OXJj_qM#N zvQ-CoAlr|V<6H~8PT;F5ccV$};OO)klPys4d^g!SXj*Wo$K(Bp1cBTD zDG%%51vWlQ#4R+sn|qS?0{UVjlB{7@6cP2{98t8~5w8+lP7p=fspWvMcy<@T_golQ#_v-k<(1K1@W!y6 zAVTNy%(ipPd#A4e|16v}Sw@NFECM z9?Lne@3AXzGI{^BHaBUySpPX9B=GMDFvk2R4v$9bbGEgGUEBR2p~I_K zo99shjY?XQpXrTja^e!K|Lg3b6+abH^$>36F9#FZfG7S^bdetP_I93`4kT#JhU8dN zY?jaNRuhWU~^<$zRBn@d{^QMFXkLNrOIbT zwB7V7yV-w`AiwVDJ!p_`jg#eP# zwYQkceSO^AK05y#-wz(bH1C}Av;KX~NBH{_B)`qX?DD~1#LNELxlT-|A>QsY?K5m` zwwn4gzp+zvl$RUSB`CVsoJ*(=f>yyV7&ovnAT^(o?GX%6 zw9Vr^F{Th6XJT=MKe-QqCtKneCg?enzPN8?HLkXg)QiwE{=SRLq@&_jZBqjZg%c(( z;h?urmmBi81UKd6bEO5}SHxf4U2)h{)N@vj8yZV}o}kf|kT92#Ih$vVua(2OWw{hC zygo~fFuZ{vEVmC_Ti#7<_yR2pTH7^KArP=!$Wh|$oso$4c-AkQsqY5Cj4$t1?+M&y zQ;i=7^-mRbeOj1w#u^7JvAjBI-y~ZiGACILo@OsQr_jgeMaWIrw*-B+EY8Ppj$OU- z%_6qwFTSmqYGKcXU)X@W!67pr4hD|hTihp6rFvO0&7&=n=J^Lz?vQ+gK_b~9L@m{M)=wM_ zV#Uh4?I{|leRoFAee7-NI#UHnyk)<2PJrq~XZ45VE?8{+al=2=f3fNcjH)5iw{oxW zal#eUiLQLtRig5)>|roksG8RY&UPAX5U?6kuHhykQ(Qg1b==gMvXEEVsAwGME&=ZU zqhmW$h`JAO8GoYHdb?nL&Grz??o(R%);xJ~P^6k@qdx!3?Io%Gnc&XS2#ha&${c%= zm@(6}7M|U|Kp;e{{Z&30rvh_UfnCdq3XMh5X{w~r(&Ut?5o$jlsGY;+x1CzUCX$Ww{I#=81EZ2rqNM} zE~*4^s4W^R^Py6PYur+(=L3RF>B;lWX|H1ET|yt(e3w5NbE4D*it1$f%)=ZTRa(mw zTFkA4pC|Ly{SyDOFO>uTi2LBpbr-nAe7P}TB%{{jHJThgpDd2K8I+Qp@*rB(Q_-@4 z>cKb^yO$(xzNcY6ZbtBSZPB$1+IOkSh`J&T_B0Bza$G_SOo-rwH*_u^K*@hXOSN}o za|SJ#CKGbhm6kQN7P>5su&Z(_{}@Qnnsl6~9r3xtmG!dsuf4g%MPw)*;!-Tt$Ip5& z>PpU`4PoNX$%^p-+Rypth>jgkQZiPpYjXyn?0~9A@VM-1pSQl*?0{Xs?-gejby^;d zGY)ueyuLN(0KxKR``pAML1qcii1dmieP8V7^1*L?V0kSXD0MD~A#zzA$M1;0r1zp9 z2f+G{+-J*vDr^MN9Q$0eI_>ZBi_H@OC+lg`a)Z+HTC4g6ruLI&x2C>MzM2|xjX&ob zy0rO;=XgGuB|pdMm%z9Z{z2JvC3JC>wfw?uZ!uwP)fo=yK(_Az*S*=xqR#QoU(0Ql zw9;(`uQ)e#cLRtdhTO>SaY=c`Kdz1_9F9AiliX;-zGz{mEC-LrJ>=W{s9*NTk@@w( zSdKqq2TVywYr1t=cxa=5K*+QH^ixfS9WW3Aq_qz*@slO8j1TKH!j`XSbw%{ZTHx^? z?wzrXHqy+q^B#AIS<6&mLo$LZHce`e1C|!+>C~ojuXll4z2X`lJz#&s0WGp~rMrQC zwu#{q%Es&1RP2)k&o{;VVZRK_caGm>!xYsEZ`p=WNIZ}JMJeeM*75p+sH$U>Y1=FO z8N`9)ADvUj_mhN4KcD-3>f%qC>F4R_e3|FPPg1E4itTy;{^^<3-Y(qQlHyA2o_E=o z)q64exeMQ3&pbq3?+jk(*0D6%+F)z0PUd$lG3jQN_59iyPCP%y8T=LDAc&G;H1e)I zH@bmXFH85k#`$?BaKB|uO`j1({99h--@k`?*x^Avv1Ix94XYKtVt_j?uXKG3@MI zZQaba*S3Fj9~90z>-+XQiCvu>rUl}0@MNUzB(xzVV3$l=+uKOX#^O>h1&n@aexr%> z%dM~Bn%s0b^7ZQT1|!zP&&h=4XhPF2G>cYj#@|hwX(dmb(*`PIX4&W7$-O8$mqkNZ z5mKZa+C}Xz1|N4@WI7q7Tdt{PKivTn^05SHuE1ZScxqlQYU}DIr>BcRuf5)z3c;5* z^*9x;rI!6F9mNI^uQ4Q%?~8WW^4AjAf*gw`B2?gEeU$=^!abmZIeXpj$J{5^ zvmI`=n2Y<7{z1-O$FiX!6>$qDpi}sp5in(G29maMN1UhQ`Y4&Z4L{=5Ya}*Xey9X# zCR#K8x{zo)rkK7TWC2+{a0Wa#no6+3r_S&6j=ymo4AEZK*sbuMveNZas zzD{cm} zTV5(VNqIvNYQ*tdQs*8NP2|$iC%a=)f67LhrTllC8H7!6h71K|TsaVu zYYA>&TQ-d`#@3T)H=E1ol_!&LCMEF*2W=nB7lPX*0R|qaTVGTj$T!Dq1DJ1UUfb2DuCp#M}It%fJcNMkeEdDkTEg z;FWg`q^V>Kd}Hq(yt&1~oY5IE{z6onPVYc(s*vyl-+?*x`W@)}n%f1jyt4d4t?-3X zJk_^huhgKR68e3{9oL5AWxlQ|42*R(Lbwc0??OdW>K&n#57}nq#xY%Oqc;dCx|Qwy z*ccoagu-k)^K1(e!0ZInV;vLfaM+DB0}Hdx-n0W$8xy)M^8hITlukZ32e4?9yB&Mi=OJ>FW#}0D9`2hM1|c20 zw%XH<%R`*13njF#x3UZoX)2R4Pz>>F7SB(dXwhO;=Cm&L@{vgql{AKuY3g$m(#LEM z^;Tss-1{NI-x@$8yq8QL=3wL5+NP*TBTyR`fiRxYe7S&~)ZUXF5{(qT#h5ShT~k** zUXdza<>hG$o1iV$`hX)#6}m;@x5%{?()Au99pH-2tyj;0iVEH`m$JLlnb#09F)x=- zw_N|(-eNA=h_%al!FTFPk7Of71k@NdnO@m~C)X@+cwXs5B6akQ!=|%B192!jYRG0q z$_vm_Ac1!p%CDqvKix}W&zul=$;Z|nP6t+2GRSUHe>E;r>Se+;fd_WATh**igs^;z zS-Tk~gL-Cpiujq5D5FwRj|Ol4XCh*9a`6zdy14e+R~+%J%lx908S>5Ej_s^ zczvz2ox`Eii5z0{oqyu#)|#@W6kyR{T@J?6C>a|()W%wqw8nI}oAC8www z@&-t@L*JGipY*D6K8h(i*fX{jl!dC8yJ`|C#2wVr8iS$x_{KMqL?{xk#o$qAm?m#V zWjy4UQjEtfmYw!GWu0TLGgKAkd_RMmeMz4A=jiTU;rVMc?5TkNV* zaI-$`b^mkfL<-69J}F=3d>NBA zAoK$}I+lo8ls{cShb_5F-zU1v6OAd1o1tc%YoocA#NzGliFIzL$ERcyUyhm}7CX6j zWzTeBa$lcAt|*RIb?@-(zI8OdCS=_}Lezi0$ASad@!l9j*htUa2i>AoW<`^kGrPT| z=JU-*Rh;AXP^CIW4tBT&1I%@alM(q%YVa>NDb3}{a`_W)k+vl*E~;lbaj?ra_;2_v zJ0^|eqoEZF?PJ(T5MB?t_OkDa`JW?CmtMVar(10L+#K@bDE#RLC(*Gl5S_;lg9=LL zwWF!WEz=7A-=TB^Ef9{s^rJhaMx@sQUh(y;M>^#A|t`f1wI z%pkK78_98RFTH|6mh+xHw(H9kuou!)!&6-{irq%fgVX(#Z@ZM5#P-no=eyvUCejU0 za$h)fWRK2TN&n0onjNm=<4+-GSv>Gzr)w_5yYU|E02!Di_vYE(4_GDWPaW@98n&}f zDL`_VwjmMce?0yf+=cE{kdTs7*P{6jlYEnCb=WnQnSw1cDHAuZOoZJSy2cn65_3$; zK04@Ub+a#d8SOB+k?VB36fBOUNy_P9?5M5GAc z43>GVsx`>E1rxSGH~X)fo(;fE&h|K##`re)4X8!!OA08zoq+v}huS~VD7;|B~`z?|XCKg>}f^)9*zEl_fkF%*gsM@vk z^+P|aX$MaIwjZX~Q0M&<^2LRrH|ipjg@h$@^Q9>V3f8eQ+!Yx1$vGpV;=XcL2QKxgQkojo)g9gw+O!0)W8obO4eI5nDb+n$<<7O| zCt1_3A7NHJ*aUO`9!1< zR#@tyhJ#8=1M3-gqNT@1$%wmYibAU^<~KuM`K}y+)LCZ_kNe_8sxJ+@#Mwvi_~I(X zQd$d+Z+gm5*Z7wWuG(q5wr;x4z?W7lg80dvz#1^QW}$C+nUC^QFNJJop2054F!AQ6 zZ$_?!Q?ToN`F(nyto!z39=e#2R~KyxtQS65bz@-q>_>QO#{1d_#9WIOZKL%|))n}! z@yP}#-b?c!jNfT#bLbG$AB5DK+t~ay)%GQ&*c~HHDl&D3Z!Oall^9hk_)4;1o=(}? zJlv{|kZ)96=z3bYR*y@;L_r-s^>*AW=<5aG6GlaLfTJkUCu;dH|M+f3AHd4~r zf;)c!By`K-C}%bX=3jB(sItuj)+JJm;|njLxrgHOyS;TOjl|Gx>~|av1o#2q-i5w?tvx-_*N8i<%8Y|;yrxFh+ZM~LT`9l}+_TbPL-s(D#Iin)z{WWwgwgUA| zFU#oPu_PFCet^z^+L5|u9!kYUjtEI^mJ4l}9X2!~_*Q&~4ts#r)B8BwUG6bylx2pn zD^VoZ*6LVWe>`4OqU)d_DXGnSU}Emw8U23dWNC7DY>3?gpU}U9f#C;Xe%w=0>1-wy zcXs2vJS5-$^yl6E#(1Z9^+ng4kL{}73_)v^f&uDQXp?MLC+mmVk(a*s%{`@sDELN( zf3;prbdX-+VMu=HM%m_=LqJ0oB7s7#NM_;)a1^zp1eqG>CH2-7smdzScS`( zz&!wj*Gi3qK`B601GIKMuphQ7n<$ZOExrbMXf{8Arq)C`{pl=yO4dG`03NtCVQK9T z7*f>I%0wSDiSEu+kcB&I%QM+{1)Mn;pB`v1*6|rI#MWZLROg-D=78?Ahkj~NNv*Dn z>|XsrHY_Dao>OUdE{0a00m*LIs&b&|S$|SxLcLEGxD-)>@*hX@EP>2w188R(OXSz# z&oRNUXLiCNyIN59t#gMzjnWK>ghGMT5U7NIM;pwT^4FO1ojfTZ++GKGd;zM5{bArW zfS?NgLKc*TeeY)W$nQfwb4Q6_uwci6DGzXrt@6WCl??El&wv!ij!f!;^aB`{{SEwF znF=LKzttd9EA;Sp>}?x0fwyqPR8N#PQoq$*el7NpylLRNw3KHQ=D>JL;-{)vjGBZ> z^Xao`(uDFJdFHR~T|8?4DIFBiQK|(b{+Cv=9Bde+4~@|w zg$`fXx*^jGToj~o--x!oLZQsB!LC1)u_*_Dl7ndhZ}3x8KZ5rxMM1zKy|f!4X)R~^ z)CqVHTdGCuj$*0BEiC#J;3x4u@isody9!mA${Czd9dg*il<60hGYdIJ>3zVKOMx_V zb)G2n>2lL58u)QiF##6Rz_Ni1C7%IL66@^;aUa%cU|`g8>e#|=j%;gNNaF~?N-K!* zPcE$~0r$(Ql;JD1_%@r%zm27;59Ec-TOF~B$eShQ*U2_^%b)BJccX9DlGovZ-d2d^ zt4^&(eXI#uD{Fmz6yD0FZW%Dt&u-VVvzNOg%N1;#PIyy6BrMc!FvxKg=XDbqr@-e~ zHb5dMs9|&!XGMG*1QrDIfAe_59OCiQw1JtoD-96R*OyV`gDl)bVT2pvA?r%m$B!)q zC(ySD%r=5iHc>w(syS*R?Zbno zF$~Bmg107&Y>e}Tv0d?4`vxRjSt&V`rq=X@Lfs6Z1mc@qA{f2)Zu5yEX1xh;zS1I; zlu#=)b~s;Y=_v*8gOjuj`Y`xA=bDgKa;EqfMenbL^0%*MCqtk3YkguY^!tS1!N-;y z!l6!OfbbLd#=6{(rPi!SaBEvdG$+_4ph?1}CUDh#*<>TDPHzqkWEg#HM6J&=#Ao*KNI8P1E<0dZkjlrn) zKhArE>=f)fliIkcvA?P2N!j)@o`VFKq4*aLCGz@L3oJHLZo*Bg?k!q<5O_G_Gt*4_ zVZH)-#yKvlkA!8x6modw&*x=9y*uv(y&$tARRDrxW!)dD{gp&`$qrc<-#nnRp}E4EZ%L@0@%tux(^ufmnXmAXf0g$uODq>mzhlrf_}=v%+C zu5bl=(%Tlvi6Uw1K;L#fAe-2K0>`JG?Y{T`lmkc~949CphSGDIsn&45GL z5le&}o{V=MYR}47gUTyoFUNyrc5Fz&UavRjK?z|ckm-FMZ$OU%Vrg$b)tu157!8#1 z;zGhv*BhsOL!+-U<-G}Y?B#b$^|fW^siB;b)Z8KB=PS~O^5!pqR6vlK;))BmpSYVs zcvGzBuTmiE%m~Eh15?&FKCmER!&LAdFP<;?@{G$9v%Z!;?-|_r~vO_yp2loee%UUf_E-9fbZ@w99Ya*Ux495pzJ7Z4t#RoOvADip~->l*oI4xr|} zUE5t4jG)x)kD>z|W|gFg@TXY>qCJ2k)PaLrL#4(rnBqAm8po$tSxkNT-e z%Ap4t+@wF|0tU>=4+Id-nVV>dLB+$>-7CIN1)CBFa{@Ko!%DX<2Im2j9RPu^?&#PX z?ODkL?epdSy_+THv-w-SW?$(JCItIuccje>(fHgzK|6|eQZvC=Y*bG7z}YHgD{9%m z3wiEZa2eUk#>&yV^PXd;FjI?}pu`Db;ywTZ`7WcdkI~1I>f-Rq*{)7!UZ9Rs8#N9v6;`O^ADYv~sG$zm5&AJL_})NkA;i=Q zb0U4_Ern{dbmT{?7jzQg9nY;Z+AqSOjJeAAz{~%9!rhcFkaG^Q8h(@9zsWF+IbRo^ z2N+G*izHK`N$w;0jS2*aA|y}cWjCy}Y8%5&_M$4iipQ0w5pk$}6?L1X;!u!NdY1Fn zsf|y4nRgB#+d0OkXX7TXWD8*_rVOmb42d}Ns{(r*3B4FF2?YN3GG3MLL0h+q zuf^Ur?@yum{2(&-g;A%Dq`*EE{Bzlnv3pqw`DJdBi#PZD-J#$=9+^) zU;y5mE_ioPzU#`0$UGWk4c6U;B9xDoFHvEBv*wHf9x@sVT$@j?9(NgEZSCU0Me)z} zBF&znd44Z;bDFmsxz{l7xzDk{f8IC~=fTdv;-DeB%dJ(+12no1kwUP&p{xeeXMQ>s zcr?;(l8QN_z!msaGKgf_(|%RLhvk$I8EVwr?;@ab5pRXs`YPXD#j;Kg?en~l(i49y|N1oI^~ zEJ&bpMbI)h;?1w)^3|Ymq&8UIc?4LEq}+$AOeVhx&rki_+r>9&5u080vyA~@7PbRu z^kSJ5VO&TSa|!Z>fQl?qaB8z8v@Z4<7@O}7P6sVU=T^2rXwt@~il3Hqtw-450g0a{ z_MEjvHE!2$vFTcwg#=ZoR(VCHS^Wbt5RU6EoDLUkG8@)pYl7d((T`+p zO97>o=?(~w=JI>bd(J}29(3LJrer#2k2=xv`ipph#*w)p7?{`D=}>u8b^F7AGNrKI zugo%xRYaA3+{RRJ1hH4U`BJ*)6a-VO4UiD>_p#8zt}~Q$*%&a10tpm|v}(&m#KWSI zl3j+dXhPdTPyK`yKZsC?P6hq|`e9vwF07tsXVBT!dYF2D#NHvu*8L;;va>zrJHJbf zu&o|gO?3-n=6ePp z?PVWf5lprQWD>!dg+-RSfaoFA%cdJ}hDx}JsWZP*QNg16gQ3=e3m+&K4Hw-=G@peG z^CqG7nT`Ux!D8u5Av}l~21?nmKL3qJimJA1IGh>w`Xa~?I<>LvkkOfwm+AvsbW&%A za$J7T=A?{S`G7-srbMSc$TR|D-8PHwW)Nz)lXy7UQo*kL(3iZ@Arzhf_l;`G0I)E( zPA_wg470KtyY+MNRyHW$we)+c&vA>(Fa9Pe*Z4bo=>QaZF#gF3PXoGyzIABZJ5rU; zkuMy`GXO){*k2kV32?DNI*!2Tmd-`Ws$9Fuyn?@lOEHpA|3Flh>CVy_Ov`N&RJdQE z{#(4ppIs_AHz7OR>NVC{3@bX=lr5?EOBQt3)<3DRLrBox$J4IX8@3ide^PP*+3q8K zh7`t465c+ib6o~-^f7QDk=8&8rm2>uJ+>ODN7jEm$xM+%DtM zUEgrteCj_-U#1FgXGoW47u~KL(EMjxhEDhV`~}sG&v}GC_CH+;a+h~qX{T8zhgjJ` zN!XE?+cg6>J4;JBOXr8)u68!IZkv?<^}oZHUES=gt-KCvJ6pSX-gC7&d|u+*xj#$4 z6~Kn){E<=HtJzoXKKXwDvTnDF literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1231-image-1.svg b/generated/webwork/images/webwork-1231-image-1.svg new file mode 100644 index 000000000..3c737c284 --- /dev/null +++ b/generated/webwork/images/webwork-1231-image-1.svg @@ -0,0 +1,408 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1231-image-1.tex b/generated/webwork/images/webwork-1231-image-1.tex new file mode 100755 index 000000000..eed04bbd4 --- /dev/null +++ b/generated/webwork/images/webwork-1231-image-1.tex @@ -0,0 +1,150 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[ + xmin=-5, + xmax=75, + ymin=-5, + ymax=75, + xtick={10,20,...,70}, + minor xtick={5,10,...,75}, + xlabel={2012 Republican vote share}, + ytick={10,20,...,70}, + minor ytick={5,10,...,75}, + ylabel={2016 Republican vote share}, + ] + \addplot+[domain=25:75] {0.877*(x-50)+50}; + \addplot[soliddot] coordinates {(60.55,62.89) (54.8,52.89) (53.65,49.64) (60.57,60.59) (37.12,33.25) (46.13,44.8) (40.73,41.73) (39.98,41.92) (49.13,49.06) (53.3,51.2) (27.84,29.44) (64.53,59.32) (40.73,39.41) (54.13,57.12) (46.18,51.21) (59.71,57.22) (60.49,62.54) (57.78,58.09) (40.98,45.16) (35.9,34.79) (37.51,33.52) (44.71,47.6) (44.96,44.95) (55.29,58.32) (53.76,56.88) (55.35,56.52) (59.8,60.33) (45.68,45.53) (46.4,47.25) (40.59,42.03) (42.84,40.05) (35.17,36.84) (50.39,49.9) (58.32,62.95) (47.69,52.06) (66.77,65.33) (42.15,39.49) (46.59,48.84) (35.24,39.46) (54.56,54.9) (57.89,61.53) (59.48,61.06) (57.17,52.43) (72.79,46.8) (30.97,32.64) (47.28,44.68) (41.29,38.17) (62.3,68.65) (45.89,47.87) (68.64,67.4)}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1264-image-1.pdf b/generated/webwork/images/webwork-1264-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..64d57966e8b0508fd63d5396dbe8a1b273de4053 GIT binary patch literal 5614 zcmbVv1yoe+zW&hNEieoS(o7&Jh?0_0(lx+P0>i+NihxowbW4YHcPpYus7Q!_bW5iU zIK;*O`mc4qbH8)XJ?pN$*0c6+uf6vB?&lZJ^YZB`tBOL!B*^(z_IFOnr9lu7($mANXoNDt4rz}7Ya!en(M}+!q=eMnyX0PIPXyeJ-0#OuhBlH}ined> zg7I#1L@V&^kVS$Sue)GEKwef9py@SzSL4OW#dJq=UnYPelvzdZ(7xg7qRRTCU7YU; z6iLMv!3FDOCm(T0`&=`j3tT?hJVvE@>GjvYuS;ca8J)`@mK>F44^2AzWXSe5ZFA3j z=KJ6j&CcAXq^3@+kz-D}<5!elNC($);q=|Q%!xVU<<(2C?T^O8MSA6Y2^Yu?YvZ)+ z`SK&NAdjpcD#nQOi!LkQ9WS?N8r$sF?aX~u3IVnvuYpI^#w?3XP;g1UVaiadAx@vB z=Ck-w=Pa(kYN6O`XUFwjOk$adan$kF1sXn#(r{K>ewt5j%B8rWt6-V_S(0{j=y)t_ z(VjrFfI8-=X?AEjU6X@qZ(Uf$)<6rcd-1eLv3ZI_p**W+u*e2;Gh>F9TC&%V5%tWh zi9)!q*K-io#x0WNJE9RO*A5L!IhvC_M0hHhwDnTe%x{E0jJq5!`LbH#x5)PJ&ax8! z#qzdEHs_0^=D^%9yEd0|Cigc(3Pi0Q7}=Bt-lt<{h#-*omK~KoZ9!l2P1jebj%jf9 z9+h{AMy<>(8C|7W!H+z;^eHJfT=ohnQY|P+TnDpXSffZc`l;Hy-c2XV6T7X3%ek7n zHZ3Kvd#=M~-c9tnu7=OFB|^W8y(W-a86ul9;pJ@Oo+o9v8kf7fhgyHSk(HJOtZ7*G z=y}7x4r``-CN=-D*qtw;iokIrgeBG`i9c3$-^Iy=oMXf&NyE^|8Z03rL3Gkx@vcnJ zL!>&xuwYj=u3Iy%xO4;<&Vr7*!I|#|Pbpul`v|VpPs96p-=Fh&nTwxt7?Dx=b4S1L zwMG(hOa1u5kKVg18r14DoqwB2|Jl;Mc`WN#N{LzM7CepjKp7MOZT-CXH>H9&)}MwW!C+NU!JnU5<{QvKNJxS zJ0DTF)A8HF(KvT&ds8jj^1`VR&iQsE()*maE4Dvxf{yLjW%)kW`wJLuz^L$Sd5%H2 z5K?s)*ZAt}`AK%bMPpK?%6Wk0NHtQJL|5^qZm~8!XZ{DP!A$=HTdP3R%$B&S>W0f8 zO@h7VA+}O1{shUElRPoER)v7wc9k5n$GYT3b8D;Ks_T5V@$WkIN;{*Cqa>41MK9=* z+y0vp{uYt(LgAx}xJlQW#087SU}W>cW^V&B2sIRlJ>MiV(RXbwQ*8@h@wteo7#rxe zlTw52Kdv*=N?oL}{^oKpXIe!nD`frM<#Nt6rq*0%GUx4>>oIr3tWxGeUEb4J_1*cE zLsH{t`JW}zE0>~Uzov}#yEP}NQ_pExB5&ho~&l4cb#naevJK$$vPaemu1TB{%EOs7Fr+`T<$z?H z75JsCNG3DBZg=x%#U~Ro^1*}7Sus~74OY|b^U%;1Wm*w+vG@ch2EDIOh!{1#iAPAR zUO9YP;FW8QSW{)|n)eJb@m8!fwcM*z!(xp|CAbq^qHf$;?wQnk#OAZeYU_UQa~B&K zjv_y!;pJ5O4Eyt(aL`LUn84T9-d)g?;mtrycd?lO#+ z3sP2oofc?Nl-^~VW}!v1MAwBl!w26}?N;-_tKoyg%pg4PW6ln8guDI!*P!dWKdJ?V z{U`DT6Nmg&y)1R>+(RLhvqwz8nsOnK(l{P>w)6K=V9IpLMgI?93Q28WWpvTk4Zl%i+|0Tkm7H3f9f`qQpMFT9~Fht%-7ii{_hxk z)e41vJZUwc7-^Qe0Lessis|jk>^#y-S)xH>+_?S)Rk!~^m4UY{+TQ~KHoOL?(!UXY zaj4{9nX|LEfx~U=?N_KgqR}+JhJ8+`VjBTulWUhpX>ik1ue)0`&f#3JXa)OM&-$TP zw2lc+e)b;bMyRKTge-P}BA>pe;jtzSe&7vyCm{TxyQjZj&urY;#-`HI#%3kUTGurB zRgPXkUP5J=jgzx;b)k*DjZN7?dZt|OWk)-FCesC>%wc32-E21exUNC86kXDtf}jm zIcbF0h2B=~6#KZ~xHf_pc?${L(>63L(NHHDsTuFQPm*oevsB9eiOy`hhzEtQ=-KjU zag9KnSIEWT*<>(n;QZFn!Ku}^w6Af~P0<7(L2SU;mj* z=wJ`|R;Q6O2%=dZSPv1j%pu}3db%%diMnxHvf22k%sT(ferJil@)qx};@3SBC()A6 zgsulg49e%R1{Xy)|dVhOE}_U zfA^d{Cpu*ZoO0lua^jx?2~JrEIROB`Hr*dwrTK%iw11HA8t~2N0fcyXq-ehg2t-^A z0)Z(ZJ^VeL9i7l10XsnuR7OTh1O$UXWI*@b5T4Fk?`!uC(!RFsp6&nu9svw+8z6bTK?VR5fCWH^54iqcfFIx!p6zKQ^=S!a7Q1)6 zQ;FAoj74t}V5_kCX;1DK5-SMFku=jtWPQV`dEs#=OLMlAdxAmtxh3T!|xnUzi5@g}qR~Ji(CS zREhtrS&3z-JhjTj!;XC?I0c@uGv#8Ee1Oj|QBfg*xUq9PRu^owG>JoG@LeFC8;bB+wy|*=1Xv(; zo(VIT_^qbDtm{qVASM^n7FpnA^)kX46HMw>5Kx$1np)Rq)d|IM;ka?EI9FUDmj1yu zumkfpy|uZazN!hcM2ck*SjQ0nl`uAzXfskQ6Ye{c5yqwVX$iXZF&GnNnjVI z43*=Q=YX-om|S1hzK$qMCx2@UDYZ}DsZs_OnUa)8jDx3=RFY3nl*Cg~PGnM20m%ua zaw*BMCybI$C@KF%b^k{PGMKo`e;W(#p&;6>BPOqzV&{7_Zc@-&G3*S~J}urCfY6ad zLr^5i)tl!IH1YuP_fNokR8^;8?3}@KohhQktc`j=#}2=j!^@>q1D?8O;f0zlS)n%@ zT|#k;4EPtD?i20Lsx!hJw~IO^ri2AJeEI{uNzX=(G&_dy_}z&qRPFT2UJ@eL9KMksFXx`18YOXY&+_bT>CRFruL zRrJ(V9-aM4Ol;6dDHpL$FpHuhyP7Wvl5;*sg(`#`n&Jal4%Jjn*B*^1)PX4O$6LP= zyp&5{ic=EIQCD1$j*ir2e2fZFL|IkqG75ZQCaFRWLzSClBl^KYLZA7_8KZcX5)%r3 zxr{28K2TL-Q@>@J5y|e*lw-@G>{AyD7hyR(-1h0ny8UCtp?%R-M+`n8E?baTuZN~6 z;@0?lE4)?phM?_U$5U{!Lt)XYd?!W3cil!Zb)lvLQvDa-cogxT^YE-NZ7$(>&)3By zJHI4M!7cQez)v$fOa+rAtmAArc-_!}W zsv*w++L@sL#Hecv6=m zX`hoU(Q?WRe&!`I$jzlxq4*}|Ych3g2u*n1z=Tl(Z zI;)A{1` zStDe{@7IO&?si06(R&Ack=n|IoKkllkB=M4Bg$X+lww;%W0Y7w?vkjzfT#DYiM((x z9x*j%n5bUtsKtp67Qqu8QjNnE;Y0nqn(_G;e(j|mW>0NQvhIS0(lxIV>Bu5`XVwy~ zxSLtF{$BCWzf}Am69;~1>tcX*1M7gG64!Rl(+drfk^DElPEzWhd|g$tDe}iX+Ti&k z;Wg1b@{`&LHAe1URo>G4{ChuMhm!QoNvC!om*1Pqwp%~-ab*!m zUy!u{h_gb@J2>|F2_X>IC>Mxexu}I}G)bQ)z_Qe)8w&KI zlvdz>);tvD4-BOj*Go^(OBMPSClFqS2RWDQqwH_&?`bpCgmK39E4nkWA)lN4 z?&zHsT^_gbc%E>0DhOUuLsKUws%V5C@{z4>k51qDZ2#(!ih!oamNI6-qQI+?D@-e8 z*L*9mlCsxx!wD61x?oYzH9EB7MW3Qsc*4!V#ut+2md1#1Zw(wH(y4mWq_E2Z(={Cl zxziT!8~;eChB3gS;h{=yn`1Jxu!23dT$ef9?Kp8|WXyqrhVyvoo5OMb^lYzsJ(^Wu ziY;dYK>B06gl$J}D|vKmR}5jfZT`^v>{;G(Z{|Y(4Ufm(^^D=oz#KC>afoHRY-453 z%g+^RMMPC_`-ItI^>h__)|wYV4|#rBY_gT@fdw^I6-bD5Jvlj=t#k)vCy$pd4^K9e z6kg5W@jlHLdCxOI7nf7?d(C`ferNLjGrpu#(~woxl>*9NO?n0LJ{?e>^iUS(&eUN$ z(g|KKKB@Wc#(pQx4}6o2)SDZhHrW^hs)3$sq_u9okET@z%$8Pm*9_C*#3#8kd8-E9 zclP-v`;EBNn6B8;o$wlBgI^Lu;csqsTteFp94TF5obcExV_c**s0`sFebpuNrtDsc-oKm4)qeUqr-k|>M6BH-^W0Q<{%=Kr;T zi9W&stgdV>XD2BoBP}UzD=jHuCoTbjic27%wld<9GVt3H5(r!IyZ`$emSBA(5={=h z))jrQy1N4s^yi?d4>oWPK!9KpU}JOW^%I!oziBaP@xLlG-vE?47{vHuk(r5*$ZjdW zE6foEL7@&|?hlSq!9fOqCRtv}0u~)zW#~&SkzjHQa7FB8-65s*oh5g1NVePB`?dZ< oGd?J$XNUNiRwr5h-wE+T!#&Y{o(KnW32_*NoR3fCp(^=*0l@h(xBvhE literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1264-image-1.png b/generated/webwork/images/webwork-1264-image-1.png new file mode 100644 index 0000000000000000000000000000000000000000..822b53e3b826c454abe59346bd7dae36645740bd GIT binary patch literal 16704 zcmaic2{@E%__yjv5h_JxYnOyUkxW#Q$}%F9eGJADBV%c-EtbmGVwoWvsf3VZ8EXjP z#mSy+QpYQMOww5Y&pYpQ&iTIo^8tLHok{cJ8fB|AzgE%;86G6JvfS6~zI(^Ms>#pRu;g7|L)6Y4+n_%fhaytq zs%L(9u48nTZsE#)?^nWTW#w$x^3Z8%xVX5?gC%|2lk3k3`!47r?&?rlldfR&YBkaH z%4>_yii|5m--0L;IxUq)79M>aZ?+8F&dVhxy&x*E8oElJzJK}n?H@faKU{l(^SJD5 zLf!p{R@k;tBEIRzw*A_*2hRo?eP4}T5`V7kke++i&7<+)>1&?4CzwG3#}qL~|F9;` z)Q#&z?>V;`)JoK{F8|w?tB1_Bd$GcID3;4}YVE0okES?RdTxwA*NLg`N`KH+eP71% z^KZDh_S>oTht?YZxzY317R(}n}``51du$FGabBIgJM)LTE z%u}+CTzY?Qf3iO5Gq!kb&Yix`ay!;X-`cEuJO0!kgSWUh+pvE3>kM0szYNp(E5vGJxyYLda0%e=;?y$pnY@OcW9tQ`nmpBnbp@uTaO`2&Kt zQ2%Jhb6;8474`|k;~f$(H0OTw3Gc`|)^{tgTo|re{QLG_%GgbOL-F6m)^6TgeJW;) zWpJeX7Qy{H?w$GLgUPea(%NGWO`eI4YEJDrV<*hIS^cNup@SDi?C*J;W2OFjdAY;n zvO*@W*V@ioV_E~6k!fdNwmK@BNo3zCJjML$);C8_;Sw&(D7UjkX}PIkA0L17*|90~ z79I~_TBy5q#T=Ag>>%%G52Z)!e6MQ*zrONnlL?PY$lSY7e0v^u&&7!g69*BkN6&Xet+iX=6hD}uYB>g<(}C#v;O*%U$@R4aC%()Y_s8d zC;8Fiql%+@M)_&FUAy}p-Dw#8yZrBiv&OSVvwjK*)Z`twBWJE9UMaW|*7Kl8hBx`C z{*ZyPfybfY7^N7sn5);zibIc^Vs-tqZWR1dow+olx-lU*H8||-)3bSJ^WW>;b2=QC zdD`pGa|XRSh>uu88^}WY>_ls*1KT>`i_uaBgT;Q_YedxD z?p0jbZfU#SoA}p9KYVux|HL5zqkd8P{PkyhpZ`2xaFPE)vrf~6zKhxyMJ^Uy5Y!2Z z!8acN@V;pxel&)%H~dj&+}xw)M<@6C>3N$($q}XN<7HyhWarOI#C6Bb$1#os=~h?= z>Yu(;bzzTIs^L&(^{c&_A59eW^K@sPeu>zs*I*`XlA$~LsEzN%{>L@CHYON9+FpB# zJ#l!`@QL9y!`$2B?x$Q#zL?l3WPJOu`LT-yx&_+C&t5!(bK7f9DEvZrR5)FBYH0tQXdP`1oGCLwZw-BhhqZxg`zD zEs@gwUWiVmU%Ni)b%4H5Y(iHjw|ff}dzIEq$xZR}*LP*+>Y4S8uEE}r;K3%}c!M=M z88laYgW_xKcXD3&N6Xw4GnIAihtp5Wh|7rim&G1jyL?ShVE$R~4(3pB}M2*mn5MQIErohck~lpXfcjaO#I{ zgq|)1x^_gR@bFKsCu9rPe*ZG zuKS^{4_;SiKg}}C_OKqdHP-EPFR=77kMwU&PGHq+2@8wrk#Nd3ekq%|eBbc(6Eb-?M_GYwkRwM*eT@g8!`milz8-+_h3v#oW>^``4` z2dC%8#EW6y$vUt4anr@)0r75E295`{e}U=d6Xr=pdL*A8i9d=~9cIlJZ;qa-7tdF- z^l6^<7R=dMcHB1S0P+xwNUyi5=#6^ie&tRC@m)g;(xi} zE54UQi$an;k_+nV{hye9Fncd&;@`H=vHWR{*<<*ntxe<1kp~rz$$FPT;Ix}?-cXN= z%l9A`*Uexqt_ASu<~J@bFANvgm<<=#2`U$ti2IY0OIR+hwOt0fm$g2AY=$5xEiH{o zr8ZPn>gwv++S=wj5MN(4@9k+=U8w%6wP|Fy{_mms?ykm}soJU00s#R5QBl#haJOe- zqRE>CXL~b!yv)Huv^Y;jR%q4{sS0zQEnE2a?%CCn`>a?;W4QFuID?*tRbq`5Hu({r ziHi0l`ttGd@$m4}x|kXpoz6_ZbvJ}K(U`2HC^hiuCG;V&v?$8KPXFmH(X*OI3$3oU zmA}r`)qh7z$T%$ZSB!t=;R9(qcNS`6TU#Np`?j{Gj~|;_TA+^hCObPT6O&8g;-U>7 z%E=zr-c=Un=A^ebR6N}&3gtYvwzj@|_pYm}YhGSnP*701;<417Vl^*bw6{04G&l8q zYWURM(Dt$E^WcYTE|!9V{Pdy}3k!oU-KE(oM_ZHb=i&6|&JGzzkG7WPb#*qjwlod) zefV6J8WR)K+zhq0HdVhX=xQv_Ja)A4X*{F7^p(odN{1^C9z2MPi>s)pXp9P|d-=S* z4Xg!y>S-M4Z%ErE+Ew_pz)1H)M8F#}v!1S!o|31&zP=$LAv9_%9&cSzQk0qbGFMN( z?AFbapqtg9L75t-`PQtl=Ghp?yXi5X!0xr{ii(PA-&bX4XE!yzd;aWc3sgy?(cZkt z4G#}5D$MQfC{IkJWn?^UsDJzLLGh|_l zYHDf*2Kq#z)s-v8u7oQwl9GvAw|ScAnO!xuwYApQJ1Zd}dN0i9=FOX%Hf{3U$fv=* zURYRo&z?Qc(o)RL3@pqq-BXYYI;*aJ^4Qt4XW!<(mXkYFT~%H}&s9}bKCi9$D(k6? z%%Q{5(itxwNlQzss3?PfqNBsY?}xoTe=1j1>7~-K%BYA^!qwWF*Zq8*--QHqrroWe zcvOJe&Out%Qt!_2UvH~8Y~$zfpw2xn=6$rzFNGIDcCR|(U5e0AUiI2 zQ|SKJrtX^@CMFd8B2@Co%}116)Rd#DM0ODBzuG17)nC{^=MTQAsqewK^*Q(t!kP0M z_6J-1{RMaTN5{T`j_j|h7Y#d2r>6z?&E*v&q>mkIHHucxG&ARW8JnR~?rXd>av-X1 z?bsVpc>@2iO#PDSvNKhF11o-DPvIQJ7{qH7I=uh<3;JflVT87x}F7x-d2U;tNNcn*~kwI1%9e58i#;K)nz`4tH&z48b`%of^$|O z#l##F#!{rt>%*3#UZUw_YBVi|v`IQwnp)$(6WV8a!q6;>zz3dNQ)2}8G5WE$kcqKJ zj_Dgm)1Bj*FhxyEpL;jVHV1H%sq%Ly+vZj#6io2}qOdTHH0he0r3ogIhH#xXDXZaC z(Pnzo5eKV`WbF!i7I>Fos5y)lojlSsX}E8~y03eD52V!5P{%!-IF1RU6QI7y!AZI= zqpVAoN~FbEUtSsBPOa&eh5N+ssIlIp9V0SzyC+0wl~z?^u$zMXy#C-^)qH^)My2(# z2rY;k!cPe^CzHoschnEGow!X=fUPMS?1u~2TT8m??_{vHK;a5O>%g<#PgpV9gY?9N z8h>79F0YLMxf_NDPITT8og94uV_gq&O7zUG_r;E0MogHV+_NaMcQeI(KZuAy3VoP{oOi@GG zC!$2sK=>U>Lht=p(lum0FF2FXrf^9bndG|h7dxo#9)Q6BQKHLB>VD&;l2-6;XIUYg z;&=hF6ivpfL|izx=Jh>4DvJ@0ddo9TH>ZfQMuU&vx^5EI=v?3z;QoCihDM&1Ye4Z!E+y3H*!H+k? z<3h+NRmdKr@g!RedDad2ya;B1JEiK_s2@YtuCd@^jt;2+aDIf$=M2;ORd3U^;MjrD zydB_KDuh#e{`K|ZUw3WXB3ZA#gR`kHfSV1aqh6BY<=yVCXO=_D&Y2U8Dc9eOeH^=wPIOCTW zY2u{c*X~iq(O*9k2Yo>d7-Th2*%KIS4jWWjgH z$;;AZS(y1F+iVqHDN513IKHAJV4PX9bQ#BLZ@sZ(dWFx;qTteLo98r9?*xkSjh1st ztvs!LjxhZ*w`g;}Zl5u!eWCX6QPRhSqU6=QuDzvWFCrbRcC;Ilg0%e?CyScXvK%se zySD1X@GJ~}{42>jy9sIzZ*Fc6V-$1qFl!YU68Idv)m=&```|vSbl8AXJf?GGJ&dv5 zLFOjAK!VIe;L1hbm8nx7gqf+>B&X0XRpf#zAEEl>_CDs1h<>XcN^ygCBG_Q9rDf@B zEO%AhPz<9u!<5(>aA!7wz1G-sD!+C+yTP%MF)nAoMe1DLKlV1QPg~n{fc{bpmZqg! zu>i_Ub<92$Z1+0xu}8KEpL`#zMEMgeMWo^6sOh#9@h903>z9EG);1#OTzrQ$MbW7Z z3GzvUjpC=&nNJY>iNS*yjZe_da6qj1sg2Bj^kFiFp*U0|3vf_lafrDVeQ1SQT`v@< zb_oIZbH&w3En5ZT&yzNRDf|FIs*t$Dikj`PF?-Zu=52ITi;_y?JRlMQat0)LKa_0- zYAyZ2j6k9mctwmzWB;L`1s`C^V(=MS;`hG96v}4AY7F3pMiOHOFkMW1qd()ltwb=` zqg*T{XqdfmLt6I&y2Ke;N(g)2COh?Q*Adf{d&F`_-}o%NPX@6GjK~Lo)$@CEkc|#2 z$~IF%AimEC*-wClin<4XNNIqX5+)@8bHUFddo5 z9;4FOy&XNJETh7c?3cL3>S$)u3!hEkaW-ewSZrp_pfBw~Z8eaiva-XjUEX}f364v0 zaN|({5rE!X1IWIoSnGqW?b)SEHS4Y0(q#w(n@$YZBWOu35?(|XyNp;q2@ zBJfI8<;rmS7g;FfYI!Bg%zZdebo(DB33S_ZT4@zOFy*aS+ccVJRg-c8-MB1k@RLmT zR~{!zFBf>aYyPKf75PVaFuoBYJ!3Z<0o2-zU~njN{3Z3q~VYX2{d?9n0Mq*Mes zfSNaaZ;M>mB2*T48VnI2OVUi`ueohP<`ianqDQr&_%8(8Q3NhFV!cp+q$AA#8|j|tS`Wk`wk?aRLbnMm*|XjghxyebeKM% z%s;X`$}!%odhC(;S&|2BzJ|mLc+zO?_`=*TTy|K6&uH88`MHbZzfu>U7X_6!%`}p5 z8k*YaN2ofl7|gMG9S64WF7o(8aP8G(Z2V^A!jUMdZQyLQBs9lK@#N}?8hgU_lHp4%=RH66kff( zGTpL3#U7=ktyJ7V{|>}_xZrc_s0i@mlKTQ&hg?!Q(Jf}0O{Ru4N!<->>Wi=bH#(8} zCpt;l%C<7O+mu~w;Fdd7Xvhk$LoA1jX@)*K>7dSWCPAHXruFC3Fx;z zP(H`(>&g;T+3(md4~Nka$BUW2!@~bWx1UQL*~mVz`FoT->_a(eWA*97cGQ&k|8<#w z9(Vwz!|0b`3TlY~c&jVyW2vW&2%>X zDoU0U4!=dY2ut30caHm3tb@8uI!$oBxUS&<5<=knIqaK@aw$h?Go^&@ktt=CrI?xTB<_k5& z8Zodi`car?#1utl1gq8+2S}FMgz)EK&LUqq-tiu*hhtN=6&h}uXAxGrO^s&!+=fge z2B)E1HcQRL^`OtT!9pr3Z2AgjwY=v{Z4;<~1_HtWhec13ZSmP#eM)*mS3n4U4OEXp zN5|H!%cq|c{76coI}8KVl!LyBx->csFSF(_ zwW#lE_4h?#x(dm(b9kMoz7GYvAMAvpG?wD!fEot_5;T%w8;2CwvNZ?ldgvUjkP>QE zdzjgm(Y00-aXZ=DwYo7O=JU2I>8txk=<)ZyTWoG&U5-NfdqTms6D!(pG;8*M%2 zZDuz9g0jnL;OY*rKOC~Zl*JyWWCfD!16G3r>P?Xa#?CssE9c7Jr4*v(XHH13cul8l z4eT?-_&j1oW})InjlNRr`qod5FzYh~L?a?)HfglPPo-l7v&vt2Oh$*VfaTmzpC2(n zf2+lJ8zV7Lbz6yZStgO4KhV7_$D;)+ipe7_QohfvVEq1+aia)=ruFU*51oU~X3mc5!U5+%IJe%~bp>=N4^;D}!j3UVSaU z%nePOjklA@4P@r)XK4zI*ovF}V>Yf+H0!bkjk&l7%3bWm~V`bEu%>%x{j7 zSNASyOmyD!+xv@1s!fKnpo=bp(#3Zy(5nln*SDtMf67v9<=mj>#M|-Z))Zwp@wQT< z3fm-*iOqc;Ie+WI(ukIb+YF;W&uj0}Jl1M|;658D$u1LbLu2DR#Awqa(X~iKo$|FO zufEL0f=yZJG!YK%-D1gTp)`NeX2nT{jpNcL|MV*;xR_8lY7C_D;Ziu^MPhTLHOE?1@~Ai-4y!bFPb1I;RN{+9DBv%f0Byv0&j9Cvq@GA1l8PeY zAqdU}IR1on9^J$#I1Qzl_<9g%|F%etA|&JxAYTW_PGmcvSjyZ|f&sE@h(cvCYRy;w z6L4PV%<|rU0uAQ56>_gdn4y|94b-5M>5F zyj9|JwtE>tLdDQyjr{QxlT)+;ePGMe%9Tf@>QDVE#+#|y*`#GodUtcVk5=X_H#Oc$l@VC3$n zRG4PoXi#IDj8bpEHat#46V_!yAXSGA{2=Ny^&hl!p^#%xMTbkbb28rvbd5lT)-7xX zoe}{qvX2AnRw4A65V#0V08^GAk46`wa9AxnfAC3W;akv|t*IjdXk-vfK?8#U2M}FI zhTj5;oq2Zq%43lCPAkX*7mG+=FD-K8PgpYg2Be04H;}308jkFsV=px<@+DtshUn1s zeV6f;MF4xtVfO)wro@J7j@e-ZQgl?a0tqsQ$Yswq}N}?15 zbsCzBLc7X|eM!n>W-Kw#@dC)*@fDWSxH?b5UVUmV;tH*+EO?nFN|dAB6M#@od^#cq z_DLW!pG6g@gEKbPSo#?z!wY|*b{7bMJv=Y89S-`tbQ{R6<@KHbsqZ-ul8UC=F!A+e zsuUni8rxc3s3z@#*V43s@Hr|CnF416D1d$zPO9oC3)lqWd-W?I26##(`S=oN494HE zd^gA6V(}{G3Mi|D()Ysy;d53&Zi0^{F1_#fx2Fk$h)?bwrT$4>Q&KP`3mR!QjRXET z-9|2wVjiD3pwWlFI{(GJZ}`P+kZ?!>fLUy1zW#XksmJ`8@=u4fE~Sw(Z7QBZ+92Ej zvs*Ii>+8H_PPpm6+el^sgL5V5R9Wn09zBkiOJ=cRQ2~ztFxb-N!%KrApyd0n$~K?kZFZ0y8m5&x`v-h! z9a+Jdmy`u~XABSrwQ}YPZ_{_PS&?-ZGacZWEm+=)Jr{`2Ur+UG21H!^8Ymv7w<**4 zE*m{-`geCnZU*5Pt{)IxRuwkFc>d3M>S{=9dC#I@(DZyj&Z&Ua`Jk3_%e9iP+aT+O zDiG*GLMcJj=ae*luqpv1Nv_61(CWW~vLs{GM*^*u`tiK?`pUo4Mh-M8ya%ijSJ)s+@D78t3gDp&W{<6i3VR(^75 zk?P>}>4Xk{-&lZe@E^1gV>V6&0~Rs8$2yUCe5aNDIOx@{qoY->-6U+V<1c}ukbq1z> z8O}#TfB+_tPuun0iHCXjst6#oqB#%2nk*q@RU`qE#^U4S9 zAD3tn)OB$B50q4 z>0-@El=8|v=a}?qxEg2|s#VyI$Q5i>_FP_U-eXI;4D>kd<9H@ko1hOd7K9$6!!FGP zt(#OLC1w8{NF zmrbKErkB!#+TSkT9;>a_M?x5NMO@T559M67D(u3&cU$1Phgs0qp!n!r!{;&vJnsNG zWnd>G9co`2oq*L;j&S_JnqO6u5tBN<%Sb1F^y}!PlceBn(63JE) zY?2+8U>fulw_7zm?+1qyRVPOEsc*a!g~5q&n|ho`P-0FgILB*}q1h!q*Cfg|X&#!k z@}NrVjzDre|qpdi`r)cW-UFZ1iSGnSPS*5agSOn3NBH4*C;socu!d5@@F z&NGmFMM9Bf&|WbUE9dPJYO2*FH$}mq*&+GEBGg3W9VfTIzMz|^XQDt%nZpag1?!<@uj!d;1?w#lj2f_?sb=!B zL|I1etZI^g;R61)MgQD+i|1t9Ybcg=xqjspup# zh2beobaY*O;DR<;oPP+4Op{=G4?SjKsVP52qCfI$tKZG6PMh){FQ)s$lSA`8v!Rxe zvzfEWl)N+rDR_FdNVWdHE1Z`m$Mgr*X_4gZx%lnnqP*A3ejD{27e79t6@}J&Nssh2 zh$9wLm8KvGv(OPmOS<~nOyc0=yO-_v5_EQ|3CZ=6qs2#r-IW{XV^$$opV%_5Uh|K* zAZcf_ss`^(Rtg#)KqgNoL7$@VTK~y{T#$eY5TD<>rP76yUC2V$-1|yqaT! zX5-_KOB@>M#ZrdQK(C%Af=WO>i9kH{MF|oFj(%g7|7a^I<^G@+EI>|0@-C& zcOu~m#MW9v6eV9xgQ#zWs86@N{@#nZUB=ZCNkZ3Ydx!ACQQZN;u{Hi)H1fx(S*=Bb zUysEIpN;Anpc;V|PvUEq-1no;B$yXCg%F_GqsdPQ zCij|d?`F2Xvk4YK!+Cp!|EOr~|GVU2l~jUeTdTCz9#rt0|4;h%xOpS$kAP!Fm26NW zaN$&8H^D}z57WY=2&3n5n>F|rl^)kXE>5VnPo>OVLyIz?yo+k3^U3TG65O(my}A;r z!H{8)s314AjEKd*v#NhH7R0L1i)G+ATA@7C49#Pf306An+9`fYmiZPI>ETczt$}wB5eh(tSJjb-AW1D)M;?%T8_`sw9^8|_1OBGOtGhogk%wYv$HjT-_SA%$k3fxd}OGEj+H_*|fl$Y2#V zASWUy(CYPdTn3Kf_GPQF$2Kitie`Th(kXRi2z0?npEqF)kBFp~LK#(`$`Awae+V&y z8N886C&Qr=*7pSl>@M(P+U{9-pax)>$M;EwOHs&N6hV z`hSW@?sH?UwT-KUTMlkf-kq(LuDSse6C;v~Ga{A83&pdwygT(oo6G(h&-~)m9*1a2 zc5W3^I%Va%dCNxYo8r85eK*K?sq!~jKI#KT73Np_dz=7+@(-{DIUhJSdRus{9gu?n zJ8mciHAOPf>gZBj09(XCGy8oV@H%k<(Q=9)>@->%3&6X@tls9RAID4r8Fx@v zg1Su#A1YIVkj|Q2T>MWRRgja<0LeZLWTW89>RJS!je(1_`$9N|GCffW!G() z`98LBZdYsW=XdLIgdcu>7|!qyc(tTKbcD}+e-mj+&bnb*o^NVaFW6Qx76*DM{;mDC z9#_eI5nYP`Mc@|M8>M2n~S z1K9w9SY!WtTXhdkK>?@+6`Y&(XI%4J;6kh?1BeWhX`s~(`7?5MY#qATbS`RThnZEV zFd3q#u>*}kkOl`7T#taoWN{q$KXn!6N7P*S>;hfw(5lNUHH9N-9)}A!-XGv&d5Kd^ z|Bp24BbS3Nyp8#fHu|41!PL9n)o<)1^EKiW%5wde;5vY77bGIM04kYLTkHJI5!=E! z16xd1(vU??K@!wQo+(VnAV!>MmQp;K!GNO2w;&u=EFeX4n>faFP({QP^=C^7TkFfN zKSOA0-WeFDx>EEQTR5iy3HVW6nO_5fq%zak$h^EAtF*9Ts})4+2Bw_YnCEdV_g9bWCA59Z@zNDaOsBSc>SL>w0dhQRc~8X?5Gh{ciBHXu;W|nUv$TjrvU{ca9Doq;t3V`PN5&{ zsxQ!fw%Y?rmAMD;jQmstzOE6%DW?ZBmMYtmFNcMDQ*$xQj4h_fA+4p7^f;BlWYE-= zxEUV09-?k69(>dz2-W8S}C>!rz1#y`tT+qSBBnT3;Q1Es` zf=XyB!oGi55NnBkSqT}TVE1bR4CTrZfzvtKV+^8abHWuITE?1^`d2RfmuHnYO$i^c z=IFdZkl-86vwExx`iz8xufGevoGNN1RRrz$5QG<1QOPyH)=K720EaO8f0`Yz*617> z5FbjiSY{HZDX9X}D$d3^Xh~W^O)C(SI{~A`i}_ouzTE97u{goghd8YcYnY4I&_ls6 zr|-u84pwtTBG@GlpCnf50k*wcB0@4pYu7shi-WFCB&ro1ULBrfJ1zql)J1KIW7BF~1%}nzm1R zf@<)g8y)AABnttx%jadu#T4W$yas)AURhbU1un|h<46sSo2OAL} z2mKq{qs^28WXAu|r{H6PuFrso+7`&2`?V~Y-A8XS1@tL!ERD&jr%EVU<3!Mw09j_u zHPkoP``7=2DI@d<{Xl2iKS>smlU_Wc#G(6?q)CmdO>v54L{1~cgMaJER?R!mHAI;; zPf+u(Uczw9jtJ#*@`l>7o2a@6Qmnu4vVlH-my(075shf@>$sU^gmh#1V~t6GT>7uV zAf?Ahb~vSbig`=!0#rP8hXtuNS0W1Sv**OBv!g z13rj@mGSUu0rF|t(HJVkL{3g}EmL};U?y92iH=Qt!^fX6Oh6h5x zMx$lgN-U#Dy0nZfC_Qgx>;eD3JuB3 zua3I4O0$!xDjh#+X~)TAw<_6aAQ;5*!X|?TpvOtv4ir?UD>gB|5O~0mJO<3|XUYXW zrROv?T1CtN+64g{wZz)c+ z{|6KFW+K%7hMG7h?!zJcdenu1c8-7Kj9eV2X=ENEy0YuK{id)K&wyF3$!gKYO11;7 zrY5$=3OeilW{RS7vC}5(De;g?8nhbC4sIhT_JkM~+gCV2S&2wI7ox;q^WhWL&_RTa zhM@UYB@H%(hO??D!z0}u$oHVxaToe^kupL-d_*hGFqR{``|vs>UzS)%_j_qm@hVS9 zKuH69J&DnfjQsQx>?2;;ySJVQr2*3^s$N{We4_Ef9!{wS@u(#sp=k8y%-S<8aUtU{7)4Y%?G2zuTLiyxhAoLyGK6l-hx;O zHPG18djk1Ux#CA^q*rczRM|F*zBXjpY2zg~vjQ;HHl~q|6aoR7BQzK{P0+S^QSLf| ztd=!c>eMZ;B&oKRmNFKmNzD~wvh$N2%L_6UOYM{d`qnFyV?K98L0Yr_#`|+hD!1cZ zk8<*p)%n$0L9ITI>?}9XJYr8QX?$7YvT>x>3=?1CulnbcEVc55SnGR|$9H8JxvWYa z>9}>t{X7Z$_j6pbo?2Ht!T){W1cBsE@o*wKdU9$0{%iMX4^Jlt`y0EBNDiJpcn|yC ziWmij4JEa2!9a!I1MTd|1P>>7PbZQa*Z&$oHq<_|`+}zCZUYaJgVQy8p-pQ-!C)a| mu)CeZ2@`t{GBVKM>^WHlxnoAkW>w&8t_xcFXLHWj-2NYomgl|z literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1264-image-1.svg b/generated/webwork/images/webwork-1264-image-1.svg new file mode 100644 index 000000000..0c2226ffd --- /dev/null +++ b/generated/webwork/images/webwork-1264-image-1.svg @@ -0,0 +1,261 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1264-image-1.tex b/generated/webwork/images/webwork-1264-image-1.tex new file mode 100755 index 000000000..e72523925 --- /dev/null +++ b/generated/webwork/images/webwork-1264-image-1.tex @@ -0,0 +1,138 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[] + \addplot[firstcurve, domain=(-52/9):(46/9)]{(9/7*(x-2)+3)}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1266-image-1.pdf b/generated/webwork/images/webwork-1266-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..3b82b46042b60a991fdd90f8d5acdaeb8fa5359c GIT binary patch literal 5597 zcmbW4WmFVgx5uTskpT%wC8UO7=#XydMu!*}x`sxjkyN@vVn_)YIwYh)N?=H7knR+? zK6l-4cByb2EHaHt&A(%A~is|Izlfx{U2!62ZdB)&V`4QlR)?>)PhKIpUnCJnoIjn6aJ zZ7$L>0+yy_6r09wa%qi>!*n!dTtL3Qxhs%!D_)I?rz;9GNA-Sw^Lhm;yr!9-FUg^e zEmaV&x7HH-q_|#4^X2Nvn+w-Occp@wj_SmXlyS)te0^=2Jz<=)dQRY*k*_~qG<-^~ zVc2f95wmZ+(9d5i9^|5{aJ{hix0=K8^xtcHG%}UrD3mRo;#-oQI{oGBjgcA1v*cU) z<&yAZ6r%6)MzPUpdsBg9$B-60roWs|sR>jW;h<_X%?cq=X{8M#*L##D^HWu%#UN$7 zbkWa@XRUN|d~Pf#Kc2%jfxLca4{kB)if=A`>z{{iyJKpnDVCL*TFVj{V_QU#5bT+T zAy}L9O~eyl^5k-zc99b9iJ5uy%6m$`D&P^xt{qIGg!80o-k9SnTV0W|qI`t+)*Fld) zSF(M`pvCWqlW8Qi{c2~}j@xP~@&#R;Y0CS~ul4D@Hxf=cyn83{3#+wWbe<@$P1By_ zj*1Nv)M&X?l}BHk$x>+kHiW~E#+wZULOX|XDKp_CKc&Rer9UuPN?IDlL;(zx@wGgQ z<(EnnWfNfpT@<2n5$BrX)iOyguU4NL7Ee4t=8BVaXtK@ccVZIgwX=w;8kUmpyIKW> za++5k;u#jBOByIr-0B<9zqA1r?H=z&qTSW-sCA`aipV|32apwOVsZ5<)=BrpQ-M>E zAF#Oh*kK$5zI7;BolJK7#JC&J&}XIX7L>*=3?Zg`I8OjW9Rt!5hBU=ETq&rI0XEMn zAdEyGd#__a-lxhvH)}LEo}ik%JsEm^ zWaCgrhB0;+<<-mG(9rhlPRMeI!~o#j9uLVUW*g9Bo!XHy+tt8!YMx05e9>(uxkdQa zTes}W#-{e`7Q0l%4yPYxvOlUH8h&W&VI6UHGYTn-me@*}Tf5?( zSf5EgYF9-OBQD7oBqtmgZY%&VJLhcDi9-E6d5;_~h$)Y_mwW~-{Mwh<+SL{C*uqy| z%FJ6Onv!dte(7zXXhq!Xl}Jr1jw;dZ>?r1_y9tY?x^Sez7vWZ!Bdh0BiU!`g>1(fC~b|fk;P>&5av~~CH zM$*(p7&dWy>X(D}iv=G=5Z?|V5t&9}3f*xk6;`$qA7AiX(y6}ZIB}D@UUTU#`7o-r z3J}OQ07JiVV#1Sl$I(v*eSr}AbnPBQ8&QBAU0B}kZSe5qGibOXKSXJ)eamc#&1{%$ z7>D9UcP|5Q!Bw&&Ef0^FiHfGESiA1Visr&TVZ18=eCE*z*l>e*x5uAS$3nj!q(Wo+ z>zuVAy=jq^-zkmO($SQAZjc&bp0|7fB-@jTekBhtlA|prRc#;`yLMx=r&6Q&LfagU zyQ-$ooavddC4PL~nt>YkiUC^RX1`*sN=-9Nd{@(>drzMK%TFagbCRv)PT;uX>93VJ z?sX}+_|3i)ocK{XKGezTe@y51?Jt|+2mU911r*@_)2_@kJM6mn2wQKs;Jega5hRx3 zr0g8#VYRgGVyLm|q@gNQk1O%j&1A(-a~!9Ptb31HSnVO9+hG6I1%BS zxAz-r3ydmsYp<7E|HRvi9nV>dmVR+lIW2UDqVK*KoLHZq9?5lW;s2JQ2r1maspclL zaZ9IuKh27lhi6P0lD*+?y(Wq*MGqYHn-2rTnWSrx5KBlk5ltp5qQt*S3 z6!C~sW7u^Y8|>`MOD)3u>$s16G-XB`RG^ah)@HHVFqeglw;$*W*LzlO)fcY{299uG zzm@n8tU&)_rQ=}%_i=&p>i$0XW&e%U3-E*gL}zIQF*k=;S*@M96P_F%efK;kibX!a zLv7rhq?j9Bx=Zem*oHj@MoU?xz3Jn>C$)+5+v+~f30F=I3|wwyeCOXq!u$$1K*@u# zgPrSZduLytmhq%51X6ATfvjb|(lkm;%ht-vjVmvKz-(?*<@v?U|`tKG8Jp|`nJA|X0docBL9FA1Wo>jd&@o969vn~=5UyX_y9BWLU9*KP+H%Ne6;-HIxQGpK=}2BXig8?_ymosAot73&b~K(-8Zvmy}2StIR9Ruw94B z4IL?&nn~)-r{zyr&k!MBd(V;MZ#XlhjNM(G{{fuPzrg*+LI#6`|K#}ao0*c1IUV-l zMMGtRhN}HA-$<-(0qaSFP5>+z}{i_V{xI8`;P&-`0SC|8{6(0)^tDeM+P*gPTmOE z-U&5&8a#=&}QYC(+HXIzlj0 z1`7peIcq(`MYwsgJ&Znv4sUSzgeUM?2<$;WNLV9}%7u-%^6KXXMx+-ztY@OU6=@zM z%Hf;`Vgv?e1mYJ5f^apZ@k`rHwvkK36#g)@`qTXcEK2-9Cbd+okJ4mG`f^_ltXwKz%aJb(Z#%hf^x zE8hi5qoAAkhu!j_2X(iu#2* zw!I&e^(yJ*x}5dGvp#(K(AXi(=djf4*Uv7YzToVJ=4WAHzFeh+j%GH^&SyxX7Em-B zIZ*^tTtOmy^<6&K5PN!q$T19dc*l3GX6dmDjN|jwO(i*jl{43r(&>)BUBpp*w3v@} zJ$*uQG9fGe~S4pL;K1|+IsWQeO_Bdjo1;uT4jZ|_QBEi z^Jv+aGAD?$hwk`7kK%kZSX7|x!rS?CX(QO!r%y7AjNX?zDVZ2hBO1uJQ~>HH_@Hu` zuUenl$YwO@=C>+nLOi!my|Y-HS%2mc zeC%BHcV3=)%*h@1?Kh0IIuT2ouqF)J*2(l|pyrMH0`%B@?WWL~v^+U1Oql!J@O zi(qCMrl7pyhL-ZCOY0+_P;I01r!RQquNw{yCZzUt?^ru?WUhuN832d9p_fx0%Tih@ zPl6b~0ONo82=_l$j2BOlj=W`OmeVTl#&f%pPczM~!-qF6^~yW?R_U#xl!H zo>a^z)01}-(k7=Km5%)W7*s;9@A!xFe2J=By)s$pTx4^;6WzAnW-F|u>yf0+R+A>7 z@7M(Ms||U5Rrd7oOIqixVh{HxtiZbjOUeR;iCNSYNbC+g%iRa*K?bifd2PxHXuBs+jK`aPBc22*O! z#&9qXVob|~VxoN2uj+!kv(RG2I@tg-nCV+P!@GozVDI!QhyomDm42uL9&=)xFGB1$d}_?f|6X3_vo zYM1Cr|t(X6+M*xel;MRj{Vcn{g43%@}((pXdWJ@Ktttf@W0w z>B)q8D*M~|XsHRmMY8^?Y=v%S*s6bX0lWJk?JR=}XSmzj zd0xHZlg~I_b|3c{D?POAPnXU=L^7Rda9Ha90MSjyar(_&vT?5z4O21rZgrP*ZOMN5 z5|ywq!}`5cNJJ2g6ltR&U?aVbSNQbSYyg3(&rUzS)ePSee||E)?sjI&@yt>}Jf_DS z;s^JD&6rZ8O~0STRp{ztzRxbyV&HpHCffi`v#)K$GKq)KKY1vO#1jRl2v)HZiVLg_JzNrAEUm`+X}r9)zwtfsy#R_O z15{S_X4Qq>Nw?^Ie&ju&8a$w%1(GMPhC1lmzn$y8q(!8^X6?BTOg-Jjs6mtc7!$J} zA>#biiZ(R_4R9%ud#UAux%a@msr_^$Q+%0Lu@pYHGG(_>S=~tQut)~(F`1{fIaqS4 zI0%vtXskd)zmc}Nb%tj}xL-Oyjs1d#T(Att^h9&&w~sO@0+e3^D*eim{$2sVKbD#Q zt^h-As5P&$oQXIPBp_@dAZQ^31PfRL1uen+0wPdLOA$*Eero}d-(COz{$a+e?d%N4 z=l^Xh+Pun6*3OK7t(n@qI<~%0MxY?CfeHWb2+-``RG6^9p9RL&-$4|>NEy6L^$-)w zas|n=%-7?6vI6D_1RWf&JLBi-hle~r?Zk^@e-amgr5U&bog9LIW6hV=qG&>+FRf1IM7K%s{klv+)j-f>bL8OTYB!u!%1OY=RLIeUx z7f6DDNCy!)fHWZ#L%ruDzUw}By=&cjzq{@q_q*Sx%dB%|_UxHG^V`3@&&(KkS4WeD z`4lq)0|U#gn`-wM81@0+Kl6k8z>=$FN!bhxhb}s+s@`?BwPj$q8JU>qcsId>t96kg zKlALyag~U>s_~aDJbuk)D%4&0_QrJDU3J&zANGp8crK;>Qhg8CgT%8g>L(6IJ>ARc z6MA`XdzB}Lp0bI@*vYhYJ|)5~I-P+ROt?i0lQ zV?xmxNexRYKKsUctB)~Yn*yuGI;lkmkHtH0O@8=5@S+Mo^OFhQ3 zc$>(=a`5rq=^Knfdkz0O7}Tu< zWDXYGxZ=d1{TD|nW7;?Q4||J3hrV6qW{i8nru8)W#{SVK2M#_hW;>asEb;Pq&e540 z8f-kJ;k>6|pLm#`s$@C|9cq8+;m;tk)vhQ8?mBT(qtM-nBF=j@5mXX@G%@Q1T z47I<<`oR5tUn2&@nEajvkFR=2r7}-_r2eu1yNl3m;QPh#TU!3e;ql~OynETsHr#l9 z%sec{;~49C?nssWpNw+Ygf)N08|CoKs4kpQvE!sZZurYd=;CcI`$#VhYUb~^249Q} z#0!{x_x3*dsXn3_ldbl)(@DaV9~)Y6gYx;ww382K6@z)KyBaaiB zHBpR=I<3X)AeqbU&D|4T8GW)z%L*vI|L%wplgqQ!kKxumB?tO%&)u54FvsS1e3M_Z zNq9WkBld&UgX33OjbmL}%$r!ob3G0}@+8QP?p=Hm_YvNZ{Eg?=$Zxsdx3=~wKmB<2 z{;{A#o?(1p%qk+|r>vuYh2K11z($I+XuAL1&*s3=i6utQ)Zb6kE;uKA$YHz3=qx%T zGb1r`W`+f=_4V}7tI)QYA2mNNE*mcEF9(XJpwhW@FRQp>?w8+>=zrNS!knI^Gk#b4 zu9wi{YsuGguT4A&AHrpf<+XweAC>>UzNEKw{ZL9+W>|z;mRgBgS(A38^QEK$W#7Lv z?o#qw-x}w$ONHg$p6LGlKJ7i@;I+Z?e5#IJh!vTDW801!4Cd!=B}yf&BomX~B@s>w z+nqkb!Wb9-QOG*gAzDoFw(w2QoU?D%ZkFF>xkb9uack(d=54Oq_*<-Zkgu)VWj-}^ z%q7pfMx2d$)tj{Xiu6k1Y@oKEQS4Pm;nrl4*K$|ZZ}KPgC9NloT}EovT88K->($>n zqn>$hyrALTS=BBhah(#arL6DK$Fbnnq z?#FiJi<4eCEZx8Uj`J((qDha2q+yu_rIu483d@1>y+2zd{mw0mFWVd&tjY9Myf?D< z`a8)F%r)&nhTR1lhS^o;K|JNUM=gBdSMAgm)y3&ACf25e6z_y=D{gl(USvGP2+kEt ze)!dJMEFQJlE;~6DW)K10e2DS_n^kjx9?|Pi)>4Li)OR{>a*6cwjX8%6(n&@k^@zd zLPq7IpQZGoVoGORb^XLQ!>qbYl3t~Dtv z@r5DEFfH+OiZkEC;4h|5*{JLJxWHEbx`4Gwb$_KL# z$@9ko1vgbT_b?TNW~e4M>T7beI<<0Y#wSGfIOKJZog9s)w#nJ@2lzAkn%FB*m9CyM zz85MtJ{VOhdiVITfAFntT)4WxG~D{N0H z2sh81z%aKwo@_l*@0)Pu;+X?y&dr+4mi3JdpRPV*-S{l#b&NGnXIPf%J{62ASEGEx z&qkmDF@A0QtYR|q`L0BRbk=2?i`|z>rMxb+Un-D#C_8v*j%8o!}A6+bw>3`^{D2p?`BG;cJ41tbdSER*>@rL zg65j_+J$+Z0eIQ2TE3dI&P>yu^?iKy#L|XRJQIwcVB-G3BJjC~(8SB)25eU0J*=1I zq>Z6guSdDLuUSkGDLsYScq}3!wx8b_Yxwp`!S*AmwdqjVysFrMn2lvylr6e79rhcx zvfZ?kpRZR@@^-0bDQBXAeN-?2Lrh6YnZ(>N{gxD+gibgkxVYoG>up@~73|3xNX;W`#-io@D>~)o7u2U%=cwwVmt+OS))p#FnbaC}3-v=OIzQwn7)_7AU zINAOFhzzplJ5XshXO@Q7hWoEzR`9zH%VwL8XKu9emC2d=lNSA0i%t?`>?D0A)3&XLoY8^0#_6#>zwB>?zEm z(L-_TN)*J}qo+zs%W?0OI0Y#g2^q<{!rIrgd9*V!JX0$-ZAPNnl7&UN91ici{#^zn z0)=JBft_Aa<`deroCJ%F>fbR$%mh4UY>kK@D+DM1cDJtmRYTll7w8w*P-F~BHT`7T zbk!)Rd*jRYz$&Hx-uLcqC6&uBYZJV+4Zy_dwC&A%+6)W<7a15HhcPg0fJKj|85n%U z7#Mz9F)+xY7#O%bQmgdj85s7K-qkWt@9H8EiNx&eY!nKmrKM$KV>8&_w!6{rxwB(x zvh~M!Yv0%QrG@5&newAYkMi*F`1_iHI?+i!$Bwa_J#*^t;loT!OosZ(1$j@NKXa6n z5F8nJOZYC?>8Pnnb#@ZL4!XNLy1F{ZWa5{e4m&#wBO^UNKAyHuHQrvXAL}Y` zMR}H%md~F*cXM+qDJelBkv%;fWKze_K-)lHTX$E-x6x0oF6OMPES30-2M_Lk?QQQQ zcZ?2wdj0w}iA3z|>}dE{{_^F^q@<+U+S;COuq|<*zkOu5t^aFPKtRB=XV1{6L~Coy zsw#Xz!CO2Y-`rG>#bP_!KfcMyA`|P-Xmn{QE-EStUxDlUQiH*u^YgRXTHnXNj4Q?! zK6Ex`Vq!EjxUH(Hs;#Z1uCA`DtD~r>Xm4*JCntCJu8yOl#r^w+rY43qHkLZtYW(~> zkrDonA3r{Fyc~SY$;iIR0)=3`1 zlgH(bj)garzxP#tDNL#`(0VH@RBdb0{Ma)v;Nj=jLBMBhm5YM`<3XhZj0uO2a_rq( zqJE<{`+2jk-TRw2@-GS2Ioxl5<}pF9^7n}wM}FE)2H)h6%TAFcym4td-(@dnE-)Bbrsp8>xu7?^x+X{$3$A7DSoa?I`N zom&hHLQc2TlnovY&W}(-dHRC6fALju2o0U8+yVq{O-Q@(=AAdrJc@aJ>5ku7>f4^7@`RO}-}#-pH+2?8kT;*4WEke) zx^sE$%5};w>dwbKer-4-3Q)EfEHFwiG3#h*Z;JFJ6m3z$%1UHQG^p+3yBu}LDT&_c zBV8|!nQ&ep@p*Qq5*MQ5fz^Ro+nh?tK-Fj|cyx~e)f|4iG=G%)9Es;)uK{+>?tCf= zcZza0%$F$MDsK-U5Xo?I)hH_@d%Xc*l(Ll$r4rKCuR%29+yu7e^L zx`3f0{~MafY#S)DTVTmYe|3Rg<8u6MeWp;hv4|^ZuFRS)=FIYwnRMw-^6>QvQ(FK# zufya`6xWSM=tb}HO2Q{cXroXD%&;dZGnfJV#t_t=xt!__1a4Bve~onH*>1YSpi!~g zDso5XO;zbr&a6Nqdym4giVP><8`)N%@pDtiA1DjtT7DH>x7h1|v$&-|A$drDc$#&tZE$q01ZN@+=Ve(e1$B#Q zA;5U@MenEBf?8Q@(<8)O`3h(?JS{QkXY(vC(9rrI2sh}1@&~8-(*eA^nPBblgq&TjQQJ<&mE9aC(!-@ zy|JX(%(C7FaESVL76$E6Vnj1&f?pBMCRmA4I*v7M+Rs2m0-p%wgv2n~m<%|IZDLD_vX zg}w%wP3V%=hy3^`hITtF@aHsVfHmEs#0-pxcMm6wu%9E14Ag*+cqjc4 zfd(8k7Zzt(L&Y@U(t@d1;D`=b+_|u2u>g#L{s`AOQVsq?zI*OZo^#_5I@BI|UJCp5 z_2=Y}%5UEuQ6;fRwp?6lfnj9%+)A4G*^aE7>wJQw&;eEPZuwLN-g6|?fpg;BsV499 zNBJAF6O!u`!^VCM!XsYV>8C};1Mz0V1~C0qtMjQ3DJu^tHho%w2l>iDb15q*+@du< zG-GN?cWLwFFbXYNg5zi!#}C9o1qe0DK(^lIk3B6sXIyQ-RXHR9XUQuZV}$uZtAsdC z#OGpK!_s;5Kv8V~i?!xBvNN+l_9o+F{>YPp>kqQ4lNIX__F-*VI2vL@w{9M_>K(~Cxe zWC#)8T;JV&XD_%&M^u}P^HsY%kXYrae6G7 zSa3Mrqkz;4JoMK0_|G`cf5N~1M~fiz(Nn$rx(NX&+zT_`= zviSwiH$}29a%&!(v|jm&Fird@-~2lU9v}$Vz`R4(elX4>9II2pC7obS#+bZzg4uhz zM)-;;q;ZfqG?Mkg`e3%wCiLn>jB@sY1Poj#&PP}V|Ly?ARhC!S|Fv{(XN?&CbELr1K${!{JSUg=Oj~XM zJX&h~OJRPgzh*;aQekQ(2^c?Ahezs=tN7H0*I8)L#U^$*6a{D z$03x$ip^^-t8ha+KO;U?OC~_r%QL}nHD+C#B~?QpeoDk?8oR|J+P40^Sadx|zeY4n zb8LN3H2fc7Xy~iKW?_KXadNp<4o{xSqun_Yl`^Fy22xq$G?xASrII9bnUm*AFW*i% zY?6}!jDk!VlX$WHx}yKIRGn}bs|9lXM}lsM6YrVV5nV$$DL)^i&Mrz{>n$m`hsk?@ znVP>lN3k!j6qX?-Sf{VAnasR^e5px(v3&Ci!~xd8xyd2glRXe>;oEt8>=Kw4>At2K zAPR*bzdVm-t?Jp-4K%8^hLRN~Yne!kX$T&NBm~0Dt<4UB(k~H6{IVY{KPr`0f|~vB z4rTu!DasdA2H~8Boylt{!HJy8mua!_z{1=Dx3i98TiO|c>l@SRc^F7tqW?PptmRFecBGAW9htw;&M=pm78KD ziA|awTKrW%flC5LE&NmxL~me@g`Z0P^Au&`Tw6|yZf!v_A|xgZBw$MHiGTJhAoxFPT0idQv4e>%7Bjqwj>2exB3!H)@faUk^b0RQ&&Pjc1-UiX@^ z5hyL}u}NPKQ13d)@&z9_pEb>brv*71qGl}&CNi{b;~jWZ2lO*S7m=$z^53i&?jVH~ z`9Y_T4BxqlMP>ZHf)WVSZ?+D}G9oH1+mTQiGJ1I~Q*D%To1KqlVoGbdFB-R=n2Iay zn#Dasq(hTTa!juar)q`_t-?7P-=L8lcs1&7!1r@{-CJw$81MPi=S^GwYL$L|*9=)u zeTDfUQ<0|PabFC^HmZ{x6bA^w_w;aSNSkoN$f(5zQ$VX+))sPmfuh$PG`-IgrNkCV zg=tp$IcOUm238*yj8Sf2QJN16Shq&s>^dv?e8sVr=MgnOsosZMgM$$k6Bw(Kex_(| zakg&UngyC-6M;*3Yd(HnRkd(&O z&SLZ{fHkPkAM~8v)`hrupjv5-$u@_YzPP2Tr=xT5kaVxWPXBt-O-hZx=BVtoUVpw@ zy#m7K5lS1a8pPe-83$FK@1M}siADbOb(z{-$;h`r4rTOWW=X5S%nC4OQ6I9BjT3R9 zV3nyAefKbhU~bs<1xPFT=@;MWFeoy800I^>Wnn*cp_6oHnum@gSl%apNzqI^9>UrQ z6uk3gGSMGfJ96r>Wx82U2%ERdU|e#8N*VJ6E zh(Ug!`a<;>Gy`_Ef^+=dcThtL0H;*qh^_k`@4$M#kHZ!o_`mY!{)H=YW>dA? zl*yt*4!hCCiompSjhLb|_qT^Aq@^;kaLjd}tZXS5gTSKvRn1ZBb-P6u_(t{uW{qnf z#w!BU-B7rQ0I@OtYo@j3qlf2xiG^iLQ=_~4?xe-o+Sc+vYzX=>?IvK^-L)h+mM(wa zw}yQ=MQrS0UnoMcCOx75-d{IaXQGK)op_gqpgAXbC(HJ(63Nb?%mO!! zyh2NMpixHw#md;Hdr!{Nm)_!@W~V~n^3QsLkCcN2@3y~H5U@R4U9(Oxx`KlHF9(PT zq+yB`w)wcjMaV&3(ks{{|?KCU5Rg@XLC3_Ue*2(54;&GCT zwza`&h&pNFZnxRkT~;75dobiLza~aGMnD~J$PQp_WYVPgwr`9W;z)&M;xnY`ax051 zDgeu9!tQcjj+aDg(&@8d$wxQ+M$+dutvNT~JX zfKNUJM5SYAeeNGa#o-c9iGw1dlheP^_`GL2sF0^2o2aR@d82LA)cl*GSv7PWrR!k( zIWbX4R6y>Oq-DF_31Gfh*HtToFuvN6@9k2QMpe?mJB@e&6lBWG>K8Nj=#-0;xZPh} znsa)n>(a>U?G_A$yN?-laa_ctAKy5@>fl&@dK zjz~*u)a6r*6i}U4-+Y&rTwJUz8p3iohBTFV$`h08jIv<8{at2ApIe%T#p-;V9jDrR zI$tl%HLFSrT(bfA>%o%)j%!Vc z7f^ye^H#A`H`ju<>zKubTKMxRh!S%pcfaIiTo>cyt$>1(?*b_!aP7vFF;oYz4&oMg zr?;*{A#jOkhK)?;Kt$3VX5?x@ya`pge@9x+E*lBc2j=$3;9hURCEq3N_H3O^51Fq zmNQ-x?YAU6huX#=3Vji}kvm&<&qL;y$%&3rx@0J7Y1`B~Ruc6CTL3Q0;mXpnu&qQc zMapq75R9G%&XE%;KR8+rPhpHj=jB(oz=Imxm3}DG0I5>RA^@>}y)2_0Xt|R?9gdr~ z-ELLdY%k2{R8yFsWqsrp=H|iSMZ(5%A#A(P^7nTU(M8+sJEC>VKe+)#|S7Y?sgNNZquziPV8X)Qe-iR7@v7&)B}Ccv)EfSznQY= zhhHO6s*PzyF%A|Sb4<8gD3if-5~8EL1*ZTG+9!M?Dp?$PjpPmEq`k7EJDJxI3Eo_g z9p_QTUq#2}j9}3Ej1xOhlGlx~G$s007A?(O>!STi+m)*HqS+2Y^wN*vkOP}Nx_qnj zHW!J42R2iCrF5?pr!ayk9eh8As1Y-VyJn`Wv zsJBRBM`Q0kqn-!NP*}7AEjl0_+X6oGkOqY;+A3+_IZhOrrNCl$XxomU>D-3w{tVtgjuW>-I(&AK0`EoWu#ZPdGu^t%)5jf(^Fx)tAiRn2U0y z80i%C-Q=;{qNS+EA$#KKl`hD6USIk^&h(L&c{HIuw5$AD61Oh=qv_gJ$zm-k_69zO{7W)2V)rAL8NIb(KIqaNd-TQ!cqzr2t8ioq1K7 z<{Pr0l_5TWJ@0$Zl`D>l+(bfzAJ_q^)f&X5H-F>K@>t}roV;dSn*54%?l;wb$mh>V zWN^-Dj>v30@6-AFXLbp+u;*NPgPU9&H)Sb`kZxCSLGQTp27htM}PKwUs zQ-H)yf)525Y}_w6S~V+5phPro`tOEPraF;Uv@drv?PBm#Kdn}h_A!|^k%JHm6w*kE zD2uSoK?nuf?ZA*f^Xit@cXkxNBEJi4HD>JODWYuWMm#*tAr-pd?#`j7^i2Po*2n^g^d`r@0X?= zD)6XqXuK#*8MkpB3+Dj$i29x*PI|lh1_JHm&XDogMiDp%lpce7D)vM(?Xq7jU7-s; zhFYD_ZUOEBu z13M~t$r~54?ZKnj32q{^5-LC!n-cp2W;=_Pr}Re>$abZ4AoDa5qC@E{kpS`&94pgBL9g_{A z+*DF@!inyajAA&O8G32180elIa>hGt<8T(w)*K&-3P5L7wG8l-FWjt4#1V zhP0eOfz)a*&e;^bjBI`yFu z9IyczjlvZv1y&iG+-tayJfAR>u4I`y}WAYKb zq8yxvZ7RASOET7u}5W^cbi3hZya^ zUbsx?$Vj3#ZD`Y36hk7s_`|kPcq6nH?{0}G1+q}~p)!JPqd%1jIg?v%d0&+Sz& zY{6rNwuPv8fP7P%L!}4vb82TmbnCm)SnG!9P85=gi(Ts(Hpb<)lhyyWqa(GwJE<>P zvpgUG%EbjQBql4q$zN|V^;?dmQA=uc0-)k2tC3P&R&ASeJxu}J$sNds^|C|TK{;Ip zyBlL$L%Lv|Ft)vk+l|=$pjP#74l`ol1$y;6l!y_TL1-%%hRfgS`2!|x*MjAJ<;`?k zdD;ol&*-j#ArfA^;iscA5^)j?i=(^r=`n&W{p*F8?x4rK+|a2g=`Qo*=7NMESEw1xUxFhwFJlmCH0F^Hg#(TCKB zp}w=S+!3cGC;L#k@09SvSa*8*U|iBfUup+&QQJ{45uhh+EFf5h%|Daymb`?VD zm4e!2qjhQiIPnxhj_zYn_n5!e0++=$4yyipOr0eXN<(Ju|G^u`EMUZ;r758Zjyr!4 z9*Gz*u>1`NV+oWjz6bz0ev`H`~Y_%q6ayMlBAdGK|_ z8IN43*hQs8>#J*GFefqy*U-VBqdha@Qu25u9WJW&1Nn2dAk2D)s-)(`?iKKluVP2?!1U zsUPr0%D;R0V`yR`K=21dgD;2_zv*PrLF?tc5KBfQU`GnL)SYIbU2b(?n=Mx#zWo5Y zx+5$hz$*lCh3;d-K~*Tna0a%;DKQp#ef+M_z-~#59m%}E=oX`Njg_{2dKSFSq@0%; z&>;sMplK3y$LOb5J)Yh2D1{T}NR0!H;KBcEh2!ACzdGjU{F$&_lbS6w+fJWL9p67v zS9eeS5v|_Pc#k6cv*j`z4dw~Sf-1kqdPFmnkURUJP#HZcYgl;PW#Cx??ea(EOZ5%# zWGo&Sz3-frI^iAIGt99TDZq!E3Oy6#wSiapt<;nJ&$)VRcD_6QQaQrgg+j8)5E5Bl zH5YjMj7P+ic4Ed(y}<(6jz-Xqw$c$WpNqIp`U+Ox4yCQ=!ZJ& z=dORXp=t%}TyGiQy=>!MBg~%jj33jlxK+eJo&l{BF$#i(L|2pUc5zTy~BnkiE;} z^8%hcd{Sjib!Sy3@vO>}yS>V~&TTHI9OdmOWeU%#Oo@Ayq0Vi7Ugc>mjh4eF8i>xB ztW7s9HENIjW~JY6C{Lq~YiaXXu0GI~7gB7q1AsPg13Q#KZk%N*yDN8YWjAV7exW9t z32zU)iA*Z%&EAnr=svBty~@N%NoHjNM^vWgWAn|{o*KE4C%E^}NOJc|zu?W?SJ3PT zG_ou*QG6qnefu``%4@c6`}QHo<@W8q*uHJrzI_NXad{%SZj;JBw7T8Mab)_*f#~N7 za+y5??h9*X7Z%Oi0R-0M6`UD`$OZsz2Np9vJZCH&WXjClP+l=G*=M$%weRhDN_+Y7 zk{p8UWM}2hT(?cswIrqNhqH3a4pk#F$)ZNe?aKH&KuCK3qi1i~S%*7FvBdyKW2uJR5P<1h$Gr^@m;2ttHg zjD479se|qx#0%@NP$zKp^z?U;7?-KlNn9Nw90(!lR97B{L5OgRu@48W)v8pjh9GVd zSBIeGaUo$YTCHZZg$Q#2EBW0>APfnF!(0s^Tp_|-l#Ov1grrktc^m{G!Y#%=v=}qm zLWH?U7-MvLjxWZn3$weN^^gi|uWd2W=nGe)8yh_u3%9%bfnVcu0Myi~gRQpGy-T=9 zHg)=8Yo~B~Q`mAl+DbHvR!-sDjg8eh_&}9AQ<;)nS^JZ~%m)Q+;{NhY;ZwV;>&E?6a;!!u)5f)3|zg#=1z1hsnZ&#h4Ev v=~Q7Jvmr#d#n^{BV~oG1^pn~C8NvMr%>~q2T-;;G00000NkvXXu0mjfb8$*6 diff --git a/generated/webwork/images/webwork-1266-image-1.svg b/generated/webwork/images/webwork-1266-image-1.svg new file mode 100644 index 000000000..0af8aa0af --- /dev/null +++ b/generated/webwork/images/webwork-1266-image-1.svg @@ -0,0 +1,261 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1266-image-1.tex b/generated/webwork/images/webwork-1266-image-1.tex new file mode 100755 index 000000000..83fb9c594 --- /dev/null +++ b/generated/webwork/images/webwork-1266-image-1.tex @@ -0,0 +1,138 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[] + \addplot[firstcurve, domain=-7:7]{(-2/9*(x-3)-3)}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1268-image-1.pdf b/generated/webwork/images/webwork-1268-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..1bb22bfb41fe22f41c52bdcc49eb446ce91b8880 GIT binary patch literal 5316 zcma)=cQl;q+J~{lh#H3I459{Ol;~~3DAA%uZDFNH}|+^{utf`S$+Kcm8?Tz25cwb*r>aS8RvTPLu{e1b^BTW-R5^C zvR^gL;D_VyG=`cA=3%G9+=&)`D|?FZ*r$2ZgKPZwB%R5H6v3V@BUm%8<%i+T*!Jcx zmpALvuwQeReCw+-9Ni?pzVZ1L*v+!!A2Tk0rXtsVW|`o1y81<;Em+{xq;|A1Amegt zKR6dJ=Wyb--8P_t-bAsw$@RQ-cHdSXa47CD-mi$8ehw}rok~3YXMyv4&4_Y1T_qo0 zC~fspjk_0^7D0&ChG47#ChXxOBwKMa0p&sz^AB}HW`fcpRDiG#aF&iPZi)rolxFwe z9zBaW;!tw>X0Nk^tN)^C@vod4Dm+nf^o$x7OYpP0X7ewErhya$n=`i;?Utwg|`Wz-j z8|A)WJL0xdsN#d=);$}JxSYonbKWyAJYotznGXk=M~pA6U)Q3~76WpOy%wFz*|1+O z-!$o5l3yjf^JAGEuk6jUIU5uv?1SkHhPL!%WXg0z%{vT@u&(v&1|Shs?)LHT#SHl{ zpRx6I?f_Y|MXXsd8!mJCrpQ##*b+PP_P#RTzI&Fq7u0dT2FJT!<3nP&3q;?pf$&L^pp%Lrvu4kqo-P)&YR=$udyxI9 z-L#srXb*un8Q;_~OQ@E^7|k1T2Ef#d>gz8fJT(%ZSk-{81{9#`z-ije`Bn-VyE5a_VnIdaSl9aCH32o?V^vqCN*tQ=d0G^A{OzOQx2qB~c{YSPghY z(4Zg0>*oSLj<3iBo_)+zao**orl;ZY_h>K0~wM@%%fq}P3BPz&EL zJ2R+eiUcZh5}qzvY(K7(RVF=#Hl>pD#OS>hO)69`Fb?L5mKL!^kyfp71xd;>S4(E$ zu|`$qBa*~&f>?u|yEhoJ>m~!Es@{;|NuX7?yG^13Nj4}PEE!{|sX}NNxxCd4q96-~ zWxXKv7t}@Onrk(0N$E?;&4v=)4?9EgC=#YQIK?3Eq;_sPe`1mv>_K1%)-2XQo5t2Q zVyguI8t9xBTOjAS%;Llm(^b-iTty) zF|Rv--6|PZk)p=C%GUdFye};*Rc9lMBjF8Vs{E0wn0nb|P)$fha&N?w2MQ*!y{tQD zPiEan)?0llAAhl!^}n-NaZkU{K18Q-q_@-j`qzuLLO%DQxdemb4^(oSOVnJA)9$3} zDFKJ&JHuK9WtMe>XHztqS8WA%yn6F9QXWED)~*%4Yh32Yt@xi%&;1k!k-?un{-2)z zd-aDUAdvq?7h)1p|7r^>{mQ2yqPLcR(MuJ+6Vy)3qaYRF%Cpy4RIb8^pQ#AbFNz`7 z4r1bZ-*?vSFE@LQg9Ub+@dY6p!^oAq5Nkm@kICvyQZ_1sq;I596hn#oNfJ|XsuV(T zE;6xY^wZV`i=Oj-x$Z%{Gh%K075 z712L=!@PDVv=~@Z4QZFoLn&)MByPl@dulrdMdQ=o5^Cx?5k51fCr-PVjR9JXKYzjV z3~Qa4T9&W{!)&QGjts+#;xxI@JKt|8Mk)XlEt<_KoWjO^a}3tmizoW@!cG7kMbDo8 z4Lk8a*%^A;W1hLg!A8G-Aou^1dByJi8%v%fXMLb4qq5h{JElE=DSDE+LX>;9att;)bnX1W6W8YnJYa``&BquOy@ptdbAwn>Y+|{rOi7U7sk0s?Yrjt`o&%RL_hTPn#?6M-#WV!HCN*owE@(N>XFk(Ni8$(6L{2T zRTp}6e0q9h+rmR@gQ(RiGaf2H@~RFAZU_oEv(aYiFlQQ!6^ifQr5DQHsCO|#BHtAm zrR~0*==%;nvNcF64W8%MOLsQhfM{31M#XmC2+BAZp4D(_LH0wuwfq@RC0`BcKDnix6qBOju#3?Y+G%|3s~&qW9v~!UDTW z%m=O*=*C0+1^zm#6M}Z!#W2(w*RhKxR1B*(px?x&M|Ye0`ZTZWg%$;PXLx;nX`DkiqR-6@s>Bh6vZV@WD+I-;-4+|n};Z}Y$MNG#O736TXP^}+U_ji%&d8IUfeNt&3ln6Zd zBb^TeBZZ9F&kl=uoVOn`tk$+pX|#`AJ=Qn3$Ppa-h`fsHIyf12bm%r=Nfu-GHYr+fb#Spu>`)osGfFYN?7gH|I5sQ#`(D+}Hd7ez?`@ z4N=O;>E=6DuN-CoZ)3NGoG%-{j|`f4hgOta@`GQ8*gD@~2@>sH&FuoUc@$;3epn6V zm6-9o6j>(hue&ZNjt=9$81u<9{8XndIGMlx9IvTfVnE%t#M!}pVe{4C*C-VubX@7{ zF|J+SU4XI@K4k6**GjrVQYfL(1|Sz`3)^kKI8Av6mkUYC*;IUH8WVkZ*QQmZ%~`C0 zDX&zA00!Uwy*_ux&iM&xT=d*)7$K)o5%iAF}uzKKYa}ao!`bh)&ZT&TI z)+>YF8)*)sxgP~N!NN3Vtbn4da~T;*6KS1%wmpAsUBkm{vuna_DywII(?=!#?4$qp z$|5Eq`tJ_vd>RV+ePf|VFBLkeQoBlV^2V_M7ZzN1_5q;S^a3Ik`HORU3EyqP7fdU@ z&)lk_a~dWHtMuI9F@!#@sr`dv)?66rC-;wuQq%1YCWY14HB2PC{3B^NeV_Z{+s4j~ zKOEu|Ey12wq-2+rC^J@Q<@wJk5yOt8>=hyhV$h{EebE|&bPNCDx)ttaE3|x+Ai{aN zP*YCsJcc`qAT<@9JS?R*DTu=*Zig`?ifTQu^|Ydu__}IS32CyvBGsGwa1vdB@t^t8 zFx}s?Nb`^r(W;hKeO?k#lb9;PJVD|nhn`V0k$tWj>GtYxpxyfu+J7ydds2{pWyq7T z_W(%9+J8`ImJy#e!O!o;KqD}ZJA+c(v^PTHCkG5BW$UuD(R%eCWTZIiJ_D7cteRww zj*yL$>3M@>(IBqMio`1-!(^Wca_)U37}+hA171BSIRL1sauKUN9Nc^WA384KOA|D9 zYU8S( zsZZyO3W9o=m+8~-kKP90%RfL=>trJaz=DD!JY;myTpw{sd8f!p)nW~pI+G@^8FSU=!)3Qip%B&S3Sf4-*Ae&(4p`MlQ8HR^ZKsK zKMH%meSac!s5JKpFxdzRqbVSJ3kxM8R#P3gJJoxxaWT&3;+?`R)h+bHJLa;6XQK_} z_9jT3x^BjnfW6BoyKz`Igc2Q|RWv_;)GIX$`{pgPEx5?!a|(M;JXaYVnu-N<^05%N zk4w>AwlYT%-a8sG6bT7Wr*#c>e}~y%+a@N->PjNSlI6`|KD{2+u%MJI-}CUJ|2*?E z$ZijZV5->j!b#A6Q%)H9t#!F|#ph$*D8Y+98I}}67TfzC+FkWL8dxWXs7<{}{s=z` zVe;>{6%7}Zh!h(V;_9gy-KC`E6v|f0d?}Zy%BP*b)J*YeQF%7+zXIRbHBtAoCIR=G z*t6u1U)~puY84$Z-9!TW6$I$zN&VY`CcYfR=p_q=IM^lK=JKxJGw4WF;-Txa66*x%5Tk?K8~Komd(L#-Dmau8R44D zX4Xt(VKUs!cU8i%pT6XEX&$}0{P<1Lqfh%sMy!Ui(B~QIpMdLShLyiYUgpol|1oBP z_3V*`7+0_^2qO0Tz=c9%KvEL_Ve+~sD)X=5#~Um@T~NN|e_%Rdp-jKqd(ICfgB3_< zW)ryP)3USril%os;~BJmoW7LYfabNFcJ92)6tA!5Bc>7wrI7{|5O=DCQfVZXQ4_J+ zB(mTlx!4=Th-F{iQYCnG5JIi8_K-+bKCoK+c1hY~ilAl^62(}ZD@gf00(Z=8p-*My zqkzpfmQ@jVkP6|<{y0`KAv*|Jx&Cpv`ebFtKaa?`tr&Bqyw<}w#lx~zr{Uo@p+wO4 zs3cCeGQ7e`!oM`JqftR3y-x)33qq-iP#`aTrV&6gOSk$+v{|Ui=d%POHqI>R+w@b! z0}Z~fKA;;1O~;As&f2k)!VTR5D?tk{kPgbIFWAOe`UwGsE~mw?YMtn&+Tl92G~pS! z1~(e5W97q3tglUx{w4@A$BbJbS^_4Grt3yr^5Jr7&V{q>av=yhklf%$_4vX;%qN zZml*Y(b4m9scbK%H{=mJQHNAdMY^ag*vHt%l!{Jp9JUw$ literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1268-image-1.png b/generated/webwork/images/webwork-1268-image-1.png index d093e35d84ce5079be1f00f589fd313db84b6824..e082ef9a39600200cf3bf42e7d6b2e9621133116 100644 GIT binary patch literal 14955 zcmd^mc{r4B+pu<}k`_xsAu7pMmYOI_Q?iuGZX{b0%GicMi&Th&%os}=*_pB=%us~E zFxmH!FvxCfWBsmsrr-1a-sgRv?|a_k{qyTM>bT~-&-1*_>%7kWzI%C1558;1!5thN z9J{Vu)->SY*hB|^oj5lE$cgIk-5ea-B^)#~t~pp+b8uXK85L!JEy9GaYLWpPd3Is1 z`txfV;U|thiriy*tT7|?!f@O*t@|&&Y?KOmaa!xO)&@TFC{ahP{anFMHuAbXJ-M-_ z=;49u7fl}Y3dAi4sS($)K2(=AB%Ow&G9PR>{NkzYw_2Y&21Q@HszU|%_ft0a9t%l$ zTQ)uGwyC$Jn1>_lyH8Osm6^@S6Yj7w_~i@Hm3jP$8^^mb+g6IUzDXaxR);%mz0BUo z_@ntx`Cj9`VSm)v)%wztOKn!{Z?BF;lKNgyzJ@CmiE(? zqgTU@Tw3?1+G|;o2D~}iksODYOT7D{INT;SURY`{(swE(M4~y)PE09m#;$v(M(x_g z`Do+tg{{Xn-g>v)?J*5w0q>jr9FZ< zfrkz&kp*`=QBS;kYm4C zdceai&i_(>NdR0M`iEs9+S}Kvwi7$iJ7>Vdb8Q^rd;|C z_l+%&e{1ze{XAB{728g6rZHup8qs_D~~cI~k%V@JpKcS+<=N7N5Y`pm7tEe## zdi%owp@1FgQhf)lLZ$*Qi>L3Ydx`jN{LACcmg)V|TOY?dyRbT~hr8Z9tsWU54M@z~nz=FKBNu~-=hr`}ejjI?Z~VOdb-UD#`1g8!*A%b0 z9vh5Qh&&f*@{sr?P}vBkgU;~EUsRdCKCQAnCLl53x#oM#Pnx;kbzeH1c$1COs8V!bWdY^Qr-~S5Pt6Ob)+%QdN?zqfc;bt&231!^pKxO`swBqm9>MC`s#MbRwfv$F? z1)d5-1@;8`3pxl+ze;~KL69JLn3J5`+kUrIo~;b8golV=e&}@Tdf;X;X!$b=2QZ&w z>M?29XE809S2-j1^*!X4164Y^>(MwSiowmB%saxHBE6Q=oF7tf|8p#&+;m@IvQ=F` zRM0KVt+=QkF%Clae49=0;_vUXyWzs17`7;ueQxjw!dJZPj~yW9D16-4p5Il6l>TD& zC3>94M{-$xdBe8!rwJNSiOm#aYW+CmLp=LCZoA+z1@e4k64vIdlmW0N{|}xUSpFwPJ^!; zR`$D`UWV#B(* zn0Jpwnmcu)iSU)_#Sa0G0zPY+-?{s%Q2d?&+Q22}j;qJa?I1nvi$St2qI0*?FaJDo zSLbPCYl{1arf9W_wVib#76Sd)R%{D)GIu4n-sY-}r0s%BBQ4Lq>1L>Py*}SJ&Na?Q`|GUYGvZ;jtIJ(`NA(5e~GjxDDru83$$xhV;i= zJ6jA68y-H!-{VwRJio1(`;*$vF%wdXD|M*sMVXn_R;@v;;EI)BW@-o5jHmk>X{n@5 zN0X1j=dI?Cjth3UsG3C`l{FI`QfK$Q;I^pUoFj8W2~`N z@k)zIv#a0N($cPf{**f1G@aC6wvQ(1jibcG#0=uDn*Mwn@)jF$L~?TN{yNeq!`ab7 z5T7-?UpeaG?+Tg3(vVn>^F7Ew-{iT>rH&X6Qy;5s&7aaVngyDUkL~z<5H{qRozj(4 zylOqh765R|@he*(Q{`u37lS$SKSyofZ6rWzm7koF0!P$ z6&-6zHvN9e5Z$=cyxK9xXgB!P*r=v{@^wiBQuih(oDN%GHqhnZ@Rs1?Sm zhB-LgWjQ#0-{#;ri{aqldk|Z69mc`2@yRuvn_BDZ>j3B9{L|1-N2S*Fbl0pemG$>k zcXrec4U*?)DyGKClVfE1Kuyc{FEy1h;Au-U1@NR$D3+F%1_lO#f`TO_B~DIGnwpw# z-@a8;R3wo|k&%(FU%!6#>=_!3#^dqj$;s8Fr9Y^(R#uk5LC;;B@8{)yR8msJVxlU( zfBTU1{#`gp(!sUOO5*g`jCzP&v`S1;qj#YBNa@jJOcnwS9@ zJl)NBcy>|9I6E7?;(}03jnh3HsiH>?cD7fK)x-}LziAA1uJy4lF}wQZs#?+^!3sws z>|3|=l*bD4k~5v@w{BdV7|DO}%)TAx{RyV9efu^wHMxwFl4Z6A%-^5mckStV=U17N z+nW)XByo%s@~W(?OiD^B*Umm&?c!H#Op*U1U$pn`-Mh84G$kb@s`GPlOijJLy^kM1 zE+!_{-1xohYyN2k*+$aG&j^$E2L+29?Ia~7rza{dsGVtTsiD@_cI8CoUA~YcC|Kcc zGum5NTT@w7R8a6aH$6R7T3Y&RDY34sWN5Iey|qY0M6j?R;l87Vx%suU)HkZ<&onjF z*4Ne6g!*PFolZS|tSr!fps%W@v#h!*Pw|W_ImGvq)`gT~$C?@e8l|A1z~A3rR8;gw zBs%Scq_&#;&8?i6Jv?{;fqbpAndaIjHf@f1X!7-$f0p{i!h3e)cbK-;nxaC&?%lf{ zKSEuGYsAMzs;DSZsgybj<>KIEjq7P??`>_ZG&L^< z1qGcrempt({q9{mfceF4$*%;qBJ5_McY{MSeS*?cd<*NzBl<$Y&hOoGO?RxrH8b;v z3D;02J6GYF951#AT?Rk4Ke%jy1hA8=zsMIOppoF%=5|F_YuoUaeVn`a z_ahkN930XHS2Qo)^y(b%Sx&GX^5vVla%hMy{`2e=vmIeNvF*ZwRk%iR+=aJ$?!bqJ ze+v;jjT*HNJ{5SOJF4t%cG>%>ga=nMY0taUS02x9EWCIl+o?RuNj))gR3u{U`n zxLfRbDt2LE+vu3WD0S3$tz^i2b{OX4qka5Cq`^P_sUae!Yq}8b%Ndr)m7z8oZ&?On zVTp9(QltS-ix8p2X?P*jh-4){6*yIMZhjttTpoAnHaqt|GANbi)#W5cbSXL?!re;X zA_y@`$3LH&Zw)#2K{UOsmlq+|$sZdPrZ~Y_K=Sk6nJ|nvcwM{VCLy-Qw|q0hv0N&} zo^ho6legya1J_BS0V7!jMnpf?n+XW~TY9Uz|XyY?Ki27_a((#k)nQT)$4z zo-TQ<^kEC&S~UmP4tDqC^q+BB*jX@vGV@xVniD$GeQtHc7e782Zq-v-y8iWP`O*!% zzv`FL^=ntsMBC1Zgx*nFAEw3e@s3&?378aiSsU3?Vm~!Av4^IxI^yMnvN4_{c}@FH zinJy$bMeu^d}}!e@(-`uRIL46n<>2TqbJ1B#UT#g>%Fv zcdxqpl)9tOeQmDTd=irpf%y%-%xDviYo3d?l990y zVYX}2=SDpEAM5g7H!^%hY5N&{feRU~6$Ja{lTJJ&&1l`okhP=5Ymi)R`fdy7sKr{J zn&H(FKFNqb9u|#qddcMQlNTBZsUMDp^K#QWMC#@EP{E=E)umz6!D)FJ894$aQF#3U z%4Yg@{A%AJWO#wZOy$`@qnhgqll39<>ccxNarko&Bf{5yhgUpB(VrtJTC^B)RogMO zlT~%-2dy-Imwd^92!WOP0sC69;B5>F_YfwScqUvwJ{fk0KDaa+LOiRAiS`+ptz4Rg z*NSm%W0Yc;tgXanX3}b`?P!;q8sIirKSp%9JL+gQNkr#?#oIJ)|1{j(AG8_>4dL9Z zk>8UY=6T1Lk`W#e%-=y5#RtKPW{loOS)NMpp%0?yTv4YpMrK24KQH%=1G?UDrEG$-mF3M) z=F42>@2T?niRP8Yf=*dJEP)(F)<02&PBfj9o2V8UJ)#^FOnEr)O?%*KjhtmFed=dV zoE*2Ls#BG@O}r?#ouyHs#~=Zrrlmp$T1xzlI9p-CC7E0gc2NtjA}5@M9b91!PrY=n zkxL*?Ahl_6jXBh)5!fY=}883WHV(eUf%owRMH4AnOmd^ufZ zK2Xc2qSWW4Fn~$O>}O_mv3Ul;F8PhuygD>ydlBP;bO2WtEjjb^rS$;3u)X$)tV_Ik z=>}kf$6TGux)jNBY9>*~8RD5lp0Gl)i~GPzX6wqBZMR5jOI5D*8#f#zyqypV#%JdU ziAfOreWg)HaXlMIst^{7oASUANFg{>VfWpm7>K8kw>s7vzOo{k3W8-Y1 z2(&FFHYC#T6*;s%g_+FbR+H6_smp3JD>JuIBqw;U9P3^0yYBz4by#v$$j}I|o}Zg- z*(pJ^^4m%;E{Vegcjtr-mBEnw-~5&7;FmD7rIfdVy-p3yrcm!mTZO7z&dXXlS%PBv*Clow`=GcZe7rv+3qLa!c65?}w0&cI-njmws<|!C3FUc2TqqoB#sy zpj;oiQslO!-VT@IldJ-upna%Z`}{kAaX6f|A{PqbQ^M`p?Do412(Z!ayn-=)4L}cL z!v#<3g{sgwD)dT42rwiMuXW zR8tig#o7X`3#z9xr+o3zZwyS@%BiaIT5F$%h(h+#lv|NxwHZB^s;!i`ymLaK;;wBK z)fPDGdYgJNmocKU3LT?>9R+vhX<5LJq+lrzZ4`qetV2~I3SUv~L5H-LPYLgkhUC(r z%e!U()36Rj!NQUIXz{8t>wOq&zj)Cw;a(3zDLTelSZko1im2sfu_ci4Qek7td|Qze zIjZ}>#6FsA)ggvCyk6IhL69MfW21{4v@i*jY{g!o7vtWpLFr)ve77eeQ-M#KW=(?8@TBUeTA3U_2_ zoD$4~ssY9eK%C(zEcIflL{JZwSTNW;8K(+afn3*-6v`d&UW zKOiA0f?>!=y`f!{!;xfy$X^@&0rH3A)Yv>vxZ+rM#&{vh9p6r8tL`+U zIz{ip3Xr<449v54y{kykB4R@B$Xn~~Qxl~WRM;ar2 zKg9T@Mf>6>9n8icUb6elqK~JNcd}2;;72u8DMNC?Cdh>*Sz#h5wq~9i6(mg2Ar_pX zYHK=`WMEbdv{rrHJ>SkhCoP(3&fZ5OMq{kwhsbAwBYYsrW8rWkc`TqL9-u78CACUL z`{XAz#e}w`J8;JNO)k1fhu!uoF-1mC44JEIT00Tn>OwRn?gPIdKFf=9-6{Wslg3%TtVYKn9K zcg`cPmX>$0NMS1RllNu@vMGiU93c{ijL^ChA7~@tMJ$9pVX8RPMFE8ht1)U~2k0-R zW&m&enKp>`3dQUQMb$Xt`A65y491Tb$Ub1&^epF;rZ{aq^;GHi2;pzUsVr0{2nJ!V zQMos=qI_1}X0VIfo1<=tVfct=^etl#F^CGKjIwHPqs6s`QP1Z?rS^+Un~o6emvTMB zK1ag?=V-xvy$fqr&qNN9%!!MSxlC@2G(OABxVabAx|GO$9(ZP(KiNmXY~Vv94R=*? zlvXO&v8egQWzp4=b3xBvk%kZ!XkA@i=7uR}PE|M5uB~Ytz4d5m7*4m#_da(gv#F#= zUPXy;D*k<&;VWeanX#AjvP?t!_Gg8P3Adb93bBq1WOt0c84&dgwylcNb=9-8Ya%b< zX)A0L*e|?_sEYfbYDJ5|?QGgUQo}>xYLD4w?+4=ECpnS!-NBThA!3%`H{ye=!fJqD zRY`Om`z%s6T%20hF-^HC&T&YYn}P z2x;wxf=<6R?P_CEk6xqOe4PBleQBTt?It5FygeQf_03P+NN`-TV*x`?VV z2~+cj*h|wtnC2oq%IwW$j76lW_iRC6cRP{XmBM9mJngZVf9mN$?2+Rn|=-G+zCYyeck%+v-wz$ z;}z7{$IcB?WFH~yQ<$PDo0WOLu|(pw-=_ z;o<3VUY1N^BlvN)bogPs#UI!(zK6X+_QFAA%J156R!Q}w+IhB;O92`#`eKTnYD@0h{1gL2c3Suv8t5H zb$aj8r|yXl7_kJ@e$>jC8 zVX*Se2%F<)wus@fy?$&yJLt$<(}dP>Q8^i523|^k9Hc%#7Vk+vC&S`7gkwtNnz|QX zrij_ZgD34&ANRZa&w^SuNr5i##Ym*JqbCZzz|lHR3thrQ`@c#(xTxM;t{fzSJB zZh!;+cF7mzLI|}F04aG4#MEX-9x}yx{c9(5`$gCGkvY;i(F1|X`u;R=3PFF#seNK|ZNP}cqC@b|2A?nlFZPOcL*@m+--)u6u-!&# zmDi#5VG1Z9TcBfaW-v{%qVtm>mMgOr)ev*My#Ws6~Id&+#0)w~z(LL&yX)yAu8jNI}aYD(Uykq-%_uWv=2vLF{} ze`x6!S()0;6cbtU>Ci0M>2us*06S{nFj@B@!hN!9LZGBusBDqC@a)0ltbPCArW`vG z4w1$t)ouo5$1k_KU@eX(MMTkTvYLf9mA zs)K4{@qub9!ekD9vRtaVF|a-m{rTld+tO%Js>#4mZrDn9kR!uczF1t%zKw35(DK{c zz^1$4XJ_1uJY&8wZr-DBB2Pp$pHWm%WMLytX>yj&e%$*_%@MC(Z69HE_J}m$;LA7^ zsK`e>!oy|tlUrit6A~D_Z|%N4+emp*3vy-8Dfzwh!NJ^V`k2{W(w?OZxTsxx&#wW% zk4TC55sUbuUlR5iwpW^QlC5i7+SL@L*3pQFxxw(uSQnLyTaG~|H!~1E51eQO9WY>M zxFK$ucmtudSB&wjqB=jB%&3IvIHEJN&oudES?sDF+a6Xc)*!e3v*ms`5NkRvPsUIx z%`>8MMC`hp+G2Ys(kd_ca;@9|j#4k;wy zrsn%LLF->~2?;Gr^;|3z>k{RR{D4B>8kH*ppiI}ggOH`yf`S9Mow}T5J}7g62kl`L zR#Xq!<=+9&GFi(4Hk6Yw2O1~fibpUBP(3UPMZSjql=^!W=Hm2E3Z*0L$K4d_NA_bZ z6D!>R0_@grtE$vDo(GyZz|k0+fC6<0hz^pBr@3mCJjifN6&dYpINGP1&=LfpTl;yr z<3XDMu-h>%D=a~U`54k{Zen*zz$IXUzrzmFk>rH-0Z{ydAA9LtGJ2ta97LsB_VbPO zVo;j^5R$s`bIEy>7dmIGf?7LUfOtpxHX~MZ27(p_G6m>)x4hb=wE;c$541e>0ZZ6; zIK8WmbP`9qRZ>UqI^Vi$&`OG*Yn|nB(rYH- zErOz&BW^GdItMbeO!tx_NJRE1!wg4ccy=A0w8gWPEoBmgy2cNOEa4~vq|hp$Bp&(R z`)uP7V(5Y{0)WEhdRN#te1hJ^KD*c*kb@M#zyLgMknQzgQ*38|3J9W*lkIh2Uz$Ii z5%v(){%ia@0IS%SKm!6`D!mj?l|q&#W>eC^wk{#YQ%QNY8i%4#Pz?_BkCs;>JF_{# z=~YN8k4uo|8DzBI!(DH!`$i`4XFB#j@^%5c|TdACL zEY@S00my#9OCeHwdKKBRigZ(Ba>CluQey7zb1DnSf0s=AsQLb`y59EZ$a}HC&T0x%x!M?5Jff-BieiIIQ z3d`F>K{jMcKqoY?8KAh7Ko+KTsSdp+7dl#!@C0ngqoCFWX==ZKG$R*k;=@Iung#Mm z1Dl-`eT0Wg9-Hdx$`W25Mmoqef$lGfyrLlDd&+NoNpq=^y#Wm?BGju_>Wad#sq3*K zQX;LTolv8q6+}^DXbqaNd^8;RhXFGm>M>YHF41OILJn(S6N)7mtPzy-Dj@#yrJsI> zta2c;j^`*-WGu}RSJE27?zNl_&@1&^iH?TLf}Dq>qCd>Q2|N8G1Dx^Z&Puqb5lhg3 zf|`MHp3zbxB9wid#@*6j=>R(9ztOy!=~M9$OCpWPq2K#BY(|3czd&g_=07|Ajhdc# zj=B75&!ugXpbZ1*nW$iyveMAEv+cCF!ngi9HZWGf-#czba+_Kv<%)ACCuqq);-|+l z`|zXpyy~Hxy~_9dNai7yBub21H3Y<}lI3#2y7O%Gn?3h}!8-+(_;o7}!WW{Xwrv}9 zmEK2Yu=ZpuERRar}HjPWbX6G={ZLzO6ZQ!jF5sZ*3bzUOR2T_=XQ5?k=` zG}kH2mK1=uxEZce;bEtD|zFqWqtr!MMq_Q$bzw452 zjVfvQ&Oi3b!p=L6-@EKR%F=hgmFlDTo1;B8R;^d11VT>hi9i<=@|+p^3|Jxtm>}ry zg2BWRI4zioiGq<12)5SXoOePCTla$|Q50ISP(e=!+#gKLAbgO1krb#9m1~OuROU%w z#RtLu6(6WN2443Mjub!yARrX|CqAIwg}nW55C6M8**gV6sGt)EVd!u8|Bg`cd~|k% zk3-pvJR1|U{CiN3oS!&hYizkz^RdhbE_3w5%~SU#&y~EkTAcy8V&(Vn(b}Jle5K&U zR7oUwy_7^28ok;9_iM;69K90m{LSirq&ra5{~Nmd7k2>Znte%70ranw%hDeU!BO`g zH_p0;KbOtYANv*^A(gVOR*R+F|8fui!3VhiUxvZ9A(%CN)tfVsHYpU=3{Cc$SO$o< zLO8T`kuGKIYixBpw!P<4VYjoQ6x%rQRy5pi;X{p*W~VmL45r^axYWpPCKCC*nZe;k1DY|n`%u{r+x2j5_Sf|sA1bWp~^ zC{k4vM{9bNN_E{C7NVR5PIB&e-?s*|moqJt0u|nd z#?`+(B6$ro7pzc+VHWt?S*)sat~)I*^=g1}zTF<)1zx)yP+i7mX3Q3jiK!+0u?GgQ zhwQusdSgtHQa~|W5b7J;9uUF1NznZ-haDj79KFyU>FuYNC0hSU3iS;FuSf(?^QVCj+)wKG> zKyR5inlV9NuazC##qJ<>AzN^Va&Y)phQpNIza;-M0i2O4=_SH3dDXP|;G2$p!hi6K z6fJLMCDC*brm)fp$kP#J|00t7vyrkw1S{zL6_NhQ8~#mW2TeJMS^ooL2k8G1k^X7V ze-YmPZ4ab!F$(-nFm@!=!*XWTL49c_VzT$RYn8X$GWnmUN7{q$=V>#>fpg_*@|jb3 zeDukuSNVi%i?l&;d-CP}LTif})tM)b$hHX4o+stz{~OQ#TXy(=^ZBRj@V|}J|4xMa zTg3m5^zk>_KwH!E^YW*v?j2N5={@K4CtZ^~&Ii~yAO6oF3FOPe2z_46f`CwsBkAA+TNlsFc8K z-2-J~<@w&6Qf!~Fix0SFrb%LXnu$0adQ$*$xr*Hyy)hjM8Yx|qoPD&B`Tz7y#b46~ z1MQxr&SFMcy0<>(WQ=p^xqyX}&N5*;M@f7#C6lXQzM<@0=rz5GFDJe61DAk5B{23vA} z<|UICvC+6(m&7^sVFJ5Yi9zj+iO8BCSiZ}yM8KyFADRS5=cZDOK$Bz0BDRN0WThOs|k|f0ZdQOo{*B%M@s0Jn4JaxNrFq zs}$LR{18O#zEZqT7xbHdYOnv+=-E#DxAFj(*I)JC|1JCf_kopF8vZ-Ze{wg&G)t~$ z*b2Ps|0Oqhw9Z*6iyK|ARj|&P8w6V35ELO|=WQePl(xy)8j^D=@Ne=HxyigrnZ%gg zU)svwsDY}IOvXy~Opku`lyBxc1U`f?k)@N|Tp3yW${u9gA`~ft?=n7N3W{!cB<9iy zx1plEj%bJ_hEh8~HD$AbP$${Ko%jmK%vDDeRA=oQgDRSDgld}{eu~tn3`=#v-(Xma z7^QHi!f!0xP+lj)8q{Ukq~%DnV^pafdeQ_9jV1`8vU^QGBFX&=%wl zaF|WKt848Z(j!KOYrELT?Ip_oHx<~5pxQi%0`f{ zJ9gHlViwL?ZS}20zknq{+mpZ;&(-#@_eS3`bO3slVF}D?9=wGuC%n^t7lzS){X8k8 zur)7Q%L0Hjk@kU{n04e=;&0z;>EyZ6wS}KaB)P{l&&CPOCIs(L(s_a6 zY(7n=k#qooDU2m4P^Wu|)Q~9VaOm;$CqZksxY$TlER8 z3KCLOR^N`SFDs%xIT%?;$ZoGi=0pqlx_>yU^r40CIwZ6x<5?NNyh`DsbZflVq{}sG zyqEn)fAT-g%>Gi#sYcLypMq4s^8wc;7w!z>oyp6HR@yW=#l&UETgI`4E*U>5{)$)< zou7AxN;K5f7`ADCHx!Ao{SiTOLHkQ4WvwWKa$Rqz@SSE1D=H3@4f79ELR zuksmRy5EB50R#7KrLQ~d?6xqndYi*1zFk@VNpc#DV=6VqDiD$!Z23Fx1tOdml?wfQB)s1sCh>$IkqR}3sM?bhZ!#hG?AA;n*^^3~G> zir-xwv542T3$Bk*?(@gyE^+Rn6!^lQb5V9;xj9>qb;Oy^J~MShMN=7{{U)P>9x@T( ziPvq;Y^0A&|Nf3Wc@upKYZQuYo2e%<>I_P-MsT0#g|PIcBXMi;_gaxuqP%@ER_kdz z)sc}DA88L~IL8AAH>k&k`Rmi;m_NIf+ywit2AfV`3+NL&+j196{b?yxZVX>1gu}dR zJ=*PT^8|5xpF{o|YdXH-VpH+er){x*0cWW&Y%%sFbAsY?q|LL_r>^h(n7P6vcTsMO zz#&P&xYq+1rWeo8f-|G2Ra>_mj2-l1IsKW{Le!f`Mm_o?vc=Bj^w9$B$jj^Vxj$nx zhIFPO{J*Q2{G18d|qOpOffs9g`S5hM`X6 zJ^w)&^)V6jeP?hY|GE7{K2gEbDZcO9%sy&cjfx#@B2w%&&GfeLObt<`sOc>6>Fc(z zYx=S-R5=;z64FdPW4$2Op6;L|(Awk4U0NI|o{yfemEJ)ge@zjMKu-+5q}JIR@y0c1 z@zBi-^O4lyPkZx;^-r$8+0L+w$429Ct1|P@`#Y|JVUi!Oxh$8nbcsC-eJG@7X;h9^f^*+{v~ENeAu$FwF#c+DtJelilA zwMl8k$ch~Lj>9YEM%5AOk;}delMjRD3JK%(Z@NhLn>FTey@z5C+nx(M-EZH$bDrQo zsT5(C`?i#%6WaAj6~$}**MI~w$v>hpZn8?#nJAV?8$inluKxC&B(Eag+H*f}K04=Y zNKFJ^ZG{)w#M}P8ssDSgeW`*dM8I z@jAvBi7qsQMx4gqrC%gV`d7FFZ{ zLb<;PZETU&t_}~}99*0^{tE%p0Iq)ciiXDFYpyPK4)<;M?c5Lui1$In4{Yqt8rr%d eA;N2#m!#!Row=cC`VBnhxT2+}nXP{N$^QYKh)Dtf literal 1237 zcmV;`1SmU%v1);6^8^A5yH|yCwSCIdCTTk+jyir+ zllb35j$0jnlNmLiU3UGw{@nik8@In#b~&^D_K%O-Gh6-sJ!zQ#-&}jC>s2qd;Xk$g zRTtZ;glD^DetoT_vqhLcx!YerS*^{8)jVhCCE0PEJFH7Z?Q^UaSoLVF`$nw}cCGf# zv%041Y87mGQp_etb$azyz1bPN7E^HP>tOesx;Uql!*fT>sfVqM+OFr)Dvu7tm`466 zvw2+xspX*Qq7b8Hl%->amJGT+(0AndVE4S_;-ZTh^Gpgj%w5K9k8FjtRy#gJP6v!p zA>{5$g_D_zAV=-4O16dDN#@(!=W^tzXh)H4+AG-xcap_6_j$kMsTF^1 z)%_DL%#{z1?#5j9)0TtsST9%n!9bpR9b56y;!{4&rVr+=7KIf1ypnnu<01DTi+b~1 zCmCP4bCYrHmmC#szRi6uMz)-kWF}jbgWPp0oP7FKHNGp@_nviL@lU1Qt!lhlxXICtWj{}HU8?S`UC}1pw5RU_k9ABQ3 z8V9&C+ZxC#3ux?eo6qcLdA_B+{CH^`qU^@#%AI-LmQmM|HnLmj%3XJ=nlaHBT~4x( zZTyTox7+9$cipLK#zf@cyj&i%L+Xpd@$}?{49^3dG?t*SdXm7^l9y!(ijP3${ z>xpII(FQ4hpiff~onR6i&suFFgXag!iNWp=ow_fu%gI9RHt23(l^*OP1TAd`+A>cq&GM%!ax7r}V5qhJ z1T)oCGMzR!kI6GS5tx2%ohsC#RE{ITK2ivns7ekJZ62GX5HL}dfLo4DQYJ9{o^={< zz&-0!p_YYmEDGf~;GT7x$A!Q?QV5u+O2940CMgs6cxyE;wN|so-}DOv`$!qU(V-3n z2dI+iw7q#uS0&(MX;a+k00000NkvXXu0mjfCRt|O diff --git a/generated/webwork/images/webwork-1268-image-1.svg b/generated/webwork/images/webwork-1268-image-1.svg new file mode 100644 index 000000000..f864b8357 --- /dev/null +++ b/generated/webwork/images/webwork-1268-image-1.svg @@ -0,0 +1,263 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1268-image-1.tex b/generated/webwork/images/webwork-1268-image-1.tex new file mode 100755 index 000000000..b6292f19a --- /dev/null +++ b/generated/webwork/images/webwork-1268-image-1.tex @@ -0,0 +1,147 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[ + xmin = -1, + xmax = 11, + ymin = -50, + ymax = 600, + xtick = {2,4,...,11}, + minor xtick = {1,...,11}, + ytick = {100,200,...,600}, + minor ytick = {50,100,...,600}, + ] + \addplot[firstcurve, domain=-1:9]{(30*(x-4)+450)}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1270-image-1.pdf b/generated/webwork/images/webwork-1270-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..b3bb225cc00ff48e61749454a53c34de0b7c54de GIT binary patch literal 7419 zcmbW5Wl&si`lf;4?(U%>XhY-f1h>Y5TjLg_aYAqj5D4x;gS$)P65QQ2xJwA||IOA` z&CJg1?9_fa_13BP(^cnwuJhagbs0H!E)HIF!0yHA1G)e$C#|!&9lD4JP|nuD9pVa< zb1-v<$UrQdEg?Wbhu9X|vZhE1?~u5e7SwW*v->AZzDdK$y-23gyFYRCM}D+fJTYtuuF z=GPfLJg{~FVl@0F?}bT*j~ra~=7+4DPNPfvern0JB?SLA;QEpi{P2tvHwT?p^1rLD zNim?|JxcPq#!k8S`VA0sJqd>ToHtVA9K&{OMS_@bbsE?q&qg^ZYqHWxo&(9kr&`YB zVV6%hoz6!G`pmAf1_b^;x~$Q@My5x!Tcn<=b}k-y&7 zXs?)yp42?^l{wWf=mXpr-d=}eE<+2l6 zo}Ygh>qX_%zS{u&PB}%stcMha(LeB3KU{5fD{fhE#Su1EnsM4Nhn;cLQRQU*srz+` zWupfR)TnoJEDZ$*)+m7aYLqW!K`8H1MT_M3;gIT{n1?BIO=BnNOAwM)&*mA=BmhnQ zsgcW)!zZdPdDvR6zCY5M-w>^C@c?vl6IheKZAQ&;H&@LKzqEWr!XyyNG8E#n9}cxL zz{=*)-a*Sk`G^nW4u=LYBRML6A23zL=iL`MZqtE{38gx$Q7V>0vlTPx6|7Y`3b_j3 z{~Bd3;3@j@E=y1_cOdbln7ep#Iu%swzJF3l?s26PEF79To`h%R!gN31GTnxCMrbFb zE0^A#bXE-x<)<8F-gPhiB9JBAHZdT2nQd@~M*M*VFx`$IpJCdOqbp&d3UwSEu8a{h zDATaE_c*isJf6ii*`hh8npDcSU#^wEo<|r^ZuNaqX_AcC=&i*x@rr-o$T!@(p2`x! zPe&==kiTYrJZtP3slit^km}-F`0mx)(;z1nmqD^~hC;n!r@RDs^%rUnKgAbAJ4M`} zXeTBm!{JpTOF#w8b5V=GZbtC{I8<9tdVMd%U=3m}nbjLGG2Db$y-Uen_WGWQgjaEx z)`amU?hmD&W}tW`_U6@d2Gh;W+8*oe$lgBr&r3odti76dnwvG7E1s;|6RLO^8=ITx zo=W%=Y6&3(fbBykg{vh6DFGjnoI54+u_fCeX~)r@()fA)37KV#%tyP|l@V@@>3ZMd zF~dajny-oaDO9XX7{p98CZXSrR+us?-dR$P*9I)W@0rPZ=jvL(m~Ri#(`5{jJ*mXf zey7v#DjzMRn85CJ{X++t9rnK%5e8e#A7;;x!!0eiN`jW3NSh-9$@WP7+4bnrR72e| zz8JpyTBTr*Jw#{^O|KZ|1Md-U_fVuPn}(jK=-XQ`Ly87+gMsz~P67+N2~4#(&eDdi z*ZQqxfc89Q1DF=uP#zjK(fr%At4Lyyq^!M&?i!`?=4cNT&ru*#ip@IVC$C$uf(Af}-Oy%)<@>J=5{G8BJg=!MRV#PFJ0dTlu`^jVsM%tkK3M$@=C z?Y2H!U_5Pip3G{VUng@NBK0Hc8}N_blu{?ke3J?~@GyB6m)q)%Qk!4$2}DaiVphrvDK_$leW?f&-1;Je z+RIoivr}O@sApU*k^-8qE7(R&dW;#c`6yw(aJz)PZDYRF7QO%A!KI9AIYG5kJV=Y8 ztz5+KEp$33WiBU;WBdKRdRI?c?L|qIP09|IE#lf<;fo+Fj~bR-k+Epd^giJejb?(h zO2iS(#!)N(^THU#GVIvr;g|TwiXwY-h?C|2*5hBNzX_EK^xv2j#KZfKVl`6ljt!>8 z?YJg#bEsR2;I;MPWQ-Y@m2S&e4uI=9OemMY#d&!~$H(!7139m^d$$)jl*%cf3^&cS z%c68>^=ubdkKhSW7SR7f3)$t++1h{0&A@ypvA1~>RVif&Q zVaD4^B-OBw^~zZCu8J5#$E~zK`e4SHhdwX8Y{PXUKMUObMfmfCnSMfyIfR{+$49eB?ZQw0X86@ySl8Q;G zb>zh`S4W6`ITfR$YziMz^J4qE^E`hor!hj$M$;Z&R((6&6nfUP zX%(iKz2=s4wU5QI(3k2*`G7HdSg-CXyFxB?XlK$185V<62;*}B*pj$A<9uRiRy)1F z_&Dl&0!BltnH2hp_LLZMycHSNCDZ$E7$D_AkX1-DDbU1(24}#4{t}{7x~rg=rZpBK z{~^>60I85hsQd61{By~Dz`rF3Lhzi-rl6v7n^>A4%BxWG5-pOd1t>G+8~(PZzx{K&n{p@omQncw3>5cG>pKBAVZWLWiUkJH&39F z#J`i__!#pi=%mK2?ifD=O|ChpBvY_oWXpyzBWL@l7=Z)h9(=*q#(yR^O8_fn?`o6i z+h$X7$Q`HZ4D|?$TNV?CU8ybuY=Ykp+-b~CI(wmaL|8u>E24GpG4Cv`T~G?;;smh& zDy5_lF?>^;3+!cJUPE9t#d3fHo?0b>fC8-8UvE9CU6|Mplo{sNNj5VG#QIiXr$|A5Td1aHPn}B8;!1UVdSqPQpV(i zCm8TzF-;nK1H?*?@%C$?Mi)#y-^4oyg|Agd0P9=U_mm|fW)3H)Z`h4KJz0$*NlYv! zKSh`Yn8KY8scjiOq}A0gx)@IIS!qB)l1rzabBnu^*ROycNF3MwGxy3i%Q`V+Q|CMhwRXx4V1jR?9jZWBTG>4=xxG|Fh zS$gVaThn5v->eza1224<#r$4Yf~W7W$u4hNt2RXa0l(AE$~Ru}6C65xT3Kwije1I9 z(cpo@HF=^b9yH(@Uu|&cfcN|zb}$WSPw~b$(T)36Y+6ex+cz>g8-p|EHP`fT^ORdR zuF1rq6Pr=MM7<(nc8`^gqNJ2NmZ}11`v-6pL;~DDxTKxJ?dCThdu$w~@)u@;sY5b@3Zii~30#5prJ6#KH@G zA3`sR{<4tnQqSVC!+MG!^5wOVV`0rMT=Hi7o9aZ1(i@T)TyRmjhHe4juS(hLnI;D0 z9;?0;v+jk|z8Sgj^+^|_rGyXh8kIo9<;VxX1ha>%t7xX&!c%$xMLS+=dRR>S7RBs4 z+7Om$v#q13ldIKOuLtDEC;#(fDwkV2EKlJ5O_mH8%+f;pcAk=>lJII8KfeE?X{|Tx z+l)lZ4B_(B`rs}P8Z7IQ=J&mk;=dWAyni2~|M%4n#K-@SWAiToIR$fKb{w$!o$#nI zsB4#nd}U-|z&eyukP^7K$3Qza3-6m4zm9&+YZpmU)Q=Dw9b&DYQ~+SA z?l=fYMR#b0^wTml8SFcY@ubTt__C6Rx6KYz zn*)8IRGf?XT7*d3D%3rkjs$f!ceu2z=%Dy4S4s`_Fn(0zwSNwN!-A1$nU-Q<#vq2C z>E1iVQ~EwZGhI@jkW3Qr(s3&Ne&zCf;44pnhq9uP%LD1f)Yk514F-02D@NJ+{P2r= zBamQUeprliX#mst7`My-r|!Ywi$Sl!5thDZJUUI~u|}2A%O8Z$`oNNZ!;}B-JpX$w zaq<5XU)gwHbuACV!E=ZS$*zCPTjupO4Ho%cqahSK`fZzZ8y3EI@Rij)t!;r%Uy0=< z^p^H8k0sD!D=RQMm+?6zneeeNN(SWIisZPjrr@l=Vqqe>pZ-i4*%y84qBTGBxJ?iSsMA!~E^J+9;R4xn=&`%|1xWG57btjkAU|`9` zp-1px0_6gu=|zcD6l~LLw3}y-5*ZqV1oP`ho#si147AaZB3V@Ij@mzzS~du2Cpk^0 z-6AdHYZuMymvh@zn44)izGI&}E{AIOICP&Z?d|O;=zjy0%o^4Ud`vDibY|D7 zSoq++pVcZg_t(crj~*xk*g4!O5hy^;9szF`T{z|`cXdJusj&C;v#LwW(+6uiQUTynfT zen%O(;ie^}FGc#}GpFkcUHH_tXTUQIEF<@^JzEkjArnU# zPWp*S3bjLbi|OPw_GXJ8CF#UJD?zLtslomMj@tffiq(>V>(Ff>5a~uNU0kOq`C9aJ zOWemTy=8)AbuZ$vZ+s_9u@ku1*2|$J% z+FYfF@Rf?7PjO34<QD^Gx%2#qtXf;RoO#Cz)BCPc1%iy5qsYugIPrNfgUV)H+`@&jT8_6`YyAV6Z0 zqNZ0NA2~xGM2tb5m}&yjXqXEXy2HRwC59i$jgX&DQ7PQj7b@*eX`1sVhj*{hj|^eT z0LrEBmN^7K(8yW-Zuj{%L&hJJ_=}T1^u=VO?A}d=D}p9+W$D=9JF#Y@@U)B^5y~x` z?t{$A-?ce@-^I0|i6x_j7HCFA(y9>I$tt!0O;>-EE z)txYJm6Pe<2!4>RRJl6E!0On?!rGB8xNiU6?Zv-OqqM+9`ERD(e^m^Ecm)0rd)S3= z{?!Yz?l&tj%aJT|Ab$72n>IH%I)gZKk*!InNPR(h?Iefbh%$I{dtZAwCj{j*Pr)ZR zJ40gN^&JxjyJL;%c@}M@r!f3oFsX4$>EjnLy{Hs1}ln2#o* zko~^J{g$DkryDKOI29W$p9e)JSx_?n<(XF{J`LTcS zM@frHXV z%E8xSVS%C}d@MB8xCmM+J8H{_uLFfYMl3C`_6tPkMtp>Y4HgwZl_Qnm zA8O{3W4@roN9cXUyQII|#m7*ql^OqzQ+|2%xc}t;JAgA}2ep^8-z@8;`EHRfMzwgY zWE}Y9&~lhFwecBgTA!h!x8N2O$@W!eXSdkm!KB4OR7&13{$K*){i2tBucDsU#R0)<9~5QkXBuZBg9vsHu|W zLC%JGP}q3F?Urdj5lclKSDf0oKq$P#`qV_ML5*Qu^fQ$_tis~UtXXkzzUnkI?DuQO z!zY_es4n|pb>g_Kr5=h@sspfe5LZ8nC7yZ<6hcRt)#Y;1GFyauzW)=Y{|$^&gOxBs zyWKS`i&I~;e1>^3HxYT`lA^x zsL#c*cw@ZPktD~fXo!r|m&EZL6hl`0In|egB|HJ%28={YiH=e+8&6JwE4Lx9gCy1V zHJzSN6C6uS9pL>=eg#QcrV2cRq#0ZW=FlgC@R&&^aF{?)R+kF+|~xVpLJ7 z)RJ#6^)eD(FOlh-O|IL$v!1d96vZ)TCZ(lb8eol9Zc+K%j-G}nhJQ}>BIfOBTv$iq zz+!5!+U?Ve>`Z1zt7yDr@|EE0>6m5uxX~O1%PR4}M?CT^&3dyulsh zgYBI7oV3X2e~k{;KSbwmOFvM}+z#yS2vnuz;{9v=cXe~873BX9Jpe!GpHkG3xZ<=d zfg5)IMCeAMj^ukG81*#>4@)JIp4L`B0b>wz0Dm#Lqo6=-V+n~|8oz|Zac_p&%F4R) zc>&&~n(m?-!3+)89R>f;l@iy~J2R|%pXVgV(VGVYO4|jmD@2Rr6^q1y$l)!In^`16 zB_CK2*vP(h@+I^U1RRtle9PW%*7w`H$fkXQIrbMulo@%#YRS~E>SIvc;E$q{)&*;V z)pvY)vS?u|YMxJIBJM&o?=<5_qe0#bJMzqI-1EJ~JOfk3XXB$zDFpF>0)&(VGcb8%;VP$P2hO5rkp5!HY&T)4KcYJga5h+0;HkzHZIMv<`u)*sz3!&>DZQgl_jc(QV% zU>d74E~uUoo`Y|Gi*U*wOzf#9Aa3iF!ZUnk5Be1zAC>GaH3JU6d+Z25zV8h$+51NC zhxI)`@Xy5(Y%Ha{;DE~e4{YQTk@;KgqH`)S8ATOM(%B(iH{+v9^1vgeq7>VrRU0mpYza@)e){PnZkhffSV89zi6{Qc>xK;Tu4d9NUVmO|16$je!;X zoA-71rAfjA)Kw!oXLKtsT$aN)bc2s}L^W8xm8^BfczZ7xf4KL@|1~k7f2hm;S7P)u zAyz;I83Pe>D_$!LOMZw2kEI2#8K zXPQ6-Co5;#ztx(WK(MV3gcigL)HUGx>kY{0KO|D%pB5m%@EBSkCu#6j!Arf4JrF(U zjX>~|mkVom3}r8K=~kOaI~|>VXxp$B{ZN?3PmYH;WQ&kWy(xMLZ-~e&?IFlbD1h4bEeEg=2_-4 zhB)S9I*xNVyz6<+(e(TM-}nD{@BO&9w&&Ty+H0@9)^~l^^YrQ!$xRz}Y`|bJo1~@A zU&mn948o7edIo6O&7vcQ!Tf!|@Z7nphPt{KjMP(qe}k*ODlFxM8L{EVr~f_s zJo{bzHme?NjC+52DD>(DW4FxJywBW@UwD3D6^pw6UZV@!Ho8Apz5U)pzSULP=FHd5 zsF?Mzg_3q-bC+K_)SE6@4t87Clg(Cfx;?!0soLS@_3XlqN-s9nZ8d-O9P|heD4LqR zx2C5hXA35w#38$-D|KP)Bwzp@FJ^wfIpW)LNH}6bw z-Nh+fv2(+Nvu|!6{JYA<;!Wt)EeabH&sbgEI=k`SKiR9h9x=Rm@>%5g8tNv;M_Ypa zYSLMIV@>OmuLA3=A3OFE&5nm`z!#DyHDS0Xaz(o*ndgPX7&rC?PVQR0d2i8azbzV% zJk7Q+aj`x<`&WfhzTkels=sM>aasGY%?PIF!2@gM>p80%#==Qzs*%%G4yXP|^lHNRN@Ex*YWpur- zqf;YL7sM&MOJ6XG)snTzHOqsoL{@w7!;N@GrFBM+iG{8@P04G&TpE)e+dsD1`rie1 z$&y389%f#d+UoxvVN&)oD$^+0)Ei~C(bhcoWcO;qeeXikqQGxEC%fiP%}Xz?KJ%b( z?~N_Ezs(=*ezf5%Z|@Erk4aZ4uGr1BPc=(!^jY6rJGE_!!8~OC{`viezL}AmuQM3( z4~q;R8{RX#>7{HdXNS+j%Hi*M-w#YHPRma_9DNlO#wy2m);Rb^`i&=Fo`2!p5cc*; z?^U6z76%9X1pQ9=shHQQ{)8|E}u1$&l{p!)1M^DbbJ)e9&wdC?sL!N-x zGghxJUY&_4f3F;~_4uQxOZ|=WX`yKh?I+r~cAqn7u$vXJ-$FH@U^KX-{f`IC2G#_` z2jp@d(&J>@#Nhp+@SskJzK4Lsr9)EYk$d0GNu^)fBwZ_0E!}ZR@)FCX3~44AXFr`P zk&2S)vA|(JyS?r{EdfNIS|9Pf4wtQ!yp9+gDi7rKJ9U^O#U9WWKnm#LbC%85a=LQn z+9&Bf7v5a&jV+4bd#*w0=#^yIskeO||6Z7t-ws^(1o_KWTb%t=-^MiYhNP2TQ~&`5BU zT-SQEryegbxV=P56BAW~JDeA8KL?-*W`pf)sN%92${8vsNmYk>()Ov;k z41Y6#tTx5C9&xpE?Q(VAX}ELBGuCq==|GaTdY*|@+elm4$+8z^k{%pEpJh8QTL;eu z;f|dUWDffHsxBzz<>OZ^L7pkY#&Xt27hFZ#JL_=4WV;!r?NQr@w?`jdN@-5_$T0pG zqFJoEH7iP|_L2WH#URB{|Ie=sci(YrR=pizyj5S`bf!kBMX2awgY~nB(x?3)-*Zz0 zKWzPy-cf6LI8!Y%aD0mc_rlqQRqJ9OhM)5gja>-1#wxNnh68@tmC zniRZr>*Rv31(SmvEc`8PzI&PW8pm{OFxk}8o_g6zZRum>_>~U1G0ri!o+-7kGn$6o z9igjE<;*ZoXW-?=EZSSHb1HEjWbHD^%9&f&%#@5}98<}Qwy4J!xfQ8hV7M@F!M${` zPYt_c>BdxlWB2>KHT$FXOU~)c?H}Ly)iibae9U>nE5jwLNNaZAs!1tI$yg`Mo~v?K zUDLtM>|p=%4@C)YbeEM~Jm}m!+#?6fvIi^_8`%gU-6<rbmKxD1~z z-<^6&!={#C&6LQNE21ZOZy=1CH%hW2nKC8qX5Tfn(@tQ=F&^CJDQ${QQE{91F^6Nc$2U*c5}85uv63CY1r-DGvU7BfpUi~pSZm9 za(K9TNY=v5F89j7L;NiI8<+g%QwRMA<5Hc!F17G)*mgO8d#)xWXWp|WxZl*XXRgMs zPTVnY{;p2`WM0j}R*o+m6bS}gh-!su$q^-7Bc+-8l{oX|dS7EB_AKA?d|%7U3g9?# z>PlU|jKSC+z+hY+VK5YEav8#4tOPKa5p4|SWDo|!ViuBp4ZJH1!%Z2<^O$Ai-`9E3 zzR>cwnUsnp2D6F}{jqc#2CvKoBYpmif^GX4)!DY07)n^i?lnJnGGc3aD{IGzl4e6W zme4g{*|GNzUo_h-$7_6U$dEWajIVk6&4WSgsEaM0^*6`e z{Dk7%+}t_%$$&=9y6S4RQ}^+v1@nvd-ItB(tE=NGtE+i~zLAa&I3k^3f|zx}0t1zO z0s~1Yk>S!O_YMc#zkfdke$=>3{1kkly@b_4Y|rp5>9syp3=9m@$Pd?vOYNL7;{5!k zl_L3p@ndU{u3S{6Q|vfq++QqAmpbP(HxkFb2i)#qWo6C0rLFy1 z_WIY-@^o)2rWxraGwn**cG^~(XAC9lN-ibINwB>SA7gMP**HgQ3mlY=jgN2UcP7hF zsXfbiiCKfM(#k_mT3zZ{n#&;O^iZ$#kn3!|PuVs~6x^!bKHh`1J+Rm$hfRQ2sw14b zo^UwbNGe#eFY8ZDuQ%hC3u_KG%|9f4(qWzbHNfHH|+pB|T4=6K=4t6&-zU6kFJ*T0p?d#{{ zJUhCI8M%xuu8N)iWH1Re4kIG#Ay3*G-lW%unWJ@WYzUF2vzCkprGcNT@9^ zl48hYo)1h*A9{V|>JG`h^j?T|rbs$Yq=w7ek(XagWqr3y%BRF0E#Q{I*WnNGn@*Dc zk`I3u`FFuYUAWhwl&N~fS*_N%*EdBUh7|FA;(NJ2RWCq0pvFg~Xp+#$GJS7p(S^I< zJClS2r(w08Fr|iwKT>LN|8j6ajzS|_akpf+^v`lqB8fERO z{EgKs*WBM&IlK)oQ+LdRujZI@cVUq8_G7rpMDL}c?V`VO%W(@OvpmMBQ^!t~j?L(bpG5wcoO|5F%V&-$>HT*Sv()5xoJHvaNUyF}hu3hVk zy=6UQ343||galcg=d;aNXE`sU<%5~}i5FbNU5rLO8WQQ`)tAoqGa@l)#oI$q>?(_P zauai5()*keu{)}(nEpyhueWtNd`?%4ZMe`_Ijz&*+E46XMnh8HlV(j0uFs&|E-$wZ zUQi)lbY)n25m8=_!#OW&-($j7g(f&2xM}NV*}Tj0X_KW~JC^&cxz+I#$gme4>|sD` znXFAvlKQm(e^b4;l5yUXKuUEQ>rDQ4l ze_V-;ZAQ?^!Pc(o4~Uu$ub#} z4ES_Qg(Yi`Dt9X7>DQE>f3Rmfr@DNh9$7NH@6TIxJvxo>g4|n32%w# zQ{-2*q`23GQ6R*Ki&oP{&0-XnUWZQDhZ`P64xb*7`~R`g{`lcbLInE>xl8%WI1r`n zvM9{zd9^@b^DX;R)4A4T zozaO~39;s*mOLLi6s}*7%-C1&y^uCzJT9ny>?1Ox&_e?1wGm9W2c8{s>{qk8gPRsL zxl!;;d8fWOKDNL2%lx*@nQKOXx}3tu=~KYrsck;x7EjqXHM`txzSes zi2Pl4aNB&Pm_fB}hgF7yuQkC=(WWOy`j(B@@kEDIXIKWMEMue$$0v{_U}G?#Fnhx$ zNt0rqaPrkFHXayvy=9G9Wn5*!ZadA6@b36cqsqe-4Q@gbxmL~i6L%etST%@yr4%#v z=UKJehI%@cTm@A<9i*$zmU}}(ZY03;9BlH*5z8lK^S1mS*U2_3$Qd9P%ITo`+n#a~ z(V)1|3077_kI^h-m*3*jIvQ8W>+l{?0AVn$h9@~|@L{`8kz!sK23RL&E-eMY?j6c3 ze`X=AlEot-umc}%-n@(U+-vGOuiN9VPBpb!6AUkFRG7@;8^ac=UNo1>Z11TO(20FJ zaOCT=jnTQ@3G)oDf5S^fy98nXWsXhQI#du5Ruuwje0+Vvx_YDg4yd62AvVP_) z!^)Gc#JI-pong7&SzHMk@<&G&%Ius)Ee@+?TtT!9By}Svm|x?uNu0gct>ra*llMqS zFiH?RzlLB=I zEojFe20!=s6mx9+{yNwYv#(VKe_8T=Lk5;9P2AyV7B6UWc#G%g!^sXqqXK8@JZMDm1<} zLE2GfCn;|^GK5G9RHvOfmSqp}eFLV!ij)_eHjq5tKK;eIt8X!i%dnWD7*p;AlhTeV z%W4=VnB-&3kwXUEp933)T`o!DYT7oZ_f|+~@Lg7hWdA?pCr|Q>(9e zoj4-=(?R>9eORnIz-p?ocSzjK<&MZ3Y*;7axjQwiSdqs@W(|D-O{SzU8Pa|&yFha@ zQ=DxaWkDA)!f-?ckJr2uN&2Gbo7~MZ!A0-XX2Y1!VU@Aw_7lwQOS6L>^&nEY98y}W zp3V!Wxh8w&ZC_Y-40?dh`!D3Vnmc`A(MHzHS)-GmuE@*-<{0dLo!i7=Z<97y${nz3 z>C1iG@Qac{0lbL#-C_|!w}X^P39>JPtcg_;Ys9DKFPgwES54ST8O8BjAI%9~CuiQX zG^?LjdKoLvGH}q&*4o0ZnF{`LpiznjR{Yt?VkA6IVY^JmTz|j zx?2<|U-3Srn->uyy6+2e!+m|WZ&2_?WzpUiF$(5k;Vlk)@`NUM4bchQa)!biQ0WWx zx{125CEZl^+?CkqK@`$D?bv&Tu3(rNy<|0HqdaD=-a1T|D#l_9x+Ph*U5_KrA5Xk` zE8%8W9wTz5ABkduDtmeSbKZRts8i!GR@l44KnfMnDoz%233aVZ`8QRnsGh^4-}ly4 zQ#Ff4u(+u^9>`n=#+ z4VxRp+Z;c^`!C~BPH2W`O8X;7p)@Y=R(u9|H7e=#mY9mi@0^l$C0>UO*l2TX<;er? zU7#!vbwZx-7(Im$5_$c(yGW94)Wz*NCrre*cy7f*D`yQ%-njI!MaFT~GIHy2a3b)yc{6zyv9X3wnAqPVJ!j=fuf~IJ>(NLT2${Wjb z83hUaj>9tTDYs2*9_+utrkcT zs9v;NgZSO7uJ>O|7-*(-zH@NuetuL&TGgV%kRkAK8{*>A9p%)!UO{Uzb4A7P z!g*vH1TL`a$G@3VPy9U~)#p$*v-6(uG7i>7`dj~OX}5*KLGeucY; z!94VdPGe%4YQ8q(tW}8|u;*60j;?drQ#B$6reCnN1Gg#^NBpUGDPIz3s!$+4(Dt&w zob6wDhqjk$W36e#ndBfuP?Vz3l`|o$9FigN<^B2Nbx8N8gVummLzL0mio~O!BJ*QQ zOYBraxpnjSJ7eIq{m^qu`_34(m`&a@%?Zu$I;wV_GdqRF{F2%Kc6D1(+>PP+%fZg2~qJ04^UH}S_nx(dbv48Oc z|Lc4+?jH}+)?A^a9`rHHt!(8Mgq55=(wE!!a)-)TtA~iGty;xcLngORIosi${032i+`tJQ0Y-u) z)!SKSN9sOzXCJcL$b|3q;jURJB5T`=G#x1yFvU%)_;*b`|0woR16=L60&D<9-Pu_d z{%i!n0UDDfTlNGkjAK71La^6jC5%(F3rIRxKh2E12Z5N72|;~)4My;qN~k1VP0$0H z8B&ys-O=_UVhcq5@e}zr@~Z2#Y>Jf;@2z?Mdn)3^K;Yn+-4;;(aTHvpm;!7R5(}<* z+IIZ8?d22R-ATJoX8yAeKd&v9>b*5G&(QbDr6w`U)Ujw{`v)g~n69n_maT%9z_XJ$wX?N{ zTe>~jGQ788!Wbl(dbo}^EapY&-OFln=lfT5IPhGMT+4)bLQEkCaUJH|6`7vonYB$I zk*{S+>IYF5fTMfcRl2kCC@XoNW3lC~mCF#^=Dv`by zaJ=b@1 zx6gRH_T+4vC)BhhYaM1Kj+XWXIz^drg#Sn!pX>c_*^fv=;lEn*i&KAqX#?mcvr-wd3W&#Qf)KJ&EF*7%L{HzP7`B>%-)o z@0k%S@pZ!y-dDUQ_lhTFe!>q9XdRNy77&;tJ(4L9uHji&UR^pMTK)EN@I4()Igb2E zZ^O|7yT+uAERmxV3p-N3Tm7wHiNKG4rG^|JhilpDXGCt2@48)S;(TzJX@4mD)X}*7 zk2g&7$x)ilR4Zb^@++mpgEEeFR%&lFogF9p>-iUN$lN_8KGK#`X1#|v^|{=@C|p)~ zBoUi$(wxWcAgg*q{dIxUtao`(tD&ZyuE_u3HC^1UFFH@i`ARmhz1Fm519h&AjU>37 z5xq!hOc_;3>n!E7(>NwWn`=ps7xm8ly3l&v0F_ClHeBel`s#7Tugcx&T*Rp*9l@V90eQOIwGK0= z8p2diZnt-bGTj3XO3-lvOXX)keum&0jSh@?`1wC!Ma(w~kD20N6uN}#M_cGId1=i! z0);hxqgHbp0hs73+4%$uU?QeaeejR-hdlmyU+p>mK_S(@WQGaCa<1kO=ST;7-SpNQ z!q)=Ge&h-5Fmm@Jadj$JO(36jEMR&Nq_tM_-`z<}*5K~GT)GG##4(A2kFjTcu1kKC zOj-7y^S}os!(q0%;;n{W<`79BSA(|Dqbx4OGUkwSPhRvRMP9K=`id*)b;$M|g=2Nm za|CQID{e7@DbmX;7IG99F@_ zMSkLq;P&=g25YIBqdzNO1^~(DYX0^#fCRG#Hufmg4lbnYCCgJAg623x-Xb^TTNWN<@%Mmb4tQB2iK|W?$n) z!Xb{z|2N);^bAprgh=a4bLCw=_ z(LMX>bi49a?gMZWARNt~?Q3a&0bBz7 zZFG-~`#zhwsXsR&=H(y0jwXi!7wgT_&?jxE3A&?Qf<*ohiBzmbFP&e26V!9ro})Ms zzyN?6y1KLe=|7&o5TYz5zmM*zq8|1yZro2E91T7vySULyot>mLWdS&1?h(ff4Znn65Hk5 z_Vx6$|8P0Mvw(M8&DrUQn@$~^KnSa!X`J{GI&2>F0Ik^g!#g#bMsWm?C1%i^jw5`2 zrxrtZM1ywX+qy{UzXA!!qxwjM1jxYcJ4&=oL$)0B1ENv?WVn$L@zR=jtXm+J1AaO^ zG(z8A2?y=y#t&(PdA{q&h2N}%+@;wLh@cREQ@|I+r~r2OB_;%&0GAk-nQsHGG}9-yEgObJsYxGil-cX_5*hD4xDuXq7IJq6n^nw~&a$W0 z?9~_bH4xQWo*?WXCUx%!J;f~NKBJr~H_Eiev#Ox1Y9(nI)Yo(P<*m6_l0MV*q1CBj z&+2veZZ(_MQ#Z}GE1I-So@dLON_rWdQ)R#AVeo59{7i|F;3)=%624S7mAXk4r-_ZK znM7$T@oDi;Zw<;B!^PT+dt(L0*DB?)jbg!`NPZEy^*`|$c;q%k|Fb;Iji0r=c+{#* zGeg1V$nw5%D%qQ@Y+GK-2`%NrnXiNIE+Og^} zTT{(^t|Lnfn4n_NjO~ibx8fxqyz$Up2fBR(PMP)<8UhHB+Rt+fo;rdB(6K-C5Iu&Q zhlc&pVGdAOjGzkY-uys1&H<`OJGquWLuM4}2lYTg{hxs{dbr?P0XqPxoj;Cii``1} ze2kM8>uUq^3=z%A3 z3KIdY9>2m<=ov2|UoQ~;6`n%vAFpPf?7+SfyI>kLuyT^&kg|=w&~`(jdsMI=-e+Gz zZ5JKmq1mWEU=O-%{!un8gLEefcA`U@A=s;(gX2{?G=MxuhX$S}WC|cyEiWBOqfOPW zRPY4YUsvSL`NJ5Y*#V@3D8V{j>lQRxLykNm@t>DTh{K9)+hs;KHgpsUeGbqiq!q=W z(1t%TDEd?UsuNW=OlQ`*{TSdsvnIc66-ZsKyFX+_Xre3k2pBsW#6f#@<#Oz%Xh;0p zAOgyZv}5$kR}iU{TLh83fLen^AQ*iB+Q(2B4Ou&)m4HSg=~Yh{4ShZ9sU>n>Ab6W5 zhu&!#dP6@@KfyKRZ2*^kT^QQ5k+v{2O7;bR2ob=~VxqsgBC$0ce?;^c0g_N1D$d6MT1gWFAf{*z;dwhE_VqI{Lj-Tb<@?qL z;3$jyz;Wmp)#)I)I(p*Cb72YrZyHXhDiwDDGP{vw+~JDO&U|>9f_le*A5UvztiIX29lAg14@zhNxy|W9?d*H-Mm^B1Xkp%DNh0oRkWcv_wX-$pI$%_gDK>z>0wXKD0v%=~8a30w>__AZ zge0)mI=EFy(PL)n;r-m{#AD3cJa)|>c#GlWHoJ(3uCEYcxxDYvPN2CbSXD(EQRKU24tL3c#~@Fc3+V0$61@{_2@I*5Ed4 zAYZQe*z0`TCwv#>UN6V!m!u;J%S#Uzo6F0c(hDcEB#Q~}Bqr<)mfXB$ zo|W=bWqYMW-Yk+grexQfh2y55QX;tbOXKT9<@uIg^!9fzP*uj4p49N>c8s@O^5Ij~ zd`V%AaQ_`t3EH0;awrjjRT4*a9p{U%`X%I-^|)(oNfe%QeO3r)5%liRe*sl&sl+xS zb{X%4(#~87Hoa#DP5Tf6*`g^PDoPA*)6fWVY zYMMv1E~B7?GUvG3I$pC~3=EORxk5)N;o<@^`l30Gb6di>Zk3Qt!)8Z|)(&ScD>H{h|*vwl> zxugZ0hDbeM>YnX6XASD>f+?jvz|~L$%K*9MvUo>JYQvXD>;-zsP0c`W3>v1zV)W3Y zQvlM$g*7j*Et>$SQ3Oo239Og*e9#-^Qyi2`f1AS%5ZwDT5=5Q84^15<-0$wevUU|G zErlf@oh2aAnd06Z;N$|3VSA^zi4##TnE%9vILv+Wrwx?m&IAE&sVC$N#-f~-QbKp9MEtpQ+9|f>L zp|RIY%1OC0Y7L|(?r5~Vhg4x9IPgJd5`(0BYT9EzHPw+gZgUXL7qkrPZN1o#bFB7D zc98885-`p9{mShkdWG^cN-O~clJcP<4O!IqZ_k%? zY6w#`AltyPYqTaa9Gte-Wa~J%6pUyv}`g%7z*2wn3e-qSKn zbjP5xm!}r-y_)k#NLGdJ8=zTyogXX*@K^r_Px5zP<%cIp^HP5DOpp%1ll(QMi3;S8 zlqS;lUs9T=FZS`blqTv;{pn7kjsLUP^(zU7CVkZN|LIPmi~q%)M42>*oLnm5JOa>* zdP?9e351$!CdVB>+4HiBc9g|u3Ft-QGF zAi6Oq`3}i2uZXn4LV3l%r!~F*qDTwY;bywETOdh!?@cW+%FnwkA0_4h;N zP|)b7UyMRT2;A^w6b1T_Wczx{ix@ynw&2lF5fmIW4wa7pXdJKuCi0#%Uqqh+#G$j~ z5Aq2z(tjhLAbRj8`2_9tKgcKOlYi!Hk;dQ2Cy3#{VZ(oJHLCd-{gf4mxmw( zA)XB_8l??O2amdGi6}&YFsXldHHa=FxYIA0O0`mxoF#k7e6J^yuP1`tx z-G+pn;KPHJ2CQsl30ZD`1lOsWPP2aRCL_EVf=d1h5np&JF?8zIe3pE4-7s+<;2$v+ zpS~S9i7#QMbmb@?bK}c2+vBaGzXjohSe)E;d&-Mrq=Hh=$Q|8hXudRJ6V~bzif#69 zyYX^=65?9Cj6i{e6W&|{)VWYC2}<|Ej8lnJ8G%abX+wzb9z+Vo-rvt}sH11ex(q*& zLy%wqjBQ(}sFS*eH%<}kZc8Qrly(N2+eLOWq0H7D(&<#sPM$<+ zS;-U%?nX9Q|K0E8Bg8`dPCi2P=s(Cuw8wsM#eQaS>013e`3O1k7+Hf~@J~eP{(}=k z8|hC@4DI+qgohFma%&L|o(tRlaNupaWbFWc8>zXWIpkwlZUOnT8Tjw5IXTWJ=BL{E zV{T`;uUd_wFD@}a%0IZpGx{d(c1V42cv2JLLJLsN+VPt9cPE$KmHC?xvax9~1Lm}l zIS=BdG=VJ9pl>;^pF{91?mZZZG{OoLe}%fA&@>CxIHTa0dO%CySF?t ztOG6~Nr>@~r4}lLc$WE#kQ*YyoyTIr4*{W_ll~jMd&kEcvb-h2W@oTZ!>>s;u>*G$ z74L!ml6tM+Y4!pm@v4=l&l4Y?@)xnPuUaT^o^gP`upOjXvLLrq?tVcFe%jpNEk^r%FVpmM(r&+MCb=9+s%w)S82yx;F@s4jbj7i&jNB-x#*pG zbi}p|jRZ@G599ggW$)jO(=1+@i5{d24=*G+iw@(REhxGO-o5fx{97PQw%8pYc3Mse zLJEdw%Yd~Z%~EJL!gX%g!t126{7q0-LdaB;qU8^O2B2c-nxm?Ed0Ix4gB6tUM{{4R zO;=FDkGC1%5@~l+3J<#L)sXIQ0aiC~#C^DScM^@H8>g+bq#V?mVIz=2>MTc}J=tiG^ZZLnzbIJF@Z49cW?ixt9m3U5KyE7j))AatF5grPU^ zUsVAT~v4+FjTuBDB`eAouC zYeh-<8IQaXJDBVItKr*s*hxuSY0AZ&+mI&t^fs=D<+efJJ0SZB2H3g>=GB8OIHS-s^+&u8lDBV?%VW>T&Ww! zzL!OST7Z(VgyY$Xjy08}i>K7zb+U***{->Gruu2#6aCro3La~=0-O9=VcRZ1itm3Oh@;z+^i&&bZ%qr2h4QP|{r-e9O+=$+@g6!QI0BWW@B)8rInJDX=ljv}TEtVe=d4i z4Fxa&m;gO2=?(<(NXZm+T{}XNfM6Q9YzqJk+*Wr$IDB0Wnh1+*gy1dh6ToME2MS{l z+69yaSY$I{w5_FFzbSZ?U3{=daV(Zi{KL{1J3_VZei$|y5z2Irs^D|)yJwu=kezZH z6o-u5ZjS)-cbVdL&!YZ2lL*weMW-2jB<;K-q*a?6Ul#$#TS~z1BV=#+)TTBGQZCIv zpszPBJ}BvRQ32|`PFl=n0=%jMls}{l3!(i1IvWjO#PD{!a5~g@=uRO2&0grf3{#Gw zvS1rd_np|pr72$stR& z&-r5EU+Pa1NNKQH_Sx8H}N^liMJqiIx5Hz~6J$8eMv%O{6wJ87*5E`Hl ziktWQ)_Y|tPXl(G)QA+9eQbjQ3wzPD%hZNXNo`hd%Vz;@NR<&fYjTW*Yn0vM-IU?R z**(!`f!YbztNc=aN*>51xK{bN)(B}w*wN3Rudm$OS7G$81|bz1G$P@cn`DV_d;#J5 zaZ|^-$)mu>LCRc^x-KYJ*&z`HH5`DXTTP9=%z?d{dnx=Upm${IYj=|e!CZ)*+t2R?#JDb-RKErjC0euaWU>5Squ(-1CK#5HMAZ5rgZ|3H zecb`H4hEb6!2M03;PLvZT!85&kM{rF1C=`K_T}r%ZYih)%iF&w_Gr{H#Pm2Y^xtXbaM1FQGk2;DcT7M^Rk=xB-&*eD& zYD00o_dPDhwD;^zkDwwKXsrBgJrd~=gZi5LA=DfG>SK!h#N~ns<3Qp=Pr5nwS@`0fGS-jR1(o$5aG8U$N? zZym}T)y)g1A|+YNVvEKuH1P-M&DReuyxd76h*%251Au}32A7}QkqTeZzerUjq+2Ac zI?{}*S`CLn>i)r`_TUy;9`@Q8_es|Hz7T$?rYI^-Vol`cHx~NpZo3KluPHcnLE0t% zU`@n2kQ>8mV64)1#HjWJFv#fC3?$QQsj!q&u-yFCH##Mw*VK986tz-ZOHM6ka7VBh zyZ81krz9uwq8=HqJo91h(Sq^hp|;TQ9mG-GJxXK!w(j{B6%EJQXvhxeIo_i@J6dp{ zOte~XCg@f`2sXSNr3Q?t9Jotn@_7KM6iS=3Xf&>4zIeQ;hZ1!YyO_*1JUQ1kaAV4U z3rB~|w{)3n1%14fuB@AeU!42vO8Y2*SVt-R&4*j;nwH7ov+JCbogNgtz3Dg-wbN=s zK~@+?jkAbXzdN7qS~s0ASRA-?&NP*W!0UX{xV68-@_hrTX3j}hX6beRB7XVcXi~(> zNrO&G;l-}WU_pBVe%WseK75XV5Mg! z76=7Nzyf8_U83|j5Y~t<_beNJFPL?tG-daDow8l)HOiD*wn#0WXLM81-!sb3gTLm30iT^GVEprlLQJl~5L}X2Z&qKTCQ~C^S0h5U(M;m~3>mp4#Vqr)A1n zA|8mH_DP1pmzG9y*t~}+z23}N3k9-J+>*xYYsx)l>AeLVSj`*nZJsR7kZ;JYA)LQQ z3}lfTa>%B5Ycy&(Dr;b8jlNOYneFmlp4_TE#D}F2)keCee4A&GF0*^FOUbNe z1eqmABvB$1BH{R$A?TJ84mY)E;KGSBOh1L({OBge2B1WxMHXB@W@&` z5h2X%+n5?6Z0gc>D?tCyQ4L&vVH@h+dM`B6K8TnT! zgaeC4l%fdUSu+1AaNmZgj?0mc;{q$eshiqZ>JxdtXYo3OeE6H@;5aKzkTi3n7s|JG z3lcRKl)W$Ch5$#{$HsX9B4s>5!>^_7bIqb)Sd!XuISIh}+RQ6w@XOI>fCIv2S1p3d zZ=ln9OnKwI2^^4|9%N(>jw;~7rY*@A+LRREPgNT1mp)~k|9t!4IbeluZEqXmA$;vz zhY`r(Q_qt%$#((Y23$mL+|&CVkX179FO z19V2JF>`_-{tAAjxSRSE;-|)Sy&3DArOajvq@vdMXTD26K3|5%6ApQ{<^3%ITcQk+}!gG+7 z>}a|6+|tc72f4UNN;8&;##T;;Zxg)NN?q=AsC?p_hop6ay7u(XW-|ZndhaG-51>KP zPbO{G;KU&MYs28+Zu;8Ik5sQiXa)%mA^cr`p)e69aH&d= zGC}~|2fs6qtomkgbiihm0HymWr`Q{j+&2_-cv8MJcuC34-nN=<)u*S3z7l_QEC)$iXbQe?jHX(Oy;g)9C(i4 zw;{Y_#Ie1IgQA+TzJ)5QsVS0ySQKQ)kl&7;uYZ3nPZz)Op||tw{pJR9LDb2B;vd); zL+qWfMT)Ou(kTW5*1uiT-e*7C7^>m%OP=1P`e@BYdgV*9rPj!^(ThkO5BQSc&LdSP zu6sfjLuicbn4#Y(*&=-;KTpKt&;o1%$fp1xL=A{%LJ1k@I!{a$ASb9P$dKo%|=x<#Wy~Wg^@C2uU($E!83>CrzDcg z1J~8d{!Ka*(tsp{5PCSm^;S$JJ3+Eh0qSv(`h4&?W9?>FwHW6+HxmvaUY&jcfJ5NA z$L2}gw%zdIB_L1$Pq;?p%7^H{rW$!KJkB@l6PI zqkAUSYnM0@@hIN!ih}B9)ttdPb1gCsYMS%l%LO11`L?GvEty17<}W5ZhYmG%yug<` zw%b9S5x9DQ|3lF>*b@KJyS2mD+SHdhwG*Ygr@cDvd&Wa-!{lWJ?8}!Qfn9q$zsvq> zCt)}EtUC9@64bnbV1MT#a@g`9?uXbesde2>=D)iOUzDi)JMiM043rd_m3=z+Yj*NE zc~v2!&Dwn!VBFIRbPN_(#T)CWPNImbB&lk+Gx;>u1IiZTK@| zxgyeEGM2QaKQfjyY5kG0q>cM)#&YVDLkKAs3cF}4LCS#9TE)fC0&+f-Rzo4?%1cj9 z(JQWL1V5TClp}~8kgOdMCMiu52GJS>xMk8INsfw=Q}_pk0U|lzSJPFK#+zmv{lV#{ zO*)~I(2{O*q@GSwr0-I=V@Wh*zJ63=(guT1M9Blm*r!g7!eqML~}GR3{6sGkVoomUbtZ!^=5S zj#kgq98;rHwNVoGi&7O$XWszvg*Cggmy`TO*nlmwthoDg3f_scI$a4((A9nT|4LM1 z1Re;Md&iBwAPfUh{bT;ORyD=xW^d#>UQ(FBc-&~q8p7c4)&oaZt!-0J7#(xg+dSAv z^fH>|=z1B>yLenM%&nW0)Qqck*^oGJ&&#-G`kC0UVB5@&nle7}Obt@u_DDK)arUeR zx#A?1P(beR&?ZpNf$M!)1?l9e z&xu(Fw6rvD&WdWLZDpF-m_2-PFu-4+;HHa;{uFiy%K8xkn5Dj5lUDcPg>*c8+X<>4 zsWSybp&Zm315W(N7^@`a1+-Y|AmPa;;yAb=^m0vqpDD5>qQ8^EU9uh^6xc$K>;HoW zKsFks;i9Vdd?P~QRDTCFQA>du7n)7}S?~Zk^J~B~cKr)IIS5sN{y_*D$rx1@m*HQ^ z*JxToBZdD+!=uWK)NmlAaNiZnt9|S@KKPFje&Pi5CZvD?CF#O>1x^HQ2so_3J6|c~ zrzt)9C5+#=F7)h>);sCJ_SKrxw}XPGs_Bl0`z#LmHyd_eLuh3fgTi;$Aeby z00UVOt#P3rHQ**YnXVLeFwjL+qm*m^vP#koioAdpLOWFm>i`mfNP!BVAoSJ!!FNYz z@h9IMT{3Vt=zMpm6sHgU4gDkuNC7#cE25zpUrsf1op$?`TErhyeP(b6($kyz#ZaO< zZ8RFuouO|RO}3Dc<|jG`BE|0|q9_9ez8eCUyn@8p5J>C7&5dp-og@MuBl)?+Uzsm7 z<@nsuAJaJ#!rwIrKGg{i%ZU|9 z%0g)*&}4usPwNrX@|PMZnh4MS&H$$^|2GCWT~L28z-bylE2;Rw07u{codHg3`hx+E zPUojdr7^%)1VL^z(S(miAVow!KOpnt6#el58M?*(^#d|1szLjJ49aW&zpQHIo}hg@ zi8Di|8|vHx|}`-ufeRXdI0ZQQvq_On|5M7kVQ&|mHVqBP`j|GYk3$U08nZ=cT) zDTtva?P`G9$l+qrlC-I0SQ`M8@Hv1eHnC)eOhXu^Y<@U0?1bh;uT;n40pwH1K) zCcxuc#Ow>K$8cv5mN}H2ivl$olF;72f5gTTB_AU%zmg=hAx=6PxPFNFKaG6}IF;Mm zzSHDX8k{2-a7Bcg)7vKaE)?4Cy_@dFOy^9qe0-~(G3Y(Kf8ummXokS;&gW#~SwlLwsp z_)^wz*>(s^&{zDiDijl3lP>>8hRVWtc|Q|-0LsCFbR<1xUV_u%o=^qh3Z#B_!sh5} z$ZE^E>VT*epdO>8Oz_M(-|yqHqKQOh|8&Uh7U8&ZFz@Dc#v>+Q0cNUs3k52CATKz; z+VPfkN~-k?rxWe4oUn8x(}KVMYZYSO*ScEA5!XTl;sjLN3R-HJkmfORf9|GSw3oJ1 zlfTMp!hQ{TW!b4Q_BA&!5mEiOxGT2I-U024_uKlAv>xCuLC!z{;F>QXgyxiX7*Ls5 zmBY{Z=0Di~0N_dz6TlE}khEUC>gA%jxp3{;M$h4)A(r1JKgJS%(n01O7g3zV!0?{S zi2#<8!vYV0Jq$8eXS(V(VwkJ?skF*lQ@_@pVj_!nV z3E%OsxjO}u0+X{_B9z@0o3z;)n{AuoXQ&q=X&qL&Cu*sk<*rltgx)s`T)#QR`*r-3 zjtZ4_G`ceLJn`&)D^kkCaq)%n-$}w3kLYZr1Bx(adt$Y8u5L|x-TWwl#?E5x-pl`S z$d)!JBYaUeIIW8uL=twXvlgQG*HlcqzY9`=uQNFjBsg1``jV zeL!xU;sQgZlhpzg4;B$5zIo?>4>94BeHV z0IObAxBUlseXMjvq09)~vq8}IN7d9hLSt#E->_ZJ9cUj&2K`ap&ZFIY9drD2*H!72 z)fIRI>P;}Rvp-2M_7SK>)ijz@fKvF07@(@>&*o2vc>3}4XC0cdF0VAPUkI-9$^KaB z46*^g1#?XS-q>(tXdkX)l z!{!ST9AD=&efIqV8jFa|abR#%PsZxbc4w#{wYiaF(=#5jB`LMHn9ow)BSHs|wXkR% ziEUu4VyF>iBApdTV4#*U`<*5edLI~E3g+A6YUMD&GkxCE-5nDaMK-p~2zSh?_sAqW zF7+)i{mMb&-q8!vfhtityCtvd3XX99;h+BD?f$8`nh`>bRJSWc23OA%Saaw7nTqx@7)0#ot^Fn!AK=qocMTZ$u&WG% z@k26sZ=sr7E;f-5CKFYr!7zDRM^<8%PX}0$%hLgP{ZL6?1WM-%{TpEC-2Ltq1^NVIZc}1`MfXF)G|W0q-(+_xDnuJ>!~Nn5zNs zUSMU}AI|8sh76S;2mP^#$cYFDxCRjB44yo~2dFa)HH-oM>2Jc-+A!P|z@9P31i6S1 zH6zyb8wl3d!H}R2A@yOdEPNy8%HoX*`QKbwyk!s@Pb(4FacSK__}g zbS=Z}A1uFkBaQ)1*@BqX9=P|6#aISmxNOr(dzOU&Alm>4b9yI`FrwF8F-H8$>sVH_ zsXmM2?rPbeP#?h5r4XE*-Z)HW5f~Df${zi?b72yQ&edPGeB)g#!2r+|NapX_x^mS{ zen+});PMM_Oy4pu9`BNW@#2@i-*r$lg>Ldd2;~Zl0$O+2AmfP+V8<^*w^t^r#>m@p zO(4VII&j!bVMs$@nZHPl^eOpwQX_q2LlFrT5zlbH_8*f^PBm>mQhQsOHW0}OBQ3iw zdjIx(`iJ9(f3V;RLH-O7~Qet>xMWA$pD?~z`7OFffzb!urqe%RX`M-fF! z2h5q<&Htz1;{6I;clV`}i@S%NAhQ$sp49Y=vSYVbL1EIg^ z0mew9i(9N)gUv~>>64jIbcMYJ*-?Bs{DWn23IX7Oc(Kf~z!YTY@1d*Z(G_G?+PJJB zgGn_mUy7%Wm!%r?kZ-@y-#{hI%2U z)`$cR{`SpR`pa=8Q~bKJKbNZn?TAl~f6Kr4ua_u@~|Uj?oz0;qkSld)PN9=pBtN2e{6ni~Mq zqepgDxfq%s1^dIgho;5xBj_E*{J38VwI9=A&U15@#yb{u1>Ew2W@7GPgX|Fxl8z$B zvTGe0q9ZzBe}g*py7l-69;^T;x_TsU$RtHuN(KqrKA|^57|J^K6Qdo9^K;Nz@tD1M z-zi+FPAHAdEhS&|6F!fdWw|rXhuzgg}JjLf}~peo*Wv)1*rs+wdWf zJ_Ph;(Cu%5<`pjstMWwlf`)&zOzwg)!Xsb*c)`DFI{fm1T^C#|5A5;>_fIR_>(ufK zVDM{m*7o{GceN_Txg~!mD;CdQ0#Et{c$yd z6#>n~K71wl9O&3c;!igA>LJ(V=eUX-aR~TZ)(U%+{`OuN?%X>aw@5q6wVy#nJ}jkspr^!8kl2Wsf!enki!5>Meg^^Q4ux{|`yEP2fFsW7v`CAs z15C~66fFg~7$UFqB2_L$d~*?-ss@yA!p*FTvFj$PPbJ^#q$U{aq$Pr`@xspu1n%pu zayZ{!_V?PTN&UPboJIrq*Tt7a33PAdYwXHiJ z%w$w$0P<)4W84Sc8p_XT!p*N3lcmgiH zlbIsEcTT|E9*gxrvie`B?e)xcI^g~f-7^dyTJkDFNQPS7(W%1vEk``rtr3SGJaI_p zXEZU|Und(Bc}FNt4lDM3xDM#NHrj1XsHVXzR2*~>JOc%NDB~jqG`BG?iQ@_t!)t84 z9_7M^ga;8{hu{6eD+^2Mp3@GXM&>BV7ihHx4*rn-3_CHV*2ksVJ;_SL&Gl z-o7nH``TlN`)!?v_KxQj8)q#|Uv8sN)}bGS#ChwxbmKzYd4$b{-qfCYp|ZLIq{+LU zU))Iow5HMtp8yj9g)0-(pD>F=rrkR)g7An_--MSK2T2aemx3tZmtxKyft%&m`=h5# z|9g=O;ImL<>)5znk6RPj#eZXp`2^*by&@a{Qw7C9l^;5LHf4NE!Zr$CnRT|+#Z==y6~sk!2r9e;{s6cAZkZFZ&drD&d1;)J z^>r*O?Htf*AP5-E{wa-!GD7T;i>L0}4c?mSJ~mtA9sJ(r&xt>G7A<{lKO-`&+gYB* ztwS=?G&OIO9OzSC`uNxZHfB z^!Y32>+-U2MNd!r>cX)QIjUE+^#C-hwr^wEU>5gbx%mK-%?DmvhiA?f1wRM>;O3oI zxo3|`cP7-X*4jid|CSf0?hmJQ93ATRj0l3v!p2xYVF0&oS9l6rK(uwo2|iXbHQa*C zUL||LPi|uSw0YYt#q7y5PkVsR0O}(s_)L{<@E*7;7mBsYu<2Bm|Lahj7!=r*)u6>b7`(SA1 ziXJty^aWp^D@naK-&HdPAYjF2yy3VFpNy<##ga{PGh7;Mtl_S(bW% zJOjr(L3)O_zJ&Gu8G#-L@zmdU`YzbfUXgi~^Q&Esg2CjSRw4&l4vIF;MEJBe7XTYQ zqKX|Uag&k-BOjTEdsYo1e?vrDn4^?6th_9ANTuzYM4qifEHaMjKLehMw=hpb;RB-n zAWO=NX(=v3h>hrcP2zJyFhH0{lC5=x6v=YVDqQp3v>KnlFe}l%OEuolg%FtnX}x34 z+i4Xprq%qry2sM1)D)ujFsUe3tPOZ7GjGW=qF7ZhTLZ zT5MHyQDqm%oVwEMMyGYAv><&aS!8i&#h9Fo*7C=*-dfz zq<_AzH-_jdIH$EaUla>|An$22T|IHyb!>Q=6sJpr#VN=r1!n>k-Z0 zytIlu6OlJ5@M*Yua9SXoWorDRz#%vROnerldrMt&u~aWoh-JTbUt=!VTdpl(KBpGa z){mDaI=KuL`k7J+g!VXfA8JmPPN@zKGXi-Nqov%tLNq>o*R;UTsEmIjD5OMdTG#DV zs`V08xH5Po`*rpOu+S%3Daq2*LK(k1CtShlmLT5JfC}Z1RMn0g`>VRFXZ%`&w@0e0 z>>zr-V`Q&)eT2vYsAWtQ0_kCW3 zv|&8Y@eQ(^@ER3}%eG~o+gsZW1fl>?S_LQqFjs~mAoG8Kl!%=e1%Vcuk096&3GvT8|8ecg0dE5SFix z+#sV-rFK5zPa#PpuPDy?i^yRTm#KyNpY+qE{N^*MDd}-irHLpfEhzCxywy|p#t|0v zG%Y;;%CD1WPWshnjh=tB+F1Zd*iHAmuyzaY0XM&zb$8>3(2d-SZq>xb6DGQ8;HCks zTiCsTdA;X>hhp3$6JL;U0S1%X(21HT#0r75duKOPQ4Pk%p-cx7+xXV?ytwmF_+obyTkOtAM~w@$6TCO?143fOc^oQNf7XWd%L}O1%R_QZWhr1KQDG|JtO!e+LdA z^^g8P4*0@0uL@w4_#bp{P4BUI3%)N*fE)K~#!4N?Qb%|7cHgnG7fv_s;(zB){0X?` zy)(?s5%)Uf_qJ8Vrg1Y0JQ9tVDX12OXs2_FCqyMt`A3948jU`f1M!{&a6~uBI6`3> zPNm|VszxwUIC_Ic+Tp)X`7e}%0yMa>;_En_#@Kp*q4>RXx-hjW{jh#%b_eTtD$F(c zbDf&xhe~I=Tk>XK)wc&Yv1cHh={*_!*}9nS6LlO;u?2SQ@k&ciq6Vs3`+J5naonfz zqF`L*&abXH_EjCpTMl~8fvzyA^TuZ4>ZTeNJGj%k^cEUa^ zw-id$qMQgG^S(_-EY7mVYma9Z3#xiYq2g@8BvLj+@e$CaySGrfP6LKJ7WQ$@_JfHK zC4q0kwJE|J$q!1+G2TRPt()co;?t|_VvV%6m`1}u22=HGmyFkQ4}C#AD?X>_m-;!apNJUsSv=)12hkj#=AGahJBT@&rR;)15u&QI#Fc@(G90J? z=S3X1Ka_WLsctbq=1E9GTqxiV8(d5;zlTC7!8+1E_|03}-(?JbRmkY&@p5?yP<|S|=aW8CL^0of0O3F?GBHt+# z&qVTbCTc~mTO(H9LsUdzIUy~)GW5SUiW5d2r`J50V0T8BDVA(rqX;9I%}9Q+sF z{tCkMs9MFG$90|)pkYqdBx@uj4aFDHau;t$T2UvU_CD!yy4AAwYBeSp3XD z^;&S^#@uF+{v|ic-grEZ12h+Oj4wO+Idg~dfm#1BHYLNzL|^Y2zexF+yOEm#|HagC z3rt+Bor(;W20Gb@a_f_sCZAO(Ju~`^E=~Q}kdXqwFi0JA?J(G#do8lY%=FO5!qr&r zFXS~WG01oUYtmFBy($LwXOP;4Yv82oPRmtzm1fm!;e`kssJ5Ryuq=sw zAHpGdx1EU5ex`U1!e#HbO}C>RH&DFoXTYy75cISL6mov@=7m)(GwFDqRe=@02d3JI z)S0j8pXbiU?KLzn*KW5?`T10@6$yxHmjdwtt3=_wCyoQUBb`%(jpYL6{t$S4M{QRy zN=m-)Vb_7sBt!a}n1xj*4#&EIO9~gbeKvUrANATb421=08-Q&ANx$Fv@w|*J=uVur z3W95G=I7V^P5*j5Xe*!`D_#HEH5)+QcqZ@2VKg`&Y@lp{s<->G>>7iyHWDo0_5j@I zQ(OGPCmT82VpBlfTVflYmh;dl&Z;TijEX4unq=ttbpB(Q8NI@V=bQdj;i5kZt8j%{ z>&@N>gQH>aU+N2E%00^q97&=qmxJh^2YOd%#-Us?oI``nBnAfY9|-Xabj^4qeuC%q zAaw<57Mr@mIPrdA;9$fQKbJ+VSFZX?)Oz_>xTrMskDAlJ5!>?XPJ7B1tfuSk(?YK@D%mKXk2^-?j}-$&oc>SDgesO z9-xoWL111%KR~^cuC==SBmhZ3-3#)ZSn)1_lDg~6@}KA^ZKBgeDboYIe1L%TAXH2k zK1FR+bzevCAOMzPuumD+PWA?rBVX=}e@Tok_b1eqQ6!5Wo3q^dm1A>IwPX2pc(vnY zLfA*__y!Ej{E*(1z5N@BdI?UV+ABCo`lF0P_3Wacs{Rzpn6e0ufgOpm?eC?l|L;j* zXw})40KIXl(Nm^c6WDc0hn$AwSmYXngyFa@^yh=-g-IWW()?zb?ok2vF1`ANM=s>1 z4s+#^c-{*tRA?X=g|53bYf;E+2zF!4690~2dD%*|^TZ^ok$wl$Jq$#QsUR7F!X`RV zH@BV-J8tQ&Pbos4F61$pZ^uvAMF)mqEvO!}HbbIU^XChF74S~izkmb$=(^%~HHA*o z06Vd=1pcWjPQPrWYzhayVY=e@WQD5FKXk?MIR%6T_dBX5(NS5<*j%L-mzK|vTOL?? zaY?ue&(Jd{{G+b4@`k_EyOwR<-|I^Fm@ms>H!;GtjNb{ePFR_21(JX_jzNTrY;Cz6 zQ4AnL!k>r$&>y-!(ZQOdMmy<7M*r%6RFvqiL*jp!^}s>H)9)YBZJuC~b$&g$a=GM< z4+qAj!C+p`&yl9V*Vg{61NzHE=CsqL;7yoA@7nodD#4e)iX0ahBzK zOVAFstRcPpt$0oN#^ECh^z(wM;+M^=h$#IiA+R8zbWd+7T^w6{@5*CKmsPRFxtDhZ z(w)EX6;(re+S0^4H0wN2t_aj>Ozz*A*l|m?Q(70)hA;YGxbO&ROw^b10jKNK-N$$r z>bL{&g13>X3woDWUOWM&XLmb-m4d9JsiSQIFvvES%pArJ3W=1`h+~tLfwl%T|}Z8=?2ud z=~DnpLSmKxYemP6qQi+l5~!|+z7wqp*Mp7hS}_u>BADv!~S1iLj-hj}bhbh0cEgRa11B6u`4M@fFNLmlfPyGB@TDhRB%D z9G^XQPG0~JS9D=ri?+3wB5bC6I{ya9RHU*)K z%>*% zbZaHJ4@(Jh7NRpTqnc@7e#DPJMk8&S%?6*xx0}H)af4~{^^IH=jAC?hSwzB2@Nx@q zED5ewUDQ86lLUh6RY(n~L1$BFR^%nau6(kDFY(>pQhV`Fm5+z~ZwZ-jf36UgIC-?k zK~bpm`s?S(LU%+JCPbNUj_30H>cGc$uHL41TfOCnwY{6ueU0^ZDvWHkcPRcvzEQS5 zCrarxDevgIkDESQbi42M%|86FXsN()o$VgRJ&e*3TX_8`dF&V3-G^N|nr0~7E{1Eq zUzdL|y)a_Wf9S}UXI$bB^|3KNnc2B$D>psdZ++Ysv=aW5P4O8Z87fyej{A>f@QYZ^ z_6^x2jGpx~?_JAGC3E8I;(B{_QfT1(O<9gWf_>#x`PzYJ%)NwHYOOSL@2lFkZpbV7 zCsat8=4BlaXO&R9d^N7o@E9|}r{&O+Gn!Wn6@?!o7Uc5WmrUk=jOvw9YrJfR&rqx7 z*oR)#w5F%gH0E#UgkAM(coUtj)GTg<8hljxvZZKfy;I-$vvKbhgsx+a*W5ue4yei- z-|I$+ce|P(mZTo`rOEZOQ$o|gM3&=Z5r~>!z4hpO-lkg?@hmij>$!WTdWR(Q1YhY~ z|BR@I4DU#i(;37Xs$)u%nX+=C&8SpU#R}$gmuaec8DPO@-(5#eMv!VU=eeaavzqoUoKz6&WXRN^jL9m*qLD zyx(2ivQ7x;O6ZF1IvN(~G%~n$OaI6hybq?YaF-mWbXW|1TZ)F!TjT4e%ThCAW9~dK zOh}OUNgauCx@g zSq8H=U@kZ(dJ?(N)h87eMj~a7_E#IX=gU%8pE<;vwDIuElwY}@EQk+PJpY7G^k0Wd zqP?!P9h5g%JNbL(8XNid>e~+yG!Fj04n0GE-CD}|ooi{2kX!H+q zZa0jGu+M)bL00(sQ^4xwF3RNN7TT3X?xrF33w1mtlGJ`j!a@1@VPbaZbyA(kl$(Dv z_XSnnb$@j>iCFJ)G&VDYe9=)cpxw3ab1t#Jz{+c?H8oYf^pUrXQ3B7Ch3{|Nsl%PD zW+eV{d7Jv+;Xh_FA9=@GZq(E2eG=!_e*v2mTa&}qgL>1stn43B(nf|;ygBc^HVov< zToLKri+aODP%3@sgX!wI^g>CslQZP~DCfs;k}t>gXZQox>hQx?-i#>I#CoDiHAs_O z#~CL-OtUqch0bZCtONrGQ+q7g%hgRiQ?F;sz=ubxn$5It?2>+7dGL_W+NV#SzPdka z=I^d7)HmoW@$-qe(B0E#3;HTUQ)q*SibLY;hZ|Ma%=_m@!}6#N{`f}O?N+!no45BI zt(s)j+LLVR|8bC%OB{w}b#(HJIF}S0&^FHaG5Y468@(jXzPhz`rx9(Lf#V7+3AE6k zH8lxxU8*iFR+R~x&iFl$75&xY2P})>rvZ#pW##vTEi`5N$>XBoPsv>|FH@3UlTMmf zl$Cvb`^?baT~6rT4Z-lOlUyhDx*F|i)X=G#j7+1ZVEHXp>biROU>xqZ--*2zr6Jq5 zwThTBIpn75D}vjUlOb$(_AAf1ZF2tcFjssHMdZ|KsX}}XX=#S%-_)e@*mAD|6$#Et%jNxooRKGM6NcI(%P}S)6;nN6W=Ayw=HU%YW(8BSf^isqrp247fEV_(!V;Vhhn@U_3>J_IbuVna1kuM{nEp2OX|{ z(rjAgY7PtqUBXO}I@c^WK-wvYUzT)t*Nqg5poCTt-Vx2TTBEjz*sj+)Fqa; z!(vowH>=yT>-sgB+27w2C`7yI5v{hgK@oTgU(kWycaEwpB`a*W|7ONed!fR{KFfaL zg~U-Yu{SQFZnBMuc5p!;?3VitPAynDp32zFen`by%xEG>^sRcKIxPpWaOvQ#s)XmTe;6k<;`|^JI_Z(`WwuuY4v_BS(q+s{&p($-2pLH ziTcL&GyC4%oM-e53|Eb9Tiei*>y!q>2rfsfdq?j0RjBA4ak20Q+q->(imF!cC8K9& z8WWFR8)IvtRLi9N{AroNmY_nQJrRu^lkhDQVTx>NaY0WS2@_`VRHm*bS7tj>BkgBI zZ<zNpK0bIL$R=x6-EN}Cp6EfKd`*5wII=tW0w%)^?_1Jf`)O~ySA3g+cMuJCfclt z-Svxi=hs(fgt>pSdndMdTx3>w~B~xOtkGHyyM}7J;I@iCtTkg7;xb$mj`bn&pj>e&rjP(cfaKo z*oiun)%9-@=fBg{W=}KmDE?YsAASG+n%;h)mV3o2m#=acjm2m3Xf!5Hz9f%#3>*zr zJatcEtGoqw34vHNn6a2kDp2Iqm#&2m2BavX#qjjx6A|ylMfr*o~;RK4o<)!x@#ZC{UXy;1qX-ELs))o=C@(yuql zh*uqT;1lcqQa1H{z8;iIx+bS?&%5hN12u_Q}DW!3(8SPYDt z_gp%CdR1}3{%dFR_smih%Sr*U?qZeUKu+3xqOFNIb2lm1ZL|K7#?-?STRCG}&AW}M z)HB;wQS-KMR-uktO4M+|3YaIHl6oih$2F!888cUqn`LjGbnI%z&1+6)!F2SUx3T0ikPn$Xf6U=3cgX5r186MI zYG)@M*@8W9PPhNeYk>XFPkH0P5W}Ymd)nkLDn3n1JyTT@44#For&ZnTT+O5}e8ClG zSugY7|1{{BUDH?5VYYPJa+mb6W5>#SZ3d?s4Bu<|_x!Z@;mLfx&s9O7qd?6;pCz>^ zLt%d0r0u}#W|ujq=RRZlHFL+mO?a`vVm_hF{`_dIp}f+DJrtSHoezw~h*{i>jEoI3 zf(LA0KU9(Db&7Bx;JB&+~~_m2BR-Bs&%w z*q!#Mgnd+G+^MhN561Aj`!X_!eBEDWyB7~y1~xagM?IbyeQs?l#&Ol>@O`sC#68|^ z+OoyW@svGBpa}mqm4!l4SMZiphf{YO-_$P`3RT?aXcQ?{R74BY)rV)zprtsqg=z4U zvM(5;>E7Kl`?IHXKLM_NOA2Y8|{(?cWZs`gSTKcdi#yudoo?D zRd5p@w|CfDa@y8DD+6XkVryN4)BO1##(mcu+hU5mwj@kxRz`0H6CNv&I?!iZ7tT5j zpWQ6T6gf@qNO6RYahm_Nu8CP44EYfIt}f%PL4mi~`nNHuHm82Y%}UQ3fi3gSU^}v0 zSZ~1O++!=6f41|TU9yCUcv5)ibO4%VMXw|!;uP-PBM)ZvFDW)AZrZ+9g;CZD@6;f69?BEFz8U5{|A5Atk>9t@F&+BIM7wNVOvsWp*H#|5e z8P}fKJUHkg(KEq8n~Em}3$cLp_nGhtQ;ky^^b@s;n8*h633hnHpKDch0uOt0+5P&W z+A_I7=XMd}#iw!PQQm#r+qIp3f%WpqTrQ95mv|%H^$^<5AYkGKdjJ;p4Q=Oo|B~2s z6w58#=lI@SQvD`nmH6zyru>B~`i}%v+nM?2a)e9ecuX>dj*hEHhY6$xhEyH|8~Cnd z&&B;A+WrCcGJ;3DyQ>f8Ted~Wj$3|uAink9tM7uY0iwR^OB#BE#DyaM_(juU(eKnS z`>qXH4&NI_BewPt&v(A_=KhqgFQ4W`@4;3{?XB_uMNFfY%eY! zw8&h0R`T1(7qzRmyul0}I?^KE&9n)20=QNC+%LsNuhX9Ggl+QG(%pSuriW7VK8Sw}EjVpkA6#KB zTHgGPIh_ozu@QCP0V@U_dr9zY5_g*Im(#wIEoPzW;QB`8hjmz=^)Z<3fp|BG+FKw* zv+q_=jtQAP`U{Ozso$@r5v*+*&&$N5c#qkZW1WxRyd7ys#5nEYW5ahmT~`SE2FAY} z^0ITQI}MJ#%y-R8K|KdxKBDM%(x8oENuMKEqyOjz$gcZZunpvqwEqT zOwfoup&W|d0IU3IQtnA;pKvmK!31>OKJ9a_%zH@_A=C3{^G}FYy1@M0yYkW|?=h!# z%>;0m&ntW9}mVbukd;RRoApgK=ZVldb8PP#PpJ93=v_?f}xF3@L;yTOL zOP=2!^>niL^&tDM%dHRDrncn9*;?sb28&$Fws;i6kRvQ9#HsaecKkAnbKe4n}mBLSWe59UhiZ-_9I zW-LA@f&4N_H>+XTK%ZB}z9hqpuZ2wj{6KKERIxCxV7pI9^b?c%vt&MF@zY5`zn2g1whAKS8gL+7 zYJvE^NmP0L7@STApMDCkHVQ7sF_UA&^yNWyWS@^M#YAY#9 zKbxWICe|e$`fOasy>81^V0W#d@MMcW5V?wdJv`#Mj1T>PLLRFJatJTpmE@n&eS}>b zqAoad{!8q3ipHcPfthw`-i=f3`jWI;e+HvQ(r(tLld5lpWY))~EdqPOeTzgC zNTH3Fsu?c1IZ^#A)b2sJHF)uB@MLdN@r~UT*$#??lX|kzK{KwiPumv=#afG0s!XvQ zd;q`QIZZoLL&Fn>R=1sOOigauF`QZc7sn|ZJ5yt$I~=N3#&(VdHbxx6fsqIY`G;i_|nSj=w3f}vnMo`z~)KM z-5ycew?~g4h#CA4Vk89Dx`*pFdQ`3VZKFtW!T*arsA zz&S`{kXzN+o`SD!ZD1kW!r6MB212y%z(&R?%^Eh1a5F@>q78`q1Y5W<*KTb75^>`B z?Nkq=kSlu6ln;xKk`MV5oKaHNs&hp>A9KWjB=Y4RJn(Y=5JG=6Z@dXIO~UrG3E>I3 z6DBLQTdUd*c~d>d$H3MZgdy{aQSzYPGle|!K6Xm*U(b5nnX*_3H1|pui>9y7y&)^riF3*LkPzTCXwPMgpuSfDR!I+bP`tg++W{b)L_}FQq6Qc&DGOR zjdPdPjw*JH5<}IG)N?}7bg_YD&r5`3l!%x&th1oGyVdg&VV3L7uJfGPFFnXgv3wY- zZws2cS3UFWndFu*$xN!*w;8n8m;{yxXHo%YuuYvgV^xCkg^{IaEFqjj$Zjxt=fHw} z1*}dozJwiLvf6K3(O3uy*Q-r+24S=8sTkT9Xz{f$NClwFFK>MS)Fud!!k5vnce~)~ zVF`=w-SAi|^y_Sj-I%Ai!kqyasBQu&rv!aH-J_LzuRd&%a3-B~_K6jT37SjICFSkG znM*Dzc8nHHP=1)6S@UOB99D09-nMVEg)?bOAI6zuB!y=K!_VFs;7y-Di10D8EYq*L z+^w4+)79CZ_Bp>#&F7lze|N}!t>%%vI%{MGZK&sZ?|OOfhG;tHlKb&tN!J^nDYnjl zGnYK*ShYxae|k<3Q}{JpMbQ8o4gSCMwgBUuN zvaTIx%I*L-hpOlpq%709vh6rSmIrM0>UZn7jvW_sAg3KKq2CsB6*8yQ!kFB9Ndrd% zfyC#bpos<Rg%xi$Mz!;D}U*8Wc)*ni3x z%#A0?@3OZu z!TTWRagwC7U-eIm-Dkc(5+>Y-$6n08%@{8NPU2zxS6k` eCw%|Kko_NMqj={ose>;70000 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1270-image-1.tex b/generated/webwork/images/webwork-1270-image-1.tex new file mode 100755 index 000000000..151446071 --- /dev/null +++ b/generated/webwork/images/webwork-1270-image-1.tex @@ -0,0 +1,149 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[ + xmin = -1, + xmax = 11, + ymin = -50, + ymax = 600, + xtick = {2,4,...,11}, + minor xtick = {1,...,11}, + ytick = {100,200,...,600}, + minor ytick = {50,100,...,600}, + ] + \addplot[firstcurve, domain=(453/209):11]{(-209/3*(x-9)+124)}; + \addplot[soliddot] coordinates {(9,124)} node[below left] {\((9,124)\)}; + \addplot[soliddot] coordinates {(6,333)} node[below left] {\((6,333)\)}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1324-image-1.png b/generated/webwork/images/webwork-1324-image-1.png index 1531723fbf3bdd88d0e4cbd7012d4914dbac1a2d..c681966d9f3be8fe924abf94b4116d1c2bd15ba8 100644 GIT binary patch delta 1602 zcmV-I2EF;J36u?xWq$@4Nkl1hOh>Yiqv3bUs%6bAY|6&4nD zPPuUsC;k)5iRIMEOcXjvEGs|TvSc}SJL0>ie}=PoiH<-F^waL8OD$hYlsNG5?dAFR z?ojdPYo-3_b^D*zuO`*A8k0xqio5qdzx2J2w-!H|-`6hypnvMJzqP*O^R-gHdfk3% z{g%dm35U?V_1J(wu@lmpvSFpPatYnrZLv#A$JjWsu{CDw3>rE%=reYQgzoRagf|s! z8@8jrhx(|e(86rX19vXiy%h^+u=md3!M*hf!lWrX`VULW2H7cWj&x0Ixg|FE(#q@T zUO#uUCp4D8=6^}g-5ycew?~g4h#CA4Vk89Dx`*pFdQ`3VZKFtW!SNrj0xnx8Q2E~&cHcHWRP3c*`9*0ZEav7+rrs;o(4j+?!ZRIDa{%-jc_wWxS|b+ z`vhCKG1qQv{t|KG`R!B>qmV0l&y){~kCG4h6r52~)~a(wJs)$#fF$ze9z5`J{}4id zG;h2KGJj3N_Ol7$3Aqy{E45pz+75YBJ;%qu))|B$^NLaOpx!fuJo7$wO7LILdfb_^ zSP3-uN*9Z!v@lWYt5c4-Vq>bE^53!1xEJ@f3DxWA)G_VZZLZ1z=C}RtWGk%gdJb9+HYIYSO^N&t4(zVVYBS17}^(T@wG5W1)$3> zZ-0FN)Fud!!k5vnce~)~VF`=w-SAi|^y_Sj-I%Ai!kqyasBQu&rv!aH-J_LzuRd&% za3-B~_K6jT37SjICFSkGnM*Dzc8nHHP=1)6S@UOB99D09-nMVEg)?bOAI6zuB!y=K z!_VFs;7y-Di10D8EYq*L+^w4+)79CZ_J29QPtE6=?0$r{`7jqz| z9WSBZ7IPIcr`5uk+_!GN_ie26lEz>(%o)pUH?@wi|x095$ikbi}wMC&$>tP@pxppWFBS?{B~SxtZ&SY43l~FF(xU{htBd`2TSI zD;uwNJcj?${x=)usKAZ==K0;quDm_M{MozzbwR5Y&7fjutbeT=+`~>4HRzj>Ywp?L z9^BKc^E~6hcXiEfzByqQxYWya+nZyeb>hFdu?1kio%`CkJ(m{enDBV+iMjN!wIhz} zdF8~00kBUaOtu;8GD$6vt}{b}d6X;93{4qydq;mG&pWs0DT|8@Y0P(p=rDKfmo2kI zXY2g?`bMBHWuJsE4Rd4n?Mo98^0 zA(?WCwon<4xy{Lx;e^|4nue3!pnUmKDD&Iwz8)*RT7SC)_rdlvZ*s$Y1m2&_;eE-t z7rFg#zwj5|me6el3OO$_pJuJHZ{K#k_H`S|zI_UEx_!G#wr^Ku-#!Hy_@xG#@`6Fh zKDS}iet+8ICGGj@IXMKoliA9>bKR~~pCw(&ez280?^G4Ckj$2o-S=(&jC*gl*)#6E zQ&q@9a=Lw6Rku}cm%4otvMSF!*}re|SGW_pt-w2z)7^5)@gCikylaIvZs^0WFP{X*z|2@Z04WIH(l;aU9;WCeXD79K(XCf@caFu$ve^AS)dSMv#Rl-4` z9-jVAlH)YBI!dTRgeRzk%RKg>)M|m9iLewWt<@$~t;Qg3lu-AmWz;f{6A4REYPG=5 zM1NQcB+FVxjUbE>gu_COA>2%arAV9O5h~#_k9{b$T3}})EJew&z|KTiiikNT&q&e- zZ>@)LpK{hiD!AU;Vv^Bkp@vseN5-Odi<1ckUuwhz)8d+KpyPx);!_JRUt2@%Rbeaa zWGl%iU|8Yl#>QeSy4Ejj^(L92qw_fiN`F}&30E_3^#;ogKF31)%<~u%)oN4CEQapw4Cpfi?8b|nX)Kv+Ghq_9XV@#A|=Wjr+qs-$( z!cvr4EwD2Ymg2;<8dBA2GySGtps*CtLLDmnfJ(T`V;@Sb7TB2xOYstR{0`tcB25W{ zQaCi!k7ult+ta4h#Gk@TB9(gnWQ18d`!2kfK z_3>fyf%viYi{%HTALaXGgz&-I)07PeByE;%!iGrci4xLe5_UJ_lsr+|#>SnEvoYgp z(6F*$`Ns9dVg1~gU@clCHf0yCn2k;JcPpaMQbkK)A8jsq;ME1wBr{S+Qk(s+4gi+q zxF=NtJ?oH0^?zYrS?zYaw2^|PTVnH?*q};JN=H}vIlbMZF%Nc_^k|XZl3g0MDH)Az z@8=QNs<|ozBqls)6c@*>2-Ntu>p6OyX74*kk?ewoqJZ$vxJo=+x5ze@+*5D6chHkQ z(t4@~+oQ2OY@76Ry#p~wKDjM451YHCja@gm)jZgyWPiAH5$xc8{;2|yN1>b;H6S5u zCX~C^sIk4sAQIg;L?gS%i45rC)wF@!f_+;H{bP{7 zOP%9<7!c&u4MiZgt226#HD@5dP<94?#;d}2A_K2gXL$-9+u9I^oTX>WSal49D5XF~ z#v@OP%727#zeTug4u~HavT!f1X>9(TIWGp2X+F?4#t3-n^VDa_-K(s$6HW6*3pw zM#@lS7Jf55vz$6ZmOY0E+lYr0+e$xC&mqEDt~a^PCbPG^u4<06l36}1t8df9%jh}J zo_`6qGY2$QBDzY5!@R;-#eOqA(?q16Aw)Qg45|uu2IFXS24QuM)N_zw1k#|S(}gih z&sah@kOnz{*`36Aa=?6?U{rHxt;929I?28WyDzfJtF0(31R3>eN1cJmEP0n5jaV?8O!w} zJ#0@BcUSb=G?C@EW5+5&S!31Dh&{97Fq$x^%08Fmmz#OS2%AgNVn^yZXoB))dd`|Z zXU#F%jn4)9_RHuwNZ4BJoH>S9*d8wO3tlqjl6h+eS#_pdvv42{O1h$++pYEHt$z(s zG7DGnVKnhFdZvkI*4gWNoz5U+Vd>4rDoUfeI~b*^taXy_6mNfuLWwME9rQ|}_0~?! zQ1AHV*-dp5S~>cbuR@i8yHVfrwE?F#_3}FajdKX>tf1F5Lvg^IY$H}TaKbqt)B>0HTnY`nFijzQ|x&(?7pJ1*uxPCE{vHom$PaC}o2 zGN))@be+A-2?RXv+XTHe5STize1?vpm5wu;Ivx15<3!--rYh9Wuc3iJ(Ye!Dah*$^EuMD# zo_4&zw$U-7y{TiCjT4aVz#$Gq02giQns)32-kbWgV*~KB?0EYxqXrJ-v6Jxprq*g) z6!;nYPDV~$J1zvgABI>)cz^m}z8~IH7$PaorS3O=h^7X?kj{Uq!0J=>U~d02FBCka zzyQp0qu}VzmYP~r^tP3*7cOzyirI^EYLl(T`$1R!vE|=HjrZ2nfkiQ=?SZ{?*!Yog z{PCCt71gK7%DIvY@i(d**Oz)$9P=VSAS=Y$0>V@bN{>C zCk&WLRM=ZA7Wza}t9_@18pYK=ZVjZ>{;_i@A||{ox)~BT9!C;7bX?Pp$)69Y)GX#Y z?O3O80Q^ep)p@{4`-*EI;5cMdb8oTT8G%;&P79rOoCF--)UEc9p%8E^L8-0!XyA5n z^?61>ruVetc6*Dh>{>XpshbA^PY->Pfg7I7Jni^d<8dP$XWMZy@V8t8dDn9wrycK{ zsGl7bbs#Z6v`QgsQ`bEskojY}FWaA%`ZK2Kqe}78Z6+ zxp5LF{u9fI<ByjzA3b)9$59EniBMIPmf9<@xvS zQ1Rz$rT*!4`=8dYCe^bVlSk=_yZ1i7^u3R_7C)Nb*DnB|>VLAowZ7x?wNk%&-F|BQ zmd1YxhtR$C*nmK>6VjWqVWqTk3EkUmu}ezF*f_GWHD>G#8ag)UGj@lB?(e{aHx+Fg zwxhp?`lzSS!feX}cP`ky6$@yv_s-zKz4Zygq$xZ44@=4h*(q#}bWLozB{uld%IoM} zKXT*arsAz&S`{kXzN+o`SD!ZD1kW!r6MB212y%z(&R?%^Eh1a5F@>q78`q z1Y5W<*KTb75^>`B?Nkq=kSlu6ln;xKk`MV5oKaHNs&hp>A9KWjB=Y4RJn(Y=5JG=6 zZ@dXIO@G4nvkBn|xf3QUwOgy&4tY~O$H&0d8H6G8ic#{Q-ZO&>q7oY^lu$V#z%7^`m!n!8s$^X!@AmN3aos@b<0wAYvfmI!B30cWsHojGGwg7Srt zrDrT5oI}WNFnZ^}f_(+7PBOlP9bdBAZ(Gq=2nyG$O?3ugv+SuD+81c?wJ=Bppvy0B zeSZMdCJ2zim(j0xyWr|!35)LC@K`JK>uieMn5VeHodFoAZUQN%1bsc-qm_HFK5UV2 zCY^Qmi4}(lnoG?k%%vI%{MGZK&sZ?|OOf zhG;tHlKb&tN!J^nDYnjlGnYK*ShYxae|k<3Q}{JpMb=G%IZ&cQ-igEDuWn0ma?uLXUgsXIESj}7^E!Ixw7pzLzV|@_3C%)xQ-ncb0DW3 zFQMNSa}_eD)xwzEdr1RF1A)Znp`eKd0t-G29Vfp^LJ>#D%%;uJLZ~G?m+&0h%$k|k>*a-bbQ+JyL>fP z^3{m7bX*9yT2miK$Ktf(WwOI>8C7r~{Yt|7n>wiRyMPBeq^~X(R}Y&Z^a$@hSdN=F zJ%+h8`%=SMf^PaoKS${5UzKjwvohb}OHh-5UJ@Yz;Vn~FY$(ly2GJkF6O3n_{ZrfsM7?#ITQ1vj>kD`i*5hAJr%}h5?ih<77Kk=Q>}F;O-J|D>#3HG>)3I{_guBs z7O~d)u{dL{OyC^)U2~_J?|-X#z**~xT_8YgGTL%&vECkmTI)`lPCL#5Uf*1`){mhO z@LGZrOZ78fAPMO*{L>0?@3OZu!TTWRagwC7U-eIm-Dkc(5+>Y-$6n08%@{8NPU2zx zS6k`iK2?Y-?1MpO(%T+#gP3UXrp-NE~$er00000NkvXXu0mjfFSIhr diff --git a/generated/webwork/images/webwork-1343-image-1.png b/generated/webwork/images/webwork-1343-image-1.png index 6a8f52d738ef863f3555cad055b9816d7fda636e..702bb75189ddb9415fc306f1fe8b4334a65babdc 100644 GIT binary patch literal 1182 zcmV;P1Y!G$P)9&j_lqS(-Z z>BFe+Wi`&rAha4ZofTp*_p%LSXi2BLDf&W|x47j`om@~*eJ%&re%pAq2CbbRpJxb! z=S7t@R-IKD#j7e~?hY#JeeQEU~o)AS9xfq z!M6KG8>Dh1YugN44a$AJUFf$f%0sVxD{UX!#TVM~LW+I%zB|h*^>s@MaVsTVu@qz+(G{=d7axOc}W=$_oZ0 zhs@Ta_Ea91wC7*X$sA-SduQ&<^R|t;mc*39@Xp+_LsiK{vUfVkKDOJKu^nZ$_l#S1 zs4AIAE;qKL=Ito6>%4ssa#WoCR^B^Rhc9zF12obI^4q;fO zX5BxC7uLK`Cvf%f^mmaMmzk@RxSAu(gphRFS03{qM7YK{gy^}N`&1Eg0_lE{=iL31 z+~qFJb2WLc?tTpDZ@jP}b4Otn?P@AUSMxL{VJ3v6)4uYU2O+{W#vz!6{AMH&HVK4z zt~MO55Me6H#+U~o>9n&vWjd2Jy#)PsEVJZ^F7@VHNi!tZIY?5;xQZehT9VX~~ z=4xDA-{bTQ8@BSFW>t6T5Z2H=)M&4=dUOoF`Oe`NckI5OGNSJ9oH*rbCEujd2JMVe(m5B4PS7)@fYbJ!4%Y#@%A!!D38@ wkaXH%9+M$NxW+hyea0C6Oz9V~{5yjCACz9zCRtQvYybcN07*qoM6N<$f|VvkG5`Po literal 43477 zcmZ6z1zeTs^FDkO0Y$M8kVYkyQbLe45h>|z>FzEA7DPb0k&q5)X#@#L=?0OKmTq|G zIfvcv@Bg|VKf8N)&U4q?GuK>m4L&a=gz@o6@lYrfzR2?@(kK+x4E$9)ivypKnCg8* zq0Z46K7RbtP)7%adhQb%s{b-riMVlPop0`)(A7uYFCTxrP3sYIS(%|LGfn7E?8~Pn z|9m}t$M+wOr-4sT5vzvYG=6&R{QK9ZiEQ35oo+6(B$RrnWHC+>yG)|})_P`~!7o0nc6GxBYrMDo3M%`%bJ=*uZqC^& z9}N$tzkaQ-+GTuggZi|n+t>O+QfVToMeI_*ajm|B;uqCl_r?_`{-WW)0GD(rNl68`gpbm;G?m9;q-m-CxM!25tiJQ#d zz*oZw?cfZJakB#GuhW={9f@r{vXUFax`cE|~r<_m&)+p-z zDyNB1l+E&Kp}kH8Nz>BY5Es;g<-hD_B9EQYLh

    dYt|vgu`%J?$bFN&ns+LVKqeD zDB@qITpnH?KO8gUH3{LDXzOScq-VFa@xsDUy zd0FgrxX_s~Pwca=zg#9se0VSLYVyT7p=X!L^1aBZ?>CX*y?&Hnz;LeF!zv;6LUNxlU|aaKzxv*6jXO zhvB>>YbpK`?r`y4I6o2om;ChQo3%nASJd8qu()!G=DN?LGfnczml=iUKguVQ%{^Ye z@#q!Nu1D=B0|vSm#JWCK&vp}b(qwz&W$$L-*`Dt8oPRp{_(Rf@v<`!N%9Pn}iiOsH zdj2u6Au2J@8^tWRZ69vfb5vO5yK}6Mh!{Uo};)vRqQ|H*Qe-jz5uNao)kQ z;=$PI70-Yg^V;xVWPe9@cy~k&PCtBIb5rq(%Q?%p6mRh!-I*ZO^84%coF?OPn~(Z; z#ou$YCZ|=q%exz6*^>EOeA9GWcQ*>)`Ycgxj zccUWWu1hjKGKp3!QuH1O9JqrQmnbpulIx`v!*mFH2ycjzWyM!7ZUsIumrRGE9iCOG zRi1NEZxh~nKS_L&_oU#vxR2rOu#AVcpPs#3Pj5_9NGITUoAP3^Yo{=_5NC*Uh=$^^ zey8IGx6_qF{XLW#jYud**hYA3SXNjCHRCJl3-~wzA8Q!2V)Xpj1YR&cw@kkIdFy%6 z3w)6_(H4>67s4-yUzCVk5_J#JYUXbG-m(}z7vgyHeNb=MW>8xY|4nCcJ9&Q=ea6P{ zJ0ZNx+s`S(`op%v#+ls3sx;gr9!h-^x$!hXdLpAX>*nK5`MVN%VylV2{jQ4tP-c`* z7n=*}I-f!lTqmX}&*mI?Q`b>ffBM4ogXvS#*o0xe@h{?DL^l)2y}qr&`JzayNLVg8 zC3z+%EVD0h>p4YkyL^C@VDeb1K<>A$vp)y2JJavB1?jCS@?;V9MbIgAKVz3GP+zYi zdiFk>FxOUbA%`o=G?5}nE!$NXx9@JRPcKgv`&YcmW*51xj6J!eG8(X_6tnlsUnw7V zh_g!O%B+M|MY(-Bayt|_?7*SJIfny#g`e)l;^pWy>g7&mNVfVR|Y25iiX>)nZK%h4PUzAOuzqV{}gVJC`@a*EMwFUZ&e_tI?tPUn6|nBMBlv1tUboD$j62fU zmys*3JUn-b&w&z`FU}#KPln%pv(~}UNzR#nn`@(eb9p^s*JQ(R%W>9tR^|8pfMhpo zU2bFiZ>rzpQ-V~QR4+-XFZ5g*x$40CoQr`g?ap&|UN5wyu zB!(pysTcnlUsZ{FsBSnm9D9nle4TKuWL9#~YN%J5TArHW`lxAX`4(=^r98n4i%ONL zRvojo|J15H#d$jY^nLxoZxun(BgNIpuCcUAEZP)W;VrE#+9k3<^Md0i=}!zL=DwfW z#-h+|&9BWb!M#sep=742?);Ah!(`x>+U&$k>1->FX>B>NUW+0%Ta^zkZE;b%byvK- z{Rb!wv*ps5GY%a%w*I_%kXq*N^kHvJCqgH(G4B4({f)!#N9pNO#d&F~-K)uywFG1I zPSLGVQBl*;qRPL*{K6uGZ_uwCnH*ayWSSYPktJpSxyBu8IbYA3Q0*6E$3JTA<(jgY zwKo`Lr|hhS+mo06xQC>haS?BxlyBB5Cv`Z#{7|7%BOqQuK*ROn+M^#&YwHi*i-udV zw5dsy@;TG)H9zTSh-*~9${kzToTvCYJc#)eqK@=XV-*CXT;JV|M!)w9SwSYoF zBKr0C;P`M!MsZGYGV%C$%b3!A?a}u?!WmhK_0AuK>=ev|&SVz24<7Zh;$0K3BC1f| zE8qFh8a-+LVSKCAv7O&Fe8)_y>ThN1J^|GL)t&&3ON?@p@^=<_m#)2@!@X}H41f#3H{glpE*Ef9w?C~4`m&O77yKBqNihL_p2$M z`v0RN?ECmjFxQitmn&r4xZ5o(h7Ip>r@kut>8Y9WHa0lS{Ps$6?B_2QQ>K*s7gEpb zFe^PU5!Exvw^>`dTz1#T_efCQ^S| z`P$`TGP|UQN3t8XG`k`)^A$JcS{NjJTX_QDPy(YzhV3(}t2G(}s1zpK`x(xnsCjGk zBnK`xzi*{0z5SlCo#8CN>8Vdbn$(PbUwVqdW%d zDqTxXdwTB73b+#&%%D6-`g=NA0|Y}jinS+ByB^NpIeAWJ%EXZ`i{1DlD&ohw40n`O zvCHu09Y)Mcbm4~=(BI}4EZX`(EH?HsKPBt=eP-rF?lWLedI6n!Et9Oi5uCWFR+I*t z_P}59tL0w^T2G-OT%Wav5nsDz@a;7=vBveg$N;BonJe+$2CG~syZwIpOVr6AFoV*C z&V53>cO2Z@X;B~Ki0gQfF%73yy?c5(<2m#Seo}tWyojn{x)#k`GKg0+Ti+;m{KWLY z*e4!6afx&;m&{z{XU-h2c~GbVQIp~d?IqG83I`TLFTRG&KZi!Ve`CnErdy-1P$^G! zoL`Q`S#&8DRSLCR-KoD3frVl#|5~yVX5A9I`innG(IQu|+zalD(6bMHlx?=`q673s z0<>@k_jyW2?&?uXUQ?N0UbmOxMhP{1RoTBlR>jkjol#8uGUl9i1MEF{pczAT&&nwj zHAR)?ljXju?j)JY8)V?ZZ zD0^7WRpp(3#WsQ5c(KHCs`}48{7b16%{p3Db_;fRc&LIWYNEL=-=3q_&#C+Lnm6n$ zD&=0t{bd(5Q{1><*oFK-yy+1uZ<-&@U)Zy`Og`{x1ayQ9o4mV5h|=>E;?8K;`lMB% z^_a1~^hu#8tbka<1{+7VB>UlhRJ;UUkHC0+pYhn6YO?KFhpm!&!5l?r{CC(MN}s?= ze^R!|DKUBXidnK;VTXlMw}YJ;amlA^FGVGAMn&LK> z9AHy1D~OoNU`pYZ@gRZaUcS@=)9S98NyXP#U)*jGq%_AFS)ZvNwDNZ4IqW$#buBf^3% zE!i6(NN<_WTQH5XdP+?mT3#f%+$S(;nN6QHG-JkD*`LdApB-VXRhY+bU=ScLrL9%$ zf|nOUfkL@lttpo)SS8ZYbZ5K~VxC26j}x=da%xJOL8E$KvH4}3jmx@sq1$#Ij{=$f zf=C_rQfCSgvI{~%Q%F)P8D=?#ZOTPI+@*

    lPWCNHHIm+~^kz-wl{Z}wJc(oEkz zt=E!DDLQxxRg-pAzr~nK{)4k~4l8b;sJ3e<8CM9+yR|31Tc?ZoH5J5_q`Dk3+{Oyp z49KcFPj;0(f?icaVCD8P&gKw`l;=D9d+FJl@*M?6J=hCw zAeUz+tI|D-3@G$Mz+FN5cx_d$wG^4NAuKo3%5bP8H&>a?++J}k?n}j}R$c$`<|O;x z3e^U!TDZEDwv@P1XBzw3;O$M5UFo5uE8v1^Y+l41O3=TuhK*FcdRBP){4=^jpU5_C zY1iI++3u@~6^*4dm75cW{Fc8W+-&p59ZWX7r_7dQz%zhTwz~PL9QHMOOm6u6j1ASc zm<}T^5EhK9POBd)+ZxOH!9enlu5bVF+kTQ8o>NRr^KPLmOx<-lG>P%?Zj(n99CTJS z7k^)eD-D-R3zHs9N$7=7%Rr|2^qv*!`Gwiy7O@P|WqEK&nMAY3k%Ni5Dd$OB-(S~@ zCU|UaYF%>s1+x;b^|INLs4_maNV}DBJ99lfxC{>-Z)8_Vy_p5)tOEpThILa4vC;?$6oiZM=+z;8R_z^Y&C4B z(Wj+HIAgdkaKjdv(}ddBYVcX1PYo)42w~t+0yR7Zd5F%m()) zc@+r<5hA@M*@#yQN^XeFy#p?EudHMI^ClUk#I7qU@Vp68uolmrftDJn`s?MTi;_k} zJX_|pM^>Si&YSqV*Q7n=uE+EQ`-e`LRu&VOxgvPlj7{uAjc`8m z;vc2c5^`CU&%i4Mf8PYx!4WfFa<9t8H7n90S~qUXRGDYaILDMIn3OBT==8J zw@Hk-_WH#xPIq-+>l$Q_pJWRh{h>l)n@iSAoQlm3U?xijc9f($_>pKT(UdGKbRcXWHbn(^3JMU6ldB7R?9A9 zxq9WvE{zHyPNh2wf{0qtOs29`eRywSiS8n??J#SxNTz#!*zZ$9q%NK}dP4}?|F{~M zTp}Gxk_%CsVk%ifp)aFf)yv^sWtUs`MMU}o0;UGX={kI&pHXjN z-^wU0HfUIWOieC6NckZS9oYi0U5=KMJ2K6)pC2Z&uzi9UM~ly@Wj?TawjPU3|8TV0 zElb7S?z#KUTCNPv*hkL+C9~L0`B&hy>%7b%#1#rhKQuYQL7{>MOx|TEmOpD}w)b;t za~Lhg**#E$V21z--2y^sZj^OU34Ips_~Hw^N;$b^Y_cEpTP<(4qPBTwdfF7#IXH1bIjh8u!11_rXnaZ~h{1??Udl zs~DgG=jEdwKeL=Lqi5qvW}o?oE+A*%!8n6#NA%dvETM7kkI04GOmme~KtjsZ6I#83 zQqdebV_(-fOXG)_5EmKn>yxoJu^FRfYVa>Bey8gB`0Jfqi$2PZTP1Tag&$fP zaRj2(u`0_U2=W>an4U+mmF2qX`Q)&(cC!jDchBwW|DMcg)~+j>VXq$@q6dKKAag@y znFv)7>Vy{dX{`8Gj&Z}U;tMBJx7?|#2Jmva{v zr5U8%9|Q={PR`#69J!9znBrv1=K$;^v*gPJpr zm+Tt!TXepWp*&0=cp>=2gyAz9c3Ww5kCptUQW!7xe~m)rw;L>Vf|I&Ne;)DV9_L<+ zD2-}qP#zD=u1G@;G0i`~=`uAOK|T;ht)A)h`$Pw`t;uO{Eh< zG58&mD@GyJ=_Wi1Ma6Mqdn3o+NDjMM$NJZg551jMzN~I(lr9{nDjX-MWoK7+sd2ie zhesV9-@g5#I(%1{O6_3#w5Lqc*mCNz#R!{T0kglmOFf||kAkOl@0oTvD$0)VA2yz^ zX}HPa?q_z$QF-BP{yCax>yofz?)fRw{q{H|e?oP0$(8DQqdnFf_yC7?vC=~yY2Q}N88F{d*gPkO@@i3}=xj#yp>6J-cjAzXoR8RSD_(3%rFCyhqs07uPj(R<-S@=vEwnAy9#LtQh zIvw$XK2^KN;XI-vBzhtz`^5BWtN_1cDvnt5eT|>fqCAxJq~xvLKLTd&sI8%D3?ve5HV>s@OAnG zuXA4e5gj_4;r_r)*?mvYwJe`aL$G|oh*_O2FO+7`BIg0;_1@!E!Ljc2Z&SuC$Bk6H z8}A6;>+vbHwUmGAx~e|u7sfLy+I4!}F;ul(#JKRL6j_p|rz0sjhg9@6f%(|nbt%Kn zn@2g%eB;=Ssb5Kddt}5e9nkuKQ!{AL#dFBd<)h0G$ALDzXuF&!5q))9T9xVO+m(J5 z=KU$BjU&cN4L);j1SayFz0@gO5(b!iqi5(?WJxl55juG^V^ftp6nG#6aVS%4h9SlH3vd zkZf9{13;Vl5;eW*Z3Ess!+4H!V_AZJWdHRQOkO+niUMFd-_gsi2!Muwo#SZG$$zeB zET{cIxXcsO*B%{9pQpmbImk~TNhwv?hgaS5q8&|I#l8jcVdfxcMYdr2trddEtZzp+ zU$T?ni{J$+gG~<5+=Kkh+dys*o)wns@5+R*f?v>dt5iku_GSN>ji8kQKlel zgEo5Aqefg~_a;_W7tKXU7EhtV+x70I10wY|QOkzsD)_Yk-A63DPCGO&bbyy39e%@8 zIpw^w=P)^s!X((O=fFdM4^N`US+;?!{^6BV)L4+AY>k#k1A=A*M494$6w)8xU$imf z@{hQ!cqK43014vGB1@ma=}(IOO$7-f$$GJmM(&_uJtrV0Ey#!{2FsLlntvxRbdU|d zK_cTYLHiu-I3z7F3lW{V{$}F!Uuc4aH%4XIZ=241WOon3fa|E<9a)IP43sFb;ECxa zg&z=rsLAC_t?gY33RUBosTjBD&LG@RE2FcWAPabOzWrg@a$D!-;U2S5764O&Chvy& z4>o8^l(A4^evszWc}W8nkQ|y~?rktE{glTak`tXn4U5N2d~bRz=D z0|0-boMh%7>#9@*+{2@`s%YVz2#WxSSL+-vs?|bh8~awil32ddOnQ2f27q-)9%oxk z*8h5$)8kL#lf7{CuqH>sfvPt5c>h7+7N;OH%oE<718`d&g4Wa*R>?h5l;+I$Fn72Y zN$T!#OVVv?{UL?ih1c3Jn>GAO`8#)cfEsBeB(>(C6(i?weoI}H@IpY?U6iIo+DLK=Z=;oSE>tCe&p8GM-I2=z zC#@|0s;dU?;kpHfeD83u$co!o3iFrYU`0*BJPXju{$!m*DKO9f=n)yNwn zIJHmN-CXL-7x|Yw`rqUJKc-P+XNBQZIQ#&iOnb&oD>fr9LjC^;AIx5Ph#?u?zexz1 zeS=5uwYh%kV4DGcho2Mycd$}J0JD#s3w^JiFF>aOz>=h3*LFI*FL_cZd_tV1c$OD+ z{G`kVP7EBa9ipbWaJcSAbm=B`N)L==vKJYD~)!B=6t1FV$86^jEYC(N<+2`5FY@$!mYuU znBwmSlnks1&Od34YcgyFLm;$%g zSM4AMtJN`M3C{+q2j83tauU6khe$q&3XDB5_@z_(e}lFqIf`*1qqyZo`D>7+*TfMV z7^MNmh+mcXUVV?aXXj<*E92?>m;LAAQl(xfA0%zo>2GiRo|}h@*<7oK;z}QZRA`Y? zBB!!0*#ZXRC`)~4`TVmU#75awE@Gw{Looc!W!&}7WWg*lP5zaD64e&2<-TCA`xhZ5 z6%bZDBRZXKFi1)E)s(i7;HN zYHe5(2Jqmvr$C(|gvFsGPxR*h>gwg%{*yr=EUhlc9}5E+4ZOS^U4QM;?4U&(Bym5% zo)f5R8Cy1@1B1dCdH5LN`Q)-YW4s!L_?$V04W3eyRFAn3<;dW#)5N z(F_k0^Z|7<2OC5-^jbXJ_o^sgjI(ns`_PNF(INeaBH(Kb15QpXJ=ln#0IXQ8Xz~kO z5&c&@j^G=SUB1)^F2381%H`WX;NKXHibCk|PsZspFs;@;tk?g&w{GP8utM0u)Viav z`eC3^${3Q?l42gaN7x>ok@#IZg8#%~U{d1yo3rg!ncy0+Yo+QIZRsixJ3X|*#T1f1Jq>HO{wDesd!Y#Rg?wBHfNWy>c~pe=Eba;uHU!4 z@M=SO{g6)mOumF?j8+e1LCq@>O7kK>1E z$jHfa#6u?YzC{+Se=qg5CY-(9F4ygv<4fUeA;s}vt8_k7fq22Z;^AQlR_aDk2a``g z^s6rWqc5`2IgEX8%FiF-f@o?hwGlTKU`=O(domcGc8XKsm{I9NHKr6tfIKM_hm^&lg8i`EdSKb)3U?gnH#B`{dZnPUN@b)tdz( z{VXg$b0xMGdQ}9I2+tGy?XsIGM;#Rn$1QuzR|r(@ekPN96*pw@b+cxFulL7P?7(rN z$vZ9$xmtqLhS97d>*B0r}K6v^*5h{7?`m<2K3m8gfyL@l17TfUK>+{N* zGb^{H91P;Tyw*bE-5-Xz7zG=!-IILt#_?VxOPa@|Wsr3zUn~!AfOTl{f}l?P&;FUc zyHT*Y8ITP!S1eY{L`PJ=xb}covh&ZtY{vXk>^*1qswlzv-gtF6zIoFOh0yf8B!v!J zmt>j8b?!-rAJc$M)_A`z{@{UwvTn;v>cP>{iVgD2vDLP%a!kB>Q`P?bP5$A?T%E^Q zds*2Q%|=}ZT0bv03sy2{xLeju2XSxF%w$yXrGN2=g0yCRXlRJ#8nf@bQ6se+$+!LG zM? zYqKUt^$3~xa*Y+d8Jqqg=mU!Qw`!;|a6U95#9l`}+W)6yS&`0O>+H>p7aMthxwIKw zAW#8(Pr0q%!1YG7*SKOx{nK8Cu1LQjqzT}cSoKS7YsoTui#A?FmEjMbo=5|SdcRP6 z*na#4&_)imeob<4TsC!7F`us|y^>&wLvRI~E?hgSfwjqp4oqC{MSygdHroK}EL?VU zuQ+z3Chx6eJyz#x;j7cLG91m7!}+xLjP@1@|CJ>B_!p8 zT!^wXi`0FuwNH>faGEg4VZbvL=M#nD^MJA1cUET2_znNOuiDMH>gpsVa7h5;LjNp44REwjhlUYwmkSL+heEuvc9_U_7<%NxG(!Mn%806HcIuL z!gQCEew!=WIm61y^cr-4%O)>&K@uEppo9R&*Yj6EO}k}TxY=uZ6BBxB(zFX^e6s5?MViLZqR~+&>5G|;(onZj zHW_qHG*ME&g!Y&yzxm%;yFYdmVflc3ba2d|HU$6<<{^>6KP2nRTrL^H+_Hd)`W}|F z?NN<6-8leVX0~NR6nnlTaNHS60;EjVkgw&)uNSZzf}E@nU42**q+o~vrerZ>)BYN~ z+zbgFOlV@8WaR_4=1nAF032Zn{F>vz4FHHSctz-nGwjL z3k_pAMi9{W5AHP?U! zl2b_pFjjvJ7ZbCZ3Ze8<2jzT+hW&I5kVX}#gZ1e;bNmZ-JT8!p{as4Nlul1!VCg+( z!0UkT;Epg{#GFu|NR{UIFB8#FG5#07`#~LAi&KUi^Z(Srm;(uJ82EbRC>oH%-%psvc36uEEZh+5y>TZlW#srW!I>w0- zjKdnK4V_GLaH+a#m=;hoAQJdFCJqH5#lM1SjL9{bf%n|nk$533;XQz63{K!*C6(QQ z`r%_K0JPJKIEGO7^dd6v-Fh7}_WdH}2&`+&PwQ{2<&DHKLuCcfHE9DBD%Vgaiz+|t zw=bs=QOT)%bTN~M-rXAx?6EV4g05xzk-!Pmx&@2t-Rt!5ZKZRa9J*t*X8#vSR@PL_ zwolL!4^1ECl8SD;w3M4!F@ZJJA9g<=F8+)EiUF^3dDh2l^Chh6&Kn}*5D$R{`7!LweowB(-W_wtR}y1ilE;CSsWr> zxWs~NL@OtZfe;j46agQKUXc4Mp5yuX_ zE)id5t?})K?rR`N5y`4oxfxv_Qb0DKwfvhDoKcFcQS6 zx4&3RLq(!Ng*uHFM{T;r36AbRc<|si)%GSy?&pK8l4@p=y@l!{lYwxuq=i!u0B94BLW3i4yktj)kFKExe%X9+J@;Rpt--Vl$0Oj*5;d zwk{r)3sy=?&+BOY$hTe%$SwA!w{6Y(der>3BZtG~F!-D`|5w{3Zb8Fkid-Jxeak|u zk!m_2v#)gEnH7p+1{=LGzqm|b?!%3Y4{ zgn|D>R66J|Gt+avZmC?(S~4`lTt)aUs`V@QFlNU+`)23-Te|yS25$m+JZdz83Wj$w zBP>kK8HY=D%>~usP>G57xt%@!*P?3{EK}i-py=AQnq-K$EyRA`hrQr(PJNjHnMtf| zwXvG0dGnS5n(ajFm94)=P%5+LR&FzBEtpaxP;`({(t|3)^wL1+ z_EdebZ(LG>5^-xKfT&{J;06;S0*q^CQ=s))SRE){k=PRmtHz3G)*RdNQ0w|@+ehhc zu%pO+UO!NVxzk>b=sypo=LN2=?Vcp$azRK&4a9V4n|73K!8D|$tM@6ifjWTnnePfF zGO@7(a_Tb4L-U?pO1sV}QOxlls_)d~fYra6f2ylVePp?)Y>oOVOGq1*EIzK62{a*s zO|NS9X`!pIzxfB!g*1gce|c2kx`$tDAhWL$IC#AmY}`ZPKW#NIh`Lxa1x#S`=+ONM z{GULGOPo=`h~Bh|(e9nizswXLih-h})&t=!vnxR$G=NW5Er&jz`g$#=`E!|3qZULk zaf7-NI6Aos{U3WC3jiw>ul&kj^4CzfggASl7oej`Ig7fN0+}qVRivf7e?=hWmP+mm zXYVyoI?MxmAX}M%7zV7<=^!CwsS}_;0ZBg^&nhup_D2%lkbYI%TB&R(JcHg%S1$9(bcXJG+3oBqBa)@ge>V=YdjR3KuNI8db|fN!>7ZAmUtT&V$q)m?><_ILNk{x` zF=Gg*ZUMkMW*KAQqGOT=`IX@_AhZxU*qg_YGxrQ9OMKd6c+Cql^7sSgsu!R#sd~TQ zEd?g_v42P7IJW6~W)OC}=*$St%_3{)w?b%%uY&&vXRukjd`q~Sn{5}SLQm0t=LPec zop-_1Ho6AF!BsIaEFE_S6SSMm!mmOR8+h{O@%Na(0ff~Tpzj$~ZmL&iP-lBAF_mO2 zopN@u<0oq8Kl+(AK6(BEEGvH12_R$Lp^ajjNdHeP%pB)%*SB$cdhW7uej8DzHn$)6 zNyi+Dtj<(9lDu7mFU)FtQ}J@U7s=fJK__BSbw**c1_sS;>f%ZX5Z1fPmYOUm&sRY$ zhA94r`TU>$%3Sge-@hs#K6B_!089>;FYY;J1ragxNM;i=Dpx8iPJyMvB1**?<9u*q zbidgWx!tQT1-d8hU-nOlY17Bp8449n)VhQXT)dZB1AU7YCvaULPFpUlbJOOwS8Q87 zh9J9CZvHuDx{oAR1|aOjwxt|VX5#nSU~~ZkUXYTK^%9IrW@{ZT5vUi_ic9I2R*+Og z>#@)23E*{VauB7D)?-8Y4Qx>C;1^8SEp_@ju3<<-O}y8;MNb)>tu=hT@IMo%&!_=* z@s$>Brkz9!QxGtFme!I=T~# zB+~*r;b~#>sHKBF0dW7)hsk@)u>KLPIQ;Pwqn*aU2oQ)0nu`7ack^$_$0bGJvcF#2 zRZGaF68Mz?$I7d2U&I)!5(5{3YJb-qb4dr~*SSz#19is1gJq~n8#UwUR}fWczTGmw z(0LK+kQ8ESQno8C)$v^mZ89PUyMXS1EJQ9$*sxv&QBsj`bgS(;<|{%&+7sQxeg=oj zIwbhBKKwP6qgu%A1C*s#ao@p&NKh(EkQZf$R9}GdZq&PL*fAR41^N=N+*9z}K38^Y zlkIq(?gR`0sa(u`<$~XPOFc`SVO^B?7+dL1O8lM(i8t`Cam5yK%mDDc78Vd4AHaKU zRJ)$&u)B$6(xp$&hk-+Zoudi6|%i z($If*pefEmeRw3c$KIv*FZL~~k}cW~kXa5TokCv}kR|T|&-*?1E@It41w(#SFk_!( zx-7Bl3_culO!<$1i-oI0Vy-)L{#oHsy$+J61FqsZEQh)d+1M)!#kSTk%-y3dhg7i8 zyt_HsZ7gqay)0}jMr|z4sP6tFo|CspFP1F;rxyaZjl*TB6qK1zwV556u|U!Rl4CG2 zq(=C~Q|f0UQbdS&tLU(`^H{{R19lTmSkBtC^Q|3KZZR=JPeVMB19lEgxjzAh+NvZ@ z7G?{#LCN)l9Vn1?pFJ`u;6iEB)n=U&kEau zI5gN!Gu}W5q>(9L#eBk#SE=KA=nq0CLEo8{*aOHid0s9({3o0=Ql`MZ1;>=Zcqb-b zcLuHO0#kwn1T(uH$T`yU_|p(?H*%OB(oz8V6F71P`-LCLw2LtCiFh#H=A~_z$~05S z?Jo&WU@~i4F`$RK71mOeub6mTi!-2iXWx0{P>5+6o_2 z%bGL*Jt-)GCv^XSqOyx~1PU(3T*lFOBK@WG6;d@OC&re-;j;Joqtty_QF5f~nW@G)a*9PKQPGiEXgI;?`@ zi^`R2O&!_d1pvv#wdRb5g+>qD4oGI(4|GpI0rX2+Q&i2q>8#f82ul~fvU`Q^w zJDcSJaIz7cn}4W$Qx=2Uz05H#tLeIasrZMbZ#!LULfp5(EGGs{;1g-5f)(K7E~-=k zk%!pwOxgPE*nZQO??CS(vM72y%q;H_wq^%;@Af{+N!&3G7ltTJKMJf?p1jHx5M|Ep==Mq27H)uHLs&gv206O<5lBwt}&DH^3CLl zHqXd0#U*rBFlbSIa^|=Iz~D=BwB`+vEOVQ&0jDq0AIh6GLlPhAHd*({&9-ZR^24AN zCQ6Bs>6gkuDc~c=W4_L+QCJk*A{`c8d#dMg?>fBx?r)T<74z-WYY-3@t&D; zD2X)kdp&d=|6*GbR}#0ofMigQpr-T&`Q~$u%lGP!-7BJaN= zvxIJsdSK*xZQ?T}-1s!Gu+nG02pBW>v2rPIcBrKmftsbCSc7X~edvM8M>Ne)eqLEA zB7UQY_<^7V-jaKv%SU)VuG4xp*mQJ{jU$3%OSpwNy;XZUroHs`(&WV=lKT;}>9c2^ zlHwO7a0Nfei;qk7>}Gl)Lh`7iRp!ryA`#lF6ApY%O^IB&)kk}28qUkR_b=$!k67YL zk|SLwbc~ldL`WCP3=8@Q6_eMcJA8bO4+NQ1Z_jNzt=`81{g8-Dh2Yg*cixZ#tXKtC zr)TkvjU5-}SH|}T*p8ZtGI#cYFLq&fG)kfjND2|eV&X`PY1-0hVX1xzap1Up$N3mB$j_rE2 zcRS)FocsmG`?4qXS=zyq{@2;Ya&P^X^XC)! z>@PG%Pef+AtYaIoKRPz-rf0R|R`~F42u+q3&1Qg%D&0UzZ@lnpl1Ty-&?Y;ifZ!q? zujXbqEr{57!^#@kcF#KCVW6(fU-L?K?) zTu5WQBTG&GVWoZ9j6br+iwz4e) zAX@9%R}rO`dZV<67Nroz_NS@8v!;OXFBnMY_0YY>k^x)x$ zbg>`q1&m)WkOaJSpaVh;^1jHW%^0I|dKP0`; z0ZO@Hg_v#F@&i&sW_zRktzWAxewFLhtz7%?WC&3Jboc)8#yc6KSFGe26lvi^F!W6F z=PlH|Hwd|bsv##nD~wN8B{I0`bqeF6*cOmg4<$kQ-HT%K6oyR`1C|N+>2FA94v;n{ zPf|!12}5T{$IdgGlbUYUzg<(nzhE}?RVX-0z>o4r>OeCi1mmX@^y1vZwQT@tmlAP1 zDC;vB?|^p81D66b>A`Pk4o(A>D!=p+YSGX&B=T?M6$xDX*Z@@|k zKI^80zT-C3QUs=-84WTN<55F&M!KIaJ-{qIziyTN&7Sd!PhpsS<7vmDO6rCC-elSQ zQ47EdpRDga-|BMSM2)f%-^|)FYy#k*`?3Mqs(6!O;X?O_DX zr_g+TpqzI9d-KX=t8B%u)Y+e+&y!W5ZQ?6r|FSaj-~XQ%+|SkEGV?$R){J+W_eNW> zCejd)ZK>Dt-^dhb7s z>?}6rvVR*TI~vM+T?w}*|lUbQYpDlwtIn-T{<8K)VU5r9k1KEZ_@)m&G-+881s`BS<9?jfphx*6OY8jJTmU+kwF+y+@{lWg}j-@bG_*aX;5cQlzYEU`?O$d8|guU zlFEVq$vMoBgEkIWRu<6AA?+nJyDn;LyXDEI_*OvtlulnPrL2tCKBd$;On_+1pcFh* zQTC*(G^?r_?kBXXmcBPe{Q1}aXzJ4dbXSE=Bj-2ck8~nf&Ke`~zWMmH z&^GyqgPXg>wyPE__^h^wZoZZ$upW{no8u^2C0D8wBVR>;kO$&-JW6j}zK&?OK?lw; z!#TyJlCV{=<8WXN+g4h#lHwsmG-Tb6w^VBUv~Rpaak%8kT%iXRV#1p8mgeKk(M4?8 z3k?)o(;0(`gvS*y6==SUfTlc*YG$(d6uslyW zj+rF=gQ9Cigkka;ML?5c=G9nG%{(^=>-*s7?CrYykLEG2?^(@9$Q?y;2`WN?hJxfm~>1e0nC3NQ3R_N*<+7K?D>E5l37Dy$oAHTi=bFO^YANjhFeK zKK)!2Xcw(kv>SH1o+lR>$HZT zP2i_@=r`r)X#kxweXgWZYaco1FlCkpsH(x0h6?y#mGoa?SD@_DkB*48xHrWt3o ztGYR`?RNEeXKx|5x1{=zc{Ao029mxy>|X7*T0RzXw_Caj$_);k*$knd$?W?2b{7`} zm)R+qjtWn?D^#iybr~ zPoF;BTMcl}BD^Cn_D7$b+k*ONHC-n4g2s5UX*A884}Bb)i|Oj``qcn)5P(o#ZpE`z zkLMdbCDJu|Cw!i_UxVk4dxx5^(*|625glkJ?(&b$oe@DH%V%Vjyw`RsKEQ_M!9jb! z3qcVHqmzInG0V&7ytw=8Ga~D2X_ATS#A`N3d_^RgT;u^CSX8eI%uFep?(OeyDo6^D zT$*;@FA?ksy%R1PHLzmQ#;H&*co9ePgHw5CZdEgzUXl8R8S7Ptwo)UzxJ-dc2En6M zkjvjTkK6cFt(0)H!0<&tV~QhjNcjE7jRDUw3ZLpj+!pKqqUrxiJq z6Q6dH>{39S>ZKAnDv6zzAKfmkAURZVTlw-7U^%x!TOuSbGa1l+NFLDG6oWKN)m$?k z*^PqA*p>)`yE4+vbWIZvp_B?noL1kNN-8ecQ^`=1x9yI%4HJNBlMNT?@{0KVQoY=^ z{B~bX=f`cgM+~@UX9?~pI~S%aLVW||Z#7x7`~z@62*J%vD#RThi0CH|5|H{y>j@5< z^I8M7Fw%_;-fJW+VFb;iKmCTPEq$0VxQ4i>CNwniYn)x_g7)T2J)KZeJHC4}ra;)| zfM<~5(Dbcy)b?nrw@?SR59%nls(>4U`^{H_`6ewi>qB#Om^?$cmwq3#mVL6QxdxSL z{0SH6cXV{lci#p@=GAdsfh*AEiInP-uQd0YJ1D|NBoj8%{(zFo#YQV%T(QU8|7rrg zt&yJI#VU{rK(>!d7;y{ZNm9*J-q57$O)^7afJ`u?dM=cPXYq`+CLdtmHFEqP(2L_F z_i%_exH0E`uIX|Z4FGk7HUv6Q_Nfo;f}7;OUK47F>ojO8Nt3Bar6d!)YEUxL$hg06 z5Ra3*y4^bxdrjWE&R7WA%w&}(9)#DH?hD0hWue)A4<0WBBa4Ao_i|u}XV7c$%tP9a zhAH|$s6y99+8FjZwLD?CNp~sFB2K(z2n6hCml~F=HiHh|LU>3SX=B^Um4{7&&D~rG zL>CBz0v-d(Ka&NGc^0uO`lX!dxDH(J);p&>+8vh(v*7=pM@(0T#PqOQ?cSshG*w9tz-k$6&1Bx2@4lMyMuIuJ=DKD+1do z@(_ASBdm%YIUt<}p!Oolv3h$N0#Up*T9Obyv27>EGJOt(;e8|Jpjra**t)L;BT+U> zMdd}Tx+^Q4ZhD|F0wv|-#0asU7{iF##ng`LV7k4AR4R}ISRcX$G`7B{<{gSwmo+>|KP`Q zi}G6;I?@3yg0{%K{ZkmE_bOI#hf*9ApTfoJp&ZuBm?jXUj|7m#iOQET+?O{?58y>Z zopTBx=M<9wu3+Zjlg+&g)%=D<1xl4Z0jo(4b4l(hV`mUg)tT(pYnlWn`jMqIsOpHBB0Y7Vm=~$`k=$9hPP8PaO$fZyhT0b?TyU>VQRwPaMQVPsi5h^A9wn5m2%1x~o4g`T1fq}TS@sOzX! zTWi)XrcsWM0TqPAWB64O@+P;qMWD1KC*>3g)rsT@5>pVHBK`eq?nqz3;>UIq9uHZx z$fzcZklWc(2YLu^LvQdA!af6m_>4=-x+IRn&P>lX(Bi1s& zO(Tsupxs<5Gv=ftSAbo(}hx6+%nC}#a{KkA675T4f2jIwwfkpi<-y$)rKlxPiExG`PS+6E3 zY;`lSAI%3$O7+C@#5j3|Nn}rlQE$2)+B$!A@U~bt>N4pmOb`o1N=*Fo5P^=98)Tjz<*^5y^4Hz$G@MmFa1LPO5=BMB zXAiuy;RC7IN-Le7G@<>Txf|~vx!v7QTrSBMjOO$$b@xQj$^tl%)Bivg&0% zG$D{Xt#p-J^qPTNO;E6yf#?CjP-gKftF!hL`)0Zo!e)LN3knHw3vbdFeJ%WMELily zW?jYIm@R<0My}Y_R>}jPhIOl}?Xubi6~4Xbrh;)nFPO;Z@31rS$7BfKc6<@1KYQIEQ^OxGeJ`}l zJ~3d-Ny!s`ceEMl#*zkO+ z`47Ifco`^~Osb9Wl&)^^DjmMBEZ}IcxtQb_QIoGDTUc?>VXPASVIi%D&@X(!JYM^9 z1iNa4UV8J>>d=+}eS_`Syb&L=&ra%v(ZeDseX|~dj_Kp_dTqm%X%n7?Zdg|Hc(Gvq zfIiRO0O6!MZY8}B`PT1yhlBX|rOUIc!AaZRcP7;PbTEF!_YhQz4#$e`-7mUW8cNSO zQL;Iv5@%i+O+EW^wZx@-{xCyTpxkh5Oc73ZUHM6X)in<}0VC1MAiyDcqtD+QF}ce? zk;aMtekruw?WB?Q8Sf)3zovB~T1|hfug~huFNxGv5XRV3r;@9)rV8!cc&+;g^+};N z;s&16^*>`8@GbfH_fmu<2ZyP(p8I*+%c?g@eL4fQ3TLg)S}PiLZm5XY`>L*n_+V$t z4rOGi@?IuL=WhH=8S7khzCwr+XIAR4SgY)zTb%7j!3@kr(q>N0mg4iebfRiJeb@^#)VYT_D_%+9~ z)>jhR6d!ljUDB?m3br5Sc(sFr6|6D}&7K@X|2I_=^c2XjJdhj3WntxPbtZY4QZHp2 z%gs%QwkS>C{dx3jf#SObn^}rD=2@=RZAY&HFWcGxIn#8A3e*H`==t15Wt3SwAxSde(@cqPs zo#|n9j}g=02^m-qH6I_B&{nV<$SaXJB05#SAFl17;z30+m?l+cwW^-&8G#vEY6hbofVba_~MR{$p!1^~__fs>lY~^&&cL!MRBhW-k!s zJHyLuoXWhpdSf#Lqka+bLi;E^N%`7K=j31P#>esO*AkpF-BniJFGIh~6gNxh7RR+! z>*=hpypwk~+Y=~bV53uIS=uH(T>MxYHzvlGV5dL-a=W5Nrnehwrf7rJ$X8XxHHyV! zSHif}*1UB03q_AJc7E}mrf=5E5c<$~FB?XTZ|mwgrkc1hw`@Kf*wYe_&d;Nn)9AC{PiX?bnzPDznA4P$ka4d#HzB)f z0p=QOEa!TJMKwZAI7LCOHoGrnM5M66cG*9dc5jG>Fy(HaQMobT>Ne#F?Z>rq z<~|=OcS?`DXXnS&IuHWIKMOEf#$2D6dj)swlj!Ji&@!s`=sU&BDWWT;wsEum%!yMy zG~1jcU(MYoaxNeW_PZsrPD@GlLW5~#)H~WnO+?f+oZhB+2khM;X19=%ov-(xLQrL| z!pD)27q!(RnhWFFY-}!Uc_tQ$Em4HI&)R8P^OHI|i^7vOufK}By)UKrly72E+xnSJ zb=jw@SxOGm^gIG$E&6#5MZPSUg2y12kb;ALa7Ka|!KV(AG&sBa=OjB_hwvFg9-i*S zD%X@jc+&k$-K7PckkOC^5XW!G@{r>)HEQHRw~*!NcngtN0n^b0=j*YUd=TKXyic0# zCnZwGOoDI2j+c26Ek(EK=*)h=rr%qyl2=w3bF?3F)t>%sL|jdu$6mY_pqnl*8!$Ah zr6SF6u>lB=&_;^1zZ9d`c{l!z% z*t7k&7X3=D(B_ydTlH_-$ReArgI#ft19K1Z{?S|rZwQGTBD>nGu3)KN^=tY4gxYEl zXRG_}!T#z669*fuRDoX}?92oZ?f#ZMW`wx>TdSE&w{sSbA34pLHka_$98}0#!p1s? zB3HPwCk=d@(0p3LFra1)a*A?_`UDLn)&MX1wn zq_;P`o-){w#@VJbeJxnYSXHImg=ujEmWjD*+V#O#YngT+i#`L6!-zsP80X*Pv!L?xdXcIHF~CH|SJy2PUzHp38R}inMl$!@t)nGO~2bIpMT8e)2K4>~k@`O{FJz!6CC}xzA0%gePZgpigxh+iL4aBDp)X*0E><%;po$+Krev}!v;&tbJ)!7&fJBgQMEdrdCjtD4SR zIEu^{Z76EIm0DTDf^GxBcX70BV`ua8qg}@wR;i8d57)h2+i4DRam8+%u{e@>K+yw9 zMnp?C#t!ndPBotse}`i+%nIadyoN zJYW5IhA;ac`+>mi#aM0qcrtZUw9fEHiw`9cj+iL}c}ZjHGAl5lR6J5ke3Pb!(A0o$ zNnFTdC40b8Hv@8M5AEKM5Ut4dD4rPh;RhBUsnrdQd0zGHAgfO=YA_gCdcib#STgLp z5{l%`9Hkgc=EYc7F)^oyoiNE*H)r{J#BKF5rr<_|V$l>Ddh#pIA>+4b3P`H3o3S!A z#EFP?uemCF8hS0OM>lt^Qvk^f>$#ky{EbIQ!9ApLBTjax<$q_lW(WzkUx9V;Ts{`| z)X{zV4JHn&cW2CH*{4V4K@o&*Z1Qs9jrP$`)*-^1p`u*LRPc~M0%&CK{nHH&4y*cg zODk-M`EZaS!?Ql5Fx1t37*i$nCsGP&Bn9-8gUj8^eq*8Wa@A$AE7riF1P;RRxMq7f za0^vFQez8DxH}bXsRudY<=S=NOb6+cj+`!`#CsaEklKfb8_=Upl!a%QI~K;1=vH9}O3; zEnrerV~Cr!ezY*~`P4u2(hNyA$k|M|_}ETiIAj3#tv-JrkxWblJG-Al42jaf&ba41 z_dZ}L{re*|Lq7^eoOu@nX`b`FyKr)&!OFoV&!R4VA7{S(NZTHlxd)jHM zyO?)n+;b`NNy_f$AIE}SbF|1xgFnYytJ39pv_Gro&Uq!nyDhbpF`(fO_k6XzUc369 zE;nJ7*2og$=eJeQk__QOh+~8UgTu#Wwy}^6gIQ1TaB|oK!$e~UHDm3F!x_IFRCi%N z;#XHP-tONw@}f+TdI!0LDf8*{xMd5)jFa3>#1iqc7y!UpCBDVBZti9N<+2&%xFD%+JbRkd97G$kdqR=CV}t z=Okkq0i6^DdJPN-RoRAiqmmc+f;tAouGClC%=o74;4b92!Ps}=-FRh#cdP1@tegBa zU=+miMsD#UMCZE^?7*UVCGRjeKl|yJUiDI?OK1};9Ad|)0w$O(Ux9TU17+O1jjj|7 zUOmtPjUdiPm=6Fr=mb-k1~vH9*ysuaT&uhf^Q@?H$hIk6oxiqxiSBN-C1 z^cbDl9Dien)qSj*a00whw76O5j$kti`i)y&j|twjqh>ItD;{czlmsp9_4Fzq>G(wd zkXH<=t*l>Sy9lAS5J`g5c);Fj)tcZKd1Z`RL>Fg4Fj%jDkVYB%zo8nVNpt9q0e?^kUA8 z_e#0`e&wT;pa1ka+?7%M*Ah4DT0f(p(9;>%U z+t#iyHJeI!&4B}jtH^uLx%0EZTp(4I!{a)RIKvimB{w z>3!{-qg6gID^RQ6>iercwC=NM@>D+O8&$3a@FDjz!0GY!V1Dy9ZiGD#6Xo|;eG=wJ z`5WL9Yv}e_84ls~_WfkHl<{MD`u4asaKIsKb-+{eW5j{UZ%6;mdG_wJYYIY1yi=Qr zh^!jbLkwg%a1k{sMk9!YImwQdCin4`g#kzTAvJYr>^l0^#Tno;yK3skP$wRPZyq=^ z0+e|OHLLxuPP?yWjd$}Li;H7%(dQEer+l9a4PB*0utzL_CtH9cfZ@d8G{^(8M}Q*$ z_nc<^O)s$f_w}q?UfhmRkA)mnl=>IUCJy6P3E@?OX_N3nI?ASJ1x^s;M=Tl098d!t ziRdZ_GQn0AmuLsd-!8izO(Yv&QMyrUT?o|or38RJ;7S0(JF5|ngcwM7W&$$-yv(bw zGMeHcYw8f0(=&*oLD1R7H#L7jK1a=_L4p+-K0_)*=7B3H6=>ZULyq=5I0FU)iN|vB zI};26t2o6a(e_OI_=g=B4hPw4B$vi561XTG*6W1JtYkADQtR~4Nz-ZNUuaI0>3BAr znzX6|_K;UZfSqAoIpaou>MK)Mw6^wfGH9m%C}k&3xKd|2LN(d4Ix{;(`yk5O^ZhnN z>XB47>*8HJiaR;7(b&zqk4S_zNT59kjr)Lg+emMnO~2Jpxn(Izi@Q#Hg&1cIK=v0dWuAz(y1SpZXi%}5m5MO zOKnaBm6vSMX!(`_xd( z(#39B6C`70l@T#0;OFT3;*pZD;2sIdGds0b1^)+vQnt}b2hy>&5ikrL88e8A*gnsBslre3N# z>hr-PDtb6VYT=cf$;RMO%6{CDn&RHCH6enP-j5F*J9g|m&Yt@fO=(-o3yreQrzeNc ztDmoxOOfM2i{wZ-MDwXeD?A*=qQ{P@WVQFw@P44GFT|biYT?yRQ-5Y7wVsGjT_eUc zY+Ri#A6a>?1M^+Gk%*KArievgt!)|gp;15^Cpu0aM|WF(N*jjE$EbU=(vI$YlIGQvUyU6 zY+-eDm#H`iL9WSjHvg7!9VpuaJdsswP+x=J*Pj@uovyuFn8o-#f6e_I0r%2eeJKtb zCJ(XzA)2i1`6z!iG*s%iFAy~9A>5pDWo|CB_j}ii35)FCROd(IzZADoy-+kAa&ugq zcTiN%E1)`6`0H*+;>eiy@?_|^+I-9%S%Z@{S$)=_Qu#8ULN}Io9y(nJ9{3{*3vqgN zPw;IT0j_Qf-=7rocWzE3PJ^4fh`C0$@F(9!y#JF(@QmAT-s3%g7jNWMwz*RB=q6)# zLTGit1|PSvR4BenW>q@0W+ngTEi$bn!EDABxfVz6khpU45gXz4UP^X*-O|H&HNrS| z?$@>GD;By1Ga}S1-&=yM?vKw^z8${L$&YP0Bd%yhVUQ$>pIP4as`y2g<&$I|`=60v zQ&is^e%^cp)Bx>LEUG74c*nfT_7-W&<)^6b?8Rlb)cR}03FVpMwkxo;{0e*Z&@8{` z#e8F%%>%vWc%LVQHX#Ljz(+qYZxNT-8heI^eCN?#HSGMeDy{jTO1XT3Lx!qdm-n3B z1%sHJEnrUdpiyp0_aVGaY0qipBxnyMTV0%Qz9& z{ci=k@FM4D<$9RnUS*szqZ&@xrc~ab+H0-UZFR(tld(fNm{Zs!O!^Lcq)-&As`8gL z-VN8p{9ZF?s6dlfWG^FcbP27n_7Pz9p*Mv8%lSIXU>*DN7Z?(d^BU z6nZmJ+qoEQbJ+7&cuJ)ULsirCyZ0eUb{uX9&nu9ZCW`U=0L0Wo@T5HL-nw25E(id$ z-xo)iM6Lp}`h7@_<8dnf@&;^DFkAGxGy|i#BdB{a6~XB!IH;7Oc~h)GI(uPl*>glX z&oT~(SY}FY8RvAwzo^as%+HvqvXuk0VbD)^KDSrVVYDal)A4hv6Y;&JQnxdPvDds7 z8ZY#O)0~>H8T_(KYGGVrkI~6f{55hAnjd3Grm}-1UW*?8RkjQHb4%mcyjk|F$n1NY zUDaknZae&f+7mhk=dIjnzl+`GTjWr@Y9+k*{Cu-&I1mfpr+s*!K7|#OyHhemPf>3o zM50s191{}=B^U2$7L3|-r$bSA76-_OXE1w}# zu=5^pHbSz;C1Px`CTk7fHceUkOHKDOv)c|cG5tuJtY-oYRT2^sX-|WA_MO^+2eXie zIgvMi;#;M)mV4O38;$BvH4e=af`KBq(r;H+S62nZMU!p0*>9COZhn0vS)nOK3+g1c zM$WAN0TUD}P}{qZ zNv^>oZ(s3Rl--q=X96O7F|)QO`s+RwHxOkK5~pRv5$bRDk(6QMuYcrTb7Rz3<6h?B z*up(7uzUzRQ+RG3PLh21`*^=YwY2nEW;Msv!Qw*uiv*g=R?en+icuCdGR$#a_yUj# z=v4;T9V+?E`Zhg})a>cUXs|u$3&`Wre9d~{s^!ZC`|+@#-Ge*Hsbf=9Kad@yO1E9q zJ$&e^ELp(>zrt1&d_2x)@P3y@Zw@#VhV*HMy)H?L>gdGmCAV=G;Ruar{kGkw;EC9+ zT7PQ(onW|sRCqJk?)L>+tN=Fc?t9S)jZBV(W_=& z?ByQZT*U?M9`kGU58rUUu|bJi{QJm72N>#7fX#QXQ=g9*{D0?-<~Bo;gnYW#4q|v) z>oC%n!DACi_zIE(q=Uw@ogT)aaqw8>fWLk_*W`>ce1N4!3~mp^Z}H!eo`&%MNk22b zqyuS|C|fgz81MFn6bVSjAx8H7&c(EcLlyy&9e{@Zzo%{`oq-unyrJPVk!^184&s|6 z#m-6tWNcTxYC;;JB-GArB{C4F0xw^TAX39XzE|JxS0nXa2t_zW4-mW#-Fv<-G#-G; zd=&4dO=-K{{gmW4MaK<5mm_JvlWiNBNG<<0b!*2B|UZJ1bG{kNvs^J*%&1J&oN7c>^9^BJIResvHKq|4~`YN-m z8Z`I*vx_YxC!{?AX(0lK`0T-YS>Ow;H{1Qrxo06;w^|_$xW1v)hS^H#5&(ma=1&&9 zSP_Ko8DZV{jENA?0&<6Ab5K;&_v4tadDY*v?y~0a`dMZ3@q~=AI>MUp`?qBUZmWmp zK{A4nn1i%lAl62x?B^&#zI+6BY=mYMX~LrWsB?$^vO4oCr(T<0$FZG=Wki&Jrm)XAwwFeg@3zFqMVO>H~afW-+;5~R9#Su#EbF`u$89Dp+y!daPG{_5 z?Fquf@uV40or47{X_K8beF?#E&k8Wp7d0e_u34!Q@63Y`(vy}`fo1jzZA71(^cB7` zDSOOiEy+^OVuTMgz}{dLXjLW7t<%r6s%;)llDP_R(2j0RLsGGoJ@s#o+dWWxQ&-ac zDiP;G!TD|`r!`NBM-TqgEj(oa@J^gA_R^HS`$ID)&9V>2xKQUTT3|oqq>^npOS^O6 zIJhXg`bK%v)R4s5q1qnzpFQg3N?wyARc7~oxl7ZbK$%(FKXjHC| ziU4O95UH!zKPd8;&1C5MVE#=ahv5+BqC?g0m*EiRI1R6jY&i>bD!8*|Bbtt`2d(~< zwAa33iGj`j0qAn~+r{Wk#b}g~hsUl-?WjHAhBkvhlgAYlnRDLr2g%3hI-lH5&&A|W z^#zieQjszvsK&pZJp5hngEZFm)phMJWK1D%9cNx>{u;Dt;2+~I0uB;JRoS8GdN2jb_;0*B)bVOVYX2=iS+GT? z-rS};i19V8f=b7XEC^^d#m5*RZ;QM`rV#hbFn&_kh#gF)8%;4 zU|WEyE*)6lA-*Tau~%-JbgXM}vTQq# z;oi$;ktL0FG?%Ki+I24+tkRw}8;*333FQYTKR`n+XcQ2_>(#g}9LJQljuc%W-owxI zR~1zqSD$Qb^bpJ&L5M9rhVUSwvZ^(5L<=7y2eV%_+xVQVtxM%RHFk6LgIeet>fBSj zx35kKTM-qIY;V{+zCF~TaK^&NWj06rjJa0GE~#4YHv4M?$4bUK=RdRn=Mdmg#1CYt zPR0?HYo}J{EiDu?o|G?dG`W`Ktm$0N7$(?GU!r$N@(Av=I#5{Kl<`CYy^ryVtL28j zjFD!_VbyxJ=gOKwYa#s8>q8d-9O&HCL6Yb9saYBIBm~&SGNqM}r7YVt0N2kSIzwj5 z_$#gAXURA|YnCl<&U>G~aG@b2RNz!#wMOQhtwc%#pMLJWY&&sJJ;>($<-zrIM#ZQi zO44k8=X-5z$4g(4q%C$F?dbcU74ISX1Hv}tLA|eoCoT_xDrklba5~=ZNWwyBy3TCh z>xco~_mBuYw{$xH+5?o>`Svy&_aePRsIVjav%oMBv_kc>@a+ID3Vdff8QMNH8)(wm zl;*LU`2Smb3|gm`oLM zp_=z4rfO5Oj6Yx8fgeg&HyHSDLq&G`q`s@Yc5bD$c+&fQ>~KX=)w z)JH{9YgCkpH)#ilotiqV&{rh>YWIQxqIp^z7>}cxYI6twVr6-=+e!Juqbfh?ZBS)% zA4&mE$LJWng$7^lwjEKqth1ibNy;Rb*=bK)NAt`N&Hdye6-_IEc@?jC;UAG+)d>b)P-9-vDQy-`@djYN?-S#yfh_pegk44TdLPCO16}i-z4`TZAXw+JdZo|wON#Tb{JRD%=`78_1;V}Vy1kElnsg=C?tL-oM!D* zIcf=|#LnkC<`~vh>dTHMzffRN=cBLm-xjP19@`Q% zb?HO3he5~sO|4itpSaE^nkQWw_hUdMe#a-fCqK8ZRDi54dd|!2ws_X8VQ5&SIP`O*( z{T|jfucT9Ihbe$F((UB)ma~3d%H|D(S?pNF#|sPjN*kcqR~H2KECc&G5_1LLJ*DH=JWY5j9O1-~ zBP!ncaRVr{-aJn=tGket6XxNu-}FGCy)9U&9?Y~D%y(PDszR6Z-6>S~W9|jK&6#Fv z+~U5fIQ!HW+!h>U7C~3S^Rd(Y5y?c_#4S@C@>qHQ`g|4Y!#*qi1;OKqz?sudw=E=@U0F_gemgY~T$+ z<|^p@AYWKCpJi#`_x$vTumRGc3u=ifTm=Z^UJ=Nb-~!{*XE571&qqbx z#9Or>34o9ig3xNBSaEG%znYT6&eCn9^(E%bttyVX$N7xGEs#A-#7n0^JE3IPwlh8P zpSB(nwH3@b?EeQ^ec6Imz=wv0Ahy0U;sY1no@7B}& z9=VR!Rw^866;FML*7XDv`1i+SLb}3{a>#za{O`U#zhN(lt-k~)L(t7rU~Y`CUN$MP zh6~bet-w}NK2$u*|Nb7pqkpH30YPbhZ!zuGnQF8sa_e4{M*upyxbWI5-TpmgV9Mcf zgIZ3=C)wJbkPHS1^$)6l{}d&HVuHU`?{8E8_e+6wo0iAk`}ev_zWr~i#MUk+Q348{P z8F1bm5b^gQ42SfPgC=khNzfdMjiDy8H*yNh7|F|GUa)JRd<5E0Z0&*YJ^!FhI{|yU z+fd*@K?$-8F%Tfod2ebm``q!OEmxQ(V|%pVhhp|YXT!gnl{rTJfV1xXJ`_T>#64((`WnWeRYd7i1bA0 z2DnxJ{S$@c4XCF6PXPw;Z!Y6?cV;~y6e*tNOixp}MLUNX`i2l@IJugDKNqRfu5F2g zF9@wHrweN$YA-aaiX0d|`*F2-u#I@5w+$aBQQfwpsC+-B$^>cU zb=><=RZk>h*wi9@%&O&YJsSE1O#Q`vXc-ym6LOM-;57fH2_W8BY?x12)iq+67NKK> zK8ncUK-Av%Log=m3)~Qig$80lVO26HjWNc_52fX_vGLOB-WkhT1W}sPI(e?`ZifUz zqsghiR6nQbRlcVz}7o}#~$7p7nz_YQjfd?m%65c?}1%Z4S z*fo)jz;EC75P~2++0RJ!0t$EYm%Ed0do4|@y7ubHkN2&VSF(lLwQy=z;>P@Y7+O5L z31@XC2$?x|DME7t0K77K-kE1R%kF6#f&Z%__RfUsA-aN*((P&fKY`@PT(!}qU2?so zrbiB+w%U~B_^}Q<9~EdAu&}(j$W>b=vYJvW6sMPXt^qevf~}c$v^tH}5sT(_txYp> zy0NO-bNaDAH02wPJ^GW;Rd>41+j(|c+3Jb>dV-k%1OLnCW{1;G@JsrlD@1-o`Dgu^ zVW`aj?iOCn|5Hz^mF;Q&Pnufk4dBLHnXN`#YBw*36PAtg{%ko%8Ng- zE8q4W(c-z=sGsws$Wsh3#=;jpv_KccjBwjC!FkVfwSnM128t2qyP)Y72;ExV)fZDd zsfS}{Yvp+K?jy|lr?1U?5EpqbP$NfmS>eP9psi8nHUwMZc+2I9$`nHvgcqR6@YvA` zPK7PO2&64wLA-mZ0zF;*!qC(SB&-5GhUvIJtgRz_31?ebT7K(ixrb2sg{riu0FZeQ zd6kF6M)2M}o+wC%mVQ(A!C}=JNHxEFzq1jd1khu1V>kJ}k?u7Hi3=Yjin+#{GUV@O zKCB}8udh`6t^h;)%mw#dkkVp#IEgy1OvRC4YH@Nw78}2U8aK>X2yiWu5qT z)#GU1Fs7^&RT(2iKZRxfDWS*^8|QM%HUvID%yB`K>s)#J-=Ykaj8~&T(HHCEP=aov zPUkqbioIF}(2#oU)oWL;#Y`I|g5C;OajFMY- zsYgZVA_?>07~6Ubq(Y#vp5G&v^oCN(|E{zE(88m5hWITmB30tVj*awxUhWN9C}03X z!Q!7+14{{7&jKZkq|f-`PYL+e3Lspr#P+Or@{EpSFpQSsH?G}h$$(5f;#9qfIpjb> z1BI7ph~}LRLiXfrgu0Nop0i~j3n7H8zsI16E+x^%Bi##MA84kFb%7{g>$^Zs5rX5; z&yKF-e^P|Dw#_r+Mj{nHX?H0iI0xNb&}j?s3~?i&U9lm-)Ey3m=I@2>M?XC*(C@Xl zm=F^l3Pd7K@jRgi!7gZv6+S4}#qLI2{r#BZdMEFjJ``u6*>?P7$9IS0`CDHRg zq5-cYX&Myi`3;equail(&(?6DloU%NNsCy>$CavmBLG@2hDND;fBMC}4)>HH?P%#M z@?rub{Q#8Geg2%YLWsHZt+24S{RSTc9a6=Iu%6v%3}17XXeG?0D10|(Q`?A+p+qHe zvL+?l8&Q;!Iu&7sQ&Ck?>eyKUux^Rp_?Oe838fyq&*@PY;q}kGdb~aM^Fi?qNIS?M zR>HSvvw=y;veTYh`XuJ6e8B&gEmZnA@r`%y*XAwT6|XdpvkeE8;&iVM$G-9n22ziC z1>i9mK1VwUMSZ%axby7N=UIMT7El}tVizZ8=TC||_Tv7Nf53-)25D{bzhE|~&+|Be z>(>s_PxHhLzkDchB(&j;tgiPL_-`nH{Kk@5&?EN-?QjwlcUoogEd8P$cA@RQTA3<(Vlz56%$4e5AfLwMK{a{TX0S}VkL(}1)8I`Hr zk$YQ5xM6Ge18M8;qba*pic$nYTGH7h@#D-A;X3g>m$uL_kram{?7m#(C0!a*29Xl#Eh+s!OQ=Z_4oav4Op1A15ZZduPe3gOg?uO+-cpg6 z@2LP}fK~kWOGnYwCw7@2jq`VW7bO+>EpXDr^jh!UhTI{9BFSlVH!#?pM$~HoKwFYR z+RBB1$||B}Z1{xs*oH)L7MG?i`{aN0up`><7-A1QQu;0Fc?iTt`MV&f4;zNK;=dV* zY=6H6fWP)rPjF&E(~m7Ru~iO3@g6!MQzhBb7LypsVJEHrBO zo3}6qY6Ub{1{~M(r;&v|2YrM5WxzOtn4ekR{w=ctu}IKq`1j9S{n%(EKXg5Y*Pa+A zC$jH8mI50M*tu#~uRxc%!O1!fW8;piey7J8uO( z+i~62+3!I~ul~eAS>6$} zKQ@tL856;Q=pf@!4T;v}+}QZ$_t&B%8zfCMix z%&&t;L>mlcKI?+TkS-K0M zBSjLM`nyth-i`zrEXWL9j)7TdC7bqUdT<;NGo?i?<&U*#haOvWTwGAKC#*T--%J

    Vu49Bf=fH)2y{j^A}4g%^B1U{9PP6lRRGU z`O;?nmh-jaH*z~@C~N)ihpgZ6OfrnqB}8ZOLG$%QJhN`J@M}Xftajg_Ba4qf1C^y< z;~hkVE23K7A*T#rQsB%*JOC#td?##UuDPLbMf{(VYF4{@wDgX|Ez{z`$ozazwj&AF#OyX7D(nR_t$fb5zb9=Z7qm zciLDb5UXfAT>DM74eVqWR=;!EhH=J4pg!u3Yu<&!A7z~*1X@f|w{tzEbiUF#cy3h{yM5Zt#3t%wa1(Q}EFksPm6tPnOR!&_Ny|BJe$sze z|1IiXQ))R&@J)Fp2W{C8Qdhj&8B?D!NV1AP(oK%?spSdu*-^myOK=ENU>D2F zN?oxHk>crvKV9c#SQPYhbAuD=u%GUjp_8t^AInlYGHMzoRG%SD`iaN9&t5~YU0vF4 zAf-A0b;Z7mDr(fwPz=%gQKBxLJ%q~62!oXYF8uQdjAHB28aUeG{EY?&50`eJFKAwmSF57zfImXQWR=UDc`z;!n3w?Kr%9_xucDChswih)J1%& zp~5y5ccTjuvxuuOv!eXryaYEap}x6Q>eKz`$OGa*_5$9sW<18n6a0n3{7Dz+7Len| z4K%ff!a33&+^f_MyhEH-x=8Ng(^C#>mr!54I{$T>WaWgIMay(iBkefS(!@MSEJIe+ zP!Z*@`AXqjt_@4G%dY)Ys6jJ5I{r&AC#r@yWeFzunVXFGx?<+xgK6O|ONt=#fyeY+ zdR@a{iEZkXhY!t2g)V8dw?9T^jn_@EpiD2VyNdhGT%jA2`I7|VEN&?!E4%m3wlDi6 zbc| zZmsRc=*Z{Xt(hjS-8--*tQgq9ucZ0uLL4tTSDo{6J|}GnH^YRie*%9GvNvRTVZtW0 zxyas#hb8%1eCg&5<-B^DBm$qqp&tRXnT)i}PV*Eez+*8>Xs0d@5`yEW>`Qi)1x(mW}qbqmixSG%9HXZU!x-@d<*syk>=BjSVluBit zZLODceDYU6*V4D{b+*%{6NXKJX-C+fX!Ks!yu_>Kz7#(`tLPi&E^}VH=;30bFrU>4 z$BnhpfOavbYJD3d|s>0I$F>ljWpwI48?6eFI}Ia!g^N!2;eLSTjTc#zXdlda@y zKYcop_W(IHZGYlWoNY?Y`l~8_LaT|q%6uo|rb^h-^=X%d@`r=*S+BI4LPBl$+bvf= zm*N=}@5fQpoY!t@er^*!9aw!anzsg1u>T^mUFWi}t%quF5T5Z=Z*w|!n}f{hSlIoC zw>xK8PoCV1{Q}*^K|AQsT`_TdVzsJ>#c#YeBPfu>-X*?!u5j7{6cf)-&1z#K{#-01 zLpP!0nQ1Pwb){_<{i~SL07apWoP(gCKyuuwt)S@9qFQ{P=Rgu`>9F_ll&LrWBNUhn zSB@MoS@@|+l1@q>N?r&~zOUucBc1KN_ug_^YUM@w=>5n9ekhQu1(shj7$#vuU&yW` zy$eK(WX*kDM3i&>{J3>mKB9-|1zGOPW2i8(`2QmqaL5!ji9)-;^0KM^6&bpbiY(&W zuC~JnWc(E4V+K>IqXvnB;d6JY@>!C!5n-+o+Y)%_I-}FS7xV2J{e z4F8|g1Nq>pV*@J;v))>$&s4)yX@wz{<8?%p#ABNk^uV*Zt4@O`J>R9vJ~@X7Pz%x( z=^zJ`H3St2&(cT!D=a|0{1FzSy!IqPbNmT>E*gF=HAIY)q>(y*W2g@hx{~BPtQ9BU zy%!>x{1|JAdStj&#cZCVYc$OI9^CD;z@MO#^%EYO!m%BY}7G zkz~-isQog(n#U>W{D0x^x@Z8uAYtF2=c2>8&6cUPQ&YH%q2$dS26dooHQ9q-_fqFk z9XJrqo)@o!x$4fuig8Lf_wh(yavuB5b5kBGk9YvxkOto1rCmFF?R)C?^_#pPnkvXUQ0 zPiZlCeO>u5ou7W((^iNY%wHIN-5C;4Zp*I}PPiIKT#$$-6OoMZozFI?v1evm(-6q3 z%#J*j?vl;e#On3V^ZEvcQUTlIMfmJTu%1j)EY2O6#g(~it|A> zr;-^G2FE%~<}$GSrfn*mD^riyf z%>_pu8G_h6mMUqplvhu8hqMEeb3T5O#@6uk-g0L7>6-nRuRh2eoQ$M`Jijx3iRrrc z@U?=Au+sACi`q@m1tI(`cupx*{$|b(RDc!8gIJjDlb*ia7;5t=Yjl`GVA1+Uykv(< z2wmMR%qP)%H%bO|HLG&!;_2vq21Zmip*n7^B8i&qQkmpPt>Rm1ItA=v?!ED%S0B4U z^xCFu+M3}`82Un~e^PbX0zW}8?u~ZeDEnrmpN(dJ?OLPoYV?6w`x`gh@>X)K=YI6O z@pN1$28j)Nghwdu_MoJ4&SU2^`83de?c21QChfcG0+;s=`O@_*>fC96-91t*pOf~^ zHRH%dTAA`94W=VU7NbHg$hGuWZ`P&c@h0UiDaPgxGYieSS|wRgvzJeAgz{3=+KfbF z4LHCUBlS>6mnx zYQ8D5oq2t9xxcptYpl{zB<(myH9HoUlsmuUl|sLo@TyYFV$V#nXIDPn1easNf&KMt zjVeulsY~6FSv4tGt|m4R$DmD{Br5X4v2l$bwBlyUUffug9o2#4Y>RUY-_@EwS}+lo zD;(Qh9DBXA;^+JOx;U{X!g(#$XV|A*Gkj?^dTQmv`$gm^9`-Me4*A)H^DU(-^cSx! zXeIdb1|Yv>E{6=Rtkf!8dhEVR?!RI(J5};?&?6DZ+1Z2}3C@7Ob7QdNzywyfuV;x=hD(!~S z2suIXOpwtvld&Mej%`kIQ%|>H0>L!&_~GCF<8H^Rw%0#|+vZxTtfpEXpfYw@79@4B z&ocL&Bu4=_x$b21IUo)}oz8?fmk)>Ck8$uEy{fF+73nmt#vNezoROitz+27vRCQhgyvtOFgb(n~x z3Uo|J^IXPGy;}xb1CiZPU}}@h1$r^cNeA8t;YI%uhlTgAU(-0u4&KS{PAfOVb#cxu zYG0y3B)dA-V$`6X$SQZwMvd>O9-J_o{l4!F?2QCPkjSt4soqykdI0WYHN|z&?!z>G zh0a&a0Il9?;1utH`Ropo*>!lxG+VQ8fVU5XYx&6Ap0|CCxDi4pyBMr2A|jv?9g(WY zIA~V!T&hLr>U{MFT?-8YUgI0)(t6iFYO1}3i$GzJapL)D5fL-Mzg&KyJTD#Z zWrgZS9?sv85kOVW@bHol{xt>wfGe!Q)f-@|Xe^lHqSC)HB&dLM476!I0x}Ws$bj>4 z1BV){Q-ls&Gi7^;q$NKj0C?e%VI1vXbO1u28LkGr8Pr9vNMI5G zA6U0vN7|kWf>9Q~r9>rJ%Rv1G1dx}=Cju6y#F(Z(Cj0DMFV4Rj!ypLdJu3ZFM6fQs(ZlY@4xt0!A;UT)|c;g-< z=24Zfm$v%9DmkBOD->+xZlC^e9$i(#GSueMZVA`)?We?KI#_%1VfdTz%kFAP(;JIB zt5KUI^PzWxZ~B4_AUSx!rGL>a7YCq{Rl4fb$>at!+UP{b!RKH`WkiKqLQa8qSAJCA z$+<(cb`6WFbKC?q1|w&Rh)U*&!D5+_Ue`q9F=Xz)Jnpk77};tt^g0m1rxJN%hPvWk z|2veSULOnwL5X*8z#jC(brlMP8h9w|$DRiN!6hn;bq}}BHVi+TUr)*Yd4RvxgEL|$ zTf9JJ{YqEY*k9lw%c2l||6zOm@RDWv-a$-(L3s(i)3GOUypHgp%D}v$_y=Ri&oqRY zh*!^hwd_l|lsdg~SM@w==gctd|5azymhSFR2hRt)2Sa;yKcN%6EG07v`3@qrj1IF; zdN?p&cUVJnJ^qTtDF*OPXhbBP0+Zh1_-5yeS_Y3b#&vti5d1iDeBeYTuI0$;oCvv$ z{CI^9yQV(FHx5lrST78;sbd2dos**JCw5M4-Wh^;1h`jmr$7K&;)0wy2>;{Zr8ZeBn z$$q#^j!1<v>D6nkttAY+B@~&V=)_HzzO(rdpLxpUsz{yHicX<4%q1 zt37JB8b$F-rRvga$6DUHM-7qQp}RR%G}z>6#_I29{f!GOW-EQ+X6daP?{7RAt=BzE zINoEkVVe*d09Gzytj6H$0;$8Z!_RCkjjqARa;#2^u~AYA)ZxQSsNXqofSR4V%=4|) zc-bxh<9(5cp;C{6@w~P}=w}rdy<8&77ULW?2yRh*^*>GlAHyCtZUs~Jh z%7BYy`Sf~!=-W0kf4^K2J5H{YP3b&-JRYynlBhBp&UN4bIk%^!`Gg68+aNh5fjHGr zLy!F0OFG}%yOj-% z#lxi^vzvPCrgL4(_cI8L1d?G4&8n*@Y6;k3#v7qOI7f4UpPAMC<^yvbP0cGnQeu5c(K7P+g%iQXb>67CZ!9lP)t27mE>%s1CtpEOjarwq}fB&5{%)d{! zz0kFm^K;any8S~JJFB?ocH8*+dJE?kVfD$q{T+9c^kRteu78*vw{wSgsa}Jzm}AAG z_gc4*;^3;)zVobA(~BQgF+YH9_N07~Wak8!55s23+2 z;l5;k&Nja;dD;q;<5!#UerjBL6nIfiUfF>=?*aDVdcBZ9oN**+f|JX>qgbcg$X zHs`J@aBB#vOA-hJ9FL^QTLKIvRi28t~*rC zndpqBlkCSfe#M=;+UON`-JxpEMCWqHc2wSus&>oU7jce?Gw$p@w(%M6g4K@DwVca6 zaw`9f?yjusi8lAx2C?5APgCu(%q^ORb1*Hu*?!_YIb2Fjd5^ zUY@g81THq5PBxA$lfWCvq+mz@4(V!`TYPwoK}>gf7m6`KD7cMdms+kS*p+~(nr5l} zpC8DSU?7AV1qXHY^!9g=7?;V_NxC`&Oc4rhICiP!YJyz}n5q_I zf?Wxis)R9CF3;7QG400eiPK(E;pVM1&p#NrR~~E(%Dz;)f1db3y#`QBU_Ce43xMZ+E*UOej{2Y2YX&yqu}ta=A~jB38t!ij6H>d+c7|*)b;cF@%8s_+ z$7*m@k8ImEZnfC7(mPLTnvE+7l@-6-RWC}tlkl;>6^kj@Ezn!RVaOc^QRPqo#{O2$o(pw9m6T-5%&qWO+|po-?`F;QG8AT*htZ z*aBI*JU%br34c$m$}iNV+w%MeVc<<L zJCvc^IG3sH&fKY$-FRN*AaCnB_c@y~aZ-|4nZOa1>3{XubgQ+eMsDN@?h*P$KL3)D zU(P4>{+oXNcyrfdVDfcwT;*@Z>%b^w7jc?IW;LSzE~w*!kAAD%Oo4l-rt zZYZx9n1AduThH3}^1P(I{CG*0AUWBYxfADY^Oh@b6Pdg0P&G1>>`W)g$F}>7JGI%) zGw!lO)yPb8zOfyYw}Z@XdHW#bpg6n9{;>_t(Vfxk05tMZ$QSgkW0qk@8%O!%xtmJE zbFY;`A;Q^Q-D3Q0!Y#%=;PoFSCSl*bV^UbNg@1={CZC1})z3F!iL3Rz#|-Z9rRYE5 z!e+Y4;|r@mSMy^$yIQ7+gheQ1I#o80MJPnL#n^{wm0EQFAY53>LLI@?)6?HsV$99e zYhf;;xmqGDL?P3uu6Zm&A;K-jJ{&Yxt6aGnMBFH@mY~h!Ou}5WT+JvKBFqJ>f4mbn@v+(Lx8$QxrB3Ykuo&0`S?5pFT|p~aX{E<~7%h%rW|=lEjGx-h%TSr4he zc591?dSAF2-Pq_+U%1@e5BwUR1E8W-9c-1A?p?w?vZ>P#TRVlzo5Ggc(O#lnv~mjH zZfvajvV!RqwnP8ed)&J_#`wC|zA(Fg41b%Ew_4j*&+ZsA%SWEa>=_+Nn15~^%hix8 z#(~0IM=W3lRSF77^`U!=(h~{dkumFWjr~2kGAB70F82j)LW}kIM66QZ+9mmzfGdR{+ zVmvGs9xTRu6f&JEoX2bwBHUu^L!B|kKU4a_EdLDP{s%=})Xc}rPhtQ7002ovPDHLk FV1kQ!FZlof diff --git a/generated/webwork/images/webwork-1376-image-1.pdf b/generated/webwork/images/webwork-1376-image-1.pdf new file mode 100755 index 0000000000000000000000000000000000000000..130497b56022a7d4059802230130dd83e36a8d1a GIT binary patch literal 7530 zcma)g1yEc`xHayYU2tEWzaCc7#1SdEN7Bs#y3~+I{=BZq-+P`*z*#d-}WQJEz&S6_lO;d4veqHoqUD2!MP5J^<9#i9k}4 zR|(<*1AFo+x!AzK3Sc`Z2+XSic5{F^0)V2TLekO%UNBFvjVpoQ;-T578=VMQ=MJ|Y zjKsEuS2i&sd>{;ZpKzr2NAH7S5;E?vVAT^O!d7H-*lpA)ZkPegm#ADE^2?aj*~X(D z1!&MdD^?VgGh>f9u?>|U5xDAD`nj@YuJ|OS2%5hA2*RkqL{+9ewY@pG_T1*GIbw~m z!Z=AFG_9^Zt}&>t;@-sI2cxs|mufP_q)naEbH~yKU)^p@e7=&y5UOe7JD0wk3A5PX zvb|UWKkUQJVi8><)FGYlIIq-qRoF0?!9;k(x(Z_8W4-CmT|Cux`taQ+&d)6N z;5I{Uz7yTl(WqsAVj4dsv8-zKYFc_4{~?vqX9pcPu^qj{&?8tswfb$PO})ebVYA;N z731PAqnT^tEDo*rHD?|@Q-~1@UMPgUcvL*$#mhN#Huaqpi+2yj>*)r4Vx;4$UXDL5 z0L?M+7!&82OukX##q~#%0IMJ?kSaz!NeHg{)6K*1qZ20oD zKr*?a=LZ(d2imOtJ-ByWKjo@0ei<4aT9$+2N6RKDr7kBV`Mj+#>~X{Kc`)>4!{Hab zRA{j-pTz3WafJ8C7w|9xXBL8)YP^$KSsK5^2F;-5MXnT5Zkjc}ki&^prevC_HYIZXt6lJdP7%A(QzN zbK=4zYvMw990YqI(=aSfm0=-sGdvD~w~!grBf+;35%~E2P1y*$qIjOGF^Zk|iB3*w zsNvDBSDM6?`Br?0erqvfqj2nna~#d#{6ewf!@vd=y99B!18^H-k(Ih(w#m2@1T?0a zAbnj_aj7;T;C+)C)foMYSdUvej!t2A1}~X2*hej8fNLu_)PHuZYis*s)1dg}vDrhv zW%0nGeC&8aXLjNTy*|}p{M&{+*mM@`2GmAxxB{=^Ex*Z&)TXzq6?DHURdtMJriH{I zmQtp9W~ZHhtY`^-E~8T+?ha{p>pjoO&OKkU-nbaW=lv^4{FeVv@&$wK4>U|G#x+!LdqAi6~x=@)$KD)7Wl=ozkE3#b{l^5bO3!j*bJVBw-l%ZH2PHQ@b^qd19{-UheaJgb8{6_DlOwyDxQlMJ0#rskaKNdhAW%@Ts1W*$R#{?l@Mg^p zBCTI3;hEtm?wIDN5>@81x;RXsF_a6)gf&c6f-ceE334kP-UTXRLn-hdb0V$gbvrbg z)AmO&7$xr{ESliS$1az9NuH^*30%3B1jHDb%_Pm{aSD@TP(Y!}$|qYF_AS(mh@Mk(u}0;EncG zhog9d`4>)cq=#rA&dhtV$JikRQ*7B4>Fu|xfp0eW)gvzS(_!g#bt)zLdGZVc-W78n zz!H_EUj@_>m9pbOloj4eE!I$r(sLeto&B1l4>6Ri&q?wPPjh>?#833KrZH?fcrf?^ z?OW9bZyx~cMJ)7nTg}m)bDL8Z*X$*Nqemr5Ub5N|@D!=^ETXs3SgTRRP!9ui)T<#e zZ7l}EHs%i(KO;J248A;(DyVwl`IMbqC}l$o>l?_UzM?36WahQD8v<)bteq>%VB5r5 zDginY(zf{}SDnN^sVJJa1qLB@nmxiUO8o?T_i*46Q18((QZ*|K;w|*gn{S^hsScDd zD;vdnk~_om>Llrdn}g zg{H2YOWNvhZbs%W+eG_CxE9I-J1M!q&e!5^y^GY%&V8EHGw_{c`z4=yNv}RCn0Mz{ zsl(kn8a>gv5$le#=`ZsJkFT4HDsbL1=yA5!6`S4o?Fu?g-$t}6-UP3;LvJt%Jv9^w zz;2-bUuW-Hzbi9P@V}I3egVEeS7;0ETDJiJ(aI%3f{_6wzWgjxjQ&uC;nd$i#)fAu7r2emF@lXE^yIJMW%Z*QZReqO|HxIExj=mASLb zQm_+JdgH`K8db8*jZtDr+ZqwFdt3WoDxySgm!i<>>g(m z)!**JBF%Qd?MO5Uh<~=6K$1 zsS}&h_Fg+PcJ{FsVYdG0YJc9MT+87{nUVM@PrDyK9vuhiZlN%tF0K#%0V?4?pwjcU zh55UKdG+rCQ2t-UjQ~*i&&b(%nOmdqh3aCh*eAWAhoJba@Q5FOM{%4~%frs56g<=`IZ=^Gi*F`b22 zTUR+)TW{uCX`7^G_$2d7DJI5GfbCUyW zhuxtpU7yQw0hF3bVf<}BBO_y~v)ugMv=7kvfNYR1{#{=NWe%y|{YuXn=NTT9e0DBFy2Hv)!@zKq zXD-p|JHEAnZo7I+s)$C{2hRJ4{Sf(MKmK=h$uIa9acQCb%=rc1owzg_^HApuIbfEG z)~4v4a{aJJ6%aeRi!;YM;P-t^ zGf@K7WQrDRpSOJjiiCPHbJCTzMYuv0}#eEA1M_B_4U}HSW83 zIBgaJCh%op9s%`2+;pogdp|~zT|~+(n|M*qpI>UWI|dKqmqun%C?={4D{FY<3V|If z=#n-)%w;X-v7e6K{-SSw6=E&oq!OY*FA%@-^*f}WJ)xf1;i;WNpisdTdG?bbQ5hQ{ zCWHtZt8`O&+J)Z6Pjnkj5AT0KPV5il{&OY^3;l%y&_gl8exajbVWFX;qZ64JDrl(b zbIU)|75#lLF>Bg_B*oO1q{=B zXjtgzcrd>xK0W~+K0bbVsJp)>#K92;;IQKa0L8>a9s~IK_{0EDUBR9ZI~zB^GaHyA z*wqGhcjE%kgW5sBFn<7tq$3RGF3!vA>+8#7mQtd<^h~+*u7>uou|V2Mhuz zLET^gO&eD*;P;Yweg}D|tGoA|MFBj!^Cq5dXlUq|{Ahw`!gmY2k4A|`jfRDRcK5w# zerO-iZBfxAsB-uU^7!nq3ccqH3ERU&)*!7ZbFv$&abFbVaGjF5o`!tt(^aDLBOg`}=RSr(upr{^$H441T&1@IIfh7fp6OnC71^iMk) zCuEH%u6=ZvGy}{DXFWa+J}?jmUngR%g5gR>r@v!dJ+ETIl|^Aw{=~jc#=U!`4+8a| ze0it76utQXneYu0$$(s=gs&#gyd6Pwb~U_*7r9+H}4=Ngki9&T*`GsQzvUe8e)!kyj) zmBU(IO~7mMj(TSo7&_qyk9@~Md$=Xs`0ldub?~ro^tXcJk@ys%D5L8<~?ZmRKS#a5}2=q6D$J|8?SY__9d3%(1O|oo_8`Wvxai7&|;z z--uy6m(3eg}Ohwh? zrH5e6aAVIXUwk?OT{Wr3Ku+ha!9D#iYk>?F789Xa=S~6UvyklRDZO;X_ssOnoeBZ) z)Tq`atlNl$?IfatY6~RMWhBj$EVO%_ z1^{H8*S~R8(;#uBmMa8_STn6m9WVq4imOH{4QAeH1E)TQ0BwgMB*XDD!}|Nv@8vJP z#qNIUFKmE`<_zM$U{EeqwG7kmJGN*W9C$Lim%c>%{fD8-_sai_kI{W71dg72M z>XNwcL>|-WQ0|r|rKv%R^FA33WcAeJ!Y#5tM^H+7;Pap=OfA8^W!54a25;;ZeSIOT zYQWYSbjlh#@YypwexF=b6@m?(*2il6i8OMF4Iz_RPCD5V0&o&*RV0`NYjn_)Q`*i2 z>eq@=OO`~Iz$T-o{MNFBgzKg*G1YLO#s7esz#pjncXAc^OE$749ID?(ryD+QGRcTf zo8;htQj>Fj&b^Q$!nZZ%s+tr{HtC_xN> zx%OS+4TfHFz$78dkrS^|zeoY^+VPuXTtx+DEJYpFQ6cZv%FQa>Qmzk1CF!BY*0UcbV{9u~UJV@J_e|1`BZYetK{W3aH}cMv;@p*gkX9<{1A# z8G}(exK~2#9q@8wZjG;i-H!KIP~x=lxMubxe=`BA|9`$_US# zK&=l&toV<9S7rlLCjVw!(`Jxs{-V7q`p3N=;&(5`9GpzAai3a!1@q!-qj<)()6B z>r5KWCT(Dlw}{a-6@N3I@OT1j~fx?ySTXD5t?5!Y>qKH6Bi$J$=k*mtsHoGLH<(kL>Buz>TB z5nG1H_I`E0+cIX2=G*q>{Sf))X@^r2iR#8zQ+6vyfo$Ngm7{3h_v>3IQy1q0|4^E; zchp0Q-ZeZPi#fWz@}G7XX>xvb{bWV(3z8^3_{5O&*2HVd1mE}FbJ8*->fXRm@nn

    QIC5{N zsLcZ;@xF3$QpdYfRM@$J))u4EjntTax z)i3PSeLPe;&#ItH>xyI3hb5+HuMK%V+F$G9%<+l6>_k@sQE1_lvQkx_`;jT9+{GNc zwo_-eJTTF=3SN0bzv+2j>KE?o9P*+^N0n!K`!(;LjQ0zTyHy-KYA3Ho$Wc@(HSSu@ z#HLRcP{`Nbp15d@;0~_f1K)rfl0M3DoH>@~((3Rd8eWP-$(b=A&U}`c)}?AuiRRi- z{G#d4)}=63+1KBhyMH2DWp()R6iA1=vbYpwmE9Lq!%J7qlX zVX7M=qW67w&v?E&>!;54dhX>lHpeKz+*Hb3vPNSi-N;+fYf`fGw2jOcxhJow&dBbW z$LnZLM%Nf`D`6ApaCzUSa(d6ZPeo*Uaf)3@=lo7dQmVK#5}2<7&X)FaX@f2J`Z?L2 z(P({nBa@);mG`K?gx5brpjSZAEdS}&vVXl_o#S`LJ}JT7v2s>fYWBESqeB}ZQp2hY zr(fo;*ce-BHT_ic(M5uk=TnPGwv4>K#~a}OBt`|b+Lhi!@RJ61HZOLw)l4s+?8B`> zrv124Qjm4SHzoetZNT&|Mmti7Z?)`V26Z`^0mO5+Eob+&oHojQkHcLD29x6W`XAGj za?{Q5pU-EktzJEr(Q@BOn<6Z*aQrSy5_vCv9TN>p)v zeE`J8Q#TZnOS`OTrJ*DfvY;HGLi<_H65z3*X39jg6oSq$Cz?9!{YJxt!d)h)0Yll) zd<)(9M2_*tQ*Qsg+gU^4cTx9u_yL!=52N?Bwf4|5xX+}%A}^uvI$r;qxz4_z+RQbN zwC?4=qQh51*=nDhTp9O$Po6JweHOFGD5#g1kh_>UpzE*pRt8;=;> zkpBNR-LvFT~s0K=|;ZIgkl?j z7%h(%4JbHe9U=Wf7>j~CC%wJ2!}k`_U3F2%P@j+eWgl0*!P4I$ff5p@Ik>Ut*yyob z?qT=@W9XVcaLv`M*Zqhx37>mdGlWR4)4gEvpnZ`TW~PRYl+{JQml}l|{QUtoY1&l< zUYZv|_Vq*kFPJHQW4-Xyll7BiqB;Y^u@iU1Bng2Y|NAL|d1Z>UB{BIypY=j239<@Y zTqC2+nvrw?3+zodQD(2s>JdNsp#bgBt8<5rmbQQhu2U2z?-ykXk~eD9oDcRgK!|e2 ztH5vsPyMsn88$w9X!fyZ+*GYUrFbIAMy7z$I4jGFJg?J@I%% zL0NTSqA6UgpZ-=3HDYQx^6SUb*Zo}F;@>SM&V5IO4Z0B8&*6l7mVCETU=3K@he>+` zd|Ws{^<)>)8OcbW`b8gjRlHPv&T1k~#J%#MON5}}Ly%W?9YY$FGkB@`ZNm^lZrwI% zAD1~8=!VOX0T&71r=b{bHqjw^q2)1PP-x!fu25LfOeU$F%lm>Tl>$DP6phc3<;uV< z0OXDxtA7ir8)J?Wb2#qOD%xumMA&2IQhs;NPMNc0#_4+w(+ZQK^e(l}Sxt;KBPJuL$=qUy2 z(L)VwYz7?K@5M2cYh@i`dFUDr)_Bo_6WY|fD3W-R1&?A`&O{MD79E@uMgguVI1-a$ zGvaTfkeP|8%Q+#J&51~3DXt}B=fo+fLhp!q_kb`a-%vUFKA`oB)ic{L)(Y(vvlIm0@5tRwX7j&-{|+R{GXG7EQNo~$Z+rq~i& z)2P^04^c)s%GsON8Tl^h4Z2@x84B*h>+OR#x-HtfXrL4{Uk^aHA2Xt~QfM{ak7iDu z!rXUnE$pi2N>EU<`!f33-^aP*Bo5VXk#=2m8 zUR4D%Nn25{hyb6E2#C+#R?uDm#4l`T!!N`yA|MPD769HINdDg=EO>RHP#D4QDRg;N z-Rz-&-^8QKs|N`H0|fYajm&^|J^U8`!Zg6Y>Jzq61bLV&nZl&oQJ@$Wo@9OpUWD;? ztmpE{I8;r}V{g_3(7JeKBaIc9$tWT#6u2Kwl9OnMPuo)ROI6>qPY9*R`Ix%3;ZPsA z3~r9K0BwnO7>W%qRZ=wWioNY2RqiGDhjT73n2jgQ&l7A+o?$P)6dp_^)d7j_regAu(Hy^kAoco;Xy3Td3bFHTk13itM90xhr z*w}Vn(o{2KW7`CQU(2nVK+ownvFU7V+m1V{sv0K_ihipzj43UgOeK@ z%5LwweBtV?!2?MP0?L#%60X^O)e{=IBqtzp^w-uj3GynE??4T^@P z#8ywwdTknPFW=3UUxOw|U*$EwL%~{eF2io-Z5l7r46EejnO$bmNAP8%wXVi?XR-7n0wa zd;XLYo9^>{iCdDsD81YG>Ot?9)4W?E?{Vqek30X@z`e~|@4x0ckaj`p$)1c|qvtPj z@f8K}ACmjX$8ldJ)k$<)!(ETmB!k^X9L5*C4Y+5wd+jRQ*dMqhHTaXf%%+u{{(-xR zf3@0dHrvz@+%3WG^U!~2?v_j<$H;r;FDrm9Sif4}*S_DfN;|d>#r@*n$aT2-eDrS1 z!0=nUIgj#&sQmTOB!f##<7bpf2H&Xaq_B$Je&*fk=T4%>wRr49JT5X*e`g!DnHWjt za(HiSzxPwUUo|{kExXxC%3Lu2LCJae)4d-~Ui-_~EF;|1$ms>C!JnV~@a5eRbkD|y zF406hq{HtZ{est%w>79V>_Cmqb?CL(%N-`{Hy_Tu53*@3+}x!#erf#JIG4|!B|(iE zv7xYA5$~>F+jE-JG~#BRWzEi^7q_ zX1np*ZU+hka;S(89kdDi6{LAImx~c%Rb%$m$9D7d-svs36Mx@RJLdfCT?Uuo7H8C` z{HWBZ@aRsGPRF6%rw{5!zf->-pD~`fGJ}(hCnodipHy*4Fe^3-?t0QC&XJs^H)J4d z;2}C3Egh{Gef2iwU68z~k`6xaPVw)v)0d~uZi^304GdOGQ!7+^Q==Q=d?GgYg7@=_ z25=57+cbw;Ch&#Ux0c^;liqIWIn#4gK-KXxc2?eR_ln~po8{3xW9ObSo}N35)AcclIPEA#ixZDlJhh-H7~2`U5IcC1pi^NTpm*VN)g@u| zRKua%>X(OAKbuJE73xfCAlA|;Fv}OB?qt8C*TsM)xkq+Bq?Hz}A3@Z$8 z7~Z@u_Hl|`Uo)9MlH9aZB!^aeCQNAWtvcLMPt|sy6GQ^b4*`#Z<%xyj~4hL9aA#Ilg7v7Qm~W zIYFm`utEJn1U_fJ>G0g}$%5krKG&$O-km=?>lEsu>NLWHh@W)&bbS(LiFm0q())-d z@lC`W(!=<6VtCQ0i@uNKQqb9+z9xJE6AN?hf3bgbf7YqhqPF5NvP(&#)dzF#(ib+2 zz?jFzMB}8GPw~zIH~iboozh*n9j>^;bdz@3>XOeskJD>Iz9oLA6iL74?ketOc%FJ^ z@h)y+H%??pWoZL@?t>K7m=9Mp_R*YZ`!%ATg|s?kH8wdpnvSeAr7LX~OzEuQE+v+_ z+#dBlR=W7kq!jJh>cjocyZpoC=}GoJT1Reyu6gh12BkZK>`KXZij=OLBg|Fb!TK5F z&MnB!HqA}KsZ5ty=XvbN%@K>QOI`Y{s1F6Sl&?Zx2fr%|T^BMqcxXo(XWyPXikh;b zvf1LieB(lWyh`LIoBk^AXTryYHwzybyE^u!bFlAFxv$CU zC{_Rb099j$&I}z!#}OW_Nma|HowPmPa-vAa<3z)WT$vjRJtr2=&+3Hf>R{B)SKluO?KLLBK8aJEdy^ITa6+xP5KF!6PO7& zj~Gwi?-A}pE;+p%t~&>N-spN;td`VI==JK49~yr&IBk)9!OD4{H)(@nIlOO%Jfc7D z(bH~t$mEbHZ@+74`8<0YXQA?r@vGDkS2Z;Nm|BRQV=VRvwFM3`KjQPR`|$!tCkR>5OmH+yf$h3H12*_~8U?^DnVsv7~3hB2%j_Yo4ZgZZ|FY z((`}pm5;govlf+F8J6gCuHQ4r|Ha(P#qM|?bDRx(TValB+ksZGagLt{l}0>XW%U-7 zub5J;BU4P#*8Ue}RO;2MYnMZ{<2+6?EcHs2aK{!K)SBy(X{MVB2BzkI3cQ2-ChELt zW2Q^;{&8+*{qls?uTZJQxJ43K*WGtEVV1n+Fk`WFcl10>;EkfCFJsDw^VI>0yq&by zaPkUuYyrFA&RHNJ$Un`8l{hGM0?##lebJ`+LuaIv&?%(>bdaDp{_twXa?Wz=;$U%r zJ9j(>Dy%nhK^ex*-?^_mJrns>dA|sp@+>7zUrhImE}w2n%I(C`CENbc`ZzHZkHhxW z=-+Rk7%1-zp?kF*#j#hnVn4-dvHW*9J>i>s_~1Mp+jP!9?zfvw#V;y-iCd^kXc4^y zpJ@KkyymnCzGbm(rF#zUGW^=oqO5ZANyRfyT_fN)9l~fD>awx<9cN>^8_34C2)ge6 zU}N)^U}O7vosCU_$i~KVE3xdd5*ypbuLe3s>YqO|=yZB|dODFv)X~wgwYBZ(s$W~I z{?yz!GEDnEMCXE(lbAvf#Z zqlb>tQX>7`+4PSI<>UwlJ3Td3ndW9XK+w|C`1x~VQxm&AA*4B?6J#uw*Ei5b~5D2ZUjZKWk-tPMD&iany4l8U;O0Bli1kUii(Pvp4{4-rj0mE@(#`ne!juHpb86Z z=FgI@1iwNh>9N7mmQc4((fBe`op;*G^_7)nguD4VdL0ea3am%&8JWD3B5yU%x2Hc! zJ-|QSko?BM@xzN3FGNI#%b%9rbZAJ6>n)0IZ2{!bySp0t`|4>l+N=E3u&|K+p6dSY z*Yh(KPa+?cze~=|j!AtU^`R>LQSgJwpQXb?#ciKol89kNujA?US+#d>m)>y5J#j4K z;DHhir5ur?=?4zHmX~;c+df}TrmLeYSNVdUpWnlW5A)7nAQ5A1Y^=-5$ho=M#g^vp zN%4jHx;Zj3&E|2bDv{Y49 zb#-;r)z$U&_0VXvy}gy9qN0I;o};6cnVIpEV@E?IPutpB>*=Zq3i5>n``*2KcgKz$ zUfZ_s-@jj2SU4j+#lqa+n#JXCaj^`(gK0cG=yPY()YRU-d42k{Xmu5}thC_lSy@dD z)t7l`;^Lww#Kdy4pNffz$;rurf02>Fp^t;p4)Qe>WK>sIYl{$n%u)?FK` zcV7i(OZ?m=wF^dfdM5gpFxR^Mxo%nW%=HWW<)u>=pPRX^13qaFTX)?{Hj8q>oJb(iU*a3L|}(S)zh_wYIkKE9IM@}WSx$cVF6 z6Yf!oInf-UULC@GCy5G{xOWyJX3;4|+fpB|zx5Mo&P|due)zuBlnit`-jgm#qdcdV zhhw85e$YT2hpjMznP+LB5n4}aC9aiO3#~V|BTaIqew`BwfQl3gVE+PRC{zYp3&PQKzmklsVsRhul@Fuw1S5T7*WKR?u zR=O*ny;===xiJGyUV6>C^Ne5vgoajI^yl#*D?;y%3pXRqLLrg`J_o1j!<$vJ42-L|n5DbxAKQqXvVK1|?P4I@*^RN{NF_}d?Yz| zZBXIICqSE|&{gQc`uHe&E9$oc3J8+#q~>{2oLqCQ4VZgLvN?g;E;-!r{BI<2$5r24 zdlk!G-ob)!0m7oCFDIYyj;;R=z+=tO0ODb$(xvF$zVQ*RJjkJM6jt7MK) zrG9LaN6@?_nl75A>IIo^1BB>DVT(J;F|b;86AQ{?dgHm_;ZG-r5s**ma8n?*)I5oI z*bt#Gx5vuNCbBY2AvKj>376TFSdyg+7e5>=wb^<(BQitXkBhTBl>i}Dm z55slmjv>q3CQ-TI`3c#&&yH485-GXHH)HBfB14YCqg)N##iXW}%p|w+2&TN$T6JgL zqm!Xeq3z!?m}%otyv|lBRP@DojUo+92G-@1AyqhV@K>$-IK}T9Pw?Qo>FI=l*%zo$ zwXOZSORk2QekZ|L&QhcDoW(Rv`AECN#dnkfe?bF{sGUP{bIRxnLE%~JZsTuzG>W)6 z*DTuu(W8oFSVF6 zguBy5TU(KByibQ62y+$~M}$p8kt+NP0!%dU6g~kMTd$Pj98}k4bVSDY1nbmX9}rB5 zOJK|h2wi23u3nx*Q~_)1|Kvxc0DOhBmL(KMi85Jlo=LV)PT?%VWJKQ3Y%{PDhB+#l zKv9AXG%M7KNs0)BHILVcG2?rO9Q4QSegq(@Ou6e3r8cFi@K zo;TJq{cs{GKimw(+jDrGlf`s${BoNvRy(`6fuG5}{Qn;?7=zBGYkLDtp>!L1#**L! zrrLYt^gE!hRDr+Tjsq`kfy_BerzD?5MD3DvN`5SIpx=@@Uw9xhouT1+MBC`cW6K*^ z;>n;GO}r$0;D^zkQP-7L?<_QPmEe75e|+{ucIL%ON5H|j_e!3)GH{ML0y2w1k-9pmE*&_Z!lL#pG#p@rXm%qSwBq#h z?SDK3*eFumiyjO-dKIDWDwHb3vL*d`Oa06LaoqoQfS5$dl!TNIthHBsGGAvxz$@=v ztfet+^V-d8z{e{N=s^c`BtjE?`C0<=EP})a3eA70S0I24jbkn|OaHvbGFwWwGi!Yc zR>+8|&XKD8WZr-(PJjnli)vV2W-3k1_EhdbCPp(2Zx*hM441$t)}glx7`j$4GH}Cx zGs21EFbq>=sqKqe2*OK{XFnQef$L?OB7nuDD!tsOQv(jG?TBu9Oh+{>lq6X#O(Cn` zwi}pwv+&BA#9A9BU{q_(c;rT#HZszWj!G`?iBi}AnP{a5D7HEZ znVxo4^cfJy>_k?}^N6W+)4sP5E>D~7LALIzgf*I+AGvBUpFA0!i1ertQP#w%J~B4o zcFtN=XJuZ$vwLI{k7+FOaWBz#Kh+y!KgMpVFl!M9x;#12h@nE>{%;U_ZMQQB=s(t~O!yXpL{W-7XS1eDW zg#uC8~&g@4O?{nq5Q$XIx=IZdD1yb90B7;H|>1&hllKy>!9Q5Db^NbuKMP9!gYFPqmHhY-Lc(eI+#k?Y9m|=lVg4>)8f#DfJO)BW9k!;PwL~0S z&IioCDMBZE9Opt8DPJ~AOPagbfP@@3&O~NroJG>_Pt`3$0*aXxlRnST$PNgdgiGC> z=Du;h-kL?YbDo9Phu4E7vv zQ{DQ88aK=p4+vpykg|4ijcf$nbT_F2L}b_-@HyVNm;6nISKMOZ@DBVQupi3uP^cmi z#Xnyt?d%ePjbnzeW7J~RFL@r;ix(*p0Up%}+zsV5rchtiGUtRye?2ws=MiDb+z~Q? z_ivA-A~NY>zAr6II<`K%LG#^ zW+>MX^}+cRn{3_l?xO1;PvX>Ym$S}%fpY%U&B1`LbfpxFcU`gp_BO)^9H&LraJbP#Wrw4o?!j+X0#A#5$TE_R0PTrIUE1o!#A=trq`BbTEdvs5xoJ4 zOX(?ydEp*=+stq;&1&v@=$YRkcM3U*&fyHxl5QHKf>Od-ZvKOeDIwUse@ayvQRQjUYb7RQeTe)eOTe5Z zF({UHDWi{AMTURe&0PFM3OCS~DR;{qO*5Dp&iX2od>*HfVB@$EfYd{>U-Y zY)hhY=M5^3#qwu174VQ2Pi6hXY%u*=)=J5*Y_)&08~^oQMKYGVHc1|COv^aCUxd*; zt`;T}X%`8c$HTbe7vrxiq1t;D!4Y_7o}&k%bsZb6_AB19dzy}rcvIwCL%7Nkru(-O ze*WlMc@}AxCHQ#v)&0j9Ir7eWj>rMDm@fgos6!-}c|F7M;S;+YLH^l-)zNPR=#?_= zrCDZ!o(9VsMO$Iub_>&_`@p4oOR}c20&<|yp`M-GH(1KGy?6u>IiQX^3;yW#e`vsf zZ@=FF_F^`$k@f9Ie_8JC9S3qx~@I&116K&FK<1)-LUL~PErq{Nppy%i!o zsN84AfDt+Zl`G+ecSD)xr8{x5ZCb6xhF0-AcCxahVgYlMF;d*)uurBHA@ekTx1S`hP`kX`aFIHW$j ztYh*qH#n{K6|~Z~m+h@mEC=To#{ob+I{O^}5ChK(K$>{03CHQfa#lH z9`|9)2Ygz$bFH?p5pGH@w(&Z0&Vu8$`7T2fO@3L)+Z$~gSlnI<)nPa(Cax!(h9yq4qShN+Ly($EeE^sKa;4Nod>B=yVEWyZa7h}1+LcF1JY$7G(DTE5} z<%cwmNUmvLjz~I{d4dg2i@p&$<8yNV18jz&|AMt(JaduJ+o{o}(M0jzgw2#voaOPO zDW(A+BRbJCDZkD5AsU$fRFu|qF^8jyawyZh`vqmL!BCi}aiX2J+VE-3`J9o6YK9v8eb|PAZKOtkk9j z#@l9^cU3KKgvQc;z52Opb{AZ^*iH5s2%1`{Xr4S$v9_elcVxAPQ|%JmN%$2E!rKp= zmBJU}g%*b@cOPL4F5j!%XFn47#2{yNeiuY=R2~iZ9!$v4%v`MU&si+D>>Qd}S&@Ya zzc?Ror5{u-d%%qB5nK7KG1g01-mKBI{EX0pTcNJ$+ z(es^^-x(%F%1P1pMmIFh8sqSWe^T@bry5+hus_)M-4BDhC_;wZXc6j8CIc28KrcaxthBSJkf_ zV}V92;(w6Lf6C~8^M*ah3W$Zmn73#Ryix`FCO6t4qk9oa^#)u}Ym%dlOqMp$eLP$! z;q<=8Ogm8ycrj90DI;C=I=c0x7l+&xah@Qin&=j|+voEir;tLHTOF2LtuerTdr#C4 zyf5LdxY*LnIr`%Zsri+%nRnVjt<5&*D`Hz?4R`*%Fu@Y^ze?W) zr}7KP!}Y5Vj2l0+O`a#}>&I@dqU^n~a@*uc;HacyMTML~=0OK{el8fbQtu&2^cC+sd7(AP+x_GgI5~Zi zH18*d;)m;2eOfL^xXT@^y^b!gr$jKvkHda8#I;_z$`W2WE`mcbwWRsJ02@4GXT=3! zOS_ewdsbDReM8GmO@z;N;ETBEjmo5bf!OBDmr1?^lzw||u@(kQ?X{e*mKao+Ik;5F zj5XBVZ|8!x@W)&r7AX4l>+c->ahIeYY>=GI@N!rR@|&^sf--NMLkYr$w_fKySZhizBw?uwidom`W2)!u6pSHdfx|LpyNSLLF&ibPQk;bvK37Q> z%(5$_DnAHVBgW;v!A=w7PN)iWhDZ@cERKjqzNg?`?tO%%T@Gd%M5IX+1S;0FEq2~? z6Vz+}aMovX)%$oOGgZ2Bi99N#Up%y!?JTvn+&o@M6MTqWFcaf~!=)!a$Um;emr;%g z5-7cj$z63y!0CSjkKvc6$zByrMyHhVkYXjnLXe>93_Q|vG)kphz|WzSrqK5k-AiVvl*+ini!(BN{F<3AmrCJAF%tj z2fZBk&5LWfBXPCB~9ge!3~wi1K#(G3eEtpC_;%X(TO6TMJZ zB_XtG7GjkfkuJBk{3K#mGPeBgu)`ZAoOMJ}xmQmx+NbY5)=n#Cz`ti~IAy7O#$|7! zAPx9MwRV%Ycl@8Kp?_6eJavV-4xO4JdcV>sj-z1BV2&3AFWWHJP{Epg( zK8b4!LYDPN=U*hU#q7hzI1AI$sAcw>Ai1Vw45^>m2mD0=_@5~YZhN$at6q%5@f)d9 ztdvkarLD8yt%X(2+e?x=$Ty&x$tvv7fQ`7~|4Ueh#ed0yU@iw7s!>=5K{iWl{*!ia zj$kx0JSslN>mzn{o77;~rS;$qD<+Smxk98-336aY4XIOA6vUSU4gvFKVIPY`otqq@ zDj4aKElLN)b))_9eg)3xwOhY`$C3*w7JrJY=M)l*a`b3tS10vzc9|Flm+~_1$oB~k zP=-+y*b$^3Rydmy&%%XzfuH){4o)tv{f->8R~Fg*Z%RA3oPeQIzKN0nfH0v>!*zB9V;$EY#cN7-NqKMf^#cS$!{LQf&`G3yG|>8u#^psrB$`mNX@` z9vOj*9MpnD&{2*Rz3V~MPrC1)G+Yw%m8UZ}ddx@CgZ`(H!jzCJw2972?;Tl>)yTDd z4X%Y?(`y|9CQKrj1GcH&xaF_qtJ^`m9mmvu07h4&Ur-*bqkh6s-xmD}&F@$SX#<{U z=5+vj>QRg=9K$fA=CzjE;205BbbOaWL;`T`)eR8J8)Lc}BErSrXF>B6zF6pbvn~iX;2Ul53d-F2$!c z-dzS+7>(=b?|_h#+|%96@nVCX{UUYu8n>rEbVVL z_&|}MyF*0U=vyJ39FT^t>-tZOh4MJYk=cS;o5{3CwCq(VG-XVeNuk^adP$@nKe`O{ zO;)qgMS^f<965i8RUW|v^xnwdylyPS$p!VSd>S_HCfI^VroGdW1u6%$Ui!(gY{s3qNlShWGxe5Z?Mxu1W zH1HlGuzrvDQiwG(CW@J>QH0^FVHE_;5@j6UaD(Ab&+R$F2pK1VT*^d&(NuT+@;R}X zpP-)GJf3Q_wnD%+Z-}{ZrM+8IKAb_bcR`b%l@>As>OhLz+?!(;oR)OI+@pL3CQdzI zGK%?{-Xa32IBD;-0Uja^!cCnZM?2`?jHXyJLHzO0E+WeP%4!2Y^g+6EA*ldlY%3(2 zQIw*bW!2ELkg7Cc{&zTF^mb6o?KS)57a&H%JApJlk~K$(!b($U@h@MiJ^V5ibU?}m zXh*AS5C~f)K-~Lxp4@$WQ4^`;Di7j8B`Abkudp3)+J!B z0~C^xw0pvW)q0X$3FhfY{*!I7u8c)gJE%1*v}Yk2;)gWBr&IkE^6DflbbURu-t*7& z%#`1p7=VZVN_Xe4B(2*~YpVB|byCpp*P&=i6PzcutmDO@ZCO_M9`n3u3SuCR(D7f^ zF+7s`!yHL{67%QXKc5?y7yqs|v=>~!oBvp@&oZ9Tq{`O|e-3wRQ_=?4^fK`zjVM76Uhz0?H z2k-SrJ}ZlXKa6BILG3e#Rq>K`H<28~;AtYt4?#hp2JcrC#UJf?T?+_p10b6B6 zY&Ov*P6@~EH(!yj1bQ{SiXt0>LEoWM`+CBffktGCN}%*1Fyu73q%Igt`HM&{A4l57 zB{bk<@g!q&dakYoG2c(8_7ceuP)tAQjX?6w)qI_3u!sZgFZ8jb(~8goTJAd8l^3UJ(saqex@%ef zGKv$e!-{I;0<`g1AI&?Y>hj-RU84JxaJL&EwYn>IR=I(G3+l&SR7ne9wk-Vk|gEC-)HqPrf5ze|!a0zM^juyEgopcp8 zK&nRo_O8^+6Ulz!#ved}mP|^yBf%;ui+s(%9>xJnmsOMna(PlXg^Tge9>o4UB^E^F=&e-Rc@*8X$t`s-INWOEFYB0QG`)N&@t zCou45fdu`yDX#wA%$nG3#jib&u3vZw(aA=_IZ)0)T>bAP{h*(Vd4^lhAP4dSGy!7q zsdl5Tv$w$*qR4wC_j%LGGd@VMDFVBP?zkuyx$O*38t`}G{{+)#R6Mg-HG-Yc#573v zoE#k8$O5(x^5y)0PdxD{2ed3m&4?}7CgrZs_As|;F1M3YVu{l@8Slo+X@TDsB z+E^6?dLWF-`iMUsC?TM7Hqc*=#VSb>IRTzSLL!7yjvX;mm8_Gxy@4{A7>68N5Wh(m z9z@oHO%es--YLDj5Ss&O)Y5g^RrT?~0HBDHv{?|&ChLoi2?HAmY zi+TOwR7v#h?`xZ?uO^&#eeP}x?$)02QaAGg|E*5}<9-Y4;q2(-#isiA+o1~{Ud|5o zcMe@~ckuGH@vuK6B_S!fwXF6nm?-)8L_2#=jED0rFK2f*wtt!6X{e!c=#r}HAp;M0 z2WJ<1?j0L~z+`S@@+~_D1rvJ@Ph_Hj+Qn0nr_WrGHLn7n*)FN;sl8IUe*b>~F&K?b literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1376-image-1.svg b/generated/webwork/images/webwork-1376-image-1.svg new file mode 100644 index 000000000..bbab8f7da --- /dev/null +++ b/generated/webwork/images/webwork-1376-image-1.svg @@ -0,0 +1,372 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1376-image-1.tex b/generated/webwork/images/webwork-1376-image-1.tex new file mode 100755 index 000000000..c875a532a --- /dev/null +++ b/generated/webwork/images/webwork-1376-image-1.tex @@ -0,0 +1,141 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} + \begin{axis} + \addplot[soliddot] coordinates {(-2,2)} node[pin=60:{A}] {}; + \addplot[soliddot] coordinates {(3,0)} node[pin=60:{B}] {}; + \addplot[soliddot] coordinates {(0,-3)} node[pin=60:{C}] {}; + \addplot[soliddot] coordinates {(4,-4)} node[pin=60:{D}] {}; + \end{axis} +\end{tikzpicture} + + +\end{document} diff --git a/generated/webwork/images/webwork-1381-image-1.pdf b/generated/webwork/images/webwork-1381-image-1.pdf index 279c1d7ca90957c444580fcc6f8538544b3808fb..7a56e776eb2d595ca9d280b8126712042e7e6d88 100755 GIT binary patch literal 6111 zcmbW4cTkhty2b?&B(zWiQl%L>p#}{EX-aQFim+*+Bp@B67b()Dhu(V)9R;LISE?vT zk={E7Py~+q%$+&jyZ1f!%>8H0`o6WknQwk?d7hU`OF>BhA}9{vT0Pu52SC|`*&Hp< z02oY231#nQL@u2)!S>e(5CsQYh<i}8%h9&o93%`RX zJcmJIhc8YbAy4Ikamwhg0a~q$IcI(6%jV?-@WAJH4r_b7$_Eho=m*GGumDpCUK}#fMH?czp#+@UMz~*xeTKXeuWJaGxY5+FS6Eku2Q4>>S(|< zv#h(F*n7tKq$f^g#o@y&SqW%1xZlWF*|&o^FZR|}T)o`DgXhs2PhJa(b)58X4n+<&g-18sRd-tBv{BtfZN!cb z@=6r#dKA-V#RDXe{AVVm-1iMX=xiRxh>v4J=jJ019>o6nebL9dlYz~l@)>U1!x4@G zz6vi!wB0|C8nH!h|2*36YT7?NcsV|ZZu<7t71||b!-VMlen~==YLi=l*df<->V7Z3 zd^MnEGCl}c@QC%QD}cO*Atdjfwn>E<7UqA*J1|Y^B!XO|ryVBHo2Ghc(S}TY@BUCh z!5IKT3yQ84N2ZZN#wN<8@B#5t2?Gyf0IMbMu7 zTE%_2dDTR%w7O~3Ag?(t=L%yC;ACl@PrNDlWH9$Jt`dAWWF*z7@+AHctSU#1hKx4* z@n#{4!7BTSHa9O9Z9dh6GQwgAR#cQ&j7#z|(nu@i)C*V9mU^M0#}%T2NGlEb^{9VO z+m4UAu|Y(8umXr_(a3*dz>N;dQV^(9`*>$qp{QOCA@U%?2%T1nNsh>G>ZYz;BQEZh zEl&M}pVpw_I#sEcz=w!8>7|*n%75~QIeLkE1h-5l$(!(z_*$kGM=4x$n<)G$S|EIncD}t8e%|48d3^CT z(-r=<#fQ9H|D%N{a{{Rf{T_VY#G}rGi6f`dkAsr0i+i<{NPZnoT~|UEn_UX__JvJQ zM-dIuHyj<|@n{OL0aHSbFURpvn*3~C+K_M&d^svsYTyBuKAtaY;Truury?ew-ZW-b z*L?*j^_uRz&?t+;h22j~UO}!MBC?E{B8lU!7h>uY?*4YKg`Jz!%?KVLo0@U%FAn1% z#7dAt@{8S8Auvhp>u87FTKs2^Z+i6OJ(#|eQV?OQN-hwSKAIvNK7Fe2vC?v6eeMVT zL`(6pSe~2{@{U>G`psLq#()x? z7qhJ=Qwt7PK}nqNf@%`BbB8fMp3ZtQl}UztNi3W9Cfeb+LmvO0Z2xI~zTmrCRZV6# zK~;}mx2?5bt+9*!n$f=-loX%y0{)r~46B4$p zHxmukyMNUQQp-@eI9nSM@SLTc0$HnQV3QiX69?Dc9~juy`drmh@U|PlXw!-WFI=Ww zcs}$5&G>VKyk7UvRS?ALnObio6W&1!TK*B^d0IVIo%n6PR`tYr)EphMVD~j!Pc4p) zso1!fE7iCPEMHrLpGD@h^}(NW!6WwQ1XxR>OKl!&Qy^P7RYPyUkLN^LEzcz4 zA~7M9cNAD$)N^~3gE3NC{P?N-?CE~(m_$D@{)Er2fr=~>6$VAgAA76w5h}Wq;uC)Y z^F-Phl`ZqaJ1UOJr5P~Lxq?hJQT=YB)}8R>`1txIf8watkz{fXV|I29EshT6itia^ zxbPr%fQM(GFv~{sY2RqHP*2c3pl% zd6Ltb-F`t4o^>M#VC8`PFERYJ`%^w4BL9hwiiisTE~6$|wdi-kr1O`wcdC}0^MVPyy)`GO*Dh;cbK_qEG>D^x87gjK9Y_G? z?`8~iM8}kRb=E5Bwusk)zvZ|*%w@d^H|Z5+eTUmybJ^e4AA`;_Y|L}i=$t*Ot)@&_ z&>&9_yD@e!+bJH#On&g?)p=5l?ihV^){@f_JBP)|6$sQo$ve^HW$Xu6mU>l@0tjvp zO(-rmvTea~+2T6>(2h?05uf+#7Yx_w#L5!dUOs3Hb(+lxh+{5V>sww?U-(`)bVTHn z@QnL!P>KBum9D#mo0pT75aQPYk^eUiL=+yc0Y|J{p4{<7SnKp zczXTnB*onD+*M|W0u{z85)DVD`42#@fj03zo4v<5FO`#nf|lCY!hL%v*iCN-K6Yp8 z0`tA=>>e1##fe^*`iR`0yCtUS=nN0hEY;lW@^b| zhCxDndWHf1>MEFEbNf*~%9=Aeb(GLChgHuE>bwiY zK7SRr>-F^cS?}`P(8Gt?z`)JA4^JzbdzREoS#pf`>gk83FixnvSuPnKr_cM~+pZ*Mg!%^~I9RK6I zmV%1@j+Hq+Kr7LmsXxbtaaJ-Do}1k%%F6PCi>^C2oRt#HxouKP{PH2u8?zdLrX|lW zHwpwATvrMfw}OViw<8{E;0L{vmjkND=>inDUW~oDJ-XQ<-$CKGh5c|b*S5cK{T5!` zDygo_Nuxe_oV9hZuQGe#ZE>6p){Ggrhq)#f#P|^JfR}C=83<7s)GppL)YSf%AN$g^ zLORoEWr@!r+{cfHSe>K&yhiRwDO`>++ZN0ZK_5@SKuPuSBvCOjr0d(km26 zM>HEJJREg>fVoe*p-q0oHlEA(Gen=E87Yq)DlGvUY5l;a+lG;OKT8{v z)D*^(LU>EUww`ZV=9A?+bTz3srDv)xXtdbTNsdH@ioixXsdq|Gl}-;JrX~1Cq3_kL z!hRC?0Cz)~G1aH-e$K;L%fn+NXLc8KS{I&c{L$>)iP}mxmB}ab-EXhwvGy&Nq<4h{ za93n3GDTxP@o5xyTTc_V-hH3&LQRMpgTia>2v&vev}kEs4hvcMx(>cXHL>y_;X}%N z0K!6yc+U;vXG+6Ub+8PsR6V;Q6U7aY9*xXy?%wLTeL9m48bH>IBjZuMhjXRN!T z3{Tnnt+EA-LK{a+H|}@j3BkJcPAuLpf2B{A@xN%@sq%xny^@tSZV)f8-+d`=rr6Rq zA!^$6(d}odJ5Mx0ob0kDV1Qw_^^$Uy=sRmRDgO}T1aP7D%la_qrJwp8I~0Smsi z#EO004p#nl*ccSo86Kj}znV26wzZ?Cce-R;$K5E5rvs!kKk`ZKs{j?V)(huv$!O_s zmqMu^fsZ#~1=W1$APF&%Pv1{gUo}+mzs`+yIf9L`pW&##m@7X*& z-_hk)Gj5jfpsNxQza7yyt>u^hP$fxRB;{D4u~7z=Ap?9;zlYu4+n>HQaamf&)ycttg~9 z)6(9ac{f?2$aih2g$Yp}UN!RiS#g;f3r%K}Oc!~$rCj_={J@Y!lR zFMq-97L{pLAD%@0xM|0@VGGKMwrY(JB3?{<^Na%`__6`|y@Rc{$XO@c&S|V|?v3+E zcjgJCLs8at2hjY}&d1r0nvo}*(eDObf?{^?l=}?G4Gq@`8h@UpI7DMXlKYQOPutlU zp5`>lIN%`19tv8w4Oz8&cjvb63UU^cC?)mvk5twM(T=hr`W>tO3Qa`#|IuG1B>x*j zuGl?QYCNX?gC@h27|b{r>Wi0=}z2l%E)lld_5^8$=b9|4VTPn zkvcjWGDfM6BX21J;|&ZR2L$mbw{3<}K5qmyE@h0S9Gp#^1+%kLzQ`{ve3Rc$zS-wh zq@tHl@mMf4?9KJTBs`nuinBXM?mH=7EXns3cHaubH%UR^cLJQ^-uIQq2^&;E zXbd|$O>eqMu9k5UAB~qveec7a9e2|BvJrL{62NhN1S~W^@Qp>GZCkn+C2zvGwo-y} zlG1XrVO0q|QM_TPCH-(wCR6LY;JsH8&1u+gf@e6?9wTv)Hc4#1r{FJ0H@o zfxYShW{B2O!<+U8%=fa$ymR%}gW9uY9B(>*u)d+TVjMmcNLG_a&%HFe!r9u=HgpU! zNRpQc`xN5Kmm|nbf0LEeJn}4 zL;rPU7?>5V@TKBD`XlCvv{iYiMCr+E!B~0vKBZueE@3tElVm)}$;HaD7}Vz@>FEo) zB^_a9>R~?lJYgFo#k;|=hcLY)xPXipr^X(r|CL z>5zpqEzr7d&xADCAmYEgvWu%5o0P=AG0hTC=j~ewKrZ0)5E~1Y z11O0Vazz9hv%{d)D+c(pb)R{*st+oq)Q!r@c{^L{MIZD_221C%b-H%pTSLswanQv_ zDrROS|r`avM+!vPJ|P+U4)MSi%iBTH`BFJ@E-+aIz!?2{97KMIMamy zIJD{G{nKq>2fA*$gQO-1MptIZEb`_LlTMZ2lIBi-AoHK3lO{hb{E3?|5ehaeaea|p zw<$hc01_i_c;Y4Jy)8{XwQ2XDA4qmw&sIC5jT0!+l&k-9_@&&djn(``>Bnk@a+Ix@nfK)VCHT@j}WR znXtg+oH|0=)B?Z(M{!Thxr?{0GeKVhNL+GPeQIVsg5~r^uwhppCeNebOz>vRMK*6Y zYy_B`R*5D@FP(W;-fKlTe=F|t5OC2BDUL{@kHR?w;`){psRU>m==152l~czoA?iP( z*myhhFq2r6$#DMI@%sT${TC~q^(voZL4XZU=|*n9%mG5fK4n8OzTh3RGPFCB)lVC< zI>Prd_}p{S5Y$1u2^{CJ&nopW>1+ zTAWGvM?@j6L-`Zo*ir|xZ2KC|oMm?ACok+kKi+F2fmEE3YGnh-U&}${x54qB=xN z#ma4m6@nJyXXjT;4z~k!nAq=+1MYzdd6Nf;7IkQm(i1=S?=qz){DnR nijc^(3!BgP?&9X@Vr2~wheDtLE-pn)CBQ!bStLp^ literal 7986 zcmbW5bx<5@o5hjfn&9p(!{F{2WN_DzL4(`iu0eu_;10o^V8MdB2M_M<3=qQRZf(`Q z-+s65ezpH}cfHkJU3I$8d4A_*P?H6506F>48Me<)VCX{BT-1)HHt3?F0FbpE)WQV- zvV%Y^WG&1b%`E^*77kX>chuZG{9NMV=&n!~3y3|s=hBIRaU6lbE59|9sW{nk+v7`* z!ZvFA_)Go~`A11lgqI&fEX zQaPqAMOnc9KKF#5ieYy%ee;VGLrfON3o8ALuVssxJPeU^wm&D24LLq9N09_L_CBAT z>zx#v9M%gE#a^z;-Q4!v{zT%eKEKxdndF*NW48PVYC8zHb?`;zbgu3%Tc@Z}A+a~4 zTG`J;U(8RK^nNZci)l^0cm@<1&j&a*#n^$K-wfEmv`OVrV0bQ^yVI@2t9HvrBy7M! zohc%Z;ho}>``di+$1>hR8`ZuIp%9sg{_^7nH^ip@$tx0jokP?*6AJ7oC>3 zH<%=3?g?Y)2@Tfa?!-Dq*YegFHY9@l70q_|48@3B3pMNVaw~$_lxx+@b;B=+grGAu zo1Z@V`B)8$FYj&YSqM-i)uyyRf$4m)sH$pivCBJ+Fdul3kL=-Vl7}S25+lC02$&7% zRo1zA&*K@24AWR;HhI0S{Zb%0_a5imR_gZ^f*v(}`%!cy1%X0Z>JNuEb|)pW!xmXO zlU(jfjLE!d#;SHZ5wr;yImfh1m54~g9^2F6=XR8W?9uN+7KMby5UsMNIUk;CelL|o z-msiJnq75ym8i@%`nE?{EqBq*ccEv& zH@-%-k1S>HG)&)R!_c#&IM`Tee3rsNZ|zB#Waqs5GXBDwj1VgYmtTdvys@{|c~h^6 zAvOkYV3@<%E>qCbG16?Sx7Hg>P8Qb5z82kMM=+STj^RX%6wBZXU#H#gEk8r)N3}&MX}tv|5JPXODfqZ;%}Y^uz| zm(Sb4W3^<-HO*7r7an^g)NwVKeRDI16}D^1GJ{C>Iz33b6h&p9x2I4KC-b8XvSK@> zWa%8{#|DAG#A{+^Z6CNIOCB}!J2=Hf%iPJ+AiUu*E|xV!rar?LJ}y_7x8}QOA8E(s zS$@C8Ei+G}k4Z~WNTw)WwM5G`T#+M3jfDX-S(-afdWM7r-w&avf=46jAr#%hs&MiK z?c?U}2Ty3{!w$#n)~r_>iA z?_96DJYxNOSFp81iRbo)6{rrDYCz28L1cD2;$P!=mI=Dn(E1jpqc?s9Upvryos0!f zwxB0{rhG(LrLSnc_=VIHv@l{v!y1F$;)Xsk>{BzMlDbZkmKZ8t5+Pb4MJ*V<>TZY^ zY=%q$8$%kM*!OQW)xrdJ9bQm6vGJp{wpkc=I1A?ZF(d!V`$8@I?Gs`y97Y=%-_3hnb6?P<$OQ)rA<})T_KGH>P#A)86gF?aYg|#cXzMC|PDPPjzLUAd+r0m|^!q zQ|S;;(E0KMWr#CLf+V_uU1)7w6$Iu-A z&9Sh(@>t7c?w4bPrV?<`UV|uL1#=Hdx;sqEGzG>grP-R-w~iB{s(LR=n~Ht1pZ%Eo zSNtUOgdF;rym$B-Js8GMp4dO#Ig5CbR#R5b^fISX6twW_S>lSM&OJzCR4(Tie(dQ! z+g z)H!%@FK=5bl$&wI9;{jx>aQ_e&~|#y+JBPy_%w$E98B!3c|R=21u)EI6|fl#rEfbJ%>ZkwCGFDPA1j3nXf$>p+V%G2e)jwsU<8 zs`+A^rLV#pw_q#cdewC@xwh5fj(+TsV-Q{`5B!c(4YDOFW014*ofvn%tE0R)0~VXf z758|aOhH1)>CNPCK|9d0=fuS}pB8($!P*?xo_3+CN+xEEu8GA2Ly35_a!oW-@}|P0 zaZJDbrLdBexFA;xEWCb}hbk-*GDDK!Vgg=duP=2X=2i6&5|7mLiq!QQ zbODfu)ld6cp?WNbm)^JQe9r2n_j@;DiI98VAIcHLoIx@YCd{?FaxI<~}0DrI(x;-cgMMnH0b8_9-W=TdxYMV+vK&(Dul0?#lBevc&w zbPEUb|3RqVSARekkmo<5FJAt?!k4kyH-}Ja-1aL1*OkxKvdS5hABI3?hmfyhtv&}> zkOq1J)Xi_5)zlZ0i1A)myJK4n+J;Sv!txyzyA;C2g|7&$+Vx^5x0neOcyJTH>V;*f zj>Z?Za$OBEh8EsCRNV}?y0x@-ik>PzAS1lbax%e^COm39omXJ_^(D&Ku4NcgC2}_B|pPbu2l|^u!WaNFaXZrx+%e3nG&s z7^f--@>2a9>G=MXPSed4>g8ks(E5D<%KQVz3h)d5RY`()TX{_gdH-7d*oi%-?Hns} zlZPgE7NLZ#P|IybReJaV>~f3TKYeAi2b{O{q*xO8V+ArB$5%dTlK>;gA@@{pz!t zbg=BCVpwF|i_pnH;o()Fx<4t|&IeTb9~bf<^;8cE%hY=X%%sGguX-ItZuG@YYj5nrt}XC(5i%{8h$#emwngB>5MCog-PBlEHQ1ZN&77;c_bbAkl(nwwfK^w#4`9#a!JmU^8fRyOF zI!?QGj6d0rcU6WYe+5t|$X!uh{n*SyZFzGynk=g%3O$fJTBZs2Zo zf`_#0jPWp6Fi}*+0xl~MAApfrH*d6}%aj-yo*#lV$5IgHDYt{?f3F zzBL@PtFy!sZuOxhi6*Zji4a0BG0eS`YTNo-Bqp|mn)K5vfIpzAbUWG(3WkjKJd zXNkF%e%UUZX)m|&xkL$Pj(C`WBLO=tIjWbAgjDWS5DW;Vhh@TTkRE32Mpv<-%>>W7};m!Y?p1 zAPlBHMW?g_8>h7Qn>PfdhZ5ub^5Yw(^p?N0yxhH5mG$!C{BnuwCq!p+J;}mB5o_s! zkk|Ld`V|v(^0^Oz1JDVa-?n)hf`dBsKuS0>6RdoEZn^)D| z)Rx-eIGouF>W4S@a%Yx;=F<(ZhFkG0*4 z$;8e?Z>-D`6ZfQVHec%W3yvA=Lh(1^3H~YGe;i2syxe~cB!=-bFrG6x%K4f7UMqC; z1ARf8@hwUfC$7c92EUGV-Q^ z8vOSZS;Zf^PTOyMZ6}Zv%A?|QO!4tG1p4sV*K zj1&PEGf733-OF?1UG#-8wB;+IyDsx(`GpsJ`}P|UrZ_@%XL0W)UeJar9;&9B zlKGGUOPRwX-!WYqbG8&!9CwMrAmN^i0(&$_`xF=6*>TuzFxbX8WFVq|#fxcR&$_A< zDc|48RT(RX*09_Mg9?Er;uY2MJSn7Fg(LEe2r$&@A44$VSYxjsM+UKBJUL( z-{1Vz|JnW!|NkNsWz|0-;ZAqdIR=Q&l_ADUk52Qg+2)fm&T1}2O=N@e0VV8in(%Ht zFH3;OvEz*fzji@k$aas-YJP|mE5vzDcjTovc@eEOvb0N+*T$ZvmAnR}$km)O>d?rM z?vwtHlmSgeD$YIOA0b^d+~d(wgZ!PPOBU{ed^VF^#PISewmplBNbaiAcbcNFb!(fpJ~!v?uHj_QfWManur|qlS5-e z*{vX^8vTq2svTJO?{abj{~sX%{xt1k)f&#lA30g7d$ouk-jOgCLEK@$ifbxGuOpi_8h@}5%x*4IY(_Kw zBK2*z<-&Mn3On12XVD=am2sbJDx4r$T3OuQLqBUdP3+Uv{O=Ye+F+DgFW-}L)@LG9 z2gJ=)`esykIgFN@eAH0fPKQXq9Q(z=5<>@G;CUx5JuT|Afj_8d>9ba;X~V9@`=Xkz zLGA>v43aM(Jfgnd5-n23Me`c`-2*UjRyH50k~cpWK2O?in>6ClTUrld{YG9I;U&v@ zmYdJqMT;8etGmqoI)kKXm@;YKE=y^06o4r;qyFtfUF;%kDW!ZFij@`Ly7x!#k-=!v zp`I8TXsfR;n1Q)0GDm8ti3K;z2CwTzHA*kJxxDe zFx{fHR5oXJ8z&+YHUDrgQ9CzoHk^jz%)mM(;9VQ@0J9yxt#8Sd3PMaOL^yU3XeQl% zTh)Jf@!l)qC<)eG82?Bdr^%%*2JuZ5E4H4N5k3}nR?`2dk%hWUbHH{JlO=%HZ}0II z)!+Ukhv92~hd)ARf-k&;%1#f;No>n7BsJ&^@&W|FY;l?_Xm#HYSAWlTY07vU$}1l% zmRwrM!`kx^(R|(G+@*chUmyrfg? zi~XR~4_SPTXFs9+gd?%c_Rxm_Y)iQn7S`ED6M`Y=u{6!OSMw8iGXbvXzCOnteQy@& z9gMTvH#7xI3sGBMaXkRNFX)la3Fc9Ay}b;{TYZ#NSyP&1e*2g}fdc&r|0}6xc(}`+ zgvDgNFj1uqY?q*wp|a%Lq7scH;%8mz+R)|fWn<< zpLH0-kT5a?CZ|Q38eUFHfyf2Vv7J|J{U~5gY^0O*v7pN^%;sI zfD($l=)W>Gfjw-pF36cN+;O?1{f;rfbMo#AHb}#PMRKoy0Gi$^^gn2KB?1klrTUa9 zqYWvNlJB`Zh?hw$+FApcb^sFWkz=9qmzAAsCS9A0Ljl&!I$k4_R*p}{re$#}g~jLT zy(*7^Xt(c~m?)-uZ2~e2H~i-ZXW+SUS+VaRMRBn4PVN2qO|-HLw}NQb#AHS?f))0F zc!SdUQaxWz=Z-4Ag~N@8vF5^GZF1dIU+HhENHH#Fhs@0!7Hzzfna4SN5T9Q@NGNNs z)c#GJaQ_)6{{U<}z`sJqAN{e18uNF56n9T&{9W<#|J8^&F!d@!TE6e=OE|zzq@SrP1 zHwke5DN0VYoCLG{KC`n@Yc(mj%Dgv5cdl{gQ_JykZ#5SCFdc3(?x>d`9LpWBgawPQ zY03o>i+nqsj`|Ss{m~?I3REILv+aEBp5#JG1Ry}RW}*`&&v=0UGT6+_ z-$Ed7&CR!?QEyRQZzkPuv0{07;UX8dpD^xkHumi1sHQO`%!vaAxu4?Ae$kB6-c3gD zNZ=%~BO><5N#WkdQpt9_f=jTG3|4;=5fqk?P*CxCxvNWWqN{6As57^YnMbTZR*Ftv z^V{U~w2|Dm!LBZiZ$N=B%sfIs)o%i|^mKdzwJ!#T)KSM($CWrrwI*wZm|0kArsn|w zz&v=AV{QbX*fcY%&OE_Bq=u@)k~wEX9FI)IfrZIvU{T^*%F;~Ulo5%fEvq& zPcJW9ezH!x_5c-Xpy2Nc@8Sxj266-csmu%g3p=+b+p8{f;SQ`_v7Ay= ze%>5dK$DhlCT5bq4Ez_@l4*^G}In^<3&8 zbu5I`!Y)ClD6D3UE=d?>vD7B-40Twq_Bk&Sr=&9)C38?X%F>mT6g7bZbc6TZhcKYv zE$w+1%4gfwZ48HZ)C18 z%jYi1xkNF|;|f+O-dJcLGn5KXMG4|b`KmiER7EtN<{K+_FZoe?d*#DsP1-05 zC$BD4m22JT6s>k;ZCL4DQU&t-XR9h27CQsIJ8YLYZy%)-9@Vneua_keN;(;Flv_}! zL{^VIR=eQ$536%W(Spw0w(sX=mmWFcZc<&omcCQ_0zPo?b2d*qxhfj|X_$w_G(?<& zLdghQF&baM@jx~@ntte%$Xl>sM~w2?SEQsgZn{zW{yFehv~kaOL!qJs2WP9S4NKw! z2Hko=MG*3if%H9L+_B8*tO+Y%+-2Dtl_y?Y2qVQzHQq8cgWG8D^VJu>%RD##gOrP9 z#<)7cJ`cAW^(gnPpEtt!s;Ww&{x!M2Qn8nVXevmFf>Z(yT)Zk5ew+L6__K*~3vI_5 za10X0>B;wUhf|e{>%!G0*D2jBBAV{@YmAP1@k01Im^o6!bW~%0a_dVcrqSG5NuO@X zX#%o3Lp3$qr*uvo>ov)auL(t}yJMXj!##dOYydKwm%00mh?Q6N7zzbQYT07QsO(9%=@2;t_p5acrF=L4E^o0(dgivQnt z7y~pM9iixdSX%?2;9%)U{l|pT0BBl!TTt_G0dx(4zgM`8|H0Nme19EQ3_~qgfs(i~ z>KTu3V&PuAVo<{wi^rQJLPJXUW{@O$KLXbQkc`xmrNzY$E0Lu$=!Q2~_gz9yGiiK9 z*d&qO$6d7Scbj3+zi}=k{z<5KUWdhILw10m3uhdJM)2?AxI!T=P)`>NOZ4AcgG-LukJMf diff --git a/generated/webwork/images/webwork-1381-image-1.png b/generated/webwork/images/webwork-1381-image-1.png index 06ecc24654257150a596f890949e49adfe4db69f..d0d9c36175c5b05948c608f75b16c2a34bba3e05 100644 GIT binary patch literal 29510 zcmcG$2RPOL|37?Cl#xUzp&_FZQYf>?ie%4>WMpsIq)1lTTSoSlospTHd2sBRO~&DH zxL@yc=+o!>{r&#;bzk>=U!TkM>3W}Y-t##g&&T8Odiy+kDn>$hkq`!hkw}P($iZN^ z!{AT!6h7Eus30c=gPmr4B`o~xm98!fChij&YVa&rm8xlVgKz$w(78w6&xAkSxb7Zu zR*ktkGfn7M?6b!v?@MuSet3WP@yEwFR2rdIjUQ7G`M$xWwDG)!+fr_MUiP7?#n^>d zB(31L!$_x2^8@SIQR~iK3!KaEJ@vk~I=zr9uR$~iT%e|CJ2A%W7avx;zG;Iu)>lCW z%dU4SAM4!BIYsvA)&5Lr={KugmNz!A&r5p!ZIUvo6G^R7q=ARE2FfaV8sm4yR3?78 zIV?TtsJ?{?np$js?s}dOc8LW=bLP;tT!$`3A`eD=+ z&GAHx-=@Y7^f)W^CS2&msK?1uZ}QGwNPKwb5E*_f1ZP=2^Hj2kl?aUAZPj?`^yHr$kX({jy1sPQ?%W=oSUt;xpG82aw#K>Jq{;!tjnC^z zCQ>YjUR!>9Fp9hC5m;kh8$M36HnPLJBe9SB@J-EC6*A}3mTzg_5)aZ$%_%}bmi z<7~T+X1&U!-HVg!6zlkwF*_b2*IxyfCZCnVf5kS>Gk<6P$~;M=RR3i}kZ1G!bmcVT zZ^hs8znwUuBI2lJZap%IRw+{P9{4zLlQ1sv>BKXxXI9KJA)Fz+A*z<&O1*fL`J|jP zUl;A%Uzc6Ke>&=I!dq{VM3H=v!g^_+S2w~k9@>5ueYTO_l%||ce)nyP@-;Z5Dl11Dy z`RbQ#@ghkQiS{S05(r5#Nh-+_3DPHSAv!HQKk8eT!skOAuKEV`g>4132k~EZlD1O{ zxNX4F6n-;=mlY{a7X}YQhK=2Fld95kefm)LyTp~p333w|wOLn%yOcPd=1Z+7PWqjb zZdPMaN|%}s>LyBI2(FXTR$_OGysGD*XD~xN^I!&N=H&UX5Al+5lF=>Xif?YH-;*qo zDiTvnPD!543CrwH+!m+J?NADo6-*vY70CVGed^~xc2_z_dyxLR%Ka?L{s>0Z9#Kxk zLd}gTN>ShJ^SQPvi#c3brirvk&$C^`2>LnteERNZah4KRwm2(xXP^|5${9dBrI`7y zl+p=zO0&u2%CCl2MY-l3xS|Eno%oFSr}4qPB1!kU?d9M#;^jv3ie}wE!+#~0G1pF` z($p5d0B?NI_^DCMk3Qn3)UdQ&^k#(foqL?;BZ{LsBGMz@MfFAa7tEW;*m3N6-5(n6 zaE{(}*dV1$p`52oWj!e9E%GZdDUQ*sQzI`+(P@7h`av;5F*fvP)GJyemtHl)BolId zdGn1nr9Q6O;x4-nNewq}Lw6+courG0*OQk;l$lu0+g>k2l`549SoheGm)cg;t=wKA z7;fs%$dy(@%;WIArX%2sdtJaM&+oQX``W=#(TN|)wb`+?vXQWBviWM;Va|9?eR6L= zriZOAw<&&-esXMDkY1bq*~QDmy`;nEUh|4`F>|Heq^4P-AExFj>ChSZZX0}s@yf|7 z*A`V53*lqKmn*L5)V=c$@zDR;VdCDzRe^A~_lI|Cxe{-^ zVC=q8aM$Wa%Z-e?Mh}K=poBK1{G_EE+DvY77vI>njj`4+9d`aEw!gmf<*obMViAoO zhLdFsMsm(_<^?aT?0##1cq;ku1AE`qE$s~P@f(Ixp56VawqJU}1se{C+x?zjn2PL= z?2BA2+%N2S`Q#;&9@4yf^s7OSLV#|EOtftDZnTqCsI~ocfcb<;I)c!YWNfHV+E)Fb zxOw?0LT2gm()+P>^|*(cuSOBEIJ^}b=YN;X$t+n7^~qgUy39;HVp>+QP0&l4FG#$k zTA6CqIam9>R{b&l&cnTDp*dkf~Nk9oD(iJ5ZQR$4PJ6s7tsik{o5`#ZPCMeWv+ zd3y&8(7no5Ok>SJzrMTu%kx2Md4Quo>bGu$Ze&v&_YU_ay8a+NUA8zsZM|nbd8(Ft zl*uu=Eh;K%Ci;omc$i;UWbhTH)dQ15YvoKc*b=J!V(j=wti4=P zwz5!zQFdxhIt0D>>B7AidRUeS7cTP6S>>c63M$abm0E%E$^u$04}U*ueq7tI@B1X& z>UR6{r)7Lj*HJAZosDr#%6Pe>t6K}SrL&I9jkZmT%6mL6;btl$JZ?Rcvt{Z_>aiu# z=JuP>nh;%@jljhRWU0SDcV83|h zhV$8VZIn)J9X#+3JuBa+fEV3O=c@a=($epxDnw0KeafCVK7C<)cbPG|V?knb^k;uV@U6jFmD=)oS}vbU zC12(O^2-a$bP2qVidd)<+b>>y*zHB1{aaZ7#0Bcv16^2G89wa^+EUHV zrfa0l&fXroZCjO9t7SWFwyL-u?uiWu^J;V{#p{e}!H<_C`bR6Bgw2f2&BGbKBqzW1 z8?o->ryzmBG{`dhtkeqjdrV74wwF`{?WFGVcO4uY=mniy@OWL3!D=O?K+umKljcJ9fEXsLZ_*PUE zrRSp0V886OY+87NikD8HOKh@cc$BkLZL+pZ-9k>;{T+kxW-dW5+U7Eeb>6OdWdv&MrYYF_0S)lp(vokGmBW64IV6Yky zp>a`#Ou@nAcJ*yy^x-V4lY(1;Pc4WCTyAzszW_t)5E>ZlY^eu_#d_EB!%H&&p`S3* zCib&oh)x2S-MxlH!^1+XguwFXdp9lai{;&K0ng#oa1ArDV0moN3;1HL>7|QK=puhf ziTo#%%-kSWE%1`=yqbDHY)RV`f1b8frgUfy__PZ8@cVe|D7m6pOpJ`Z^h0! z!O=d27`V&_GvQi@G=n%z>*hUyjm?Xd2ScK!qXpr$I579L^TMakO5bKOBrO5^Jju%R zDGgj53dQ&(c_5TmFnSUX=5G0&%q*gtT&C)C1!4cIrp)Tz=k~e5y5FxEnIrvn_?MUI zMh?TTmmjh{x$U3@uJPfe39tDqht2yE-2_E~pD312vpH?+WrGcszo)wM#+0{*?Ur8` z$pw8yO*o0VyrhM}B8bFFb#8dl7dTIN2fp(o@tU{D2<$!MB25H_9&qva;0feRHqn)i z7o$+Eysgbu+!~3k8 z)Bjh*V1Ud`-FDA;s^E|v3Ybf2)+Tt~IX3tX6cYoj-#&c!DyZG9j1ezOq{S6@Oo zZq*^Z(4!M*#FLfrQ`m=a<@TB0Ev9HMwZ}@s;es!;xX=g~Ok3{cN*tZYWC+N z-_7C$1dFK6ZW*s1nvMx>ZY4gBC)pdQ9zLu{$Q3-??bO_WCv}>aY142N#>Hu~?zaUK z6_XlWIfTiCp-;_Qtd4m?;zm23!dn)Vwc&Tbi+1n9jX=7H;SjPU8i ze|J2WUd+j1#4F6o&^*>Dpvz-)WtX3BmtV8tOXnS#@qj>5W`W|d&D94L<=G$dPlmAf zX%v;ta28&>_(XD3Y@f5k#?fYf$5NuZ_Fa4rQ+KMw9oHCgrT|hGmpR%1S@gI3GYT%Q zzUNZ5Z#Gi0Z}H8~7iFY2U5#e_{7SyPvh(8`vGlgmPWH+B#`2eI9Ir~uIVaMhw@As`~wOnKPru1%=R{B)-HKV~T`lyOswEkK+Ny#%zGfl(6?`5XI7%-E_VG z!kqusr}s~Dr7sjZN>>JlB!|i88<_<3h3vS`&-#2?NIIP|Y(VWWH zny%6wF`DY9TG}V3_R9{kbuGn0DfZ?Q2^rBu-zsYeIiiIZ@+M>~>4t>)z7ys-!*-Z7-p*4^$z+hAqQw9;u2P}AZx2Q@S9 z1q!6n<|-}ZRPYNYI5u@iW~;>I_K}Y9+E6$rlPZDq$z%!&`()8gHOFi{pP0GLOJ1s@ z#=>F;@?EYkUL(oHTp|77WuevF*k_Q$&m=!Q4>%b}JQ5xTPBAC83mlO=JSOek%sb^R zB$GwmW`TaHF~~km3w3leqauBJX!1u67|g;}L!wojrzERVX1nb!I62KB1ivu7mxkpS zRh`#7C;?|&DkrPhzQhd{;vvf8jvq$HKKDFOQ1K z5l)-<$tL>b*b^{!e$Hmyr@BGTo}8ikCH@xRhXpSj4xSBE@kj}Rr>3^?%UaDURantmO9AYS z5Fb{QKt~+;OHOs`uEBElzfTjTRXKfT-MI5u-SU8H`u3SzxghebmcS=vQ4BCxu#|>XqM54G*u6u3P&&V4 z4bIkuNje<)gz9-;YY92Dja_Hw!8hei zv}LnTOGP7n2!jP4sL@7s+n-h|ogW8UwhrwjCWgMP0X@wSOH&6l5CKKnjOE2X>~#T`VM*BpshpK1~| zbNZFw%^<;6xheN!K$?3*1bP59PfZVfqPVLKc_?`|9;m+1oiR45nx1NN9$$a1I!?nP zGh{0u(Iuk?jTcDrdW?H~P_M6FM?vLflf3&J@}z`Z4f19Wd{V^a8z}be?F+JUiDSJ* zhWF2BHqNR@Klugsx9*5^MWhKt)d=X;@UA{pj32w&PoAECUkRX}P_9B{P#yGhTSqh~Ig~s_5w=Qgeg?029ee697fjyky^* z5KnSl22t07P)P&-3!dFQ2R#srzTWcT2MP*0JYi)vvb`F#L701gdhVI*JEP}jb@>j) z;jOZhF(IANm=7#x7S~!TOFf5pn&N^2 z3CmFdWf^>xtXPZJ%{U(vTZ#{u3m1Q$p@op0yu0ju4ST)!z&R^^xeW@g*Hpl{RFKq9 z6HY;Q;(bVJ!sF1)mfCIOIrE!JsL4x&kD=oUB}Z+;)2#GV2V9rG`M-1K!QSdFp5j-j z8IsSvbU~u9+%BtvNL&kxGzTQ0u)j<#);g- zH@3`adQt9hr+fP$D}!70yRDZs_qEn`OJaR447ugG}{s^TNg1h>{J~Dy8j8jccS}`7^kFBkGC`YC;=;zY> zGDEy34@M}HSk`e|?$jBeXi*>)jyp~wUvEZx$I0~^M15Qb z-SNI;Op6_*c+~J#!60ss6eXrMR()%d`6aS0*aGjpbVt8aO8%h#MaNOB8iF)Mx%jlM z)qP68Z>^ALj5l|bB6mUTJOCjyp;BBZMAS;eat@PbpMKlERxB%M%hOU1q70)tvuW&! ziR1SmqvprUKjcvrvJ4b?90ywxzz?*yGCt;B`>N%%K*DVpSrKylnQpS{@>e}ArG=Kh z4^L*V3SlF$9zN0cJ(2L4jL(hW|H_ntw(?-I(^xJc2X-097i=2cW&xO1u1j`D}>PFG+ zl8K#fMXfW;mU7IPguQB;4$|rOBfZYCNh~Ab+w|c{A!v4$~6no5lC@ zxFt=-p-skmftS;p3}jnbtfk)`S{&X^%y(KTYYIV#I(bV6`Ab|)FM&I6ymi4}l&Cj< zR+04p@tBn%UGQ^S9Mi~ZbuMWR8uGF_T`MAIWn4DnZ92LoqR$vbVvF+Qljx2t_Dsor46ePrJR5A! z1Rtm5JO-x#F8c0ufzLAfJ5trAF=>GY@f@G4t~@<(^;MBc?pKN)>Zmtt+!Xkdf{J9? zS}#z=YiaT8aA_1fJTR_A*CS_m$%k@d@0bTrnjXeYrUXh?Z;w0Pff_HOfu3f{Ca3AD zgtxV|wK@G_Q(()r|IA-;w}O+YrnSAUJ@BMV-q&9{{?6e=RcsE6=(66f)^aj&MSkCU zyL`1YKzs_WRO`(MI%v)n&7WtW)G`OjLyckpYk-2VxVX zdWLNoNfYh$r|s$$F-Lo20RzPi<_OTGCrfcN{LN5!cy$ z&mlQJ)J2f(;EF+emIw6?Jp^WXfE7fK;RC1;pvPZ%j0^1rb9>HEDyD%tR<{nOfj>LR zzGwTOH0SD4gUK&AKJ-=Xkc)QoKSAk26>&J=+NU-#y2Y1iMZ{?uuW7|A`nTtsq~oPM zwut_i++)h|`*l`IPQ##CG($7oSO-B}?DC8#y?U(H@lK6(Ki8(YE~oWNi@obMiasVExZgI8+u^5k*>o=J*te>l*vaY(G;GAvQZ^1umM1Tv z`>%97qTp5@7yiw_p-8sK;bXTYP`~QNBC05t4PNw46*6BR%-tDqwI+CP7R{%b`9c>@ zQd&RmYs?3|c#iSTRNX%Mk18^6#ELuZRPV7SXl=i8oo!B6QDyCJ%2}jccO9X>w-{?X zCO=of^r-Pg@HyMoMy625J=5S9{1N~Wqm~nku8CHDPa6I^4<+oX=e72q=9*r%K;G1) zB-c4x^bwSXh4n$35kGz-TSKm6u-7X9FuO> zRyhH(`c-nX6$80g!}vL+yX$l2#vNPmR4)w=W2Qu>Vi2nnC5~_0%O-eG21|NY`M`G=>^{>PeYRG(@hL;-W|m+h*QT^ zkJXmd^|yXt%mYOs;%G0TN&PcQAnILZetBQ(0dumOI>-N-4Yj2Oa=6wAq-dzV^M17v z7$%cYqtFn3f&MVJpx)(SSSj0&{Bg2S{=EV)0#i`IETqBJqZ*Xq-XsWsA5J4KTj)A} zEphrf)h9V(i+lk66_r3_<^9B_*I?lHsw!Dn^CjGI{o=l=48}hkKnN;>p*#&G&f0(v z(;x?zGGs0G2|N^VDNA-Ed;0h315S9*=&t5ez?;t@-+5dcasmvYlR}{U_EPH&h~abH z(S#;`Fb30KRJYWs(zG{MaPk(jAE|r_;3_j8Y&^qgz(r_9Z*k_8;pq#}j&&rcHkrdv zQn5ER{_E>)J;q)EqSb(%WG%bKJ=p{?kxX&U4*`+D4`|D}y_aCVUhrO31QefMr&;P6 z;2lyjdcU-U>Uk*#w1HBxGJD7xJ@Pa`XuNUfM{GR*Leh{YKHg>!??`|$7q<>KI`gV=CpKHR_|7|y|C;(F1yOf6;ALYGBV49nLLmU(l0E>Qbrw9k<5>psO9czMs(*qwCQC zATe_F2X6pG^*biEw8#*#XAM=>iGfYF+x<)jVxPX*ReTG#oR~gaC#m2LeUZ2c=4x6uLCKbI!kmFE}n{Z;Uym@2TISI>`F8;#Nlz ze-Txlbj^7@ju$C^rQ|wOJn3*aDwYw?t@^QtP+kh9Z+fZ^3XgINX{<5VEYdw!^2}GK zNSnXRB2H>rZGU3WDt2^Z)k~M&ZEdk-r<$HlgP+X_!CvQRS+;n(h)0V^{^W`({i-6Q zqExTNYDPcpZE6*0&jo|b)NjtGMKNx}fzRQo>gFqS>j%~a=e(*q%B$vrkR?ppq%N`1 z#PP0HK!>7KjB@YjOJ}KpY1@iToxT_vm=%f(FDJUVZFQY!7{CqJ zsB6=<0MvB1ldZhhP_7htDH(c*myc{Do>U4Kc9r8b2ThXv0k-_90BFQa!FZyh&cBsw ze;CmxHd<}NOkXPAAs+5d02_Y+;KBB=>`UrmP!oTRa<`sFL_FiE^>3w|mtFWyG-F{y z1G6YQo{k2NrUr(yr5t5=$nlvWHd`O$#MG0OA75U>+{vIk(NIlELEpQgOSNr`M^4Im z{r!&*GrK{zUCOpy%$43b1Bz1=pr5aE0f~7!6@YT6`xa zK^ZY0>~a>sS;qzINqc0{c`?moEt(T%QhY%7SawR1Q|*D z@)zIH3ouGRjGkBab+E%O@FB#!cJftvy}^SK1J1(+bQv%T4tC&z;#xKtM9xCN$mdup zJ?+D!^=z<^jV1svpM5}Tp+Oy9w|}sSGNvI}t(dHGt~`i(ZcOL$G`@U&@rLUX&P+$A z5-79a6#5}SYJbNQ+WO%P?9{~K+Jd(1JAjqgFKK~6hUoOE;fM&pB(_J;T51BSk-qmp z1e~2Lx>-MzzvMYfvz?yamqpK^_@AP;zHy2Uh3$zN(vm9g(?Ps zv4bJwAMlH5w?fIH5323J91C#oQHO&VkK@U&6m(2LhsZNyn(I~22xMn6Ab5L>u05a; z_iQ?Nh~Qa5qo=v<6-?&=z5nhXTn!L%%S|@SlK{QHi$?)sdz?mxW~@SZSk7TkFEl;R zXch~YRNHAfB!G!A@fikviewa}JOz(*9*z%vPdf|HKUnq*%W{G39?<}4xOaWlqJe(l zZfTKzM3vzZo+QpYf$NdBtCx|P>2~cE4x4Wot)MI4bg?&`qj=C}_Y<~vej_$L(m-0x z30gsp!M$bBp3c(k2|BzkVL5IK6;Yt;Q4o^%O$-kYBbrbT@7IFsz5N96l9>)0l)Lhd*b{9ZGOj#6H3yzhYs~wt%e-?AY3Hlcfu)4JnQ*WhEN)R)`uLGjtz3hkGaB zwtU~*zuC!+VlkG(6Q@Ac|nHq%JUO0j&SFGC~@R3o%D`x(hn?fZmod9_Gfk&i)m^p!MAY z^fuv1C~}07|C5_|#tg(&80>1_*J@F~L;iu0{*Q1VV0ShhD+uWEge~9tm{i~FeGLaf zhwnpj>oUQp1`(y}M?Bp#%$bWU$o<369qj`QgJt2_q=?+ulKhXpgFxE_d^j}YoH?&Z z9RT=?WYB(5MVLKP`yVCu*`{*!<>bOl?M@3}i;g_C0S3MqqIbOHS$kwW? zV$Sb#V|I-ChKJv`zYFN@I6h*p!_{Y>xalm;J9y z;a-9l8>Fzf8G~WxC|t8!vxOg@_tKsB1*Azfn2)e~_dNPXqX)=Gfar|?fQIlaGe)8S zbr+lUpm!^jWk|(&g?;2W%Cl13|KJ$@BQO4N3{bVdEC(o#C*6SJ`IDvp5IiRcwfKS0RQ)4kKJJ}0>2@3@#Pd@=Y7zjC#K(KGx zAgL8|d`A^}4lZKHjo2fu2wYN9-Fz{%1D_Hw4glSpAsv`j$hX!(VP|DctqVY-J~B-g zfBm{B^?#p4ndNcL)2Q@|RfpH=@*$>omYZJbiCd!DLar-o$L^cX>Sd|6p9h}Ho-U1R ze=s?)EjT)wERo^Au}_6-6!D<;g30`^Gq?`-@=A)kl6L?Cqbna`Ebf_)CKgu;a$kzy zt$}al_XVas3Hk;vrWA;OEI^O&VmOKWB92&?wNXm}e`QgBljZuLIm$lv2yy~Z4P26S zp&3!xa`YJS!QM~2k_)WqDH$^VW)kC=z`zk}=Gz4Xd4b3YPV^ZxXl5+P3!)1&lhdmu zev6J)+oY_hW&5hsZC!lyP!G8+f^)2!>0og#cUK^zs4oS$8=^QWWzy4KXC+*|ne@9C z+yRn(4wH4EBWPIvj|2ccXLvA(jmk1eDT62iJgUO7?c4^8D1c~O1CvmUmJDE-Xg^`d z|MyD&(FZzmNub!SF}!pm<5_mADGwKDUqCgWI1NQ8hOmGj_V)e|(4GL*OhJg<4)GGS zDaYI#AkAP_B4RbF5BrE~re!k~;_X+?@n)E!a%QrFb05X9b1Ewuq1u9>Lm0l#s zaOwcITM$;l?7j;qW=B-rBd7x@Vt6^=l7V{lk#GjcPHZk<&mSZO<{vz3z{il6_K2i{ zlse$}@XX+JkHm=PF{La(rDLxH#`5}W`71a7r*P-bUY=WoCz1s_V^U=HABO=b7R1kK z`l~7&hl}r(5zB-lad7%0oHr5g#+EiMSAUU^h%A>ZRr&L_jk25v59jxd9J>aOuZvLL&jt zJeK&uY@Pss!f|d_Ip-HgN+ zgocHA(FQxVpa0rfp9b^>$pWqQe5lVERG9vLo&CX|g!$g!v>PRj%H8&YPM!y%fx$dQ zTCX^nkOKAA{V9ddm4FN$?h+O6005%qRj=*ERpLZS2t)7-L_-QhvWPWX9Bb+uH!@#ek~~x5lV;XSoO0H)eiG`< z-w4c=7Z`Gtb*-8{^dI1l4= zMT1E$d~DDY){#o8>FV~5IXM{U`xG%5b_u!}OF7nH;yHrih@%F*`HJq*n3TN=l4WSM z6+U==+ww@o62pcR^oIY37O zC6ZA__{on5Kn?Gz`+_kI`$MdQfLQR7L^T)Ye3U<9pXbE@AW(eG`}XlF&SeWNB7S~K zK#8Qd>HvZ|C>gTo1*n}U@tHupkKEckVLV|Fr;+C(BFyS+@C~&*wl9d<@F*c^niJe$ zZFe0vvvq8mnlGUfGuO&nBE71dk5qcEe(jcZ4nRJ);!mlDpD_O2k-DEr9Z>F z-M#&@!Rh;(lPSYoV9JWTjMuTHBE>C}=n&t9KG;L8rMrF1O~3!LO~2&k;;x(A7zsDf zSQnf7dhHlRO%2lCG3@Sv{v`%A){bvGJMNRZd(wwFg#af|s|%f$Z&&nH7vSVg%ug1n zp2r1llSH7%0pp$&zl@_}Hjwjw=o1+ZDOj&^CId0gxU!R{a#i`kEp2Bd*2Viic zg(5hF`IYlrHt^$mz!rTNVmDN0HdHC7sD6g<%u1xf8GRoY$ZSOavE>?6WKTJU@ZP{f z+H#^oAlS_N7N8c0T*6MJ)`d$tO#;rwc?` zNFy(R=&H0=MGs9K?o}R-q=X0evcTM`FSRP)n58+}+;w6w3+u~$aIyCm-~gc2?^-ax z{I31W0=8BVftaZC7KUc9Y8DC9cT(50cZ#)fqKUQZSJ%@3UsAM;F2ky5vv^vm&tUAg zyrwa}FVc&lVx?14cpn0nxwkS$=1S83%9%*c;H!|MH9AA!`G^uAL7>>BRQ^G`1zVi&=WxhBN*?QnFEciJP4~(WwN)%nI^@iyL|4*572VCq> zO*PW(UJtWfWFDsYW{ofqz*mU>2c3>DDfI zyFAwlihHMeg$AJX9v?k3tji*kwQaX(qlGB6%!cgvX@XJ>V6EE)fLk$-7s9O=dIT5~ zvC8sMc!4nxql5!Z7n@J`dqwI%Z#K27(05y`Ze1UU1zqSk^#54MfXx2IB~^qz{*Wbd zd3X~8jy3WDO$)^;eZZ}OqK8cfYaG)3H^p!<$zXQOA0N4EJP$Vrj7~z$utpC!bAiL= z4_AWaVz4QTtsm_B7B4}cB(yUf-^T*+W0cJCqMB>&=+r50SAf8WvR6akf2>K^MENhW z69+oMzsb%czuM7dA2(kJ*<*n(bk;)2ka87TuCQoD6L9ocj&D8$sgKa^Kez5*DkqHb z%75k9|7^27yLD{t;^9L3`fm9g$}&Le8f<*%K}>YC)1HbTvlxhw7)W#hUmdnE0Z9l4 zlJ+wOSVVo|o&PV1ODfyyc&v-Th^Pk|#lfRu*GE|VAIZ#2_$ERWD_fLr&6;6SvAOa~ z>t@kwOczmYKM&47ktY3k*aG#_T>{iaZNL8@h3_91u|M7*z+_@8D1^ZNnzJrJ;S7~l zG9VlN4}%sIBz~fxA7scW5BLsJEDZ@A?}-Bo05JwIXsHMcqOkuSxS$GLb2M-f7I~FL zNCT$^NUMZrm{b$Vz&gNA0eCh?utE(ORQE;G)JA33u>B=R+98<-3 zNGl1jGC-o-;G(qR0K_f`Tv4bOw0-dnQPva(*8rVyl4V-jRg>!NW2FZ3J!llg8czSX zyr5+BO;5U`J}_jIt!C+JbU8<&atR91Y~&r-(8l~ z^CkWO=3%^Bn~@M2Y#z>{19s>@e@iSLw0CwElJHrgU|?w!W%Wv;yl8YyB4WrKRI0rr7{y5#&# z7t1rCiel2jVDv%a!eQAJ&{Bba6G)lN6xe)xP=sbO8r@HvQ28T3VU7uzPf%!kEx8Su z!#WG+BOMC+qyHX{usQj!@#wZb9wyQLj7OqJk*+*`?Bez_cb48gT+C%Yhdq~z=~}#8 zEVA@&(_}8cjkUsT0$$+MFyyWJ-;j_~QYr^)4+3Au5AP0wMP$Iqt^F0CM8L(q8f|2n ziCupsfu7 zV6{Ph+L=+1xw61!1MDoV>?TKcv7)V%9|);zz?OI~SFC8J0>M(jcBdg>T*D<<3HVBW z0oTa<7Mu$*vka{-uz9cajIMNnbrn4U$~Qy7nm4d+!VlDsaG-HOv-t7s#=aw%nsy#0 zWdoT=GU=z>11y!q5 z-A96OpisopMen=Jj-GOI%>M4?qI8d|>CR08U`yt@B7rrWGA(WV0^*Wj{7|wWdJ^t6 z74N!$K-GU>%d~u3_%UX=SR%-|lCs!e=Rd8eR!Z>+%bpHauKZ<6MWypSC8Yb#Q|m40 zx!h`Yr|s0HE#&5w8t{knYi9Ck-gG)4!7t#e_B7HsHP3!^G*6e!jHI%AoF6%?+hUqj zS;yGSb;S*Nq`hQfWG)lC3P||;&>Wi^*2+gDf#s-(`-8>7>V{)uTm(~{h%xJE#)E3n zVT-9{a@0NFmKsx;CHU8LS8KMYN~I$E0SbDmXDv3?v~dnE6@RHROFns07a&nNOE~nG z|INq#dwBoYx9q7v{_N3?pRUWvz?0JNKi1VN(^j4Z-!dtY25ee>KF$}$=zaFtn#5O) zE?%Dm;{z1k+_yvs=>`Q5dVOEtx`K6RkxZ#C^oqpkD{|l%Nv2z0)PU6$SurLVWEN?k zWVX}WYx#Uto%cmAZ-7`?(j5z?LA31Nq!w$eCPHX@MY-{o9ieC1yWT!i#lX%Kh!tnC ze8CzA=tl+H^j3>o2-VACKGx%PpTdz3N@57u(~;Q@l+6bC{SbaNMNNkIO?y=zFchXFy<703GlV>RM>jcNU$Fh=|Yt^*tR$RaSC5#d=oRT%sVafg=S zmX=R9AQ500l+vxvK-8S%lhf{&K*d}H1?{+gPQQR6UWRY?4OqOUD!1m0z!}^9$;anz zp!d%FJd2ZA1;{Y--jZNYmi>IezyNCXT=w(ZVm`rFaol}K%rBJju!r+kJO?Y{`0U~GZG4k}$utqLilua7FOH+bHUjSkoi--lkti3(=JL;OQgNZ#~vshmHfJfp+ukO3_^#^B~veOrrIGq~C=lwAt z{$@M1pgIQ?EG?|e5M>e-9esM3op`vty?yGy8gh4IY%7qr2mpsfMvW__orVYUH{X4< zSs>P zA=HJTrIZDgAs6ucW?uA8^d%%h7H>cf0ky}9@f=L37xZ{&p!sw&PoIM$Pk-9pwojn!~Jv;SY2K;S+$>VSgrJ(_eQ6j)uc%9!2$1KhMVOw z*QM&Mewwcq69z?dV}VjbB-0v!ibX|CHaUBKLx@xkU3{ro_uVl3_HlKBS6f zq|2`jJ2*I?YN-S=S=A2|-FAeNCS1%ami0%fv=5S9C(vlhJ~;Qyd-qg1Rlz%fy_=@+tL;!SUi8F02=$JJKpEg$D;C zKlq=HD5nIAQv&D@g}6sHSsEJf!M)u-7$FS}3d-`c*}mYmpZ0vNW6N+Q=Bvb^hTFQE z>lac1c7c>Sdw`#{m!k?b#K{z z3+ZNmc#{8quyiUs1uR(426VN0yw(o_VFJ98MX@n17gkr7A4y5og6ZCLrP2~f`|=I| zdO<^B+;29>KFdY<^u6yOu(9_YRt#pxsGIjHAE`Vo%L8j9y&=Yl(I41>uTY6=0DE0-b+Ih?ThEAf*L@U2`IGr| zNr{uqp;lPlbsRF_SxW#N%}W<#Asl3*#|9v88Q=L9P8OXSBEMnbO*AUHss?)tH~?^j zMBrc7RM7->2w>O&2l;OH-uS{-g2#%aCGGkZS=KSx;1t2%Xae0TX!KGot}F9BsW)%m z4!htYJ=ySzM1ggV1swN=SU!jt>z3eLDfl`^B;Z5d_y_r#3RT*`ESj; zCYi!?kUay}-ozv;UeF0gx$#c}#4z4zCI;V=9jZ7DIuveYE0>CEn7qtZW)H;Nq%aqJ zKRCJkiIv{o-oo_s^vO=mnqS~n0bQFAu_Ho-4Ua3n@3@UATSu;6mLcNqhlJ6KPIaZXaD?&HD6)c_9@41+jPsLSI0 zk0c?)yj<;b&3d}Zx7+!zsT5Evm*qwmt~svKOV04>4*$jP03t>xS-=XdKr(jt>3G2r zMDSpKgI#cBWC0c&9pNSrJzz>cCV2i@a0K480Z#o$WpEkTFrG8L6=MSlAYf198OQ4) zmdFpcHq`dfVi0lxpJ0@`G%l&??KXm{oa)!JG#{R9$}>an-GE&M2Gk?s=kWKhI?0{t zl~29G`C$DvEGq=|o(@`Mz;Q@cmuCvpPB6kts&DHNo)q~E+8AVxKLvyP{cCaB?YXMk z;ZLg3%jR4C>gFUP+hv_DPuesGSUH;KZ)fWAdLJ#&svIM5M*N^tLm*o}DCm@za9T!7`K-ZsbdG*CE$47^apF+RfPtfRo7m{C%1bDeRl1} zU%~2hB&!>H_G3Lm)#w0$-DYB!<%3+qiQ3Ne4XM;K=vjNzyo_7QSoL093e(`$fbLi? zeC4p^?FbCG>A+Ih^SVULU`1*=*FnixI!W2GvDTg!!+feB8vZdKVb&hE)>$^OC)h2P z6Mr$lrFQq8p6g;YDcTJv=s@i~djCdWvE}-{z4JDk4eG0M_1=j2mboC}-mBR|Lr4b& zZ@R!J7azA*M!s}+uSx`sx~-O^WHsE6R|KveOG_MSAI`br^(H|sBbR1QM%R99MP;X&-Ec(K$@hplT;8rjhCl z^Lcd>ub*{u1}TZFk)JxjCDT$93zqrO{!_1A)P%H(zdtbEVe56}nPo2hkF zt~)9Nd(v;;Z2%7t@Y+R1KhDj$zCsj~3|6cJFF5>T_QsG&V5KK`ISk;Z7B+|OLN5g2 zl}Oqh!`j}#D~2H3d-UB6x8A*j8&unh{CqaMO+{@shgUwY*364&!a|BphY{_%X8#n~yXoeXX zuo6pFkPQzvH_}d`K;>2g-uJx&q04zwlCE*KiRTR(Tz0SBPRuo7qm8-+%|T#24{o9? zXn|N;JebRD0a(4FgJ_F5eBaw6BbYt?a>%zW3#16dvjF}EdP4^o0KE3XAV@lKDOd}< zGy_^11d$W(kwhnO^@0`%9wE=?KTVb>bZDi6Y~_G4sT*ax-jH)1Iit14FZ4CtwhWl1 zH6deNK6tl)n&tH4+xm6;P3H%S+rj&y7)Cb6Rz6cR7K6Um3z!@5sQQ9yLp5$9YB-*sI57@`+d!@rYL2;INK)*!O6cV`Z*e^zm44i=nMgwz=DvVzw$ zKY*&S;!+snJIH=$`3}*VS>c+p>O99wQCGiRDSK}d-=#UWOlK7d)&)W4cZlKPP~`QA zEF#US2B!7On2J=thRzsv58fb&yy*aqVn&hJ@2<-j_6QiBdV$H0umM~F=Diww%jQN)-Zj3!sTO8RR;^CLWhsx&26yh>OPizFvIwmfE7Anr~qCrANhd zqTN&z&m+-tfmsK|`pK@p-|VU03NxTEVN%tNs$o5OxMY2QVKLf^wU;|jZz2j>B}Gv8 z_2d85*qH}HxxRh8le8QhohaJSNy?tIV_#>t$QQASz;b9bfh5D^%M6xYE_f3 z`rKji)2Ix2G_i{Qj7YY-*sO5k>+>r;yNb!#$V0_T>>xfhlJ@-cp9-it0foKk>O`og zLyvk{Z$<%_JJC@JddM%j@J%F@K+H|gfT4t_G3sR2v7wk+> z#DpNPVuUQoJl7OmBPQr(Qe>KZcaKlK} ze0sXT{THsgZk%CT@SC$B13Nz1%ma%#3iGr1XD1h`j)$joWJmZ?f>-w(^-9A7^YIvv zmmCh8YBsrNV0j2G#Yj$eacfEYn@rf>@odoam-kv#oKXu}^h>l&TcC#sQiJ7#A7YU_ z&NjfsbqnVWO4bn6XXt5CX}ko$*6 z0WRocv=E7F^Cu2^SjAFU9~PbWFRiyuHJWe8q?B|TbvW+aD%-`V+ND#)tI}E|i%0ua zBl`|J(z2J;&S5XXj|~*NhMe&)P5XSvv&V>RlXlp`@yu5i?d33a znUIh&K&!@1<=vy50zN}9CwYLA8GX3(K5Tdi&3pgA&#=j5Mfa4&maj-YZeg*0r2V>U zu5ki7y1V?bFpo&ZeKosd?ONJ5-P_`yJZeVu59Kw23+AiHt*FBB*68c2I*!mAOrQOGWVeZb{)mVkQA4=TJ;S9C2 zb#}C0Mqe7{Xlh#8f!|JuJ9%mte0n2XoPhhzE{sg8V#^-9$DNEAs;#J21vO*E-n=E? z%be5`zoTsbjHF~*NNvD2y)nV5)Xi5U8iqXb>?v3h*huA6zXT%WMX%+QSAU7%YDIUi z!#G)5m7v(0B=HFyLsYA?W30?Fl2~f=(LZ=Vs4xMQztLSZ>0T8dap_F z{_T-l7Xq`2hqJ00BlpAk0&m*e}ijt5N zrkF-lK!4&e#G?)LFO0!&O4E$hv7voz6Eq>ryd2i^D%ZJQfD$MwgVHM*^(kr`xJN7T zcFFW*!$MMyi`pO1ZuVZ%*B>-l{EJ}{w|N520oO#*AKbV;5!_3-$-3G(gf;+MzYx zzSi@0!NrKi4)-j7aW4%O*Ax8}DuK*V?-!Qo@AVhaoH%#H-p1VWUY&IxI#G8YVSGPN z;etzD`H*iv4VK!(-CY?7Z<=1XhTrPcy;n$-jIMa|hI}T;D+tR4LX(q|*@2IEb0>RT zC&$0tY7z73U5AIMF;94lYlF^?z`mMcPY#((H_6Uv7o1vUlKCk*6eTqT3V67beAYE1Da@6$9KcBqO77>cI6B@$67@(}%hR61N?q+X3?dmLD|v zCg0ZWj#o~3jb;o8PL1zsS%o%u(n zr3$qrWV4^xkD4Re_+F2YOWNUmc1lT)U#X&d^kc6SmO>0S1}g{e)ic&av%v4Hx($1E z1~a$>{K2zQNby0Gh7!I%Tp(SBteL;Odd$ulHSp0NKh3E}=bDzN5qUf3*c>(oEL@?$ z3^s>!o(^)yU_JYl7bSTbk3fJ4v%}^!Ra_t+Y0w~(+`I~$T96h3I{l}UV7LAwPI?8h zP&j63!foap=LB>H*j)Yx&Iw(9&NBiyCzM2c`-76y_SuBSM$S;vhx#*M@gpN^W_%>Cv?{iKh}`2l!F=-h=XR2gIR^3 zcvR@_9#kGj-WI%Ym*5U$@wBtQ$a?aFxNe@t^Hlx$`>sLw5o%l|trc0S1nermtnaMe zyzFK*)B4vX=++jN!YC-wPSFBlvHI&(AjIZ1=^tYtt?BJ@3)IkwiD=3D5XBomO>&hTTRpua3-Ie@9(^%) zZyz0`dV22#@56Y_j?8OAp z&=8I=%&&2S24v-cA*_d@-c|HY`?QxN-$Y%m62@UIhLw6X$Y@adCoBwrrF@WM*Hb74 zB1fpN`ZVAgfOUnm_b393kcFDJ)7N-c(2V;*B?nATg$J9GG0C()p;-Yu*Wv{cI6@;U zYr!7-2n`p@IhmywLJwX(u=NrM#W!r!2U*1Cb+!6e6XV`>-vz@1C~AUtqHIvE&G>_I zV8@W0y20&;3i(Uekeq})kg-c{tZx-{n-QBrM>7_tb>Lf2d?HQ6tYI5)h+J(Qcannq z7MRe@leJx?s3ZmfA~7GRl3sw7J4@}MxfYLfs3|Z>c&`=>)E>1{;p#ZBiR>pG2b!6F z(s4L@ufpQ|z(asfz_12YxMt<1=tBClDGq4WyYC=A65X#bOl^gL6NKvErcoWZl~)P_ z>#jS1eGaO~%}O2NM_i9o>D*ZjLCUC7HZNka_sa&0(1My)#lyPPZR6CMr)53Oc5K$= z=UwV_itbc;XLET0-JQ%#$k-ho7GKsQ%Q+dS^Ndj5_LRn-Ex%svtx1cIVwOV`7USDS zRd=;Ggi20$$m=QT##uq!>SG>nSH2`U^tPuqYw6zQ_U@;xVq3a0D!Ve$v~B@_65rjJ z5xp49*^)7GD-w}g-+WUC%79G`ED%r(rMa`Gr{{RbmCFy&o%s^-(D_|hMV zuQKYh>w-Fe?89_?;e_UC{WWmR+yR4e%+neqd4R7DVuKt5;eSwJLrogIW6*?vjinjF z0of}}3rarFXF0XYR%JffXj@m+Zcv%KtDkufsrv^r{b!W*g zutt=sxZ2urpnNV)a=zlq#Yu8C=;o->#Nyam7y+u;;x zoI-3^A9;7`g+v^)PS$5^3Zy^-4Z7q@9UPC1lg=EcqO+RkW#t9b?*hSbtmOdfp6KYf z+#x(e795}(M?kGUE0h_#=jq$#B;#sBA-H#QPw?DT4{S5 z=W0rYJ#1D?gu~ix^@#Rr`=;^}?v(%vp_c3;dFGjVh$3twf>WqQ0poLmA$Yl`SWv3A ztx|5vtp7nWQ`hQ$wo3Z%t_&H4?XdwTb&EJi)=W78@@;H7`hdf14}=iYKY3i7iu44iv5%}TG) zjDIkYgkcV#x}F}v*r1$IUH>D>xG`X0r(=p0lmM!lRuf{8&9zEAoF7!1_*SxQt0Au< zXYKT=7f?3iW);aUEzRkBX=hdQdHzQ3WHHD!4;?Hx`Cyzn)$WIjA2Malp@DEq;wc;I>IF!fkSzCL@|RqBm7g150EN1K0%VAHp{1Cw}9)G z3g$1^ZXMnk?Peg>L!Ug@VELNs9LHr`xWPHCX%^I4!PPYX9ELMqirh#>t_xhs+hS+_ z^yVA|@R+jjJGMdkg?Dd2ckM7%lQG^$HBeAha$ThFUoJ#iY_{?IE3y(ibf$3sNl)=cm#GN7%2p2Ep( zi=qZnfp|)(iCFXB*{+9ar`HnK*8RPatw&zyn=Awdt*a>$l=~q{+94djb{E$lIzyGS zekBikGgct7{^Kg3Gc8i(xeL~L9pU>vdnwA6q zH{-RV-;)Z3_5LwH0l73nHIOEZfWLq1g!T+<-?8j?J2U$v1ESd6W~+5F6Ko(2A$vyj zcRM~a$`5&F{`G1R;BNuP#C0}z<4P*H?nBEm&|QH<1-vB?y*+J`Y?+>h%xjPG_2uPA z?{2Y5D_rY<=J0=!4=~VF^rzP$$a`lTuwbNVbOg;fiK<~3A{NZ^=8w(lY{4=aEl1fA z5C*UyApQDPlqo_amEY$A2e9u0aXe;rh#hkql()0SiLkm|DOCH(I1$27qeCokrn-WZ zyy&eGw~bxNF*lx+C}2`@V2KZSN&6mWi!P9lyMw-tVqD;Kt{n?Gh`lCK7UZ#r1uW{zl^oCK>PxcPff z<7CD0047TrUSNt4kHfre$Q?-jfq&h(e)o6mcTm)XbP)ui;V$OJ2jDLdAy;57)(o#Z zw6|(8`~hOXk<^d~K%i()-4^uY%^zF7bA)#DR@IC04I3egi+!&>VzUxP`8;pdq1=~j zif?{3lz!3+LY~0>?_$Q`SZG=x`8jJ0j)~arxP$1a7xEcHc48roG}|EbKG%vW*>?{- zC*Dm_E`Go9Cwlh4X=Y(Gvz;Aj9X^@Sr&wxt0^Igvbw7@~poXPvi#Eu_cjTIb`5%b9 z<3MD??8u`W1+V|58A){vK6rw~r{hYgVZulh(DSmeVN6jK5L@SvCMnX^Q*$2aZJdI2{L5?W0W z@$Q~p2vaBoSR{esC=A47&|?ick5cN5QmecYFms*PQ)aO>L@=s%Ipq?EYdcnx_qdx> z;b6|J50}Ijs4L)O_{!?*la-H?b8@aZKZIh7S6XR=M-2V{=L!z$@!eZuD*Zx(mc(Xe@X0dBAv6gx+q~lv?jR`oUA0a zY*tcQoKyN7PE?vX(bCG}YAd`-GK~Gdl6;z60_~3fl~}8yCUL^YHtR>`UHz`s)7xUsU{r literal 51758 zcmeEuXH*njv+gJ=C@3I`1PLM-03|C*R6rz&zq&(x3geEkUVN{}`)aK>>BrLqSt~^cko@-J zv_TDPXBfr5UlMguSi%ILY2w8V5;H{2t?Dn`d1@nld)4(64<|eK5pBce+F2=|3->Ut zErw#+Wq%w{z1ArD-ExPK0F=$bVd0%N6Q41Xuo?S&<-BOw@vq=1tr3EU1(2_L(1 zSn11go0p{Qc!8DVyr^@(4>{c-=|yePA7eax=*QtA9sDaOF&#Qyf)wFvMktvtXTy(0 z{N^t_oc_B1_jQV6J}*h6Uj+#t8G3n?;8i9GWy~GUPp9KfO$*;6q0V+cPk-|V^@&$^ zqm8Z{uXnSGj(A9_ctYup?L)FPLYq^?hX>yri}q;c<;L4P>HLN?_(+HDQB}Ndk6-Ng zc5j_W)}L{QpO~!NUebcwdB|4LES=rD!B0du60~&wFbPAIus^Bh8}DbN#EcZ4caQu~ zjw87$KJ!^Qj(S>T;lkY~Q(2KC<%I3wu9q`ngCgFs zqDd`Aoa(e`uk(dhTVGBY*^n2bG(Vb)7Q|*md$jpYJ*OaYf2ph6$le@GFMZyCD}lnA zqQkw&i?T*qXChPeE0Hq3>D%>6cioPxqrH-I5_3#*BzC8FX~k={l2^B7e-7ZJKwQbNa&c$q?yo`hMTn zKd1kc{kgQFw4$)$$PpSGNg>O6_i31Fo~lRhr(Tv5kueWO9^QCpb!FV2%b(9*&9b!6 zomYil+9~x#-qx*Ux#e5OL*GQd@eqv>%@WP2k?}NU4otaY`{mxl)#SP)m1HvRHwlts z?OVAKxySl=`WR_M4BPD2cpXUh40lkPj1mFdfonmHfnNhl>90PaCpvk|=X2#1-Eadh zc0tLj_buZX;y3Q+NuHEwl4_9XmlT&gCs`mtEamF2ThIHWreQ8<+TWhxgKt;hx^I)O z0E459o$|-)hF9x?Sp4~}ZQiF1>a7XT Zv`H_^WuM8q>${Y`} zq?cnxy-v&gRKKd6EIsYpPME;xS1qlh%nnM8 z@Flg5dt6F6TC3&c_dcYZ&9GISO~3KgEQTglGtF5Xzni1Wv+LGZuEG;#^-fCdDLYEB z#f)H23FaRb3TZ`JWY}ag6cz)@LtQfWUG@a`T8>>hcKjHu)yZV{>+bgMgYK@>#?;H+ zDc%bimon@gm6_T0%=FaWuKipq?nM{eDm@@$7q%Ac#L2^THaI`DIXF4wZD?1pclPvC zSv!th_gj4f%}!z4_N&C?3FOn{iP!eCJM+8>p5}*ZRjZQ~CFnN233#s*tP~N@8fr{q z;@qij6#JCSK*3_QQMv0zReqb@``GWEW8r^Fv$--a`Df8o*f2>4b4V|DwBK7vHVAAqk3w{ih_%tT29+C73~x%7YSN- z*pU_5mQ*iXU%(%z>rTm#QSYBV#Q%a8pFi?NHot;^>w47-dj};)fz2Ch&Fc%R(c4eg zj5q8jO(!)*cY9?!*s3$?qDJXPhyMuC>Cio-q9^Jk9ytAi@BWP|Hr%J8i4noO8zyqDn!jqMdZNmHglB|7 zFv#^ChO_EM4C~`d?abNSR?PLxDcmNv`EJvI>Rqv`a#EyxLi+%Wx9@P+{+E-m3YadD~!(&Qd65r6>V4A1? zWs!psO%^qNFkN$K6OTr}F}o_e0RJX!shasCEys7yu8e)ktV)YXl~1$M9)GMP-SsR_ z(^kXVsVOpayPDL)<6|$aahg)nwUoUV+#6G`ZzmRibnxC;(F@iKsf)b1b#ra6W~krw-Xq&c6Ux#|GM)l)K1+|7r!$rS)`M)sjMPt`(TDXoThfni))1n=5I%O&E6E2(a$i1@JP5Vs? z;tU$v^UKus>{Oi0P9)1ZJGxcXJwBXt*Z7A7-Dz2z8201P$>@%q;XD@$vd|N3Uj06s zLaX*0FJ1{PulVE&kzZbo@{0LfpUWV!EscIS1ik(zwwi{UlKcHKh+!=^kR z0`ggG9cM3OmG?`Q80+U3tdx1T7j~GrVV@pSZ(AbeqpN$J@__uW~pEM!`(sr}5Iu>$pz_qEm;k zpQJro@vt9)Pe|1CB}ZxktBUf>#jS;Lo{Yv&F*xlLW{0K;Nq*gs~kjQV%6;F z$M_#u$NAu{CXrN)_Rytx>#kdRFmY(k#N3fSpFod^`iAkIwmp_ubjDL~RcZYEhM%YQn{Yhg% zwl;fc{***tD#F(;!`FW1+r&ePG~;AqL#!&*i+ObtdIk)=OgZ z@?iR`sK1O)cisqL>L}F2lV8%>YuirO+;z`)t8TT9u5POv7||9M>zSyYrqFt0&1)PK z_volL$6?%9srorQOWRJnRZd?kS8u8tY*C$6!yc)Oky3q!Pbsw2>1rSzC3xG*<^pZl zHQbKUKW-qOvfMa^w%7hN!@8PW#W|)%k55d?&nV%p`xO>_TcINI)2B?At`wz$qa}-> zJf@5iEGp@pn$HKn8)1jnImEV=qvBTwuPvFT)+>s>LEuTab624l_GK*^E)mwdOxX%U@{WzR0H1&PRvY zy6wDa;c5{mUTiS)a>K62YMq#ICUSsUQkFVzj`$HqIA&-;J$)@#ZqU!My4}ymRr-v1 zvB9)U;ft#h3k?$WrOu+W-5*tp+0>ge{26O&wt{uapOpWhqH>aE*XGo=I(bs?QZQQx zTao*N&77!=)+V`&DZ|y_Dh^)wv>B#P&(?9}oPW`G(`V6AaC4AF0<*$)PHDACy2uPWXfUH9Vm=jaeTVq?}df{pb)w+z$3UD+}phAqL@=_^Zmc{msDCQ+W^pnaie z#Ja|7PEk91!G*{4DbxK-v#TCFmz)?XA5(44-}Uo-sI zWAHRmIUG4c+uqCU72@)i|AU^B?GLUoS7*!m8{`;kB`=`|4XjMhM4U4eRP4_$EH3uA zdv7L{&b{<8e{wK~gqwO(9OVqL3{`b=gn4`4Cs(^99gNT!ixloz&(Xg-R+gWQi+QIJ z*c*M!sH>=tmDTHiwjt4;t>0jS4q&F)@ylv_d+^}F!(P!IEt96;L-*cPD0?giKbPuD z%$nzPWC}24O)Bi4T%8XaZ~XM-u(Cb(N1^6ik%yc6BSD>LKfU^kihbHyqtWT`4Q?-YLb*2M(!0ePDeB^-lx*wu-d9{ z)@b41wWV>5Xa2Ky%5_h_t2-)?QORjsE&y#u8gU%4he`&iFN9SJV8>lugQ@2#mhn5v zCy2T49v`?@ZAXE%n80Jv;=!jd1wnb6#F61P8Y(Ils+(r$;U?eO!#-X_)CU8dre7lv zX}YPcbkE(aJ>MTDNj1E6rIFw}-HZE=H3+doNxf^mNaT$v5gXU9;F-?v+vyZ6okjQrd2ky3|jLbM8;;>~St2N*rBx z{)o9*Nbx)q$)r7YAQFK$w4|`S55ha-R{F)b%+*HtaR3(M(R%Uj%R9Bw@g_{otJst2 zLkR>J&J^m$|GD$mH_D_Atf&9_M=^l_xrEN!ppa91xb6Ay8pc4L*)+`A6&+Jt?8uar zhdtzGbYbn?!x4iz(NA)#O^dR-JLN(yay-LMw(^d+CtM>CnE1Wbpv27PD4e&RxZYZ= z-GQ95yT2tiIKTCZF?p25M;0vv2dHLZx>&d z`C@fK%&whGuD&B~Q`+uKaS#{XQu9-`y)UowJVU+NRM9K*?v*_CBA=t_$*QJ``C;YzBYqJIv3BhYuMwWq$~ne ziW;sCqb*Ft$B4Ni?Kfs&ol~CWh&W;CwK^hf`QW zL0_*rO`0I$y5hC5xBP***dnO@xgfixquF3@eh^~oil*47q<1HA5dGa6#UJ}VgFmIO zxMOCO=t^~D!dQkP;>pdplWO6lD=;}n%#xH14%RuE0b&zgTmXn{3{d1)-$+SGDXUnf zu84Nt-H<71sr9C oYIjj5`-dr`fm=}rRn5i0$qbJ{QdA><|+4!pKGQzx`j96vl& zP!gExN-wwhXXv3rUy}ZAH2lNj`85^z$HLFDZ^0Dz6-N?G0~b6#1o{3c507i?7tb}H ze;2i4IZdXZhzJyX>Si{Z@mGD2}b#HLqRO$b}17Vl#zv8|KN&QZE>&22zbexESMMg>r zvnB8Oof(yi+w;I8p3_~#Ug{>MsH$otI$IPJl0>)5n0KV8jPa^*hz}<4;>Ls&ZRv9kn*bEJa z%(B6}M|*gl#g{kTT`LXS%Q88^on^@jvwY)!>k z2hUxbRFOQb_xS&!RmbSa8iN{x9*HFG{%TQ~nn)t;3ZM-?u*d&r3h}?DhBcG_?=^a| zwxVXc=jOdE|{3#gEsNpERM$D*gZXRy@pOpJ=+gLqyiMel<_5Ti!`2GEw4P8Q? zk{qopRCf*!!Y7EAECObV5HEG4i5OF!b?``SR)3e0u+n}YKQVt%y}+)f-MP)mC7zMv zQtf|+#J^pjIBL)7E%=k1=JdCa|BY&ht^HS~*5OXe_G^w3{+{LYQ&>xu3hJ_6AzHkF zwN3Ku#;0skOkriB?H458`mX)@Gu}R)dV8-^SHdHp%sKV_DPH}C=0Ag2YF1eT+pFiK zUw1(w!<2uCAYt+kWdynhBRAh=Te^!KFF5$BvP0>U9K-t#BJwW0^Ww+SQ;X`EELt*h z$A6gL!-3ePCxHR9j%N*ilO}a^RjSe9u2dZGpdB-OF&8ll(#duZ)dms!Y3vbpO0KVK zy?jx8#GBe3+cOULaJlj&vK~VRuc&c3&f&Ug^gibEBv;gpb!zAJYJG^Lf6BR2#w7|^ z`F>k?F($S6CH;94p+?_0 z&}px_+MRJ4{7cFQ4>MbF%OuH_zOCKa(%PScbZf`MgObcec@ZDu^fQCcQugU!0g{_(4x zQ73V@pvQ2Eo$nATK|<3~YTCGaVBx@P|3aI(N`Ri8^vgzSF|zPo_F7AI-t6(>TlK+9 zKAx~1^1f*Tm(Un%(=L0f8JHLa^{`hgfLT-)i)Utn1$BJ*$esrADs(tvI5#7P{Nwy<93f6UOc+RB6+Wp$mabRVmu&clG#`MM$$53RzU)fv%Ph}OlEu7`yZeaCK6tU(? zg4ammFz2^s-(;O99_KEAr+4RmNAGZd>g=WRe3%TlNPtp!3jvP1N{*HXwkl}JM#X7n ztqEuvmwXwb+w*F}W(~q;LSo!j;9y!@_2+@hSIyqa3%S(AILiuLZecIR+WNbgBRQp? z5=$eUeQA9=xitE}jyjoPb}{M0#wp#Q==BdPdoLUZirej8k|Kp)l(0Wz@bq{7@pW#= zzOPxz2Fb*utj1)aJZW=LNzh{KPy7`K8smMM+1d zp-J>NcfdDWBo?Hi9j^Gz!sYNFQm*aL;*cWBo2)_(JD@BH}a-CKkmK3 z#SGR%cYV}YZI&*Ox?oSsJOg~*O7Rk*bGFDw8oYag7%IlkyQJ z*=fDvb-r1CTq*EpV1$GRcDvbSA^%*$r^^s56gOSzVM*l`GHT^eQnoD5e!psXQs}nk z$r&3Sl7rcyApgt-i-|1W*}F9uw1gePExtO0R_>Ehcn0iNPvFH8Ln3at%wheTfJwTQ zOusz#ER?~$DN-@mR=Y^P8+uzDbCeGqZ6i!>E;698&qIIs4K0V;!H93gr$|K(SNl91 zw@h@$trR`)T_1`+&jM4>103T@3*kKIr2bN=i1cFzncP(Zfyav30y1Lw;W_SfL$n3W z3%4Yvt~M2wVt?4=BnY3ho!?#b2l<@`oXoCQF9N!U0yBa%Kdb<6%9c=?YJRwXU{ zkf*;wY_No-QnBiU09v>1@bqJ*x>3TL=qVc>z(ir(sMc=TS9#DAQTVl;v$FF<;ttRmuSo*cVXyFIY6SFL}^`AxY*Z9yKGQaLHBa zX){Kq=~kG@dZHzSWtuXoMP;l)w^Vvwx0=0}ktS%ZSeHUfFe=0JLgZ5SwV%$nezsez zd4JAwN(JOL?MXD$ED+ij+nW`qRZ?lP+-rF5`sMTfbn0zn_UTsBQ5EYI{w1l8=rCx* zu?#wM3bH-Ox~cJ%7d-&r$tpDR*wS4JS*OfcuJzIFbC8;p=Qf`I=0v%{(`lktRlho} z_(%lw;|P`mAsNF;YX&*s2lNAXIfm)=5j{wJkq@L7G?d^~0E|^8`nRWN4 zLE(AI&z+VD)%T_@+E9_yUDq!E8ioFI+0VRWv3%ENcfO$G7OYor0Zx4=i(?T0d()$6 zqT(TU04I_=UF&>G%2$`d*mvRh$~valYFu*Ax@EAVT&KY2E$vSU3H87!u626qGW{Qu zey7QtI1MPx8|I96wp_Q~XS-~!TDk6)?ELQfJyQ2&EG+t6 zm!%DeU~?X^W%`sr)-`rqbqoyuenH)34LbwX5pfFz>Jcnkrk@F9T%#Dymgn5>IZ-T7 ze$<`rSAT)~Q*oSgze|W$pKd$1H2Le)rE6YauEqQS1VRuC0qy1U#I9SGb}e%Djsv*+ z=}53jOpQSGh@0GrpX?rG-@x+lE@Rgc`S)^!BJeGvzmA9}rrxm)dwu|hfi?29Pt-3a zJ0%=**ZngSJJ06fET~OqA%~{N`EiXewg$mj+cHTo5peT9;k^)3~gK$tE?tz|Vk z=PQByxUn(UH1Kf`d&f})oh{vlW};u$+0hgmM6H&{GPoT*t2_m#U%?fn~A5$HyNji*t4p1FoGxP8ptP%%|ne2CoYtCxe zB_J@-|A1e}bZdF$wWG~H%}13a8h=z?+0q3p^q*GZc9hhCK3$B;CA@z#h-#rs6blq? zM@bwi&!9)@@Yv=90Fj@RECjhG6-1&n*U{tJ|H2bDPf~|U#q0u$u4d~x0R9D-|K_7% zB>)qM$~ZH0lYn#l&o{TLRr~0+xb>yZ;ufbejzB=i1cx~~nr~<`kHD7rlv#>?#5_zF z4ZCI(R*7tsa3X@(!r>R!-1LE6tcAI*`QIo-3$1ZXP6Ase>@)A>`jX@LG39`uj8aA8 zvtDj9P}oqxl-0M;C}@p0e;V;9sFT5YouyYRULZpA+YU{NZjnj?g@I(42J5hvui)4u zvCvp!>5jY53tkXMerp-IE%L10P{4{qn_3E{S=hW9S9&Iz0$u;bq) zrhikpbP*`wlak7%>i)5WF~n-*+}QQc zPI{b0j1ae6^vB>TQXJWH+Y+8S)c9El-3vhqs07xoCYXZ&3~{l7l@#U?35Gw<|NLq9 zQfyiAgFM{WnNMKsqo*Gr1QHho+?cW(Pq4L5`t9JKzP5L7_&g7zL{GW~-{eZgepqa6 zjJIc=epE+ac1!Z^INeOt{v6XuS6+c$K==sP!7t&L>AEjRUQ<~~$xZS)o{D)&L`;o^ z+gDA5ka()*3!wo9td=K1++MzQ=_|kXgYP{4rc<3o2m1C}%V?Xy*5i>9RTQZ4to}Ae zGxGqXOs70VwD)FI5(g*CcNfM?yv2BC)@Sxq%41W9F?`+*Wf%IH4AZ)3w9&7Wg?0!c z+?k$lujudnxhxcNLukllcisjeYZGainXT`-t;3At#e?Fi@0hbm$Y>_ZRD^KN6HXh) z22wtOIf9@W4bqpfT1-o}(9@gZCpAkVzvz1kraiJWU5;Wmi{~ug$TfEvKVCMulRj{H z|G~}O*I&_7WpSJP=y!r!&Cju>xV>c<^i*1PdXhdkdgRDSVrt0;8&{dO1DSSz_f@X- zL==w(>esouf_`>=8RT1T^V&B7lTtz$%z)L&aVjd_ze?DXG}2x>Cd3_I*9f)`pR6v> zZz$VY_9-`?{`Qa3d0@HlCKS=<)5oER-X&0UGZ4M)A%J&VTQlQh%mcQv0bdE#x zQ?#i}Ua_w8S1ICpf_RMK=6NMxr!%+g-LXN=?H2tSDyEXoWe$w7w(C6P77v7MRe|2C z094{Uj}$ot!9U=nfCc}S>W}p(?8`uAv^k6GFe3^S)Tm+)Vtdna(BrpY;3{ zjP$q_66HR-QF{axy+*`aesZNivI7&t5m+}!cHAfef#NG~wN0BG_yO`I?*Z}t47VAD z+x*MHY>TPZAf$Fa8gI`ubi+QEj>H>U;qz9!QU`$>7u*y}><=(Z3d%C06u7|f=HAqy zU}2pt$V&j^lMJUZof0*6Y30;z+6qqK!!YEP%V+>UAEKB3!VnIRYcn@{Notu z1uO$1VS^=Z^b|s0x&0fGUmWI?6@PqyfO94Q6sl25Yy3e#6kkGVeFT>~U6a!FF0-Nd z2P$~jEz*&Ipd4HP`VnYAknIBth#0i?$6|M-FIZxx&cRJj!-NRIR{zNaROkj`5$Obp zRF$v5S~86wm>&>N$nh=X=U+hvodNJeh7}N0*`c694ms_=IHucCk*PyEKaczyv(&i& zvr@InpOY3?YS#iED24SB$zS!L0}luQB*a4eJA!C@x1^uU%{{Pj;B$a9U|Klab$bUR zyt|D9vJWp?@q`-*h=l$Nq;dNKYg7{s1Qd`)TI1tfBnR3EvdL>W+aL*48xh-al80gz zR7t~P5q;Fwbn?-DkE?a34wOATRTn(|0WT@5)Ol1+&vya?an97a-E*?uenr=L^p_P= zh9u$cEq4!Z&*8M&m;Mc2kU}X`1%TgMF{f_g^|Ve~AH_+m(^21Jo0h&;JDJEv)RYv-^QSz@ce*kjl(aZ+4%T&1y}Jl*7dqrR&Ic6eOtwr+Q$m-I%L61ulg#}! zmCOF9+(}Jn`=@L|lTX303tt2nh?_kND=T-*`hWq@c&6!RP;8M`P%-JSn2_GRi`7GjE-!1&HB^+caMO*?;c&NL9R@jo)sNU`_mfg_sX<1 zUAFmV9OrrTF+(c;fG~zmjsF0bU!i)Ve_>1f#1Ean@-6I2w6|b{3X~%*7rFmii3UVn z4`XGAH%>3zNBXnJwf45ewF}mC9HBr6_>U9_rL@27;-6lJJ}nSbkv=Ui&zGWS&5Ox{ zhB29j!jq}9UDGDP%lyQ^H&>V&KvFukxAX6{8(+QX)ZLW*iScfUg1!FBg#KYUeZG4m zf0YqmLxl+vMWoy)KcFRY&Y}E?w3qim1-xT#Zj!~uy7S$PB=ITEqa;C)i!piXOq8Q9?8Fgi7*%-~x9^V$;lWwiUALA2AN&y!4@An5 zgVnA)j69ZDHpmNqXG(H_r=WEf8;yud&5CK#72jH{D-yI(FdRZNSs?9|qlb|pvmxus ztnvYR74w#ka^q~uc4}10rtkA+pEl4PFnd!z0RhHiG9dE{KZ3yT#&o)A>EcbpJWfvK zqzMh7hz?2b+MCy>fN*v7`;Wul-xF1lpboar-)f=sR->}c%+em zb=%ZCs!;DixxD~*L2)p!qwgb99v2V#onmGe)nuPTVqO^4o6LFyCzs!W89es-Hy&fY zdL0-Og9G{@%Z<}yWcn2ML^qKsKcOVcg1C_u%( zw4_aMKF|uEQ{cNuYtKh83VkTjz+(v;{YNRVsc;1mxM7HTv`wU-)2_D7p0jLk3&M4=;}FImf)E!6d~ews+jw`FxFQwPnFeAB4Rm3-6(YWI$ro2Y zh~hy9e*i~^x|8r_SV($v*mAey*Oa#`%ZnqM5%U}X^F)ym5(#!&? z2lD>t{BN?rV@OwsY8f3CeVYc(LDux+ri1DJUYDy(g;S0OrTwgo&_Msped+E>QN+Mrorn!wX#sfCR0 zdPf&=Qy3ocBT22yx*7XfcftESm_l9X{(T~?Y7H&G^;iw>iFELX5QD_)5+1Jyw0uL0 z<-=UK3~~K8>v+$tJ6oYH3H=R*zYYQGXa!6LFz$fwSof+I>By9KZV?z5@!o;zt*@W6 zq{NNI;KIDeBSVt`Ecavt%#rl*9FyKDW1ExDoX!CEK&aXK*J|$Jj^so*RpJQVJ!CNAvz{hB? zkF=w_8OTq0#P0W__Fvg-%VNM;rvL7NsG3{V=fzf1)1%3Y+}-*9N*oV*pGB}8Am#Xg zohU-^T4S00s_<$=4(gmX`Jw<$NU{oXxRQZ9#U`zwE)9`Rz*~_>>PC^ZxVGys;Sn_T z4w4}_2Uu<1?H}9stg)a(LT#cha2q{wwky}6vsd80#}UMm2!HJiW5x!qlS@Rmlx=um zD`AL8$_b?Q^epwx6j8wS76O{nTZ^ht;ryy6v+sb=Ql?+Z?#@wv)&iK*x1ImxRVsbg zry%bCrKnV%W^R-Y*|H8Y3_f)niIn7i(2@|j^-KGz<8F6mUtocEUfw^Y7%X~8UC|R` zh#;xrMDBO_S7ZVrstzTX&Frw%czaa%HiddKds~&k@EhqW3LZsVSDU%0nx-`*u`jI zZl)u?7;bV)D1SWuSw~*xf@wo@IKhKH?Ihx&lxjK`z|{IwR=;a}ejg!1wW@v(sJ76O z6MLz7cH2%rjxnWyD<-sCFt0FU_Pa+<^X@J60g;&*YUS-(8>tNr2!J|T=x-7h3jL7Q zf{X{00ogenduNeYhtV3bbxk^@)s9Hd{?hPQtiT> z^^@Ky30<*$A6@baJrw113Z70O?K5siKOfW}k%XWyAiksPOL0@lKUBoba6^|V<%akR zG_Ghpzr(~5U=zM{Z^$F%ix{_@ddZi& zxzA&h52lFvzzF#xz?=d+^nBMkq+)~EVp>;1JtuYr)L{ksp6@ur8oCeUTcMYIk`wY| zDX)5R9XsbST*{f{%+wPZSCx0wWR8}P$aZFiltm93X?5g4m+iA(2!?7o&C2-$g%XnZ zpwQGu(1C)8+?6&ae5EtgR7xX#to2Jf|;UZkB8c;>+=|0*cP%#v89uY9FBXIAO6 zMbuWYYdirG>Qxz_Rtu^prHac|H8CdtoHD|^xR!(BBNipSsv6H|?)6>1G)PYI$;RG& zdd~7IxDxACOuY{zniufAD+B1Dr^bc~g<^l(_iuSCE9xk-;+ml z78{;cD6Y|)!)B7NFXO2=ux?m?C(AE@D*q)g{usOmW+|*l+YBJ`tH?`a)`L@k$4X65iZ)kYdhx_ z6~RB4ZY8H&$@EcrLdj;@4~86LVcSAw@Fr#Bv=K+kSx+(SmbNT^N()Q(Q|kr*T=W7iG^P;Z&AB#Yk0x-!)Dt2UB1ZEWI)1jO=y& z%>*gm9rt_dtqqIK6i$Eo{4NQ&uIkpmxPE&-C1NEaIpJB}bmLnO@5m6TTg#@(a$jyEG2JKY9vnp7fpx zp?ehCv)MLI8S?z+zzYQ5-`NA_+1N66GeKhZjDYo1M)X`9t9wSLSa zmM$Q1_NX0?P(&2NDt(Lk_@IjOz^Mn)(*sCikcBeM3q9=EwX?mNw1iw1{h#onzWQBhJ zA}mPB`v8JSNGM)(j?#vz4w6dVlN^nE_y^`c>0Dd)w1rYq*PuR5iQ(Ep^p~zjhw;3> z=vz7-yPS3`>oCw@;Q_2P{&~e>mteBmRpP;n=%?13+$n}GycMqEXABag`-SPQVU|3<+NA<_)tdz-NOq1|q2bWwh+>rnrr z1al>0OKsv=7X=X?9Yeb#`Liot(odtNsj0fih8->1qmz{Hz>cwd|JSNW;Gf}`(+ej) z_=a~%q;j3Y{HZUEtca-;oMvz9J7uBTZ(ubqp{t~MYps%oEcF&*E{Ms3wWr3=>oxbA zjv_&z(&n#%3TgykguHv4@>I$E(5^+;mbKpI?^0#QCvl7q`X0B#5+I|(+ZQ={38+7S zDJk32=jY2@y}>p+(Ka8)^YgZlMsjJ>>mzmdv2_gPhy87IcKs{f`9D*+hGwFAtQX5% zoBniNv~VF=);p7~$+6nQrBHj<$9p=EMS&Wj|2hs3?ZBJgLG_IJS0s6^w-A~)XMT8k zW>Y)11A-F_Z1C-gMRt8@Q3sVjO~{)ls__q1Pr_<5Jr*J`iNVDdxD zHZw-NiO7&;lCL@`bZpob33#9Ye^>%WOm_9ruH2MdN8h=Py#!h#DhZv__28Qk1iHoi z7Xd-`KAC}?c7YR%y&Uw#_iaYfj!NGIce>~oj)< z=EL58J%g1UzvmD5&2dqNciz&JZ)d&6^8@owiM%Jjer+QJK{XUaWq=#lJ|xwvOQzX{ z9ucI4h9vQD%fu%JjVw92PaM;WtI4CIL7#n-#`wCYTk)P>xyU4!N3&PFOMPywm1S`z zT4ktfaJcXb0>}^?GhCM+NYP2&J_oLax}S}sL_3YiLAg6^;&=0BL3w~XA@cA$1w9X6 zm)yq}eRFPsrTH}<0b`C@7XR7A@FOOAF3~3vK${4QT~)t z_l{%avwi#)LQE1wyK2G9u}*qBHG#o_Qd}0^)3A_J8iQo`{@e9|k7xr;6h=qV(S^|X2I@QgUBoF2U0xVVMCE|<7>C`T*BHuN)bRDYy*&A= z886v*K1}g*d_q$-b$Ja!Ac9~1LyO@-zku#4=$}c#we`5M2Uz63&D+!Ci0sM8j}@Am zWNVZ@?x z5%hY0LHQx&14&^85%|W ze;$#%I4M&a@^wetI-!xKo03dWd#wV$`W)f#_rtvK=7VQkPvVf4l2ZJ@z$54F2JJ|@ zC~-Px|5P5}pn_;Qzw2G+JDx2zI)1irT_G$U`=I?Wrs-ZH1itUuc8~mLX$nrAok)3N zR_iOU_}%no?w+@6=+>_5kWCP6#cF3!{0|#V_V-r}wY*8s<~9?d&(^w#SOKi`+4H9W>XDZwYiU={v`1L6_AiJh%HS#GHt_KGVm< zzWrLmzVHx*#KWk|at(gY^{TcV?S6i)*oqtlyee@lgXxOx__bg1;@1Yx!E2>~Vbq3H zWFyT%A>$f8)4=8`9@(SSQCBaNI+wv89k<4fyq#lde^USN zS5KQQ%AGEKzuv=Vo=5y*t)zf$kcU&@xWkP_JLlNFY{qGe61Hh6dt>zg{?(L-PVJB0 z+rv9AW^#+@yECFOt$ZYZ@~82O%4cc1J*xd&^IfNDx=V&eGNWcOLpD8eN*r%!SD!8Y z{w65ljeY+_UQ*I`;Tn-L`H$P@QlMeqMvqsnAgw%kjeAzhAEev5%U?9q3%dI-521lS z73rpL|DAUPdEDW1`>r)ymuhQ8<5tIuF24Q{S9!MQvLRs*3;*+p(Ke+8duV*J#HR!bEv+uDM=!gn-#z zGNijF6QtomTc}sN`>6a@e)dQOdc!0w$EW5y&l-d*CBPX%86BY(qB6W<%smQx7G48E=F(lSK%l=&%QDq8;4*HS=FTZ*X|0}&)*av<(lhnuCab1@lc?Q**0 zLhL^e$3wG)!^o7}gTFE~b*_W=1TTb&z( zTzH{Dl;t%59ajPBE}5+uLAH+*2@3gRYB70*0Lf*cZ3RbZ{B{*_Vgd(N}G4Y|rZ zTzK#7(T$^+A+(`f?m)#ofUX%VUdBCg^gY3)lC(V?&NT1y_8Dv!*@d>p_RAxAuzYoY zW-kLs>Uc_@6oN*PSNWh)A>Gww&MXo5C{+@kG+F^5m@siFpfF8w#7n^65X|;`77bWK zT!vnVGo$;E#?B>Hv=QgoMZ16P5~P60`2iNv+P-OIFgXtB*%W+b)}eL<4sjGj=rcEC z?}?M3n+ijbV(PyO(5s``Bg>=JrnZoiCLE=NwdoM7%rEpSCU( z(Ksi)VkBuucbO-<4J0n%I78rjA%1ltxV4GP?aLFxWXMafBur4BGXv2VcHM+-c;mD* z;3vr1pjl^>PS(RO*~vpFh6(Kvj*c5gpiN+S&-?kMd{Nk(fOWO#}xZd*4 zM1A+tDT07nWg`D4i4QcK<$WV=%Ox{~e>v7Myy%?B#x|w5Q=O?QkjZ@pvC+qk2UY#` z8+pllZ+9rGeL$Xou)&;>?FGz^J+^#SPeTG+ql$T19f*J>!Gx3Yqg>dgM&j#`qg%o5 zDVB3MtE;ICxlYOVx5-YCCg?=B-}UsslM;bSI<)3MvI9kIS182;+5+JOx6ru<(F8F& zz396wUtnwJz2-%Zp8kx`rQULRxlg1U_>(K%HxN~9ly(%=K~j6lX+32lZU3m6H<>hK zExyW2YDP%Jso@2!qS8rbuTs5ZCWL1#czXZyQc#_@A|y#Q7ZTsl`hh9cs@T3bw?5zp z>Qbwu(DIwt(vbdGv2C@MWk{E%kTT4_qGE7<1XDY476dG*eSvcZ30FYq2vsJw z_)@EE3$f_^`|_?dB!5Wg;xhehi77r?U!u1i9%%vFw+2Q%c6;d)!jRta{Wsm^Fdp~v zA-lPj&uPb4e?IW8-_!((TK+b6ID1hOY#~naW{%R6pLv&!kr6=%vF#2?Th;#A1!*tw zE~?$jU$ks%sn7VV)9BfL4TOSXr4f11BcN%I&BDh~of;JS;f?I%Y-yZ6#A-ArebTXa zevdAlOa65A?JXM)e+y|kzSgyC7M+(poS)tT}UN(S&I1K>yS_Uq@vOEn(TMljoCa8zKUU7ODg1}yS|WgtG946nGC zWyeGKK+IiEhd9*&t#3#SQV%GfDL^R+3P$%T+aDs!#eN*n<5qvcjFW8{^kaY`1?}Be zGP0pOa22^#0=O6$IhbRt7H7WGYZ-`vDaPx;*-mM$O5(hRHuv`fZsxyZ9yv0mz1;zV z1|;*wQg~z#8*~`{1r)!$klwWegos3}+}kz&P9X~cIUs?JKDsrnHu6q~6MK|Y@-N?T zMPxy?@AbbbzYH4pv_H^g*9#!PzjKLte@s?+LbFi$^}S#lgdBA z7)i`APfZoUD|B%{M55tlLaNj2CV7v7>*R>fveH8`2R%J3urBjsEJClAp+E?QyTPnH zVPn2XMTh0ffOna}q21+bEh|@kN+^N4s9^CSID>BggShtM8h{L$@-Bb|^9fGy-3TQb zLL}cr5oYNdx}S9^1&+KJW!%-}c#~z_$L>lhZ&WiLif}24hofJO7UH|f93|sga1884 z1zLzMDL+>GT*e^5tRPoF=yxvisoxYI>OMM8cwc@TRc$)kUdeknTKIu%1RrPeSxyIgNDu3scT1-Oy)8+PT9 z#Sd5|*`uo12jiT2C6g;N)(u%ZX<$0%sG3vPTV{D$t*#rDJWzQJ77W^XCIMX%UiDj8 z?v0@e0}08%?%WL-^>Gf{w-W5iw;H*_%L#JL$E}K-w=s95Sad~}+Rv%eaddA(du=QlN^Kv-Z7kY|tsgR>ne*-AL(0w+=j|#p6 zdAPRWyCM zzk!LyR}ng`$)L0OZ2<@&z0e*2EXk(4SPRO<#%54emkH2^YQ03=p-ivu1Lda+nWLZ0 zzc=nuW^gGj*Okzll=HX93~|{7h1CdMcD$nVDQE|<3s-rlJ>WeQqUsl#bZ^glQ)vXq zou2libk9=JdnxV>Z#rCBGKIT#vwM+CH2G@XB+_Fb0Ta!-nqid)s}})UGK7mKxVSOv zk82BQw?s*&0xRP&i}Va2!_lx%dsm%P%FH}~P%9@_*|Z^FPf69@eUq$my)nM?DTFkz z6?u~z&2LNE^@d-tw{g;H`FnYs)8!(%sG`SmYX|LfGoWR1mXj#f0G+id>EK>K01FX_ z>p{2yrBw|td&x!pP>EvJB9h)GWhiqhH!8(=HPh%~XDU=~2P? zvEu*G_SR8VZ)@A|GO-X70R>b*q(MNCRum)@>5@>C4pF*A1r%{9jR;70iGUy_AZ$Up zyQI6D_1<$WbbHSGJntCaAKxCs;oj%KTECd{j_bazYvOcu{3^JVjwy|#i>Tk)I43c{ ztY@_%iUkGC;f(fO7@*ri{FH;a3N!>NX_HJ_$sG7<{B3o@Gxl|}vF3<2mFP=eCs`gY z`ai~zeTc7HF~q)1k3Euj4FGcE1BZmmvNE}Ln*g<#Pa_ubC1GRpk%)8#kB?z+OP)od z+&a;pr1ApI;#C71yq|+m>cm(iBTL^%ed~_+7kM{eA;rQN5GnfdDlTPWQ#bYvNp|c( zdIYEoD(q9B{t;6Ln;mQ$ei|1JS4EvLw4pvTtYT;to=tSlils>sTQqb2 z!zANYZRB(G@K$tz($9p*DS+UBa=|4ILY#ij=rWy`37zUsl2T3+t?(0uE7PZS9dC7t)N1fTNNkx3vGM>!(sn+r{$C z@y>AV&ly=m()p8KepK%&0;hd?a_Mr|)cM=Ka(J^}Hn2HnIceu=x!&i-QAhpyWxBcc zq7zx3Kl|+0ruAYhG_g4f^AC$toh8GPR(*x6vJ|J!@TV~8ssF)Mu+Cc<6RB1xr-MmVYBISL# zE9)9=Ro-YpznU>wEtFTtZRO|DiXN>qwaZ@x);h9N+?uvn)jv^HX+-yQeG#*6^NwsU zbQlWUQ0-(47caXnIP>ipkz+=tEpC%!0oTO!d}D1gyC-FR>TAuhXH(A-E1##o1pepN zw+ksiiVdwb5Lf(%&&R3#vd7~*GC#P?(jL=ld zuyuQ^Y!YAA>VD^Yez+@fVMQa`psQlgJL+j-Jq8R)gymaIN9+qXM7#plEN0ui(x&ED zLbBs&nM9YnFBcfIJVBY-!2M%lqMe5H}RO zeU~oj?d$s?s?jf7dQ*&~>t^jpobRX3*itadAlev-3lBf-cI(c#;3_oIj(S}8@7vqGj;H9J2yfFW8Om7ZGER0 z)X9|(QyclI`avvSMkB}GAl6x>%@;SS={1oS2cE@%upjSeHOTUA^*D?xuXTT=9+4=g z*c2r$@vg0(kDgKp(~0SdUh-XP8DJ zagfe*-SF{%@nADA$I0Rq`f#Oeb@p?66G8oI(^z0ucw<-i|=sU}P;neA3~oOOay z^qP{^{^4DhNyXC+T+RshWgB9m=ukgE*Hm5KP5pXdmUIV3+XQ0Ec@Ftrj^N^-zAs4z z=55bNJIh?;_S9ERsHLfIqImHnj~eQ620R50>ej0zp`-LoFw6nT4gyqoi|((bU>S98 z^t%ci9;$N|jf6TJNL;s#F}`_6rLRr9Vb8J^E1bE}I^ug~N(HRCu$Gc)dmBJbHd zz;1)gGU{Nt3AX^`y(UeKR0zYxBt`dg*W89J3XA+GHA0W_Jw8mqwm6ELoE=HUG33c* zY%4cL|P>mIx5R#kQ(K!+U_Tv1BG!BEl8)+ib>3{!85zk0%0!wM3do&jqY`z zI#eI>!QH73=*V9ScsjF|98r`qM?4Bcgy>!AasHPUc$wp`$)(&QqZ0~aYh-+Lz}fz{ z8zSO9vMWMy0mJF?UmeArQ{KVv1ZMRP3xsIEP{K>_83L;k9btqGiS{;XR ziSVd7mtZN-f4ytumcI!ivo$CM#Olo@*eKj&e<3qP0jC#bN_L--Br6Y!rn6_ZW=i@x zW0q!TXvpBX&kH*z{Xs2Ak9;wC7{gBU$bL_Ux~qDiTt-p-ekP=qds~(jvpYnTHeaaw zwl{^NP><+0iZzDq8~HN$Dk$oLr<<3i`%3{<(+jA&)R1WfMO2eS)4J zifX8dfldxH-Lzi!rALq68i;7U4tSqzFnMtOj+}FR+pC@Ylpb*hm=?&iEYr_2v4J~b zr(i(>R1o(h!^$e!yQ}<3VZms@!ioWc2^#jY1YC30CpmPzc=M5&F0@6PHfo=!HZq&m z+M6v*)XJAmmbyKxv%3$nRncXr2)_TF`(QE)OQiW%YV175hv6R` zzt&3yWO9XWDjrIrz?-XXSjODl2YCSRgO}I>LmKJM=2k)dpsjfLPc(f;Q|(Fq&8?)T z;!s-y2rG(kn{w-<;3wbDXs5G8f^C?K;sZRufnO5gjO}6P z^eY2YlSKUs$SCH82~XuR5Qi31gk)*IqZjDw**S2G9yUUO8Bp0m1Lv+`co`M5UZV+w zF8V4)9wAAXKp;)>E-y|SjPyqvnCugFSs~s{VZT?@v|}dh$x*ZLatFyOIomy;>PM6; zS71-7hi#bT&sPns%}i@H%((ArXa`#V?+|uYj2p07fYAN8lp+>G^jc)r>|ogfu>FOv zJObLWe|Q@B{#Iz*3)FWMD=#UyOE)GjnvTfysB)u4Qp-;&5;|5V&j8=xy-& zQfPRfIfweu#b=p*vU@$pKXrpBqyz{r(_8W>&# zER70ign&kL5(aPp%0ErKpy@nU9oW$fED8$TR^Cy+_k(xjv=8B5#Tj7IxIYPQFIRTj zJpeQou!!(ebda>I@k8ODdRGK3;N+>owy}tIQWj=)b1m+%1eEO@85H+fqjeEDLCfz4 z;Rp;3;F9LG30fCWO2MQVoDA>{`^jMtTgWN|21UT=Rf50yd*|{H8@GDH*#1PpXx*o6 z2sFSjW1z2l0t5|N6LIJ3(Yj0rCK!L*j&5W=vKK8~mpu?4KEUS1SGJIeKp6ovZJ+90 z%+FCYM3Z2Q%5H=JgE|9H-^Q(LkX*!p&I1Vxu19QTM_?eXw(-4^Osl)MzuD*bh_f_9 zhgJ6U;{lzb^PqiFRx;lrNPZ#HLo#}>0S&pFVHPCo!PE5&+y|Kv`5KtW(7mV}giCKl z3_48Ob6^<%)lP#f1m+)^zo9n|bC?AGM}9mbd4L(oSHJ}15wFxBUT=#w4Zcmj-ykDA zqAXvo=(rrhbT1S3S(*jvsxc84Kg<9v7QVZbj1Vc{RkHNFO#zgO?}*BI{2Smn%f6wX z@)-1_?8(G^L`NI};OAaC{vtS2{vqyCBw{yR<5y>b4ECE$2(AYr8U2iKwwh^ssq_$oxaj;lFPKz&MT*kXPROQx@;Ax57#(d98LIj-SJ1d8unLfra{#Kn0yP3ss$o#PS}+% zZnrm?O|Ng-Bi`%>b_W=B9+OCw$pi?Ec+oJ_x2D#%uYeIKpPM5e{nut^jwW$2G`ACQMs$=MN7MUdmou*r5 zrD@^d)Cd)K_qXfBno@VhD%3wkzfsF&>kKI0qIc!sxQs8Ls~!>OgG9t`Ys}^!KWZmx zCS20v0y&|93KATipZvPXXheA7@JJ>VR&#udJxP($y4`#6<`Uy(ijLRRXSy1DP=(F~ zJp;4HY3H|5>IEZiRQ|2q-fzz$-y~1q3+&J3_z>5j)N_U{_Teo+uLh3SvLGYmG&^wd zqlHU*%VOC$=h%l67tpzHjQ41J-$#)JPce2k{Uf+fO0PT^$L*@KMxX5Uu z5eKO%rw0$ZQJ+|I2HE}fj$P`a@R}k$h1kSE;sMwK1|%@J{J1n`2hIdnJ5cT5h_}T7 zlA-)3L9c32X~@LNlj@o=QmB zf1ZUq7;-s`QDDuIf-n-^ImeUz!k-G~o<|K@Q2C#U*4^=w2<>o}8xT-IhzG7cF_}U8 zgtv09CJHTkT*9sN)S0VT06z?tzUCrvmDAYEN$m|4eG_Ay5fq@1Hkq`;0;B~RYU-UbfSa;i2K`%+y}Me@hRzS z)HXu&K|^LF*1a$`vB#N3A1`4Cr7c)3K-8hxD0&WHYx2c^dhf_#<_Vvz3T%>i?B*8L zX4$mg=w?9S>Boa0uRz1deZtJu&KDWr(O<_Sy_n)G;Qk;T1ehSjJ8t~Mg1#}yASGJ& zitmY{`}o2~q;y-v_{n*=;ZKbT!&S)liWWss%wGu6yF`5$J7xw&2FxTpB&+<*(#$V1 zxZJ?Mcb6+bqDB`C7qA`YftLL{FW%b;+=-9CU3Q%AA}KbMiC4wc;q#h{?O*OWM`OhevlkHaPsqr%@yOY)V zuhH-H0Ch2ZpbGdAA^ASY?zkt&1Dh|fZ?wV{07=-Q^ZW>KlA*8TPIiQ=pinVf>W`+) z0SmkHGngY^33u`f2zeEK%-d=68h+zg=H6!5o>OIG&;+cNGw$~iMO7@Iu&@a|`0*Fu z@roeUfdTG%oBX-7Lf++ z=>G(OvWY+i3oQhE42H0T%eJ-2_|@ewD9*mE2SC*hQ1Sf_3lkD9%vleN3lT8b6zKMz z9dSn>;9ODB$Vmm3Wiy*eFfN_ZSh2$qG5 z{)?#Z_T{iJyzFsdUtee~*a_>!{KwBead?H$NdQ#Np0otd!lf7g>W;L)o+r?454VGI z_?O~TBVpumf6Nkr<-ei(gBYaug-+!JtXg0`a}egL)R&MTELdgr=;Vyf-_3g4L1Bi~V&9t_M1$1rk=n}8sgO^J@Y->VT0;;^U z5p1m!l#gJ!X`~H9C5QpTZiv?y=5Q#Opd5oig;Uwgz#sB?+lLbDztKnqvBF{Ry?$V? zrh&Hq^K^ zxQ%oQ9h23Wu(}e9yXCYKXWRv`VlszPZWO(rNw@#~U z^^=*tLMCN5q?WBUww~r#zM905Q^S6I3w4rj5z`O;AhH`_+VGe-QyTbfcvs~Fc+(BZ zbt9(ceP@v`yGs3CsVY;|bfrYTL&W(w>=@;Nf<>f^fte{?G0gVxb4pmX1>v$##e+hK zS~k9Ij`%Vle*v_aAGBCcnkvo8pGMw+peR8D%?&ILA7k+Xxm1-d zOpzgdLf;Oif6#P8Mlb~>av$hYDe)CeAW*g*IIK$+<_h#pCZ=9YOLP#;1^kA_j`cs` z3*@|r_M~()vfuGHU#Y8cLiMDzFwOQym-4S>7=Q?9yzvrU zSkiZ2er_}-jEEJ1Y%&lY<@sV}7Y0oIQJUqb*+$d=d-i|tl?4P3drOU4D*_S+enbx; z9G)R$1YNkr#3dFp=%d%9s+3N&4m(~Py+8Z^YRq=+#GEIj=)hwJ@v6|iK7k48t$zty zQ0zfX4eDgn#=%>-j^`v`IKuz#!Co-K@1UW&4dGdE6G(|D6(LC1Yt$14o$oimN-^ga zl(Tr&09aiRguv2!k71q@>KeL}<#e>opnXVXV|5^AE6YwTHH&Z4r-4~_4j!T@a^=*4{{aAjzeWfDk$GU)?vF2gUG{LaAZ(31 zhTqqM1Kv55XCTXlnd=25C$MTsW5h4ZpPq_RRbuMA2>A!>%q5!eO70VCbhJIy7Vp$= zgQ@t19zj@p zV2dJ8xS7`og_>O2fq2iviTO=%A<&;7f6I0Sg!lZbgMwxnjHpoZ3Dku|`5VhIK7?oh zz{5rV#}R}0HbPVF``+uu?^FaC8j)M@;FIZd|G7|xo>Y@F~3Q;B@XbQyWL zzxF-Kcc~gYx6%&apS{mGp*n;r1&_RF2lM)VMn=6_;cQi~UkO}2x$VM+#(rH1sHyOk z62m(@4&ra+0@4ctC=2=_Joc4r^dD;w#UX%ap?so&7-KE4YHP3>VnO8+h;J*j&u@uJ zfB*t+>vs+Q@%w+V>z<)LUTIS^==0KL7>7&aiwgFq2tl;p%fRiq=jdOxIx^_k>Vq+! zqh34^3|g0Vo*lw3`cekTs{85i(c|enbA%9?pC>r=aT2SmVr)>3jtOxDoC4! zITHCN%vl)5t7>zA$WwBi2a7sEYKP&IUJE4BMj)QQ$k!V+fMLFZ-j55JJRbY`2Ae<$ z$g~bRe2IK%QKrD9Z=Y$r>^KuL4Mqze@5zkdpE5F?MO`U|nVj(3A^?*o0x$hvJ~4QC z{*88F{0SJ;36z#Viw4`F#84|R=u$yKi5UE{?*o3gfmS@%D6NzLpkw3yEh{|(AsC_g z{;1);c5wVdd`^O}<6!k4NU6KW?*9v#e|n_EVlSyD`$N&KN3 z5%mXn8Vp<{{(lCQhM(XcRnJsjLg8dygmr$&y{_SR6N+}U7hlpX_#{Nun8} zU{x;e&M5WGGhr^K$YlR{(|TV&oxzNnt8sj!|MWqea6e(9?7jM!wWSss?xuo4ML@e% zu#*tsfE&r{Ye5@>7p4#9rG^`sHLS^O5-keZ7XI#+bOA1G^5anl$QUTP(jcuPBM~HA zpT&RGSuVC7knaJlso-gcKD_cj5CF7~Ex1bodFpn4t%P39{*jC z3zqe1D7}8;E82TuuY!>VbdbAEU?oFi0>U)0;T`+%XWnV=MF>RcSI4z)GCj&2>wO5E z4ovZ$P*6P4Ix2GZ8YEBfK5-!QQovE=$J69;WkTPK1Kco0i4k`N0GY78y|lJ?@>w`YK6wpJ9WjtV>qcBQ~OlJ z7jAr8a8N_$^!f!&tS{ae&iBnOX8g6Nh~ZUPsF~`TGrGLloSNLsxt!VC{c-?0!gi%=}=an1SCWSlbru#MX}?>uQFVVVX2}8U`sULEF>W{97F`smX^=cExMz5=pWkD*rKbJ~E6d zJ&~V(#tij;D7mK-`7k%VA%23fbZU1@7A%Y3tV@^uXi>Wao`g~pAcP}_VV1*({~+nv z-}8!kZNT8PFF(ce55PQ#2yw{(+99wx3@oUQ&=YXEpu3hMq{|ni1`F_e^C4DRdt3Rk5`nEKI z^K3Kz9_WCvC1)Zo3>RBU3`N1e^k_Ujz&Sx)I~Y(HS>f6WhA;6RivJ%V0hn*NYH zDvNIu?exg~8*m!X#kqhxPXgf0kOHB)!J7_Yn73MT+#$_^X0&)lZFsr zL6!C#|6oi1g>HVLnGC{92c4qtc2DO=+*TH?wfegs*#4bXv;iCwwSYQnZ1 z+5Toj>pATfg{cz@x_|b_9AAKCp>5OQgNLW`1{R!p86N*h$c|_EJ#_**yawB%^Z46o zurYl-W%>?iI;T;4P zwe$>HV2jV2LJ=gDL4fxOUT~21q5LJqZynN>L2~k*f@4mITJaSyaM%-=dmw>~05?7W zJtF|?VHN}$$iGc4dX9gO6$6o15r&x-(AGrn5qu<& zPWjWz5A~#93#-Z#hslY`m}1Qo!$xbmUH7c{I;N+AXYs)_rUy}q^p zg`W8^x(DefxE`2OqP7w^9?gyr?}77fQXd|*xod&Wr4^uM*W1F5DufI|L+v=i=AS26 zC%r5YZj4}j(-x~K>qLmN0^h(VWjZg_wkSLRGC2UdX*SEO_KW_ z<|XV$k2%a_90Xty+tCvAzj0n&j-VCmA^(m_lF{q0nXAa%wS{+G72O4v^(Me{<`yV$ zK2Mg`_3+K5RG5u!kb~87CtEGM0~~LaI#!@=GRl)XUGgxO!}wx}5dI{U4iX-*V~xWS zjdKseE-7_AkOt{xo?~hKW-q05BZv*7Q}XE4$KrsH=V&|OypGC;^Zvt?Ep|eUMqO0- z&GYwECV7%o!^Iy@tao*y@nEC9Y|fyd!y~Rn4*ANM^hx%h3%@t7I4PAd(M2UBuV0wtWAqepLau zw9h=Zsf+a8`AJ*v+Jx^$(bIuc)y=BY^6TXkH3_1f?9qMQQfq_Xa1rAmhH7=+lySJ0 ztanAd`mxDX~t zb0x7@vk5RMZZ2;^HO$mfyA@WL?4P>vPOMBu^ES@QUSXw5SYFOCaUyJ=03^J{A}M=6V=Y#OmE}utXhliD zQX+2n6s?3ob~J12vLLD5^?L<(?Sq@0$Ate`7w2yjavBn^-LlJ2gz6EUqXA&;oZ8zjT-yNi4s?u0z81dzE8B^z z3aM-iQVl?Jj<=Yt1nL^#*1z-kwsalMwP1WeeurHUZS)AF76Hy26h<&G+n$P~eSm7t zX(Kc8C=nI7X@;om0sp&(T|MoIkhMnF^!}qm{%=YL9+}X734?pg&=AlGOAWoaSI1!} zc4gcVid>&8M_;FD+vFcwW@AohdVnW*xEGr36V}#6RSEMuLTd~{JnBC)elWbLe~MwQ z>KIUh13{wFn<%1t?46pMC;-HPqRpaz8m5Jp9ea;@%rLgnbKU1o^kheG}`9QGU8gwe!e*)PH;=xDMSs!r}ApJd#kT>1o zx`=XrTG4-xOaB0VFfKj25A?F|766(HNjFI=xt&a5vwLp2#25j0_#qidDL;DXq&K6Y zbqS+VA)wP>&XVE$J==Do8X96MZ$JDd?QddGvWJ)%fb&`ws~N`g;6oU84~3A>raH0;C>v`@tFx^DYtCb085gT-4gQ297Sm zM!oNhWa@$9R@;>7cn#QdG&2|%RWKyJrC|6DD3A!A3rTU4$XHKtxb>FEwTuZ0puyocppnq~rwRCyk*4!05>1a>RLa2>2~&K6fGnc1#LTL) z(MpLY4E2sD9)v8-k{vs`(BPgq?zdVKIEnP@(=t;HCrfT=RxirL;WAK(&Rraj1q7j6Y!{U100l}~|R0Y>xL`8qz4 z58jR4?y;eLcD`I2!py z{!nO3_2E_zKdFH6OMfWe;^q~gNdLOk@I-QDQHI0gs60w+WG&!nVrFf(*dkn*@P;Jy zg(Wo_SQR_BZb5qc)a7@@$OdR;u|te4r9Iha+VUTC+#-OiW*dFDx9j%8ce<9Hw$cBX zgTiHpy^~37Ly^kP+0H5qx_hIh=>Ud&i{sD8IS_$P+m@i(3PZ!FU5}lH8<5X794mC} zQxy=Jg}a|v)E~g6WA2Y;tH4=Dshz?&v)siRTwG$$<{Q}7$Ng^qJmIukqwiR$NadBy z5})CPH$SQcu5O_AQ3YwO2@oAB|A3xW!GuZkKyP9i>l6dlzeXUy^SQPQ-eyROyiawI zicg_t*p_FPEE}Csp>>AP;)|X;Fs*ObB0IR}-}dyNU7R7;i_^$SJ7}&*xs)0c-{V~k_}!YL>DLxa|v~oXKwC5M~+sb zN!VTLI(I9aDh+-x4Pyv*=w&8)RzD+)Gu1K}Zag4p-Wrk|u5O?$8mI|-y20nz;ujr1 zr|E7*^nJ4Zsr8RxZwjrn?Xk6Y;wHVg(M)gILEpkAVFUFMAZMnxnMik%sF|c*Hd{GL zHn$Nb%zv&%rUsOCP-WVcDHjA9bdcPq1}wDb^N^h35)kee7&=TCEjktK+AoIe!^pzf zT#k)dDbj*P8Bjy_lT6|}ayhBz)P<|F&QR-x8T6J`+To;)YOQny=Xa+wJq5qdKna*i z*KEvqE8)JgfpVCwioynbY7Dovtxe#WQ}9bAd4V=79fK)b$u>u}=JyB12nX_jp~zGk zb$K79IaebFBU~eK{i*`<_UQ%OQLdI_8-7EU`GI81^Nkj$bt6Rny%VnnVV#w%9XBiQIqO?*b7in*Ay+_F1>}x{YJ*Yhzpy|8h(#R(t~wT8oNzni zFQ8;r)UbjG(^>YJTYQJT>kA~QYcO5Ex`6t~e#t0zZ%G|Q#EeR$rT06$8^i2TG&4|+ ztJ$AZxV)>WV0QVaqrME@@f;0|ky&4dP)?d&077r{)?=a6KY8hq4H#3QGPZ6V41MbG z?pO!^?(H!}{m9$Tw2YJ(_vUGE~A zx58*tjx5sfoFYgTKyQV(O}~ry2?Db{RJRl_1<$5kZ|9$FM9)18HmHBOQCo0NEB>e&&Nailb(*TK(~*anIM~kF`7o# zrq@BB!1_Vj1-t;H(rS3`2Czif4MtgU`<0@j!Q33?V>dSKjZ#y>Rx+IJZ*#3?FZ&XF z$?CtEBBue zwG&91+XoFfxzb7>i6t<}etNrsd~U)|fi>QILm7?%8$XRxbBnqrK{rrn=pdv;Y<3h= znoX#q&}>5XZ8(t5sz+es%Y70w`B9Np*1?F?{(}w$&1N2RI$JB3x-dg@8crIMghQ9E z;=H7LcVK2cAtPW)-HH^c;63NLV&E-+27JU#?+&1jH@!g-5WuqD;tNc(s6o#)wQWze zF_U)unP6bWMEC;uUmmZs9tT~9xnRx`~1Q+*f4GxfV*6u=x{39wLEsgvEcSlw!f zY{NWudJslNiVf|tU_GU>%mBjaRW}QOE^Rp>r)W1Q*(&%~t(wQU9fB?tjr%f~60V*(_KkVr#HUjFGWR{$#I3+4L$gHqy56mt%6ygoNN z$}&QgG~P|b@Ydt{W;X%rx=C?RK(cUoqtwbj$YEl+FeU{z>!V#pPYQ6xPk~$w zQG$40b&!L5<|1G3A)^=WxQ(~a_i{9lJg0se^;};t9gco;yLX1?P^9ff3!DhWXd{Ku zHg7CwD(uAu0bdg+|FS@^MVgKqv`C(C&^o*#XCk8GL@hg+HwPDxm=Uf#5y zl{)b)|2st~*Hvg?xm|fjOZjx9&99mI*uAa~vKr#xx>B{&X*c!keMb&&!?@@fYC)R$ z$mSB=+9i8$9BJ9q!M&wE-Fo`S zWuyF1tUJlQ>CDUa&dUMQYU@?jj>2y5YG8MA#wI%7Y&x~BJ287eiEebHgsE}0QIPM8 zt?l(_lO33%WKKJO@h=&qc`TELNpTeB9d4puxq3$xD`y)L+O#-KEl>Qg-v9HHt8kMy z(dE(O&g(V`SygL}-x&|kY{a!`g%39iUWpN87bcG}sO~1=eErN@l5UdgRs&`K)_IAV z3gu#}e1rE@oHLwNGEQ?sj?<&gnTw~poLR&~MdkW;%BQEtm~`CJ|K|>lTQogguHW_a z!pn0E2JiAa?rGQ7_I?asv60_D+q~IjGvJsS%G<<6gSDLTu{InJkz(ngCu1XLyW-|H zbzF&>t)$NS`?9s@oA4!GH@6P@h1@@gYjsSDzl zwN`9r)*EGT@yC=?-VXDaRhaki_1M_N-|8x~l@D^xR*w(cYOod5)LdRl$m}Q2#}(ID zcJXgf(7t3+zPOo+yPIvYeyF^b;mcz>v6!fh4F{c+&=TX+QCSwPA8OWd{I?pCKuD}q zpn6t(FnGtM@K2vUNjL`!ZOnWMU7f8d&&SbFuX=2)U~#P%=uO>dMXh+aEasnopVB7^ zyNY`moawx*C-&-{wx?5x%?}%iu(V;D9~{0y9Mol%Pn$2(MWvFh&cnG22`lC^_QR3b z5jEpiDH{8yXWfXXynmVSrau4r$+E>6H%()zwN}l)wkb?4*U~nx)@S#4>rFs9gQR0I zop{~w;x}xl?68b;%DGhQ38S3Gm?7ttn5{drIxM&iE&b6*?4v^=db*Y|;wmyDL&irW z+V8*DUF4$122u5=D)10d$+grux)#c{4WugI7PC#(va>t%ex-+Ycqe1|PpKBjuxnn@ ze0+$wi+&|hh*NOF>JV|Z%+F?nq&6V zMvYeeG2FzdH{r)K2Bfi%6i5yV9@yxSkupyD%$L~H95U%JzjJY+Y%G$A$oG@YKUtHt z4W|?5s}A~pest4XPcu?xp{#w&dO%J+uU+`Ha>78aQVQAUMkAx8Llqh=(G_k~-1fRd znJlVDA}5_s(8Tx5Il9hz&wA_Vj|koSK6^rHsGpxf=GezPoAkEJTSOs(4#u5}0&CSFEgiU=#_WH_m9v-}zC%^GPN>Cg&#fJ2uTAk`vo=NqyB3o#_!ZW1=sb0&-Bi6m*9kR*(P;4F z#oB{J3-E{5KsxiCy;PR_rV2~B4ECaIaJpsWC(ncQU2NPuQoBMmD_NF;KD>(|cDFJp z5AF#~qg0b=8Xq*VU!2RzKh8SyDu>4iSwhesDq$xs;g!tTRrE8#WV)p~p#Ag0>1q98 zDHBe$_=mj8DbLFGV$Mn)A`y10;r-%pVdHxxTZ48SjbzAjXNsB+4lyqsv|~ zvN$dredrW_v*NB~`3IehV|IZtszzjD0a1bjeu49gsAv{x>+`{1f5!M8SZ0It`T237$Pyk$R)yKx!r zVW*$0Ry-=vbB!9(W#HyvN-8%?vbnNiseM%b>pnQ`bz`)qakU}rhx*+(Zn}<=(5uga zc!HbMJH20VU4^VC7GtWpC1?6x_Y{<^{E5aRvV&Jr8Bg04w+l5KWm687PZ22pa_&Y9 zjb`d*QwZbc>a^vGG&aTSN%X+Zt~)>4Pxv_xMdlsFV19}oubO!)ma>^{I38PU_p{D3 zfK@&CVPzI268v+qPLJWP{Q$8J1RJt!#M&RcULFX_fuhXv}M5anqZVL+QqqACCvG_=WkCE!C{rEI*v&;9txlO=$8^ z4t5xKOsf4@Uv1X`p^`^Z35z|Usll8ALVE9Y>CA{R{G(&k zau1&=`wiO_r{u|{F-qJ>l!rHUW;U9>>J)#IL%O{80*6LZ`On1TrU5yx#$`d6{{Gi% zE;?RP_srh-U+RTA7BAahG6uIw^m<3Y5(&9t!Tjrf3~bY4>=pLJaCc{`aw_NEKJV$r zr(zxY#CgXZ5u%c{T#3TiUbEH=zPq-EbLKu&SV-p|s~qL&_R?2JkY*BtqkM|mo3O)1 z9T}0jw%6q*zq7*n$h^OI^1RR;M_~>)uPX8<>fUcMP*W zg^OoSZr+=e_w|X0X!@Z;N*D5~FGIi$3{lv{+&e>iVE}i~=_amU)bNUvT5@Vg&}Z85 z8C9zNZ@#M)rUvdkQ(iAlT{juFdVG-*7f!*HaIHFqoXRPwM>z?eE5#J4NBDk>wiWEZ zhTn0e<_F0IVu4GP1((uK6?}-+%TE6Au1J)^+27sZz0OYTk0roO=hfF@I(r|ImsGYK zZjtSY^1=4pea!JKUFAg$wYF6;ZNHquw>x#gJDuTEZBeLA(KS6QFu)eJ4sxvaI^ZReBV4#vF6GN9=-3< zq)s;K?fZpIpYp!PU~~}{HQ#Vd5jOF?__iQ+hqAKbQeMHi!a@Ce_dhtTFRjyLsRsn7 zd}yj%%+TT*F~lwQ7TkqX&t14pct1=1AkQA1=+m)tTz>bKx>MEpzTM8ga@aD9&3!t2 z_A5DWJN6V~Gu*KTw|-c$+6P^L1rT^HL($NxqQ~7*i=7` zPQ9r#IIq{#KJ?H8)(W%ftsjSz8QQ`jVi`J4O3y4U5H;Wa(k&jvHx|4AG1RYsLD{I_ zlcN~5z0khr)NS8M(B%wNkHLSi2(S9FbQr_{wo}BlTe2#5JhxcW*8&F0u{7+JIs=>5 z3BBKDVxssc4-VS(3n>=1M>Uw`RT{(h8lQEPpJLf`3Hk7j^sac+qC!4I*fqWPt%=8^ zAB(+54;#W0{UKqpe|JwrR!{3=G7w84p#kG7oR|HKqrzr9fHuup=?RXtUM` zcrW4Leumtzs#Ls zR11lKaL;uXm>$n}8oh+9%St6|(viRs*`bp+Q!;g~?g4-)+GN%t;!P}$Fg;qV8Vl^Y zApyH)V(uOef|@J!*f8V9OxEYb!4(6Il){fYWKz93>wVwb2Pm0z*;Izr7{l8vmq}dA z4&@#T{>cJ{CvL`_faL&2UbJ7ljwey+{NXY^D3vkeq8 z3wx&%?Tk|ogb8a0^}=}Hg=LRq71gSjxS8c$Vxus4Ny?$m_U?u)NCzL!s536ct)u{J z@ExbH>BH7q6bfW~iAr8@)A5&W^ zVcUNrKXUr;utLjE~d)4S9M*3)-=E7#K! zoBU}Ojhgcq1u6%2P0jO6_#V$X->c0KJFENt`Z1f`hivRD=U&bo z4Ji02{u9plQuQMiOb{3*wwJ9ZJVX3~h`0;e0N#0A_ThELd`QF|@IKV!Kk2Ap`8S?9 zp?WHyHp;BRkCxkM`r+X4u=ox8pVw862F}W@Y&^A97evRLMFUsKO5$3QP{@!1)JC1134~l@L~bU>j*0u4SFy4lN@+OSypfWcbq8i|1u;g?^a zUk;&9nndIkjLV~t&7t(8J*f=)A+Kde3z7w1*RgF;R2Sy%{}eL%-s~G_OdYx5iyES(`eQ*~c^EjhU*l%WHB^GJH`awSI8n zXDr%9B8}ky7DM`x8d%n(nsubj5P%mAz`y~gLgwX^I0rA-Ow}XCB%``Bvu(;GHx}2W z;6OV@c(ctxWQh?U9zbF74ZmbuZXPP+0-(#kT-_b(71!zwO~ucP4k4V5FQr7p9ep%= z9IwN^c-6if2uEdpcyAcr2EVRr3E!@*25oHpg0AfYKh961XBrqu_rS?nATRZSR`51wd7j31CwrmmG9Xwv`qBRhWjulY9LU z)d)bQx|vrj@j0rJRm6pJ6(g!t@9vUvzJp+sK-IxyCiNAf7kmGeJ=adczuy*;O*uuK zgpIS4ENk|%%X<;a7?7-6&6Q$pX0yR6V9KzC?kp zDCyM$v_u?BBR;}+RucsFc5H~SdkkO2go0XOy^#D(s|GK(R4MTbxM+f+AWn}qAm#(~p{2|3 zG<7e9zFCXueeJ0jx%1_n%fcE76_zrn*?KMD{_FMUw*jLlzN||7u$?fl)q-t3SI{BU z7ZLZgmu6H-D`bAT?NYe7s%ntDoKxKT+B(rFV(@>h`Ajuk#cW zqfJf1pW_F-a53xt|(MvOUoSvQ9s+hp)fsryH`TEreu;`KX5&D8Z4;EPbJk*~)h z3z(KR*c(P)65Fa9=jA(#BeZN3?mN4mOfpP6??K*cs$HH!j1p>(mIdz>S`Xu9M|)e+ zfBbk%I(^2v=jQtLWg;HjA)DC%GlS7cYw3ap$~KCL)fev*stwJza@)*G2dRDhouz+_ zQyuF5ELSgL6uh^+-6C_AlK69$7pLIUTAK3@A3kjIeqEP6-BZ|;H<|U&F5y?rHA~Ts zLjnQq{g#7U68zd4Uv;tEG<>y9Q?Pw}#7Jl2Sh7tNYDc(A{0i?y@h z{vGR1^D6(zqyPE3zmW)59i(N5(24Uams)$C3OH4D$?T;k@lMX-@Atm%-ao#)nvI+J zoKi4y>%-Xq=f`la9Uk#1ke)pVuidfLCze8}xA6L2}pHIBid~u<3R?Uu82qca6SqtL9Jo ztB{e3+5tthHUqo{bcvAcAe1u!mIhrJYO*1>nEt3)_)}LO`EUz09Q~jCwFE7%12Q)V z-&-DZ!B)N9kAS;*1(IJ15`mHJDQwsUD0>7oHboPRXAm;kzOO{xT(@eE2E5jPLBQ`gA*hbTV5L>htN}0iwaR4rOl2j>!-q@TRjQYoK_<w$G*p; zMU?GK;Eb(BBhQONqVRaIHc5aldZ}_XRX|tb`925#`Y>h3_X*JU?2HV}7>-_OeA{8C z-u~=q`So!_sCE<5`yWDX^J+EATjbyFJIj*<>+x-V08zHO{3n_SsryMTV=OM;6)rD~ zJYl>0>Pw^Xu>3M1;d|7Ms8Y;Id6#!xK4Js42Jh&5ipX7WUP`vc4Cx}M)1Nj!&m>i| zJ4+b>qj3VqMqFBePBPm!_6Mw@vvFhlbwao8X8GjJ?|8PLWLy%kq8uO zmP^6qVn1j0-4)XjcWB=+!~y*ophXIf>1;`p?g8)fJ9BRe=BIn7)V_spDmoE8r0b<( z7D>PG;E;w;!--CzPD6J1(sAgLzO_*R&_d3$un)d5@%`pW8dIr9%!id6(@$R`(GHC_ znO4RZjiG5$Rk_eHR!ES`8ZKSB1qZ%je&+6xB6ZIppcW&8+wC;f zZDw_Tr#p2JZ_dwLbl58LaLgcpn9wFK8EMS_(0p~e|!F*FT0uU{~;XH_r`;FG6 z#q$8>&6zv%_(EB|3w)S}@4TXptt#rYY6|Rb?tk;6WqF^1!@DSiU3d1#_Hf6 z@b$~vvLz5u4%I41R*??UR{9N}Ne9mf^n}7^R(B;p6ART~V(~DEG?@cbO9-LyjzVoI zvq(EEWfTGvza}xXUG3#K5KhES1hz~5s+8~nMH?qo%m#i94BE|p>w;!X@WGuvC=`Ft zSD*?NP_tyXqy8<3$xvHT9TK=^0()7z8KA&*JxiqHtc-|4*O3YR*Le7;!@HRh8Yfa~ zKV_a>mu6B*Hg0=AT?afnkeh$&Po=UNC+KR_e#|RxwiQaQ#r(EA69gx8k zaoqqBj4(@`a3V5dfPZ2xeEssMv+dl|8+U{ky)?dAUYh_7QJ`y4(qU+=Alu@5D~xt- zJ+(?AfZy_KyNpD;dmu%?Cdtiy@O5$*3BbSm)B#Yr39l_lQm$_71isnF1Rh0k6IOio zVQTrg5CfR%cxOjzz?y2B^U;#-#jklu}eD82WX1jsi#;Cb$S?*H$3?)=)xp0;PNS$oa9-dO`~ z2?Il41n}KJ8URcs)kCq#Zi8;m>Zn0=z@NBeMx^;6fTQiq{5ZHs;tv53L}EYOVTj|# zOrRHBzN1m!q22H`nIpI06h9jFsj(gk?@`8RO2xACJTz|jUfS436j)WYM0J3WZ4E81nqx^e+H*AR#*_nL#MUjRsV zbtW+Hj_+GdpdI%IDUqY?DKD@&HPFP2rTJyPK8&qTGl-0i?3cl8R0MLOdnP+Bd(}y3 z5w@!tf;WPiHFlz!nd_!`812%PFUHSA>oOx_-ZpwP=+5@^qrX}p&JFPz=Y)Ew>{xGF$#= zR&7>|%-|pSVDW|MQre>y^%yO9ZDf4ab^f?6#qP@@h#Hm$cK;>_>S^2eT}hv-6w`h0 zo?l%gT)22w%+qs3v-X4L$xMbVT0)^$e!Hj9sJRT=}qyw?~XQYl1A@m`q z_%!=ZD_34rW`^dFS1r?u?vtbN2=S8+{uKBr(9-ob*@#`rwc)m(-)^cpv9u|fH~4h6 zNF_mO+ploB$Q15iMU$&ayWG(O=+!S*2>a%dNMA z-mDIy)c{}$_|AUI6WCRsC;=wRlANYK9_(?1^EwKHyoSI40H22Fnqt-Vv=xq7Cjb?_ zh|U6Pkc*X?sg2d!Tm^W*bi6v=z?f`hr)wk7{jtKG683v-LXSX9jHgAzlIH=O{LL*5 zbV#%n9WUIOzg>QPuNK4`Fm*C;ud||KlSTJTyv#i<1R+4+Bv$tQsj$wI!SJM-6f#k~ zp7$<85X1|<@#gP6bOd5`V?+W-3GP&Y0Gs!a&K^h%wsUh7X=uT&Qfh};HyM%kQ$Eq*V6cHsx^+s#D4>@4<^b-3Ixb+9=JV!d-WV!eZzF zn7_siL(V`)kMeJnCW26f@2Fop2v@*}6gl;;2BP8#KriuRv|Q2KN5ECvi;c^b0C=@) z>u|A54z2}?SDumRR0AOd=#>i;q9?jWk_LhMeqU?o^Mcaj;GwIt#lb^UvGbToV2}}~ z=;mg#d8qnn>*@?!6EI+Phr9S;VC@F_W$0UADFbK^#Bhb%2v}tEzJB8Oz)CVAbMgRb zW)kCizBd}dZ-LruQqWDC-$TX#-t@G~@((cB<&z=HbPw*uVx0^?!(c%JysG@(-22ry z+Oc|H&jX@nRe*@ZBm=ycQ8gX_-Q^WG zRdw0yrY69?fw09JoZCQb)tJQWqIAzfk9aBhNAwBTX=VEW_7xo)TobxQArT2M6zTM?$f$}l^Vf` zV4~_#eW5ILA0de=Dgg&Q9E`gTL1ON_9fpy1&JP9Kt>=9>FHU&OM3vx zMyOXnS3_t1yc8|;Yveb4Qc{=6ulI7q$*}6)?J$nf(#=Ej*_EE6hMp+lyF)(bCcqDq zDlSy|3L!LGwOCl%?VkbPGb;)B8p)F3Y96>#p9c5wi^%?OF^K6hr2JFGJRvj+y0h0GI_Y!|C*z)$Z#@9@;zQvqu#U z9oh(mYTl#FYqO}dq#nU$OHQp7g)h_eQiQ7N@SPN-X5@i0=y$&9@T+&zAHVlz zZFZsd^68t!R|Jt%sfNOm2pJ(&ow`E#&IM%3Sh``3n(~$3_ujU?tCkn=Ix~`Qt{VeL zEZN7YVr70co)P)Qw^)ogu$jl~k={eTvk5#ufFlVfk=BcJvF*Oa)-LIGVOhvz^8sbR zK7oN17{5sk?Ng(Q7_Q();novSi!*w%+0Wx+)|r=nRj!k*$4H84=WlCi{4wuAt{#6obG1! zJ8&SFp8nzCy-N0oY6|sxL_+okDRn!nQUhWcOHt6>*?v(pHlymWTJ0=-6M&eu-JecY z@&<@kG}(bGFymh+I7#0GwtShJysK+tL7BXUYJ%}y+qoA-creEq_t4XQmfyd0%zYhX z`v~3Dbp%Bu5?y2_8#r7dnNTVQ_RE2n5q0)(JD9`hzrv(vkM1g15ZqV1ef>jk-Fl^Y zuuJgXUQHSSREFJo*h=(06Vv&_CBjRYDlbYtAKi6SG34+IQ7_BC*zJoNZmPg6UyG=? zqhw-Y+W(OlX=O@{wG4{L+}*KtSef5;Bj``L?Te_nO9e)T9xLYQSogm&szjU>@;9kW z?YQ4oE4DA7_96fd)2`1QG@7gUUEH!^aQ(a|VobXv0^VK@(#ex=(?Xi8y&G4zs2@Lb z79(Ci+b<4*ZAPgb6B2~!lD2{RlI7I*B?LAxzz^|;~z%=Mk~SjYp~71rZS6w4F3=>?;WTx6(x&EI5)E zhE=H9?Q^%*%SBK*DDR-_fDgn#Ed$NEt1PtS$zHN{TyAR3N7 z1e8MrQ+mC`s5kE9pv1)hLyiKNMb;9FFMJzp(P6P`bClit}cXO>*40eeJZJ zvH+R^`|4Gn#mW4vM6SQaRIU#=eK#dmb{lUG5cDSMH%w!XTXC)lSY<1^0(9dSieOY05Z%# z&GN@LAhu#cFxVER3g8S;(dGP%3=55PR5%^J;_=3jkr8BlNzx<@lwnfkeB*AhamHyV zV3i%RzHVlgUR_f&y&qh!Hpi6Ct!i&ugnP4*(xT8fbTVneE( z-Wl-rm-IbsYta0>^D_N{i*9u`S5{e>Pe<;t8)3o0^lt9%_>Vlv$)lN?zjnx2ihRVW z{F)k{j=z6BbR;^nz^cn~1}v^8+mQPXyYK{c^<@b_EH)%@CRbfBSz6xK0kiDF!b)aM zjk2$=uM4SG^n`c#CBSkq;0-Lb1a0S6Rp}~YH+9=8KI3qw7^W=*FB~`!REc50jOmh| zj)m~~-QvGh?YC21Y=(VAcK;xQxCIpS_?DGY{uOH^IdezEnQGPJ&I9R7+v(nQQq)xD zZwuzleBi^BJ}EJp;<7R?59WJH8WTKxN{nFoyMe!g#*}@2v01JXGL2(E(joPWtNpV6 z(_yTz2TN4-M!=b&qvwH6(COZ4=YJG7u+jk5`0Np;dU{i_v5KTEIP#`ou^C^i*UxHO z+j-PdO5z74!YJ9Psj11J^3E24rJ!W_uXr(&GYzH7;<+{?4t*6B!>qbZ%8+#UBgyXqUC1%*-_oJkjBy!2KTK|S%a%@VCsqo$6 zzOkK=v37%~Qtg>&@!9V)c*S4egEV(QaCKwP@lSMnGkw^p6G@nfR_3kR5~r+*hq`QX ziyZY1pI)ZyE^D3>6nsx14PyP)8$*x!cMlI&iS@Y~7#Osyt*`q#^t(Tgh-m*SG<5jN z<;%YuYA8RNT3Qsnel|z^jY6T+WiTNj0wWsfFWA}Hcb&}5zf)fGu)Cvsa5x;B{CkwW zy}h2jePP(1By4!fN=Z#r`cNf=%eFRlBc#5*J^}t@EjlwHeYO_rFxZ_fvuD+q*`90+ z*s4F76p~arGj`Jz|h?+U0quOyAiBhHjTk~Q!j0j`*Um_ zMVB{Gf)8F6;E@L8S*m;?(jG`(T*-^au-C^ceP}*|o1B1e2c&>QdSWnD&+}O1X@A(Sd|Zge)I}a5j^)`>lKI3M z5C!VJ_LaBg`$%}&kM0k1PD~68*{v+t9#%hfB3MgP_#p7hQXgrxrOmJgUf8sB<@rM5_5YnaZL0eB6QB zT}baw<)c7DeFG*e!ju-r6qVrOjYcNC-@B{&V~vXK{1+|Sli+MX@NT%k;eFunB6H4Q zmmyybiUY$A{=-8FNQR5=FjK=!YU|MS7$MK!zn4GOYesNyAFO8rth9qxzVq6V5S#(0+^ASC&z1Ga?t!En10kC*%5}<93(lnSE0BCtQkgGvs~QVbT4Igv=SuqGLBza~QZEZTM{Dln zb5;mJ{)7mpB~sQn>q7$6h}O=obi5d;Ua6EL4US~K8Xfk)i;z)dl8Y0w3NR72`oE|# zL+&!Dgye%7zupMSYOy@!1;WxA(n(JPos38-^J8wE6a}vg2afC1_#sFvMQrf;196&^knOpyb$Q{$3rgvfu^rg z!qC!f9u9KZwlwDV8jg=!0+;6VPcFBC;Sdm^IdG&QFM3%#4lv0F+wSVU)$6(1fDrqY z3l>DyyZ6&qk3rL~${hCwV3kI>r(@iv8oQ=cWx?4VlNRR+o_$m21Qg)Z(w*GP( z<|5YhG)l4QLF~YwIVb=9mJSybaJzcyp&9g@TS^VsasiH|-+3W5k=LpwK!s#~hQ1F_ za|g6@20ML1F!dss2IioES)0g}#w)Dhh_(Z8-{C-13vQ)CUF|kc#0(EYIRZVyT*N!y zSU_K2!mGT}gmHh9Lgw{}E2$1NIQQ*@yo1(0mjz|VJuc?Ze&=@G>b!ag3E}(%XH3@O z{=i@SJeAbIto_K41@HY=~;6I_wwm8DV_j>l!^1FRp?IqDqxdK!Ds#qe5o0;U4H-rKE2avbbUCcnB+U*x0qK;S&tR_!qQXfUku%liF|1nl*#GTB31T-n?dPCpU#?n&G665IQZ1douD}?%eb?U0K ztBtjl=czjwYgZ2oXRA{e5fT!N)lKD~pv3Qj_pMwkooyUlZ7`3Z|5Lz4Tlv~4GzxX< zwll`s#@^~E6Adsdp`&nN$NScDdRERZaKYQxZ;DEYN#Bt%tOvIt^bL*c#n;TA{0Bv) B@sa=l diff --git a/generated/webwork/images/webwork-1381-image-1.svg b/generated/webwork/images/webwork-1381-image-1.svg index b02d9621a..483260f76 100644 --- a/generated/webwork/images/webwork-1381-image-1.svg +++ b/generated/webwork/images/webwork-1381-image-1.svg @@ -1,41 +1,35 @@ - + - - - - - - - - - - - - - - + + + + + + + + - + - - - - + + + + - + - + @@ -44,7 +38,7 @@ - + @@ -53,7 +47,7 @@ - + @@ -62,248 +56,147 @@ - + - + - + - + - + - - + + - + - - + + - + - - + + - + + - - + + - - + - - + + - - + - - + + - + - - + + - - + - - + + - - - - - - - - - - - - + + - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + - - - - + + - + - - + + - + diff --git a/generated/webwork/images/webwork-1381-image-1.tex b/generated/webwork/images/webwork-1381-image-1.tex index ae7b8e226..b29b77295 100755 --- a/generated/webwork/images/webwork-1381-image-1.tex +++ b/generated/webwork/images/webwork-1381-image-1.tex @@ -1,5 +1,6 @@ \documentclass{standalone} \usepackage[svgnames]{xcolor} +\usepackage{tikz} \usepackage{pgfplots} \usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals \definecolor{ruby}{HTML}{9e0c0f} @@ -129,15 +130,21 @@ \begin{document} \begin{tikzpicture} -\begin{axis}[xmin=0,xmax=15,ymin=0,ymax=20,xtick={0,5,10,15},ytick={0,5,...,20},minor xtick={0,1,...,15},minor ytick={0,1,...,20},xlabel={\(t\)}] -\addplot[firstcurve,domain=0:15]{16-0.8*x} node[pos=1,sloped,below left] {\(y=16-0.8t\)}; -\addplot[secondcurve,domain=0:15]{2+0.6*x} node[pos=1,sloped,above left] {\(y=2+0.6t\)}; -\addplot[guideline,->] coordinates {(0,16) (5,16) (5,12)}; -\addplot[guideline,->] coordinates {(0,2) (5,2) (5,5)}; -\addplot[soliddot,color=firstcolor] coordinates {(0,16) (5,12)}; -\addplot[soliddot,color=secondcolor] coordinates {(0,2) (5,5)}; +\begin{axis} + [ + width = 0.47\linewidth, + xmin = -2, + ymin = -8, + xmax = 8, + ymax = 16, + xtick = {}, + ytick = {}, + minor xtick = {-2,-1,...,8}, + minor ytick = {-8,-7,...,16}, + grid=both, + ] + \addplot+[domain = -2:8] {2.25*x+-3}; + \addplot[soliddot] coordinates {(0,-3) (4,6)}; \end{axis} \end{tikzpicture} - - \end{document} diff --git a/generated/webwork/images/webwork-1381-image-2.pdf b/generated/webwork/images/webwork-1381-image-2.pdf new file mode 100755 index 0000000000000000000000000000000000000000..1605b7091540f36814333a48aff4002870fc1f7a GIT binary patch literal 8839 zcmbW7WmFyQwxuCxaCh0b%icIag1b9m0~?1R!QCOay99y-cMtBa0fKK_g1ZIioZCId z`EH-@o<5`h)Tptl>d##7dfr*HicVEhnhnU#gF?4+esYfjhZ_(!G31Qu(3!WilGA+jX>c0Ewlk!_seaRoIW0 zW8&yyiBEGHaowt6Hdtb!Q#X#n<8bTIuP?GkT1O(2)vpUuWTD|$sYuWTt&y)AD4t`8 zf;#A;OEAZ@@j&hQI~FN~ZnxyO?pClLt-uq&D=gy0OYvr^2E&Y@?$orp7`Ai9hiCRG zuQKSGz`V|rK(@lw$i~Pvv`EW^-Os z)5nCFk+Oxx$ubfHLsa|IkrOM!aSw#@2rl65R_UuA3R0i+yg9LcjY|fqg8}FH&A+)?|UaclJfrs zm$Z)4@M%?;g$Jf;PhhKtSA5H4{EDTg%A@w4>T;gKMx!g2 zl^A{sov8%Ardn~oeLC8&_$%V32!R6e4p10^+Z`fi(M)2rzlD;7%d+Lj3Fp@9} zB}~ix0F4S&<|}H-GfUDHCYAIjd&(;?C>=3fHDrU-20jd^AmNbm|ed}R@Nxay?Ga*Z3)QMywp@eUkZsTuNFVh%=;^euSq**ixr9(v^(#IPIO&kz1M^KdSKeJfCO#L}Mq z!T2$UtABkjCtD`8J`q}UE-nvM-(P?n(aS^)y?kj_k}Lj+(#+)62n;2mV`L^!cn(9E zq(~x;iZV@F?W*T{RBIL=`k2YJin3WEVUywr8uXlRUVRkm;)VLe;q~7}+8cEVH~2sA z?bho|wpJS%y~32~%$ab36Vc>}cA?;0w|X}+8fb-ZVt9%%?qtb3iM_M8OBzu!DrS)s z6mi%6TVfzYJ0FktqZ+8l{d7H@qWpl92n1SwP;d=4R^okpfQS2Fc=qA#3xny4t}C8} zPpiGke2l`*sy?S^XG`1h)CHSFSl&mofw>r@V_8{>s^#y7VMeS?(rH}r>6cDOq;~l( zQHI?yNH}^Zb!VO31k}yZ_B*6%yAvJ5vFbp~msLxdtY997>=%py>; zaAn=es7pY$fk%OtTm;ZS*GM;YT8~;#rtOP{st+X5VVYqa9!mO9cP-ltOCfwU?Ddg7 zD&`7VU2?tzUXESSO`ln88AAmDlRSuxsUpLV9$m}>hHshFj%q&|a+8cmX3A)|WFc8P zdxKc*`v29os_CJ9e(urb7()*Q|R=}P%+fJ6g zODKLnAOuYxF0!6aTWT!|_?$VWn?{rKEPC+sSf8SJT=vtoy=iW_1K$0qG?dXmm;^~W z?*hHe*C(ZwGLzP6n;#ufA%}h7)ESUC?0Z@2r7|A3IS>5#6D5vz&joKzl7VCm(bsGI z$ujz(?|MH0J`*I(2LciP9C%h0W`i&heJaKzs}WVnuF{#Z$RxzCUqVDMoNagbd09A$ zYdxx81(I6R9Vq+7Lf85DI9Ar;8;#HK<{Jh(pkb(ic{ZY)dbO5$bU*WQOyxQs^NFlo z^`PVSWcN4x2&C84{4#rFJ`#+2$#u2F73f}{8~xB)tcx1>(sjQ=D9V%%s{Z61`Em3ULSs} z?;?Ug8UCcZuy@TA|1DeK!c68||FG@(AS0GzIl5tTd5|kTl(a}n+8*srL3h=)>X5US zj#`Gth}!AK+)*TT;Gs)tAPwxeU*!S+Qwy1d?pl?Ve|`Hzv0VoL)z9RoR!u(J>~3W9 zwbI}a(oQZH;*OI6R}3MgIg#>j%C=@rf9Ta^xj(EY62dl$xxLx{*~b5z{bScNs<(Q~10@E}Mi|B3O`?eT?(3^urc?&DsZT_&|&o)1=Yl7ju zyi{g!?H-%>1Yw>pwVlv7tqX*z+^%pS{d%P8AXWX(21Trdo5*&`y}scHff#GG9lwS6 z82miF1u(%6C@976Oc%-DjJZNDebHO4E$URY-~;@ih*dpjpFDKj`isMTy+RGMh%xp* zGlV!yf*r+hzXDvp45r*A6>7BzQq7=p(UeB!e=a++36g$3G9Ta{G1}xxs!gP`mo8vl z_HIeax@+z@L>f^us6rQ6->HZy!}AXuvt#`msd)b>m4>T{i{~eE4$VIYpv1rNPFz6V zzgj0#Gh+zE*vxDP-SNsXB6`~G7Cp8B=I!F%%T*dg@7`JD2-EsAB`8|dEW>XY_=;tj z;C(P~nHwgT8W_0Q0SNc$!=yGs4v=*P^f0pYclQnttLe{M8yi#9gXXL2m ze@&<;HMX+0t|~A#Gd3>WNYBzw&d5$L-pJBTOvudAeY3N}2*1x@`7KqG*=o@&IT=s) z@F*=U?TBuV;AoeyXvO*$W3pwc86$iVK`}u}(Kp7H-Tf5Cd%C@E%mHgvNc3N>oKfW! z)T8ixt~Au6E(2ms;J?R0h~aQ)XrlJwFgkPhJJZh0d@eOIABtb^oMx6#+X&g1PpVR1 zof9g*H?$R+LGEllJGTZzPY7$CQ?n-7hX#a{ZbUKnX(_5GC{Di^ylZuz+}(sVSwF*+ zf&nF&SN}~v`2N|C|Jl5Pc=-R?fPXCA5KgR)eU`nJBD?~E3qbFd_IZun4ETVDz5soW zR#N!>jwfPG0yCpM!_&dSLv7xGgpM#=q;ij(ol};Mb<(SmEN@7S8HWKtYZN;!BLbAj z|1~=PE(J-2YSv z&ZoTRt1px_a;GnPYTjnu7~d4ySe@cuyvV)PP5RDbjgx+D{5j{tF`CzXpU%g-A01Ka zcpq&!lKaSdD3rpGc&0b7OhcEQGB;3y+d-plc7;58m^Z}xg&TJn0_2a#_18Oe+P2j*tOg3%`cz{Zv7NWovAQg8l5Rrl%t?^X_KNn(k6kOW7P@OvFWwRwrJ|rEb{U|&Vx~#6 z8P*}b!IM(J%hxIWS;E+ykGZQLEC)=cY)nXuV9ZQZO_^hSM)Z|zkeXvzTzIRT{)p42o2B~-%;7Cnr?*|bO*7#iF&TiRcX=k!hYPef9}Tl zm0gh&<|yd_nzRB8K=hkpyQ$H7;eTuLtzJ*L*r8?zMqEk{T%avtyfY>Y(VI;;TuMe^ z1KhcPdRq+MC$YADkczIZzUQpGl^8e<^UR>@323O(CJGv>v})AWHm--Sx2bF>eq_JC zSqr9kxMFW)U36>b>ek>2el>g?@vRQ@>ta)Yz$ekH*{!)IhcA`9ER~=2j+MrHTI_oojBA6kw?oS$2V<~8%n8X zJ#w5&j#YLE51(htTilI>C*fbCZeHbWmC+U*omu`031VHFs@M7x+ZVM#R}&qR;g@Fkc3LF5VjKwLqh`LqUZy7P?1RReaX z+x~7h6ZOu#e9Vb5&dkQ>!i^Xo=z}n@tC@;P{yx)!$FMGP-fK-v^ zcE+cUs(>GR78@IK@gaTiD9wVhj`Q2$8dUYPL=X=cB6~nx;J|1djKR%2^L%yPIBrI; zJ~6UfBy&-!7(=d*L|PF$+7q`yjz@SU=Hl^to|xbpJ2-$rT7uIUl7ZMwHOvX^bI?4&7sUK zi;#?TUL}4X9`JqT^OXrNfA+xKW>yq9<~_|kOqk`;qimbaa}-S(0wWU&lpTZRLT;mi zSf+0-fVQ|HpuGw;NdC9?6B<5v%0HdT^U|D)sd4a<>3qaVUigEl0D88d2%?DK@Ns@i|gu>Vke72-O3ca;Nmp7gYqXZ>cC z>82tltGr6y0i+jUlC759LX&NhMeiK+cxcr8AUK!@Fp@a#wkG=C$PlI3XV(go8JC$4k1sk0em^f-HJN8Pq!M3`5R?m-(> zeygE52fk(Kv4IiR)|`i}g2)j{w6mI9suGe|81IyE?mqTACVH%|c)v;Cqt}RfeQm5A z)%L+sVpaT^?~Niz|8;-w>{q}HANRp?56mFO8YQezA}0U^kl<;?cR?)@X48#~C=H#P z9{&;$N{Y+a5s6{P800x9!I`X=^PqPy=DaA#fK&G=GBo2cBZzXAQYt$^;?1#OI(fyi zMi*R@gr>&u`<)a!4SX6-L{T>kVKeqsAAb@*3t5p26KAU04|gSQI0TS4ouNmSrsM}+ zPab>*AxGb118_&M-a29-)HXVB&Wp+YLerId-5>z91?xVhw!A*|;4T?kWoD zWx2eFwA_m>&|1Tp)q1rO6a{rj3cy1(KU1S ztF_Rqu8k0i3`rcXI5`cts?sjd&3@RriGw%3Vof9YFc(4X1OYegS0%xLrI*6ER8iR- z$Q;v9hD7)j>fgK63*oi*@-1G&s%Tke9=7{I`63t6N7U0vJ;Jvj?YtYR4EJry_|v&F z(Wez-wlqmnmD|VH%f-k1czU3OGrs#`Qz8wInUmh34r7l!ivpUm6;?L64hdn-QD)m~ zZRH`tIC!-U0G1YQ@5@Ot=JL8R+@0+Cx}I-V<_st}H)z})4;$i8vGDtza-YZ(LWlHR zq|hHdeu~8Cb)R)o#C?s8?Kr!uJFHMCYdW2>YO(!ZxKy-~^fNL4n4*zKd-s<>m2MJ$ zUQBg}XkpJv%EPaaYp)ak@&j=884mAa%q_dgY0z~kp!Y+Cyf{Xs3Uo-T_rs99h{()?#n!>fa_YpnoX7srI}lRFH2=UOwm`2^~k^zU6;D0 zEy=nSTJ5x~s?=?Jd%g+#)9fhz)#JVBJD*pCpP6{Fe?u;=f0FCp*lupVzmf{9D)#4% z5v%JspbpLsu6LTTh3FUW+z6bR^$D_x-0)>jS2bG}nO8h_TPC>$@8)hPx8QrH zTkQp+uk+HLWXA_|;=x~dX(!%W6E`uh*Ad3jw&R?BxJDjH@c$zEz*SuaD&qL5kkA>_ z>sDky9~N)t>RDDyWo(qW@WvD#@bCpcY=dFZmvLp|$L!MuN(V3baaDWl7cZC7Sm8{V zms1s#=HX0q(~gV4ixPI+fzsRUf^Ur?yu7b|^F+?<+cM+^1Li|m@_yYPL>!7zNv z{J}xPxzyFTMW<`OTii6KAt ztKdjB^jiFmKZ_X`{jG9bcujg$77{VFl>z-P5U|Nud6}!DqWMR?E4aUQWd(Bj4+OsvGS_ zX$`HaQI0mH)`Q%(uIH%!Mp=Wfh$A{`JwVmqh>Fa4m->iTNW5zH{?xALi*CaH9)hfe zHZ@saxCLg3=1K~)R_L;gbM0pZuh`{$w!Bdm7WZUzd}b}W#;NfdL^iuTZth~saqebS zl2Q%r%D_#7xzshU*!mMz7H7dP^2^7n`b!Gl^72f@fj>0nqk|z?#N@8iHk&1;4H~4}?KJ+^1p6p_Jl1s@<~WPU!*Dtqea< zHrgH}KgUO$2xR(S?_zV6{UVBn1kFGDfBAXNr3FXk}6ZwmRpOaKS|YBqBGp)4NQJt z#&??v{fO1mAM3jhm(N4a*IJDXX5bm+z{QT&j!6wDdg`Jv(b4(=*)N zt=gS0IKxx)zZ@N@YTW(kEkI}s!S@fm1<7cj4(1W)y)3n;P?ajakhmq4*BLE zb83uJtRt$(@@&d%%nC|spT?$G)Xo*gm9zp{z?k9!`}+jRsf7eg_j|>K1iP?b29RAS z`#^~F($WMIl;l35`+n8S;~r;2?t8hvpyiVL<0mv|+o6<^Ffez97PbG{58z+S{y!2! z4iys{4Hr8OWdI2HCp&a%23&+o z1R%oOhX;fT1qHxqSRsdn`itK+>c5r zNeK_BMHY{c`$A40*V=M*4H~91>z5Ahk`A}((VYcHXQ(L}LGewH8lV62O9i-~chWj3 zZmG=hURc4&R9g?kdW+hORBHgJe{WQ>VeM^@(NCa-c{;JWDp6a%J4wyS&R)`P{77CD zc+f^u?N(aq^)%mHVIXXKcU}~xFPl2U@$6^Hz;dZ(qU3NrN&Gd@y7@y1EIL&Eu?gC8 z-9v$~+ZTtMWgE%E&OuGV4rQe^JlvE~fwoXcdQ2p=rEk+5I@<0m5U%#6{5p8ejnhnD zT9(R7))z6#QV<2L_%_;ab#j6@M0WM`E6~0*AYFSm)x|Zk;N^zuX|3D?O9K**cJ#>T z8!ra;wvT4^pK`admiC9UC^G~r>u{*5blBMWf^y^H`I*I!O?IK?xv zxa~!(9iu-<;ov^Mr5+y7JaGVsgVt$}Nbqi?!`L?wb&KHGYG?ImYdvmet7Gb8dfcXy zGf*j6N@be#I?JC7H-DOa6x`>JxIYUl#i%)ePY;jnH}jBmd$o4Dxxxc08DXIwbO8u# zF2-#G$V-A_M6flGOdzDX$Z+gn2MxUBLUQBpsK+WA_=Ushwf=ZJu%yVP^}-iD4V+Dr zxB8LUO-xR1^0g`M>Ww&OZf<jS#qG1`EbO$p3| zQhf^|+Jj^xdwTZE4@-il zKLBe!VTgI!Cw<%Iw@XQ9Y+he*J4lQcdx{O`F=d*69G=KPWhQG-{d6S?_iG?<6bD$f ztzDAxBN$Bwxuhoh0HRByEAdLze;sD7zkD(IuVL0zH@D!Bll&mW!v!=mG2!9kHRm!j zgFwuAfe;fjQvm^Pu0KU{5JcqveFvCB-NC^H1^9>G>Kto5dX47py08qWUcC{CU03*OMUHXzb!d@3K}; yS#&D$>XaqC%}R;^vFg_Cjk`Yz%D>9x>;iFe@o+M?K;h>B{`v2OPD(`@<$nQ1#$~bq literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1381-image-2.png b/generated/webwork/images/webwork-1381-image-2.png new file mode 100644 index 0000000000000000000000000000000000000000..2fbabe0a982c0291bb3753be8cf135f52b4a0411 GIT binary patch literal 35076 zcma%j1z4187ww3OQi_U%2#8WjOG!%z4Bg!*NFynYilC%)mvl=vgP!H>vQbnwbc zMR`dW41>y8NJ!S$zyJml`xq8xBpaee(6qYAGyneSO+jy2p`b^U9-+AEG+o&lPv_!f zpP79qN2B%m!20advkL^8VfRg+;bHl{MZ>p!M~C*a^3`oQ0X3^JqPSfW{_2w`=MKwb zo0(CYj(w{O_ddLP@wLTSU%s*q*&IMbh}U{)jK(h^yncPl_TpG?)eTtgH|NT+j{Ur= zH-d~0r_0N$t@j_kwS^@vz36Kdmr zuH@T@rwvYHzF@}>s?m8*ylaSbjAL?>%_JdYb@q$-Ts<3 zIGwjlJ;~bbJ$4>!tci%VyJ)!g>z{_+(DL-Rx^bO?@T1_RAId4X4@DP(lv9Z3g;vM~ zpX2X))F-~Ap%N!}@zGjjKWQ&Ru}4{vISbPct@q8svk@WxWZ{gCmygv+bKjLd-Td`t z?xih$B}^;8LbxnBKgqi@WX^#Q+v|;HiX??6@EnDBz0%LsODfR zUom~ZUFW6OU2u6odZQHWW6TnWH1) z31#R6&0BhA7UG#AQ3X^bliI4 zbmPeA0H#F&4`U7Aif9eb39r8Q@cBJ#9Q44TIvTxLLq8@y@rPosQtqejh?R)rz}qBR z;7D;%aRTu&_;m^QP`#f|etc_LikJ^|yzd*_8@?Uf7R-C!S;}5HfWhctQv_`&H~p>{ zX?TD5Zul6TyJW47o3w!3S2+2zB>9P~`kebhoyyG81(NHZfBW5(YF2-!oGCdU+=Z1! z5z-*3tIXsab^nFq3!`c5X|CxD)0c0D`y`0Ri^u%DrS$fZ2Ag<^WQnL!N?OWHUU+uj z=N&PU{C4F)IsTN0j@^#Q{~%h)|lJ%FZ4HZH3l_` z`jJKck{p(@kJ*ZJdCbOgJF+yoJu)-ueROZ6f8o5Dj6E~Ti*snW-6dwT|Zy9A-I1_jaEO)s19MT0^$GL6ejX6@|K1A z*7h5^KGNa>K1{v$w{^3`#vi?ueAm^NZkO5}!TQdWWtOamdgbpa-=iTMF|VlFxzcmJfFFBFttQ>NW48W7y~Z>2 zXVcGozaReA;JkG@S zaFAB3S!ilIf3TvN{Pd+h_j9&mzzLI3GRf(3JhQCW&eDr?94R7y&0aD{!r40J4 zqt~oEbMLs)D+8ST4>k-U4WgRjIrccVj=mjdX3CWoWUP0ur%cx08l`rMX^oDKo{o`F z9}o8nj|w5DUOhHDu~EskFx4VX&Yi=168380J7W^UFV>!S#Ky}tZ9C^+Fxp<-S?@|u zL8ed-QTM|o%!NBVv(|a($ik{4l^UJE1Qk9VSAh+|=4bWa4}B#ftQp$0q$_xwDGz=M zcQnQ~sa(t-UEN+FDW7p#ZnSG!R6#v)jj&K1dE(yvd!|BTNh7XI%F5UvdLB{rljkh9)A2_&i@T1yme=!1Nbjx_J2KsQ{Ky4&UH3q*zM((x zF&RD2D4!Q8t;_x6l*7!!?t`%sH_Kbmn2dhNz+L`L$DP-2`PVlBi}~@XHxohY2JT29pBEJ!Nc1A_1?4!($JDPejk3{HWV@;v;puG|P?6l2CDNb-#VK zk|CsNQtr#^-4O?}2p$?g-=^Vq!{oK|#S$V>qLZbSDf}_r=pQvRxw0 zy=J>lBY|kX=^9QN!O-Afg}i(UWtM^hl>^~{l%+YATyjIs2dhB+^itx568DFTMMoCS z3pzfP$0u=1ZU*ZtL)c0QjaW%0b}@Lw-ANz5NDJjt6KJ$rR|?|c&&X*?R{7lB zo3cY3ev8}6`I!b6_Q?!KMeE3$Dup7$JK6X}h9bNB4%1P0Y7Q2%gHzm3II@FhC^XlJ z8?$6=BtEE*7)mAd<;~F;tnw_j!=+U|)Hm>WX~-O2S&=p&pLY?;;Nq5ClA%!Q7bl1j zdLGsFwQvGwnB0uTX;miUS^t&pQugi)Ubhb(_f*zer4#rs!eD+^E?HbE`N_gIRB5i$ zfhe;;rh|Xr0Tzy1{w8SLR3t{t0tS0JjxJIcR)o}UwFaL-<6uHVk8U;$Cq8|b-Wo^M z@^?B85Pnt{8<--A%nU0oz+i8P3Ez$uAqxudYmOIY;tTq9MsfyJeJTF@@#bplb|H7y z_QVzbZt#ES#=L6(^K+q%$BneR9=_o#8mw_Ns<9z+_xSUd0z0G|IG!QzQH`BS&1(0U zMmluhT;z_z6L2-ff)6jeZPtBjZ`raUw%Tod^8>h>tehM4U%o@#*EEu7iul1=E0BFb}h&XZSZ!hKHh^Iius0( z1o*-N2IK3QXSvQ>YBgKkE(nCm+LH9LQ-yHw)(@c;Z=RxSU{T)_SrGc{-fNk4-%~{} zTSo4rFR40qbQC9oqb4nqwbxmslB+#ObvzE&+abgR@gj`uRi*E$>_WTkn6 zz5%_fy1*UcIIgvs>3gmvU+6z7-#mR1F5WYbb8B~Z_{zX(Q)Xd5XqyfWS7Ki9@~6|6 zztKm7c?e7zdFL+RBlL0$z=an zaAvu()RN{79;jqS-N=xMd`OGg97kG3YBn;x=AHT$dvM8Ke%uvjH|@fyA~!Nr?mNWd z{yvt@-clrjY?&mDujNBfNvVDB=XcvF9UB7l6oKAHA0v?F3;X;njnmWb)HO#FWPS6F zK1`tP?t96`S?X?#5NAoNh!8HYk?}h9tPaT7*CJO--(wM6n_oR#N-cA8i+7g@ALClx z|2~Z%;@@Nu*mP6u_bPa6Th(Loc-X0){oN5f@eYG$)&yF~r<4HcQlc6;tjvaaMepv^ zHP5IuMWGJ7>4l3XKz?;5TTyTgJ;i37v*}IQ8>CuYz;+45ugy)$CNjkDXHxUw-tfj&F0W zHLkY=R=i&EVB@o%B7tiiA?GtEgbth^@ZRd2dTiR)$}VI{L752vI|a8bKhi>s@2M|L(S0l)*)DVbb41S-ELpB$F{(QYD{)fV z8b7=M(_O+&J-bBuN3FGK9SPv0644=vqNJ7S9p>OuiuO60!3X{qU>>FhPr#iIS08Ju z-JHZ2u6qffXXh?<#la}@$ZH-T+U{XB`2FEs7_5KbBYl-pv?Xz8lb?iC-{E$4-o7ch zRfDV(8Mc~z&alzA1^DbR_3rwWm(<`Qu#34E0`?3TMq_!Y@+m#bZiy5wPPfeVYM1sn zeA3>XD>5-#;`;D-w*d9<4h(jy4jd3KBASQFJq7@mxS8KGxXsN_R%(HcleGf3NsY4Q z?z{*I=mo|jRvm!oKIpJILc*1=K{epg^9MQ0T=l$&y$YR_3}pstA4@`yKpmV#-Cnd( zJg_z~g%bl08|HyAuAUG!B@XaZHvV{&ix<2LEZi66{7RQh`&G)}gp&z_*sx15k9)E< zzxeJ4vbY~TqI{lmv;2)INhO)%%L@Bt22i(C2O?=8=*^?TfYn{PMS3eWrhklkrsEwz zPI0i*!%bv~xFE2SlC%Xj#QQfkmM?Zx#LldK&QCiIH-qDWi!iEGG0%MXgzVY{}CAU zDTs?I!jkOj6)7|rOt-c7ZgKW=sjpvzjyC)Y2vg*>FCb=K$-#OE~NcHDT5 zK$5zw3BbE#+%`ho3^r%ll3#*iVK8fdELozVv&M`^-%SPJSj{EY4nvq{vJ?3?0&Hk56dBZ?S6F$-wwY zw4%m>#N$tH`-P<&(vs61cEcxN;TM3GhWt-yYmDN-JS#fHwfO0@4|)KASjcAsM8wwe zt+dh0%ERdn{UJRP>C(K`A~LuxR!TH7>g--P#4V}seqX8I3xb*7M#aVeb^1akgD}~g zZCj-1mt-n;sf{@}0-MJilzTIuo4;9DzJ1hKZA4U|oW6}p)DLE0_ZB-XJ-YYTNbk;P z>u@u$srhmk(K4_()>mtz0RjQbi?Y@z2$;g#zbm0xK=Cr~8XQ{sx^f$ebg}7D{RD%} z{(O0-css#bi_`augdr#7*MIuEp-Q?pbHob_r3U(CPCt@Cl%azRCd_YLuDF$@6Lb2o zO$`&>mOcFxU#$FyvpO2O1egs>^H%CUd)z=Fa&|y?rJo9V{oB?1cmW_f5hNO*_&s#`2}k&PY>sJDYWEW>=qG&6%C(^Z93P zaL6L;eb^*?F^h1t$WE`?>)W!`yp?oN)(+_gr^Pon^Rs&quBj39=D^#piS(C4m0$op z6h>H@RpV{@B>jga!hU8)6J*;%yf;t~ys1=2*gM7dez_ugvmCH3rso#k65vk4kUp*f z>9cBVAa^jILD%&(5g_^_HI4By9c~-Zyqk~Z{~RKV^QB62G#cV)2s{NFD~i(t3%$~B zu?U+%x&%<>sebUUs9rDyTbbva?F-K&ZD*|!M}z@_^Y@a2hRdO?P;Yugs_94u zh_@y~19w9U^Tn+1smdV}eIakcubYtD*+vf}*j7!gI33{-2s)RN(UF}pS*QMbyNzPE zgt+XTwBw^8xy_kJ`^b~TQb!xRO6#||)VWlXmsk3fL)10e%x+Y^a|mfsEqQW`V!7b| zOvJVR7qPe)3A(1tfZ-Q@j8LR`Fj9)1qIvShBAjtlops6mnNb4mmSaLj@#W~F9)}*| zf*+#zO|P+syVFZ(+0z=<61cKW8}K+Q0AVjBS}5s|sw^2ZA3IR)KOw6=QP!za1h-Ym zF4;y;9?e!H6%@jaQxnh1)yg%7n+3l?>S_{>TqE8B=?Il(d9~Yl0n)cnx8U2`hHRQw zdy6xdeS8;nc{BY!aehlq5iTVfJ!%F?OXB?d3Rl&pQo^pp{)~z1OhUipoa;(LR)eTr zI5*8j2VqaqPB}9?iibUt^iIk63&syV#ts|rIlVU8Z_A-JW69HfTmZV#(jHH51>ez@ zDWZvTJrEvK@T-mwTMD$$q5h~I6Bs8_-W;Zq}y@rRsl%bA%@Dl zjWfgSd6v+C!hYZy-!}U-=v8L(&!M7^n}0&(PfDqt#r?yZXEAHdt@%hzVZ2C3BN+d- zp~J2zfNX%l9z94}+t)%tW4?pWmj{Ir>}P=}2gt~^S-j@uxv_lSeJ|qT!Y2vmsp6#% z5_vC7M}9b36pyswna5S=YOj7kPIOJg-d9@YowDKK?Z|?+kj>Tpe9m$i&=Qc=U;dp4 zc9p^EH=y9g{S^Jq9%TI5U_MDibPm9bLW)KOn$xsIv-^_}V0l90ji$%N&R(CkFISAV zRH(=unB4;f6#OYw_7qYsEtr6m!YT?Y*T;B&|FojWZtq~;&L5sD({Wy(F=H?kIzY9;Ke{TI1Mf!(7s1W@xgp)hJEH+t0DJ8RZ6w+q?9}nbFTiHG~ z5`!r7=Wvk3pTo&TTXq8i;Tiz$pg4mm=KhUx@3dxPWznjFvB9o)Js3cD;GyVyk*E3>Hk+i^($J zqA(M<9v=;zC36EKfqT}qet>ioBvKL*Rpu#V{4_@85AO5$1n4K+5d-N*8wvXR>grshiHV-iyBjzBN(fyLJ3n|FR*yZWKb!L3_t)0I zIAtV$S13LNG;hmkASmxpBGlyyWmA+J})R~ zfcO6sOtdS4+{{m}Lqr|i)OT8rgQfn`GnhKB@7Uj;9u55~9|REWr#F>cw``U9T;F`g)owJ<4&d$lR9>Hlm2g4#}fLlBzpsa8!ogy z|9lJwB?3_H6Z$wgzrSWl1Lilxe_r~0R_j6LOUBlXOtEGsj|^B!nrevkpNx6QO%^1WQIE)8-!OpKifD77U#ZnLzGs=DZqZrJwx z+1h7(`Tx@2?Amm_yd729ozOUt*qPbWSw#)sj)mpz%=?#0Ux;66@#3)Z#;$J5;C$I{ z*^jrk+}}PNPb;BcowBYabDFM;?~IW&rD0aTC;&?{HxZWL51_)ziZ7!s9e?kRci>3hR#0T5EOg=Pv{wwNT)v{ zMNzXwBrI^?gO+NGCtE&84G$A4XdX{{k+fT1kH+M@y9X^>&q(#wS@CAV14#?$V$Z~q zyVY~hF+X;g>dMDCKjjS478K6}gNJ=|pdC&VBUB;}ts4MxSC(&~W%jPlzq4~-t_Sp^ zwzDI|&`;06oZ@uUkO%4HN9u&t{gJL@#vhZ%C7{8f-CakO0KqVo(M9z|)6a+OFkMN& zHHs~b15Q&uiXMMt(~$g%=-WE~AO~=$=a?t@yM8-vsq|t*U6Xh`o1Zb`9pwUfYy-6q zjB9}uyXBW#%b}1@V5ywkkjcF|nh1%dq%bQfz&TQWhY?TbWG!;A1G2Ieb#lyndf?9W zt=+!uRuKeHE$KC)@z)F#<1;&N$?)X^H5BTD6L#Te0)eQ%zw~XMqLBKFpbr87Bi@I% zS8e?qa44IyWA9dctoPsnCqFP3`1#4Gq^A0U(_okF_r`-UUEYmf*6^-Ab5!IsI?Rs? zAkG=Q0z8HulmEINkUN4?Qc^rYLR7UqUo1Vsy9Q%20Ug>S?Fu<1=j+?VSb-nAwD`wT z_7_#;X#9wzu|5dUax0Vm8WaV$ov|T_b43OXCL#q|qA~p7@}B1}EIPs{nd1fZB$*rh zJ;$b#U%Zs3`K8N-<;*W@WJm_8UVtXw zv>AXe>sT6S_&V0NcxdS_Q=>W`}X0Xo|3PjtI1xz zL9q#QUbug0p7a+HZ?!i1#uZGPib`#FjhK%wlKX*XnbI!3LtRt0`fjBCO4Y%G)rFDm zF$E2U6a3&zj<#N@m}Bh_i|Fi18x7U%YraV;%wE?Pm? zhUv1gdCZE1vHiV{JU)bT2};L=4d>ukG?29B@K~pOeAf+mve|Y_Jea8f6xDL8B)xF| zO&k}2-pz?T^R2#Hr9_j&sF0Ref;MaIyNHsP11CQ1d@6G`W(&(i2UGkrKVB64VEceY z?ck&;1L%G-BhBsrIl26YW`jA7#>pTqE~&>+6gEG=qU=t3d5R^?yg6x4=VOytt+v^B z$hUV7$y^<~Gk44}Y~|X~an1Xrj@Myj`^dugW5wO_{GC{;w*k^2Jn?za(UM-;Qa$s` znMDf8mGOSmKnykM%*J1mImn})tiX47Nt2!3J?ferpeEg`G(5`pu6FJaU?dh9(80EM z7o=zCua|w}JVQNnszX@1oaqn%p@9z3C$~>LyTHJ|V%Lqd1M@ zfa;lC*+wXHE?F$e^sqUrIWS1}e=h-w#kvv}$PSRGM?0pVOAga-!+eRlzGmnXvIXQN0Z7qs}gaU93J=V1%DQ2LSqXpfK$c zFCz1C{P>GQ@7)8UrFiAR^w9NcwV_o97;G(dR6EI4!sgIxIfpmqDiUPV6Kk?6NUl?g z&m>~9wHzr$EFK5)*S#3O0CNKus8ns84AShV@wa?P$-w~D8CIkhbY@<(3%10{Iq;N&t?yLb@xEVCUvnbCbtmumtf z8!@l6f$GRdcM=8UU@*%ndGy*oUEhFCTwMH@_cVX9|0USC!rufKAXD)tFmu`3`609` zdkI#Ts0fg}!A)HZEJ|1fuS)p0*O-@~a`rS4o$FM{0gdu`*}Mvhle}@Q6r@Z|f|uKL z5I%zEw9iN>PP`BQwCV?C1I_8rK|tYADom@hMC|--8@ziL9PN*(ceLh2KvmI>J@hXZ%TOQVp z+a2NUr)|&E^$Cfs^H=JFLFhl! z%B23!AmNIjgp(*RG?vrbDIcJUAPEG+on#FQ6Xrvw-y;brBg1BU>MpX(a47Va#bh&Sp6CV^uAYa<{*q_O805D@zqkcUqY z>|l5bgPpYp0^6n+A@b!RV8gA+6|0Z}7{AM3a@}Q!@cl2j?(_(6{+8=N&;J7Mmo9ty zm@h4S3m<{ne(0_q@^<|E8g%yh&wtccP}RYo#FN1E5$Gg6y$2-(U?u{XCg%A`&!_;X z)lgV!wxXY2WfD*{7saH5ItsqsmMAV~iurSozpu+2>BNm}(9m|p45>f}GN--9 z%V25h%9X(f0Ct-mA1FQ8u{)Z;c_z^91+{>_@G;ru=(l_w^LybD+xlG`d3UPmKdy^+ z6hY7F-L-kb!z)Dqrd`~aSTyL=_qwh>d2*BR!Ch7i*aJuw7L%5f$+(;8vAAgBVQ^#S zn&sRAeAg9|Gy4oUo`w&uUvwJ~D=^GlAn>*WAP9QfVs<|dYmHYOZ!%SUYK`;wwzpV( zq@yV^{q-*xco12@2+5nus_6hC&Ez8-gh@NplxEA=#?`%-0g5PD=C2`xWkKgJeTZxs znOH5!=9@WmFING2GWb;MTNX63pNHF~=|DPKOC|eZRv=5Gi+y<-8=B*C#x)+({rV4) z7ge@|DOe40UZ(Y;Ttfn&`YRO3A(l8MG#+KHZzfmV?n^)YHZFnG&n1~jU;)SJ^kC+gA z*|F0w2dI)4vg}$he-geH>PEwV>&6aw4-=V_duM;*P^JXjQJt|x7qCR2vK7zy4rQ^B=jj&v`TN}*^UhXK2$vq!J+vaDmhj-mq*X%U-zMlofLs%|pKt_(TsRm7+vcIb3_2pzs_3MNft!@` zGtv(LAPFNk75rdps4BE}V9CK%;xOO63&ehK%J&YQjPjU#?TfoQF4ByA)D>=<;WL(w zy28#CV`syoTh{LKkiK*hvigq(qIR|hC>shR6-?hdUUuh1ZCU*Tks;$uNO4RIj*jp3 zM^y9rn{!gnUx9Af?^NXj3bAU59E+rCN;V zz`E#k2$}~Vab?XUY8ma2{>HxZ&QYdxX~$XoJ0K_kqwacQULvmzLHE$ha$MwvYpJXf ze|$=5rxp904s^W-vDy~Jd17;}nwB^dF}v7Lj_bkHMF~^^3?ziIR4>O@uBCM=tN~jHtUX+Qn7$$AGDBP0!&l#fnJt0XGCMq0lMuKoBmnpdz(}>5{LDGCOE{Y# zHMe(sLrv*HP`w?u#-Fx5}6%6i-QgaRTSJt#ga zYInqk+XGAkW1|~T%_4uGbfzL+eVvBnen*NO@rBG3+&)wkAVDQaVgO^LPI8J#0 zied(stakJfOXwedYTrS-_nNr@2iCw8YVSL#p$&D&$xfG>Chk~ZHHgL5y!iYGj3mLx zI)TkcLMly%IMM@pj;F;JBr}+Lawm>CgW`rdbY|^n;yDVR($o52NMl6GEcgS|>V4NS zg&Z{jRPGgsH(Yx5fiwr`Lh6je+;bi?u1i}yrU|kpQM^v+-M$hvcgwx4-IkJg%4nUN zsBfNY5d7ch9_Hm)P7FV}b^^&AkwSp((M0=xlSEtyVE*=T< zk~y4H^TY9*q_@6P4qdpI2xj+Q0;o{nO9bz=!_<;VFL0UgD ze>tC$8Zd4!RrWNmdpZZD6*#Lf;2oJWxw~E?23lamy!bwEaAR}lI_gpv@YuKt*1oWJ zK}1c#iwt+L+-JwK2{$&ud?y>w2r#}|*&Ut$K?e=;x&!ZuecCF`rg)wr{Qo6{XttPSTYXa5q3MO9HY~#;`I(CvGAN%V#;2j(&duRDpma_Snw{|v^wt%iYVeiKezqf=d- zskD@Etg+$bUvmXfnda6}D07x9x3vfFMMNRufP-2mS-so6e-RPBIAvTR9k1Lvi~q8V zdXmj`kl^-a#bqI<)?^8lteWQ3t(sD|qRobh>X+e7545t6n$?i**$Y?rH#qST8xrD& zeL@DMnwr+$KbF}a@(d@#CmKh~oceEtz=68KbXvCNXiQ!ePzsZf06qJ>BXW~%w<;@* z?in(fnUr-{v#=)0#&2?Wr%EjNks(89FZ7Xt#GoeLQ?j>sT1UCzSjBcH0?`S6zZPxo zwVBNIo&IFwLl5;-cS5gS!f{UCCXqgxdo`9NiK26ego0(qC+7v~Y0~!w(xT3wDBEV) zFS4^OxqqMC^8-r!BB`TlTQF;5Q%~gxRmI4X24(YP(8Nxz3JUja=k3rp&IiTz-x~>0 z-O51c1}3uq-wAh{)tH-R=6_7f3=TBY+$PzN3{PmSPj=|rwA^PSdAxyeb5u%1rX0SQ zhGlxROaYIRdq?n%(J(lmM^k3S{&cRd>O*dBDr6xF0O&@bZp-gxtl69O-LM>bm zR=1iF2lbJb$qS2ZCgr$U#tO*W1Wl4NplOT};t?0O1!k7>@ml|Hncj&?U=maw5BRXQ zw(Yfx^B7FMGv03%RG60d|1}sF_!2K{G?K$_N5(V%h?cnM;-r%$3+M`>S?R^Qjikcy zG+TT?LCk6!K5$EE}4R<%H?zt5isCBTJ7CAFcIFJoFf40CCxh1EOk;i2N*YnL z4WR@RgQF3rw~gBN1Wj9+@$rH4JL{ct)-fuO^o2WScQ{s!08Uro^DJ%hES^=!ulotJ{nr9N%Kw?D!l160qItWs>7W4%wLR;=C)<#D2i}$WfNaMj%DgntK4DwBZZ1{PHf2W_M4p~m9%wyyILGK;N@ zZj-x@62W?a$ISWU=4rQ|X{wA;ao_f2=!-=xxH9&|_UCDkZE(lO8}$MnLi6af@z(|w zFQcTBGp7kGO4sjtGhp!%6gWNYn$uBjx)(|5B#;7M+prT9U4Yghv8=Lz+IGjkP>)wc zIOo0ZAgr7U>dk zv9)0TPsh&T&65T7z8mm`rROo7&S%+m?jvBJ8mm%94_Rm@GJQ6Ha7T<45R?+^R+;Z_ zOnYgvgXtpNQ28d@+`Lf(cz*nT6?R+1I%%0*>-=uGlN;PK4;g=Y^tUd*7%vsJGx<4n z&0v)kDcCc|M0MhAmR_K+!z;NefBn z8}6Yly*&;@^NW>h!*S7UA-BK!s*3?P5tz-g0VBibWWSot9Pa*Yf7REK-Y=4=t%wa` zZrmYazHd%h#&eL0p_9Kbls9M&?LfF?Iety=e7X1{0}RJ!swjb%?iOnP+XVlI=9Ic^G*>s30Ub z>i{xMR+IGjm4_M;ay?5jvWjOhp9!R$24WvY%ltVGLrmu8Zq6Sq($SKY^D(12J@#*qk#aB_cf^+cHPOe6y0@3gpt8aTG@d$i5;%HKA4bVU+COG2d?w8ECi~sOTU3J z=v>P^I}}ssm*v$^chC|7`%ms^Lvos_Zy=@*i2GTIp=Agw@`cb0hWq8eG~F}Q@Bs)9 zq&Vn3OkH_>`Ej7)0NVZMg3plehniCV-}BqEH=oaMT}bc35$vbuZjk-AT?pKD;Cn!| zfSLfNZXg(gQ+BRFCY3pLDksJVB7H+cO1@yn3tx*UQ%}9?*VHk$&6Bwf@6m0}+VOcr zcPwgEIb`t;9kn^Rmp0$9g+j-PnoQ+u;F&+$m~uKs1G`t+IhgiWaFl>46?;26kd$}9 zVTqa>wOUQUp!pJb5GvXIZl@s69D=$)6^i9b4U9h^s4z7_`GZj|CGT#&31R|0F0Y!s zBfV44aQ{<}6}ch~!9ba~yeKu&Z#3ik9M(@Z#{2cjU_(!X`+}JTh8z%oe>68{eQ!&M zy7fyH2qhC}O3vPJV7$3)vP0;^YYg@ZnP$Tci$oQuh@E4|?b_u1x7jBT-{>s1?w&|3 z^QGk_&$RU#`C#JRA)%Zf$}I!_{|0mF1DAVtf78PpYAJ>yNvh_@sSRr-Yb!E-)o!n~ z%rJ<0)m9SEyoGKeF%ukD!L38g%O*q}XM1RtO6@SfRWB%?+2AiRCwkCu&;L1)O8}A@ z%X>CiEq-Kt(e$9nHU<(Oue@96uPyDc1QvfGMlv9dXK@px0jk7iYDm1;ye<|cG+?*| zc4C~4xGD$CzJN$4YQra-_8DK3amN0E&pQ}3<5;~*z^;j z(Jxfq&odS1$x!@zJcWFfc^mw z^X%iG$NHCD?(7i2E(Z*=Gn@pJmXQvylex^v!=VE8!oq^Xbu(2s3oJ zf!+XmC7|#_m!M1Y2XUA|%MYOf;r+9|7Kp~XoVCD7=VpGImR_F$<^`8r2CCHQBwZei zl|VK^mHCfB;!wMl%Cv*(&w+g29F%~jQXMq>LC~Hb7&J@z%e)x-=eQYXSPDMu_18=1 z^95vn+@J9jggZ|Y&$1ra6Jbp1<9R?-voj}L4M?GiPy+#ibDmH+FBjL{%IG!Ae|>6c z1ms!ylLZ)bK!!fUn9hGR4X#Ff8(;3PoK#95%5{-ZTS$SOBLyzD{Yfi#QOa|aNhMA=9`o5BMuXpSKf<`aOK8E}iKKM+ zFl-T0-@qG<*Y4Cc^|UhaqVm)!N?|E##H6dkG;VX7nt^F ztgX{$s7mfYPF?XO2V{TciTjw!E!Xc1e+#64atiE^p?dZjti_%@m{+;o0e%d+eqhrEy$Kk9 zfT150_Sn$G5afw#KxaM<6c=^4j;NAF*Dcc+di=KisuP*{z!8C@vtuX%5>IK01u}4v z!w~cvhh>{vA20W}yqrxtReW>@fUOT2E%82i3}QEKZ_@5@eb|GuRFM0Z$+zlwUDV|A z=_wQc<)1&D1Dz%c8W=bH0)jE$^#E)@mO+}tf+c~lAt__XSqvzF=u7=Ghv)b5H4#;E z;q-{(;D}(TCJz|H-!{W@ocPOT7+3K?1xR#%(Bf=I#v8E6fBd{>m+;?y-m};Lm(l7B zO8)8RJwNHc{Ja-omsU(cfr18-)!f0LP==8Nvi6wrf$e2u1t$lqfd?bd@D#XP0mQ2c zR00Nyaj>1DcR~xf<~j;8%ZzA_$!rvF>&x}NDkXZ5yP`?RHlHUn!U?fduzf3P?!7u8 zD`=90t#>K}PaFFIsIiX<Pp)qGWT*b0H==w!%|Sp&O$$Zgr{oE{5s#j z_R0TkoSL_Y`~oa=mCXC{zj*o2d=kCm>E{#Tz7O;b5Rg=~m3JKk@j zy(L;KNMNR8HBBqn%^lsQ=vnHd=O`hBI`WS}{nztP#e16%1t9SK94>B~YTTRzqxA@Y zXiQh3*Zxgv+RsB*_OCANPn3|KS;+sI#a5jM$kV5wP=QSj_l%UZPJ3CP??S0^T4CfY zXqAm?+g87uf!_7=pLm1d{(E#P=oIlVpvQb#o&Jh3s04ry>wY*ZfMVNj42B_p#n%ms9rLgVPxPpJsszQ0(KvPZt-G7xddc%dTZl%U3<>V9h8k&73Vw;p`4&HM;Zt z0u$fqCQr^dauADP+fZ=IOgJtGNTwds{A_OEFXs@A1#Y9j1PiD<*ya+0q&XNc1D#;5 zxmHPekI9`ar{7$T>SR3YT3T6o%LoC#fUb0R*~Cw$TRaXK$V8Banowz)Rmx_NVZ)OmH@KiI)`vyH=aXK7GC za zv-1OAdS%JjnMX!z2(p?Q#sK~E4xpvBk#@D@iXr!T4tnTzJ$8?dP6m^TP6}FP2LRjd zdz*e@G0Z3fF=>#m(F2j!+NAD`qMR~Ir1DQXkd-|{X~Z*bf;z|YK7lV@1oN?(#%mwW z%aQQAX>w?ir6Klw%bDGe{=?+Xbhu$v`B-oUH!%3-$l0FO;cbKadkS`Y8SU0rUO-|S z4VRDqTZ;G5na?bH1bNLVPVMX0SLqOr9^3!i^8~0ldtnEiRtC2a_|J7$WU$(4HX0! z>uqXE>3}5iX^7Pud=C8sCjFWQ23XG^9AK)|dOaKY^LUr4g5*dh@t-?U{60o=QJ=lZ zEe^TQ(re-R^&h)Nyy)1&M@^yZM+a4G3$=PETyk*k<989TnX|!J4q`tVbQ|C)inJA`HuKJQ@yW_3g-a~+ zt?vHJmo`~9|LAHF!^skjhKAnMWC0T}2fKm3%+a|1p77qz%y)%%QTn`>D>;MMIC?(IOy45864jiSX}9nc{@x9z7Q0AKlMO$-77! zG0a{??XhdY4gXLk`(*qP*uD?h&!By)LvKU#WRDC_&2p)6`@Bg`z|;2VwV_dm>(2n* z?BdH0osPW&FWaqwEr?ge&rH=WUgx_Nm4q~MG)sCg`<2aFpVMn_?HOEt`S3y>Y3XXU z^SkhopgiDt-nukfXfO?Js9gasYP;JNK?XQ2MntboLZe7JWURBV(foYw;RB&U2yJ2j zo!YVu-B}&fI(Z_Ri|dld6VeQ#Aep8P1|5kMY`i6PoQvlxO9y_up|^ODdk?ZNYcWR5 zHC%V=yY*wWia$TSBn5nzfGk1;jl&~x0&I;cOnVB%YrZCe+lI8?qn{<=_LwY2GhFe~ zFp|U7g~{;jS3`bkETndg2Hcv=Jy8bvfx}E<1GxxxQ+jCstDH;kdowF3EcTE|yX6cF zG^^lr)}YFwM$ivU&1_Aeh2%Y`2LoTHsibMwM#!T$w7sDLZ3a$9lm&pWyqI~$E?ss^ zof!=q9-c5Mt22}=Qga7xbKXuw{2H;&)LKnAw2>KDapkjsxk6v)&=Q_@{Ry|0eNMG$ zoW`fFRX4~7J{I)~1nbu0IX81%zn(t!Z49Kp|1o2)+n!ihN}lh?C2L(uU14bZH0rg_ z*!n2_1Mo!@Wc~SUV3%G4~HyW0^AQW5&hIE?(p7=vG1q!ebs`eELaN6w$%ADp6Rsv&gk`i8?mo2V!eM~ zvQr^{bN}^h2gm#`BpEyye`JS$u!p&9x-_%NCQEI6l8s`mKp0i+I0<# zxLfUV^q$j(k2kBMd&Kau_PBke;FRsF8?AX^UsEb=8i#0>#t>tx16W93)W|=!XKj|d ztk9=R&o33dKQ(+s4(fLTM7d^zJzn3Ox+K2z{f_HuL46>$ zs5aL`7zq%S)62z0KdxvP;BG6$W#ek~xw=bZpH+~CBGX2m(BbXx@4Gdnup3-n{>Rt0 ztUG*-I4B$IZEmVfKed{50xLIt3?jEVA(k^^cN?tepG_sqd;R(&+)@uLSn>?TpQrye z!nWitptFg0+a6lA-CN5KW2M$Ey7`*|E^SnlWxAl`m?qtwF1m>2%a=Xho&3(EYO4BW z-}f0@sgFycVEe~na|P4;$dtu2@D(x8WXh^Puydno?5K26Xz7sHu39fS$5x^mTyJ;U5X#Y-Gb?o?fC>!KyawNA0h;Pui=s5p(%Say4LQaGpXdw7Cwiv- zS6vUdJB-qF$Xg^XSYx{0A|f+x81ssExjXmwe*FIMI32TneCEAe@9TQKo)=MySH-ylTT)=>?wvd- zvVXk?!|K?$`u1E;j*D?ovs+MDX>O|$<&@O1=n(dzb*e}Hhrc*5KEi$KNMk0G(FWkC zOwD^04sOO07fgQecQ;k3iLu{qc_p5jt_{JcT$TG;7IkOAIy%fPb$UUS`f!rGq6&~u z@_`1?jOB)PLsJ_B5~!$+QpewAvF@vie>%qNFvCk0W|_V&y5K7My2qza9lYB5@~)g- zMA^J=smalnNu8)OGy~u-h#dYWmi=X;F$dF!23l>9lS8X(xj=TdKB3ryPW{3>?ywfH zTfjUadj@w2JpLkwkv?UnEB3b>n~w3VNByT4{>6h)uYhIX6}Js%Gy%KC*Zk6J74?<$ ztgw#;bHG6ft{Kj){Pr8@%Jxg!TnYfUh;P%(6gNf}W`Ea3=>-@(XXpn6XIL9wXX!o% z`E*8vZ#2HJpRZe_is|ZEFr}c4rtdLz?3aBlYgnteU7YT;7s-3dD}ay{ftw^bdFaV6 z0r5sa#cLvE837j1O@TGfo6oJmzP84I;5GOC#wxBr#C-E)*QMXG-*t4d0I7gAG63uq z#JtBwdlK7rC#Mp|w;}m9bD91dKq|=6s%oDX{AJl!4p#izW>DaP`eeUcEjk1+@`33@ zX9EG`+WFIu2{$NG;7&&__W$LcTTjQq?az6-;-4k_6{z4rMg$oO6m>A3gUkieI@2qA z9fN$b1RY((rVg8R{f>ppeZ@D+gib$!lmqA>pTf(`8&EuFvhWTw+jpwC-=XXjL^zd! z+idMDV%Fivtcn|FgMxw%fwN+ii#{N@(6kr#@hPaY@>4~;RbzWwl9W5TZ~04Zsfd3m zPTG8SZ3`bGAtfQ$`MaPUwc5#2A2sO1|2TaH$IEvz) zI5jHsVE7_3qF(R96d5&;#*;#`>TqDJJdX^AjJ=Sd1X)mJJVURraOM$?{@&n{l7(IC zDe#eD2MG4Z3|uE#4{qI`mNmN~-Wz9RH{jFVJrK|UvOtrl6QlqNH9RpcHw?5#xo4;A4$7aerE~HK@ionz);k;xc#(m&nY8x;sYeW znil2dgN;yNfkA>smaV;OI_UIkWp0H&s=!z_FL~Kme}Ko1=u|M=0JZ~URUyN)>X@9! zigRj31;$-;e~p7)e_|qlV``AA-M`ngIXoFi92~@m<&6OgD@z@B#~!wOXtx|$Gp ztQoLF-c(uLkJ5roS}r8l5bAQTe5c>nCDK0ZXgFuy^%m=Vd_*n(^m>;x@w20&DQ;<_ z0ed-9-qpCqe8MlscZEKDL>douJ4}tcr1o%p-_u%q&(SB*G_qTq#ZX+SN^E|;&b)63 z!BHTYjePN^(YppJY7&W33q!TGjy9sxQ%<&%2I?0E z4(qtO-W!Uz$0EO?5ph&p=)hYjG@qoJWJbSi_)4r_+Cq`S&cBdej8UxB>6+qTclzN= z4aATPr?&Qc;G#7#%>1FXtS2^k&Q-H#P#5HBH<$c3(041+bye(orjk)9aESK{aZ7;j zsAGqtBt4(GPhEZe)B?Wh9?K)j*4zGPm6h$PAF>%>eJL$@d3n7`q`5neLEdx2)udr| z`3#3%-?#GAoUdaT$MQw~{sQNZcYQ~srPP3PrDJ3i{HFSnXuAU5C7)0_Q!76qbsO)* zJ&lv!$0pvRv1oI>EScnWFD%UJ^FC1m%NHw;7%k?q%uim*w9#WSsyf@8THJD33D>uG zC1&;~%dZS6Hbm}7xf`X;gsnPhteK+n>;NH6z2-f!%slQ!RbNyPhB7Ph>{9HC#=>{w z>>nbbaMN&D+Ea52ixWcr4mEL&5|gd3bsN+k$Nk7&%%?5q%dIpM4{gW3`<6?ZqNnMK zi~DisV|ktnA7r8V5wDN1df2T6#wmxJ2$a2KxSw6}v^EUJ%Y{E0a@q1d%UQs1+J`aQ zK_BF`p>Rgo&*`VZ%J=))Af-`42j;}2pru7;nB8c0%8Z(>X(PGW*>k})`i54ReffI&1XoJx{Tx0=5xMi-bJ|5fF{{)y{SyXUmytKFT!K z@04xR{#2G|A6TdfX3I^46&4=t)|VfIYD$#jX4wB;XN0s4NYS`z+&4P4lTO6!IhPRL z!~btfkL0OEJxTOqn#<}O{HSd`Y3l&^1pA30GU0rG(fxC1`j6U>uHvqR# z3$cT+z-+rsg&Ge`z&UT<8Ay167+p&co;EZS#8DqV!**G=&nnkbW9+0ixQL1E!H1yteW~vTPQjd+$@1g) zfYp^Zs)^gjt!sk!?E1**qoBwJMqN7|5zF^|j%{~+95ywTr7GRKsjBGSyxH+lb?eSg zl=6>c*i>0v%%UWnOo(T0WZJVeTYSw+i)NQ|(5U~lU973t1kP?#$qT&Mk=syldiPNy z!3P`o588ZwekgipwS;0=*fW|!SMisyXn{Vvi-u1UVU1ykW$|K7NHw~#b3FUE-D=JK zA#rX!o$C3I;Y@c9JKqZobo-IdE+PX#&Oy7aNK|*zN7JkTEU>KK7aJpwE1Kn*tm8;d z(@r@o6YQ|r2d~b!qiAG!_-IXa^@Hs4+X-+_kO%}xhLMw8?UObYiYyx12c!?)IHyXd zCke^lz-ZYp1HsLt8T8_HjK+}61`o)$RbI@=Q63=sfWJqT4i%0dUm&vef7a#DL)%kp zSlPVrb+oBJm@ccyM}Lj2!3isYp1X0@>>pPK-r&iPM#qS^hvGSsP%&#QH8Al{Y^qPqy%2pYJwn$8qKxaFn!lGuo%?0M-m!T<|03u~=#;Djrh=$~`FwNSWRCFt63O|8~8T6Ijl^21b^CUL_ z^pJ~$j{?_{m+t06mIKu z@q<_h5le_wK^_u|5OSfw0=_x=#r*Abz*E#euQEorCqUmgT{f`%rJ0h@$$|Zlo z8XZl0VZM$|gn6SI3vo{PSUu#ImZ$Js1pIS=7r_Du*`G<)0&n_onDZb2hQGfXKl)A} z8Wc}3#IwMbNRZwJG1g@9;RH~+K`Cf6=JYo(e-w_zY+^KqcR&&^F?*D+&$oUytmpec zVcohS*SbvbzDkHWmAuxUhqW+P697(dYOp$0#N*rOU2$GJ#|VSZ9mjsw15rLX-ZNK z1;?5pjug%}fum7yv@#~2 zo$b4r@$m8EH|IAO_|d3xG&%33x4wiVp0vE@%?_^kvW3Tx#ZB@U@KnyVYS7Zw2G>gT z>lZ?Lwmt7HcYTVl{~I=VLNAW0+tSiP8sVXJ`@(ijG!So5A{{Zu*32KN#s*ywYDyG#@a$E!0efv zz8QF7rKbPG)cZlNSF>CdGnNm1uPw$?8{g`fYE=`}nMz~VYPGWKNRb&Jm3@^@0fCLW z0rt(`STe1I){y}3obM5dR0*KZ4gU6d5&P~3ZG6Qu!k1Wzn`D>ACGo5b(vxy#?%UJzK?bh*&<%qgpdS^{p$|4o7KI!znRyC{`=O6KWHbW%K_k8Y)=@JrBi!@n+FrtN`bCMS=C~V*E7kM7SA|08T zUPIC}WdO964*OzstX3@|OiomTn}ry$arc-1NHK!CGN#<~CO@AV@#kL9$d22vOwQ#Y zl^yGvC~(I zb^8m?gb`|J_TdUnIv`>xRuK+M4wDQ`EiP6uH2s+qOP%e4i2tS2&T=vC^pKvZRC*h02Au@K)zJsp;7yGdw!ewp0zivRtJ*6lo~JLY+=3 zm$==VnDGNL^Ih-RfhmF38_#_KnoN0A@A5n`thKdg{vyxq<8rAuPvT~A_^w(Tec%u= z$5qzYPPlth)d%qse|&j_(s~;Pnnvb!oln2qWlwy!(Q;7E@0+p88t@ePVWn+zud5AW^ApM-KaDnURYWZwyN(_!Z>hu_Lk>b!W53 zoZ&PBUT6y}%gKSgS#nF2@#=SMp1=BjdJ3xt+}I3!!@hPB%hh4SRdAv!LPfSfc!wFN zCGVG$=o}V%Vlmv4Uuq{W`ibZrk(|6`PtjEA1|q@4tcjeK3*Xf&S;+*eo=|@0;@L_< zY#cbYC5o^UBX75}HoU-b{A`sLgcWu}tQm40`10It*8AmOj9tfH9nJP*Q*7{{?G)`r zf^_pDPsQIi2SZw8S;ir=u;7h_{N|->Y>7UP?+-D>{AV2c7DTEC@Ez_HdrzNVX^7v} zHydprPD`7j!xkBa!Iwx5{9CBfxJK{8_8&Nh@v@l=VtQr^v3tBMg=EA0@gL7Jj8Mjxgi2KQ^PfJ*MWCNyF?%#?!V+<;hTYK zMPutDQCIv+N88h4El}zYe>3TsDJ^Q>=*XzH_(u!lZkVSECO1 zxaB~$4=!?aw$ul_8>xqR%4Z0G{J5S?i#!<%oZg4GV?{=p+DUfN!MODl>-qYBQ_EL5gid5%}v z+qh6-cvI8#<^I~EY9Wtx9r}A_H{?XLl{$#5=6ru{IDG4Sb*$w`e%Hmtj_j;<9Hk$( z$lO~T)(J=|vnFSHN{OXXw~FfQn%CbiMsh2|Icx@YE{0OHr?4ewICO?u2jZ& z>>c#5OUSKm_A8}ucV>q!q}MwAHeiXsDrl9voyBssod5Z1O~fh%6(~>vpfC1>K*Cr7 z9G<4Pn26|(e{d=Oxe!N_Q89Ml3+YerK6_WS{lIrQDqXJOx{cTHJ;%Y9uLIvEvaUJ` zad(cy`ZYz0^??WkyZDP?a$>N!`S=NxZtF37Pvyjy{OIr` z{=PhHnE?mb55C99KC4kW^CiKczz`_-tLgiDL1J>oC-`& zIcr~r&g`J|r|!Rl)NIyQXSYkU8VzxqJuKfPaDVHTW_&ZIzxY*8a_LL6(%)y1lW)Il zYl|1^}P5ZV@3}UT;Mr^j)itC$h}<2fxwTh~REn(!R5oZhe;l_eV6s9q z-NkfOX0JP@dWI5Sp=7DG?zGVCv2?vc0kT3<#fl4K1o1TUPCg$141%NSalTYpWI zOQTEwLYY`p=yi^<5$>lI{g}*UGa(_N`cxU5NWG4`>NEaxjEC*f_GO$Zo$c+FpVYN4 zkG{|3I3>J=5{(Pv;L#WdUWpqN&RaCOqGQ^n-fXi1%jJ$D3P!}G=yF*r@ywew^{?!= z@O?%Y3Kl;_OP!qD>p;_d3YER{nv6wlu=tkaiVFqKkp^OkA8_wq+xVQ0Y4(@)KXA2z zWwJO_UgRogbnS@EbHd%aYg!)T+`7kzMRM0$OXqx2R;=Y}SsyRs* z06!}QGKBBtGNoM_EWR(t%Rm1%NWOD`Lp`J%@6Oo0qtb@ivCV3dFvk~pM%s8;`je9R z0@z3+Rl8OWNxP?$6?@{bn+|Do?YS6q@WhpEF8OV(h>eblgj3m$y0Cx3kfJbOkojTN zbGzKIya~q@;YGP?gkX$th-zQJBO5(@g&Tr~o=SBeH=q1_M(=1L8!^o;j^)TNj^UEv zd=~8(Wt!eaT>0YUG*40Xd#c6Cf=5}SHqM(jo@`-mfN`?wvbP$oa5xYW0wCNTtHBH0 zp(!obL|>#$?ZhkYb`|S;rk&+$S%i}BHA>d%-QlFh&t6p|7^!U1r^hF_j@zU~Nt>$3$fVKn zOW-G{hnw0NyJIke4Xq~oDehY(KYC`Lo_;mKJ)N!c6#^?xxP$~vV`&XP2Szm65mp(( z2=7Z9_*FDB9c1Y*QK#j76U5~)Y5=REf7$Y*@jjeLnK_vI*Kpetg?yo<5?kuo=;MNb zJ$tl*4rBSKt|Jeuk1LrEer4mE+N8bI(<03q%!lk^%uuOKqqDXOMfzY#)>u;^+8=F6vA}NAgx_Rr+Ky*0cI<{Kd&uF86^o{= z-!4Ly1IY9%GsYW&-WqS?o{eJSak75-j<2R3&fvQDEEZaZR#hS|P0Ori*e4o(h zOux7Njy{|};Ydo#+rOA~1K_uf*_e>Z0_-A=sJhvdW#m0<8SQZY^p7%rgZ1$ZA&>=S z4h=B)8Oi`T>*5EzHWaH-uYj#q@uS0yrIO?52Ws!W-qg~nrM9q6CgM)~$1W|Gs?qSU zf!sdhI1M7x&r+NuhzcBx=1|v+ukr+tRSG&bhVQv1krC*BnoQ~ zl?+hKOVYVp_}>5`8`Wn3r6Y(LszR&0l=I+i6F{_&3=iqJ28HRU4h+oCcQ@wfy0Yjs*lSg4_YjQ$FjC z+7h5)%PG2+mnSyqtL|nRdhTA{>&mZivFkHS?n>ii|^^Q@b=qf}dGOkei0)|kl*EI5zJ`C>sX(UL<bO^k5;W z?nE>lDv1RtJ^`J~=5eu0hoiRlOgKCj{@?p(v#53T&g$1DIPmO(`_{_}52y~%WXmRZ zjmr+YlodW>Y|oydh$ujnDma^E7GiEaAY#KwyJYmu?uY9^F`@y1)i?m7O1k$QP)TWP zZH?uaz*@}Tk~y=GSpWI+KzQD}s)=@TJ@$Y~i4E=8wF1T&yI7EwebRSVh0r`7Oboyp z-l~3HXM{-8E)H!`EN$(odnNz|+QB<47s7xv9!KDef%N13&J@U# zyuuV{2&a&EKH3*tc~0oVF4jwL_FZpp@c4SjcYnunHu||7OG|i~$6g!e9XBXBW1D6iu^O=spm0B9+Ix#UKL(d zT72PMA_h(Vn|vX8mEcU&ahGv1au8FhXJhRH=Fl62l`sSuH#%8sgIs*IC4$0573`X1 z)kOGU(t+#8W@cr*&m-oY+*i9y7+R1NJZd|92hPT8CkeN{#X@g#K80&v6gce?>6$tL z2EU!)-^=L=c{Ob0wL_hyS8RPw=8TTU679<{8bE`{X)3NQN6!LC09Y(_ z%09xaG2iS6S|i9ipE+JXgJFIdRq5dX>j+>~JBp-u+}7%7bDk3WK!jD>YcC7DM`!vY zP{e~+4Kks7bGJ1@5*PlZ1uJGa2TwlKaeuiz<3Kx*jd4qAvOElVMPL(Ud7ZlibjK|_ ziViNmJXT6Pr=NvTu3(3c$k;+k3juc}{#;PN+=faN*Vm&uA1VgZ@g8=Sk*P!8_JNm~ z5QnpIv${1{YZXVwJ3&77Rgd?1;30)Py8X+49w*F)A{?2!LUwR-jMP5e@=P3vHE_a; z)X3RK^qOcxs4-&x9Xnuh{OT`MAn=3^E7Z0<%LkMz2+twKd~2-c3V?sUzr7-Q)T#j1 zT9klYlhwZo1g6#YqGJ&KCq_t}ms<&@hy3e`FR!Q{CH;oSUPU^TGY<&(M-~fM#FxW6 z&};k?QjyQjBavJEt?Oobp`T#wMDAB^ahC%b@3w5nJRg)SJ#GJ>wB zY?3*ynfBKA0#`&IgiVL@|vMOxZ&SMI~|?UK$+(PRGu!YMTuHgIw{jv%}CVEykXYrin7+{ zKf204glvn=44`5~R#F;aPXA&&vy$?ae7w?amB(|W5>MhKZXeZ7xzthiynB|Sl3_Nb zVd|t?6{w*W^EIoeh#9cVkn+36JSTXl#o?y`^qN898|1_?eI&1QFLd&V2WWO7gPWMt zEqGol)pNLOSd7J26HC5lP4AWyvo0#xuxmhkDB?lTwgi{b-{9JWuFyxqR;P^jZ{4;a zc=S#Gk6zSdVJi#q`q#_0b7Kn$(V~`H$*gH9=fq@Q8W5(BswVCZ1NL+EY1{o#4Y65& znI%^+jYNrxU1GOuWSKf{{@wRNefXu7`l2|H_^>vl=4P;+7=-@bu{V4&J0=z=RFYZFWT zVr}TX#gk&K+O8B#)UI7n%HcA&AY!$qsX4I{cv2K3!*SM!1Yt};9}f%FwXm3XB!!7| z!zwvt{9|q@dJx30T96Ka)IjPV=JL)i>Q!x$wqHi`d78j5OGvN={7?NC5P&dv_!2Gb4}g@+esr0PGdK#QdX8r7s&j&!A#+T=Ik=SvFWn>dEOac}p^f~k=0l7Qga zN1CjuCci;M8dCk5AFUjLwFK;SH87`+KceAzUgTi1`6f!?Ux)ZtFR~oUNUr9O@N!dN zDFr0GHX`v3Zbr=AKly^02nr2r*G9)xxGvX#M)E3`R0z2b(CngvP7Yw+O1!=L3;qK@ zPu=mW$P;G&muUZlEVDuj{ z91uppT{qEHBR*H!%K9)Q>aCLUwrceKGDn}SfO;QNJ~8zoxiA>mDND(^GMy%+PSXE2(y z=rqtZ-IEJ~l|%H_>p^K!M4t`L@OvMdrU3`mG1E55tm)>^9Vs7|Tk88-US`n`^B6On zb!@y%J+tfexS&oKGtEh-s~xyvDR0wo4j{t|&bKcrMby*;7pah5wxxl&3s9wa{_^D! zj~HbTM#kpALk}Z&S$%O(eqt|HV_==~fTALaF9DN4vk(IloZQ@}s&uCpu8!BJ1KAJ#*qgiON9&o2?V+(l*q0!m znTCc25{0L6vY^-T8DD;WlxxoNyhPGRv+z?#pjzA=M{vnGmE7cIK3*~NAHR95T&8i( zaW3T7F)Xo416{vu=a+(<;ttLMTgNRe2&4#Li%*LqFH9*Wk{A*z>5+QSl+PV#2GwRAGglYo0M>}>;zq{c zi3^^P;ElYj8<>#<`2Mq6m~wsSuOSY;w&7tZMyx$>7T3B@6b{)9^~ZftvlaIdre5aC zx|o?HxoS}d1MRq=#MZD~w%joWpj{(GLt}JWitZ0XJ)_f7dN~Pr)S|2id6KIYm z6o)CQY6H6Fdwl;Y4D+eJ0S#E+VmJYBnLgee3A?)gP`9`@V^|G`Ua z46IU4cWo=1TD9Duy?@6F9O5ViPyU9L@1}PaRIGPx{h0fu^Aspm&wxl>Fdj#qYxDrr z6BylXRMx>uj+gwGQI}xHJAR~IwexlFis_rOA%)b&t{8F8Fy+xvKMl3a+$mx>ztDc_ z{JQX*oM*6v!Wz?pq~5=}V4S3UWw6qd7E<|EEE1(qdE(%EZ63@n_hsU+a9+ad&BOfT z)%)Ep23BrZ%htPraiTgVw>T0Any`o{f@P}{g3dp9sFBV`jZ6BbjJ3y-UvGp(x@LSe z!)NLZku%b>@ToG0=U?jZnbveLPKl1;lBHL@pw*LOt1@}YSyaLuybmXxccRycKzxi4 zQES>aw6a&;y%C?R0ACWRZ#Soye#qn9;!*~#gAMgl>r*zN)#~_>Uv|IT-z|j*b{3V( zF=Qv5RVTLltJmLP*WI-Vz5gf9fBA@EFSPiVk9hS){KrR(msCO$NIHLVx&cH#E>TRV zPsQEyVPB<5or!a21AWO2_^Dx@Dbl@}G);r~VIrHfygo!n`s_)^*as{ViK6D$GBb_y z$H%V%>tQ*5cUWx|_3wJbzK*jTnFp}-AFS$sne6#gpen8jounq6rO5qGzs_hK%ZGNq z%HD^|_M}lMzO@og%W+^BHvQ1|q+SUe#X{!`f!-JkO#qY_@~`?0YeNJjZUmp&-cl!# zBx!q`6yVa3yN1$0h;E1dwyuz?oz_0rB52*9HNd?Vwj$}Yz!Iwh>Hlx* zP3@{pj;>HfxbPw{W>m_vQQD=B`&e>b67n;_!N%g8SdBi#l0 z0nwcK!>@V^n1rx*17-p(VH-cM8I6D;K|oyh|1fD*Ux%tEz=wB{CYfgP+YT66mj6^o z#{x9acOi@>EVpCb?Y&|r@X}M3D&N)%ti9{>n%VHVQ#n^l8IH|xNW?7J)q>Fqy5B%(mQi;u6CP#gWB6|y^B6i{qA3IV z<^Slg(MogL_zzYwdJZpYbnttnj$CVn14%4ZSZdHaUtW_LW9gNx*1Vu=3(Fs7uGLmJ z5WvCOiMn*aJ^9nukN&PzUq7%+R=~~lOTi5Hm+{=1atj$u(MR!ypfep^aOtnknfuxU zZ<;z!^b7dL=VDhvn&z5T7WTgOxw-h{Sj#})&I4edXP_DwU2IRZiyAKv=FqpZ(-W4u zqE0%()2=XotEP#1t(~a1O39EodW@%icz(_7#>kF3Yfbtn{uv;5r6GhDYOQITAn!#;#pT3J`xN3Fwy_^Q@vSBSJSXK=FopT6P;VXn z8tOL##M27e)~)~tvK}jf5;7-dd9K0PHd}t&2gtN2fe#t(Jlb8*YiqJ5g;ofJ)F8G|! z5X;c+g&TubWow<8X5>!3cRMTbCqyH^xk>H)PcCxh0aJ^b^dk>0m?~Pa2^}cuLI#HktT6skN?}nM}3Z&JBcAQ6?={aHOK-L7# zNOHBPJ^-x@&{=`bUs~hWZU>^v>VM4qFqFA|nfVFmf!~`RL*M_|^Z>$y-p5-If^+2N zhY?9U58!B&R0dHv%z4OTZUw;|-FlCBm@oe{zJcHT<;?Q0#y9XQ^a;O5|F5_6I0l!9 z5MUF9uKB*V3{M|0e?wS#`Q+*0v<%bIwaz#gxlTwUtmZpT56;fOBGAXIr+$Bl{msEc zeC=-=m#7id2z&(xw$)YSc16D-AQ~%fV-A{{p%#CT4Ep_XPOB!80-#9&+6>eM`Ul{; zL=qMFD<7Tg>%Sm+SrW6rUAZ=^J|GY~DAj@%Il8hNuc5pQY?2pvTP_@Uuz?Tm%;~|U zQAm;dAS~#3l?0n2IQL~I0u6#c z+z_wAIF2=aJ1i}c7<%f|=;qan2#$t!GqKPp4(#)!zVC0-+vtx`T>qwZT01i7^ffe` z(Ypgv0ZEa#Iud|<9%LZcwweK@=b0waO#K`RLUGq6E@NQ2BV`e^iUWi?ht~#Ok1V?- z7qBros(!fklgSH`){2f|KAw0=9k4r=``{9xPt6zi--%D?1qr$fkb$1#l!80Jw6aezgaL7lFaR~FmZ(wv+1rpVXr9R$ZD}rn(eDd+%kJOSmX--W| zDEHr}w_6B&Km)=XP>COEEJ;*pczF<0AN1RN!nfvQhh+0Ee3-v|e}QQjXahQ#CaEBn z@;ObUhe6>p_?SE&AufP-V)8p=0xH*OUt_c9QTY;xX)`3?`Z(AMoNYH>vJ9KbBjSn`#77ZDQ?F0Ib*+^@?ESB55n!aIt>Jjc8 zuhBRS69+Ocimszah%X}hVFqk;yc)UE;39@@IzwLt!w<>&hF{O({q|Yk(F(fxqHO0G zy~h7MsUH5ubIQs>Qc}@9u&;^C-rYR%Y_;W7TX)>1|KsP!;R{M7Ij7p2r&R0=^@JhAhcGj%edExgt}dh4bq(v5 zRJSTBt^PosE#0Fv7Ah^h$ZpRZJT=~uM3LOu&=8jV>C-1)3)s$*uS3H8$`>oH#ajxo zMbPh^oA&=jXK&3yPmt^YGYv`W#Q!P>0)Dj-IY%9_rLx+K3bzeGijA7?13&Vv^o|NX*imNzY2Y@A(foWN)N-!HhSrL6FmilX9Q8ZJ&& yHujdRjO&8n#jNPX&eyD@bS+(Oq8DnMRT4gaR9sua;3GUtr*iJn*(`;t0sjxVH%oW` literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1381-image-2.svg b/generated/webwork/images/webwork-1381-image-2.svg new file mode 100644 index 000000000..9afd49e60 --- /dev/null +++ b/generated/webwork/images/webwork-1381-image-2.svg @@ -0,0 +1,303 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1381-image-2.tex b/generated/webwork/images/webwork-1381-image-2.tex new file mode 100755 index 000000000..a5f970e15 --- /dev/null +++ b/generated/webwork/images/webwork-1381-image-2.tex @@ -0,0 +1,154 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{tikz} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} +\begin{axis} + [ + width = 0.47\linewidth, + xmin = -2, + ymin = -8, + xmax = 8, + ymax = 16, + xtick = {}, + ytick = {}, + minor xtick = {-2,-1,...,8}, + minor ytick = {-8,-7,...,16}, + grid=both, + ] + \addplot+[domain = -2:8] {2.25*x+-3}; + \addplot[soliddot] coordinates {(0,-3) (4,6)}; + \addplot[guideline] coordinates {(0,-3) (4,-3)} node[pos=0.5,below] {\(4\)}; + \addplot[guideline] coordinates {(4,-3) (4,6)} node[pos=0.5,right] {\(9\)}; + \addplot[soliddot] coordinates {(0,-3)} node[above left] {\((0,-3)\)}; + \addplot[soliddot] coordinates {(4,6)} node[above left] {\((4,6)\)}; +\end{axis} +\end{tikzpicture} +\end{document} diff --git a/generated/webwork/images/webwork-1382-image-1.pdf b/generated/webwork/images/webwork-1382-image-1.pdf index 8a0ef0e2d7e3052dbef0d6f97c37a63a73332a84..0612db3b43ed4bf7a96813019638fe42c26619ad 100755 GIT binary patch delta 5021 zcmai0XH*m1wv-|engJw)-m3_Kgx;ig!O*(|g7hXJ1T=ymMJXn92_hW~9TJd^6zK>` zuhNuWL@A<=ch`GseRtg-_xzbTXV0%$d+#}T!wqqupsAM5Z7B&EYQeRG-BW7$R}2DZ zDLGkrRaNSTL4inTAL`KAb=$!tdl}%uF|%V=e&SN{t{wy)VyuH;sNE8Y=TT>q?O1xa z<7Jh0aaVeHEJeh<_#!EJ^rT?ts4OslFU^8N%@SDxpPaD^M%`Ejy^&*4Ys%6j%*Gbd zyNmlKF1#Dr)o5BNR<%QtAuPVDKekC?6O*=0*>MGZ_%IK%6Ypi@3Kbr~s zm`96O-fj?;J!WLAKFaAqk6R)Vzp|Dfkl0mKfOQzxAB2kC%g8hqJ_3|NQ<>D#@rU&a z;$tIi&aUNm*L}5buA`%^4{`y0C8yJZPl9INJ9#c8C;p`t?%Bh8CYO5!byl1(sX=+m9EYt7Kr$#=Ga$Xu6?I$yPP}=rLD)L2 zH$>Wi@Vsu3iW3hH9wOv4Xnm^}9*{VNhSU4cPMC8B4m=QlE__3XzH+usf;CqeJd?38 zc`JDtckR+jH7M&+GHWxYjilm{ODMZs(mm@q{r&(kp(OSF94tx{%)}^esaMP(6e@r* zdYy8=#y~m|(&>W6xpFyU96av=xG$Vea1MAyax|z$b?o0b0FkuC@NuLXjAS9o-$=DF zzHd{urZAMz#8X(R=qaPwmhfDTcB;<^+kk9_#(++S;()!V^iUex2Yb=dp)}OhtsFh1 zGWFT;x3EA9w&wNB5ra$-XKB|FwqYnPPU(w%MAz12jh&_<{>jTqP6OQ~9G=mz! zL~_UW&d>S4Z4$pd^5P1qd~HrI2#9V(XE9(s(#wh1Nw@t`B0Txx(1Bik^n8uV%>IEqh!xk5t5HNcNH=Ec2O z(_PxaieUnw^b5GX$K|8HK?e+t(Zr9fPDjUy6tJhI6wfgz%3iq4^EVn7B}-9Gla~4X zoK?70`M3to3LTC!Rmu#8ado}0-A|-b;q;k#e4x$zGB3Ju{We9h5rZh{BB0F|>}_F2 zRW7Tf*+|@h^4JgR>3=e%ZPFTn9X`6v_957V42mkqT!Bhp)&geEc218?+8QQ>Lki9y*fTy?JR>Cn0- zltL{Qa)$W&liL(fU7`iL-W=>4-i+!jOzF-#s!HDeSP)+Bq`IpjaGf{g7lS@`q(9IV zDn!5w7GJ|Im-bU=B|rf3!NlYK4=R~yd7?g6{)S9%=}iotUU%ssl%(@vP@A^Jis$^x zg`wu;M?pH87iYM0b-m$uk0scR?xts&3l)!=qPrxSa#Ta8$a>L)G6EgxlOZcza@=e) z7i)ZMK6U;Sh5q#Te01s3;ocoXs41|b0V`xB*(~_2u=|w!)rL;Ru3NK>ZoTM`|AflL zL%t-_Y{|8=s9;pvw!Pf zt%Ezq$tuZCm)F5=A;c%tNMG0g3&1Pphw!A9l77V?h=xl4GZ`IBJ3T)@Xj?CC1$nc6 zh`-?y3!EOmPP+a6%kEpy*&p{Sb`FYk7^cUW0 z`$tM`BP1mww4m~YYh$ZE%N)O{U&|rG=XG|LnNdDE(>d(9z&80`FPO5gT`${)Rkpkl zsOXvj#5-m(=v&;BG54`!f^9iT-a1!zj2SiRHArt{O)PicU%4x>xb(A_TR=CUN%k}p zTPINt|4rh>-~;-H8UIWFbYyUsqOO$10qK>e|>4p#Z5BOOZhkN!Y}PVXAm>y8ipLzor#Q< zb(-jDzu;ic4XI_*tP#>p@~4iVR;g3gko^I|>CJE_$G20(u?rRM@RF>Yiyji=YdS~D z{m6Ix9?VI1^?r2y(35vjIH(;0u$}DhSKqD2{btBL&~%k1DASBs=4eOQA4++6@05bf z)HDYBFM3RDjSbk0J)R}+Dr|o>S410K1lVC?0J>;d8JT}}3iGI!o>>66@9^)8rBVg* zu5GIbQPW(cA`I4DALm^Safl|0un`EW5=}TMlV)?-G}BF&K?rUL!B%tcHpI42Pp

    -#|H?b1smc#ygSZv^vu-M$pwjG*rl7P*Hv>vvb5W%QcvS&g)U`CYYtj#rWl5B#<{R{%cS! z?Rws@p1>V)#NzZJv4mfAG)Vr2lmU{7kY?+*6p;n8KrL5v9Z(nKkXe*~=f`v@g&^b2OlyN0#AfpU1oqOA3WUk(8ldbpq z^7*R`eO%KnUik6yN;+1QWMS_8Sg3xW3R`~a! zl%QuCd}8r>Q!8o4e@S)>7+ztSX%x|rfA*+@sk-ko2(aaXIo!zH0!;72&s&%pH$XWn zl0pd%f14iV!}@UfR}aPd*t6Y*lOcCEEXE-J} zKRN{Q69q)QeB00BQX(3Nqh81GNg2cM+pQJjaw*e(DIqmf=$XYj!#X1_-QsYvu)EhV zbLH}3#Eunle~)4)I&6Y^dEf6j zrZg+`!(Xj3ESy#wdkO-g_YIi>IQrV<9ylmk>cCAGCHUaA3cHbD#?(|c@u2jg!eH-` z*T;mg+>xd;7q35GS3j4%Wsu;4!i0K5l5)wCu*p6 zh(H=2?T589IatOEgOQE0FELr=(-^OpKQAAj_c2Ry>3#(m8NG@_YvqKR8+hAe|6uju zK#g_`;&&LNmpc}rTt^eX3&9}3i{GxLZMDFQ1_q#Z-uSn+qp!+lVbV}xVI|xUye#me zA?+|*eQLb9Sx&B-h%fOoM$oMya>Luzvod#8oJCq1Z}|>`*h+IFoFi-%J!Je*aLuU+ zZ~hAYs%&G9FfdmDIDSRIVXx60JqNq_NX z?yMvIj?eYbokPNbhtl0su@>W>s;KH#da?sxvE4Av60{Z2-0t!G&1-9v8O)O>LEUM# zx5i8Eu~3JJM}9wkf|9ET_#;gD!^)CZCuA8rYTrX$iuRExO~8aS`-^%k?%0k zqx^ZFJy#Go0#l_vgnbQpcpK=W>0yzb>4q(a!cFhE)l4xiDl-#a`r+fnkxHSHlp88FR4a$3r6-nlO@#W9JFms^u8Un{v-I&?E8uT)cDg2X zXk&P_k83hAw975yIj7G+j9&x7$Rsz;E=eL+e!;Xw|32YU&sGssf`}M?OA{OsiqNM; zA*G#HKcWSwSa$xA!2ch>5UeQ=m_|xQQpY3kVGxgkEVc%~j+T{E_~%GhV14{%A;72Y zzpUnoQ9=iXImB`e&#rH#&f8CeaU$H~%f+vF;u%ZwFHV%$siS?zVaEE^`w?v!oAlfw zR0k^6Pi<(yZK{+J-t1s&32hmw1W!s9787-U7H?oWO-|%4m463w=(A_B8}1rDBpjD$ zsnN(mL&lAFEjgfj;!=z;0M@lEjmxL`?EKImXIYM_)#53BB9hCbC>uNX=jqUz&Z`wg z1J_1}s9kCMAE*c4%Jgy*q{FmU^+avl3MHM!t?yPo!*yy|hB@@sk}i>Fw}KcGnJ&!+ zjK05bQ0*Qv#c$+-9%cOMWzLqd*3j_>j&{?9w7B){-tc3m`Bca zZe6UTlUnrZ_Zy>)`-5T<6Qm^HJ)~RxqSbO-=z4^+ghr`3M)|BP@a($rgz3HK)-G&L z$bJKjSYAihCe~&2#!+zOUXS zd{i9XD)*sP)%BVwT@_eWBdcTY$YHwQr39?2Agg`-$vsrA*A)A!`&141Frwzcq_YTz zOMwz%k6mgLoJq#%yP9HlsYdi5%bX+sX_l_g57o~dX9*+UE6r}-*pO*dI1`=+tFg)S z&P2|icr(u5m6{Cw_)RHf@~_(zSRvzY$l58p`U;M5tzO?>(=oYG#GSEs7$%uLrgGfz!-KTjX9S1S`RX~@WN@Nf!XG3{NP-ed9bq~J3`c)5kRB_yys zy*#YVow58@P7PFDNrmvc?{x1ms&-a(&!xH24H#X*6vPo;;>&XAywA*lY}$4{lb`1% zGrbF+W%1t(_I*6@69Y=_<{RIb=v!adzIkDtzc_<(b4kOPbn$R{;r;`x{MI&t!SO1$ z`K}6yO^f1G3HOKVpv7Y$1bTkgTK&U5I^gu7*F++qQZs3xflRowZz1U-rCEK+eswrN z%oPGR4?>ovG-%@Z5WIM7?K?C5jV9agRBS=6-y-OD?`c`c21xdkowiCTs~;QcG~g60 zOX_+2!mnih_&dwYhZ)d?-u*>xK$*$Bqp7>}r_1q@e9}C-X2skzh?47hIW`({3%$y; zu&G9`g9oR#hNLk!kE5;gNmfX^na4C|rkSW+Uab$+tOT z;W~Xlp@En7J3R)&!whG7>@?8S{wr@UAnyI$mcdX)SPop!>Bxk4Nn}qiU`Mv-hm>Go zwjYz-D2khvVJz`$#M2;9iZ*ko_)n#+G|VkJ-9K~SqJpqY0@dPoxBQ(_bOHVE-;}Cz&#n=y}N~sk~#o-fZ(39waC1;W7GwilKV@WN-GD3*$K%_?tmWv$j zjtx@2x5)ZViiu1IOgJ{#u{Ht^N##+ zxkTd7PLUeZVSF#m97F(wzk}HA_or--lKK9`bWqH={kdG{IPQHy=}FTuq&fV_o!<$f zcj!u9H@PWQ%f~x?91pnuu^YBg4!6EBoKP!y+8`R&t2eg1wfDZtb{6%1o`-3ZaBcE5 zjtLzg%=cj&%@_Mw7S}oYi&9P596JiuPSl*hAf^^ur?+4tZ|&9%y+GA06rrvcc{9~( zY^~mfUZ#2ZUf|P?8#T$Mn5J6R;1CGG*lT;0s$?ITF0r;Zl_BjKl8@mNf*y2$9=~9R z-ocNycm-6J!0uK*KsCuy1+r}$R&3MLA+&$+%M-Ue&xPU796S*QV0XphWB!!^E4kf6 zy-A7ap#XH<d z0ii5v+2xS)O#q|3x}KkYIb@*UT7QOT??~8F&e4*iBMl=on%&_X=13A0ZHyjO3hJl{ z35VMfxuNFT1LqXmAp+69*Q^0BK&04krT--^(PD{fU(TXt~!9BZ9QLej* zcE%FZE{PWHoYQ&PO%iPhGMZPRT*<+lIdYn5!ZJb_Q^i6gApPp;%h z^S02gr62oWo`chxcOiC))P+HsUu5=-4Aw1&GBkTEi`y(JdsMayEIkKZTeyT0Uj3tqeZP-@bte zD0G)W@iL7Hdlf!3{l345*9zEDRYz$Efj?kwclL}%rYGuItGn=hR>0K|Bqx0&AsdYX zSdo&Bq6On?_l_=`M2POZUcTU^P)eKrWa7J&M?X%Kw3c|5kQUy8dn<+%WRHetB+T36& zF>J_OVw1BmH=DgBsZ_bY{ECK-%1_Z*p9{ly(aA*W#9TyJ1NKoxX_=XupmGiL*`}1n z+eZ*$nU_Eh1|I&-s(+{iWw`!a8EiZ?#=DS7*O31@2bml_6M5SmPIziIPy?p(Q0nuL z<-4ad=Zlb)R^t{uA}n#Lea=H2aQwuzz?(x0%lEgr;+1tJ=sC>I*6M2I{~mLZJkB`* z(CelnIW_Kh72<~Z6ft@KVUnqTnuOvYa9VOE;_-2tsS+d2j@q~oY80`=l`hkro%3Qd z^$Hu0fWIM)t8KGg@bx&kW=MQ{JlM+XyJE& zEKmFo`KUgtYMX4s=~AUzm$P}wK5N(KS3YSORL~Y7|3et-Fs-eDGa;6>i`D-0g!5A;Br}}2c$0~&zrW#9W8o+*tqM0LeQ6Q5TMd$y z-j69&nZp__tHe-xK6j-hdHQlTpX^a6YLnAGye&O|o&=47Zq?Ow<*GknOpscxiJ1$~ zSJEme1Uad*oxa(AM6_K-au0+F;weA`{>3p7AxNQ5hzKYs2#AP?01a(@Spx+&X;m%Z zzhePgc;c_Vhj7$;DujD##CsZ~dm!>X843*o0>TNwU%QX@*RJFLwM>5uk^vzC3L+wg zmtQnDHyt(_MgizSGT2MhrVvD5K#bA#z#Ia_J!+yC0Cx8RQ zxOolg-(?q7<<#})Lm|M5`N`5wD8^BTm=G!TKH{^TZ0~JXj6y^gAfyTv!1;gha zEy*43J_)1S#Dq!!y&TZC5)+^U0}0SA0AsZmxV{1^dEm5gdN?`U30`tXsBi-8fc9jyHaFDQG(mST?#NgU;mAN~sF|^s0mdB({2R$z zsKe)=a98U+ zxOoN_dNtM}c6l-Yu$T+L0b-#D6#_8tZi$2K0HA-w^Pk9ewQ#`V{lCIUi2q-)n=;`N z!3`Kb_s60YRV6}Pu~)#bN2RZ;S=Bj0ma-t9#n8k+-w1h-5KpY}&OQ%mtK>8)w zOJYiv5#B}4Ylgb)-4}*YRV2u%X|sUY`f&`b=X3U-B*s%A2C}n9gH~D&@!JEQ77r&6 zPQqo02|UyRxE!?Nv>b(Eh3xQxx>E|W0&Jl%fsl{+qocmMNi|B@zTkv>i(B`SFL0r+ z`YWt0&j=i~$nd)xO0v!@qPQ}PnZ8TlvD-tbo~V83y=L2`DF z?N4>(*H6c6YeIb3BsyL3J7Tc*eE0;CQ))~d~W+0(GkS;Nqe!An0O~hrT?Sl3KX|D2U zH5!z+M0m;!MA)fBbOcVzF_Gaa(?ThMRVc8=#g*)UG6KSKUb*X(#=Hk%HbqhvSvSJE z*#q8qsX7tU?i%wvdhMJArJk11=8^5HYE6YxL2_6{6sU*DwgSHRhuRlfP!3tFg*{E9 zxRf>Zp6>Zb?TmWMJOc?9@Oc=enFz=fPfiz{zcli97?S*MsM89Ko|)*oF^Jy? zCCDbv_3}sOo{T~(M>3i|@?KLHmM2BxLC|kcV1KEt@F*`Zjlnszyf)p})O)^ZEcKAd z0$EA2XdM}d5AJLwQjfa8rk#*~H}EJ{ZC0nea)w$sg}ZERiSk{f?8!{+V&ZY-Iax)$s|D*7DDoxUbKt%?kft9mFVDCjlY>8<2?}~q%R$TgmGhG9O-gI5meR^w!Xn+ z+~n{lT66=Wz#k+Xk;rEr%RV7yh7i=YBFFW4{X-YxfN|W4k2=YS*~mXu+TPlE2(|d( z-sC-ypwSnt9t$_tfbceK_td|jWPkWI+v&XOos6bm&c@m5+u<7+`uz-pTEd9n{_~1X ztukc*G85K=r4=yYz0uFYL)R&Dc{+)p(UoZVU)AM%5OdyIOYdPyi7?!Q<1xrKZ4{UD zyhn0+Cd}UA@y$t+IC6wrWLw8pUck3h9TN1!vWacgf*hc7QFzMlu5K0(6=9Yxu;v#! z!&!pZ!IsRaqV&>+{&EJ;!sfevyein?T&RDAqH;{oA+>Qc-p#H1c(&#_wqhq9zv}h1 zuI1H$r9o#eFsNYmr{?zt9iWKL#ogDAlZMB)prPMsZ^fq^Rk8~geauqI2f3IG@5aJY zq+w6+_wS>EG-r%T8JEcfFkRK3$OIgPTVY%zz!XYK2rnP_p9#!Zqu1pf9e&ppiRW6G zy$qhS5Dzm|0w(&Z>azHC47Wil7Rt}9O|qGjVstYbzw`aB5r=QwCCFB~u1v!S5rU(; zPGnpJ0n|gF-K3Ubl5-wOW;9AD5cJ5%B3o1$h{w}{-zxt@f9?~6TkmD zqswf8EKeGMagy(k~49C@T&nHWk_S~(|2W3t7tcLO8sCN<#&?lW` zFXX#w|Ji&}V~kx?vhA@fNF_-(^iz98@uDl5YG{6yih`Xle$cG>1w4qMZu2;W+)R5) zj6@4!=fttVj8U#7zY)o4Rs^P|u(G9#n-Ik5h;LlbfUf|D7tPK4X~`Ej?cO4;1`jK~ zc!hVZBtE@3cj!#_X&De|pLfLr`iv+@JUkE-7;dH8{8sOo8tUWp_ z-l^VsAhM%(dEfoUd(-vA-LJ+(Ir9UjWGaW5Uqt*A??$PMwQGEOE1%H5 zQdO?fG>zu{{OyA~M*otX8zXrX$Z3*y%Z4&fpsLjeXESo$Rg;s3NT{y!1u>SVoR2Ag zIN)@c@15-2dt~YN@gbTAMnjq73Q~+af=Q3);!k9^|AmzZ{in~(_b&?Gvtwygw zh*WIUlcymf(wxf2-|U`8|GV4Pw6!<_B_Ke3mZ+wqtbrNn2}2Dto}1z$@+TK;Ec1b7 zy*^>Pk{!+cYSQnRHB+-aBdB4xn@@+1S9*nmcx(kx-PTh@?(Let)*%cRC1;d%wFH@y zYPh4R2-CI40|HSrYFsM`_sO5{ig?IH`|T@as5(eRgomYzKyLjV> z&IbR$0`q?(8T==bl_A(Tc$n8dGd{OCiTwNsu}gbT*bg_`hfd#UX91E{j&EaJN-cBqnk-Z7BN`Y=9BLe_%Bq@fV^eIJ7s}(RZ$ev*@m_-u z4@uK9N=ZTY2j!)t`-u627+$pfylBjFa->9h33|E~O!OlU zbZ*kdAs}E;*?jp2nE!(%ANPMQ7XSa(1W|SUPpx1(Xw5Bq<-Bqhmc{R)D&{2ltw}XN zn%ww5#iFOSXjvNX%lO5{SfDiYho1d@Scou#!}T#6H!AF8#ug%$3JY z+0K3EHUSUg8h+LE@T-kY?eGwOpjD<$S7(R_$y|P6WoSxs?0+q+)6VyW9I@z7*7e#Y zY8}eYan^sbuIQz7_t`N{MD54OiYc{5pn^hye4CdyWBkC3M_A%95K}poT2KD~rRni5 zvnIZoNcRjQ(9bAYv0h9$SnpVR&pDDO&T>`gR(KOV&pZM}-*UKS*# z;p-oD-ieQrlJxLu&lnNE5{ogb@g2DyIR#jWbp;0ye**x-=Urj5LV4%-l!*CL1xNrK z924He@c;lj3ALv-lov*HMI1z=%&X-|q#T+@9l7*h}tu-rh}RcyODb58RGKFUeO^X+%-hW zK@*6K%m4#LpicW+08#lwbMJHB}X zp(b3{ZK)50_y`UA#^$Qqsycpmgq@$9ow%H{xL~pQo=l$O9^V+u3KutYe9SHR)KP5k zeCKjob@pr7L%TgNpj>U!=V)NhdwsYgQ&!!sAXk8=N%XU5izuB?H;Z_&aMkL^Rm1uO7|8#~;?K*$!_+-QD^)2N%Q9ldZ66CU6dTUA1J4ji)u0-9F z-O~UMv4wL!K1}+ewVgG9qLCNBsZM5@ie~S&Jh095=|7CM3%TXB*Kc1=U~6*9jk@Q1 zNU7g8s$P2vcf^KdHt8EQkR3TFqC?5GgL5r?_uFfMA|C!Y=%x>kw37|phpPF!g{Qtd z{I@qoEv=dA2it7|xzW0b(7_-HyhA(!0)*bl(>H%RkBEFQaYAwcFF%)@y@#h49gm;@ z%#4r<0_GF`CuwWSFmzo}!VleqvmMkFG1T8R8NW%11C{furlZXAfADuRr_Vs>XjZg` z1lukeIc-rFSEZm9tkX>GUyXl_M@Gercf|jqgP?+RNPTfh!kwIkG)ISuOu9AEubr%f zs9WR~4KMOT@T*2S3r%gsFA|5~PzEu*Qosnp1#*{>6yffMq0j~w(-+SOvlbiDS~pH2 zCCq4SsbJwt1YD7KhN$*gWmrA&Q&Pld^x=GGh9cc{5a+iGNJp!Ep3I}0P9UmAe#zoU z`QggbBb}+F8x*xyHOGY9z^dR1R8`g6SlO?+a|?M!>MEU6%u($r_RxV4GaDu+`p3y1 z9BkeiY!B^?@KR+{&v|}Alhlq_(S~Pc%bK^XJmL2bSxZ=re_Bj(ZSvI7GjR4GNV+O^ z6WN=M->t~6juQ0a4Ci|{UMVqmJlsDnBbzF))j*LomCO$H(l0{GwAIPyd5Qy7<_hlr z^suTQa(E;z-yI1;HKl@73|0;wE!s3R61{|~>14?rrVB`s%-z~o8S3HIFgHfQzKiAN}-YV&5^-@))JU`)TXV^p zu;tzM4g9tP(l_?pbw2K^A~Cu{oU&@*d!ysoBP!4OU6e?j7JIsz&h;{-234tA4s?@j zAQsS&H3?`W@8(wg6`2h|s#NqPRrS26;9F@%ohBqUKy0S9Tq-Ca6Kid75#w9xwq9So zsH>;40&->e&w(I8@dsrY!zKg2amH&Av2Q~wv3Qt#zkd34=h~yED-=DjM~gofw#Mto zW293zBf#S-t)d%dZ|5E<9Uq`47xy;I-2naVqrz7dc-G#}s=k`k@yaFPj~2T@c)8VP zb57A507;`xT0Ql6pquv8penk2t77}MTm-Vq`$>)_1{WI$vnu1N>lJ|Bew` z+aiBJJ*t&F_Mto8hGb}ATs?~2F?j)KHN=)iQEPT~_eUPkc=;@g_UZq2o@{Eqy`O15 zpjw%gEUGzf)ZMd1HR8yGjra?AJS$pW+`700&$#dIVhEfsAJB=T0eZj&VRB~o`eB|L z3!$jLCD9>{v3YSS)2zI`x9G(cfY)6SKE6-4w{v8MiyMf%zn9Ur>!OByO)|*)-5Yf+ z!WUytDVxG;QnApUs8n-*>;IQB0;c1p``eEb6XVkHceCcw)3UbVQj{?i6W|lH;TPn$ zu;k;h6yW2x1e^1NEjWb0LPA!5k^qm5pu~T7`G*HAp9F}+&kgG$VTADT{wr}Xjr8Dz zn~}f!DSCm2aUX^uR}vc;Yxy~>sir9brZI}}>AZYh09vixf8K4|*|zl$^0yI7 n=*A+zm0@z2iJ{^0-NM^A0(8J6NvW~HeBAt4OiZ%sa#;TdKyoSn diff --git a/generated/webwork/images/webwork-1382-image-1.png b/generated/webwork/images/webwork-1382-image-1.png index 5a9b853091db0ee798a4e870bf7453a8ae215165..9b3138d0f26c5ea446b8a7140ba07fce45b9299b 100644 GIT binary patch literal 22158 zcmcG$1yq!6_cuDiBZ{J^NGJk|FoaT4(%s!iBPA^@Ij9ICr8GlHmvlD*f*>K?pmY!2 zFz31lect!`{@*(1taa8|&yodm-*e@T-~R2rAAd!8iECHxUV*`2*Q6vxm0>WfH1JRS zG7k9U`Q;OG7z`I~E+V36Ze#?5N%}`bm??%mBW^`)^Dey;yd~_XDDsJ(+B^J)7Hw}% zrr=_{qL|hD3am#T-m{5)6uUsI6G3SycKd3;8!RHnw~w*ftLzAsgr2>ex)Z-oCQyTp zaqqT0bC{oW=stXT;r{!#CSQNJKUc157;6i@Lvp+0pDEhF#K^|Y9Y^e`f$E#E+-CQx zsqVwP%QrunpUhQM)Yu==y>WznUNsr)kXCs1JLQKgeh9kJOhdCk=hu@d&EJa%*HxLW z`o~A1Gb^3CUW8X*_vnsDucLj@4yX^RPv7heXMVRzSFl%gQ;)tM8dKJbOCI%O=QfqZ zH@K*m$_6^(ge*iL&3T-KimVxy}rTvPYl{EGveB{%idUvf;hBTYM*f(U*BZL zj%*-$3M2k?!9(cAFznzy4i(mg?^u-vmuO*P1{6;(XA3gCfXRI(jKhil#an@u_jc?T zBMDB(>l?Cfq6Gh$e0}lqn}Qp6l7*gpyp?)=Nl^R-X_4=}`<&lNue=dXdO?fZ?ronG zuXt1Sikgs}^tl=RPoLpMP3%Fx5#~$I?>v9+zGREL zve0n2p$De(Q)ndHAUxpU#k=}DYU3W(4a!Er@SD0mK`(FOQ<3-!|MOiv^#+~9@+b9F z(j}4g2g1fghu)2!U(mv(iB0_N#SfDXGFAK4RavsHIAIOEUKX1W2}%*o?0)e?i#+#j znc()%*NZP4iK<|_!8W4hDfvl$Jzsd>`_*UpE`=Zf&7-**iIH+&_jvTtosdmJ3qg;d3zJbF7Dv9D7AT5Md}S@9CN0nr%O+k9xPppksZ807W%erX|{G2zNNOM zvgOW#j7}g?cr0uct68e)H~ew<(UpW``QM5hiuSZ~;jH02;m_=9Dtw=6@XC7Ryed85 z+Em))!bSQd`S^(@ix!F&H_Q2((?@0tIeiva+|Fvv)W{-W^GTPU={+ckFTojQAEhD_ zG3#;NdFpoa)a(eROC=S-7P%AE5&0#u=02VAeY|TpA)gv(4dYA$nfayZB<)fu)Al4w zrLReK%KVTTla`PsmM)jVmqCOZwm<#e{9`q0DcqGZAao#dH?%X9kJ4SvSv{E1jIK56 zQ8*97z9f0%P~?8()MJEfoxYd6kkVJF2VzOezq1>^P>S@Zv&a|9ZYEC$-jZw6qEpY3 zT?*~JnobqgBx|6~>>fjD;%Z_xhd0MPcVX@#VdRHI=>+N6b^^6G^xEvwrLv_GYN_d| z^LddugUNf6Wcgj{AxZ+NlNtQ^UwbeA9M0{@V(APu-PGjzLNpi+f7U0?s#dJGT}LDy zkV}~Fq`8vE@x?lsEJZihQ{vJf%Ygp?*B91`E4A$&YQ5P4u#?5g*i|)#4+5BF)JxJo~j?q*xJ{s@QIKs1I;7 zmi0J)NcrYJ6E|B^#9BZwTsqe2z)+!G5w&*Ho#t5h_`;>^w}~PVO)3(EtuI=MBtC`t z_nBt==z3wMv2faz!h4ZCaj2P~GP=^rZpn$d@~A?+lHZ}vnV`a{x@nzp{nB{rV0ONo z*4WYo-dE(8coSX~@v874b{k*0x~aMI?Q`sO?XGVp9a`;}@3}5mE@)354=eOBHRZP^ zPE$-z%?eN$P$=HLkJpbse(M#FBnK@=<|7i)Rf=&E-tsQPiLXv!58w|jK6tqDY^8W; zYW#ln1H-0wLE%A$q}@KrBG|&QBE(I+jmu5*jjq4he^c^DA>N~(G;$P^@v)~@3KN8%L4V060?XY^xSw_F-6z8C0J?>E>ak6=k=Hu;CCaUxN#dIZ=g|dgT zZP9ak=PiQ|^3p;dm#>*r-b(Rw?!{*4;kB!DTks_P15m*cIz*Q=`RlPTFT>ZENyl3aj^5 z-%o97CkW}8PmaZ3;Hlmw+$vvCShXJ=P`dg~s=d z+G03jb7BGCPNua5?w)CG&h$=Z)?!nqQ%md_?oqFij@TBXMYBZB<(HZ->|>LebQCof zm0#i{uX$#pqv!tqCGE_|g2vqB9OYbl{ki9AvI8$mb)B?>JUSDQhfO#A{DOzc&2!Z< z8M04bvF$Cs<<6)Ib_+V%GKw~eX-(if;M_TFKFi8dDl5#~?AuJ8X(X7Waf|IhB9U{k zGFrbP10!R?9?+o9tk4b`IX0HMq$#Qjwc{g6Zretl1&E#Wq?#&0J9GEw%7ox1XsyzbOT z?V{Z+39TB~`ID&KWwMHSx3w0h))kH8r=C$ZniEeEebe)m+N;{}<#M(zJFz?EXwxn2 zWA7!wRw(H)jHv*emAteGtG~0<`iBSr4 za_n-Xa*2s{ah1ol=pctk&;>Yob)B3K}%&6|sd(?kAvb#O3Jl)$XApH1aU6_NM zDgaLRjU<)jU@$j04Cd_vgB^j7ycc0GCuSII*#HLPj)uXAU&d7_@q!09&t)V;VQA>T z&$St0;1k@JlFuAqunUhd{~X>g0aPZ-BPA-N`f7A-0-+PWACEfSXFgFh78F%(7X1`` zBlbC+Z1Lc*=EE}6utkm+Mjh)(X0|B{&Fe*NNNtx12hDZeayQGcDn3*EmTLE>c`f2? zdFdms)Mcj!N3FPx-gB^vVhx8=1W|t95b?ruK)-$)*wB$ltb@Jn#!v3q-@#PpA>iP- z(CMVRUHOnhmBiOGOH0+zTTJNW1@z8}2(P(g%i!eAq%KCwgMsF5=jXAr98UN8eUJzE zf=PwJvASV0Q4a@Jt_O9$vbe=P=I_$^G4=9=%O^v|VKV5P*#fF1Ga5uz(ad%DMHmeG zru6IkR%33(R-w{Oy+>opEqXzNvmib<(X6p^!yoJy9eNepmAM=h%zHf(g=I z6!&#OEfkjP^$#+v_6Bx{+;dX517 zdcEG=ynJmqzdV>ciYfwe5N$dh(`k$aYq_`=j+J7ypYSj{J5Ocv2Pw{kOTi9VeYaCW z-Xhv%FbZY#iv64 z&yzbBv`^*a7$a&Yh&fI@N6defA5*;gbpcKQozqK8aA~DaP^A3jWRE>F&{yR-Cl|KI z%i~Yc39M$~oS2?&GFmchS_My|jcbz%Hl3@Ot}amjx8qTnZ;pNyo1j!iq9(L+dt( zel5iM(+O4UhjJ)l4Zop)0JbRUwzm)td1$xi(~51WOM9FcNook+K!$lT5nw@QLgI;C zpM*mFP=vv5NM9IdLulA;jh0kustq`WK6S*$ymEnSPAiEpWpS^m4t~TFqC}}Yl&Tx1^gO}c^&h8|1fLm#E9&jXIb=K zu1%khT~J)fZ_`k;p6c1q(*6B)R>)z@XP9>%X*^!WPvykvl8X%}1;X5}YZWz9s}aus zC;}H7=5sExE0IpnQSP~RznJ^55_N4?Prl?}?wZFKIqZh>ALFdi%aV@0=&(60rd#FM zwon$re$*o!j2t-rw7sd~ooJZ=(Sl2?-=V z?6QOjq*PBlj~wR~7Fch>V2i#@`8JU{9R@kjOh#@l9Vx>^FnLN?c-(ezagurgc z=YLR=$zjK{d~)I(8akf4)YqZSnu*K4sS&35thF<*&5uG(HDv78e;MvgF&bc^d?1;I zLHCQXSI{RwJR!RVeqtU05&ruKh~?i$Kqg_3bc08j?_wUEe;4!U|KqS!S*sNyDs>x~ z&L}SXzHV(`h2#z!Q;I|dsAP6TfDIELD?_QMga!CxH0NNbfA(Kt2LeCPKdY?fG}N5P!mdNQyiw}(qw0(= z7BcLW_zUCx+d0`QB|zzbEgr2hnfMqY1?0EXGBPWxEzNptpLGeN)cQbu%S&APe~>-E9YW-NT%+|w3cgd|gT9{(6T7LUI@W;u;x)H5I-u*AMpRrXc({bt~>z7jT~K9jcl z^{J=72y6@~ZCRiAC8|K~6wa%N$$|G~_YGTvH z-oTTzw>B9HbS*lxyBwYPj+H&H_ZRr%nr1J+ZV2xFPK>=JiJL|erk=ORV@$Ewn`C0k zgTr?T2E*%A>-*fYC9Q%*2iDfxaxEt!`h|*`Q~Z=c70^)65X5hDMyV?>K^VKA0+I*} z!mOqeCS%e{#;ti|gSH00(DSS-syR&*bj!$=WC7D>*z z6ho-PpZ9~AyFS)NRR*jbl}vjB8=MS)fi(crkSkVD7|0;J^wb#Gk&oFDzkYg(pDA}D z%sqdG(1btl(5^PU10t{nYv!Y$r=2*;{tl3*xSWPVUR56v$;xKBdvXa|r z4EmoBnQ}FNjAi4ab-MiELbjFU}a2gG|=>YwK)I0H^ z<8R5s!r8T@o)w!gbwuT)3xRKDA8mBcWt;ze%aq5-_VE_9>J}*yJ_qaW%=)JFtgFT0 zhi0zQy3uLdYAhDGMSNKd3}P`CAB8_Fc;4)F3G&DdPw`X0qDHQ6N0fO8eLrKwY?LK4 zc{URq$c3gCra}o%wxWDbP$JN2lNcH9takMZ>GR*Eb+eC~Cz=gf{Ot@4BuJ9Yqm=jW z;aJ!RCH&SGrNo@PE6KI=O;6LWA6V{)pC8cuKsVB^Vrl9IUTu_`QKk0T_uVF-IjX5QyVIm|i$~{>qKUH%shy z*U#5r@SG&#$Sr8}>UIw`Py^SO7RxH?BF^UI9s&}>r(JiFBz*T4HfJ6DRiR{RoYoh|+d(x=gR zT9*~)9fSzSJxF%}&PY=WPCm`1 z<;n)OypcBPHVkI415_;AYoDL}bSGok{0@KMXs9Xo&EO>%4l$o=esgD?8j`O+XrLY z0##g+b%K*`0dI&@gOMO|@wet4iO&k&Fk?(Bb0n}b83|*A%#gP3@87)$onOOy z=DV^p$0c)DjhQJwk3oX+%^9P*x_v+1#&^L3Gsiet0ag(Td|XqKSp>sQNRo0-J@kTBla{2G!l zahgEsA<`+0(oW@&F-m{IE6@rn&KQK>Mzz|J}19;`v8(mKIR4ZAiO!qUfKJ64UvdEnq2(w8O~`% zlgE)`u_7jyDJ3}wv1JJM!?mvnwcUib%Li*4VgVmQ`r;oU^vx~nQZf%JV)JZmZ5KUS zLf#>Len4tQ)i%BPYSxk4ftu`L3btogJWob@yL94Mc6N4&3rdgg;Md#s&d%LRCN8dL z+w+}9)389g+^)VO4|I-&Qv&%Oi+ZdakX}_6-co7x*dVDUNxJP3mAeFMO3&Z8BI?*U zB(kf=D7AI+;c||#9KpT;o(R4)s@S1l<;6V0a-HgZmSA_;xqQO@VG|$xm#@VlQ>nTK zll?r#VrowFMBMk+2jaL~R$>ImmkS0K19O+8Q@N`j8)^_B#@bSc^+w<^<)(@R7Ss6< z*2e!LHN(WO*oodp-bHM_o?4i!+}d~Q-bQ=@TOZR~;IF@(rWYbS&B}k<@>U*(%CmXu zK|wH)v(#55$Dr&|F1E>0o)wVoV#w*h2%_O3;jBu4UZX}wAF??~hfL3m8|sGn9Mnw- zaJL&RKON>n{ngJYKwj8S1wsWya4t;5DYpiX3g)(qU^U)tw-dv}U-Lt;Pe$ERf{ zzZNqO+5xKZa=>D;&_YtanURyHS3g;*U(a^{R3fU(<2BGz>_%EM z)E_ZZVel@-ax8vs*(d8Dg9iS^QE744ac`7!*_J*K3$lXDT880z=K|l6iDFSgJ$fBpeQ}REb%=_j~HI4ibM2&V-`&m5m6=+K&x}1L%5Jb!XaJ~AA4bnFG zpotAc8()CEoP#Br^YsVL(bD5C)5`)ig*HgSnAx(Q)yFndt+2d8O2ldijpVI#J!Qx( zNUQEtA2`9t{PzB4|RgB-5;w@j0{9U%>anjag3|^Z1A+Jj$RssEP%CWe*u8W>@7^8@1O$OYF(LZ$ zv6dN~-k1qc-d>{z0GbWP*wFHEa)1MbpxM!;nG3n{nq7veK&j^382+C{H>{aI&hKYJ z9}7(GY%`*>wNSQ`tgENs-XK|l?5{CqZwMOn4r*Y{f^&7(#>xX9^O-&&wS?25#(<9! z=1LjAlU{$C zOq7Ss@nT9yiPMTY-`=p1+g2Nm=XyX;&_G8FYs^!F@VLX4=OTJEuDkSd+Q)bVSEBO+ z`M@>8nB%*-{?9~M0YC~ z9&FxGkP|jr+seChYXKO877__1|4~vTo@?$ioiyCbcRY>(#|VcI=}3GU2y98V+<>T> zU%2>3WZZzT#OR!#Ii%k}eb2@Oo;?n2b?D2Fe^ww2p#AxsG+6s6N!>Y=VLzPYadKNZ zi>(@mu=8FhMrC~gX8oD9;Czu#ob~F~QS9W@?swGE8hXd&0u#?&LQW<9vuzY@vBAZ# zYZ$e#g z2@o>^d*D4rWHsa7NDJ)@Q=1Js(3!;G0j{4uoJL=Zp%%W%F+gPo?#^3B@vAzCOY0Q0?^zR72Z zStK(hWBsNI+0lBA5PMxYgwoH)>yM`$It+3j*leT0;84r~`2xs#cgl?D%l$wsXg7gS zKnls66CwQW6gFmq_~(ud^Bg!fJFXG8jr=;fd?n0j*}Yj~M*@b0Ad)W^L^Nq#)@IX~ zC8ZFHTbt8CXbjj-uI}b1cCZ$Szo^PJ;AOd3X#aFk`-IUML^Q@v+bPW| zs+0G=Ag?Wg$P#_r-@h-e@eg~YArHu6s%hujsf10DpJ>D*Hd~i%0RES+aX^~pKFi|K zI6EulRt!^|kQX)~iT`4_t4RZL9Mpe&Rc z#LAI-YzWQG&mWCP#l?Mb-JWN$t6+$lI>_Qj?0Q!%`=Y~(My-%#&OzzCXCwqT2D}(N z#Z#!!m*Eu4fdtg3Q)ypvmWo^W(-G^@ry$6Z3|*H(v03A-k$ED%#ps5$i03fs`}gmd zPke|UCgotSq~X(5uIjdyKzC-gBZ*(S|O_D|w4h%|U)jp&=PrZizOi2Z) z*TAvdHruI_7_Z$^gzA{>iOxH^2lB=00?F=)x><`p6%VFt`>EK<&j~mwHMx@U+N5(s z25rChnjh+6F8R0C7)~zr9Engd7MA(orboA144ma^s<(}3_Iy3%;LSVM1vk|lS~qzt zsn{!73dG!7jOE?{9KWL$J>GAOXH9d?+{#S#9Dh6r%z!5yx;7HW!wkYNx)Z|Pg0L>r z0R17M2C4ulqK^lOoNhEVh!X&j!)lI^LZg3uedkFrFL1LMZN@0$f5aOIbc5!7y7=C~4s!X)`f-Z9hf z?n2QjaKF%GEp~v4R8mbwHql^1K>;?T*mxKjS2$=9_FS{wW?u_vlf8?XkwmvM_jE1( zL<-O_bP9RO&BXQDQmow6l`*_f-mL)&`R-2P*Iem4%qA%s+TTR?##7G_9rN;>cS zTlO)dHC^mGI3L4670QoXl9^HiDQkO*E+s%Rf_z74DoEkQWjs>dD^me9r&jg6U9N6% zpcHAVq3ZI=f5s)@lk3d2PssbY?f93;%}`z1YCY3X6Kou!{IJ6%cfh<;Re=$Y$3?~fgH`mE;IK? z`uI04t(fvm^h^F0Uu|=bsa}hS>}Ve77)^b9cT@@MYhL@4&=|5Q&bR~l#2gVrVDGt~ zVWtEulQaqDYNLt7Yiq^V-W_6u^L3#SNPh!fbrs#HGDZB^{>4Q?VG!~D0HW5i)sBf1 zpkr$Z&(7u^S>Q_ntZuIEb_K}pgA`RSM8<+F5!ik{$m1)^JL80zJpV(yT>+x*!}VXx zQ-kJa3dN};H8u6okNOiCzRhNWx9d!g9`&Wm#onf%$Oi_+;%H~-g8p`U#3OE(^*E}m zloSzNU0wct#L-g5(PjwZiyyh0~E9EtZGG@XQRP|3j zXvJro^4kDD3gV^cxSj6WTEH&bn((_zGZ3}i{mo*{wyqbzOHM~A{GlM4Gnl+Qjp>DE zC+XYPK28eLq0}KGC0pAmlE*~vWaZ`&YZp2zEUUJ*>KM#7P9C2QK0if;avkkfIviEk zk0no|DAuIBvfS>{9Ev% zduw&PkCq%&Rxt?F##{uWf@_xcD^;TSvYB{mrCCl5`yI5${6{RsX@{D>lm;Mm@*#pN zX=~)}m6^}*q~iEjLwHLd1zRTdzvfJkd+%&OEbQV$v=CF*usL_kBTFhGlcTnf^ISGx z3h)7U8O7=Na-p2b$E>y}RCjndo%z9CzBA1-FVFS3vl)TOp>Mv4bS{N_4GM)X&alm! zCf#S{0`$C|wO4SeA&#u%y@&2r?ssAAg_cM*oYR32KBbpJ=4(acLgi$6#0HGL;uo-V zV)jbC|CvxZFSdcg)ETMuCa%#k&qR3qwEK@M2`Rn@L(^;E~j0-3NnMT zZ(c)8PQ7~~zy;~!j6($>#c65Z^`i3u3lz-;^VYv9d!YqM9m2~55~@&@2fXPYC@x?= zzN0&iFc=oJkdf&*i5j&0wxn@673Tf9Amg@?^oX4wsKTJy-KAmha>WjN^1+Y|M2i5Q zKuU8z{J96nX6&MQesU#aDwg!}scrIiKv$J12OAgxSxAmSlXzY{Ky74E(J597jbJo9Rdx*$V zkMk09IXiUF@yCT3knDv&m*w6Y3>Ymls$Z|OjR3|dJ#N*VS_&kcc2EyI*9ItaJG7gP zFW$sh9KbySjJ-4Q(zg;D+R}VpqQ+4LlvuumUz?L|3KcQD;hwtfSu5VX`WS=^=Ufz2 zNa0PfRICgV#@z5}A`tQBFjJd!M%FSX0>ccf5p!fpCY1q6X<8a>@%<1z%rU@>{uo{8 z`MFgBHsxQd1RNr;O4M{9CIDCvJTZ7|J98{(2yODo)vY{00zxM7ehaNSbzvgFVXe(eEteAbe~%&yAf0?CIraJe9?0!)R5@}Lzc7=bpA zXPe(s%!7Xs;z%+336#Vv4#+Y91JKIeVw(VNV4(Im2|0CyfBKYBXg#qe%|(mR(qeEz zTD$HuTBkcHZo*X(5V(*F05iR`1fF*WxBm6$8%fQnK+A@c@|tg?{DtLx4jC+kKrB4-zDqg4~K= zNo4KMa=PHe;7p+@-GR)~Pi=;2yXDmjB0truo`(To2lopAGoY+6rq-l~YJtW9hjRch z{N-0!-3DnO3H4{MU?rTbnS#hCXqlry0rWM3E3k%K-`=jg%K&V~a75?X1W;vtGl{5f zJ!xZcU<8$e!pU&JVp1*j&Pi74+?=U`@77nn_`bq+mTfB9Ix<3dlY<@C`5cqmPTm7_ z->nyqlM0~HyvfO*hY&c#*jTBeSG?q0@m)Gnt=|_fo9$NX_W-;BV(Cg0xf!Uix~vz| zozOe%UyJ0=9V_YaP$$jJi6@ea?@a;KihV71AovP(RB+1xnQPSI@uRL)BpZj8)1h=K zC^J-FZJxB#4eR=+JaAZp+g!Z;eQ(C$^oR;r-3}a_3Qj z8helZzHf^eFFS7hT649jBzC9S1s>miS&@j*Y8WKYk$4Gl<~1io_`Uqxyj1yA}lLxXmf1~73ySRII` zjRNT=kdgMS(Oc%M>qx~^_`ShTF+*rt441D{)7i9;f#lw|SZlVnHf29IMG;fNgQLOIa=a-tKWlb={$w=bmkS&C(Fas(jT0 z=u!rv_hNVfZqhjjvXK}S0;MHD^0pz0@DJ5bvURrDW(NW~OOPIf$o$&g#GIYonf9~x zQ@7`x<>6J#0hVLfkm^C46-Wn2H83NeeS`qKP^~Fo3Zd*N>!(6)$6s<2pH9DhPP_{3 z;Bqmuo3mLKka4f3G+_P&aUr|Htls%M3}jpW!Atq0CeEM;7+8gwf6j3y^bSLyv@&1} zOo(PT#i*dMyzQ!%mv;|yE;-HzCEfx`4h}J8#T1iagyLhcCxd$+dO!rN)Ee&m)ff_x zjbX1-Xh$-*8}k2o>oHiqqK08r%pa}#hZ@5;I<~6hlo)R}0ni83&mg>Mu~aXmwh}>` z(+SUt2X+gb@;Jl+{F@I02~U>O`M3!61t)GR&}(turq+!08w05Snu0kHS~iL!y+gT{yhF$b6)@Pklc z);lhYX9j!=$Q-}H#fH8`k^yq&@_glwGIN-FK+FVvE3;d{>$CoCik&gbaNvQ!VUpa0 zh#58XOGOO*0un=*@qWLt{;%0ph)vr7mPq|F0$@2U1g6EP!8l~Giu=axUd3?v?gd|NvQ(}`GpN;gC|cKV0@C)sqvNOS z99kYN1YMvWX@v7qt&Kx#&rCX;(}qS@mk26JDubIM%AB2^0h>o2B0LXO26Z_<*)Y@Q zB@c;{H_J;kGTzzD8gSK0L?NKq6F?KX5P@u-1TE^EPNF>r=iiMN2Zn}@$JLdTk|<6W zCA5l+{Ue$5*B$hQh2P9~#5BcTgL!MK%ok^{^)4%|HtY3?_YK~BP(GF%O%n1vxr5Yi zKc9%TJ89KUyMOYgoi@9<7rMmjxL!|X`}mA>DuZ|PSo(PKu2_qEexu{*e%&ZxQXXT# zbDBN+-eufaWCxN3ky2ZszB~op72_Lud-%xf&%CSU( zAi48TJs?4`fOhRl1*SZFZj+YkZ!s3TNY(ERPq4^|Fq_HBw-g%kirUp;Dj7Sp)Rki; zc2B#Tr9-nEdU`De4zr`Rv;9Gy@o!qi?k(vQ{u$SwWHgk(IsHs}iqxb&TxYSdKvGE| zL!S!iCkpgW&i33HJBPqc1@IYNS~j!@$kJVR z#HKZf=zyv`-)UL|T7f)jwR9RXx*Cq01Q`p=Qtg~EiI}_nYSHO)eA%D?By=27fcxRo zdlxaO;3))QdZhaNxx#z@ufQ5o@NA&qU1tUT=S&G$Kz|7{Jx0~N6}hoqK$YG{L5g_sP`U-}U`puhAZI9u-j zNk2jsXXbyG^*766j{@Egu z0F#S9uZV%LGYSj~CjXnTKCcu27X$X`*CTi?L|{vR*A+d#@}gvn%>KP9G@9Lp7>hoC-Ut6V}wxQE1xb`!XG%PZ!N;d%s z%4Z|(lz$G6IHAo~tSyhtT((dC#r6pQ&`ieX}49lxE{MXBWq;>1_ zeGWLrKk(GkkuO`#YB@N;+TmXQOKJDD(^z4F5e@)2#bwYY$>&AjFGmg71+SqU7k22t#Sg zld{nx+EN!$A6>Z(FwRJx>Q!?o5YSD^Nl&^6-*DDf+<1=>t6Vd-Gb3(B_9Xd(*9A6pC&NGzRe!d zT)wz0o}^Yj1HdWa7b=FWB&YBJi0EVW2YW_ZN|+lx=k2EwwYX(rIrKhG@Fys9Oi?)D z1eF}pp@1c|9NfADl`DdL(kcJ`7O8;RoaLqdvuCGrQ0N5t>`B6dPf{Y9@m$EUVi`u7Cht?3fA?t zspUiq0SA`e zGOH%;lxF6a`lGzw;tXXnNL|b~UbFq2Vi+Zy|3_I$`m`k~3s_iSooWQ>rHUh&J_8H^^XdHsVy^ESg2-_8*ho;( z41EH4-iVaU<_)0^t+V?P;IsSTIK+|)eC3v+IGOUC_?+;h4f zl}mV|8%zCU8R=c(ca zBF9sA&MsRP2$w488`UBM+Wk=@vl&|$o;*QDw^7`XW*lT%nt?tZ9%9R z{7edm77W}4_~E0R4*d?oROl{ZAfp5TBlM=p|LZSAGuZz^>{KvVn3p0sNXo}3rIuIO zT<+q5uIS-Rbs}>y1L1QvJW4w0OTXk8%ixx;&Eu7czo;DEjF{Rd+_W3kI*Ob z;AwdlqPE0EG+Hk#NB8FPauKcb^L)zOgn}$kiL4SH$M)5QkvmT^G&B@$xgKSDI8IDP z#%=*@-+RRno)f4?%|CeWg>Z>0FhLN2xJCbe4{Siw#h)r^Osx5zri=70vHy+jGY7Uy zMPO099RJq31Lf49EX7bW>^COlMDlmmu)jqr5+K-d zBcR~$Jg+P=a}Fz^`kuN)09?Fp{JC=It^v%);Dx_;Y)~bMx%^O20Q^7-6xynRmYe#J zwze+C=gC1JjRBahq4Cw;DNjWHb#gZe&_Ce$b=%qxkWD+IXUS*3Dna7_VCC!4EM0v( zpphYII*OCzmOQ5T%rHAGP0Vue_M*1Ts(sDAG@`;0uzZVVm$ zr=JIK`=BXH?i-UWW}&|Zzv$QNp-tB8PnV8bJ#4zcvH<4~xPFMl2H#9k5a9nwp;cd8 zRe)yp0@ctDjD82mgUuPfXvKFn2jujT&Dqui6veAyuQj9zzu#i*Yt`O@g@nn+m~u#; z7%;KdRv2OsJ$bC)MKM$A+&RO)54d?DD9r%Q0;-6{978PBY1WyD_D)wwWXy8--uhHQ z1GKD#BW5Olf<6m#FuZTDKlJJi(_XW*ZYorR735HP-GD-=Y28bH>3qB$qgvTT07r^f-JIE+08EB=!IO;*k14`<9G}j$V{ETVsE-Cvk}060i-*&J*zubH};Z2@)b)lkQ?rLA?7INCVh`7O%HJp2e+npkw!D z-bp(~^##W4uO#fqgFw*_>X1CnWdFGk2U*Byj5q-J4XKcF(vE!}VJ&D4QN0MW`TFiE zu_Fl4;8+i`Di1$^db(Co1I-7!Vj1gsHo7&hK4rri7+%=gn%%llR8>{&G}lfD$7-s| zqavlau7P{Jy@)YLi^S`=mlIoCXh_b>2rFZC6(b&)m?=?4O3b z%7Yt$+}YX5r1@u1$zr-XxO$WuntkQ^r+pgsJ0ojUpu==%kND8IMeuhC6UtHSq7$H7 ziREQWmP$1Oz-hqn!mEe#f|z*ujD2@#rcE0&H9*^^_$)F5JR>;Rb=2`s5U!1=ahXll? zRyv?43cptCkXcw-_K~^Bs`>bnspjh}@!aHjU>Is{wDIalUNC`;PMrWQ8p9EsqftFD z?jfds(5U+C@oyCZmk@26a+H&Ieo4vl$)Pkg4UOe|;8Go_vZzmTGNnW=+}e5#`Goqx z7mFw3&Rf+}y1lb)xaqbuNs6Gi7)dRqk|CcCf1n)J((>rThYzYAHLhb)w)s1U3sJIY zE2m2C{>jO_*;y);Rv|fY7oJM{G%*U#oy~%MgxOsA?ziLmFL7~keWyt~r9*l=&a1;L zo*Q2;i9#ragc1%HkB@J1$#VrCm74dZ_nlTdn|4Ge5hDu37&Q2MWCX@DTog|1zthxJ z%yc{LPq+kq{~S-Dqw2J4yOV3ne_uOptfcMu8-h(Tv~JIxZvJ$)_fArnfIxl>tFmLa z)%F=OC(kBq`)$bXaEktbPxK@xreGpl*e%y~Hs5T*Us6EN;zHbNksPsC-{*_?lLUKE zt5#BC-J#5!#~gnrlAUhVatjgcNhx(S80wxS^}auKbxncLHedi-Ki(ATO8jzEUbMg4 zIezIjh3%wS&diS30O(2jEvC=13mjeGy7ShSKutP&+cv|u2339dV^@Fh^wYhJV}1pX zU{0-vyjpuN$|CuYxu8iKY9|3|&(E|W;9N`J>Yh(Bn%JK+r1B_$S~&o;!%>gaI8!p@ zde#z0@p1PxvuHYPD*McPnMc;~PnnKBopdS7T#}B8^kS9RgfmhmDH7x_^`U+iNfgT+ z9}4iy_4@b^XdQLt`*_L$KMhij01jK~4`Op|8tR8FTUr}+dLWbu^b-ff>AAS{evr$t zTH}ELEc;7Us@PNkl#t`s=BZv`GZg-^1i}U5owBeA_h7s`l0zUf6vXsJ9k?S9R~?Hi z6_`DNJjt*&(=zyX{l{9UtgR~#2BRm)8)Iq}~9)K{IcoFPa&X1nKsSptam z@P9gESpASXthST9a&kyfIV#ZetP*x3r!9`lAw2rV&0Fj&u?#k$aOjzECj4Px{JqDX z5}`Zh-(G)aDuHmFT-?*bG=Xq96%{S_C_@%&Ch>-^%EKV_#8$!QMn?IkKgL(wroiOF zva;c!VPSG)BD4ddpI5&X{o7XKf{NpXlNe6)3uddhKH!=>7|prkq3D*c6K&> ztv%(zF0tG=md2@+4?EFu8SsbI@n1T$9s7u%NlQw8OeNn&tAY~|91g%s`$`3e`!ugx z!AJsxm)M96ewv-k{(PDSVE<=P35j?AdGbqKNl8gs^8Gw3dDHX;po~CB4}ScG+u%3A z-h!VPC-Q%T&dwi!Rx#ku6IZtrdeS?;L0w!bVL9kJ;JTsP06++O0uuX~%-(kXr8kn& z&l_16+veGN^nA zO$GhYFZ%R}EJ1Jh&S;Jp^nU~>YHFz9@kl&RysDmT;bdREZ&gV2oC-9Mxjsv#s46qt z3DUk21RBfwpy8Z5pVaPIC?W1gq+(ji76U9SoXy)FcIRYQ1xw*$e=@TB2_Qp6_3-*O zTrSQ^WqtY{$dRTti;_lMWL}@BAS$b%gensv6pX<%q5!+1rsO=(45Pg2ij@PK@A3QHWrDsStN*qJ#fP$Y5*`Up_|%K6nfagi{QAJ?Jy|L z2sCVO%6aBYlFG(sJc`D2aCqA#?YRAlBYGOHHx|2m>MYQb|Gkj+0BAcn<6<-wx-RU@ z;ROQg{+CxnjrylmLZLfpOz8zgl*YOTaueQ)?Oe$fHnQt5Um8l$o&kB17|l6I3VzAzjdO7kcEj zW$p3X9QqTq`|q%0GoOhp-LN$D@Zhz4AUyi+YH9h2PTqzn?KL{kZk!3a2b;(vfvc4S z0fB{uz4=5j&mZwK8i;uGk4#{cV%0GD1}`0o{4KD&@_{(a{33Lg4_Y$sux7C0l7v3c zcZ8(!+Em&Z5Ed}fV&wk)SNV^X_R5M8hB)2@SMt8RyDSt&U~?*Mq>~n?c=so$!5Qcu zcN@#^8#Y!0<@#k6dUO0%9n)=6J+KC?_domlJsfySZ1$yvHY(Df;Z_P5$H-%6MJ}1e`via-iLqe}*MC#=)+%*!H^H#C&i{tDKz8xIB1Z zs1jx{D2%ce&wbZFyVrW_%F2P{VCw1euIu`}`Gr@!p6J9+ED`{>WM^gdq4;fgdD7=E z`0WRI;QNWlwX{P?1c}L4t$^w7>Xze!Z#V$^D&X0)<77fsJnT z#wrR_Q>+{xHAa-5@Vsl(@sWml{)H)FL4!Iv6)%N$cX#>A)l_#p8xUF(pKFf=HY%AO zJP>l2??{5fD?O-{qw+<4zd7IP&|1v(SF(J-euo0H+A{ts^b~08sINzXb_cDBm;M$O z(vDv`e(x8WmWSv(;UQ1CFN{F9-;`R@z`PKeP!7~-?b6hog2x^K%+ zf2O$GV)tkzUnX>re_Nu8)hU!}pJDCLKg1!H1MrvupbbMdzilAzvXj%Z+n`eNlX{5x z49VKRsfvZ27b2^+ep1kqaZqTpWW z*RZWInjFnh;uJ$h3^azz7H_G*5ZJINdNVz%3mI$%w1MYIpsevCEu^`g?p#H$?>)DvbxpE#*2t43|#pcM`0aYY5Fv8?TPvm!odOrA*P%#23S@UHY{ zQO;dBONO^D=PJ7wge38%yHR%?-#|IJ;67a?pM~+Soz2hH&_W<9QDM3)EwGI-nPnE( z(2Z(MpRtr{IqovDE9=*l;_l4NC+ekJXneYFYFbRN3I1{lI0+hLl(T%4Gaz2O5I+kG7OQhKvbNsc3h%SfE_2(kdk* zwUc)uJQ1FeWKCKC;hZ?fdYGW6D_T{qC&l4OKLWOS;-$+D_w?ftKj87|A@c|OnYi({ zeJ%(YR8;LDr3@8JsTBBj5yC(8lA@CfL^`_1gK_)~Rb=u|Cum)e=9=7a8 z@CwU}^_&890}>3blYI(QR>t-l<)1t{2}fS8#Yg;HLI*xVUf^qbT{qz2$=%9x;8Vf{ zOa)Ck+;;JY26+vz7jO2D?E_KMcj3XmO8`mGYLISFd)jntzr#Z>aZ@nx2HmcVhTtuK zGb;xM2XA(SSc%#2M|~bdZG)gTRC#ZoT+8-$x=t&9i4`J_6k6HLmH0S1y;~jB>Pv&l zj=p5jBBvWrrle9brv&$ZHV_oT7~BC_5*Q|NW0$7Duma*Plk8|Woj5_u+YV0}mV0$G zZ|X>8eG|{Sq=*fpG#}kB_Bgav*c@W_utV5w@ITm(xOGbE*zvjs>+@`|g{q~hL%UcY z;)w1h^SYlIfX)K+B3bG2a2DHa@0iTyh2O^hFpZzFs+yg{UmI%__Urk3Cm!*bt{ z2P7_jB5jq)F_G{9DNQ1hD{^E)QuIa&W%(CBbV49yF;F1P=gTmu9Fx2i@SlMESGFaP z*oldx6q!VX;~`Ny$g%Mn(w4gZH7KrjI^rQ8cb$gcsh4U%K!6Nb=*Io8BQ8 zwVAupQ^eR=&+iNt(OGjy=A0` z%sThd@%H_!)1)8G4`+&t%U|r@e*F>=ziiys@=#f8BC%QF0_wQNR73N-?r*Mf&53z0 z*Jb&(s=L@vQ%kLSp5$i{G`F$T=Z}4k9oGDx@W0*~Ncm}%r2OKEn|j2xPmzUPXXwJ4 zw=Un1t-mGZsk(j~dPw_C)+9Mw%H~DWEeWfa3W6A~^Y^$ocupFaZZs^(qpr&wd$pTN z8&piYAqE@}*A6P(CPEP}SN{>mc4{bFmt=={AOu&}xlcwOyQPGZkev4ZXI`9;`5(3T zGcR9}au9_5xWbR1_gVmx^aVeD_= z;(xN<4F6`MBt*TssPH;m{N&iHQ>R~lzetrJ!S&(N*YgYFG8d_HeQ2)n{iHto`u-O) z<}*#+FTTV+B7JgJO~UaJ*%tB3^QHfcz9szP+riI6fFnV^C5<@Q{p^$`L7(p+2a(e| zL#J3r4mQ=apA_3T*-M_a@-1sW6XBuWR$ zZl97ssyc#q_OBwN2Gcch0DYI7;Qa_pPEe3p9NPJ`Av&jZ09Pc*OhY1@lY@cZg7kmVJ4# z55I=6f=9Yb!Jfc|%7(z1sJCC<`bs584p&PW4rk$=^{BEQ;Q!GO7hFmC5*j9K;SVma+wcF3Hog*Pcef+`vEXE{| zL-gV82lihXzU@3Hct|4GD&H(O{808G#ls@G3-Vqe&zksu)-^AOFNC-<1bpfZ+y2!0 zNrb^&(Mdg+&GdF-I7^5S>+S=(u>P>!u<^TI3Y7+)N)js7a@VE5Jf29a$zYJ|ROeL6 zQP@bB47{ZHOZ&EZs=~sjF5;vcpKBGKs&lwUG8nrWo6ek@5u7vYFJW@jpA>nD>9O@WLn zeG#{`x@EZ4^7Jv4S7ZV*$+I0bm$G;>tP^Mx^)ivNM17pS?|TI@xQowLGCOdg%L%k=&cA3u6x|%5rt805vh^yqIx5Oau=+W zoj4DC1cpZ1JfinqF&C~RU0Jx2%zBjDQxI5WRT!gRt4&st^sM!5sJ~i-T5M=XlsT;h zvPau2(TdDO)fUsD-pgB4*y-e-SpR-1X1Y9=`#ae{!EmbsYq3so_zI~z)4~0N6GUlm z(2}9Gs5KYo7SZIayFW@f4}higj|K1GM_uOlmoD6z6%aJ*T9EmkiPb?9~? zD|RfaU1eJ(8folH%U0ALUN|A_LPsPV=aMU|D&n1>AE6X3YI`TuraNgo7J@@}u;N<^fr<*QEmn8=3h(SV#yIF&k9d^J9&YS? zd+YtSP)gU(Y_jBr#bb}hwz-BcoHn2OD?OC(=jdhFewy~+_Z>5ZH(hhM`> zlJTKDMMs^Z!e1*&!^+Fomi@*zbmAoR&BumgPY9J^$Ty4Tl$T!&^*+9)evO%O)Vic> zhp6X5j@Y?nt%~Fq?Q=DLH9FFS(lgQl^@o!>VpoqeH>SGAQYr{;Cf$_XdA4(Ng?i97 z?^r5T%3Nuo?!+zut#M0kO>PkpA6>bYjjq1CpB?kmhwn9+3F(hBUl`07swwo^73ewY z1bMW^MeWy;`uYYB(3xkdrLd;qTzGco-v}m`2D=4eH;p2UA{*oQ_V~7Nbw{bGDup>I z8{HdUr)tQ?nB1aUqN1W^qUE)JhXsa3e!kAMc4T$zppkB4sYjidIe(cy)PAv^{YzC~ zjFZTygAX!kI|Dly<)rQYjHo9kRkDYw`}XqL#jC<|FS3${bIWiV6$U7@hNuBjV)Onl z>6-e(0QvA2Y^{1qCBp7Eu}xC#4RMVc1leP2+l#ctvu-O5j*Uwi2mHuz8_iLEukOj& z5}jq8*dj$+=dI|iqGOXyodfR$@kZJ_AwB2THKz+%ROS57xnIu2;VPDPU3YCSWYf~o zY*4#$T;;msadG1*_E}ABKZ=WR~r_yuj0zEb1(K;4o@e4CA`v z@>*}AEuE9Lswvb~^|=qcNv@tqdeHW6g_p3Sq$^@^4QGWr2}tSF&JjP_6MIyQBWE^H5O za2qr5c(SWY?Ec*km7g6Hp8()=&FH~nMFhg_76Re@7JgAsB%` zP}s$kstAK02o2?Br4YyPzxaye&)}Ogb`P{15Qr0Z@qZlr762;qw3m~Tc;Yg&jPvs7 z+eWed22*w*wx(dx79b;)bhmI#xsZa@P^RdDINXi}}A57Z=-W%1G++XQ@jM-F8Jj zkdjiUoc?Z@k5M0t=huu);*E7#(J-y|CAn=sHjkxrw5fQVpX;#ugLzztLnc;-elAEo zy02;Xd|+%DQA5Xd*{QrK%gbLw-}IVqfG2RMJ3diSAvaU}q5VWcVZ=5_w&Y+<$iyj>dj-cg~%LR{NO7po(CnNea^LG3JpUxfTFQusWbFr|?5 zY9M)fAIoV(OwvgQ1V4$|lP3jdrcPZD&0f|!a8xWu~f>UIs9S597QuZ zee}`yh!E8@hXX;luJnke~r8#4qDr)Lr326r^EE-jHr zN%?i9uL8x&aUqsqDoSJoAqa#L(!s<4&GVtCue_(i>LAMqt8F~&EUh(|^|P%l@Ym-T zVA;ubXK zZr?&|c~b#6t)8Jx4`xxnDZ93^uIVtZ^!2%ed9$_r_@AwksD6WLeC=*)>tLXzwVS2F zKBPHvsVOgyo16@8a6A*de{e9;IxJq}$rI^L3_a0m)_qqg&6@Lw*8&K;yZ*dkbs3Qc zW)9ukf7>BjG|8Ih`=_4!t@FA(!0@`ES5dDXM`T_|=IG?Z7lLilJ+ z3-M^U1JofWS7YrDQmztYU)X%vAC9-Dt(?ARLR!)Jr~l(~ zNndk!va$mbf22Qd)Bn~p__LbV1!E;5gM{`7Xog!qz`Eu)Bi`+@qkwTT)NncB?NRie z>th@F0sWlC##onw|=Pi_CFbNh+Q#bvK;AH5-9HQ?EyK7#R%PmZ`Oa z6$>S35FH}WUd&j9gt+584DCa#(bUq^0f|-zs}aHr+&RC;h8e+6n`~8>si!u#g`nmA z*xkj@bd$c(RPb1@;Zm$=MBpxK&SwBg@$06RhHa(Z+fMma58nAQ&PeSae7WLR9!}Na^c&} zoiFZi;j87}IB|>R%$~-^fz;87PZ1ryEx_e-reBrhtRKyMbl#8UT$)W~-ObkT&aXr6 zC+-ftqs4o*i!VkXFB{cs$tx{w?D*W&xbCy^-C*cIMEKvO;B2mpIQ!m5{5Ugx6-0Wt z{|E$hegwSY?`P=Ke?Q}Y^5$+w+(dX!C7K!8XJ~FuF}qdrHw3l52ku76$Hp|@g*ro(|{k)=lTds#q zgok^%DwpdaZ@3HyUJW$|;Vf4X9fsmkbzd0GH$a+5`{+IN@`-A68t)k%p083;bz?Y! zgiZ0N*ZGe%#-BvvLJ1NUKMduoA3syI4(|Dit$4UD#?Ad#dquE}&N=nv!slC-s zrb1mHko2m8?dIa*xWV)K{<;pROw2EGILX3ANvOcP!o|@1d`5qq+DHD9ueVSU}&S(%cfjeK6onL z)0dQ@$mA4$!7nlBw!M>yoe{ zEI)mbx=!?^15wJ0$(x$nKMcgKj=N9$l9Jl*3`TCudL5zPHdY>CqwQB54c`0t9q)`+ zQeL=ncfPlu4mn1yI2RYo?w&)L9;k-oR}bQ!tEJ@m8R%;1zDwd)ioFcBPf$H}SvNGQ zi!xC*RWYGO;2MKr$B#tLMcp^*2@I)me4@EBp96tc2&So<#51eoF6-8e5L3G3xEOPEQm){;OFLiwrKsz788^#~p?v zceq_g@zW7fH#WUj_dSsCl=-(!aqXVoWXg1wB0<7IFN#<^;@qTkk-YEx2oL0MfwwH~S5r{LG?Q4FF&`OZ3O<{)USN zf3>AV5P$y!P8Rh{6o_L}C8l(bIS|(vL>EEW+!`y+5r8-&0>t2HJx^1JFQ7Y1@wA3y z1CIlfrBD{FvTAs-J{@Y~=3cfo8h#0KC>rCkRT_kSF0OjYmhFo)}r7Jt!Yz9eFiGYBCf$;0S zh7WncX);2sUY%57o$-__G@_rb((`@=E}VA$o3xQpUA#mPr&;`XdY{GO78Kn67}iHA8H*a;`+oiAd3w2f>e6*Lt3(QAQQNx?`4Emr=UDvIk~I>K z+E4F0Pb(MXLkMR*@Ivk3L#f$Eb5l~04q)|Q77YNkt9^_8)bA@L1%@BHLw%Vjy}Kug zEE~6&0Us|DgLxJxSfnqZJy_!YG-QpIaxMnqxdohm)}w#K(_ZlgDv zW7y(AcJ}`K0H2@#Eh+!+nAm?F{-3AdKmW0Vf49DW|NQ^q*8inT|3_=|o=O0z9JnjX zNf=mu{Nez>0z$;g|NQY+og52fhJG2Kd2l64`2gcV^ho~a=jSjo033AZ&)x6U*MA#* zj18sAvx!Ti{1e3<|C;U61hFXRu+yEK;Lk&&!CiOvmr6dyT2l$($30U3kQW4!Em7uY zgb*J_ARekz%Prwi&mO#7pPljK93oyGK=SwtfU4EA7P0!zpvEl0r5&ud>+9=-nh*%& z)Y|dU8Y9Cb_4uBQ?*TC}F^Hakln%G=Hc;ENYy!49gLnwkDQ7Z1B2n_ zO<(fF*rF!F`sziVf`vNEDnSHHLC zPM`AvSJAWviPth|GL!y`xoMCS=05ZB^U4uWWv05MMr2uysqdiLmLNWuV&RX>N%MP$|x9wW8& zAqT860{E$L#0ELCW&YU++w1Co8c(h=0QURRY}C}`bo(;?_%^-+@T^Ct3eF3pwZ)wc zg!u6@4Bvm?U1)PyO|Z=Ftd-0xPIi~|9Ny^a69`Ja#|? z`eX0}iRMC(qnO%BaPp8&;Oz+kCjev^0NDS221xPm=RZH;KSQ4Mf9TRbkNU43@ojd)(#-o9^XivP%>k^pwCPRlpjdnBTrE*R3cxz;wi z*g5otQL!L3U3tU3_sByPeHEWPfGU#En1-Z3=a+WJK{K-#!V_Ref@>*?hX3J3$(8iS zdyTZKI^{0*Qc`!d^HXY#`VG7)5>EVBxYR=QjZbE0@JM$^e}62gY95AX&Y@I(j%yZ6 zk3r~m-ysQ13?hxBk~NywB01aj5a22Pm;Cpj*+)wDCKX7^b|s_<-UGNQa6+rCF|)TX zSw`Aen=)N+_ZKGH&(Cjn%vFmMMs87>^xV9w5VrwTeAIM8#Q<#xG9ImJ)P@Vx<;l@_Q%7SGRHuMaKQ zRlZM2QwrL86=c!fR=t>y8e67J|D-|<(o~vlV;VrYl(k%koe4O4*2N`NI29w}*;IYL zUtGz=rogCv-?;RjobLbM=nZCrpTPf3dih_<5&!a3JWl$T&%fVN#h*lU6hZRnAIJVa zA*h~pR}sh>klaBc2xy@HNkl#ez>i2M>F9oNx4bN3a0y6I^z`M&Ya31rC_})hWk5K< z>rCr|I!5(r+y1Bi-V_fKQdY@ooaW|_kqGVspRfa<`WOL?0_qf#QA?|EOvr!z{CFEd z3<(Enz&rtDsd*+!idPHK9v=41Tsu%5@%O*;jsWuR`Ah(y0kcGg06N@j$#+wj4d@tv zQxDAoQVuWp_Nnz7>+#r&;e;?0**lDhbbB~69Sk$#w%gCcw->r^u}~s9d|Jm5@!_`s zraiSgN_sO|-e~jd2q4vb6FHftzw+|(75@xJCQlbJoH+#oWWm`jvztLugwI)ky`a^z$?+7K1da z%X!>W0R~ZnredG_gCNo>0-3{Dn?`gpt-TL#sko;Zf^l=S1q*i~`jhI~M*TzlhEiHN zD^6rT3?NW5Kil~pFftX#2iAhB8fzXg%Zlv_`q5R#2aVb`di2G?W-~jV`m>D1S4etIO^b|0v{A1;@htEnV-97x_YdT) zp9DFHTA5*+`@MMz%(ExV9`f*}DdzSoT3AR~_+#_MuG5R8@&2-y@)%JuGU^tXdzszJ z?zz#-eD`jkpI^4eO);Cu?Qeeb*tz`{Q#P{_J*`Oo2m_aiQ?#qiO<4Yj-9}6S9${ka zwJnUkeoc`ztx|1B<95#Yl&)lG&A`Cr%Uky^pt9~;m7^5rkZZctBam!Wqm{~c((?c< z523`p|9AEId^QRrldfnv<4)AjHcDUwAngCBK6kriqycdx_1N+mh>FE)!N5HG84}9o z+P|E!DY~~5AfP9#{GW-rAt*KA85M|tP0_=IjbCb0!8~7P2#!HbIuT?4sl`0#Jc4HK zktbAFp(l84jR9jPls~Q4fGx< zsdC&bN|0m*XgX8CH&mGMS-g_>uRhc+f_VhAAtyHlK*vlvPe4x&mSxUPNqKm+i80Eg zB`=Tb@;QWE5Fn1m%m%(i`+Kh)Bg8D$nlvUVPdpQj;! z=m*HL=?)mKX~P8lz!m~f6C$tX(8XEj0-dr{65U3L~B__HqWRNwx`I1w#W&Dk=Lo@k_O)afV=_Zs?~=9|d~wIke;OzD4Dm zy88RpZkO{dR1*r-ksv;VUjl)tJCYCxK(=mJcWeD)uVMGCfe(SoE)x(mG|sP+63YjH z>{!_|Fz*qg@X+4gehSh$*Tf~TMgVQ4rJy$gpbX176O5@A-S2a07dq{D@XYSH>3;fC z%Y)tIdLkmiIssEJ%35FGfaIa(IcslDaJoD)@>nbvJY~K+>oT|c3WS}J!8s2$28Qz< z>kjoCfW=|$G6Zd3Rz$WtdFFA8SXlf?)lF*VsX?4Jf&G7VAK-4r`f@3XvTMmCkD^aL zejIf~%ky(h&0S4RG4H~{f<^bBHy!E6S&x_xAhs9-gc17_&tQ9#V)OGG{2t&5<36O> zT8RL2{y*B>e~pd(rzMM5afzP+^O#?QB8yaAvAqD~Oyk<-9K9a zY6*&v&~y#1f*eoUnV(UCvj_9~V3(u^dA~Osr~;4?uQn_0xh#EK?9Ofzx<39H@(1y% zPPGA>I98M&tbQ;}yXWpxgBj+Z#29$dOeV)hP&wd-5PiJm(fxP_ta5m{fep^@hrN1k zNxre{4w)me1W!fiwLmS6wdk{+fqo%UBgA_ACcIAEz@lP9CwHlt7--jk>E6J9shkP4 zH9Q2WD75~bm6Bp%9?A>OxMr!DJZE*v zZd5P@MIHWmT>uu}ASeUdxiz?J+yKfB-kx8!A8+qemjaEtY-BwXmK^l+Zo@(V+S34k ze7O4#l=PsI0mbxY93+mw7V1Bv*n>fBB>~(JeUZL1|D-2iiW}2p0SDzlQQwx~(Gf$~ zbE#MZ==$AWmQh(01^~nP`MGSq#{)Ke=?Q>1h6M8@VOU}sUVnsUd-libG=PW*Pg4bM zTw>AAGk0h)P{y~u8=NpF&mhFY2&GGfxDfY|N_M#M-k!$b8n!n(Tj`k6^4aq2Vz@wr z5(EBJ{vuy<)3vdR9OpItck!~I?n<*(jtRcsVPsprA@f9oUj6hB&HCFS>JrKom^Pst z(>2j|7h6!RmB<6baZ&bbZhIzk>k|VN?nU61;X>;GN#BzW;Eq&y$I4iR(9SC7#Nb?Y zIdBA!E}vWg!;3ckYNTlycb$v1VUNuKm;oYD9UFrl_OtGL>l6GQ1IucrP0hPn|AdPx zDk{HA5x5=L1I@-Ji}ROvrJ99j%YORH_T10?RFquB>Et4wn-<;`obXzhHk zH`|=7>by2KIzyZy@WJ3OfQI#}kFl|X?E2z_nQbecUR5{k&WtbjfGiSw|HjiZy8X6c z3)3;(K{;n3sj0O4zUY^8)<}&$Jv}|37Wex_1$1Eux`8?aQG3xpZkUH*{tAv|=egb*6@XsS_unqHdpNM7~z{xN$l+t#n zmXZBrWPl;+=7knDWAOU}SYRqDIV?-}j@ztQLuPM$ogNPeZj{}{YRs}DTlZSqc7P>`v z-t5T-^~FWi$soS5-~hJLYh6TXik92-Np)UFbHnzlCAEyETZ4d{)^>(4vkcSL`kK}Y z489@%8Y_D|0g4cuYs!@wlz=d+(Whi&ymuZOC9lE_Q%Gs20Ubv*`s%lD!wh`R9_Q#@ zK9LAi8{Ikf+*UeA!~cDJCT(|Yn0Coaw`ai!eYE9j-;X{l`nf-C%XbP>wHHujrd#^b z{E3E!dp%gcI6alXi|5OOc}x>iMvcKir--lcv__c5itxm_n&=7d>{U6rt;b6&fKsHG z-?LgStQu)+1e3`X2F|>T?;|;tTLN7TO|F09HnM>_Fe~>g_>xbZHvLG@sM2}uNV8&A zOc1#YNki~(6=k?CwuDmfrI{*6)pSu~!I?jMRw=XS>#_&tGY9tE6WFR%AxD$ zL*4?+bhnV<+SnWW{uueoiLt%Mr7UA^ODlVeUg#@{ZJ8P$`fN`+UkD5xAHV4A=_vEx zh1nzfdY=fF1=FbjFPFI2`7gc;vU=|ArFROecd9?c&H+&_XEoXX8JzYjgD{%FZK!F)u;E@Po#XinlX+7GyFBWS;|5}Tc|R!yZydcGoA$(T!&bu$i1lha=wg7TcWQTx4$#ky_Tx<@sr=zXYnTT>PJonCaJf<+s8d(Qsz%abOY|pb z<uQg(J8nmou;kD;l~OD&Czs9jUO+Vc6-*3K__@YSBn zH&se9`8* z8tLm6^0{4P7lds}ySoNfBbqChN4~hBeg1vCA%Wlu)n~?&lw`AHV*n&n5sXoRz z8V-5xnoPQR9|ye;qU5|!I-m1J(nnM97blgNr^h*3i*M(jW=!_brN<}Bpy+5LBaz2+ zrpIB{8yI)G_oU&$s59d+RAiL`+>3Y8jrx`K1G;KRa+cn)s%Y<7>F%%VedIQtysgJr zE&P7g_*B@B&O|;TN#8oS6JeI!suX%JRnF1A;9=PEITffDQ);AErvhSqo7nLo<%d2> zQduldh;8x0@W&!%u@@NAd{?WMAdUJ6c2725+4V)--n$@JYWS2RYi#$iU#xlEQ|Hl+lR;{i|Kvl1csmS{eQ z^dXh8L@f+hqcv|J3+sqoQ%e*vlv87PGPB9q^>qh~HR(j{J(^cC$5#e3cHi!E>)7wM$iveY-3%1zR8}kPAK& zP0JL%OlobKTg0YArY6+%`6Px42r}C*Wp_u8kL1KCG8-*oqzX@AVZ`l7|w-@U45kn?j@5O6L>yis;2gjYXRxxj3OopPhSe z+6oVvW_Xl#G&-ncBaOFtVz;7Dkwc7(up9yh0DU(`xG&VdK3Mewc|NQ$0I&yX>CmbN z-I-(&V!uB;0AWPeVN?Vp^vS9f@26FK@BDlQUY*UOKee3wAHl<*Qw8F zVRoC+g(R1qNP2$T4kJ)pQCicOR}pP$Z$~|=^|{$13wDojjyPOKYFMvZo|x~(=)lUS zeQt1|wt9tyvfSg=VYTS#ndR56^dv4|3pQY@639pA8=<@baB}Y^!#gQRG6uCG&q^eN z04@x4yQ>huuSYP^8;orFRs)2!+bC`~W;Yh!&Wa7F+VFfQ+A5n_`*Joy#oDyBD%zt@ z*jB}}xe-jpz2^u9fLJa^vX+Bu4`4MZW{|HA02M)zBv+1>wzlQ1cV|cy`;UxGjT=DY z&>h0q^Dkh19BiRn_<0jk(*{tOANDx^yL_Tt!r(Sgukw6dt7gU0{`blvqYeipj>k9Wp{R5%z7~ZLP4-r4pWbHSg~Xtcz%vyFA8$8 zJA`Kp7+djjD@s`i%l8u%(nYkgd+?vKwVwq+69B5^Jn)MX@Se$iVk zRV6c1v`hNSDhK&~W*FZS)rUOY0K#p;Xwlvq~{8gc_z-*eC%1UNGj& zG%eF5INe6wM!nU05UPyaERPUHmKxgK-&6rBC>t7J1t%44hr^W7FZ6xdNpo&uG;ONm z70O4SBPjO>Rs?-T=XW#p%WlJVAAq?T0FeMSC#+d*=#rj}QZlGmH7wqXhA!wY5e^$E zbs9rMS1h(04!O?~;m7@T5Xuy{_r=goVL#;%x5)24@O?n}Z9xkZ5sP6tZ@*vCZ6f+e#fb zu{jn5>)+YW)H=srqW%%dDc1M-NxJc6Fs90SXNKvL_dP%- zg?C#dVY3_kN^vYHI2}pQ!sqw;8>$f)ed*=bqF#FUbT++_#Ly&GaX`rId1KTpdo!D< zP1fLObo?Q<+{mW<&Wj{r>N5R-OOBzv8{o{||6 z9mIG|v6a<7aefInqtUVDuwjGhcn-@!f@CSG$1Gg4QZ=owHXv0-_S?P$AY%Mac7zC(Oa{#t#HvO@6EP^4GZ;EP{Ns+SnCg7 z4=D;OVy>Er;N;UM2UBE~A8GX5aInYBh9)a8qd1Jln6!6~ms=oD5!FruEP6`YaN7ZF z6a-$7xSHIY@9ytqEn>HDbr>C8did&S>38)TPSaH!K5NLe%8I!zzAN6TItqW=xC+3L z6>fEicehtfovGZWykILMkhiP=?eAH#_S9o%`HQJOhZsyhM^X=^Sx&Ea|UMx#G zHsf0hq#I?7KC!OV9@P~^MJ1JA_?mWkLQS#x!TksnwGcKU4(J-(#|3$&fV{1u7JHR|h}X%jZQ)Kxxb-+W5S~AX5PRFi@*OoE7xOu- zj!+$MCLSA%yy%vau09}lThYX*;9g07#JrQ&5zPV~)*qf31g2lN{(fp|C|PW*#k#SS zLPtB^8tmu!>T(O&eK}iqb(QXmH_I#&%o@L>q?r5VHes2EJX5-j=5)iz9omG%7}|b( zTSk_YXyXg@3_=HXOzh<}RY-lIShs)aJ^$I3Ws4brJB`gR|5~; z8SENb=@s&J<#A2=%+ZTJ6LaX?kyLrC-+j5E*)rj|bN349ytS~q8!5=usq^W=6io8BuXM~p=-?^nb55j8lG%Ce#{s#L zjOftz=YGR(jk634lOP3?=~Ki-i!nMKF0~u=6YGF4zXA6RU-i5JX2ZbeC8NvbHEX{= z7$xbfWvtiyX)71FYvSy*%A2W@VKLglH{IH+WG(K`Kwq(*Cs6&kFE+># zwhicRK4ql46bWb+EV{u=H<2U^%4+;`h}by3vC(^XMXf->HeJ*B+nN*fMKZV{c;$Ej zcNX~x1tyUI;I;hsRSVT3h#*g5`d*a_MOQn1OGx@+e5HKb@i#Ix?-AV>J?(5BTZ{pa z)w(Pj1LD|iwheKh8Lq<9K~F^`lqW=%r`5xFxBA*Cep|##6A&UR+GOWO4)=>E_I(*a zvI(DX17GGCjRuf*^0<$Uwl8J%+JYVv<+Z@q;k%8CeCOl7KN@?aX53KD2NM}k2m?R~ z66R{Mqk`{dMsnT9esU#9Hxl6vv2oqMjdA^5beRy-0X?hdE@8)lYa*z7U!`UwykHTy zwIwr05-%I26R`pUMzm!$>LxvIa2%P~oZasxSquO#@PJAOU0zp~A^Z8_-n;I0+J`Hr z=o{j^T-nCPHawTdb6L~IZlf-A`_>m%9q3e-7eQ?d_UOS*Sd-Sn`@S;<5VSz87PL-h z^)sY`5-5P2#jRcF*F^LI00L;KKZxe*S0vEh)n<2F@y5f81rXde^3b@NgHX>j3t89l zm%EIt^1qlsFaZYJ3qUdq+rTr2mpeR#Gd!n$+q;ehVu$t`E`&DLcP}1+<7!N< z!W4ZBi@Y7N+xI{3BHWZEcq-RykOJ$qUEhqhY;P~1*S3xzeAqKfi~|MMXIB#GEkQcK z(jk>VH#g=yzbycnrfj8q&J_j=jjzf}l3k>_xi-&WOq~QRN3}$+@*OpSI%J?Mz3YC& zW+#j!v#l(srFNI{`}-YHjP~t|p`5e0UBN>I?l`QH6|yLm5CG42|Jt(%kCicDfxQg? z-mQMg(jLuSRq1P~V7pvn5lE*ov(X`ko|2(&og%h}bL=|2UI)v2S5GYvX%}&%gD3{h z4%*Z?G=HV+{(Ne3oPS7&AG^miz?UF?)J>FHm!Bu zJ@}=ru0?-}eJy}ITYcvjM$r{VVUJNhPOQaghU3cDYTjRs+$V(iVWY3r=LmVtIv zmc*q#ok$WvUj2D+Cj?K6O}}q zyB9JAF2TcXzO;QTd|b2Fj$scxj&_$@7CDFsdvs))uF(M8hW7q`=!J1dh4DOSWgYEm z0B8Vs2_q)!bNC5m4J?7YoV9-QNA78xmAY0yTVm}RL=B~zQo zy2RlTyI0r5(QKGQ#dTOG*{T)yEM3bV8|^dgZag1!Q&qNM-Qcq9M!wjZ{WWEiy}Auf zfY;q{!uTz}odlwpuzPpW!E@fi3V=pc%8NO`z>I*LMPT2YGJU?3w_xPk@VeQzv3{+R zT<7??arseA*xSaWqzi#3OuxyFzLFF-e`UV09Uw~O#UC%aBfL@Tf@wSs1LKt(T`-!q zE-KtlY}7*UE5vL^woA3VPZ=%{N|S$>koMf+Nw`(8Pg+{o)g90ruxHlb@#hV|zVxa; zoxOF_^5sP`!3@K?R2+IFa1TMAHwLaHu_UME9UmS5y zJvKfE3Tru@L!+df2eehn?{x27%j9IkWipB!jl|`PT8^0`mvZmPq4V}Pz2j``ihwES z*ftMy?liR5gS2tL;}8Wq)aR4*==G94%Cmn#8SJa6F`jt!FDnOs+t%vrkK)ahzw*;P zuX;3sI@=#m3`bp0eJBlHdk+l^$d@2dXLiSjbGp7j#aTjWvxPH}f$wsbQQbF1`FN;b!HAkFrul(`BRIlEVi|-R z(7)d^7aVEd(1&%&7i|WK;#Lcqs$@Uea}*yvlK!`kcruM-!r%~lKaIOuW}AcJ2!gBL7zeUCzl=haV6~l z5T%2)qBUf|rgq;7k%cdqALjg=;<1p&rfZmkqkC657h;;>f*u5%< z@sIKe3mW|-(5;Q8C62BFk;3DvYpTnFm>rkVv^xhQA_>x z&7{Ie!Hv|D$71=BVrZGWWw$P!PBtjFnS6SCsyo7dKS=;gg|KjYOUhc zAIfC~1q-0m`y|aw&+%Eag#m|Yez7H?C7GDNKDakcf2dEhznyI~@Rj(IsHti6=Yp|F zB&0#kErD+*@&{AhV)y+uSQ3BFAXU`0(CZQQS_DSbJ zy8bjC?4VC!sDQ*Vtb$bu$O74%zJ0#3V`Hz&OOuu~eJaz>g)=HFZq8$(3uq%Bd7c!; zNsI$RcB$=jk$$CTT5E)J^7bH%!=c2o_0G5GRw**u-3;d@hFpGcJL4n4PSo(XP$yCb zp*&C=`Huogvitz_2l85`U=)z0>i>k4L#5jJN>*E0P`H2>vi3)!mkZ5B%#lut7V14& zTC_`R`$cTd+U+_aUNT!2YY(-Ik(bNMvuCb)fENJ~h#RNp`bBZEV5as8c|`#7PYtpo zg`93^<;F&PSvRkA)dW+wtpFCXQ&q_N;`pGz`t|Xzb0H4sLE3)d4V$h@KWF2J$da~Ql)j1Xr4 z!LcxMV+;V%478K80>JA)_ch~%f2&|ozkM~DWb?fqt z&R16*Fs9(t^PtoOG(5%U!TVCppLXEJBxPTz>sezr%ezroxY;2`90GQuaHAnpC0v+=GQWw!P5-ABUa*Ob9p%cb{mw>}h&QQGSE{m|rceBM%WmxY1hq5cIJ z9pcNPvT3XKCqnh;hG|5(8$B4H+YMB%*1xR=L1n|f7IUF-2cKPb!5hAHb;N>OOE#5s zIH}t%u3lY*rlg9Q?J#&zR2`^WRhSwJM?a55LR353KF=qVDJwON^l}dg>6Zck6pntk z*w7`p@eC{ujeegp@S+3uw$`~lo9XNVDgS>Qib zz~e@WUtUGHm1^GGFGoz-jw~Ipa0qP)TAxr;oEb*bp}C;Xy4x)s$-W{r3sfihq~O3J z;v)h7%09(mG);!B-&(~+93C)-&MuYhR`p?8`<*h}bDD3i1HH5>H%ABN zOUVE2bg;_pPG+M{8Tq0@`%%ti$G($I{F>wV7Xe-xgX09mfD|~P(A(MP(8>xzd+mQ6 zMvk66YQ0{xjRBREThOYgK8szQlJ;&2{Yg;ZI5#Xi89BoI}WO5 z7@u?gH$VS9Oqh(}GwTD0jT%&^%F6rPLB>jSv3~}=jtxZ%fX_5=+5`RJEB&I|u|U4! zHGs$l%zR}1u(qPKu-uVSL24;)=UzG#fgHoBJJJT+}Di`5b0K@;FM@UP(3>X z${pSF5>|W-+l;j;JlpIC7p#YpL(A!)Np*NZVZ=vSct_m`;0(^~17hfbh7b#ybj8A@ z`slxW_T<9h%bs3(KZ3GxtHG3&q2#LWM9UhGP9_26^k<>3Oz#)7bOv0+d?^oT3v}IW zTgi!l8^~B$lSl?W5Ph+duA)J^KKj1+>eL|4A{Me`R5z zL7}J)ID*S7{FkS42^(UXb3f|~i6WOfG;s~p%;^~@a7LmfZFuU{<^lp1AVTYjgUj6R zXPf~41`Jl&Z6qVWr~XqaW1{D1!+>z+aiM7Q!O>2Hu$!Hfl>BHMSUcxZ!n0U9F_4tu zWBumKmRewr`W<+FI-6aG`WroKtEj4}hYryum>x@|xw(AyR$I)k%ytF_uyC@!gbh;g zkY&+J#-ZELI6>^{ldd%cY?8Foxjzze?jKaCKm~;S7^WGwB+TEWQZ zIck>KuB`|e@>7f#t8t(2h)d|O`3J(Xx_F=Bfpf|uw6Kd~GPUUv#*l5tmRW3dS9wbLl4n2BH zDAQRR1Zn<6cT4A@?Ln}|lkDu&mw?q4h%QMT1PyC@>E=&8QgzQ)QlN8GmwHloSOwpt z@b=8ii|uRg5%;{C*{<$oEm<{UULiD;&GX=L-LNH6SI)K7et!&N@ZTZ*XeVtky3F>! zV$gx{HQUUDr0cmv$ps!GBgw*AzeP#^?$K@9pTleUaiGBgT0?ckLHqBKU#p-RPA2&e@fQE4$|wkxqhTLveY2s}r<%ATVsDR}v%NyXRlg`urk z<25Am&x*F1KodfpytCLP;QvsR0t_DtNx2i6EWpU1sG^OIeGLLPEyOCMq%8WGen2?@ zV%p6zR2?vRoulLJ?Z2~0f20pYf+SBb2|)eoqU4-qq@*?`?^g?hI4$zpRXxL{fZ<`2 zCCuTRY%CSVa_(}jiX)pn?i%9J zbzqmbau0W98(Ebr{#$Y19o1CYz3W&%9TmZj2-vV7ASxY#BPfH4hzckOCyyG^n9NX><&*sxRo5?U?XT7QmBb{c~7lByA4sqz}LrMeIxa;IVU?Y-bcL6)bq5oiq zfaCtX6?cF-A~Ixa17pCuksiDKe&D@o&er)~)(5ShN0>|W*=wAvAlnd`GfvNU70qY8 zjYUMnN|f`GRl{eN2N!>IyIu?Qa5P4HK98JyH185gFi9?=Xx^)=gV5w44OHNPHxnFT zhluHI2#O<982A^(^Qd_;AR3c$>a~+TJXMBvPEM;qA5eEfIz4@SMmG_;e*&GPv|CI& z;HSPX2`tTThb6w(AZVkhFRFi$fwTf`c@f-wJVQSlvEG z0-zBS9$aw(AGEJPj)mUvtuEh6nI&j3?PRYof5k3VsriE-rGru|pw)rRO(Y++@R+qo zw7`EzXhGg9uvreOrzHc{+xt*CV(AfVYd_ zhvGU9-O?9BX18p`a$Q5~vRZrME5efIRa-g5rdau1iyA4I7h-={@I%U^x<#&vgLhq8 zdq%agoiW1K3)MWGMd_YHhaOC>_>`M!s9qI2i#EsOP3c6GK?0>Sw|OuFyxafWBvvV| zMWHF#cZEZY28fAS7vCzOEasg(#X;5!dvb1f?-b@A0M=LmBxbQimnVDTr%LGkPlG&$ z2Y+qOQR6ExG#M}T=`Fr)FgynThL7s zj(obSnncgc{S%nim4{VM2a_nZEx_rQJ}`&VBuVS=y_pP*%=&jqfTdd##&TAc`ea;UF>BVH5Te+mpQpSDkD?x~kn+L#cW z{IWOw4^~r3uu8ItJ}6fh--F&a5WLaA5+9G{|3`6V?3Rl13lxRQi)EKnzafwMg-L?`(P=Ua!_~H-q6H=>c&e z5;hpuD<&t|>Ob4ZS$k~`y8vzy@v|akDZ}m6r|;7e-ye?bb3Grcnpk>Y9XnmfO2 zleL~jdC-~N`X}PV7lu-Lmd%ZFcaV~>{r*B69upePw>5RivRe*AYJdeV7{F~~vN8A^ z%!d=s_oS1WPo9M)X!+lOhTm}EBE0Dthvu+PLN$2JoxA*os@7D7Exy4BQ}PoT+j?{D z0)X0AA_NFl6012fVZA{#qmR>Kzdk5a{QiRmJO9Qgn zZ-%7g<|qUtN3Mf+LUYZB1yni3WC~l!si&%u|3fD~>U+%wqhen(Z1;xZ9x1UPI1pl8 zafA*HEnbg;;{^b28N}rG%UF%C#`lJw>Ptm&6Kq&&3Zr;URjf7A1Iw=v|6FrI&WEL z>Yb*hrXnbfrGGj2)v_I$@_=s|`dGF{Mo7!Q=wOECNVcZd&EK_jsElCQ#|59_tizII z%Zt~`$jZhzA8oG{Quf$2-OrXPd~p5IFalB z^bAwvs60&!$m%s-cH^*b{e?zPj+vWxQ?TLk>7yB{BHwyz*VEZ?Qts^W2?pn4w;`v` z>F2w7x57Oy=r3FknV(3mE@E?mS6NFIhD=DcTNLR_39VdBAJV&{IuiF}$I;v>FM#NS zUjF3UNbc>f^1%+q3wHM!hxa!A)@1rXRzT} zRTM{A83_&btFL|kVAlS#Agu-~8I3PHa>vemZK$4l?cPMY-Ew&nfIrcL4Q@8^<`xIv z=`_C_a_Wmm8Iz@gU)F3YA3r8ji=rP@Qk@-Mg_h*T1#=!|${w&&OwIA$CuDcTKW)#t zjhmXETuEQJ?WvCIHVJI~e16Wl$^g37gq|DGoWYA6pn5>Rg~DLGWbs3 z@qVH8%-mu{sdR z*&s6L=S&Lv?ds9Snmqk>)n%>gd{#wBot5%2(uPEGplyE8M1*YNIA~z zuZsCPpz1qfoO?nyqCawIPgOH>xkizenoXQ-Yu8vAeNy9d`{WmqDW9MmXT2yYP0mc$ z_r31Z?Kh>p?$mDn5RDQqOOyrU%4ICK?6?i`R=RgVvjwXzqoa%c;+F8QUF(xlYcu9} z7P}1}sC{{yH@Ifag2c~2Upbgz#Swd@Nb4*shyHtC*6J+bguX;(A9j-Ujya4LIQXQA z%aSH4HS)?r8aak3GY#|IN!I}hbF7^f(w}096yIB-8mO(chW}m80B0D;P zs*PU}zh1n~Jt!c`mwlx4DAOO24}>eRLdd@5TvvS;w9Sw>gwjb!pj&A8aHWdLXREUd zI>hUjw3ZQ!Z0Zmol{Br=ln1LRk2}4(pgy*~9-(h2v(z-FxGK-~gwVni6UvY%=hJ!! zmwZY2zD9K-two83yl+6nBQnWQ6?as;J_y8(_y@`lt%|%EAEmafar}U2RH~fu*Uu!H znq$dFzh5IbYA5Z+oWAJw27Q#K`q@y~<(e@E?CmL;O%*$|G{&l{WY9WmXBUDkP!zhj0Z^>h zokG}b{4VF+akn2HNR!Yq+3ts^2^P?IyJR+W+Z8-&)_za5f;rLYAI#tO>`3mdYAomc z2;g@zbkl+;?Y4)krYQ8{0Sxr={PagzRaKlH!R>@HWR5uSKs%4SU&=oQ{@RVQ;t-a; zHi;Q8L*TEp0*YK+@g{N&KhNYqa!Rpq|GY2uKkGMhO9!J`+baA3^Qrx^1Yp*q?#T5i z%BJp^>-gc>l%-jgy4zGshM%APwbA8kOD;0l-JPeZ4lvVrbOnJDeoXS4#i7C%Zd2s^ z_{fp$2d8lTCwX92J&aXGicMx<{Cbz3O#7mp~>APku` zYikj*R}hba0-EMKH(t&J6=ooOD0W^(K&%wN`$M$3ggOJ;zwPwk5;=+z1|~s3_cO2v zdIr4>SPlR@905c*Wlq1QVFQh|7O^tNgExj?*jNq0=(F~Z8XVwozgMmSwZ>{ag3R*f zd!^d#jX&oD02-D*l+7R`BDs(5x?AlP_w%f++e2+A^NI6avEqFYz$nqSJ3NSi3m9~H zgh~K)wRP!?8o15-Do-B@JKH=w=`*J}u5CzNDK%}ZoYPwvU-6VX#C@v+GDQL)a#x46 z`mGrtA3zJZUYz@hWk9`5;4r$Pm|U=~8TF+n^NtaqE1X=%r<#^ys6>+j+FXO=7@1Z< zs2+JS&l;(%$eoVz=&Gj}%^z%Ikk_T0u}N?xpqU0sd&&skmb*^RlP?0k^Oe05!aP{P zzqSxK?0Xj|&tO0fFA-qpy{Fe!SaWZw6wYLO%FDpJ z^F(`5bgJ5>sXw=7!^ycMA>+#g-lvFwEg)|gf~UlAI;)2|R13dzh8mmR?ckj$_%vW; zP=JFeL*Gg9(t~*??Usqg8M#@p@$9Ph{`88%SZ*q1nP{?@lZ_u$Zz% zMdIK7F;M&YGdLk4xIa^QGwSjVN<(IV2LeTbSKFO=^yZ>{n#?p>>Gby49@7L@-YjKT z0viP5VS29Z?|>S1zhIpP2M`IF#rfd1k6sgxGDZOkt$cG2yVkfuz^P#tGva#gg>N3kR+mE1qgx2H|CM zr~7Ep8+HM$!pFVf`B-1GD*tMFn#73mv^ua-j(XTah#l=bB~s9#;J#zmGJcV=NWR}~ zf;N!X>)ssM+7vSVaL@k>4F>C^0KdaN7i~j?nSJe+L~EVdR}@01T@NmPU0<<7B2z$^ z^E7Dq<>=?=vAnLl)Mf|za&!6-p^^oQOZ{|A*jLBSPm2^@I=7bJ$tzEm@Bh1Mo*A@} zjmTT&Qq39PC<*Kw}Sb^wINptHS0;ww|^bXz)h$E%-U zrqR8BULC7RE?9kWG@Vt0BoKHRU|DNoeQnrgW`W25z5Hibd6+}0*#AcY2`zeCmteGl zCD=uaak+=?3{COK*uBI0b`~jzGg|JStMVw7B`%IcW&z&0e?Goo-hd`+27eoo(ZgoUSoj@x46~g2#qvss?o{UzxfL~a>Wi_9rb)y$xDeG zu}sN}I~gnNe`iOphf0ACnK;W*uIYk+7lTg*nGiGax8&Lv$J?_#!9`DXS1`b8s{pwJ z5EBsytqxu0Ucp(}fQmGPe}7CR1F&Qu+4a)9&V3RtkHI`1P(Kpso3?K$@gzSkgt!51 z-@am84|r9=911y_q3?~HmJGdM!NxDE~=&yK%xQXF^ z#4EgNP0DNH{4t-{i}5M^*bH^wWsOrc7Y8aFd@2biw+FlwPQhlV5H2g7b|xp)r%Ekl zjA}8{fVN8@SKQy@c+-c@o;!V&^?#32ZA+^RqKEO9y{liRk2MEY_;aMBAW|Skn~#nY zFiXFj3#x0MY9|LgbeOy`bMtWFN~l-wPLmW|h4t)WVCeRBQ6kgvqhFYS?1 z);)OudK)8sTw_fu6D}h+OVix|(7mLw$B(dz{zox{Oz>1*ueAZ~2vj4tOqzIVT4@7S zm>`JrHZN(xG7l7>$^*eJ(|Sh#{FTd;`ci)fnI4sLL&}`w~(?@ws`Q|I6{$ zBusaVHq&>G@g8Dy4X=$5dTAc589wE1K$TfAm?I@DL?16sFk^iKV^aDtYDIg4n_Bc`U5W_r+N@0bGA{x;Cg-sfZ_Gb;JvrYrQHPy4W7TABW{naD1ijnC z1vt>d(_Z#iI+hT07)t}YGM zAkrJNw{qlzz$RL>7@prgj$V3dwSb$VtFj?=Vau!pI>LDe%^^Bb@aRA5Jzi=Oyyb9s z0SIe;`8fazWAu=>Z{Pa)ZIhIQ8f`x8E+LUM+SgZ2Vn#JiS^znAG^ep=P2F`qhgbDA zqXi2+jK0XJ0~L0gcaCzWQ>S=lWg_lRnMF9>ZBjqV`P>=Ui+)OKFN>`2I6y zaPu8{Q_W$TY*m((*s}+5spyLnv_*>@#pVxzjF;2Tmj+)g6IxgsKpkTI_Pc;j<%Z^0 zo6fr4t95D0Fo{_S3c3ZLdxR8Lko*Dxu{hwc!Uh+VzeEvGtyb-Z(9MeA6rE7UFA0hg zAb9I?#O!O)v6%^FR3E=ZtUNKcn)lhG5FVxwjYWu1_%(=bg zaJd8P)A)K}j^$hBwu-1Pg?m`^A=Z1rX9W42-T_p8h;19EinlR!m}w0h<)Xyo;>2Hz zzZMaq)U9Q9SDPz;f)=$%i0;%D{Us0&9rk7VnQktLhO(5rnFJ`CEgy8n^;eb9im%pI zVGp^#G(hcExL?QW1H_*pU>t@c!CL0+C66ot?N7bT!w<@jsn}6Q*e^0tEo$yb&f!_! zN-fldX_bbz0JaiHwil^*ngXy(8!1i63KIHLj)jh}e$kY&cZZI~6EB9&aK?V#mmp^J zc*7#&227BEnt(vYxN97jnqUWAn!nfWzz$oC@qxW`>@2==@{;G0LDv?Vft+u8!W&x% zi`jHA{>(hZo3jaCADq%k)WmVTAIMvv6Yu7lTE#}U32Y1TU1*Ypr2g5DLLPyBl%My8 zPE>PAfxsG*faLP=*Ixgoi{S-nDqz@%qvTMr^pC2O72qs)G2G)=f5tF=!CSlC0qIIx z3?s*yBFpVZ$ltdVt#oKfDljBIvlrvI?~f8*7(FR^VcuW(!Ca#U*eXrr9uNkBn1RV< z;=BUjIq=9In=5BNcyvxBdbltf##-~QE)3!Qem~^yKaqz7)><6zK9L0%c+-6Jd2Dxj zJ^=T(*l(pCdelg1G)Vj1!306!64|12FJNeO3DYU`hL}zz zKzKhww#lipQh}5?=+(fEdGqnNjw9Lz#51_gd3Xxk0Uno+V4?4e-NxE4s@M=t-8aF&Hn~JTR{@gJ)TLbfpPEBRY!6bBOc)T3oKpW6meI8Nyg`MHBIRPvI} z;Wh9aa4E4dFyxe$UPp2V?>nz&JQySnp#n(HXGQD1$<36M0Ym;oa3KiFWw<`I4g)cg z?NGaWzC%KgbTWhi8tNI7InG&Lpy&Df2|ExWnl#Q(ZTymO(0Bc_23lgEMrN;n@gdcY zjjuR6r>7ixgRXDo$iz;32-ET1TNQ)n*TW|LLWFD@B)&uTqO&v}Otyz}gx*duJDiiaIJ{9?W z42qK;PQon>x{etLDMkyTAN%;gx(YOgt$2;UMg2Rm*up#!_i#7OBuIBbmtfPaZ&4N{ zd|$RquOH8g4vLgJcmB*|xnXK%rdZKpMe&WlZ4;5-xvA)5{+1%2p74;~PStihZuYF} zrl+=8(BmWszndtCa-~hZlN1@SrGtR_ zJD+cU!6#7o46pf7c2-HrT?5Dar`oz)H^pd&jfsgTHc`&T)uuZ?SL^{58S==f49E-- zP5Ep*6MD*JkS3n3)ERsmz9#G*qce5AA(ch{b{DHaplx5EAX4;vv2VI1MFOojT-#p47m?zFe1PpnoNl>3FO$0?Gj$`f!L%-F_Bw< z)qt`B90W|0R7mxnm}E)IY8$T8RT9WL^mmuXo~EMV3h~|emhF4EFCQi#UWuj+Owm=m zY;zmX^|&w`q{B!wl?J*AtQqhZp}&V1QtQ)p$5PH6{=Bat6W$9n{xS2>ix9W9Yv#ih z;2B0veC#FJQILsn1h8%1yLEfsAcmcgp6?^JHwrBou!s4<@!ZB75KsAfAp>%k2mlMK ztVh)cR1Twgh1a)(_SHcMbA{piJUS2|1D;nE#vqpw75&GhGw0YR0Oo!bKeC4bWd9v3 z6^70Y1SRnq%uPVPUgX};nQ{w^1x671J7j)ZT-vMw?}+@xXk9ZAU(lCxL(V*8ZRjL0 zJH~??1%y%a|YyC0B)a2R*D`sTqRK0=~V^2bnN;IK$@? zSpVpUL(_+UkcE7UG_H%B`OZDNirsx;ZRY@Is-Yjj0pn~#-e`@{*(d#GB`q}?=ESU# z(P7a=0%gG8<4PDNg|5zhK{upTScI$ zl%a=wFA6lGN^nVexveMc9<&NlZIj~L3);M#i09$Yf{b#Ysjo1NX?%no4c(^R2S+zb z!5&daCR10(VkIhoM)8z?bU~Qp6ew$qguf;he!r7Pw*@l|S#6NXVexERT~Q9=>KuksHu7gmcM(n*81dCd2M|5 zZ92ZyOKLB`Uk%(}c87C2WZuK@9? znso7tdIxdmv%)#E?p^V;qgcZ>C%o{2#SJgCIC}aC2R)x?+uT^1kU7cBsM2$Fh0ua| zf}JF*T7PBpVJdKfV(oG6st+3 zmsYgzv?sBHK-Y>%O7fEy&Ajo2REOhZtn zsY?M3^Wyfqa~22)k52h|<-enlsT8-;o>D1#-KlG-i3>(3XU}yG2t6xrxs)us@QP<# zt;^bvCysCYe0&|K7xS&VH$5mLs%2MeBY#ie=1uFcZr8ATg17EHB=u;QXoP^QIna*1 zy`SfA2%l;%Kb5n|Q?E;7! zhf!VaQmznQ>+SR5t5g4vH>4>a;mPlGf1Bh#8`|=}=!IFYHX5#pZ!cO%pQ;J}PM>U* z%2p2XY1^|py)HTFh|ou=-2N*|N6pPCLLVoZ$7_iIaa>*^L;CpfFL7y&+iM;;6z_O- z$j;!3XItgiRmW>*`a=Z+zmnghhE$`RUHvR(NPa!3e-R^`N{hCUR(vHrMH{6 zRtRY{66#JN4u!G8z~T0!GkN_8#e4NAHgD|gk~zJgC-muY9*FW1AjbT~Wm;|Aso74n>JhW?iQL+`_X<`lc>98@BJ^q7*Kq zqqh}kpMH_QX+;UI^RGQa^K5sWb;%4G!K1GO8#@DJ%eDD20=cgWiZI(2@iI*1gUmEy ze@6rxLG1JH@XLUJ5V@Ktmr^ zxGdZ96iUE!eX|R?sh@l^aWj`C0${ssMwKGP4);^QZSIi70UQyRQ;GQ?@r*_aSM8E_-u(F*%gotY#;Q5HZyDJEn+d(sfc&rv; z0dze?><2YY{u_(F9r*N(A-1IDuL3SHksIsnD6$k6x5*04xaKOuSpTvr48Fj3cZBV0 z0ASzUc-Vie;Mw=)Db?}s9(`q5+!_)Vguss`y(TW>4{ZP45I}{&MymD{y-5x&;6;S7 zmLL();kgcqLX{gzObn=2LA(eWa9sKT_h<}wAUbdq`0IlZKxkx2+aBo=ZTc!s>qy|r zHF(qa_uz2<4&hTF#KIE&E%x|2vyRX>ex0~V;*>G-2B9t1kc(mc^)R8un!AZX*J6yL zD}RGVCKI3fj+fo9dx5;KsXXyEJe~+S(Md^yg3DRng1}lb4)MQg$-m%H{4vR`5XoasFfL2>+6}UPdrmTeN2l{Fu&{(&G&cP$eK2L-rWcnjXSpXdX zxm&O_WnFy_gUl-xO}7)<_mDH8Z1b{hz#Y*(Bnmu+uo8-8J-eTX9EL)_$?3H@*{M}t z4G+CJX1I^$<>{3`Dsnq!0(hh!D0EvpVz(O}DHbAF*ER%2F(f8H`5d82jJA5PyAxV1 zA{;YwAiD83XCgA8EmIm~)@`cj;a;T+hep}9{iX29Op zTR$or0%=4>eIUs0%3Rr3Uf3RNY6ccHBt?##^PoGj(MH0}N4Tdyzmix|YJ6fwm{ z*nI*J80)=r{hmaz2KEON#9?su>yI^jZL?F6Zt3kSNWQm8lR!~%_{AIUgm`lD*4z{Q z>+Bz(ObGI((5-nlzd<@_=AsNPNI6BB&V`CNtu>Q3u7&%@>&R}S<3q6*e_UNNG<{L} zY{!q6Y--=?l3#%B6TjmZ$HzB7$GLIba_vOl^w)1tzra;80u{~vm)GEzp+;Y+jr+GP zz=dj=0+ZNAI^k0n>uXtd3#DbupjT1Wa}IPbH48fbh`%w2od$qau_s?ufPzQRc3&w; zkOT4CWk}4d;~B0j&-6rJVdSWL(W|=?&2@7`(*GD>6g7+~93U>!p|y<;@*$=lCzQ;7 z2VIq+>2i6_)Tk+3Qc_h4I6Bbtl+qua$<+c()1aOJB`Dn3)dLq_y6gSq4Q=*yF?)&0 zgCk?4)7ySWBtl}(&thxfIB!=@$l!_=Y7~P!X2-|?KR*E0&xqr_yU$Y8yj|5!=jPMJ zHf}63UlIf@hfYpiw=B;LoIl@JyV8tQ`&*t=tM||2h3)B?sGbSC>c-(VtYDSJqE%q8 zjR27UUIFNLGiY_iI}*cUl~NWo2iN@^aBoj20!SO0bdd`F^D*5I+AEqzT)l&os{}qE{R?Cq zD0TbCflG$9qfIK}Y6A*A0P>IcS!TOIz^a5>l^O^edT<6Y!Z%1hVB*- z`hocyI|`o~>Am}THzDHAvVg{l%?%^>M+e!HExSDrcV2mPUh>)b@|H-&wh!+I^68Nw z`(Y)`3RIXC)S7AR;MAJe zGDmE-Yp>q9<(pQ@R73eh$}d5y8275efrn$dbe5S+O7dMvd+TP$V%6JAf4ILjaIS^7 z;MB)o&L%y|Gy1bMnW1;)?JTinwq;h38AKW>OD>spY3&z>vI$H0|5+55;3RR*44vF@ z0LJ!uap0lbRlyz8=WgxYA6m96Vf}`vWqIyuw@*5#n}44iVHvf5EsgsW#HMcW$WD&X zNU-V8Tjar9YpUmIo#FK(CjQW)+XDW7*!OOJFZ|N_-dc;5X#uBM#iv6Wr0Xa{?<-jQyD>E}iGY3b{3sz)Hmsx5vpG1GV;9_NA?jd@@!NSG;+y!&d z-Mgfv|CL>s4hKrl9C+T`+3bRqql=Y;{jC2w!1=hweo;*|HPK@i94xGC&6h9!Hq)$ diff --git a/generated/webwork/images/webwork-1382-image-1.svg b/generated/webwork/images/webwork-1382-image-1.svg index b61ff91b7..501137198 100644 --- a/generated/webwork/images/webwork-1382-image-1.svg +++ b/generated/webwork/images/webwork-1382-image-1.svg @@ -1,44 +1,32 @@ - + - - - - - - - + + - - - - - - - - - - + + + - + - - - - + + + + - + - + @@ -47,7 +35,7 @@ - + @@ -56,7 +44,7 @@ - + @@ -65,262 +53,139 @@ - + - + - + - + - + - - + + - + - - + + - + - - + + - - + + - - + + - - + - - + + - - + + - - + + - + - - + + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + + - + - - - - - - - - - - - - - - - - - - - - - - - + + + + + - - + + - + - - + + - + diff --git a/generated/webwork/images/webwork-1382-image-1.tex b/generated/webwork/images/webwork-1382-image-1.tex index 8f8a5f942..97aa525c5 100755 --- a/generated/webwork/images/webwork-1382-image-1.tex +++ b/generated/webwork/images/webwork-1382-image-1.tex @@ -1,5 +1,6 @@ \documentclass{standalone} \usepackage[svgnames]{xcolor} +\usepackage{tikz} \usepackage{pgfplots} \usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals \definecolor{ruby}{HTML}{9e0c0f} @@ -129,11 +130,21 @@ \begin{document} \begin{tikzpicture} - \begin{axis}[] - \addplot[firstcurve, domain=-3:7]{5/8*x-5.05} node[pos=1,sloped,above left] {\(5x-8y=40\)}; - \addplot[secondcurve, domain=-3:7]{5/8*x-4.95} node[pos=0.9,sloped,below right] {\(y=\frac{5}{8}(x+8)-10\)}; - \end{axis} +\begin{axis} + [ + width = 0.47\linewidth, + xmin = -6, + ymin = -2, + xmax = 14, + ymax = 12, + xtick = {}, + ytick = {}, + minor xtick = {-6,-5,...,14}, + minor ytick = {-2,-1,...,12}, + grid=both, + ] + \addplot+[domain = -6:14] {0.428571428571429*x+4}; + \addplot[soliddot] coordinates {(0,4) (7,7)}; +\end{axis} \end{tikzpicture} - - \end{document} diff --git a/generated/webwork/images/webwork-1382-image-2.pdf b/generated/webwork/images/webwork-1382-image-2.pdf new file mode 100755 index 0000000000000000000000000000000000000000..3ad79b47df3cb1bdffd8c46f26f94b3fef7ba7f3 GIT binary patch literal 7475 zcmbW6byQSs_xC}1KtMWFIwWTX7{Z}z=BgbEk(Ti1 z{af!^x6gX+_j#Z7{&TK#opb&5+55Zqcb~(eE+fYc;t>F_tREj-0fgy+^e%9F02InA zXXogNaOahCwD3g8AS_+15WLC=XB$sjdXO+bznB=n!_ysM;RNuWKQL5s(G?^{UFvz` z#Wn?sPd?F%{lcSC#1pK;D8a2Asmi2Q0l@Tl2W15+5GIJl8Nl z99!ymaUgP9btvi3pnY2hV`|Qu%ti;&nryaQ&aVTsb7rEy+L|e82bE#3EtOr}m#Q{c z_ZGJM`5Lq3227Mv?ru0hc}{+K_^Ef`fFd5sBOmz*$u4_!%=DcJk1<9{*UiOwf^a3^^v;P*&Gx zbgtQYcx2iTGk!A8I>q{rG+>VfXNTK##&at-LRcDbYb zcQ$m+k+l~65eiH5jfhtVah=2o!6}rYx=mDI9t@mjou@TKGf!)3yC?XL!Y3B8t1XAe z-1Y*?1H=XI{hFS6z0~p3HYvn+{+PSm(xeVSw{aA>lfl9F-~ct()Zjkh8fr|>Boblv ztdg1%7N*ss$PXje=$#hXP8{u2)G0~U8}7hnDztJC+I~IS$)VHdI4G$UuIg8?A1Yq8 zg;pY}GR08h-vRgjVZOTjUUd8I%0O6RfL)*O#-kR}ykENnPi%f!Zu+dx1y47FT3*xk zv2%SXS{EWGoymJStk6mphlfYO$vlBe3MEk}R)JONvt-DnFITIA6MzF*n$K)8NVujI ziyyIim!}RfDIddt84w0LGficSIVjz3< znKy=^zUL>UMNF2Y4J{yEvNDBwC-d+3QdmhH2>A)gB%@3oXX|%mC~nsc;liYFAr5KU z)WY?;6V{#=NS4G@Ltq3qmJXGc>x{;tCtG;A3M@58gC(cNx2p=!vYao;4-}QlaveP=G#Z2* z*cXfe&uC3{**%c`Ks|c`4`d!#Z}*u8vJ`x&u9YAx&ajdbhZ;&vD1CMyMV^{lj*yf2 zUbgAZTJHR$c?#LR+qXTx!Pu~8K(`b99JRKN1cFNU8&{6-h$4# ze!mLn%DkYHYqv}HXz>$OFVW#nlwX=z zYV#aI>+oH(TR@|pQ%>%66Sv^$zVY1^zOk1HpIpGb^~~M%#vof|97xS9w;gQ?z+wEv zn7>#HUis_#w*PF#IVSnuml>N=Ve9E96DN>gSWlaPhyDyy^reHJ{^@IKObzM&d}PzN zRKmMl7vkJH2Y803!bQGdzbH#`kk;-?t| zTPoc-YHxNw!HsAOk|Ngo*c?B+Va{Qz9Ciaof~_-G^1Xn2w$O0O`bAn=U!nPeo8?#z zDoH%ISR(@E3`Qx;7E*r{aQ1yMv9^;}q^rxAO>lGSqqhb5RPND_?N%=h`6Ktnv8~V4 zjR?E+sT@3TKFV7iDxcmGqeL7Ut*f!c3h?rR4OojF4Jz6O7 zqA%bP4v!YWOm8Tx3c>3F@jEpfTbwAt%1R^}*`~~!n?k2fkSvhfvw3eD>T|yZWAa$O z2Jzlx(*2knM3()f0T|J}q;-c!x>#ltTUmF`jm;mMI;3 z7qe5*2DP8aF&fWEE_;(dH@DXG+|6{}4ZuCLJG`|rDO9xc(k{tD8DjD>ZPO>kz1)3^ z^Bg6Dm8hYnGbAy#V4WBqK_tC*#**^*C2~|5R*}QYVEw&0(vyiXF^=$B6{DM41f6d zv5c5G45%G#+b$lq;tD6A=*Su&A;(uSuUR38m}e4eE6;5(811;wDNzY$1KmrOx}gOE zf>B19xG%i(G4+AES8kJVGo{-BA|v#22S&!dgD z0EDyE|7p>`@BV7)Ain>myMy^ae>Qg$^-lY4AVKR5r{_LZcQg?qig=quN^2Xkc~Mnp z$(dnM{d(>3kIN4x6WxR)+dHBPYIQpci`!^t6Gj7|j?&jZnr5ez!qM5BU;`3%)b1%Y z*%3^2!uXdw)|kHD4hek|9v+)ZmAo@bdWVuL-ZsXNs1A2jqrcIE;Z|yHZF2NHv&Imm zW7>VfuY^}*)kKw}4t$-5^T*-GQ=iPT$1vGpCM9kgcg%L=JgTs>E@jyJg3|om&zX4| zzkCfqaX;j*hOEl1qO-Hhmo2Af8760BrAy|P%+i0b zzRr%h#b#5Ss?A|LYn7Z#roXe7mX@~1vPrSGL5WMym z*soYNi#dW(6*#Oprylr`VFi&3(ROMWpXp8x2 z#nY#UImXXFy&T?HL5D9M5y_(MH6dR7Lp&gV$K(I*Ou_u%KLhYr?tY;sZoMJ*_^zqw znZ_*&%d^5$v73Lp9~eBQz`@Va`#W`Cv+n2PoPWF8JXy4oeNh%i_$I)Q(}-Zyp%AL- zbwQHfElJxIpI1lVm*Ol8kgqAAIMj;6HMqMtS>#WUMYTXt6)b+(5qEM;JcKJvGgC;D z;$TaJFMc*@aJj*G@iL`Q3&+BW^nQR(glQH!VPkVyxM_jy?0kIlx>qmsKyyx8Dkp)v+Y=mX_eHM0!#0r-wf|rYS6YhT zfeJ$mG2k@FvDMb^G-T<+)(tU?T?SJURu>hoyW{zM(hzkx#3N7=8pNhMuX~b3+@7&~ zTI>BOc!_;T9{28zb+%dBiIQ`#lU)s$4-7`kymgSbW5SWiqfx7xJ80kFsvT1OhJ%Nf zS4)$H-$Yeso;q#H)>YjQ!^UDCn@5zU%zoMyAw;KBHPhub98jOfZ z#)Kp454Y<9=xadAz|XLFWh_d1h2kb{O>3-gd46RZ0{&53Ks>q|w#dM5XQ&Vypbvvo zl)bt5WlcD4x+93 z$K5uy$AE&Lw2N7rAAEt5(t6xRNA%Iz9UB&*nH*8&zpm4ksPg`o~@HRgqRoAtL}Wbb2Nf|7m#m4koPm{WSI2A%^?$x zE>%8{K6m5f-c&vKsEy%Rp+gHkfDiG;ms;l)adkKGzxz_(F#wEDEkjq3w8N7%NyALJ zx@?SnVlonx&LVrU&++WV@Cn0@PB&GI=UYMv*?x<1i@uBUi&3I`GVNnZ9(s|y)1R}# z0wo^?HEDaN&DyRtcimukzw(azcaUJv|3|G5=I8s9tL$4N9*{qSFL7`?b z4bBYQn%u#ean^{T!!qFRXOVzaTgUgHInc8kW~$`11PmR#}okR z9#l59*^rePM<`b@VyR7|A}LmDRyGOju?|BO%^0LB=5*Tv_Z|-vq?Fa~s$fTDCDMUK z9yfFa+!*tuN!d7S=^v zuZ*l8GjF7qIayC8DJu)L>gP|6uNj3fOMZg&OOmbT`|?YZYl^!~W&`&qYdOzIqScWRQGH{$d(GSkATfoQC=HKs{ zZ}AJT;b;W`IZ*kAxBHDE3}}J#3rF}0Q7bc*TW-VTL)1OSP13Oj?R*#Xf^q@=v_y-W zHRs+24;p=4sN4Mx-ffl>zRz2i2`fOBUjiS87X(ffUBFjr9I&ldqPEYl@9xJJIFFfB zGW%w>fE}{5Ugr~>kT!h2%YurmjJu4Xco(Y3d4wCrf>X;kvD7Uq)1vb~JcJ3D& z(|emoR?gZh`f5vnM;SL)%bzcf9)}*w-X>U`qa2o)pXg5MU5e#@9EbER3gy1~G92`i zBl|6@HAOe6gxavM4tSeDJGYdNj_>*)02SaA@^t%fZzqgLWGk_c56yDiu5(v@!^cWK z1F5P@Bx~7wP2@)kYLZZ!WlU{I72Hd17fpD^)#i8E#&1XXft<6spe+1NM*^F^iTh2O zi~9+pJQ2$62=I{ z-XBd{84DQ&gopI{hLo%k=VyQ8sAf*X(}m`968f)hokqjeCi)c0Qm0`@Sj|k(4U$& zQLlCm0}`X|$OVhva}XrX%hkLwWYZ?7(@=lG_S!i{ZvyBVlHM*fwa$je>DI@=^o%A?Ca;n9LRc^l(SQqDtm~6ru|7^8J(x7!6o#L|{ z-)G}C>a!KegX_dKA9)ha`7!sECJfEv(3qEMSBY;in*|ZH{`$UVruj3)VW^auYORFf zJ9q*s8jH`;?*ptgGf%Rs(q>+5 z+4M${(-rjgnog#EznpbH3%nprBTeG@%WK_dzZlep20g%XNBakO{C~syuf3MwzdBZy z1f$v7*Nn|Hgm4kor_+BR&An&6xZdR-Zi@@rjt+h-Y18VZ}67EZP&RVD;n?iz%7 zvq&pTl=+NeSf`~i8KNU1&PdLz6gAZt`i1MG+^>FI|F{zi5U1*L3%5w;;$tfWO205f z>e`gB7Y<@se!0B7rkf!=!6iw6uY#gsy|1EN-8VSy)i{JBiAgTxul-z-+b?8&?D5}5 zVBq!v3%=qhZp<+w!8(ZQ>8hXLC-OZGB&Eq`XLT|hI%UcwJXut_mHFS2GdDfsqBk29 zg&`i$g6wP;tNLfp3=X)kWtw zw6i%yeTCQPx)UgsaT`VoFoi1&7upl#ig!wVg#;=jdy1Y1DgigOvye+hRK zlVd;WPn)&-*g0%iF5R$YMHD`dffB3iTSMSO?Ze0A*c$-KU4z))9|zKUy-D zqqT(*4L<@`i%Ff499JDS+sKWCq78D?%g##Oo4uR^6nrjW8 zKB`PNCzf|}5BpKiNql!s&(iGJyYD%+UmCoiCwNpDrxg*6OiZdM1yG|U%q$81^sx3< zLVlo&1DMk~9om08$%HmZ7X<=a*-99`i$ipS_5u@&JBn?Fdem`V2Gmak5Qi?llXyvw z8TN(G!H>ztm5h8ih3748)rqC@tYkljj|MqGBqnJ)pA^o^L^tpGH~bA+G+T*>ffSdB zOIVaPNBQGhqVfZ%$G~#EvG>tJ0-dlDMIjV9=SVF;;O=kik zyE%Mu?H}lGLxlw(p_-x(Rsh6dl*|VE0jO~4qL^X5!FFUcHj)6c2SvORAzQs4+>C^JMLw0l;-rACtHCpV0QSq(cSo(9-T-RrZH%{;8xn@hW`-`8S zSUt*Usm_M>%&l9rg)d+my1l1vb%N*xuJZ3hO!NER45RCCxqe8o)o@7|Z{4&X1kZlU z_QG0QWHO5D4;oKl&z}pNoQ;KU;k#?g@bs{|QE(m*xG=@Xzug&< zh_zsZf$OaA9~Q~~taUXEE=EqoE0w3_jRMGYxlXGoUbyNFpRZ+f)|l!f;nU$4OK!T= zuxBDWw_Kg=w`QZSE<0LifUOwHKaJpuj)^AkOtXowo2HfpKu1Kcd_w#!rs1a^C5ABc8Y< z4{tFTyxjt{&PM|;%+S9;fBn1oz<0 zTM7X|aD;`0H5?>hZ6)^qK4HSE>EhxE_)FiKyb8|NF7$sn3{74wJ6{AnACOni2=x02 z*yKNGTlio8HL?sY-~|JTWyUhDAel1i$KCPegQn%(a*iEO;gNU3-z^N#gr9r!7f`6G z$H7S?nHe8&xVD#bc;)I0TJT1lG|Y{)BO?2(M~WH+W*;}@=$Z~O#zaJ61WsPzdj2~b Z4^InsPj7dGH9!y~2n4XO$g0T!{s*Jo>PP?p literal 0 HcmV?d00001 diff --git a/generated/webwork/images/webwork-1382-image-2.png b/generated/webwork/images/webwork-1382-image-2.png new file mode 100644 index 0000000000000000000000000000000000000000..ccf49939ad1f417e17bd5d73c51f8da4a50b08af GIT binary patch literal 27419 zcmbTe1yqz>8#X$KfRv~-2pE8bNJvX42uKVa(jd|$-JudvqI8dRgGx7&qI64&bPhd3 zpZ$#Q`+eVk)>-SUb9Bu@=80YRzT>*CZHTg>3^4%>0SpEsmVNeA6$ZPM4L;IWaKSe( zuAI`tVED|Il9I}nCMGc0vyj+WbLD7l%1-2-$ZFu@8xq0Fk{{Wa{9~@^unZRFJzhyw zma=(QeThBv9k0{}DJ)98*gMuzWQ1XFE>SqYy?^OPtpmBr6YW>C)Tt;s@%oEIuRi;8 zr==OEzN1%Ix8A)q{r1!Ag=%f{WY>FYDzfg&vn=6=xR#xLXYARLy6dpwHm};*zN3;W z*FRdGE>>69J07vVafW@`FdgldQ_`Nx{HZ_^dC_97_58EmZ=Tubb1Qxx8}hwP_m89I z*L(DR$q8ULS&wP2UIbk@Aw$&#-|Ua)eYeR`a#Z)!Ot=-5SUHGKAOCZoj8UeY`Khnk z_C@R|UABx_Zs}7y#~;j3Y@8K@_x!H%@o@28HZMXE2Ayq*00fC@MA z^)-bz@sBUhyvDim=JPe`j3+!FZe(3ueJp*A_G{41TLRx{3EoJgTd?5&@OMm4RlcrH zpz*{-nRK7f`D*Q@=>XjH;C?|~>@#AYfa?jDhm3KaV~+;^;=*$c^qD(&#hXH~+-qr^udb6YQiVud{;rvIja6ptqh=QEs^r#f ziI)^d{w<#@SeWG~O+y@|kJ8b3>cg7q+yw+KmquQ%Nli;eWIoO7v*6L8FMeD3c(4EU ziiI;pElmHt-P5Yf()8eg=oL3AqM+Bt#@$@K3AYq(nmx>+a-tdvstKoVQ!rZk{Je-r z6VE#EpgG8R=nKxc+=lE1)5bN|8z=NKZLD+QuijT1>D_ohqV?XoL%)r9F6R}YyFNT|T)Ag$fystemW+lDNXr zjKgPirM;SwD&ko)xniZ?2Cwvw7Z2of_e7cPJQpgW7)@Z-9+G~jQD(5$NFg0oOkV2p ze7%If$Tov6Q@_|p25*#mBxFRW=wUTM!w+wb!GdFr%vwfZPdRpBTh;WEeQ-{tQZ;03 zW0LRZbKf(uvp!sATzp(GSH$^24}v^`ri1)wEopZm3L>^jnM+;u8f;z0*2X%7J3e;E zgfk@cD@?&%llK$6dH5cZCsZc&CgdjuCXFOSd|kCsa^*e=5}KIm^-ezW*dw9Hp;)EJ z*3@ks8~dWJzb`Gpu8gX+vtJX1~|1Il|vkIp7+a z*&Z^VvQYo^;b+qEipd@)j%wZN_|5BH_f8~Eu<#1rA|zv5)nv#!EjlS=K1PQOndScM zwJ_IOKI_dC!J$WtwUO2&)Yv$zx-iuoS8LXYISsjzR=d=-Zavt-o9Y}bD23}xu40L} z)8mPxxqlT=6ZJc2arf}l@DfGw@An>T?WG^t>{}jsEL$(@{y7;}8scm%?L_=x_%pj8 z&S1o#OmmB9m}KgP`=e+4Ec|)wRJ0omQ&b{Vy~fkuT%vC?-^RIpcU^nEY;1PwR^4sm z*1(9E2xHp5fDB1&iDXI2R*{yq)}&W(lt$^1` zc!KTt@xDShT*0H;=Dt8B+o4N}lb-FAcfHK%4mvx)KcMod-V2LAHH=oO-m3OrUpTt% z8igv#JqhI+xpQDt@a#96g~Hpx(Oj48p?LB3bE2Mbed_td(ZrENWZ7w1?@RfY_e@dt zgEQskL+bBMdX+2l_W*b>i9l)+te&nvq&6(^N8Zf(4q-O){Z zVqiHlnTqwOZjXGoYFTN+abiUEmgX%Ms%hJrx!uXnFl=0ALHDb6TVEp{|qe4(K*@~T4LMK{8`CoSoy^?Gpd`*C{9VvRhGf-`sC z!-BaU}X5=;_1(*)20`)6(E`S~i9U-EW_cCzMMNN4VOCU++#B`qe)>->%j zk4ucceGhqVbK#^_XlJcYn_0X-$IBRR_)EUz zF!d1Y2EiJQ$g*Qe?&R0HGpz>0NQ9P{q0f_Di7u&@_R}!=c*h4l`ieCoUQEY7p7wR5 zb!uUk&L9ug=&F}IH#=N9*R@Ured6t&PYe1D{aLEf-Ox>~g4?_8C+}BXnC%m3C!{&W$$LQF&rFATlhk-+6 zMl6V)-TTgY)@lCf(D7`AuRUoJ0cZGRBuaeG-h;=(XTJ{i##R3e4vI_M|Iiri1Xl-%(=C%{ zs&E+0lNko{4}igr!B_q(FqjJ$47O$jg9#_VV3e;?YE?wQ1>6_%GEZR_(BDrDxzXSo z{8!JkonSDm`Tyb!Q?@W!Sj=~m8le0*-!AXv|H z6YKDa1m-o_{-wdye%Jc$kW|%&O9_bzk&7>RMo3B5R}+pbMK-w-6BJ^ytR@peLn=EK zoFA)L5zE0QUxmE6<%Rv#xD`n<^_^ne#;8@NdHu+J;rdHM^zPY4#b~n2(fy|P@2`@R zkuhGzJ>Imp-K>^`IWKin$Hv6$qkSzc^Kaa^;py$SyU>-yXH7wh2W$RabhY|eO2XQ@ zlkVFSy3kOIhc>(w2Uu3N2hn^*`lOO6K@pk9FFrA{x;hpbn^;&LXz4ddNsqOSTBw*r zH)Y&OCAezCAi$R>cw4cGNxH@dsj8o8;JM0Vda;7!Tb`+#q9?pcFq|cO_hDu2Pb@13 zbVOZ}yN)7rP~}hIfod0AzI5g^s~2$x)dPGr%A|RRR}){*D#5$UQmRL(qvG(Y?VnHb zxyTi!OMxD#RgYM%$c-2f zo&0rj^SGVUH$AuMBG$lwm&e4~>QF1JmFL{*kFn4fMQ0k}t5yD|c*~gyA zlTBDKSUMRMw~Gt^=6Frtrk6+X?k`=CtAj6vlKH`Wq!y1Mv0*R?R;~@tN%^&w8DJ_` zvYX$nSaVvUI{!Y-KZax?CM1kyoOT|HDM5azBEvjTREuQN=MvYSyf|QBl9i|6>_0Yd zTs;IP`IrUAyX3@R@Sx9I4R6+`##$+#(bn6z;%ei`*Nlv!4GtWbzcK+4$=FQaTV!KV z1ZUCKfe_(R?)s3@%0jqyV*Zw8R z%8}0|d{f{wPNSf_TvbIyGdr7^dk5z45GI=K|0K=Fu8!h?r%s&{iS^uDC3c@txya4j ztX?PmGT&_y&%~T37E}NQ*W#hTyQtWG0*4a=FrfIMa z@2YGRfs$BJvb~%s4;IYil9$};>y;+UVzF@=RwwN1o+6cxP}q^B9D)W@2QEMKs~#po zbL9W+*Z!Xe#fGS?NtB#CdxK~3`ilbUhh{81gFjwg0uv3xwd?Hl2@U;Pt3P6>Wv(`J zl4?9O)UG&Py4AFt=9QoTEmiaH1v<9=-SCsC#;QlCw(v5iz2mAOn4QQYYpd7pLJ$7iXBpt)!yj zgSfZmIYWu8*Pol4i_%F@hBYUAoAYXZT!+%DoACX1TJDRT&)AdJV*qR-oy?z8(y~et zHO4UNMwmU_d#_ldz-6-yRvhW)D8dAS_mrqw%5pGA2Pt4jL`$Mz{@jtG+09obQ@x6j zD{oG`NRpv<)CwmLsz#UlPDd-yz>K<0lvuf1GVT=1CD7*vtU64b|EQ&pM2^=e)j5eP zx9kfx97tDYWK@rd!i+XB_r`H1kC8tsB}wTMtI~&Fh_g2od$CgpUeFnU1?#{$jJb5z zW}dL{V8dxLR?Ip(jnllg5Qi)UtJNO*X7SXo#oySrC!)hyHZByhj& z!h?m&cz`J&;&EA1H#IZkxekM^)V8ALj1{Mo`d^G2iLoYr>R|SRrw$g3cNGIq{#!kd zk=52L8hG;0y3QrM-YsX#ZcM~bTIpZB_q zdYwighAuyS)9C5ax8e&tC^1@;zX4qfeKd=oXH?C6*KFCZF6oRj?;iSw%~@D7lLO4X z7;HyoI!ya^z^3XBZRovGrX1DcwvS6WISPl%`fRn-V>D>*%Jc2>H(tZl?(dY=*y?`2UZ5AKgeVnr5v*2! zEH-a;BG})&{NhaXvk&cLsysGU;gaSz8=YG)Z2(FYUVih_$5f%EhdT5vAf{UlUKEbmq(|W;5-`@dZggZ_uxgl8(zPkO@ho1$9WAdqGP2?HX4w4sR~H}UdqI`7)y?Qt zxiG{WpM##h5%Xm?qDe!QX}_lC=Jy86(;CrebpG)fvb3S3iAMYWmU!pT-aW-ImDBRz zWAD?I;EU`r124F&Y>4OSm*C^0;)!5~#W~JgwN{5OH*&SO*;BQy7TkURPD63l<5)!f!NL}!gwL-%<`QZ$p}@TAY~ zd0ADF%G4AYpR?B^30CRY*IS#;6|Y}#f)>u$Cu(JKE0>P%dmoZNi{_y`K#Tdvaj!|_ z7h3xIRncyqqu-niC4da>SfvsrDS3%Coh);3*EBVF#X)Bq%lp62YOnxNs^ zuVa)$RsIc_mfG2!MztUy_b!X?!R=%TV2{Orpy>JTf4CuMoCjhXh_TjgTXQAIMY7xo~4oGH#s4Xw{?sH$%im{>!)a1m552&)kf=)wOs!7xJibwV;3bh z@HUqN{zZ^Q0q0kKo@j3cqI}{jFwVKWs>0o8>3y>TtWF;X1zvg{=z&R_>Y&s-d5K&E zFNHEfkUJ{@hdNR25f8=VXYYn`+wC2{18AEwmbf+jB&%C0slha2tnXgTi(%c07xI!<;zt1a zg`!!91r^6cQsJKzYpdq5rT(r@qJemqqTR@pZMN^4GwE{U}j$%7z(_RD!+ad z2J`*%($%57|wglT?vn#+?4+ZlG|bt1fs3EWeqVDf(1`5z{3>lFS+2Mx|aIezsN? zC^Il33Q*CS_pbJ?;7ao|9%T&SR0i{6x>C_7I zC%{?iH-4Ye?<>vBl*GZoA!MX`DeSLE$!VYtheu4J6cJ*@5wrqpbHNOwCXu2`htDl6 z@^awG9c@_6f^+4+oaZ70Tr4k27s`JfdmB90DX??j#ShHR=CMp${E`bzMyD|^i(ROl z`yU4>PTvICy(nE4ViG5eKgZpZ?Yhjze`LKDquH*=zjZV0jaqtCR(T>kC?frv-2lk> zTCG)8$n)01Uf1%aOU)VgI;cm7XkFBq&6Zj*$qM-0mh(G$Zto=}V`j4~_jbgnD!esm zzagbf(QagzwWIi99zNfWI%yeHVRv;6-L`T%2KBx8c#`{&b-DTc3G>is8_~}Y>(qg} z(55x|Z9F_;&Fdq{Ov?WZGe-vM^sMS&cC85I!3xhe?1*9vg{mb_AOZ>8)%CxN&Nc@G?JVivdx8Q7wjz*EpjF=vsl=_@Epy8kT=u_t;QH^n z3xePS49OA#$^IKO+7qv1xsu$@c?Zw}ShcPs~dRuQqCo9h-E8pzslQBNLx}s$tZ%^EB&pI6`9*>J8eEt$V~uI2XcQPnh96% zX@QRI(@IyX5I0oDXCF_V_6)=o@<{Wp!F^e-RF-{y`7Z{t0%(BAjOAIU^@oRPNxp9Uf-pVv z;Aq|ckop<4R;axL&)=!b_*tOS{-xNKJzY0?PL4d8=h?;#21y#rLb|B%z;?HATazQV zogP|dxHF1=-ZS%6`se$m0?=*Dj%|+Dt9n|GTI8eG9V1)5*`GEXn=|bscym4(RdxV! zhd~z|><%-k>F8S3&QZJdy)sgeZvfY;<9e#G)fVs|MU4O^rtXoY3|56j$T|@$gs9k< z=6MZRVNcOHeBis-1Yi7apk#1gJ~BSh$9>qypNmLJI#a(soGyOZWuFScC#Jbd%j06R z$q$#S32ZPi*(mb^zeKtje|=t@bclX?ygMr{QK|x5uTiz&gGq2_Wy$Xm#Fks}jx%0z zBavHjN_u$Tma#W*(dy@6BnNX7wxP+%5`xr(e(w&3&|)k+3MwR`MY56NCCT>5pFVxp zWWFl;3Pefog$?#PkW^_pY6fZbRi&2F5B7n9&xsnk2zJJwz9sr=a`7a+ELSs;qPvm} zsP81AxPu@(>+GIWz|WG&K_3-v9=D~b!nU!%;(ygg{-9x4KM-;R5eW?bD%#4#UWtUt zk%g`e4Ki}_i*-Br#k#%wO+%1Tfg1`e*6Eo{#U4BDtF%paphUM?oRey>!7d4!4iuP) zng6OUJA3%xL+#;x4$qS#EV{6xZx?(n%T*9~|A>5>Mhf#0>WL?%ZAdCqE~NPapT?-tQIq1dRXgjo-MB8VO~T zR&tuWCJL98nmj+M{hG<#f568LLXBuXI~XpAosgv+G(YMiea6iKYHR_Po2fM!$jHnn71ty-&_KKa7$~rEF>g(F|l5v74fD z*0)H52Gq1r;hx8t-$txsa!?=)`+r;W^V5rRZcV9c@a@Tp(ZwyN+-=Q$RI1L9Pa*e~ zZWd1BBeg8oa3RFa8d4ylng>!2{A}hwYZa4*RPf#(pX{KQSW2v3Ji09#nzDM9*8Vy` ze#~bc>EpG~TX%B%o8zjVhB^)n zKb1@&f-(P_*~hXTbQibd109_+xC*Imrc{)=6YFFCq=vE||ZzMxh_l3rEH zI1NWZ9*-GO4?<%#bUoN_&K1!ovng{=z1$^o(DnQ|WpZYgnbXL^@eXk{bCplNQ%ynD zx`IJX3$os&o8ulRro0Ne%_AjZWe`-fBqYPwqYIyT}A_y^wE* zUc-=A3oa62e~S-4H2IE9Lu>fi$H7M0V_5be?VhLJXdS9+AL4Y$3%?T8tgf6*at zK;NF;Zkq?2qQLeeZ$MuSb>)s(NtgRfM_u6H?@w2yq^MBK^0q$ypbmXWY(3U?Eq9Lg z*LYXQv_nSrh@r~?MXU7->+-xWU4|+j6>;z?Ba@Tb6cvZ-OKTZIeJ)-#8hCB;^&~ii zY5G=GwQBx9!t4nyL7RPuXHhMKsy+tMPNkbsnP27H&P60|8+7kI8_TNNAz)Chnb{o7fts0;3vJI*U)b1`e<%bVe|Iq+oj77hA$U1NCLno z`3rMUfC?!D`bcx8m+PzLmec5DjcSSq>ZS(St?044-d)c4W>-Ta7f~!LrLk=27@aW* z)rMnyAy<&6XKyb(BBGN~D{IoZNntf;ivj2P-l>KmSXWd?2|_H4E)24vPDhDi$= zR8(C_`+SYxh8H%7Nj2E>vloi`I_z>5AsA}?X>N<|c@fA2Vv(Y2JA9%vvnRhJx02+s ztlW}JXzg>7!cm+5rW?2&E-sGG25rTBW-|(_X2a`{57f?Q8gDYVInliCvowu8ueOaB zn>@3Q>T3YZ`}=V%N;lnE6l^5Z?*ktqcMaegU>_GJh*ZZ2S=`HryU?o={QPxF@p|eIa1_tADH;O_H zeD-+jUU9LOfY%@pZ{OK>A*?nNK{yL`cCNy(OWVIxOo{{zeEQA>??Kl}TueqMC2>8u zJEBAl`G+S-VSQJRL_v6>#||0@^78(6-Prp>JyE8Vc-N*IiI3wbPfz!c{`RH3LQxz1 zEbvHk0w_0D%34ts%=RI?dh>-G8{USVCWqomP>JF@vSy1`cKDRt)G!UDZu zC-Y&#;&;okE{9(a0t2ap=|MyRLBFo#1E?FIu$z-6IhLkcHq-E9!#Ktu6Z8%?b@+6Y zGIb~`|6%aW-k?8&=Y};EHf+GVCu&oR{UTfq+wt)1lXY`QkAC`hs^9Wje$M(OIK z_0w-q_V7C2f8wC#c{v5t`#Ho9O7`ydc%s-I2C0BD&CFg~lBzvnzK;yn-2ZeYnoYKV zon|8KsbLBs6)5SL%sJLQn?Nq%T^Z~o*sI~3nkzg*GEHea4dp5t)rW< zTrJ#jS)Nd%X>taxYQz0+Fb6l(5Rw8T@%PotnllbHIQ{%!^CJ7lAw8Y5`{#~MtF9AW zZ()92kBUEyXZ}hvKgZ^sroZy~uBJG+wY!6)6Qq2r0^!W24|abEeS=`yj@RE$me#=B z6`WEsBw0mjZg$;(*IGxgE*EW3GRQ?md6Ol{V1ulXR2te6L^@FE;km+oyrI>k27;Pe z%AOX;$$`zO;nIS3L3yMxH5KFF;G0$DW}Gnw8NdJSlRvY5GL&K>=^2QHq&pVm%k{Zg z3Qk*r_a~CrWpj|}!s_pTu=(tk(ShK~VN3?u-uDD|r=lnx%6yO}w4&lPL45$4>zR4c zD2W1qkVcBrV^GI>dv>xuq<%Jk@=JI0nluDKqcuU{2ymj6@d;ZfRgt`!`Us@Ym9EK4 zXEqJ1OPwnATA00N4f$D#XO5R31Xzvv*oq0E06*-m!Z~)cUTpp0Wf-*m6ig2~| zkN44zR_9Bq&C;C)_Ha^6V?=de;86KZ=e{4TFUl+vMNRTcK2=oc=OO(_MiqiVO)GZU zeA(TFGH0QP$TQ*+$XKbgK@^92=CeHwXX~wTmo7$i%n?!Z)pXJXPn&UxENGjjNqTbyhA?$^F<@b2uFbvsU{?a!20Hd z|0B^SfZWwe{&DLaAJ@<7931Cs?=01hYh}D?Kutn|xfpwPx-4A3D>XtEmzUY|>)?z{ zR_fyPAv`=ei@K0ra~Ji{vwjpLdjTpCFYp)Ct!QEzvFo>QFMRPc za(~;NpM)hG0xTMDtL?}ybNzy>)j!SmM1XU3sOqZM*=g8O`!TATXu4vH`)tl|X6}Ik z=9cuM{bG{+)vdeOzNvemlARZ;6=Dn|p+rO7MMPlRnnYoUiZmbeTfdLO!rq6zV*-tM znBiv5p7+Q*8aVtlBv?m2$La7r*~tC-7KM7o9ETk-6y`HgLy zXcRN9E^$6x^2h-3iqLTH3C43{ErLwH(fYji^1GAp`uB=O9O0Y zL=$RZpOMK={H-i5bd87u@=Z}7_c!$tzuCVj{Wi166D^ROXo0a8rN8)FPa#)QO6W5`n1TBrOEE47y#bwa_ck~k6q1|$@N536!WR_tdMTTe z%Qmg}VEdagtNgMWbCs~BJ9dc$oK5?)}eWfdY@uLg*7e%gfYnV{!mt3 zy2DHi1+U+aX7*{A!XbwMBO=Uv^3gacF*bdn$W(y}V~cK9;=<{X=a$z0!P8h+c*13| zp;Zp>bT{D;4>u`>&> z;J@3naLi(0R1~0-dy93k=i>CBBhdZc4ALjo-r_q;&!-a(+~o9PimF*bkq7yPW&uz; z4CEKr7)%FR%5_KrHEihU3ObsuL23(fKx0`*moqVa`aKc)y@!=CCbD!d>k*Rm# z#TjhM@SgMe66UnH=#_W;?QK+5y;N3kv6B(xelWv@UWDK7E6SXATyVeLC)0<+jyO^Q z2ge?n%nf0o&NTe@0K^F|gZQ)T#V3xH$tD=3lBh%RL(hOgn0#D;Md@=wxZ7=I%yZ3?fq^qxJ@Pe>Vl344Pb zp|AOupY!$AS~q?Jt;OK*fFz6CIJ0~8A!Q88S61r^abkZBb#YPU>Wa)VInV9KmTpQT z4Q2Cn^w7@szwQLKk&6+EmfkDkhO|soJxbpBxElwM)hcei)#;t{IQQ7fFUAln;At1= zf*-H;jVfQr^>bZV<~^*{pm<98N9jf&CN9mL}orp~r7X$)mD_4AVxJo%Frb z^(7>0`tiZ!HeUk>v;JnPNC;T8l}`Hj3p&HWwq#vKp~QM_i=8oht%+$h@*6ik<$eDk zhS1VBb-FNNy7Q(Ba_(H!n4D|N%@{-4JaQ5k)%R<(u$TM!z{Q7-oI1mDC}$^_0P5>( zu!?6y24|1f#D3>Pgxi@H^ybtE4=nJS-`P+~39@Tlp$H_my9Y8p4~$%b`Le~gvk)L+ zDSo5YO^6wA08}r6!cQ60f6PP;Y#eo2U|7OFOK)!E$XI`?e_DF=02>U7YMCvPYz?@; zvmh0k*-Xx41L2G%c3ZpII;Rd|omYMtorLL-VK%S-<2Sc0&SrZyk84XBwX;p*o%hE* zY^Pnr8?qrg^@M7Mzc@aq&4|?1U%ndzOaI7hDsRk{PRsHaQ-h|99tf;4_p5lGfG3*W zmu;a^+WYoP34e-i`Qav{a5`u{wdyWBId6N??xiCXXId;~L2k_ABbfv8`$S&!Q(*7y z@Gq`>1L7`gF*p7oOgsB!?%tdBIDsn= zfw|!tt>7PzZJLLd(#-bFd<_CfnZ|)fF~(BpGQs{~lm&68^GpP+HpCw=ggUJKDG_j4 z>Jb>8Xx2~Q5+P7jXF;CbkW~fi{fN_O_m;6|`FmgI#`W3w2l{3=usG2VaWeZr!VTC! zo-J%klyW2Tyn-I}!FYoe)Unu3jfDktr2zGQ3&=KropU@oXM0WdR#j`Z=&b4dDOmYo z?`*h?xbUotMI+}JiWBEC3tidka!^x(Iz}|wA=a966>s)e+sp81uG)Jp+3$;|b2`K| zP}I_ZQ?V2>CX*!2cIj!r)le!ZpqVg z_+tsQBOFbLNb#@9b&lhD0OZ%4W3SdHVqIfN_vr6FH@8 za(=(+J4=F?S=xDZnnP(gwl&qN15}d0F4TYs4p7XHJtln+)EN2l@rPH_V<5?_uxfu% ztn(#+WPs}GfA+l!nmX?Pz8Z?ZQRr1}m7^wXa;@M_3LOCX$ z4r%5{`dNrFXZRZtM=dCgU;OZYs7fjd{GrSW^m>)`MX9)ozJA$ODqZOZd{F^JtVU%? z;l94HW}Kj`BJ=fIG4FkCi4P0YxkuLt`9W3)VcR%TL6&BkL7mIJ073#5m!+obL=M(( zu1ec^^Lk~hme-f`cSAq}0IB|t4b-zV=yETzPQ(ibzAo+^!6NRw1@Y0S>REO-U63K< z9f7BVOdJFg=gx3ia)I^<7nY{-jVQ3CAWxQYN3tCBl^tv-0!3~&Sh{mK`u?)`VqO)= z3c;U0DJVwSq4?fa)M;(+o~%F@$mIt{^(0B=>sp?HRW#V-fCJ^hq1NcZ?g>Lb*nmVA zK3t!n*a?|!CKHI1qe;IYCkXaM?K!8cT28?FdujnT(B4*x19mhHu;a9(`dI*w@Gxwz z9f5e^x@lV8F}pHi4LndD&_MuDhVOnGAon#=_R{FxPmW5IMW2IC4M6+762(H<82*|| zZr&@QC-tG-E%!^U{$kn8u1DL~_4D&o734C;f=Ad%5F5u6lVI|Zn^ttD0bpaWvhHCa zQpBzSG769rfK$^L?UUl+k_LeOCIHdFa;!f5P-}+|UJOcr7Ng2H^HIE?T4Ueh$4;Xd zT}dkLw!7rb#8c|jT&lU=L?RLlnD~>%URqba#(UwHn$azDUZiD|WY40Qv#0*}eAui^ zuTrE%8+4xQa36`dyFIcIe-4y%YBT9?N(j>+DcW4m7Xqr6&w`O4!{4YHp^QLcfxMd3 z>Qt=)uq07g33c_4i2QtDwB703aX;^RU7!I+`NZ(RP)8?!Uzn*2!hn8ejUf$IWXB*kJ<_+~wPok0;XCf;xXJ*6@yzm-NnRrrVBb4RA9m|Se>96rs_uBP zkdk0JX%&~0U<~D-HCv8^*}lF}?M=*rQ~_xRUaf=LTtZhGhk_-0&+`mDOKXGqMT_D#Ih- zTaijDdKTE&*yQ{JKBg2@RC({-^^Pf~Pze_y=*ArXtS6yg2wCPA4K9~9?0_X1(>kIjEKRn!nl`85j^^kpAz8OVxyllA{4hIyb2~<~enLQuAT>+kc>k{VUm{ zAwcbFG*#UHuiCZdi^iAeh`L8%>!rDMTQ>YQM3ign!jg&YCk|xGh_M>Gi^D^g;^>Ke z4~EjhytPyI2e)U^x4CAgV;9~c>$4s;p7;+koW03f66=hUbFpXjtZ`ciuan?;~yinH>NAh_iQh|+G zeb)6hh)%zM54(mi+N*RRy+`fsbOyLkdpk!%Ef>@vTqf}FDQr_`_{>q@Nt1oTu-SMD zSqcH4_vD&RByB@ViU?f>m;hFw*lQD6e=sJTDZ>)zAy|?&NzFg6S!!s)e&MZtY6rZ{ z4~E zbBD8T-knOaWQ61$AWVP+X6@7LT_DK?!Zi7{4W-(%SQqc-(p=#bwmq9=s|~}yFtQLh zMk`L#^XeePc^Klo`9TOadxj(>Hw77Glhf0}Le4Iih*Kc3W7)a|IG}=mg%3buKFGZi zB;>Zm9b3NWUEzX|1;r};$SOG>sq8Xy8lHuf_CB^x;A03D_PO4&aL!?d3I}t%ie{yW-9#ZZe}h zAm4Zj4n}wpC1sn&vWaMCRDUw6XqT?rG;Rla4(H+>pW}=x%)|hZpj+QC$94Ut1`u+r zS!BbB@vnp!-JnI(Gk^xf{iHS!tvV4M91?iR1~84~l9^&QOM@)_j!rgk+`?x*gcze+ ze_dA6n`1BX<;hU#`N>jN^3g4iBcbh_F$;h#0UE)1#!?|_xNQMCPGF*;?)b-_WL7>^ z>vx@mB}f;vf`8Vb;%}CSMPb)WhzZ^q+jMboQmeJ&vx-0xJq8C%w4pN^997~aAYWnO zIN+C61!74cB1SN-p14i3EhmZ-PtMHlP(b{b1%PG7miEY;@L=0?IueQMzeo1=HV`Rv z{p+<=vpbcEjqGhe$+biE8!#L|QnoS?W@%y3j9>ydM6FY6i?K@TQje~ISMOOsb z?GL>ykVhKOb#Tslm9MX7*^BqP`PbNtotGAEOcW*o(HltBi4nPw_#fS}X29SSxIEC` zcw?b&&Xn{jJgMU!N27-K=r{kHh?4+fO4<^Tva(r4w~%!cpy zC`exb`-ljL1G#90-qzGzJvDT`?x*{`t3*AA+njPyHgg{h=p?p$g4=0VoEF0cYs_fi?0Ov(42oSyGw|y?C{t5QrE5PqTZ^eUVh``i~tyg zBo=sNY{2eaaay*XIONd-$&4>T?N!0v+DUohOa+5u;M9)itc$%TfE=JPxwpTlU^^Tr zEj->dEVMxmJYmp;|Il05ll%&aPduu*emsMGB$zj_*XOsa8~{Rzku^aFg|=t1a11pL zCYv;mIK_BJYJ4m1$@8w3@a3eGeKqtRr|y8$Bn7`MuGfKdKLX*kG_yk!_+az4`vV>j z(hq^`edW`ZsrdcZ<)5Dd694|Gf)t#a2u*k5YvY`#?c8q2h6pcE zf0b1wAp3oQAB9%Rar7(J^r$nh1IPi-hc1*yCY$1h%X9sS^1rJQOFo#y| zl3~bwpq2A#!yH*^oc|YU1422>GsgH3bB2IKgJclU{JYop-05s*P)_#k5|ppU-K$tQ<0ugHZYiwe zVlsU&2q-R^S!}3Bo^F8R*nRN9bx74*ZA-LpWM+*Kmb=I1cM1Sq9fi-FcVE_Bx=$C>6W?R^kumlB?a*iLa-p-~sg2T4&8|3jv;vgF zZQguXTt6iLfxUx>I|kP7$y$BZUa)!S>lmIA7N_Z_HrzHkUh}cAhU`{ykDtO=)>5<; zX!0#MkQ3k01Aqf5&8*?JDVqoild)xH?MS!Yw+=2A_5yu?u+mG$%dFZdsUMPOIZ!)j ze>|^s%hm7KMhD)lhuwkNR^Y6_$CuPYe3rI*dp)#YoUU*MR#yuK1_iM(I!C3is{m09 zZSv{J_{-DE5u!PrVZXvZ6YKW>jy-$BnFH_ut@!2-vvqHL zF7c!wr~cIst#i2mR0p8*iR8D6_}i{u9oP@S?2e4+@^Y8!kA`&kh~|=+|HNu}pn98;FUnbdm*SCrzG23fP>E>wDF6{;dbP#Pt zF6;k(@a#D_LPk8j{si8wOPYE30b+iz2_VG@#Qx20;$Qi^@;){!Q4e#7$^0*UwPQ); zB74$l&k8Y!JdnA4aszH1=+!iU|B38NPKnD3!)eF%z35t%l{@#vB_)PzmC73fk}oj4 zWC4ubf0^ZmI*Y{|97`SaXk>y7rkzJXn37u(q$1NWUC6Z-b$Uhw2fCkNt;W2tBOfFm z@fG6Hjrg&ZIQwps7R<>hY?4d>O#w(<9eDKyF?#+@Y0=ZeeeQpj3eLGC>1C0suDY?wV+8&UBvFxn*EMQ%V6P{n6_* z&+Y3z_Um%Yw2*h{e+Xs60OmoSjNwTIu>v%647eExTrG|HH`Rr*JqWc`en}M#{(B&> zRf#bo4NQN70##(8ln6AaJ#Ui8iy5`OAX_0uMQ(M$iJwUo1EIX08}*uJ2ciUt%>^VN zY&64&vUbkE>4YnpD{24&f-v~i?7N}KFGd3JUt{}5HhJ#aG2eg!jFMi1bQk~|1i}qw z(ex{D(~8ZBF^x*BCNVai0NKl67#+HhRKf=!833reAw5@7Tf20-!S^V0iUrC|sVM(4 z5JQiF#{zkGqqc)uk=(mT`u=qQks3HPion4a3U{FJ8Rr2gyT96Y4(}~rW+H4&9aq_? z{o;FSc+MY(wyMHnxI}dz9`*(RA3}j-zx0sC)^Ix>&`?oH34%n=0Vx3BY!a3a6e)Si z0A4OWUiV6P3=onR_aQ+gfb>8P1jN-q=1iy<$m2le%~EWwsaJ#wo@Fnz0odYG5Eo7g6QFgSBWU! zkfh{E&}DNj2h4R$2#NeFIK$86(gVGMc#re7uDY6aQfSx{K!gj?2!KTG&H|v<;l(j& z_Oz%50O}I2L4D`N=(S0ujinjr*|oqJm%7tBUP9JjBJ{u72}`cn+G7l>0#NoJBZpA? z7V3ZP5->-Z69Xp8Od+j*zX#wT2yOq8(F%nCzXvEB;1r;TRR0OiYY!CmY1%E6U!CgQ zS`5WNO$!fhil#?aA`da5QWa{s(k&*s?{o-g1BZW}h*ZqmH+1`H@M>y)&%!%$U9WDE^7@8Ki8p$||I`2uOmY0rNe~`OD$?|rJy{IA^y+|y17z2; z5P%kr<8shbP-t&<2MUu90|(A&W|iuNG@U9-Zim0<&MW%L4W(~^)MS67uV~P~hvXcV zr7xWal-h*k^h+-Xxy;lx_^Qv6=`o|cvX`^du>L^7*ig3!oQMG8d_iz1h9|weXEhhM zt{uQHslHKJae(5Rd$B+TB;7!53FOZO=-6qi)9Bz1)OF*e^M$vyoB4|bp;y#wixHFV z9ZujZqI2()x}kB0a-saLgP$R-+}n!_O60DHGy35+0Y2)&pCJR2n!&AoWPGbDA1pB# zz)x8eVsmU@ApOmoH#kvOqXs&ey}mi zNmOW9*ufCuT%C~e!SGQYD&= z6?Zi-sGOERspJ3Z?9Ah#+}}R#)TvVmokNi&EwUt`3>u+GH8cpNCcCm0W6$pBM9NZG zv!p^yc40KKWjm5EX2=$0Pj(u+=lYI%p7VQt&+GL(ujh|!bKm#ebKm#(`d;7b`h4D5 z_#RL2*~UJuO?>LmM_i$s^&gJ_D;@^3#6c3C0|#N_>X-z5?UPgA(#NwP(w)fEK3N3U zs?BFYoZa_0zt1){Ykj{I^Df%8#e6G#=;Pk+CE^Xy2XWHU4`UaqN9@AD{1W={BV0B5 zAipnWDnI_$O)gd{P32MHckk|SsZUm2R7?x9w>f$#nlp)(%<=(N9tf*VPfz}DRq;)Or_sEfhVM#n?l?kv7Z--1yn=YFlmHB06)n`^y%2Ck4n6y7Wzt>F02UQat41F%bCS zxwkg}!C{KzfX5rwa3Scspd|u2WO`xFX_%NG{tZu|>jHK}2s)=Bn}$j-+5#jRAa6zE zcRrq6a^}DUj&8iXasZSVC>H!3*c>ERYeRL@A7SNaykMZh3Y-xL?2{p2({i@4&C+)> zjeRur0L-Eqjz=Ak-}~OP8CV&CsVb{j3iI)?9P{)z*5@?j={cRvJoFu{e^HB|(_&*y z`UBJ}==ha>C<%v?Y6(OpIi_<5W=B8muL>ahSog><6KrS-Tz+fi8!}=3!y^MZ?n%)( zcox7T1C76u^fpLl;fP9~p7#(W8FAR;-^_^*CSQCTNkk8E&uHW7WUd}U+z6(MdggFB;&;EemnMzoWPbe#a zu>;;wg){+*7A447gby#g0Oj^hxmQ72urh&j>3^HHZV^G96exL5*TJ9nMl7OYKdfbl z+<>%!14x_ZmlaF>=)gcfXbL<-XfsVnvB0g^0BZxZ!b4YRhS}oyake>9hnmr&=ckR` z{$>oL4$BM{o;%O-N$tjvk+SiF7GN+fDCreD46ns5v*jMC0~l^s&+u=q45fl0b>4%x=&xDB zZnZa@RyJlk0xPxHf2-e$k+#IIt5mXjz}+1c zZlr1rVFvVfm>W{QSzD-ZMez6RboG*TL+f1=yswkrxt5664wQ0Q=s7+wooDnkLv6~{ zyWcx6`ZE}*Rh9R`r3h$CE^K|S5xr3c05>YC9c+BH=h{HOav3+C>diloFZNl~9f=yr zU2F|^V-&s=c$f=ajhCc$-!6Jpi{l>x%`akjKos*&UpG^iV%%Ay8_p6HZilXpOgR`pnk>QN`GL=-y?Mf0ZiXrDoFt99}9EEVRI#@^qzLhr4N-5xNyLjyjb z*kI+~(&x8~8L+`SWPz{y2B1(s~-H!1dDS`UOa}b||WtV^$=#cv;9<%b=h6RN*mZZmxsv~oM>OF$4PS&cz zSxBhCa+*}MG_C{H(5xZk!3t`CBv`En3A|yCzo@4+QAb@0ojQ^rqHsaSi*Nk3VMro9 zZ9HrM1tDk#P5wsxWHi%mKdq=FvhF%Y1tJ*JD@))O+RSxsNfoN#XiGnEmu2=`_4W|| zQu7gE2ujS@QKGw^XekL+ z&89GPwxY8kx&1i{jpoXucW<$GGvjW^_GGvsy#4FHuqOZE2?QdRCM{8caqu-Op&@;u ztC>`obM5jG8)#6!!%AEN9dTYwPa3?{OL+OD=$PVzJ)lEI*7aRVP*;Py@irI^_;gJh z+jJ5tifO%frQV9bwqBg-W0ue!_3iZB8-l|{xDbXSr^9vZMmks_NV3cyj*v}MOwV%} z-T82b#{91kBcz|n^cQx@jINMfY*koaL45Q3eE`JqD14w@RK#@QD|~}5TT;%$xx+X3 zvNz%=gcbOP_RxO!f8IfgaFXGRiUuZR9$B8$7Oh7#s@XSrgJ6DDFpvOJf;j&=#PUR2 zzq-Z{f4)M9z}dL)Sc9^7d43<-U$HTTh|<7k7#_<0*kYYC<-FGJ1Nw$TqN11gj_lnn zFzmd#?2dwhcXm*5A(eCb8nL#LKDDb~Z=73Hidc=ZT5; z2kW5^8gs5;`HM5T*57}_*&8>YQexv2T80=|{{C2*eSa4X{1vOssfCfQN3jnCe_w|m ztoMPCZ+4%d+q{xYvLVQPd1Q~T>w*)7)yPa-SD&c-&O^9>4{Gb(CWtqnf1`76WCFGL zC=auP)E?2t6kJ?558|3HaLz`yw;CB8J8G`5PGR70O`aGK!am0{TU7rHGygr4IcG2o z&jJhQkGqNPJ5n^lqlfM>^L|SiurMD4+&hH`NgVk5bpfvv)`-)<>*z&9qDRH=S$Z~k zB++*{)^%dxq%{ebP%*RO1+0{jOv84=3MYU#bE|6RRUkLGu5VHVvDqM&5jC*MQfC+WK78C{M)mV1My7QRW znsFtm>`H-NkMT@SC!<9tJOVn<{(aSmY}Apk^rzBFe?wN&Vu*;GGjtOq(r4=~7d644 zHR11ArUIw}d$H$(LC0ucH%BL6p{mAql!!l2|2osv@~4>=JpASIy84zazNhWz)>;FDIx;Rd~0V|eWj5e5j?y#K|6v=2%OID;iK48 z{{d~Ks#Np+XygPR+uG8X?|9uYG0aW*M45@t;@#jNbTOqq9F9rMV)h;xt3|ewB%zd) zPqO>Xzo?*LZosYHXvCVs2z&p8Ne5u*>Mv0~;dYaG%yMy{z7%J@5hTsvmbXdwZG@184+i-N9-0Qu^8()ib3 zrS^gp>zO0y!&;JgB0qw}oJA_jS@}ed``J`G)0t|+6{9X7vKbdMyq?aDKA9aUVm@en zcWoyIxl5%k0!!)J;wm62kcI#utKd?>_B|5i>Is1>`T+c~FL^jNbLj#wrXayND_fdF zfKN_``rVs1CwBHL6xjDH5lU}7K`%U-ZRKkxtk5hZye3DoJiCT~uL0zp-#FfUYpMK> z#M*ui=y_TWGo~vn)z4OZ_EJ+ToZAu$&+)sp`HM^*$I_;IU^jp}E+&Do{#c`iMh8z| z>8+tLihD=G4}-l?;wy9;F-=49a8wqa&>vwGel~fj?LivPQ(B!jf4PC>HW30FHaV z?}8Ef^o{BQNN@NFqKY72IAS_wv4G z%H^Rr_yr?)RmS?RKl_LM>2MLP*8I3sisZS~)ot6i^W?_L$jXXA{%k-`&K~HW zX|(0qz1wu&(JUaAMx$vw62{+478WMqZjN2ww0X1WiLRo79B@K<91l#cvhLPgZw^y~ru47$ZGYuXKDcm&TLPp_ozl(sQ%di zw>jG>y;W+Fkt;Y3udzK>D&qt2pU$pj!%!P2yNemmj^5=fure9{qMJYcDrd-&ZB|Cc zg-s$#&6OP9I8Ogg(7UgSuf?y9+_@R+yQV+4Qszy|aflbUrY$I6wEa**4qb{hu%2p@ z^qn(SFJ5xRCn`B~we^?WwmgXAIQ%9AC9LifzAuTkY2yo2uF9IGCX;6SX`k$DiK4Tg zKiAy6IumBBr=}J;p6Et?2DSd9dG&STYl~5zmKt)61eaU~byga_+Ka8-8v0%}b}I=q zGa7r;>5l?+A|WUt@FVbd$a3@1H|<4%4(8pFNJ}Y6Qt|QmCDzxMVkW}ZrkPyvsom3| zPPJt_|7|sH+hxqm{g%1Nb};lzE%|Sz8@gIGw^6z5dey78V;ml1=Qc~Yi_&uv*NV+$ z@ms`No@aL}j@wdA$irclHgf}#Jt>~lw;(ismsJ>onV!MnyH;NBKN&6(Xj`W`1sw>O z{12vJV%>}>FEpj?Iy_)bwct5Gj zfnkBLlIUM2T^FhF~OL3muxdv`0<+^GP=Q)ow3_fanhd=#S}9lEF|=?DN#^B;29i1J3QT@ zAvOgk%}EMMo}47wnB~R3kC9fr@cZhEw4mk;Q%($YG;GYC$8N86%tPceDv*XeypvMU zy9)^=KNF8QbNaNX=-*U25c9bIEyl2X{pRyMYh#mHR7+#V}@fGUyr6i{F{hI!uW zapLkW2OB}HLV zej7Ff_k7R||7@rU#5cZ4eANE^`-Mmi^U6Enib#!bim{fuaN+kC9~0PEk{&+Rbo#~P zM&E%32X&Ck8*SaTO{RMM1`AKfd*@hIVo7=VNww3z=7yY2lm}hf}4^i5w#9Bl&K44CY!BC2nQUsP2+&RT1 zj?Mz0)gekFP#t5wIhec`C=I)pX9KXj0#Fd?ahDfWQ+8BZ1DdoRuztW-7#GyD>YDvB z3aR8M0p9(S$AMx<)TqGp#NO(?PN#R~PgbPD+*g+zw(*xFhbEL@!2swClsQCw3AD$2 zU|#`XRc}{$jN1w&r(!T09x^a9+;Q0#K!+4g14ygel0b2+IG$T(1jH`5Xr66?THfBJ z`k9vl7da2!@4;qMn1e2MV&=1&5s z>&!-;nJi=OQZV0i6yD?;RSCZ0^<-Fq%QsU5%XpBjPY~Z;aWMoceGJd(c)vs`yML>F`C;Usvbln|IfIRlk z7=DG+FB4|(%_YB8xL5|gVX#szy@&MDUwl+0M*CV-?}SW^;X9}n5X#1J@Js|QD;2ox zDkxm6<895_#8|QzsKYh0lv8#L>gg-z2oO|D!ntZxMZE{(ly<)tImk)-*q;qWQmY_Q1A_i(}(BQ0(8++j~#ZRrIkO>2hJbBM!d2UvnN zGFYWM8sgoT{GLp7)HfPYa~p4JS;w_#fy&kNwgVQ-U=k9FwZC)mx6jRaJC*f2#Mde* zwav}x4|-WQm4$;t3>w7uVqx8D*C!qO(8(bs#fiNmp(Zq`0f)~r*5qd|bI|3qZ@rT> zv0!6(4j-Vj>sUr@;45LgepZ8%hM;n4Qqs&C4tI=8P%w3RTHV{HV%*{UTsQl~()6(t zdu)ypqQb|O6?zwcL)5=gmNtWJ+6@O8$5@2}8ufruTFNMy7->E$NPkT&HTKOLcETmB zES4U1TX^a#vfgXIu+!=ui;Rww!#@_>P8-*(uM`+{y1f0*)pjgTKXrh`kisu@B#L>y zAe)8rt*PUo_P=uH=fLa^ z>Et9o-jd@iDwEs|KuX!A`jJ^(A~Dc^qi1N}RPAaUyWocw@|ulx;>6UDvi|TGxhF~} zldvGs%_Vo?k(0Kk4;ZNf8r0w(z#=`Mt^Z!7F?(Pl6z+DKj7Z(!KVTR>(()X0+`@m! zK*l&qsJiO3%^-JbvEnW+E_pL1eLBu*nOV96_~X`7H4-AQOKIz!Zrd;SUG`z78XN&& zIZtdAa(-LTc!{WeHhXjvAXWqfbl@yY$%9uHs}=0i0R`(TKljk0u{5=f(M@ig+r28@ zaPHn*o#m}Jv}^fH2uGMUFE2IgDkoukT>J;tgk>2q3O+a7(`Y!~yK`GgBY_0m<4cg- zw)G_s0X6W+t?3B`<#O?;;c7u;Ve8&9tqPJh?x-_BwiA4}+vQohaQJf@ue>A4!lhS` z-A^zymPT*aBfhy@8!AxxU4eU9IYCn60uV?$OUAUGuwRI4DG1@9rgyfD@M+y>TJu<~ zg+&y7Pvd!F>dlTf8sx6kB8X4V^J$$} z&hYFNMv{FpJbhs5IEuM|05ICTS;5@ga zk#@*WI-BhBc|lpu2Grm1#^2DkfL~PXXIug6O5(0E5_~oi0FqJa2tJ6dTOAM=k*t`x zHzF4NdGd*ecd$ssx)^g)*n3i8t}v5&IeC^V`2r15-qz>vpCvc={r@-r|E$x&0iezL zzaI7gDeHr5Z(BsvMbyWE45H@>)LUTYo^bIqgr9)13@27jMpFfff_TZ&@hNwh8(r$h z!oZ^u4{4_K`kuxC)PvwZOGTL>h*+PsRW6`upHTHWS7!e(UFEW{cYrza3<1tAf);H;ST)D?0~L&FDz{ISTa;z z;Bfm}Eml&5M&h(q<-R`x=eGHO)e%knQdB|S6TtQ&x7`>}xBj;kxcIHQ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/generated/webwork/images/webwork-1382-image-2.tex b/generated/webwork/images/webwork-1382-image-2.tex new file mode 100755 index 000000000..4ef548288 --- /dev/null +++ b/generated/webwork/images/webwork-1382-image-2.tex @@ -0,0 +1,154 @@ +\documentclass{standalone} +\usepackage[svgnames]{xcolor} +\usepackage{tikz} +\usepackage{pgfplots} +\usetikzlibrary{arrows,arrows.meta} %needed for open/closed intervals +\definecolor{ruby}{HTML}{9e0c0f} +\definecolor{turquoise}{HTML}{008099} +\definecolor{emerald}{HTML}{1c8464} +\definecolor{amber}{HTML}{c7502a} +\definecolor{amethyst}{HTML}{70485b} +\definecolor{sapphire}{HTML}{263c53} +\colorlet{firstcolor}{ruby} +\colorlet{secondcolor}{turquoise} +\colorlet{thirdcolor}{emerald} +\colorlet{fourthcolor}{amber} +\colorlet{fifthcolor}{amethyst} +\colorlet{sixthcolor}{sapphire} +\colorlet{highlightcolor}{green!50!black} +\colorlet{graphbackground}{yellow!30} +\colorlet{wood}{brown!60!white} +%%% curve, dot, and graph custom styles %%% +\pgfplotsset{firstcurve/.style = {color=firstcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{secondcurve/.style = {color=secondcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{thirdcurve/.style = {color=thirdcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fourthcurve/.style = {color=fourthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{fifthcurve/.style = {color=fifthcolor, mark=none, line width=1pt, {Kite}-{Kite}, solid}} +\pgfplotsset{highlightcurve/.style = {color=highlightcolor, mark=none, line width=5pt, -, opacity=0.3}} % thick, opaque curve for highlighting +\pgfplotsset{asymptote/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{symmetryaxis/.style = {color=gray, mark=none, line width=1pt, <->, dashed}} +\pgfplotsset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\tikzset{guideline/.style = {color=gray, mark=none, line width=1pt, -}} +\pgfplotsset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\tikzset{altitude/.style = {dashed, color=gray, thick, mark=none, -}} +\pgfplotsset{radius/.style = {dashed, thick, mark=none, -}} +\tikzset{radius/.style = {dashed, thick, mark=none, -}} +\pgfplotsset{rightangle/.style = {color=gray, mark=none, -}} +\tikzset{rightangle/.style = {color=gray, mark=none, -}} +\pgfplotsset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\tikzset{closedboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},solid}} +\pgfplotsset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{openboundary/.style = {color=black, mark=none, line width=1pt, {Kite}-{Kite},dashed}} +\tikzset{verticallinetest/.style = {color=gray, mark=none, line width=1pt, <->,dashed}} +\pgfplotsset{soliddot/.style = {color=firstcolor, mark=*, only marks}} +\pgfplotsset{hollowdot/.style = {color=firstcolor, mark=*, only marks, fill=graphbackground}} +\pgfplotsset{blankgraph/.style = {xmin=-10, xmax=10, + ymin=-10, ymax=10, + axis line style={-, draw opacity=0 }, + axis lines=box, + major tick length=0mm, + xtick={-10,-9,...,10}, + ytick={-10,-9,...,10}, + grid=major, + grid style={solid,gray!40}, + xticklabels={,,}, + yticklabels={,,}, + minor xtick=, + minor ytick=, + xlabel={},ylabel={}, + width=0.75\textwidth, + } + } +\pgfplotsset{numberline/.style = {xmin=-10,xmax=10, + minor xtick={-11,-10,...,11}, + xtick={-10,-5,...,10}, + every tick/.append style={thick}, + axis y line=none, + axis lines=middle, + enlarge x limits, + grid=none, + clip=false, + y=1cm, + ymin = -1,ymax = 1, + axis background/.style={}, + width=4.75in, + after end axis/.code={ + \path (axis cs:0,0) + node [anchor=north,yshift=-0.075cm] {\footnotesize 0}; + }, + every axis x label/.style={at={(current axis.right of origin)},anchor=north}, + } + } +\pgfplotsset{openinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Parenthesis}}} +\pgfplotsset{openclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Bracket}}} +\pgfplotsset{closedinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Bracket}}} +\pgfplotsset{closedopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Parenthesis}}} +\pgfplotsset{infiniteopeninterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Parenthesis}}} +\pgfplotsset{openinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Parenthesis}-{Kite}}} +\pgfplotsset{infiniteclosedinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Bracket}}} +\pgfplotsset{closedinfiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Bracket}-{Kite}}} +\pgfplotsset{infiniteinterval/.style={color=firstcolor,mark=none,ultra thick,{Kite}-{Kite}}} +\pgfplotsset{interval/.style= {ultra thick, -}} +%%% cycle list of plot styles for graphs with multiple plots %%% +\pgfplotscreateplotcyclelist{pccstylelist}{% + firstcurve\\% + secondcurve\\% + thirdcurve\\% + fourthcurve\\% + fifthcurve\\% +} +%%% default plot settings %%% +\pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + scaled ticks=false, + tick label style={/pgf/number format/fixed}, + xlabel={\(x\)}, % default put x on x-axis + ylabel={\(y\)}, % default put y on y-axis + xmin = -7,xmax = 7, % most graphs have this window + ymin = -7,ymax = 7, % most graphs have this window + domain = -7:7, + xtick = {-6,-4,...,6}, % label these ticks + ytick = {-6,-4,...,6}, % label these ticks + yticklabel style={inner sep=0.333ex}, + minor xtick = {-7,-6,...,7}, % include these ticks, some without label + minor ytick = {-7,-6,...,7}, % include these ticks, some without label + scale only axis, % don't consider axis and tick labels for width and height calculation + cycle list name=pccstylelist, + tick label style={font=\footnotesize}, + legend cell align=left, + grid = both, + grid style = {solid,gray!40}, + axis background/.style={fill=graphbackground}, +}} +\pgfplotsset{framed/.style={axis background/.style ={draw=gray}}} +%\pgfplotsset{framed/.style={axis background/.style ={draw=gray,fill=graphbackground,rounded corners=3ex}}} +%%% other tikz (not pgfplots) settings %%% +%\tikzset{axisnode/.style={font=\scriptsize,text=black}} +\tikzset{>=stealth} + +\begin{document} +\begin{tikzpicture} +\begin{axis} + [ + width = 0.47\linewidth, + xmin = -6, + ymin = -2, + xmax = 14, + ymax = 12, + xtick = {}, + ytick = {}, + minor xtick = {-6,-5,...,14}, + minor ytick = {-2,-1,...,12}, + grid=both, + ] + \addplot+[domain = -6:14] {0.428571428571429*x+4}; + \addplot[soliddot] coordinates {(0,4) (7,7)}; + \addplot[guideline] coordinates {(0,4) (7,4)} node[pos=0.5,below] {\(7\)}; + \addplot[guideline] coordinates {(7,4) (7,7)} node[pos=0.5,right] {\(3\)}; + \addplot[soliddot] coordinates {(0,4)} node[above left] {\((0,4)\)}; + \addplot[soliddot] coordinates {(7,7)} node[above left] {\((7,7)\)}; +\end{axis} +\end{tikzpicture} +\end{document} diff --git a/generated/webwork/images/webwork-1383-image-1.pdf b/generated/webwork/images/webwork-1383-image-1.pdf index ff5c2985558203e9c014346080d5c51405507b1e..80d005070f1413c7b2b9e7e12162d9a7da8e685c 100755 GIT binary patch delta 4920 zcmai&Ra6uTxP{4~yE|kE>7ix@86*Uek^u&3iJ?0N0bxWyItS@aDQOVt?gnWAK|my> zL%8Rjr*og~`~ULqwf497+9^+Sgeie9l~shrMWl#Sy2bhu7V|y(3Kc%qOB2UcKBrZ&8d}!BZj>xw;c+;|2 zA4F9(e|n~zkn}u(=3K`Rc|1|JrAG|C&OL--z4s#Eh_8rCTp{pZ)nL8hzv_6ADW&msEIOrr2#{~|C*hkLr;a~}N z26A6EXAM;Rm9NG1CuhPbGJ1K@ig+&}*yQVz)9%P;vfM`mQ8^5V4~~fiVz2I6t0QNu zE2Dz@Cd#Wwf3Zeu?h6A@FfwK>Nnby2EWwEl<>sNiIKi;m3-NI><_ST;*9#$|hdG_Y z#^mc^1QhF0f6+tC*q*IhcUcEbf_6TkjF585U8yUVfjqB=&`haYF+4)PxR@v7J&A+4)*d3s z8UP^Xw8k35pO%RMVibn9T7}X`U-0hO3|YIK087&PGNL~RatQdH-#uk;(Viu#vm$Zb z+6x7Dh7SGm%L9FNvxUZ+r~ofRp%HEV+C|5D?-^M(BTC?Tl&Ih8pG|U_f8WIQeYk|q z#k2LHv3vRQ!S4e|NAyNn(;^6TYyiB&y<*|qToT$_n+i0LG<$9!Mf&p*4I0C=v3Mr? zB`z9jU8G00URp-l0M6u3^w_#=5)f|l*QvB(6&*naSVjQU@6eb`^Gq?-n+-Q8^fr85zB=tuIt0y5tT z(D#Y19@YR|y5HH!f{pl>!=^rchGsrNjzd|JzA5`>;cmUW56Yc$a-b*fVAI6&mYe*C z#^sd;bk3TyQ815Mq2f2lsd`9dPHVIN+CzB$(-gcu*R@PLx7(YaO&cu+O%rnzO|~-V zEm@i`Tn|R9@Bd&|-u2yAZ%05v+&mJVP>@eOy#)8de#DzcFPhZei0s8reUOSor;9u) zv1&B$M;?|o3;j0gnDf#L7YT1E*`thv_IU*8vaVi|{X`?1W1<*PQ?kvsLGA7e-6jygH`IXj*MgfgnWK*qt-nQf}Mn9xzYQYAr=~7vD zH1;%{Oo-piL=!4tmophAO_{7yTUa|M@@xTW1kZVUM97JDgpsAkB*71?Jhi%9?0Tu= zjX*|@OVEj>AGRHHqwUsfEYtY`S@S#zdDClbmMky)4+wSK{vs?2CIKaryMkpi%$eS` z_$>CO(?oq2-ws1_h4aWl!PDrj(0xF3Ou17h5jk0Yt;*V=FuCM?IMIvpld(r&0p(+# z*LS8)vqtG z2THH}_07+%*lbaVHleDlbIkp zN)pdXDS-K)Oi6Tbn$$b^>R5{?-a#%Jfr?b*x%w*S*%!1@TFOxO$~)Q1a>GFl~lJM!32o9f8di>VzTGG;809M3vdC z7@Xt@-c<1+>fWp|(fx zXfPZsBnd3cXpoe|e-p;^rKW2LCvEGg;CJ_oVycGav)3BTl@VdiGQSUtWI{@SrB5j- z?r(_~k0wbI$Dyk+VnPi)GY0L?NXCJ+8I)XA5oZ=Vr-D|T3}a1fU7kP> zM5KpWO2T@jwhy~O_Uy;kIK}??x_rE; zUGxe#kJ4){N9!mW1FwM(q7x4(4@z8Fc-iG{^;f(SgEAj_7f z=2z%5h+wkumv_b!&2=nVL>XUz+yKf<^=xc60>%GQMN*I1GT_NZM6G_{pk# zBNPWih^J`66-eUEg82)ToOd6eKc34I(9J(hs*snp8ifC)*@_~noK{p}OpPV=4_tw? z-|W)AB*{DMqn~PqA@G(r&U>eW48|*Ofp5@EfOW)PH_su9nntG>0_dIU9xQy26G_@- z?nd;vM|(9`4!yPu1@LX$+%mzM@xOJeEZ_k*#ugr;EBkQ4^W?REKHfQh;NDnk*jkE? z_!YzEV7cb*d-gZN*zIE9c~=I;*J7tGxoA6{(~MSQ z*OH4b9nn?(S^_K?6yeQ+J}QYbVfzXqL|cMv7hTKLq>@jI8qAO6r#-p7Z{TDhi<|Q1 zvMBcB)gX^@rfky=6{%02bBf+w`EGcY>phW%=u9UeyGd2tTZis0ip$C%xp(>*_h|3h zmk89>ifZ88 z%f#KlqicJ=N_gie{imGCYxB1dDK#gDBGOC#<%$82cRHgBb-6pyj2B_%;C12CGr zaNDUW*l&St?eQV09~;^7I@y~#bLRets{uDcw~O5!z8hu=0>z`o!^h2CH0i1At9++K zjrRr2qf*Bg>|HTC*ZxVvv{Wrn|Fg2TjAXI`&ssK>O}iy2v1A#5Y^_zHG$~}C=ayXZ$W0Y5qf`iT z`CNemSi#c6n@$X(|4^ra3y52U9Mm87ppKoi5nWNxAUyNG^n{w2XUKH~wwjJRUy~Du z5p1?X@sfYLjLbvUXVbbA^{KFKyg9%G9o^fyi|ppys59a=J?TeIB4pZx=9#yBM4Ip3 z&w4W(z_bz(8~Y*3qNCx3k0Y7O8Y_vI&#O}C{&Ko@+wxig)mNfb^y@FZ_!y3k{AVZ< z3%;1Jylo#cWY8D5$4X0bR>IT}vt!bk6bQ&JBZ-frX6wR;c(L)cdXWA9{{K(m=Rt!c z#sAA4)0bTiy_}S7w{)H>@JQaUMiOd%!|v^P7kiX_l?q?!M-`%w3D_0 zf2ENdJr=fifo}wjNhci(o@k*isPlRhSQ3Bc)|2|DxIlcO^e@IIs1`h(qP>%(Uq(-fLjHYH+rBgZ$iy7u*sWh zH8CjEA%m(UHlsi!Pl|JPp*|SGHAN!9xe-m*Szg9zH`~$i*2=5xjw<^&Wj@?m2v!4g zITt@c^_ErJw5fiqyPQ%odj6t`c?xhO>-nv+!s{N(G|sgFb3#pnStcj>MKNoaC;WVmn9#2%=Z!|8#91^r^D6mt9hlJlpMo3 zSXDubX4g0w$fKR$N!SKQ6Y9`9qJm5U&3IE79*@!!R94N4E5|}p`1wUdh943C{A2ke zF(LolVOsgqa}`x4HC{vJ2xFUuY)e*Uv{po{g%HD^(*smXwB(Yt&DVQN7j}zz321(1 zt&W#O5!dHbO;YzPNqFyfpG7nNQ5AI>rKynNiaa}-L_G%uszT;c1w4DgNWuhzPf+kN;HULDqlFicPsPSGTp@ir1vB! zpaA&`PBYab&yC3%st(-Wve8T51?e2Tx1p7$`DuUjr3F_ zRrTk?p88iaPX7qC%+SgcI+X6P0I#nFRNI1_o722IYad$3Mqj4k z^N=UAAN$=0P>?&irvEI7j|pBYs4w!g=zfEhW$3dWwLH@?PM8UOLRK~(ehm!aBBpn< z;7n~=d;-S?S`{GH+*Q?F(Mw zNAmigBedwmdTfV(%JYe@^r=RboAX9&T0z;nThT zgU7h3I>=~CU}x?MXfiN(W+|34QP84h-A z5C*{kv;a%}H@kpCQ)KTrX_cnv>1ZjbX?EOM`NZy9f-vA{+`O?{NV6t({Whg|xEWYM zJc7q2ALp^#VkPuud03+LAweV~==$1r(J8hfOrO?}_A}oxHZDIw59XMIl|)Js^3O&F Je5S2J{69)NJ|X}B delta 7307 zcmaiY1yIyq+cqm8uyjcZ3kWU^OG|fmC@rxF=z_q~v6R$OvXqk2(jC$z-3`(uf~X)Y zu*9p+JTveA`M&3yZ+{C)ux6hO(7 zJVFRchzZEZ06pODP#YM~XZid45f^F+vIz7kRXa|ct;H_JR#Z*s?IJGg*u7^VSYwHA z)5p42CVseB56iT`6`H?-LkngQh=DZsmSI@V?u-KRv#i-Ye*5y1WnNdl+{~^0o8y=K z`;QW1=oc*BuWDfG%QrOm%-&J0o~$Q9uMx78owA#DFC}(sgU(;Q;9UqZ;-5%Ot8(*2 z>gH^zHn!VEPE?ltD!oG{mg>E#m#0w?IubaPh8`K&J-b?Jx|Fv#OZQ(fe~r+C z;1(*Xo%yyuq=jc@9_qC+gH>bV2&hY&u($IeN*$0hujhG&92~0D&S<+<5$?k<7;zPv zO<7&yP%Ho2jT&8*6oZSgm(5P5zC5L8B>_rXkcq|V*rcLous=8N^rMG1t@ykNyPe&k z3XMvajqW{%R2Z-6Skhto8^H`4N0G;S+0yI z`5BERiX5R^DS8Vyi?tPx-he0R`Pr950P}s^-IWB@xKGuA`d%dt8y{3HlO45tv6q#j z8;zu@6;fU!e_`{?+F&^wAebU^%P3q-ZGzu$ZDeP&^B5B}NkO_4QmXL-?mZ|Cw&ua3 zyfJwC-Ky9aIMp{r{qVXNr|4KS^e7MX28nM}5}LBjW)8AfC(O_^8w^eLqN-2El{Pyn z;?$v{9*%;wD8zD+#ad{OXC(sa>>$s*OgO%c4=b&j(%DLKE53RGMNA);Yb|G`$hXu% zz}_YtKgWl0_>Mp`%>vo&uX*k^7LdLUcoN!p){S;)+kKDz>6r=q^wl`$yAwc8O-lHnPDuR$gG0tB1BQ z#RGtLK{+E)VO;*0M-0B& zlsL|gKe)3RS!O{&rkKRA;t`jLrT5n>1<3K4S^=ggz z+i?*hty2K4rGOmC3_iulsmKr;4=_V3hc=2Pa~kC2z&CnNe!sUY4H&m&{Jf75-5wZ_ zj>HWfk-?t$Vp8KxLS-Jj{3L-jCx^)yYvI|j<}gG5?eMd!LLTE-to&&s6Ru8anN2bc z7bYZIB$nP&;T(R>`}!6GC)ZDtd#?v)2Uo=q7M(JoIdea(IEy)V6=^%^-1I8rYb5G@ zShG$^mY$D!1=CUcR(M2`3BHo(6y{>>+Y5yDRnnrw=XQ)HoJCVl0fuUApy;lMsVKee zqoiJCAY-4+V-97Bl%;1l2t_^vnLOpAeAxOITOW9R8;+8(7UIDRr%&0#U*r0FdLM-$ zE)yp+!bM&5KXdk_i}dV$Xnc=paSX!n?DKeWv*BIIEqO|S6I`af7ze5o#vcCyOtgd- zd>7`+@pReebmM+P_YB-EIn@0bf0-fc`8$y~-<*nC&(_Sei1Xd3igm4}p)mO=a>Jqh zgPLI`6%B%4k1hj-OLi}SC4t-zc5tH#_SAt`1hY}2iCroyetQNTemO$c~LFGgn z=i%=^etDkp@_hTL`ungcX&N~#3%T&{FE$bf$*%5v60)7$!BXz~} zO`qP5%+g#uO=udtj>G<~Ti2bsuG=(p0BhUX-Tpj<2?%htkHv-cU8^Kj&#(7`t&eBG zFAmFQuLfZ4uNP=0`8n^34wv}vuum3qzM;DDm{6J|4EKZtQDub82tA5>oN#Lwo0YMe zl-cUV&u%1_=rYThUJ{D02h!@g^#?0!U$K6#nhyzfmBgJiFV1O(10KWFhUA>++*4{Z z->}t70MT3yIDU(WdTZH)vA2j$sVrT~OCsO^!*;y{PU>OD?h3{0<1xn%rTADkZ6p?< z>(85>dUZucwfnxYGRz?G+K+e~UNwHR70V$f|LsW}x9el|ve#}3c{C0R2Gq*L#tU;e zaaRi?F*eCGn?k6XBz6~-e6HozR#X{!q5S}FNZ+E8cx`>B99c?*@Ns~#{kb9~KonI& zq=|RtP6UsRny$kp<^W(n_JpFk@TgGLM1+43U_=Q2v#hby>vj$oAZy>}-fJ$T%5PO8 zWu>x|FM*kQ+f)hCt;%UD#Nyp`a11l~shv2S?EP9(QnsAecoG#yDE#ZO{EJ zYZ0qKG&}}NxgMAFb#`uxPMuwM^9;og4Hv)zO#+sG4sFAGohBETtj3CiBKGPU%5tjT zC(dRh0drP_mM=cOzQvL&cLO~87f=#^fD%H8iu^~B2SD5lz5!t2;9vm&0HS9qO3GT= z+=|);;(v3I7nt~~p36|&3kIwUM!?0xdlyXD7c@8zv9Pd?$p6M&(!X()>~CcM6YiN& zVBr7&_;8;{0RbUC0Rf1j>uXboO@V* zZj9xFH3G1`h$6WtMy{fN{Qh1J==zO@f5S#sqMuV|yl*JacsJDdxd*w2k0yM)h#|)i zNeMvAYQx{YnD3XM zRdHY4kMa&PMfki4lSLx@k@y&u`+qkrN7IyFSmglFGxOLNOqUGQIY^WjQxGyHCL|Cu zb8ScKjII>NbBGF-VS+o~--wSx_VvXfH;5jU!|`OIQmdKQ&dOQvO+cz>4k z`glQC{AItc5YTjlPB_3uGojb1k!wk_Rf8YfJL-p#IY>lHRGy<%bfgn97$27Bm!DOV zS~FnPCWv9iuwZC0uQ2&&3biAqR%B;-b7Ng?MFVmRA58<^$6zxlBA;8r&G6CGn0aaw zq;qvZF}yi?0$GE9+&#C*)Q2#7saOu^b`-^f}ezOrrZhsEhr-?Z=-#S6gQwJ^KRWY~b}wgC+>l!E!lRK8GRJX`!UjkKVm&L;JWUC1s_XFeUN7Pzqyu zQdrnKrKJO^U*y)FCMuu^I^5#D@1kxn|0U=rfThK9-KFXWDb%tqS;lJI_zH)jmD)WU z!K=`Cb~HPhKM-;-E|*Oiydw#YUmL5I)R@@{dKdU4TIYbtE3Gyf%0_ zCLz<7g6*gNsA-?|cOqUk{ik>@OkF#Gl>av_kiTR`m6Fm5G@7Qxr%Zvtt~BJFOPS{i zL63+}`&qd` zA=k*$f>c{O)}QDDW>untwqq`O;PDaan0tl&-wKG&{2~VDcgQZ4mEW`|g(&DNSwQsb!FZC#V;T8bt8T3&ox(SY`KaCZaf6zbGtNj}Eaj-dE8zGVqFVDXQ;-BHyu8zU7En z#PF>X7WUi2Y{BpKOzWv&DoGl=;Lwt{^h#hkJ*DtAh1Gzn@VhTVDvc8El0A6|-p#s_ z3dh`N6Z(MP&tG8Q-(=BhGXL~Ycv0|aGc+|T=UI?P;4C8eX>dnr*oOtzh|%DzQY|$n zUHt1>8~lqKwwm=rA;B(=9A9qx$wjkLO)|ce>^5x6E;$c-hs3N1u;_~Lk?54D-S?NO zjMkBF;^cmIxX8Bx7mp=%o=~1BRE-h_N(Xk_--=nqPoEcIMgbV?d)>5pnjWipL0`Vv zou~mv1`r{tfR`pRhED{5L9{&_%PL=*@hmK@ysgpgpP7m$_UgP$4=D-<{JwZ=4r8^> z(OD;VkNj#_oL@W*5zw;Xk41j7u-vK_D-4ayU6&OX#ROie8$f*+Vse9f8Bj%Y;Na@ZX=4D7EwbW)?^^Nmoa%-uXrg#^U}T?3m`$xY-Erk<*Il8DY0!=b{@g%`cG@WF6XW4sNR?3=O?G{NAf1JNZKgB^6jZ zD&E+=rPzfRnE+ts9p_4bz4lwNKPj%0epi^9{2g0pHT}bW>T8$PQQkb{{5iv#fomd$ zkjBTuQhm4R`Vv=-JvjZpASUv!S6Mjdp)y~F$yEY z0geg=(1krAP}!rZedR{rwK&BEGj(#3Ph2w!AFrdmhLe zCv|M~q#)Jm(X_LQS$?8V;qNwQVitQ&-fsVo1h=)U-Dyzpi<#&c8*x|7L3^%Ik$1q+ z$_R(1_VGP($jiF$>#eOkPlb>nBc9|Z#}e%%#T^_Bb*I zW~y2EC-zTw5oU~WEqvaq8K#FgIjeS*UZxBc#~g=NHf19RDn_l8ed;R2;v92rz#(Ghl(8f1z>+Ekq={BdTefRgfgB6n6L(^jrP6K!w_)5(k*=l0zfXm$<1~(X^ zDV-EOoD7W9ut;fk{zvbohuD>9$AXEAJD3!RIj1KI;DReqNIxej;CT~Avhm5^?^7IE z_~h%;^>^#j3b`h^?94N~a9nNKMG_W#_f=JYze~HHE2MQLcR*_Fl5ojPmxDSnm7gO7 zZI2*`BtHet-RDTaL~U3vC??2JXu3}PGf7YN1m3M^cKno}b_R1mhYhW8Ugo0cPb4c5 z>B$pL{8QQl*{DL@ff|P?>C!YkS=S3R`TNA2Rv`K*O9mq2TgfNks$6G!R>l4M9J(So zwZF2WLpM=kmDxD}mtR{EwrsDipBs9zR}^6okb3~%mBEAUqzLIasI!>=;*{~s)bgAu zz-nWsX>CT9CGw@N?Mc`Vm6>9_vn$0f7yd(R((FFu+YxUjUIc~bw7ls#T&hYw0x!{= zN!t~?lMOON?6$BT*5`Zot8ueW=;a8aMwd5BKm+~N zKcZO1JZ=P0TQPBc(562m6+V8u2G*3NpX^I=m7%#YDr@I-HoUMmep{m5)5?8^{d~03 z_P-Ea^k32_^dHstUoX-tK-7-mh9A+gMUzUt=~5fS&;MSM(HkFMQ4rBD8sjmzUgf4u zDyf!RjD92y^<03@;|z8TafSd?>_j%VObqa`bw1aIhKX_&a5;!*PSjq3Z`8-}O+7BZ z$zT?)z(FzB6`hU!l8%hb$%lIga-Y&xl~H+G6jO3YSpO1+JS&1pWR`N@Y)QV+yXGylSbJ_;B?p8ks zwZB#_-$;M|nDVtlOorvjavB42Le_WR+c;LXdJHD_Pb~2NUDk>JHEjqFf|GMN>73U& zalEiFR`lY|4dLb4=KiaB=4m2Xds@UGF$D|Jv{ zy>*rg)whZYdGq#detFSKSJ$(NuC76e&fHqCkaWJ1Je#Ru_vG}ng=+U;SC>JzpjZP~ zNJ3DzTdd|0n}}Eq&LCn)AAekTT#NUk(PZ@yn3J=5dV!yxe?fDUcYcIlqiJSVA3VV` zq=&D;tHaBssjdHdWP)4&r{<`(ad5LGse;7*{)6Pyf(M)zd&LD0b^%%a_;BW4$bAkK zl?P+Y4+CWPgQ}KCeSQph@8x`3FH_nd{hlq`S=*{{JE788I(!T{IcZt-*tv|MxvRGx zJU92FZ)K!GUIs9Yt4mo(OMS^`5%qA*9_z*KYg=cI!NNLNbNFj(*&!eS(BI-%N{Zje z_cfGX*Vfq(4&&DWK?M0#oZLO&AVFaWDuI*%AucNT&u66})!cPOlqz@~!#!>eBO>A7 zQs#-ls(JWA2lb(DM&w-vp0NWXps`THX#W=TaN3PWVvUT5J1VNF*W+@W<>BYxP^`~X zN?qZA2Y?rh)lLLNtcoG(WSIbD%}g_)g4)Kt@IIZezQ9+_l znrzdzlBmy7DbFn4%q|(LNSla!S=49m#vd}6<}1OQG@EBm#bKXZ^{_GlOi$nuala=x z|3n~44@Qe`6csjOcv3Su^U+i2#7&%4%xinlCxz$e`N~<->c#Ev-^Abe9*`=xT*UN5 zdo)_0KHOv%dbv*_DAE@ZTva2pCbU}B@YDQiaQn}k>B_8uy#>6lJUV4``9ZhvYtv5( zA4dkMOPh8uag2K*B7U^p<9c^e{P95l zEOokRwPI-Fh_7Vh7UVVrvEDc*s`I~d;9yJGx;%U;oEq^#JjQm^=?InAFfrAk{^?Hg z6&U%V*Bfu_agNqu)wu38kWK8e*fVv{1DQhg1(P|pYPVY^JS%5n;_@>+S#HSqMCwoL zn0T0PB8YkI_Htwc;paz#ccy4w2sHscx9xZ(#*Q?5o0x7Zu= z1tG)Jb!WCt9|i=ZL$$IjZvYB!ry!I1qqrPzLod@;0*YcenIS|seP zKe{Gf(mti)u#^EONYkW|dLU%rj!3I1xF4^}J}jNe-@XZvzj-+)^8S4r=)u&ToEF`$ z(ei=U(|syAQ3nqN=WvR02iQB*jO<#}mhU+%cYU#z@a;0!EPXYzF^|93uqY0*-9Ex% z*d}&~t&?tUok;WEf9iPXH~N(jesrg@P%1u_V8XZ#=<~k90`0?s8BX%ME2u}xhLm~T zw}LYrgAIuhF54r>vhJ3NDe_m>FN`xGs$4pg4+Wg9uu!n)Z|*psE#oykOj#o|;5p=) z&GYA}ST!lQF?l?4f5oA#N~O%q@vMxgb91NnV$0A+I6g5hX!%ezVUD$!b2CcLITLnuB@v9{2#KiyPN<3 diff --git a/generated/webwork/images/webwork-1383-image-1.png b/generated/webwork/images/webwork-1383-image-1.png index 5878a09c41ccc6a33d1b410f4b002cdbf0638dad..f2b76f3ec732560185a572ecfcbe7bf4467c19d5 100644 GIT binary patch literal 21575 zcmb5WcRZHu{|9^;8JVS|?2IUh$S5JQNy}ar%82YeuOu3xBxEKsGE4Tl5QH33@B94mbieA(b)DyN9_Mj<#{2XBypJASzNA34_sCuZL8z1zFJ46u!bCz6FGuI`p=JA$sM40ri3fOs;=`-v7M+L&+DrB!nIAlh z2b$Ni$o4(8+#D$^EOl7p^>adA{<_)mSy^3YD6v76CS<3|TuVDgfAI8x_RvpX_g^Ya z6@2TV!;_5$J_q+AOuXxi)H?w?jx!Ihi}@{frGB(dR(H7Wp&7{*`XaxDoF$@R@c^en z4fjQ#YqLAgH<@26n5AT2v~j5CzF_U7dUnB=`pjtoVPYfmx!OsU5O(<;-)3_;qp}|! zNS7nRwo&de^c-=TCX`~$t=zcV4%C;K?@VrsL+XtfRH_v;(`Ibe-56_v+HnWs%DcJR@O3Vduhh>`My+#Iz(-z0%$CR>kM8 z%!!jn`6u`XUVL@&?Zw;=Y7Z=rhrhYt{8Ik%LRwv_RvMkKf3os$%j)~s_axnCx;dF; z&A+-Yih1nYGG9jwIF+6Yhc8Ba4$lZLW#hfUMoC2y@)XBo9Ag$FAgRo&XrFZS^|E4~ zGL=%JN`q35vV!s<;Lvuah0rhw;*qc`1envYP9oE;$;*>f=bVZ3sja&$eNhTn0$Gs=0Yc?udy z$w{MG;prW(mKB+^n>0hNNG0{BNM={IkbUXO{F)}%7-}}BEuKN&5y`F7DlepwYq(HO zFaId>V79aNWR^(A?N`i+2ASRpq#c6o58A~ugbMeT)q82Qyjj;sEan7zO1614UC1Ka ztaeI0``XO&@+hC2ZJ#a4t!5H#5^@q)E2^{r{s8xYz5riFOUAj!Zyrx)b7#Bgm)&;m zoan4QTl=(DA&52di)yc$OY~x-*Xc7t2P5;Nnj+I)+>2_DeEe?QTHQr(BS5^nx5+Dd z&3%D}KAC=;KIP>0ySBWb0_*%3!)jf+qGaPn|K|@iA~j;4e~GeWzUAGfYmsP8XLikY z;j?DDNLBt>eUs8<*6^Rpvd^opDIBb` zsH0bS`s_igSxQ5bg}K()R#T$H9+vpd4|GM5Mb`G?&Rj+7g_=c@j;$_qh0Z0_)BMw< zy>%ULvek5Z#t9_cSV$$}+}=rCJMX(x<>v08;dy>VWU*;!dLd!WdeL&(eavc1@B2oV zdh4m`?7H~xtltNINU>sAFCSr}Y@_Mj?{-d6ghwRx1OwwQ)?NmQf+pj>O6O3?tDdOg^*Nj>^X^jfs1!*fU1AHlXm)@eO^Z&M9)=c+mD zZRdZQy40ioi|yC_fjPan3x<~cJ+TDmN)`^z7mTU@a_DZq%BIQ2!_arTsAQS6jpnTs z0h7w$os;)%qO|3 zn$1Tl5f1!~2A7H?Jh|5EFE-c4)oBrB_s=X%Fc*$`Ow~HqO=@k3c}Li2_lfzoejhE; z`=u9Kpl0j37`<4qV>Yk1fgP8wW6nKi;MzFjLX&m4RP2V3(@5M_+2o4*iY-kxGYiuk zqr1S7)5pDN=P>KWRn?s#r&&)*^h*Y?obWojowS*@*}6WE=VMD3wf9s|PsobYg8Q{~)y}9fs>R3K#}sXt^ga3%!F%eE8O3(kYVPRs(ezy3Z`VDwY^lDje z+xoV&&~^2DON*2Y-;?rZj%wEdII)>3UR6U74{ij(`Xk6Xe1!dpAkG2^GJ!#mvyljL z$S$V%iUfQ?VxppO5!pfids&w93_c;ZQ`B)p5CT5@Z^!%N0K$|Ol`dYm?$$lk=dB;M z5<9cAVmHRjV-b2>iIG#@MDOa}R+jWQ-B&ht(LXf>Z$CSKF1-#X-V#<5S(91!TKL%c zmdMLdp$0*WzPaLpTja^@ive*jKWG|=n!mKRh<-_|B*Jb@MSYs6_DakGsZ868A$c3K zEGxt2xmaiEOgw#wfp;(dZ*uF6MAL5UJKd`iv@F%Sw0sJW=Ah3^* zUd6A*MCTG0{LIcUlqA!V9tPjh`t75KuY!}DuM;9jkWxM-9>c(f98+;nRDQyl-D#C@ z96>O`k>mmUt}8lR1pq2dCa;~N&38|W(2lDe2J^@G%OXur!Z>&%ZsDI*C3i0=(=a>Y zH7$M?3%S~S=Q#7+$_&FVzz2ednW~3;wF}9E){rYi;FFn)vyX;j4DC zdVJQ;pB73jQw0NAY1dNbVIYBk)2C1S<*)b>6EGxvU+Of{Qd4_uEz;wEe(ck;V@vfX zvM+lf2)AN(!)qFs^C{QFcMP1Joi9vw({m}cntD2$&+n0?e>eR&Er7Lx^JZ@!UB|HN zSpyGD-%i_|t(tnpkWi8O>bL$w1V!EI^bN&%U^JFAb^nXMP+7zK%nbbH#h^NrM&Zz)}!zot&&ETPX3 z_~c26WlG0Q^Wz-nZ&ts%b>oC&O~?&O`sTkAVMnfV-4M0NWk zl6gfqVrheBHH#`VPp0&|pLzYA+`B(C){}I*+-l{*K^u2FhTKZ90G_x@1(#UG-qgO^ za>&cyh$*1|)^_BF;C{b-PDRl!x7N;U-tN=fH!I@2q(APb_Qd9^Wit_wyiBKBFW9} zOMG@xW_JmYLc!Y#N7~kNO^GJGi5@%%d7k%?e!nRO8;b-$-R-im9;{} z$#0oI-DJdgI{PkdP>#bE-)xeS;KPRxHCDT%YpLyuhNbTQz=RvSazoq}yP$Tm}^h;?Jd? zqGzi=KHU9iys70wRQA3&`8BIfRiH{nMka>oSRCW>l&i$kNms_aAq(-o~dAx`a0TK39soGy}GvxB0bnlis_ubdkZEAOS2j? zNmF=7zpRh3;TVFL(MAZL`;mW4`R9AQleKqP-&HMP!Wb;TziYU-LkJc zTi-Fm!FLKLXVV#9GFl!uNwOXHR5IR#5b?XkmoG0=#h3Hpo;@!0q*3@#?uy;Wb|+kr zYp;RZFf3znTRRU^NBpebm&hh{y_0d-9XX4p2>h5NTs%WBTKwl(qYfe>1Q3C1Gu417 zx1`7oM$^@52Qo~MhNgEZHobr%zoLr zdj+*0nc;_$dse1fM^O*g*em5youS}E*Pl0~^-NBa>)Pqbmnux!G?a*dEE1m7MmtS; zh)k#w$8bbxm;$}quknhtmZ{EjTj`jyN=oSw7WF5|iRvU`ZQ?1$^k?3mm|p&38+3t; zr1o383RCz%wFd@~{(08$au&B|Uh`b!s`kyY%|f%+jY3P4$B2n8P&Z_dk|3&_V7b92 zB?E}WiQ93f<4*~H7BTn7;4oSJ7ff;}5f)a~)s5 zWYjMP!cYhF!cVUp2+%;d!8|Yn{U(`u(P{NtdAe^Ma#2g}xP7pgV}vm@Yf3=0s|LVB zkOq=vm}>x`bO0$$6+Q(99)-aKfm2uxsMOpNk4=w~rOu1lmOebbpiww-0lxTDJUbL3 zE6@OQ8%|xqU12`Kv%-iy^nCQNgO!ks0PXkCF>XVl$rDQGt}pK|-)2G(Q_CD~h9MR# zSGX)5b+aW7k20kG>cF~Y-VlG9TZn<_H zCi{>CiOfNmB7~z6FJ?2pJAzMWSyLcbD91&o;vH)sZ{y|hmCobtHm{b;u4nlBi@_}n zLa-XQIwMl26z?_{*;)As>f=4qz-3e#DVH3y>C;RA@6Pjjp%F~i%KLM;VBB7;&|Pb$3avqj=g`<7qLAG6ZzPT@I#zfgjp(4*0%!V_s$j!yX& zvLpT0FuXe4YA>_nq{$QEMghG|@T=t~I1f4ua^PwU${mssK_b0gQvSgf$*2Do78S4dlnlLp^a35F4CGCOK-*hQ zAVW@sPym0sKFscJefp5gsQe z92AZ$%Q1LwhBEj*aVx83*?DdFh%;@>i+GrZFrOTuRSO zN|ht+KRvl;B}I38bxx(_SM;gV+xE+2I!3c}K{O;1uB{oCQNq0MjUpEG!}#6!l<&{9 zC;e;)*w}>E$(mMlUb)ENv)P(#sM-78u=;UDqLA{TQ@RJ3r`j{zhP_Gk3mMqW&6A8L z2c&!weMkpBuBxBkD(&A$@s_MhOsP9=y0UU+OUP9@s&IxX_r^<1(SE;ElO5uG z26YFR3@x^r8t+7*x-?AfZ~Yt~)?cxV16=mUx&UM|pW<2Z3`Uaz7*%ZTyW`>yrzGo4 zznkV23adjRyUd9`5bG}NyAycJcP#>Nk`Y~y`=1eJ&TG8M&0ZxLEKK@V?^XWf_dql- z@rtc<)}Oz=4y?ue_6b}VK_tH&8%zzGGj!{=)>Od$3Ieg3a4xH-kN?~kg~3;MWH_x} zvyEc<=i6LzyRKJixKqw$I>EX@V@l9ayJh*&uw!v zd|kNe^yc1K{1>X<-7XXXFef0I&3NxV;c)F3{=LZE&h|MR=z)|6K>hvdxI{7T4(fTI z4`MC-?mWH!k{DJvgFwvQ@(2Ng4@k4OZy20 zzzx zIJ~tSAsR0uN!0{s`K+J9`^g+fR%QrPkypI8$WrkxJ?eh zr!oIKD?W32d7D!q?UVCVIN?fQ65zEj0Vdp0J|u?jv2>vYUwUX_*R-Pqx1{DHm4&eP z9_0Tn4~84RJT_v8dbu+-0YktCG$D|EY#rB}THl|?Y&k0~dUM;$dz#hNLQ_}x9?Y2` z6N;kag^!1_&k-@3``@g@pxjlMjg#ZbPg~UkRe{E0{ zId#18S>8u^)Dvoz4ZK@%@ibn!7J~zK{Ol^Cs+|weAhiMrYYKkI4~IY5&_E^r5`zfT z;`6E)Y)U$H#D1)?G7ttY8x+;A44KlTWtA%(9_OuBR3~@qNjLYU`(AYXFe3j5ABQfy zAAZnl+?VzG_0*eLiTEl492L~l7QK!{WTHXRjg# zsZci`6L;q~WLhn0Hm`Fw486W4G(h_BQEc+Z)bas3y6@&ZXU{Hn`z;&hD~9Uon%;QJ zFym;tIA^pRWxDA*ala$bJdtu^ZCsqy_+xy6VYc7Z9m5Efl)VqzT7NKNhcm{Dr#yR> zMmOwRHmVuXtnygWN9g9umjRe;t>QK-!C#LA)>pY|rNTOmd<{I8ySS7@qSexOOdj(u zQF~YT`pyu=p6fj1v$0dEtF9ixIo_65;N7GaEuwZsOJ>$@4~4L)yz0k5f<%+e3)IyS zuN%IfV`G}|-Tt9ew$L~8sdf2IbJ2^Hfs`JWlTWmcbTn(_w#Cl{x|78Te@c%wGd%r+ z!VM^)>JXz)n^?!EbEJ0DF)F?*uJ=NdY)7^Z-QH;OeI{HwR^r?=IQh-ceFWR@vvK?O z>ZE7DoJGP4i?^GI@*#e0r`FFZQNo|B3Y$Z|)(x~DWY=;{!Y5d*))XU=-qmI=2Cing zM<%bzir2XDsUGtcmW?}Odu-cI+16(5oWZa8KBCoUMmvt)%bo*AzBU(a&oM5Y>-sq? zdW~eS4Owt`#y3$$QUAkuV1!&sQK8uyF^fGnb`@MJD}#uoWb9$#JEfObCTASvepcO6 zB6$o6WA+L_VjFHMe5=k?$7ob9&yIOkf#u{888l|0S2=P8M} zNZ3EAL18mD&v#RPi&|uTwK43D-x07i)PGGz%<^JvTGJ~$bQ>acc=sYfv}yvtby6~6 zNqf#QAOi0YCOX#in%C~ix@G=H*{#0{QnOo5=I>Q%>9O4c44ftY;|sVp?*W9^uJok@ z25ItGvnLcFs|-UDxS?ihZxt;9p?g#E|EME?;x}C`gM9M47}dBl{6rekf!$pq`b8 zE%!uHy57A{6%IK{05t3?c+$Lwk1RWap(L{V4XlaLsQe#bjSt#*U=23`untJ<$^uC9 zA7D+6rX3VmBem3xry*6?A=3jykqBiFC7LSy1)?y>@Le>AlHsKBMGuztZ(xn*pm*~b z`uxB08TxcAn$Pel@85ie{)jE2gv#MO7N3)o7wTrV6=yWILlHcB5RhsXFX&&hdC*km z@?oqO@CiVv%fLeL9zb(F4r-m&%tRWS&LVl3TqQokaf=~Z+28n6;*l6gLj(+wY-s8g zxBzM2pBk+T*bt)Kx}a7Em{BNdrbQ9o^^$=O-}U9a4}Y&+WihUA+2W3ko@;i}qXT>K zOGEwmrUpPF@8G%*brTCallx+ga$h8-5b|#`H1|UC7>nnRs^H?qS4rHGKiu-21#x_%Z5Vz~3YQYVF?|pBX8- zaYZ#utF>ZTCtg|w9t7WFhhTmliPjZS%bQo@vf`|o%66J}Dw4`6dVyR%v0Lvy7g2u{ z-vld%K2ydBA|3NqM@ZY&_KOR)o!XkOr8cXZoxVAKvPH~jF!)i831_n_^1ZtXj_uw= z+1NVX0aT~m^$@7qk?k)chY{L$!3V?A$Lo$#yrVC<${a>bbmL}O(*4L#(6Lfcmf(KFuD&giAr!QZYwQF7@i@o*YM%N+Z#W@v| zO}fDk59lQ8n&QqMQ}QsLnel2mshuM&WcFF$;?}?pTlbfE)>>A5^Jcm6UqxR^E*0mN zc?Okp@Ud?R9lt)w%0EyUB=uR=?s}5ewG@4k9~3{K6!+N-6+WLIF&C0sB6Y5Ftn0ny zHKMp^amP>-6PXR?rI?T}@Jh*-iZW6av)O&!7~MBn-H-m5WANP+6f{%|CNGOmxboC= zMJ!=eXyR(N-1#vMnf&c)=D%^kVA?hQ$)+C^0eqM~#%okkK1xBpguO4A#vHBZf0Q3C z-6`d1YCg}e0_Cjpo13Aqe#OpBa|A)?-F%VJXH&>rza`U=KcIKD$(vUv zHoG%q-CpwTVEB{G!|l869Cp3`EMJbDF%!bZ$>lTR$i5_oe7*0RYCOH z0U_@<(5$(liZ=;C@V7~anV%hJbsu%M0#bvJiiyE%T&+%|A~T7IC3gMuznsD zpV_u)UG|?Y*FHI7funqZ5b16p!75SSY6uDTEm|3{@c=p8o^`6oOh>i?h6r0aJ?Vj; zIyVKb-6|i7%ohkglL%A<2beMU=z~W?O1yFM-99Y-y3xJ3JIlzWw>K^Kd2SG*URTBS z4}XWC(Q?7!oz^6jy#tOw<6GELVye)>5{po>;R{O!QQ(%Lzu^^ZGyr#_cU)tzIV6Ul zV^KD@j_-yMl&$8x!Ej3I(VUf$871v9uz}4+`93rfP&le-y%c zvp6Ix^!nXoiAFy%#!S-S32m4WVtw)*?-{A+qq`&Cg?rKG6XvWhBZm4>DM{@rJp0jj zhLeEQfurm?N(IS)gw_R3EtH=iPyn9+`!wMg{QK73w2xngfb={A;0!nMJT{*5`)4F{ z7Jrq*yHEeSB*tHfFNx6!0a5WMr~F+KqgH^H#L#0VK#WI12DYB|r!W3hG=PAG$w4tR z^cmfA2~qnmJvWk4c-Yy4;o$(_#t;w8rMFf9ctk z5VUw;sz*?smC>64 zI0;%RuYhOH(wYClisjm)e-0&@{KEhf05kwN2nD|~ zl!=PiPvWus=p=y(r}BvQ5oesK7}W>NeZ6Ss3|F&STktUla@dbABHDC8>b@&I%3COk z@xOyMC454hzv%Q9uX8JTt8iph7(f9_VEzOyM`Vcye~!Oh|4>acneV)!T&$RY6?h^c zdgWfoYhnrCC5J#*ujH&f5KC08bNKDXOSU6CY}SMc7|1we_w;P*V3g_=|!>{~+&%I9_j5eDThE@-@lCqunE}-YtT&URZ*y$DRiC;#9W>!T?p=X zAB%l?UJ6PNS`(lreO8cg3qSiZA@ioo#JHCp9v*({o1(&W`i>tHKnqaniAF(0)~x!z z=z!NX^JZv7Q|JR(1I=Kn7mc9lQTN{lI zl2Vl1%&}74aV$;jV#%TCeN+860QEROEF&pSzQjhY;t^}4I1b@jC%ZucOoZT zTbqmnhaAV6#(Rqf+oitu7CFwl@l&7J%!1g0hJ=Cwu^W=eSX1n?^o>$uDct$<=P&Fu z6AVdgtCW=mY!fcI@sli(D#B#^#=b{xjjf{T6Ho2cbED7jRnEQr1dMCk*@5jdexa#$E-z zMaE%E$+1_7i8mgQO=Z59TL1Z~xWrS~agZ$1tM%zvH!L@9u*+WtBp>K3@&3L7xq<999 z@EM<#l76_=_qwO&xr$=wVyQg5-i^f zF1S`7FB#Nf>*FY6v^_7==kRPUYj)?SjK9zJqn-4i=7HBmjwy#7HWfl*Tc+ZpGS7_( z`kelfw_M?+xU*Sx!l&8$=J=uOv!XQPPETC5ABQG=B+D+3b&W@Vqbdl<$LkrLJSROb z*j6k(d(w0G`@ld*OQWcI=bLQq!}CPFBhwqi8S4{MK~iMfbpz@pr?;JCgLew*45B1P z-oM-awGhS9(*SQb>1^N-x4<*md^&*17gZm`e;ZrT&xyj60^E+$le|8vl~L>*JSX!* zukqPZzo&0YXpPYqgN2}iG1rK5SAD&5WsTNb#ujUW#sipkUeItjt(cEZNS%6Ux_RRz z0l20Q(d_(9O;qZHd3Nue`Q+WRm*O=rpl=zUN{(OUyX69!uffWr1c*HM0hTCJ%Xi61aq2Yv>{T*f<3dn9=@n?qQWN0>pm_wua;~9|KE6|g zB-<}d(9DnxayM#I^a2lw% zuydrJ)N+%phN|vK)}4>F0Re#)P#uzE>(LnhdKTPvZODVn9k>EwT6H7Dw9Uh*KnI@* zoRmM{m7-2Ft-G3fSjdcrJ6Wsco^fA-6lLH;^k)&5)nS?-w z`Eb#yHt18p)+VF(<}+@&$x&!%nr+@aQ%RPG@Qv56p4wZPzLTVMh-!l+NLW_DXTj_x zpt!R2^b`fB_@6%qr<<9$5L~2PUOq}F3?MNk4WUr9L!cqs2Q@YW5sIAul2AZTlqlR0 zsXh3M>pA*8m$_+fu;dSDdHrtD4=34**WKJp!0@F`^L*;-1;N)SR{K=4;8j^JzB{(jAwM0cclK}0006Ip*TnP3n z=^zADavacYfL?F{lpVwz>}MAL8Ir)Qesc}F3BJd~8u44Tr7X%2?Ek zl~fZm{U>cIZ}Ec7Q8L_J=Q|}^wiTJnLT%(o(0z(w1Q!B8kbo!$)`9T`A|M95)!^BG znv#HvP*9mejw(5zb^{s*vma_H>0?l=N% z3)%dSasu=%2Pny2fdHj(^MjV{&z(o~k8T##n52Q$)BGg;qm}v{Ci+LOznHw`3~Fb1 z0gLc;PZL=TlxOXtX#apqy_{jz1-u4YEywqgs4ng(pb97#i_U4S>y97Um2Qb|RVotA zQY+mHJPeXCTWywEk22>(=2MX1tD>Nf*S*9=P=d=mtFMfR+y^8w z<#tr-V6#mE1CAJNy6_1am7iYNcnE3C=BGW#SK)L|X~X1_BnqEDO)hCF>$3A+Xi2Ak z#O3~e*o>P&TAOrGT1{ZRhplxF|9eirRp9YJ9$$mrvxn#F7v%Qq-4o4LAdAB{l`wwj z13iwqvTb%i(`ntvija8l#!}(pAlv`q|+3-6c+rMoQ7+g-2QTaQ;aAktrMtdbYX-W$B3)PT;DxES}MA*+vcsgWASRKEE7lyYdYddM}j_w@bSNmP4x@Pugkh!bNCVUVe?=XSpxsW`aH zHByYPngGZITlX|!!v~%vfRKn-e?G_Nb~5ZeGy$OZc5Dw(5C=PhdOi4Ea_hjJEL2GZ zRW*Yc@u^(uNmP;gM*PR+M)ta#X(G0tx_)y>;H?4%yAmR%tP5*#ArYhB!g5Y;-WzF=WHa58BIWhs5qq8V#ISnxcWWwEWrVX#2Y>B_ts2VmP8E z@c9(*`-3yG20vYv_TNwMBqbsZAQF-cpXH$5f^L2;=1iZxBw9-H=|tAsh9epu_j zMyscNSWpS79I$bpjXHb-Vb8wkZRPdsBz*(sf>_GVvSGICb-i@Z1TZws`8l9cCbZov zkix9~qkdP8`Dxr~EoMfllQoip)${%<0=m!RPHCf0hNFzMH#!$-+H1Z}E}i8XFKb6P0-$HoaK(D1XOE3$x5Q0Jqg0Sb9H>t++-h${HOsEO zOH|Nuef{l0eEU5L`ZJm%SS}bvgS4ySZp# zk(0@Bz=DMpr945H$FH7%~iAt%Yj%cf|mw_#3Z*w(0;mng%c|pv~Nx zCIH_r3;eln0^ zs)+1%Sa5GCrjryA>`PGZ;35}*M%dMJ5|Io79*e!5pow4y;XCQe@V-bTP={E-*1zNx z_#D5>u%U1Lrycc|4sLg41)<O9q-Py9L|wIa@P_)yA^6rY#83uD(f z5nRrV8#lE5=1Q-G9ymDO-o_h2rBld{wrWqIrw=r61gv-i`Lfw zC~K#*&R0NEpE&NY=^J0xHln$(;+vX#reaR~P|4t>zPw>yi}_4P&ZU4Kt99K9ab)fi zwXD9XL0ZpDgI0vzZ+Jn%JNYnZ@m3T;1rh&}ZkwZI)gOt~ZZjP7AT;TJv*0m595XVk zqd3R!WOMB))Uv3^x_0te8xY#Sbq8;pIto(GH=@)X#KWFiQ5b!)PjexGD^mA0H(o$X z+jlfb1*OWXAMa$2ewvtmYcbCWBz=GjoQVJ;{g2QLaHCnhG`@RRO@SjQ#-XwwfPdH_ zV92o-pzVWT_tzSF-Bw4Nh(L>l4#{R+|ye;Q68Yr;FqwO3GHz~yY#e+#d| z*1QrR6<*^(ogeFt@5dozK@`JSfxT$(0py3=e}TbW|2wh(9fKmv$kRyc9%^)OT9N*Z z8r!9k3eVcV;|k1m)PDJS{A5{*@>U#Z(a`7PoVty+QB)lGU$GnNPFQ!QQ#@KisJj=k zL#K{G82!D|bk2$d4o-pwL_G?(s3p8U4qje0Hz z4p#m*dA*ck*K{1J>6R*=`G z(f|JR3LPa}~GFn@nHWZKs60`Nu{G+q*agC6Qn zLQ({`3ULLsHax5D_qWnE!la?)Rg1*K-Kl8(3Ek+tV*d|!iypNiz!wj&uvjFc{5&=j z724w0S)XK*u#N@x53XrV_-ph3k=ec_-Mtb>chpGy`Onr8&@rH5b#`=ags3zdl|LW? zB!FMgVbQ)ehEn)j`mR*xe;5LEHGY7MhbiPih{kA~%x;bI?=)*`P>DQVv`7qFNQs5< zi2wGA-$VB1{C^9y_MF_cJ4ej_A3fHfe@Lb;4aOJMJ%aQMI6*igq6p5~}7R=N!If4}pZ7D}Lx z<4$|dpd}*e?O-6Hwc}2fPym5C1GwLAY=Mg6{np~V7oELh^5B6)RQ#Ixn{x&S9tKG`3=0xT%OUA!vmxYw_Q zedp1m5B)ruVJYbfp-$;ZdK`&uKrKUe?)eQZNa4jJk6g!>ZkKFHlKeK@M3K-w78Lfv z0yE7y7V7pmFFs{W+tMC>Qm`dPYXmZN)MV}mK6@JXs;Z9VDhS>H7ZWxSs(9yC5Dkg@ ztPZWs8hIO;`}(d2)B&xG8hua^pg8JDmy;)NYuy`Lt1b>n?2>%*=k&yNXdw4Ph*9;U z7{)MMO0gQOD%#F$4QO#mx&WGoM2~)TzD?l%@v(#n_-^V=Z@@a$x+=YJLMzdy-__UM z?>mSR$Vj*i0d(W`*aMR&{7e4uOlL>PdQsoZwh~~TesCJq52zUW>C=`UN)c3RgitQs zfX;CXaB)&kviKu3H@_LBrZ>EP%($Z*Vzb?mGqZ8ZNMDfKbhWKOWTK;D-Q#2XP9nv= ziK*w@>hs4>x9lpOW{NDn);yxG4v{+BogQg1{CHvq!MsH%m zVf@^{T5v~(PU3XjgY24bE89t{wA(@6PL)SLnRGQv4bx;j(yGe2Vo+}VYX(W|jHj$z zeSBPbf5o1A(yLT8F9riex1Tpr?ep94wNs#Rq~};TVZ^X(H3b7dRhVDg-k6zO^Id<7 zFO~LQinY;s!W25QPpGo8M%B(vkdEodusL@mrEoXpfi_x++sfI~s^yrEgdJsq?X$xX zNhPD+2OLa$iiYnPhe!=iKUHm}-qMY|6tO*jN;Ew)Q!Q|(xm9bKn#KL`5h>?Ky))aE zZF!1DD~D^i?)7^*?e?|Te`K=ko>Uz8&q>7^Beb=B#Aeg>jJhV&&&jIgOrcV;ftzvz z%xQoJDD`&!R*TV(Ht!3C(=bms%eFeyH$|G^G#RS83iOI8GTN9=t=qqmriV5zk!S40 z96!*O24H!V5hAj=tLa#1^QFmwh}~`J`;`!LK0i2z0RDQHHy!29kDL;t^hmq%AUb4^}bg%H$!sh@rExT3A#@b!dW}0z}5`- z(azJvnVykxr;QjDXf!R?u;vdanI-3ZEnxp!2b-!j=O`+uNH4 zj(nhH27L5tvs)hJ+vE5e6Yt}5i0Pg|Xdn1AuTelwfqcUd^}jS;=%vPJ#n65u8#-<% zLkWsa(9|M>hZ;{~@F9T(^>oSyFE9-CkPz321oWao$ejQg63sEFl)0MrZ@d8lrIRL) zgwTv1@QrX5gVz1VU&_2!8y0 z6FQB*2q22XHOweTgAc0${?>SV?p_JF0tH(L_kQMK+IPPd%?KP>f@8%N&=yA!zk4sB zNC#~)DpaG(`b*l4T2u9I9~?*yYRPqAHO1__N6a-XIN16I2^F7bRhd@ zYeGTRkoA`q4!`%XS|}c3Bj&J%{09{?6m9>4ZPZdsp>Raa?sy|wXTmnllHtJXfC~k2 z1iHZzyN-jm5;$oakNnEUBY5
    Once upon a time, the New York Times
     1 
    nyti.ms/2pupebT
    published an article about the movie, The Interview. It included the following quote:
    The Interview generated roughly \(\$15\) million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.
    Sony did not say how much of that total represented \(\$6\) digital rentals versus \(\$15\) sales. The studio said there were about two million transactions overall.
    -
    Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog
     2 
    www.joeydevilla.com/2014/12/31/
    that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
    First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.
    +
    Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog
     2 
    www.joeydevilla.com/2014/12/31/
    that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
    First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.
    If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both \(r\) and \(s\) are measured in “transactions”. The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is:
    \begin{equation*} diff --git a/knowl/index/ordering.html b/knowl/index/ordering.html index 6236376bd..4b2c6e9b4 100644 --- a/knowl/index/ordering.html +++ b/knowl/index/ordering.html @@ -20,7 +20,7 @@
    Use the \(\gt\) symbol to arrange the following numbers in order from greatest to least.
    (a)
    \({-7.6}\quad{6}\quad{-6}\quad{9.5}\quad{8}\)
    Explanation.
    -
    We can order these numbers by placing these numbers on a number line.
    And so we see the answer is \(AnswerEvaluator=HASH(0x561336f98588)\text{.}\) +
    We can order these numbers by placing these numbers on a number line.
    And so we see the answer is \(AnswerEvaluator=HASH(0x561336ec9a68)\text{.}\)
    (b)
    @@ -30,7 +30,7 @@
    Explanation.
    We can order these numbers by placing these numbers on a number line. Knowing or computing their decimals helps with this: \(\pi\approx3.141\ldots\) and \(\frac{10}{3}\approx3.333\ldots\text{.}\) -
    And so we see the answer is \(AnswerEvaluator=HASH(0x561336fa0030)\text{.}\) +
    And so we see the answer is \(AnswerEvaluator=HASH(0x561336ecced8)\text{.}\)
    diff --git a/knowl/index/section-fractions-and-fraction-arithmetic-3-2.html b/knowl/index/section-fractions-and-fraction-arithmetic-3-2.html index f6111c857..15179b67d 100644 --- a/knowl/index/section-fractions-and-fraction-arithmetic-3-2.html +++ b/knowl/index/section-fractions-and-fraction-arithmetic-3-2.html @@ -17,8 +17,8 @@

    Checkpoint A.2.3. A Fraction as Parts of a Whole.

    -
    -
    +
    +
    To visualize the fraction \(\frac{14}{35}\text{,}\) you might cut a rectangle into equal parts, and then count up of them.
    Explanation.
    You could cut a rectangle into \(35\) equal pieces, and then \(14\) of them would represent \(\frac{14}{35}\text{.}\) @@ -30,10 +30,10 @@

    Checkpoint A.2.5. A Fraction on a Number Line.

    -
    -
    +
    +
    -
    In the given number line, what fraction is marked?
    a number line with -1, 0, and 1 marked; there are evenly spaced ticks, with eight ticks between -1 and 0, eight between 0 and 1, and so on; there is a dot marked at the fifth tick to the right from 0
    +
    In the given number line, what fraction is marked?
    a number line with -1, 0, and 1 marked; there are evenly spaced ticks, with eight ticks between -1 and 0, eight between 0 and 1, and so on; there is a dot marked at the fifth tick to the right from 0
    Explanation.
    There are \(8\) subdivisions between \(0\) and \(1\text{,}\) and the mark is at the fifth subdivision. So the mark is \(\frac{5}{8}\) of the way from \(0\) to \(1\) and therefore represents the fraction \(\frac{5}{8}\text{.}\)
    diff --git a/knowl/index/section-standard-form-5-7.html b/knowl/index/section-standard-form-5-7.html index 7db47d945..431e346d4 100644 --- a/knowl/index/section-standard-form-5-7.html +++ b/knowl/index/section-standard-form-5-7.html @@ -14,6 +14,6 @@ in-context +
    If a line’s \(x\)-intercept is at \((r,0)\) and its \(y\)-intercept is at \((0,b)\text{,}\) then the slope of the line is \(-\frac{b}{r}\text{.}\) (Unless the line passes through the origin, in which case both \(r\) and \(b\) equal \(0\text{,}\) and then this fraction is undefined. And the slope of the line could be anything.)
    in-context diff --git a/knowl/index/solve-system-application.html b/knowl/index/solve-system-application.html index 018b69736..3032f48ba 100644 --- a/knowl/index/solve-system-application.html +++ b/knowl/index/solve-system-application.html @@ -14,8 +14,8 @@

    Checkpoint 4.1.14.

    -
    -
    +
    +
    A college has a north campus and a south campus. The north campus has \(16\) thousand students, but enrollment has been declining by \(0.8\) thousand students per year. The newer south campus has only \(2\) thousand students, but enrollment has been increasing by \(0.6\) thousand students per year. If these trends continue, in how many years would the two campuses have the same number of students? Write a system of equations for this scenario, then solve it graphically to find the answer.
    (a)
    Write two equations that form a system for this scenario.
    @@ -35,7 +35,7 @@ \end{alignedat} \right. \end{equation*} -
    So we see that we can use \((0,16)\) and \((0,2)\) as \(y\)-intercepts. And then for the first line, use slope triangles where we move \(5\) units to the right and then \(4\) units down. While for the second line, use slope triangles where we move \(5\) units to the right and then \(3\) units up.
    a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8)
    +
    So we see that we can use \((0,16)\) and \((0,2)\) as \(y\)-intercepts. And then for the first line, use slope triangles where we move \(5\) units to the right and then \(4\) units down. While for the second line, use slope triangles where we move \(5\) units to the right and then \(3\) units up.
    a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8)
    (c)
    Based on the graph, how long will it be until the two campuses have the same number of students?
    How many students will each campus have at that time?
    Explanation.
    The lines cross at \((10,8)\text{.}\) So after \(10\) years, each campus will have \(8\) thousand students.
    diff --git a/knowl/xref/definition-cartesian-coordinate-system.html b/knowl/xref/definition-cartesian-coordinate-system.html new file mode 100644 index 000000000..5cc89167b --- /dev/null +++ b/knowl/xref/definition-cartesian-coordinate-system.html @@ -0,0 +1,20 @@ + + + + + + + + + + + + + +

    +Definition 3.1.9. Cartesian Coordinate System. +

    The Cartesian coordinate system
     2 
    en.wikipedia.org/wiki/Cartesian_coordinate_system
    is a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed (positive/negative) distances to the point from two fixed perpendicular directed lines, measured in the same unit of length. Those two reference lines are called the horizontal axis and vertical axis, and the point where they meet is the origin. The horizontal and vertical axes are often called the \(x\)-axis and \(y\)-axis.
    The plane based on the \(x\)-axis and \(y\)-axis is called a coordinate plane. The ordered pair used to locate a point is called the point’s coordinates, which consists of an \(x\)-coordinate and a \(y\)-coordinate. For example, the point \((1,2)\text{,}\) has \(x\)-coordinate \(1\text{,}\) and \(y\)-coordinate \(2\text{.}\) The origin has coordinates \((0,0)\text{.}\) +
    A Cartesian coordinate system is divided into four quadrants, as shown in Figure 10. The quadrants are traditionally labeled with Roman numerals.
    a Cartesian grid with Quadrant I marked in the top right section, Quadrant II marked in the top left section, Quadrant III marked in the bottom left section, Quadrant IV marked in the bottom right section.
    +
    Figure 3.1.10. A Cartesian grid with four quadrants marked
    in-context + + diff --git a/knowl/xref/definition-point-slope-form.html b/knowl/xref/definition-point-slope-form.html new file mode 100644 index 000000000..25b2b6eb0 --- /dev/null +++ b/knowl/xref/definition-point-slope-form.html @@ -0,0 +1,27 @@ + + + + + + + + + + + + + +

    +Definition 3.6.2. Point-Slope Form. +

    +
    When \(x\) and \(y\) have a linear relationship where \(m\) is the slope and \((x_0,y_0)\) is some specific point that the line passes through, an equation for this relationship is
    +
    +\begin{equation} +y=m\left(x-x_0\right)+y_0\tag{3.6.1} +\end{equation} +
    +
    and this equation is called the point-slope form of the line. It is called this because two important features are immediately visible from the numbers in the equation: the slope and one point on the line.
    +
    +
    Figure 3.6.3.
    in-context + + diff --git a/knowl/xref/definition-slope-intercept-form.html b/knowl/xref/definition-slope-intercept-form.html new file mode 100644 index 000000000..f189931f6 --- /dev/null +++ b/knowl/xref/definition-slope-intercept-form.html @@ -0,0 +1,26 @@ + + + + + + + + + + + + + +

    +Definition 3.5.5. Slope-Intercept Form. +

    +
    When \(x\) and \(y\) have a linear relationship where \(m\) is the slope and \((0,b)\) is the \(y\)-intercept, one equation for this relationship is
    +
    +\begin{equation} +y=mx+b\tag{3.5.1} +\end{equation} +
    +
    and this equation is called the slope-intercept form of the line. It is called this because the slope and \(y\)-intercept are immediately discernible from the numbers in the equation.
    +
    in-context + + diff --git a/knowl/xref/definition-slope.html b/knowl/xref/definition-slope.html new file mode 100644 index 000000000..bb6af0ca9 --- /dev/null +++ b/knowl/xref/definition-slope.html @@ -0,0 +1,18 @@ + + + + + + + + + + + + + +

    +Definition 3.4.3. Slope. +

    When \(x\) and \(y\) are two variables where the rate of change between any two points is the same no matter which two points are used, we call this common rate of change the slope. Since having a constant rate of change means the graph will be a straight line, its also called the slope of the line.
    in-context + + diff --git a/knowl/xref/definition-standard-form.html b/knowl/xref/definition-standard-form.html index 6fd0aa2b7..c9d903932 100644 --- a/knowl/xref/definition-standard-form.html +++ b/knowl/xref/definition-standard-form.html @@ -13,14 +13,14 @@

    Definition 3.7.2. Standard Form. -

    +

    It is always possible to write an equation for a line in the form
    \begin{equation} Ax+By=C\tag{3.7.1} \end{equation}
    -
    where \(A\text{,}\) \(B\text{,}\) and \(C\) are three numbers (each of which might be \(0\text{,}\) although at least one of \(A\) and \(B\) must be nonzero). This form of a line equation is called standard form. In the context of an application, the meaning of \(A\text{,}\) \(B\text{,}\) and \(C\) depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations).
    +
    where \(A\text{,}\) \(B\text{,}\) and \(C\) are three numbers (each of which might be \(0\text{,}\) although at least one of \(A\) and \(B\) must be nonzero). This form of a line equation is called standard form. In the context of an application, the meaning of \(A\text{,}\) \(B\text{,}\) and \(C\) depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations).
    in-context diff --git a/knowl/xref/example-systems-of-equations-substitution-intro.html b/knowl/xref/example-systems-of-equations-substitution-intro.html index cd9f97963..2877b5ee0 100644 --- a/knowl/xref/example-systems-of-equations-substitution-intro.html +++ b/knowl/xref/example-systems-of-equations-substitution-intro.html @@ -13,10 +13,10 @@

    Example 4.2.2. The Interview. -

    Once upon a time, the New York Times
     1 
    nyti.ms/2pupebT
    published an article about the movie, The Interview. It included the following quote:
    +

    Once upon a time, the New York Times
     1 
    nyti.ms/2pupebT
    published an article about the movie, The Interview. It included the following quote:
    The Interview generated roughly \(\$15\) million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.
    Sony did not say how much of that total represented \(\$6\) digital rentals versus \(\$15\) sales. The studio said there were about two million transactions overall.
    -
    Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog
     2 
    www.joeydevilla.com/2014/12/31/
    that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
    First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.
    +
    Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog
     2 
    www.joeydevilla.com/2014/12/31/
    that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
    First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.
    If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both \(r\) and \(s\) are measured in “transactions”. The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is:
    \begin{equation*} diff --git a/knowl/xref/exercise-negative-to-power.html b/knowl/xref/exercise-negative-to-power.html index 83688c7be..19ebbf0c0 100644 --- a/knowl/xref/exercise-negative-to-power.html +++ b/knowl/xref/exercise-negative-to-power.html @@ -14,7 +14,7 @@

    Checkpoint A.5.15. Negating and Raising to Powers.

    -
    +
    Compute the following. In each part, the first expression asks you to exponentiate and then negate the result. The second expression has a negative number raised to a power.

    (a)

    diff --git a/knowl/xref/exercise-point-slope.html b/knowl/xref/exercise-point-slope.html index f161dbb67..635259990 100644 --- a/knowl/xref/exercise-point-slope.html +++ b/knowl/xref/exercise-point-slope.html @@ -14,16 +14,16 @@

    Checkpoint 3.6.5.

    -
    +
    -
    Consider the line in this graph:
      +
      Consider the line in this graph:
      1. Identify a point visible on this line that has integer coordinates, and write it as an ordered pair.
      2. What is the slope of the line?
      3. Use point-slope form to write an equation for this line, making use of a point with integer coordinates.
      Explanation.
        -
      1. The visible points with integer coordinates are \((2,-1)\text{,}\) \((3,2)\text{,}\) \((4,5)\text{,}\) and \((5,8)\text{.}\) +
      2. The visible points with integer coordinates are \((2,-1)\text{,}\) \((3,2)\text{,}\) and \((4,5)\text{.}\)
      3. Several slope triangles are visible where the “run” is \(1\) and the “rise” is \(3\text{.}\) So the slope is \(\frac{3}{1}=3\text{.}\)
        4. diff --git a/knowl/xref/exercise-solve-system-application.html b/knowl/xref/exercise-solve-system-application.html index 018b69736..3032f48ba 100644 --- a/knowl/xref/exercise-solve-system-application.html +++ b/knowl/xref/exercise-solve-system-application.html @@ -14,8 +14,8 @@

        Checkpoint 4.1.14.

        -
        -
        +
        +
        A college has a north campus and a south campus. The north campus has \(16\) thousand students, but enrollment has been declining by \(0.8\) thousand students per year. The newer south campus has only \(2\) thousand students, but enrollment has been increasing by \(0.6\) thousand students per year. If these trends continue, in how many years would the two campuses have the same number of students? Write a system of equations for this scenario, then solve it graphically to find the answer.
        (a)
        Write two equations that form a system for this scenario.
        @@ -35,7 +35,7 @@ \end{alignedat} \right. \end{equation*} -
        So we see that we can use \((0,16)\) and \((0,2)\) as \(y\)-intercepts. And then for the first line, use slope triangles where we move \(5\) units to the right and then \(4\) units down. While for the second line, use slope triangles where we move \(5\) units to the right and then \(3\) units up.
        a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8)
        +
      So we see that we can use \((0,16)\) and \((0,2)\) as \(y\)-intercepts. And then for the first line, use slope triangles where we move \(5\) units to the right and then \(4\) units down. While for the second line, use slope triangles where we move \(5\) units to the right and then \(3\) units up.
      a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8)
    (c)
    Based on the graph, how long will it be until the two campuses have the same number of students?
    How many students will each campus have at that time?
    Explanation.
    The lines cross at \((10,8)\text{.}\) So after \(10\) years, each campus will have \(8\) thousand students.
    diff --git a/knowl/xref/fact-parallel-lines.html b/knowl/xref/fact-parallel-lines.html new file mode 100644 index 000000000..b73d8ac2b --- /dev/null +++ b/knowl/xref/fact-parallel-lines.html @@ -0,0 +1,19 @@ + + + + + + + + + + + + + +in-context + + diff --git a/knowl/xref/fact-perpendicular-lines.html b/knowl/xref/fact-perpendicular-lines.html new file mode 100644 index 000000000..104343300 --- /dev/null +++ b/knowl/xref/fact-perpendicular-lines.html @@ -0,0 +1,21 @@ + + + + + + + + + + + + + +in-context + + diff --git a/knowl/xref/figure-point-slope-multiple-equations.html b/knowl/xref/figure-point-slope-multiple-equations.html index 310da3153..4ba8d629e 100644 --- a/knowl/xref/figure-point-slope-multiple-equations.html +++ b/knowl/xref/figure-point-slope-multiple-equations.html @@ -11,6 +11,6 @@ -
    Figure 3.6.13.
    in-context +
    Figure 3.6.14.
    in-context diff --git a/knowl/xref/figure-spa-collapse.html b/knowl/xref/figure-spa-collapse.html index bb163cbc1..15d229b82 100644 --- a/knowl/xref/figure-spa-collapse.html +++ b/knowl/xref/figure-spa-collapse.html @@ -11,6 +11,6 @@ -
    Figure 3.6.7. \(y=-130(x-2013)+2975\)
    in-context +
    Figure 3.6.8. \(y=-130(x-2021)+2975\)
    in-context diff --git a/lunr-pretext-search-index.js b/lunr-pretext-search-index.js index 0e45088b9..ddd159115 100644 --- a/lunr-pretext-search-index.js +++ b/lunr-pretext-search-index.js @@ -13957,7 +13957,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "3.6", "title": "Point-Slope Form", - "body": " Point-Slope Form PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In , we learned that a linear equation can be written in slope-intercept form. This section covers an alternative that is often more useful depending on the application: point-slope form . Alternative Video Lesson Point-Slope Motivation and Definition Since 1990, the population of the United States has been growing by about million people per year. Also, back in 1990, the population was million. Since the rate of growth has been roughly constant, a linear model is appropriate. Let's write an equation tomodel this. We consider using slope-intercept form , but we would need to know the -intercept, and nothing in the background tells us that. We'd need to know the population of the United States back in the year 0, before there even was a United States. We could do some side work to calculate the -intercept, but let's try something else instead. Here are some things we know: The slope equation for a line in general is . The slope of our line is , or . One point on the line is , because in 1990 , the population was million. If we use the generic variables to represent a point somewhere on this line, then the rate of change from to must be . So . There is good reason to isolate in this equation, so we do that. This is a good place to pause. We have isolated , and three meaningful numbers appear in the equation: the rate of growth, a certain year, and the population in that year. This is a specific example of point-slope form . Before we look deeper at point-slope form, let's continue reducing the line equation into slope-intercept form by distributing and combining like terms. One concern with slope-intercept form is that it uses the -intercept, which might have no meaning in the context of an application. For example, here we have found that the -intercept is at , but what practical use is that? It's nonsense to say that in the year 0, the population of the United States was million. It doesn't make sense to have a negative population. It doesn't make sense to talk about the United States population before there even was a United States. And it doesn't make sense to use this model for years earlier than 1990 because the background information says clearly that the rate of change we have applies to years 1990 and later. For all these reasons, we prefer to have the equation back when it was in the form: Point-Slope Form point-slope form form point-slope When and have a linear relationship where is the slope and is some specific point that the line passes through, an equation for this relationship is and this equation is called the point-slope form of the line. It is called this because the slope and one point on the line are immediately discernible from the numbers in the equation. There is a subtraction sign and an addition sign in point-slope form , and you may have trouble remembering which is which. But remember that that the line is supposed to pass through . So substituting in for should leave equal to . And it does. For example, consider our example equation . Here, is and is . And: The subtraction is exactly where it needs to be to wipe out and leave you with . More generally: Alternative Point-Slope Form It is also common to define point-slope form as by subtracting from each side. If you learn about point-slope form from some other resource, you may come across this. We feel that the form will be more helpful with college algebra, statistics, and calculus. Consider the line in this graph: Identify a point visible on this line that has integer coordinates, and write it as an ordered pair. What is the slope of the line? Use point-slope form to write an equation for this line, making use of a point with integer coordinates. The visible points with integer coordinates are and Several slope triangles are visible where the run is and the rise is So the slope is Using the point-slope equation is (You could use other points, like and get a different-looking equation like which simplifies to ) In , the solution explains that each of the following are acceptable equations for the same line: The first uses as a point on the line, and the second uses . Are those two equations really equivalent? Let's distribute and simplify each of them to get slope-intercept form . So, yes. It didn't matter which point we used to write a point-slope equation. We get different-looking equations that still represent the same line. Point-slope form is preferable when we know a line's slope and a point on it, but we don't know the -intercept. We recognie that distributing the slope and combining like terms can always be used to find the line's slope-intercept form. A spa chain has been losing customers at a roughly constant rate since the year 2010. In 2013, it had customers; in 2016, it had customers. Management estimated that the company will go out of business once its customer base decreases to . If this trend continues, when will the company close? The given information tells us two points on the line: and . The slope formula will give us the slope. After labeling those two points as and , we have: And considering units, this means they are losing customers per year. Let's note that we could try to make an equation for this line in slope-intercept form, but then we would need to calculate the -intercept, which in context would correspond to the number of customers in year . We could do it, but we'd be working with numbers that have no real-world meaning in this context. For point-slope form, since we calculated the slope, we know at least this much: . Now we can pick one of those two given points, say , and get the equation . Note that all three numbers in this equation have meaning in the context of the spa chain. The tells us how many customers are leaving per year, the represents a year, and the tells us the number of customers in that year. We're ready to answer the question about when the chain might go out of business. We need to substitute the given value of into the appropriate place in our equation. In order to substitute correctly, we must clearly understand what the variable represents including its units, and what the variable represents including its units. Write down those definitions first in any question you attempt to answer. It will save you time in the long run, and avoid frustration. Knowing the variable represents the number of customers in a given year allows one to understand that must be replaced with customers. Knowing the variable represents a year allows one to understand that will not be substituted, because we are trying to solve for what year something happens. After substituting in the equation with , we will solve for , and find the answer we seek. We find that at this rate the company is headed toward a collapse in 2022. illustrates one line representing the spa's customer base, and another line representing the customer level that would cause the business to close. To make a graph of , we first marked the point and from there used the slope of . If we go state by state and compare the Republican presidential candidate s 2012 vote share ( ) to the Republican presidential candidate s 2016 vote share ( ), we get the following graph (called a scatterplot , used in statistics) where a trendline has been superimposed. Find a point-slope equation for this line. (Note that a slope-intercept equation would use the -intercept coordinate and that would not be meaningful in context, since no state had anywhere near zero percent Republican vote.) We need to calculate slope first. And for that, we need to identify two points on the line. conveniently, the line appears to pass right through We have to take a second point somewhere, and seems like a reasonable roughly accurate choice. The slope equation gives us that Using as the point, the point-slope equation would then be Using Two Points to Build a Linear Equation Since two points can determine a line's location, we can calculate a line's equation using just the coordinates from any two points it passes through. A line passes through and . Find this line's equation in both point-slope and slope-intercept form. We will use the slope formula to find the slope first. After labeling those two points as , we have: The point-slope equation is . Next, we will use and substitute with and with , and we have: Next, we will change the point-slope equation into slope-intercept form: A line passes through and Find equations for this line using both point-slope and slope-intercept form. An equation for this line in point-slope form is . An equation for this line in slope-intercept form is First, use the slope formula to find the slope of this line: The generic point-slope equation is We have found the slope, and we may use for So an equation in point-slope form is To find a slope-intercept form equation, we can take the generic and substitute in the value of we found. Also, we know that should make the equation true. So we have So the slope-intercept equation is Note that the slope-intercept equation has an unfriendly fraction, and the fact that this can happen is another reason to use the point-slope form whenever it is reasonable to do so. More on Point-Slope Form We can tell a lot about a linear equation now that we have learned both slope-intercept form and point-slope form . For example, we can know that is in slope-intercept form because it looks like . It will graph as a line with slope and vertical intercept . Likewise, we know that the equation is in point-slope form because it looks like . It will graph as a line that has slope and will pass through the point . For the equations below, state whether they are in slope-intercept form or point-slope form. Then identify the slope of the line and at least one point that the line will pass through. The equation is in slope-intercept form. The slope is and the vertical intercept is . The equation is in point-slope form. The slope is and the line passes through the point . The equation is almost in slope-intercept form. If we rearrange the right hand side to be , we can see that the slope is and the vertical intercept is . The equation is in point-slope form. The slope is and the line passes through the point . Consider the graph in . Using the point-slope form of a line, find three different linear equations for the line shown in the graph. Three integer-valued points are shown for convenience. Determine the slope-intercept form of the equation of this line. To write any of the equations representing this line in point-slope form, we must first find the slope of the line and we can use the slope formula to do so. We will arbitrarily choose and as the two points. Inputting these points into the slope formula yields: Thus the slope of the line is . Next, we need to write an equation in point-slope form based on each point shown. Using the point , we have: (This simplifies to .) The next point is . Using this point, we can write an equation for this line as: Finally, we can also use the point to write an equation for this line: which can also be written as: As is the vertical intercept, we can write the equation of this line in slope-intercept form as . It's important to note that each of the equations that were written in point-slope form simplify to this, making all four equations equivalent. Explain why there are some situations where point-slope form is preferable to slope-intercept form. There are basically two steps to convert a point-slope form line equation into a slope-intercept form line equation. What are they? If a line has equation , we can see that the line passes through a certain point. To find the -coordinate of that point, you might look at the and have memorized that you should negate that. Instead, you could train yourself to look at the and realize the important number is because that is what it takes to . Review and Warmup Evaluate for and . We evaluate by replacing with and with in the formula. Evaluate for and . We evaluate by replacing with and with in the formula. Evaluate for , , , and : Answer: Evaluate for , , , and : Answer: Skills Practice Point-Slope Form A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . Point-Slope and Slope-Intercept Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . Point-Slope Form and Graphs Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as Make Graphs Graph the linear equation by identifying the slope and one point on this line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. The slope is and the point is on the graph of this line. A graph of a line. Applications By your cell phone contract, you pay a monthly fee plus for each minute you spend on the phone. In one month, you spent minutes over the phone, and had a bill totaling . Let be the number of minutes you spend on the phone in a month, and let be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone. A point-slope equation to model this is . If you spend minutes on the phone in a month, you would be billed . If your bill was one month, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. The slope of the linear model has been given to us: per minute. So now we have that . We can use to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. A company set aside a certain amount of money in the year 2000. The company spent exactly from that fund each year on perks for its employees. In , there was still left in the fund. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. We have been told that the account decreases by each year, so the slope of the linear model is (dollars per year). So now we have that . We can use to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2029, the fund will be depleted. A biologist has been observing a tree s height. This type of tree typically grows by feet each month. Ten months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the tree grows at a rate of feet per month, is the slope of the linear model. So now we have that . We can use to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Seven minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . We can use to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In , there was still left in the fund. In , there was left. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the company is spending per year on perks for its employees. So now we have that . We can either use or to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2018, the fund will be depleted. By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent minutes on the phone, and paid . In another month, you spent minutes on the phone, and paid . Let be the number of minutes you talk over the phone in a month, and let be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone. A point-slope equation to model this is . If you spent minutes over the phone in a month, you would pay . If in a month, you paid of cell phone bill, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies you have to pay per minute that you use the phone. So now we have that . We can either use or to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Nine minutes since the experiment started, the gas had a mass of grams. Thirteen minutes since the experiment started, the gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the gas is leaking with a rate of grams per minute. So now we have that . We can either use or to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. A biologist has been observing a tree s height. months into the observation, the tree was feet tall. months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the tree is growing by feet each month. So now we have that . We can either use or to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. " + "body": " Point-Slope Form PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In , we learned that a linear equation can be written in slope-intercept form. This section covers an alternative that is often more useful depending on the application: point-slope form . Alternative Video Lesson Point-Slope Motivation and Definition Since 1990, the population of the United States has been growing by about million people per year. Also, back in 1990, the population was million. Since the rate of growth has been roughly constant, a linear model is appropriate. Let's write an equation to model this. We consider using slope-intercept form , but we would need to know the -intercept, and nothing in the background tells us that. We'd need to know the population of the United States back in the year 0, before there even was a United States. We could do some side work to calculate the -intercept, but let's try something else instead. Here are some things we know: The slope equation for a line in general is . The slope of our line is , or . One point on the line is , because in 1990 , the population was million. If we use the generic variables to represent a point somewhere on this line, then the rate of change from to must be . So . There is good reason to isolate in this equation, so we do that. This is a good place to pause. We have isolated , and three meaningful numbers appear in the equation: the rate of growth, a certain year, and the population in that year. This is a specific example of point-slope form . Before we look deeper at point-slope form, let's continue reducing the line equation into slope-intercept form by distributing and combining like terms. One concern with slope-intercept form is that it uses the -intercept, which might have no meaning in the context of an application. For example, here we have found that the -intercept is at , but what practical use is that? It's nonsense to say that in the year 0, the population of the United States was million. It doesn't make sense to have a negative population. It doesn't make sense to talk about the United States population before there even was a United States. And it doesn't make sense to use this model for years earlier than 1990 because the background information says clearly that the rate of change we have applies to years 1990 and later. For all these reasons, we prefer to have the equation back when it was in the form: and we purposely do not distribute the slope. Point-Slope Form point-slope form form point-slope When and have a linear relationship where is the slope and is some specific point that the line passes through, an equation for this relationship is and this equation is called the point-slope form of the line. It is called this because two important features are immediately visible from the numbers in the equation: the slope and one point on the line. There is a subtraction sign and an addition sign in point-slope form , and you may have trouble remembering which is which. But remember that that the line is supposed to pass through . So substituting in for should leave equal to . And it does. For example, consider our example equation . Here, is and is . And: The subtraction is exactly where it needs to be to wipe out and leave you with . More generally: Alternative Point-Slope Form It is also common to define point-slope form as by subtracting from each side. If you learn about point-slope form from some other resource, you may come across this. We feel that the form will be more helpful with college algebra, statistics, and calculus. Consider the line in this graph: Identify a point visible on this line that has integer coordinates, and write it as an ordered pair. What is the slope of the line? Use point-slope form to write an equation for this line, making use of a point with integer coordinates. The visible points with integer coordinates are and Several slope triangles are visible where the run is and the rise is So the slope is Using the point-slope equation is (You could use other points, like and get a different-looking equation like which simplifies to ) In , the solution explains that each of the following are acceptable equations for the same line: The first uses as a point on the line, and the second uses . Are those two equations really equivalent? Let's distribute and simplify each of them to get slope-intercept form . So, yes. It didn't matter which point we used to write a point-slope equation. We get different-looking equations that still represent the same line. Point-slope form is preferable when we know a line's slope and a point on it, but we don't know the -intercept. We recognize that distributing the slope and combining like terms can always be used to find the line's slope-intercept form. Plot the line with equation We can see from the equation that is a point on the line. The first step in making a plot is to go to that point on the graph. We also see the slope is so from our launch point of we can move forward and up to find a second point on the line. A spa chain has been losing customers at a roughly constant rate since the year 2018. In 2021, it had customers; in 2024, it had customers. Management estimated that the company will go out of business once its customer base decreases to . If this trend continues, when will the company close? The given information tells us two points on the line: and . The slope formula will give us the slope. After labeling those two points as and , we have: And considering units, this means they are losing customers per year. We could make an equation for this line using slope-intercept form, but then we would need to calculate the -intercept, which would correspond to the number of customers in year . We'd be working with numbers that have no real-world meaning. For point-slope form, since we calculated the slope, we know at least this much: . Now we can pick one of those two given points, say , and get the equation . Note that all three numbers in this equation have meaning in the context of the spa chain. The tells us how many customers are leaving per year, the represents a year, and the tells us the number of customers in that year. We're ready to answer the question about when the chain might go out of business. We need to substitute the given value of into the appropriate place in our equation. In order to substitute correctly, we must clearly understand what the variable represents including its units, and what the variable represents including its units. Write down those definitions first in any question you attempt to answer. It will save you time in the long run, and avoid frustration. Knowing the variable represents the number of customers in a given year allows one to understand that must be replaced with customers. Knowing the variable represents a year allows one to understand that will not be substituted, because we are trying to solve for what year something happens. After substituting in the equation with , we will solve for , and find the answer we seek. We find that at this rate the company is headed toward a collapse in 2030. illustrates one line representing the spa's customer base, and another line representing the customer level that would cause the business to close. To make a graph of , we first marked the point and from there used the slope of . If we go state by state and compare the Republican presidential candidate s 2012 vote share ( ) to the Republican presidential candidate s 2016 vote share ( ), we get the following graph (called a scatterplot , used in statistics) where a trendline has been superimposed. Find a point-slope equation for this line. (Note that a slope-intercept equation would use the -intercept coordinate and that would not be meaningful in context, since no state had anywhere near zero percent Republican vote.) We need to calculate slope first. And for that, we need to identify two points on the line. conveniently, the line appears to pass right through We have to take a second point somewhere, and seems like a reasonable roughly accurate choice. The slope equation gives us that Using as the point, the point-slope equation would then be Using Two Points to Build a Linear Equation Since two points can determine a line's location, we can calculate a line's equation using just the coordinates from any two points it passes through. A line passes through and . Find this line's equation in both point-slope and slope-intercept form. We will use the slope formula to find the slope first. After labeling those two points as , we have: The point-slope equation is . Next, we will use and substitute with and with , and we have: Next, we will change the point-slope equation into slope-intercept form: A line passes through and Find equations for this line using both point-slope and slope-intercept form. An equation for this line in point-slope form is . An equation for this line in slope-intercept form is First, use the slope formula to find the slope of this line: The generic point-slope equation is We have found the slope, and we may use for So an equation in point-slope form is To find a slope-intercept form equation, we can take the generic and substitute in the value of we found. Also, we know that should make the equation true. So we have So the slope-intercept equation is Note that the slope-intercept equation has an unfriendly fraction, and the fact that this can happen is another reason to use the point-slope form whenever it is reasonable to do so. Recognizing Point-Slope Form We can tell a lot about a linear equation now that we have learned both slope-intercept form and point-slope form . For example, we can know that is in slope-intercept form because it looks like . It will graph as a line with slope and vertical intercept . Likewise, we know that the equation is in point-slope form because it looks like . It will graph as a line that has slope and will pass through the point . For the equations below, state whether they are in slope-intercept form or point-slope form. Then identify the slope of the line and at least one point that the line will pass through. The equation is in slope-intercept form. The slope is and the vertical intercept is . The equation is in point-slope form. The slope is and the line passes through the point . The equation is almost in slope-intercept form. If we rearrange the right hand side to be , we can see that the slope is and the vertical intercept is . The equation is in point-slope form. The slope is and the line passes through the point . Consider the graph in . Using the point-slope form of a line, find three different linear equations for the line shown in the graph. Three integer-valued points are shown for convenience. Determine the slope-intercept form of the equation of this line. To write any of the equations representing this line in point-slope form, we must first find the slope of the line and we can use the slope formula to do so. We will arbitrarily choose and as the two points. Inputting these points into the slope formula yields: Thus the slope of the line is . Next, we need to write an equation in point-slope form based on each point shown. Using the point , we have: (This simplifies to .) The next point is . Using this point, we can write an equation for this line as: Finally, we can also use the point to write an equation for this line: which can also be written as: As is the vertical intercept, we can write the equation of this line in slope-intercept form as . It's important to note that each of the equations that were written in point-slope form simplify to this, making all four equations equivalent. Explain why there are some situations where point-slope form is preferable to slope-intercept form. There are basically two steps to convert a point-slope line equation into a slope-intercept line equation. What are they? If a line has equation , we can see the line passes through a certain point. To find the -coordinate of that point, you might look at the and know that you should negate that. Instead, you could train yourself to look at the and realize the important number is because that is what it takes to . Skills Practice Identify Point and Slope A line's equation is given in point-slope form. Identify the slope and the point on the line that is being singled out. Construct Point-Slope Form A line passes through the given points with the given slope. Find an equation for the line in point-slope form using the given point. through with slope through with slope through with slope through with slope through with slope through with slope Point-Slope Given Two Points A line passes through two given points. Find an equation for the line in point-slope form using one of the given points. and and and and and and and and Converting Point-Slope to Slope-Intercept Change the given point-slope equation to slope-intercept form. Point-Slope Form from a Graph Determine a point-slope form line equation for the given line. The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is The slope is and the point is on the graph of this line, so one point-slope form equation for this line is Make Graphs Graph the linear equation by identifying the slope and one point on this line. Applications By your cell phone contract, you pay a monthly fee plus for each minute you spend on the phone. In one month, you spent minutes over the phone, and had a bill totaling . Let be the number of minutes you spend on the phone in a month, and let be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone. A point-slope equation to model this is . If you spend minutes on the phone in a month, you would be billed . If your bill was one month, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. The slope of the linear model has been given to us: per minute. So now we have that . We can use to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. A company set aside a certain amount of money in the year 2000. The company spent exactly from that fund each year on perks for its employees. In , there was still left in the fund. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. We have been told that the account decreases by each year, so the slope of the linear model is (dollars per year). So now we have that . We can use to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2015, the fund will be depleted. A biologist has been observing a tree s height. This type of tree typically grows by feet each month. Thirteen months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the tree grows at a rate of feet per month, is the slope of the linear model. So now we have that . We can use to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Five minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . We can use to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In , there was still left in the fund. In , there was left. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the company is spending per year on perks for its employees. So now we have that . We can either use or to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2026, the fund will be depleted. By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent minutes on the phone, and paid . In another month, you spent minutes on the phone, and paid . Let be the number of minutes you talk over the phone in a month, and let be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone. A point-slope equation to model this is . If you spent minutes over the phone in a month, you would pay . If in a month, you paid of cell phone bill, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies you have to pay per minute that you use the phone. So now we have that . We can either use or to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Six minutes since the experiment started, the gas had a mass of grams. Fifteen minutes since the experiment started, the gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the gas is leaking with a rate of grams per minute. So now we have that . We can either use or to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. A biologist has been observing a tree s height. months into the observation, the tree was feet tall. months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the tree is growing by feet each month. So now we have that . We can either use or to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. " }, { "id": "section-point-slope-form-2", @@ -13993,7 +13993,7 @@ var ptx_lunr_docs = [ "type": "Definition", "number": "3.6.2", "title": "Point-Slope Form.", - "body": " Point-Slope Form point-slope form form point-slope When and have a linear relationship where is the slope and is some specific point that the line passes through, an equation for this relationship is and this equation is called the point-slope form of the line. It is called this because the slope and one point on the line are immediately discernible from the numbers in the equation. " + "body": " Point-Slope Form point-slope form form point-slope When and have a linear relationship where is the slope and is some specific point that the line passes through, an equation for this relationship is and this equation is called the point-slope form of the line. It is called this because two important features are immediately visible from the numbers in the equation: the slope and one point on the line. " }, { "id": "section-point-slope-form-4-11", @@ -14011,23 +14011,32 @@ var ptx_lunr_docs = [ "type": "Checkpoint", "number": "3.6.5", "title": "", - "body": " Consider the line in this graph: Identify a point visible on this line that has integer coordinates, and write it as an ordered pair. What is the slope of the line? Use point-slope form to write an equation for this line, making use of a point with integer coordinates. The visible points with integer coordinates are and Several slope triangles are visible where the run is and the rise is So the slope is Using the point-slope equation is (You could use other points, like and get a different-looking equation like which simplifies to ) " + "body": " Consider the line in this graph: Identify a point visible on this line that has integer coordinates, and write it as an ordered pair. What is the slope of the line? Use point-slope form to write an equation for this line, making use of a point with integer coordinates. The visible points with integer coordinates are and Several slope triangles are visible where the run is and the rise is So the slope is Using the point-slope equation is (You could use other points, like and get a different-looking equation like which simplifies to ) " }, { - "id": "section-point-slope-form-4-15", + "id": "make-a-plot-using-point-slope", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-4-15", - "type": "Example", + "url": "section-point-slope-form.html#make-a-plot-using-point-slope", + "type": "Checkpoint", "number": "3.6.6", "title": "", - "body": " A spa chain has been losing customers at a roughly constant rate since the year 2010. In 2013, it had customers; in 2016, it had customers. Management estimated that the company will go out of business once its customer base decreases to . If this trend continues, when will the company close? The given information tells us two points on the line: and . The slope formula will give us the slope. After labeling those two points as and , we have: And considering units, this means they are losing customers per year. Let's note that we could try to make an equation for this line in slope-intercept form, but then we would need to calculate the -intercept, which in context would correspond to the number of customers in year . We could do it, but we'd be working with numbers that have no real-world meaning in this context. For point-slope form, since we calculated the slope, we know at least this much: . Now we can pick one of those two given points, say , and get the equation . Note that all three numbers in this equation have meaning in the context of the spa chain. The tells us how many customers are leaving per year, the represents a year, and the tells us the number of customers in that year. We're ready to answer the question about when the chain might go out of business. We need to substitute the given value of into the appropriate place in our equation. In order to substitute correctly, we must clearly understand what the variable represents including its units, and what the variable represents including its units. Write down those definitions first in any question you attempt to answer. It will save you time in the long run, and avoid frustration. Knowing the variable represents the number of customers in a given year allows one to understand that must be replaced with customers. Knowing the variable represents a year allows one to understand that will not be substituted, because we are trying to solve for what year something happens. After substituting in the equation with , we will solve for , and find the answer we seek. We find that at this rate the company is headed toward a collapse in 2022. illustrates one line representing the spa's customer base, and another line representing the customer level that would cause the business to close. To make a graph of , we first marked the point and from there used the slope of . " + "body": " Plot the line with equation We can see from the equation that is a point on the line. The first step in making a plot is to go to that point on the graph. We also see the slope is so from our launch point of we can move forward and up to find a second point on the line. " }, { "id": "section-point-slope-form-4-16", "level": "2", "url": "section-point-slope-form.html#section-point-slope-form-4-16", + "type": "Example", + "number": "3.6.7", + "title": "", + "body": " A spa chain has been losing customers at a roughly constant rate since the year 2018. In 2021, it had customers; in 2024, it had customers. Management estimated that the company will go out of business once its customer base decreases to . If this trend continues, when will the company close? The given information tells us two points on the line: and . The slope formula will give us the slope. After labeling those two points as and , we have: And considering units, this means they are losing customers per year. We could make an equation for this line using slope-intercept form, but then we would need to calculate the -intercept, which would correspond to the number of customers in year . We'd be working with numbers that have no real-world meaning. For point-slope form, since we calculated the slope, we know at least this much: . Now we can pick one of those two given points, say , and get the equation . Note that all three numbers in this equation have meaning in the context of the spa chain. The tells us how many customers are leaving per year, the represents a year, and the tells us the number of customers in that year. We're ready to answer the question about when the chain might go out of business. We need to substitute the given value of into the appropriate place in our equation. In order to substitute correctly, we must clearly understand what the variable represents including its units, and what the variable represents including its units. Write down those definitions first in any question you attempt to answer. It will save you time in the long run, and avoid frustration. Knowing the variable represents the number of customers in a given year allows one to understand that must be replaced with customers. Knowing the variable represents a year allows one to understand that will not be substituted, because we are trying to solve for what year something happens. After substituting in the equation with , we will solve for , and find the answer we seek. We find that at this rate the company is headed toward a collapse in 2030. illustrates one line representing the spa's customer base, and another line representing the customer level that would cause the business to close. To make a graph of , we first marked the point and from there used the slope of . " +}, +{ + "id": "section-point-slope-form-4-17", + "level": "2", + "url": "section-point-slope-form.html#section-point-slope-form-4-17", "type": "Checkpoint", - "number": "3.6.8", + "number": "3.6.9", "title": "", "body": " If we go state by state and compare the Republican presidential candidate s 2012 vote share ( ) to the Republican presidential candidate s 2016 vote share ( ), we get the following graph (called a scatterplot , used in statistics) where a trendline has been superimposed. Find a point-slope equation for this line. (Note that a slope-intercept equation would use the -intercept coordinate and that would not be meaningful in context, since no state had anywhere near zero percent Republican vote.) We need to calculate slope first. And for that, we need to identify two points on the line. conveniently, the line appears to pass right through We have to take a second point somewhere, and seems like a reasonable roughly accurate choice. The slope equation gives us that Using as the point, the point-slope equation would then be " }, @@ -14036,7 +14045,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "section-point-slope-form.html#example-two-points-to-build-point-slope-form", "type": "Example", - "number": "3.6.9", + "number": "3.6.10", "title": "", "body": " A line passes through and . Find this line's equation in both point-slope and slope-intercept form. We will use the slope formula to find the slope first. After labeling those two points as , we have: The point-slope equation is . Next, we will use and substitute with and with , and we have: Next, we will change the point-slope equation into slope-intercept form: " }, @@ -14045,7 +14054,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "section-point-slope-form.html#section-point-slope-form-5-4", "type": "Checkpoint", - "number": "3.6.10", + "number": "3.6.11", "title": "", "body": " A line passes through and Find equations for this line using both point-slope and slope-intercept form. An equation for this line in point-slope form is . An equation for this line in slope-intercept form is First, use the slope formula to find the slope of this line: The generic point-slope equation is We have found the slope, and we may use for So an equation in point-slope form is To find a slope-intercept form equation, we can take the generic and substitute in the value of we found. Also, we know that should make the equation true. So we have So the slope-intercept equation is Note that the slope-intercept equation has an unfriendly fraction, and the fact that this can happen is another reason to use the point-slope form whenever it is reasonable to do so. " }, @@ -14054,7 +14063,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "section-point-slope-form.html#section-point-slope-form-6-3", "type": "Example", - "number": "3.6.11", + "number": "3.6.12", "title": "", "body": " For the equations below, state whether they are in slope-intercept form or point-slope form. Then identify the slope of the line and at least one point that the line will pass through. The equation is in slope-intercept form. The slope is and the vertical intercept is . The equation is in point-slope form. The slope is and the line passes through the point . The equation is almost in slope-intercept form. If we rearrange the right hand side to be , we can see that the slope is and the vertical intercept is . The equation is in point-slope form. The slope is and the line passes through the point . " }, @@ -14063,7 +14072,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "section-point-slope-form.html#section-point-slope-form-6-4", "type": "Example", - "number": "3.6.12", + "number": "3.6.13", "title": "", "body": " Consider the graph in . Using the point-slope form of a line, find three different linear equations for the line shown in the graph. Three integer-valued points are shown for convenience. Determine the slope-intercept form of the equation of this line. To write any of the equations representing this line in point-slope form, we must first find the slope of the line and we can use the slope formula to do so. We will arbitrarily choose and as the two points. Inputting these points into the slope formula yields: Thus the slope of the line is . Next, we need to write an equation in point-slope form based on each point shown. Using the point , we have: (This simplifies to .) The next point is . Using this point, we can write an equation for this line as: Finally, we can also use the point to write an equation for this line: which can also be written as: As is the vertical intercept, we can write the equation of this line in slope-intercept form as . It's important to note that each of the equations that were written in point-slope form simplify to this, making all four equations equivalent. " }, @@ -14083,7 +14092,7 @@ var ptx_lunr_docs = [ "type": "Reading Question", "number": "3.6.4.2", "title": "", - "body": " There are basically two steps to convert a point-slope form line equation into a slope-intercept form line equation. What are they? " + "body": " There are basically two steps to convert a point-slope line equation into a slope-intercept line equation. What are they? " }, { "id": "section-point-slope-form-7-3", @@ -14092,600 +14101,510 @@ var ptx_lunr_docs = [ "type": "Reading Question", "number": "3.6.4.3", "title": "", - "body": " If a line has equation , we can see that the line passes through a certain point. To find the -coordinate of that point, you might look at the and have memorized that you should negate that. Instead, you could train yourself to look at the and realize the important number is because that is what it takes to . " + "body": " If a line has equation , we can see the line passes through a certain point. To find the -coordinate of that point, you might look at the and know that you should negate that. Instead, you could train yourself to look at the and realize the important number is because that is what it takes to . " }, { - "id": "section-point-slope-form-8-1-2", + "id": "identify-slope-and-point-pos-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-1-2", + "url": "section-point-slope-form.html#identify-slope-and-point-pos-int", "type": "Exercise", "number": "3.6.5.1", "title": "", - "body": " Evaluate for and . We evaluate by replacing with and with in the formula. " + "body": " " }, { - "id": "section-point-slope-form-8-1-3", + "id": "identify-slope-and-point-pos-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-1-3", + "url": "section-point-slope-form.html#identify-slope-and-point-pos-int-copy", "type": "Exercise", "number": "3.6.5.2", "title": "", - "body": " Evaluate for and . We evaluate by replacing with and with in the formula. " + "body": " " }, { - "id": "section-point-slope-form-8-1-4", + "id": "identify-slope-and-point-neg-h-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-1-4", + "url": "section-point-slope-form.html#identify-slope-and-point-neg-h-int", "type": "Exercise", "number": "3.6.5.3", "title": "", - "body": " Evaluate for , , , and : Answer: " + "body": " " }, { - "id": "section-point-slope-form-8-1-5", + "id": "identify-slope-and-point-neg-h-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-1-5", + "url": "section-point-slope-form.html#identify-slope-and-point-neg-h-int-copy", "type": "Exercise", "number": "3.6.5.4", "title": "", - "body": " Evaluate for , , , and : Answer: " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-2", + "id": "identify-slope-and-point-neg-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-2", + "url": "section-point-slope-form.html#identify-slope-and-point-neg-int", "type": "Exercise", "number": "3.6.5.5", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-3", + "id": "identify-slope-and-point-neg-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-3", + "url": "section-point-slope-form.html#identify-slope-and-point-neg-int-copy", "type": "Exercise", "number": "3.6.5.6", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-4", + "id": "identify-slope-and-point-neg-k-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-4", + "url": "section-point-slope-form.html#identify-slope-and-point-neg-k-int", "type": "Exercise", "number": "3.6.5.7", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-5", + "id": "identify-slope-and-point-neg-k-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-5", + "url": "section-point-slope-form.html#identify-slope-and-point-neg-k-int-copy", "type": "Exercise", "number": "3.6.5.8", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-6", + "id": "write-point-slope-pos-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-6", + "url": "section-point-slope-form.html#write-point-slope-pos-int", "type": "Exercise", "number": "3.6.5.9", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " through with slope " }, { - "id": "section-point-slope-form-8-2-2-7", + "id": "write-point-slope-pos-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-7", + "url": "section-point-slope-form.html#write-point-slope-pos-int-copy", "type": "Exercise", "number": "3.6.5.10", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " through with slope " }, { - "id": "section-point-slope-form-8-2-2-8", + "id": "write-point-slope-neg-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-8", + "url": "section-point-slope-form.html#write-point-slope-neg-int", "type": "Exercise", "number": "3.6.5.11", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " through with slope " }, { - "id": "section-point-slope-form-8-2-2-9", + "id": "write-point-slope-neg-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-9", + "url": "section-point-slope-form.html#write-point-slope-neg-int-copy", "type": "Exercise", "number": "3.6.5.12", "title": "", - "body": " A line s equation is given in point-slope form: This line s slope is . A point on this line that is apparent from the given equation is . Compare the given equation with a generic point-slope equation: We can see the slope is , and an apparent point on the line is . " + "body": " through with slope " }, { - "id": "section-point-slope-form-8-2-2-10", + "id": "write-point-slope-mixed-int-decimal-slope", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-10", + "url": "section-point-slope-form.html#write-point-slope-mixed-int-decimal-slope", "type": "Exercise", "number": "3.6.5.13", "title": "", - "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " + "body": " through with slope " }, { - "id": "section-point-slope-form-8-2-2-11", + "id": "write-point-slope-mixed-int-decimal-slope-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-11", + "url": "section-point-slope-form.html#write-point-slope-mixed-int-decimal-slope-copy", "type": "Exercise", "number": "3.6.5.14", "title": "", - "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " + "body": " through with slope " }, { - "id": "section-point-slope-form-8-2-2-12", + "id": "find-point-slope-two-points-pos-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-12", + "url": "section-point-slope-form.html#find-point-slope-two-points-pos-int", "type": "Exercise", "number": "3.6.5.15", "title": "", - "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-13", + "id": "find-point-slope-two-points-pos-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-13", + "url": "section-point-slope-form.html#find-point-slope-two-points-pos-int-copy", "type": "Exercise", "number": "3.6.5.16", "title": "", - "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-14", + "id": "find-point-slope-two-points-neg-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-14", + "url": "section-point-slope-form.html#find-point-slope-two-points-neg-int", "type": "Exercise", "number": "3.6.5.17", "title": "", - "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-15", + "id": "find-point-slope-two-points-neg-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-15", + "url": "section-point-slope-form.html#find-point-slope-two-points-neg-int-copy", "type": "Exercise", "number": "3.6.5.18", "title": "", - "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-16", + "id": "find-point-slope-two-points-neg-frac", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-16", + "url": "section-point-slope-form.html#find-point-slope-two-points-neg-frac", "type": "Exercise", "number": "3.6.5.19", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-17", + "id": "find-point-slope-two-points-neg-frac-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-17", + "url": "section-point-slope-form.html#find-point-slope-two-points-neg-frac-copy", "type": "Exercise", "number": "3.6.5.20", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-18", + "id": "find-point-slope-two-points-pos-frac", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-18", + "url": "section-point-slope-form.html#find-point-slope-two-points-pos-frac", "type": "Exercise", "number": "3.6.5.21", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-19", + "id": "find-point-slope-two-points-pos-frac-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-19", + "url": "section-point-slope-form.html#find-point-slope-two-points-pos-frac-copy", "type": "Exercise", "number": "3.6.5.22", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " and " }, { - "id": "section-point-slope-form-8-2-2-20", + "id": "convert-point-slope-to-slope-intercept-pos-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-20", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-pos-int", "type": "Exercise", "number": "3.6.5.23", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-21", + "id": "convert-point-slope-to-slope-intercept-pos-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-21", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-pos-int-copy", "type": "Exercise", "number": "3.6.5.24", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-22", + "id": "convert-point-slope-to-slope-intercept-neg-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-22", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-neg-int", "type": "Exercise", "number": "3.6.5.25", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-23", + "id": "convert-point-slope-to-slope-intercept-neg-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-23", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-neg-int-copy", "type": "Exercise", "number": "3.6.5.26", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-24", + "id": "convert-point-slope-to-slope-intercept-pos-frac", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-24", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-pos-frac", "type": "Exercise", "number": "3.6.5.27", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-25", + "id": "convert-point-slope-to-slope-intercept-pos-frac-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-25", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-pos-frac-copy", "type": "Exercise", "number": "3.6.5.28", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-26", + "id": "convert-point-slope-to-slope-intercept-decimal", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-26", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-decimal", "type": "Exercise", "number": "3.6.5.29", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-2-27", + "id": "convert-point-slope-to-slope-intercept-decimal-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-2-27", + "url": "section-point-slope-form.html#convert-point-slope-to-slope-intercept-decimal-copy", "type": "Exercise", "number": "3.6.5.30", "title": "", - "body": " A line s slope is . The line passes through the point . Find an equation for this line in both point-slope and slope-intercept form. An equation for this line in point-slope form is: . An equation for this line in slope-intercept form is: . It s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is: where is the slope and is a point on the line. We have been given that and that the line passes through the point . We substitute these numbers into the formula, and we have: So the line has an equation in point-slope form: . The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and solve for . So the line s equation in slope-intercept form is . " + "body": " " }, { - "id": "section-point-slope-form-8-2-3-2", + "id": "determine-point-slope-form-from-graph", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-2", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph", "type": "Exercise", "number": "3.6.5.31", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-3", + "id": "determine-point-slope-form-from-graph-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-3", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-copy", "type": "Exercise", "number": "3.6.5.32", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-4", + "id": "determine-point-slope-form-from-graph-neg", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-4", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-neg", "type": "Exercise", "number": "3.6.5.33", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-5", + "id": "determine-point-slope-form-from-graph-neg-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-5", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-neg-copy", "type": "Exercise", "number": "3.6.5.34", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-6", + "id": "determine-point-slope-form-from-graph-q1-large-y", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-6", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-q1-large-y", "type": "Exercise", "number": "3.6.5.35", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-7", + "id": "determine-point-slope-form-from-graph-q1-large-y-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-7", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-q1-large-y-copy", "type": "Exercise", "number": "3.6.5.36", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-8", + "id": "determine-point-slope-form-from-graph-q1-large-y-precise", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-8", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-q1-large-y-precise", "type": "Exercise", "number": "3.6.5.37", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-3-9", + "id": "determine-point-slope-form-from-graph-q1-large-y-precise-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-3-9", + "url": "section-point-slope-form.html#determine-point-slope-form-from-graph-q1-large-y-precise-copy", "type": "Exercise", "number": "3.6.5.38", "title": "", - "body": " Change this equation from point-slope form to slope-intercept form. In slope-intercept form: The line s equation in slope-intercept form is . " + "body": " The slope is and the point is on the graph of this line, so one point-slope form equation for this line is " }, { - "id": "section-point-slope-form-8-2-4-2", + "id": "graph-point-slope-pos-int", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-2", + "url": "section-point-slope-form.html#graph-point-slope-pos-int", "type": "Exercise", "number": "3.6.5.39", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-3", + "id": "graph-point-slope-pos-int-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-3", + "url": "section-point-slope-form.html#graph-point-slope-pos-int-copy", "type": "Exercise", "number": "3.6.5.40", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-4", + "id": "graph-point-slope-pos-int-neg-k", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-4", + "url": "section-point-slope-form.html#graph-point-slope-pos-int-neg-k", "type": "Exercise", "number": "3.6.5.41", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-5", + "id": "graph-point-slope-pos-int-neg-k-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-5", + "url": "section-point-slope-form.html#graph-point-slope-pos-int-neg-k-copy", "type": "Exercise", "number": "3.6.5.42", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-6", + "id": "graph-point-slope-pos-frac", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-6", + "url": "section-point-slope-form.html#graph-point-slope-pos-frac", "type": "Exercise", "number": "3.6.5.43", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-7", + "id": "graph-point-slope-pos-frac-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-7", + "url": "section-point-slope-form.html#graph-point-slope-pos-frac-copy", "type": "Exercise", "number": "3.6.5.44", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-8", + "id": "graph-point-slope-neg-frac", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-8", + "url": "section-point-slope-form.html#graph-point-slope-neg-frac", "type": "Exercise", "number": "3.6.5.45", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-9", + "id": "graph-point-slope-neg-frac-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-9", + "url": "section-point-slope-form.html#graph-point-slope-neg-frac-copy", "type": "Exercise", "number": "3.6.5.46", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-10", + "id": "graph-point-slope-decimal", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-10", + "url": "section-point-slope-form.html#graph-point-slope-decimal", "type": "Exercise", "number": "3.6.5.47", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-11", + "id": "graph-point-slope-decimal-copy", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-11", + "url": "section-point-slope-form.html#graph-point-slope-decimal-copy", "type": "Exercise", "number": "3.6.5.48", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " " }, { - "id": "section-point-slope-form-8-2-4-12", + "id": "section-point-slope-form-8-2-2", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-12", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-2", "type": "Exercise", "number": "3.6.5.49", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " By your cell phone contract, you pay a monthly fee plus for each minute you spend on the phone. In one month, you spent minutes over the phone, and had a bill totaling . Let be the number of minutes you spend on the phone in a month, and let be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone. A point-slope equation to model this is . If you spend minutes on the phone in a month, you would be billed . If your bill was one month, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. The slope of the linear model has been given to us: per minute. So now we have that . We can use to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. " }, { - "id": "section-point-slope-form-8-2-4-13", + "id": "section-point-slope-form-8-2-3", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-4-13", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-3", "type": "Exercise", "number": "3.6.5.50", "title": "", - "body": " Determine the point-slope form of the linear equation from its graph. The slope is and the point is on the graph of this line, so its equation can be written in point-slope form as " + "body": " A company set aside a certain amount of money in the year 2000. The company spent exactly from that fund each year on perks for its employees. In , there was still left in the fund. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. We have been told that the account decreases by each year, so the slope of the linear model is (dollars per year). So now we have that . We can use to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2015, the fund will be depleted. " }, { - "id": "section-point-slope-form-8-2-5-3", + "id": "section-point-slope-form-8-2-4", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-3", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-4", "type": "Exercise", "number": "3.6.5.51", "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " + "body": " A biologist has been observing a tree s height. This type of tree typically grows by feet each month. Thirteen months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the tree grows at a rate of feet per month, is the slope of the linear model. So now we have that . We can use to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. " }, { - "id": "section-point-slope-form-8-2-5-4", + "id": "section-point-slope-form-8-2-5", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-4", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-5", "type": "Exercise", "number": "3.6.5.52", "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " + "body": " Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Five minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . We can use to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. " }, { - "id": "section-point-slope-form-8-2-5-5", + "id": "section-point-slope-form-8-2-6", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-5", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-6", "type": "Exercise", "number": "3.6.5.53", "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " + "body": " A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In , there was still left in the fund. In , there was left. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the company is spending per year on perks for its employees. So now we have that . We can either use or to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2026, the fund will be depleted. " }, { - "id": "section-point-slope-form-8-2-5-6", + "id": "section-point-slope-form-8-2-7", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-6", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-7", "type": "Exercise", "number": "3.6.5.54", "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " + "body": " By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent minutes on the phone, and paid . In another month, you spent minutes on the phone, and paid . Let be the number of minutes you talk over the phone in a month, and let be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone. A point-slope equation to model this is . If you spent minutes over the phone in a month, you would pay . If in a month, you paid of cell phone bill, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies you have to pay per minute that you use the phone. So now we have that . We can either use or to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. " }, { - "id": "section-point-slope-form-8-2-5-7", + "id": "section-point-slope-form-8-2-8", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-7", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-8", "type": "Exercise", "number": "3.6.5.55", "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " + "body": " Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Six minutes since the experiment started, the gas had a mass of grams. Fifteen minutes since the experiment started, the gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the gas is leaking with a rate of grams per minute. So now we have that . We can either use or to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. " }, { - "id": "section-point-slope-form-8-2-5-8", + "id": "section-point-slope-form-8-2-9", "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-8", + "url": "section-point-slope-form.html#section-point-slope-form-8-2-9", "type": "Exercise", "number": "3.6.5.56", "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " -}, -{ - "id": "section-point-slope-form-8-2-5-9", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-9", - "type": "Exercise", - "number": "3.6.5.57", - "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " -}, -{ - "id": "section-point-slope-form-8-2-5-10", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-2-5-10", - "type": "Exercise", - "number": "3.6.5.58", - "title": "", - "body": " The slope is and the point is on the graph of this line. A graph of a line. " -}, -{ - "id": "section-point-slope-form-8-3-2", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-2", - "type": "Exercise", - "number": "3.6.5.59", - "title": "", - "body": " By your cell phone contract, you pay a monthly fee plus for each minute you spend on the phone. In one month, you spent minutes over the phone, and had a bill totaling . Let be the number of minutes you spend on the phone in a month, and let be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone. A point-slope equation to model this is . If you spend minutes on the phone in a month, you would be billed . If your bill was one month, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. The slope of the linear model has been given to us: per minute. So now we have that . We can use to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. " -}, -{ - "id": "section-point-slope-form-8-3-3", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-3", - "type": "Exercise", - "number": "3.6.5.60", - "title": "", - "body": " A company set aside a certain amount of money in the year 2000. The company spent exactly from that fund each year on perks for its employees. In , there was still left in the fund. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. We have been told that the account decreases by each year, so the slope of the linear model is (dollars per year). So now we have that . We can use to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2029, the fund will be depleted. " -}, -{ - "id": "section-point-slope-form-8-3-4", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-4", - "type": "Exercise", - "number": "3.6.5.61", - "title": "", - "body": " A biologist has been observing a tree s height. This type of tree typically grows by feet each month. Ten months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the tree grows at a rate of feet per month, is the slope of the linear model. So now we have that . We can use to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. " -}, -{ - "id": "section-point-slope-form-8-3-5", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-5", - "type": "Exercise", - "number": "3.6.5.62", - "title": "", - "body": " Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Seven minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . We can use to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. " -}, -{ - "id": "section-point-slope-form-8-3-6", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-6", - "type": "Exercise", - "number": "3.6.5.63", - "title": "", - "body": " A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In , there was still left in the fund. In , there was left. Let be the number of years since 2000, and let be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years. A point-slope equation to model this is . In the year , there was left in the fund. In the year , the fund will be empty. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the company is spending per year on perks for its employees. So now we have that . We can either use or to get . In the year , was . To find the amount remaining in the fund, we substitute in for in the equation , and we have: So in the year , the fund had remaining in it. The fund will be empty when it has left in it. That is, will equal . To find how many years until this happens, we substitute in for in the equation , and we have: Approximately years after 2000, in the year 2018, the fund will be depleted. " -}, -{ - "id": "section-point-slope-form-8-3-7", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-7", - "type": "Exercise", - "number": "3.6.5.64", - "title": "", - "body": " By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent minutes on the phone, and paid . In another month, you spent minutes on the phone, and paid . Let be the number of minutes you talk over the phone in a month, and let be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone. A point-slope equation to model this is . If you spent minutes over the phone in a month, you would pay . If in a month, you paid of cell phone bill, you must have spent minutes on the phone in that month. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies you have to pay per minute that you use the phone. So now we have that . We can either use or to get . In one month, you spent minutes on the phone. To find out the bill, we substitute in for in the equation , and we have: If you spent minutes over the phone in a month, your bill for that month would be . In one month, you had a cell phone bill of . To find how many minutes you spent on the phone that month, we substitute in for in the equation , and we have: If your bill was in one month, you must have spent minutes on the phone. " -}, -{ - "id": "section-point-slope-form-8-3-8", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-8", - "type": "Exercise", - "number": "3.6.5.65", - "title": "", - "body": " Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Nine minutes since the experiment started, the gas had a mass of grams. Thirteen minutes since the experiment started, the gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. A point-slope equation to model this is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the gas is leaking with a rate of grams per minute. So now we have that . We can either use or to get . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation , and we have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. " -}, -{ - "id": "section-point-slope-form-8-3-9", - "level": "2", - "url": "section-point-slope-form.html#section-point-slope-form-8-3-9", - "type": "Exercise", - "number": "3.6.5.66", - "title": "", "body": " A biologist has been observing a tree s height. months into the observation, the tree was feet tall. months into the observation, the tree was feet tall. Let be the number of months passed since the observations started, and let be the tree s height at that time, in feet. Use a linear equation to model the tree s height as the number of months pass. A point-slope equation to model this is . months after the observations started, the tree would be feet in height. months after the observation started, the tree would be feet tall. A line s equation in point-slope form looks like , where is the slope, and is some point we know to be on the line. We first need to find the line s slope. To find the slope, we need two points on the line. We have been given that and are points on the linear model, so we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This line s slope is . This implies that the tree is growing by feet each month. So now we have that . We can either use or to get . After months, we can find the height of the tree if we substitute in for in the equation , and we have: So after months, the tree is feet tall. To find when the tree was feet tall, we substitute in for in the equation , and we have: So the tree was feet tall after months. " }, { @@ -14695,7 +14614,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "3.7", "title": "Standard Form", - "body": " Standard Form PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG We've seen that a linear relationship can be expressed with an equation in slope-intercept form or with an equation in point-slope form . There is a third form that you can use to write line equations. It's known as standard form . Alternative Video Lesson Standard Form Definition Imagine trying to gather donations to pay for a medical procedure you cannot afford. Oversimplifying the mathematics a bit, suppose that there were only two types of donors in the world: those who will donate and those who will donate . How many of each, or what combination, do you need to reach the funding goal? As in, if people donate and people donate , what numbers could and be? The donors of the first type have collectively donated dollars, and the donors of the second type have collectively donated . Reflect on the meaning of and . Make sure you understand their meaning before reading on. So altogether you'd need This is an example of a line equation in standard form . Standard Form It is always possible to write an equation for a line in the form where , , and are three numbers (each of which might be , although at least one of and must be nonzero). This form of a line equation is called standard form . In the context of an application, the meaning of , , and depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations). standard form form standard For each of the following equations, identify what form they are in. ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear is in standard form, with and is in point-slope form, with slope and passing through is linear, but not in any of the forms we have studied. Using algebra, you can rearrange it to read is not linear. The exponent on is a dead giveaway. is in standard form, with and is in slope-intercept form, with slope and -intercept at Returning to the example with donations for the medical procedure, let's examine the equation . What units are attached to all of the parts of this equation? Both and are numbers of people. The is in dollars. Both the and the are in dollars per person. Note how both sides of the equation are in dollars. On the right, that fact is clear. On the left, is in dollars since is in dollars per person, and is in people. The same is true for , and the two dollar amounts and add to a dollar amount. What is the slope of the linear relationship? It's not immediately visible since is not part of the standard form equation. But we can use algebra to isolate : . And we see that the slope is . OK, what units are on that slope? As always, the units on slope are . In this case that's , which sounds a little weird and seems like it should be simplified away to unitless. But this slope of is saying that for every 5 extra people who donate each, you need fewer person donating to still reach your goal. What is the -intercept? Since we've already converted the equation into slope-intercept form, we can see that it is at . This tells us that if people donate , then you will need people to each donate . What does a graph for this line look like? We've already converted into slope-intercept form, and we could use that to make the graph. But when given a line in standard form, there is another approach that is often used. Returning to , let's calculate the -intercept and the -intercept. Recall that these are points where the line crosses the -axis and -axis. To be on the -axis means that , and to be on the -axis means that . All these zeros make the resulting algebra easy to finish: So we have a -intercept at and an -intercept at . If we plot these, we get to mark especially relevant points given the context, and then drawing a straight line between them gives us Figure . A Cartesian plot where the x-axis represents $20 donors and the y-axis represents $100 donors; the line has a y-intercept of 100 $100 donors and an x-intercept of 500 $20 donors The - and -Intercepts With a linear relationship (and other types of equations too), we are often interested in the -intercept and -intercept because they have special meaning in context. For example, in Figure , the -intercept implies that if no one donates , you need people to donate to get us to . And the -intercept implies if no one donates , you need people to donate . Let's look at another example. James owns a restaurant that uses about lb of flour every day. He just purchased 1200 of flour. Model the amount of flour that remains days later with a linear equation, and interpret the meaning of its -intercept and -intercept. Since the rate of change is constant ( -32 every day), and we know the initial value, we can model the amount of flour at the restaurant with a slope-intercept form equation: where represents the number of days passed since the initial purchase, and represents the amount of flour left (in .) A line's -intercept is in the form of , since to be on the -axis, the -coordinate must be . To find this line's -intercept, we substitute in the equation with , and solve for : So the line's -intercept is at . In context this means the flour would last for days. A line's -intercept is in the form of . This line equation is already in slope-intercept form, so we can just see that its -intercept is at . In general though, we would substitute in the equation with , and we have: So yes, the line's -intercept is at . This means that when the flour was purchased, there was 1200 of it. In other words, the -intercept tells us one of the original pieces of information: in the beginning, James purchased 1200 of flour. If a line is in standard form, it's often easiest to graph it using its two intercepts. Graph using its intercepts. And then use the intercepts to calculate the line's slope. To graph a line by its -intercept and -intercept, it might help to first set up a table like in Figure : Intercepts of -value -value Intercepts -intercept -intercept A table like this might help you stay focused on the fact that we are searching for two points. As we've noted earlier, an -intercept is on the -axis, and so its -coordinate must be . This is worth taking special note of: to find an -intercept, must be . This is why we put in the -value cell of the -intercept. Similarly, a line's -intercept has , and we put into the -value cell of the -intercept. Next, we calculate the line's -intercept by substituting into the equation So the line's -intercept is . Similarly, we substitute into the equation to calculate the -intercept: So the line's -intercept is . Now we can complete the table: Intercepts of -value -value Intercepts -intercept -intercept With both intercepts' coordinates, we can graph the line: Graph of a coordinate grid with the graph of line 2x-3y=-6; the x-intercept is (-3,0) and the y-intercept is (0,2) There is a slope triangle from the -intercept to the origin up to the -intercept. It tells us that the slope is . This last example generalizes to a fact worth noting. slope given intercepts If a line's -intercept is at and its -intercept is at , then the slope of the line is . (Unless the line passes through the origin, in which case both and equal , and then this fraction is undefined. And the slope of the line could be anything.) Consider the line with equation What is its -intercept? What is its -intercept? What is its slope? To find the -intercept: So the -intercept is at To find the -intercept: So the -intercept is at about Since we have the - and -intercepts, we can calculate the slope: Transforming between Standard Form and Slope-Intercept Form Sometimes a linear equation arises in standard form , but it would be useful to see that equation in slope-intercept form . Or perhaps, vice versa. A linear equation in slope-intercept form tells us important information about the line: its slope and -intercept . However, a line's standard form does not show those two important values. As a result, we often need to change a line's equation from standard form to slope-intercept form. Let's look at some examples. Change to slope-intercept form, and then graph it. Since a line in slope-intercept form looks like , we will solve for in : In the third line, we wrote on the right side, instead of . The only reason we did this is because we are headed to slope-intercept form, where the -term is traditionally written first. Now we can see that the slope is and the -intercept is at . With these things found, we can graph the line using slope triangles. Compare this graphing method with the Graphing by Intercepts method in Example . We have more points in this graph, thus we can graph the line more accurately. Graphing with Slope Triangles a coordinate gris with the graph of the line 2x-3y=-6; the y-intercept is 2 and the slope is 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 Graph . First, we will try (and fail) to graph this line using its - and -intercepts. Trying to find the -intercept: So the line's -intercept is at , at the origin. Huh, that is also on the -axis Trying to find the -intercept: So the line's -intercept is also at . Since both intercepts are the same point, there is no way to use the intercepts alone to graph this line. So what can be done? Several approaches are out there, but one is to convert the line equation into slope-intercept form: So the line's slope is , and we can graph the line using slope triangles and the intercept at , as in Figure . Graphing with Slope Triangles a coordinate grid with the graph of line 2x-3y=0; The line has a y-intercept of 0 and a slope of 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 In summary, if in a standard form equation , it's convenient to graph it by first converting the equation to slope-intercept form . Write the equation in standard form. Once we subtract on both sides of the equation, we have Technically, this equation is already in standard form . However, you might like to end up with an equation that has no fractions, so you could multiply each side by : What kind of line can be written in standard form, but cannot be written in slope-intercept or point-slope form? What are some reasons why you might care to find the - and -intercepts of a line? What is not immediately apparent from standard form, that is immediately apparent from slope-intercept form and point-slope form? Review and Warmup Skills Practice Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Converting to Standard Form Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Note that is already in standard form, but it s better to get rid of the leading negative sign in an equation. Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Note that is already in standard form, but it s better to get rid of the leading negative sign in an equation. Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Note that is already in standard form, but it s better to get rid of the leading negative sign in an equation. Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Graphs and Standard Form Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Interpreting Intercepts in Context Casandra is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. She can spend a total of . Assume Casandra will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (150,0). It implies Casandra can purchase 150 tea bags with no sugar bags. B . The x-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags. C . The x-intercept is (60,0). It implies Casandra can purchase 60 tea bags with no sugar bags. D . The x-intercept is (0,60). It implies Casandra can purchase 60 sugar bags with no tea bags. Assume Casandra will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Casandra can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is C. Page is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. She can spend a total of . Assume Page will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (0,45). It implies Page can purchase 45 sugar bags with no tea bags. B . The y-intercept is (15,0). It implies Page can purchase 15 tea bags with no sugar bags. C . The y-intercept is (0, 15). It implies Page can purchase 15 sugar bags with no tea bags. D . The y-intercept is (45,0). It implies Page can purchase 45 tea bags with no sugar bags. Assume Page will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Page can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The y-intercept is (0, 15). It implies Page can purchase 15 sugar bags with no tea bags. An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (150,0). It implies the engine will run out of gas 150 hours after its tank was refilled. B . The x-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled. C . The x-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank. D . The x-intercept is (0,30). It implies the engine started with 30 gallons of gas in its tank. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled. An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (45,0). It implies the engine will run out of gas 45 hours after its tank was refilled. B . The y-intercept is (0,180). It implies the engine started with 180 gallons of gas in its tank. C . The y-intercept is (180,0). It implies the engine will run out of gas 180 hours after its tank was refilled. D . The y-intercept is (0,45). It implies the engine started with 45 gallons of gas in its tank. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: Since the equation is in slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,180). It implies the engine started with 180 gallons of gas in its tank. A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (0,44000). It implies the car's initial value was 44000. B . The x-intercept is (44000,0). It implies the car's initial value was 44000. C . The x-intercept is (0,20). It implies the car would have no more value 20 years since its purchase. D . The x-intercept is (20,0). It implies the car would have no more value 20 years since its purchase. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (20,0). It implies the car would have no more value 20 years since its purchase. A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (16,0). It implies the car would have no more value 16 years since its purchase. B . The y-intercept is (0,16). It implies the car would have no more value 16 years since its purchase. C . The y-intercept is (35200,0). It implies the car's initial value was 35200. D . The y-intercept is (0,35200). It implies the car's initial value was 35200. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: Since this line is in its slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,35200). It implies the car's initial value was 35200. Challenge Fill in the variables and in with the numbers and You may only use each number once. For the steepest possible slope, must be , must be , and must be . For the shallowest possible slope, must be , must be , and must be . Let s put into slope-intercept form. Then, the equation becomes To make the steepest slope possible, we must maximize the fraction The possibilities are and The biggest of these is Thus, must equal 14, must equal 6, and must equal 7. To make the shallowest slope possible, we must minimize the fraction The smallest of the possible fractions is Thus, must equal 6, must equal 14, and must equal 7. " + "body": " Standard Form PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG We've seen that a linear relationship can be expressed with an equation in slope-intercept form or with an equation in point-slope form . There is a third form that you can use to write line equations. It's known as standard form . Alternative Video Lesson Standard Form Definition Imagine trying to gather donations to pay for a medical procedure you cannot afford. Oversimplifying the mathematics a bit, suppose that there were only two types of donors in the world: those who will donate and those who will donate . How many of each, or what combination, do you need to reach the funding goal? That is, if people donate and people donate , what numbers could and be to meet your goal? The donors of the first type have collectively donated dollars, and the donors of the second type have collectively donated . To reach , altogether you'd need to have This is an example of a line equation in standard form . Standard Form standard form form standard It is always possible to write an equation for a line in the form where , , and are three numbers (each of which might be , although at least one of and must be nonzero). This form of a line equation is called standard form . In the context of an application, the meaning of , , and depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations). PTX:ERROR: WeBWorK problem webwork-1289 with seed 1289 is either empty or failed to compile Use -a to halt with full PG and returned content Returning to the example with donations for the medical procedure, let's examine the equation . What units are attached to all of the parts of this equation? Both and are numbers of people. The is in dollars. Both the and the are in dollars per person. Note how both sides of the equation are in dollars. On the right, that fact is clear. On the left, is in dollars since is in dollars per person, and is in people. The same is true for , and the two dollar amounts and add to a dollar amount. What is the slope of the linear relationship? It's not immediately visible since is not part of the standard form equation. But we can use algebra to isolate : . And we see that the slope is . OK, what units are on that slope? As always, the units on slope are . In this case that's , which sounds a little weird and seems like it should be simplified away to unitless. But this slope of is saying that for every 5 extra people who donate each, you need fewer person donating to still reach your goal. What is the -intercept? Since we've already converted the equation into slope-intercept form, we can see that it is at . This tells us that if people donate , then you will need people to each donate . What does a graph for this line look like? We've already converted into slope-intercept form, and we could use that to make the graph. But when given a line in standard form, there is another approach that is often used. Returning to , let's calculate the -intercept and the -intercept. Recall that these are points where the line crosses the -axis and -axis. To be on the -axis means that , and to be on the -axis means that . All these zeros make the resulting algebra easy to finish: So we have a -intercept at and an -intercept at . If we plot these, we get to mark especially relevant points given the context, and then drawing a straight line between them gives us . A Cartesian plot where the x-axis represents $20 donors and the y-axis represents $100 donors; the line has a y-intercept of 100 $100 donors and an x-intercept of 500 $20 donors The - and -Intercepts With a linear relationship (and other types of equations too), we are often interested in the -intercept and -intercept because they have special meaning in context. For example, in , the -intercept implies that if no one donates , you need people to donate to get us to . And the -intercept implies if no one donates , you need people to donate . Let's look at another example. James owns a restaurant that uses about lb of flour every day. He just purchased 1200 of flour. Model the amount of flour that remains days later with a linear equation, and interpret the meaning of its -intercept and -intercept. Since the rate of change is constant ( -32 every day), and we know the initial value, we can model the amount of flour at the restaurant with a slope-intercept form equation: where represents the number of days passed since the initial purchase, and represents the amount of flour left (in .) A line's -intercept is in the form of , since to be on the -axis, the -coordinate must be . To find this line's -intercept, we substitute in the equation with , and solve for : So the line's -intercept is at . In context this means the flour would last for days. A line's -intercept is in the form of . This line equation is already in slope-intercept form, so we can just see that its -intercept is at . In general though, we would substitute in the equation with , and we have: So yes, the line's -intercept is at . This means that when the flour was purchased, there was 1200 of it. In other words, the -intercept tells us one of the original pieces of information: in the beginning, James purchased 1200 of flour. If a line is in standard form, it's often easiest to graph it using its two intercepts. Graph using its intercepts. And then use the intercepts to calculate the line's slope. To graph a line by its -intercept and -intercept, it might help to first set up a table like in : Intercepts of -value -value Intercepts -intercept -intercept A table like this might help you stay focused on the fact that we are searching for two points. As we've noted earlier, an -intercept is on the -axis, and so its -coordinate must be . This is worth taking special note of: to find an -intercept, must be . This is why we put in the -value cell of the -intercept. Similarly, a line's -intercept has , and we put into the -value cell of the -intercept. Next, we calculate the line's -intercept by substituting into the equation So the line's -intercept is . Similarly, we substitute into the equation to calculate the -intercept: So the line's -intercept is . Now we can complete the table: Intercepts of -value -value Intercepts -intercept -intercept With both intercepts' coordinates, we can graph the line: Graph of a coordinate grid with the graph of line 2x-3y=-6; the x-intercept is (-3,0) and the y-intercept is (0,2) There is a slope triangle from the -intercept to the origin up to the -intercept. It tells us that the slope is . This last example generalizes to a fact worth noting. slope given intercepts If a line's -intercept is at and its -intercept is at , then the slope of the line is . (Unless the line passes through the origin, in which case both and equal , and then this fraction is undefined. And the slope of the line could be anything.) Consider the line with equation What is its -intercept? What is its -intercept? What is its slope? To find the -intercept: So the -intercept is at To find the -intercept: So the -intercept is at about Since we have the - and -intercepts, we can calculate the slope: Transforming between Standard Form and Slope-Intercept Form Sometimes a linear equation arises in standard form , but it would be useful to see that equation in slope-intercept form . Or perhaps, vice versa. A linear equation in slope-intercept form tells us important information about the line: its slope and -intercept . However, a line's standard form does not show those two important values. As a result, we often need to change a line's equation from standard form to slope-intercept form. Let's look at some examples. Change to slope-intercept form, and then graph it. Since a line in slope-intercept form looks like , we will solve for in : In the third line, we wrote on the right side, instead of . The only reason we did this is because we are headed to slope-intercept form, where the -term is traditionally written first. Now we can see that the slope is and the -intercept is at . With these things found, we can graph the line using slope triangles. Compare this graphing method with the Graphing by Intercepts method in . We have more points in this graph, thus we can graph the line more accurately. Graphing with Slope Triangles a coordinate gris with the graph of the line 2x-3y=-6; the y-intercept is 2 and the slope is 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 Graph . First, we will try (and fail) to graph this line using its - and -intercepts. Trying to find the -intercept: So the line's -intercept is at , at the origin. Huh, that is also on the -axis Trying to find the -intercept: So the line's -intercept is also at . Since both intercepts are the same point, there is no way to use the intercepts alone to graph this line. So what can be done? Several approaches are out there, but one is to convert the line equation into slope-intercept form: So the line's slope is , and we can graph the line using slope triangles and the intercept at , as in . Graphing with Slope Triangles a coordinate grid with the graph of line 2x-3y=0; The line has a y-intercept of 0 and a slope of 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 In summary, if in a standard form equation , it's convenient to graph it by first converting the equation to slope-intercept form . Write the equation in standard form. Once we subtract on both sides of the equation, we have Technically, this equation is already in standard form . However, you might like to end up with an equation that has no fractions, so you could multiply each side by : What kind of line can be written in standard form, but cannot be written in slope-intercept or point-slope form? What are some reasons why you might care to find the - and -intercepts of a line? What is not immediately apparent from standard form, that is immediately apparent from slope-intercept form and point-slope form? Review and Warmup Skills Practice Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Converting to Standard Form Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Note that is already in standard form, but it s better to get rid of the leading negative sign in an equation. Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Note that is already in standard form, but it s better to get rid of the leading negative sign in an equation. Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Note that is already in standard form, but it s better to get rid of the leading negative sign in an equation. Rewrite in standard form. A line s standard form is We need to move and terms to the same side of the equals sign. Graphs and Standard Form Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Find the - and -intercepts of the line with equation . Then find one other point on the line. Use your results to graph the line. -intercept: -intercept: Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Make a graph of the line . Interpreting Intercepts in Context Daniel is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. He can spend a total of . Assume Daniel will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags he can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (0, 20). It implies Daniel can purchase 20 sugar bags with no tea bags. B . The x-intercept is (0,25). It implies Daniel can purchase 25 sugar bags with no tea bags. C . The x-intercept is (25,0). It implies Daniel can purchase 25 tea bags with no sugar bags. D . The x-intercept is (20,0). It implies Daniel can purchase 20 tea bags with no sugar bags. Assume Daniel will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Daniel can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is C. Casandra is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. She can spend a total of . Assume Casandra will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (0,60). It implies Casandra can purchase 60 sugar bags with no tea bags. B . The y-intercept is (150,0). It implies Casandra can purchase 150 tea bags with no sugar bags. C . The y-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags. D . The y-intercept is (60,0). It implies Casandra can purchase 60 tea bags with no sugar bags. Assume Casandra will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Casandra can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The y-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags. An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (175,0). It implies the engine will run out of gas 175 hours after its tank was refilled. B . The x-intercept is (0,175). It implies the engine started with 175 gallons of gas in its tank. C . The x-intercept is (0,35). It implies the engine started with 35 gallons of gas in its tank. D . The x-intercept is (35,0). It implies the engine will run out of gas 35 hours after its tank was refilled. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (35,0). It implies the engine will run out of gas 35 hours after its tank was refilled. An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (0,30). It implies the engine started with 30 gallons of gas in its tank. B . The y-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank. C . The y-intercept is (150,0). It implies the engine will run out of gas 150 hours after its tank was refilled. D . The y-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: Since the equation is in slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank. A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (0,24). It implies the car would have no more value 24 years since its purchase. B . The x-intercept is (24,0). It implies the car would have no more value 24 years since its purchase. C . The x-intercept is (0,52800). It implies the car's initial value was 52800. D . The x-intercept is (52800,0). It implies the car's initial value was 52800. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (24,0). It implies the car would have no more value 24 years since its purchase. A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (44000,0). It implies the car's initial value was 44000. B . The y-intercept is (0,20). It implies the car would have no more value 20 years since its purchase. C . The y-intercept is (20,0). It implies the car would have no more value 20 years since its purchase. D . The y-intercept is (0,44000). It implies the car's initial value was 44000. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: Since this line is in its slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,44000). It implies the car's initial value was 44000. Challenge Fill in the variables and in with the numbers and You may only use each number once. For the steepest possible slope, must be , must be , and must be . For the shallowest possible slope, must be , must be , and must be . Let s put into slope-intercept form. Then, the equation becomes To make the steepest slope possible, we must maximize the fraction The possibilities are and The biggest of these is Thus, must equal 9, must equal 6, and must equal 7. To make the shallowest slope possible, we must minimize the fraction The smallest of the possible fractions is Thus, must equal 6, must equal 9, and must equal 7. " }, { "id": "section-standard-form-2", @@ -14731,7 +14650,7 @@ var ptx_lunr_docs = [ "type": "Definition", "number": "3.7.2", "title": "Standard Form.", - "body": " Standard Form It is always possible to write an equation for a line in the form where , , and are three numbers (each of which might be , although at least one of and must be nonzero). This form of a line equation is called standard form . In the context of an application, the meaning of , , and depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations). standard form form standard " + "body": " Standard Form standard form form standard It is always possible to write an equation for a line in the form where , , and are three numbers (each of which might be , although at least one of and must be nonzero). This form of a line equation is called standard form . In the context of an application, the meaning of , , and depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations). " }, { "id": "section-standard-form-4-5", @@ -14740,7 +14659,7 @@ var ptx_lunr_docs = [ "type": "Checkpoint", "number": "3.7.3", "title": "", - "body": " For each of the following equations, identify what form they are in. ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear ? slope-intercept point-slope standard other linear not linear is in standard form, with and is in point-slope form, with slope and passing through is linear, but not in any of the forms we have studied. Using algebra, you can rearrange it to read is not linear. The exponent on is a dead giveaway. is in standard form, with and is in slope-intercept form, with slope and -intercept at " + "body": " PTX:ERROR: WeBWorK problem webwork-1289 with seed 1289 is either empty or failed to compile Use -a to halt with full PG and returned content " }, { "id": "figure-medical-procedure", @@ -14767,7 +14686,7 @@ var ptx_lunr_docs = [ "type": "Example", "number": "3.7.6", "title": "", - "body": " Graph using its intercepts. And then use the intercepts to calculate the line's slope. To graph a line by its -intercept and -intercept, it might help to first set up a table like in Figure : Intercepts of -value -value Intercepts -intercept -intercept A table like this might help you stay focused on the fact that we are searching for two points. As we've noted earlier, an -intercept is on the -axis, and so its -coordinate must be . This is worth taking special note of: to find an -intercept, must be . This is why we put in the -value cell of the -intercept. Similarly, a line's -intercept has , and we put into the -value cell of the -intercept. Next, we calculate the line's -intercept by substituting into the equation So the line's -intercept is . Similarly, we substitute into the equation to calculate the -intercept: So the line's -intercept is . Now we can complete the table: Intercepts of -value -value Intercepts -intercept -intercept With both intercepts' coordinates, we can graph the line: Graph of a coordinate grid with the graph of line 2x-3y=-6; the x-intercept is (-3,0) and the y-intercept is (0,2) There is a slope triangle from the -intercept to the origin up to the -intercept. It tells us that the slope is . " + "body": " Graph using its intercepts. And then use the intercepts to calculate the line's slope. To graph a line by its -intercept and -intercept, it might help to first set up a table like in : Intercepts of -value -value Intercepts -intercept -intercept A table like this might help you stay focused on the fact that we are searching for two points. As we've noted earlier, an -intercept is on the -axis, and so its -coordinate must be . This is worth taking special note of: to find an -intercept, must be . This is why we put in the -value cell of the -intercept. Similarly, a line's -intercept has , and we put into the -value cell of the -intercept. Next, we calculate the line's -intercept by substituting into the equation So the line's -intercept is . Similarly, we substitute into the equation to calculate the -intercept: So the line's -intercept is . Now we can complete the table: Intercepts of -value -value Intercepts -intercept -intercept With both intercepts' coordinates, we can graph the line: Graph of a coordinate grid with the graph of line 2x-3y=-6; the x-intercept is (-3,0) and the y-intercept is (0,2) There is a slope triangle from the -intercept to the origin up to the -intercept. It tells us that the slope is . " }, { "id": "section-standard-form-5-7", @@ -14776,7 +14695,7 @@ var ptx_lunr_docs = [ "type": "Fact", "number": "3.7.10", "title": "", - "body": " slope given intercepts If a line's -intercept is at and its -intercept is at , then the slope of the line is . (Unless the line passes through the origin, in which case both and equal , and then this fraction is undefined. And the slope of the line could be anything.) " + "body": " slope given intercepts If a line's -intercept is at and its -intercept is at , then the slope of the line is . (Unless the line passes through the origin, in which case both and equal , and then this fraction is undefined. And the slope of the line could be anything.) " }, { "id": "section-standard-form-5-8", @@ -14794,7 +14713,7 @@ var ptx_lunr_docs = [ "type": "Example", "number": "3.7.12", "title": "", - "body": " Change to slope-intercept form, and then graph it. Since a line in slope-intercept form looks like , we will solve for in : In the third line, we wrote on the right side, instead of . The only reason we did this is because we are headed to slope-intercept form, where the -term is traditionally written first. Now we can see that the slope is and the -intercept is at . With these things found, we can graph the line using slope triangles. Compare this graphing method with the Graphing by Intercepts method in Example . We have more points in this graph, thus we can graph the line more accurately. Graphing with Slope Triangles a coordinate gris with the graph of the line 2x-3y=-6; the y-intercept is 2 and the slope is 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 " + "body": " Change to slope-intercept form, and then graph it. Since a line in slope-intercept form looks like , we will solve for in : In the third line, we wrote on the right side, instead of . The only reason we did this is because we are headed to slope-intercept form, where the -term is traditionally written first. Now we can see that the slope is and the -intercept is at . With these things found, we can graph the line using slope triangles. Compare this graphing method with the Graphing by Intercepts method in . We have more points in this graph, thus we can graph the line more accurately. Graphing with Slope Triangles a coordinate gris with the graph of the line 2x-3y=-6; the y-intercept is 2 and the slope is 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 " }, { "id": "example-graph-standard-c-zero", @@ -14803,7 +14722,7 @@ var ptx_lunr_docs = [ "type": "Example", "number": "3.7.14", "title": "", - "body": " Graph . First, we will try (and fail) to graph this line using its - and -intercepts. Trying to find the -intercept: So the line's -intercept is at , at the origin. Huh, that is also on the -axis Trying to find the -intercept: So the line's -intercept is also at . Since both intercepts are the same point, there is no way to use the intercepts alone to graph this line. So what can be done? Several approaches are out there, but one is to convert the line equation into slope-intercept form: So the line's slope is , and we can graph the line using slope triangles and the intercept at , as in Figure . Graphing with Slope Triangles a coordinate grid with the graph of line 2x-3y=0; The line has a y-intercept of 0 and a slope of 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 In summary, if in a standard form equation , it's convenient to graph it by first converting the equation to slope-intercept form . " + "body": " Graph . First, we will try (and fail) to graph this line using its - and -intercepts. Trying to find the -intercept: So the line's -intercept is at , at the origin. Huh, that is also on the -axis Trying to find the -intercept: So the line's -intercept is also at . Since both intercepts are the same point, there is no way to use the intercepts alone to graph this line. So what can be done? Several approaches are out there, but one is to convert the line equation into slope-intercept form: So the line's slope is , and we can graph the line using slope triangles and the intercept at , as in . Graphing with Slope Triangles a coordinate grid with the graph of line 2x-3y=0; The line has a y-intercept of 0 and a slope of 2\/3; slope triangles are shown along the line with a run of 3 and a rise of 2 In summary, if in a standard form equation , it's convenient to graph it by first converting the equation to slope-intercept form . " }, { "id": "subsection-transforming-between-standard-form-and-slope-intercept-form-6", @@ -15004,9 +14923,9 @@ var ptx_lunr_docs = [ "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . " }, { - "id": "section-standard-form-8-2-14-2", + "id": "write-slope-intercept-form-in-standard-form", "level": "2", - "url": "section-standard-form.html#section-standard-form-8-2-14-2", + "url": "section-standard-form.html#write-slope-intercept-form-in-standard-form", "type": "Exercise", "number": "3.7.5.19", "title": "", @@ -15226,7 +15145,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.43", "title": "", - "body": " Casandra is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. She can spend a total of . Assume Casandra will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (150,0). It implies Casandra can purchase 150 tea bags with no sugar bags. B . The x-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags. C . The x-intercept is (60,0). It implies Casandra can purchase 60 tea bags with no sugar bags. D . The x-intercept is (0,60). It implies Casandra can purchase 60 sugar bags with no tea bags. Assume Casandra will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Casandra can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is C. " + "body": " Daniel is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. He can spend a total of . Assume Daniel will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags he can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (0, 20). It implies Daniel can purchase 20 sugar bags with no tea bags. B . The x-intercept is (0,25). It implies Daniel can purchase 25 sugar bags with no tea bags. C . The x-intercept is (25,0). It implies Daniel can purchase 25 tea bags with no sugar bags. D . The x-intercept is (20,0). It implies Daniel can purchase 20 tea bags with no sugar bags. Assume Daniel will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Daniel can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is C. " }, { "id": "section-standard-form-8-2-17-3", @@ -15235,7 +15154,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.44", "title": "", - "body": " Page is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. She can spend a total of . Assume Page will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (0,45). It implies Page can purchase 45 sugar bags with no tea bags. B . The y-intercept is (15,0). It implies Page can purchase 15 tea bags with no sugar bags. C . The y-intercept is (0, 15). It implies Page can purchase 15 sugar bags with no tea bags. D . The y-intercept is (45,0). It implies Page can purchase 45 tea bags with no sugar bags. Assume Page will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Page can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The y-intercept is (0, 15). It implies Page can purchase 15 sugar bags with no tea bags. " + "body": " Casandra is buying some tea bags and some sugar bags. Each tea bag costs cents, and each sugar bag costs cents. She can spend a total of . Assume Casandra will purchase tea bags and sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (0,60). It implies Casandra can purchase 60 sugar bags with no tea bags. B . The y-intercept is (150,0). It implies Casandra can purchase 150 tea bags with no sugar bags. C . The y-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags. D . The y-intercept is (60,0). It implies Casandra can purchase 60 tea bags with no sugar bags. Assume Casandra will purchase tea bags and sugar bags. Since each tea bag costs cents, tea bags would cost cents. Similarly, sugar bags would cost cents. Since Casandra can spend a total of , or cents, we can write the equation: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The y-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags. " }, { "id": "section-standard-form-8-2-17-4", @@ -15244,7 +15163,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.45", "title": "", - "body": " An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (150,0). It implies the engine will run out of gas 150 hours after its tank was refilled. B . The x-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled. C . The x-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank. D . The x-intercept is (0,30). It implies the engine started with 30 gallons of gas in its tank. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled. " + "body": " An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (175,0). It implies the engine will run out of gas 175 hours after its tank was refilled. B . The x-intercept is (0,175). It implies the engine started with 175 gallons of gas in its tank. C . The x-intercept is (0,35). It implies the engine started with 35 gallons of gas in its tank. D . The x-intercept is (35,0). It implies the engine will run out of gas 35 hours after its tank was refilled. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (35,0). It implies the engine will run out of gas 35 hours after its tank was refilled. " }, { "id": "section-standard-form-8-2-17-5", @@ -15253,7 +15172,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.46", "title": "", - "body": " An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (45,0). It implies the engine will run out of gas 45 hours after its tank was refilled. B . The y-intercept is (0,180). It implies the engine started with 180 gallons of gas in its tank. C . The y-intercept is (180,0). It implies the engine will run out of gas 180 hours after its tank was refilled. D . The y-intercept is (0,45). It implies the engine started with 45 gallons of gas in its tank. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: Since the equation is in slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,180). It implies the engine started with 180 gallons of gas in its tank. " + "body": " An engine s tank can hold gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming gallons of gas per hour. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (0,30). It implies the engine started with 30 gallons of gas in its tank. B . The y-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank. C . The y-intercept is (150,0). It implies the engine will run out of gas 150 hours after its tank was refilled. D . The y-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled. Assume the engine has been running for hours since its tank was refilled, and assume there are gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is: Since the equation is in slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank. " }, { "id": "section-standard-form-8-2-17-6", @@ -15262,7 +15181,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.47", "title": "", - "body": " A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (0,44000). It implies the car's initial value was 44000. B . The x-intercept is (44000,0). It implies the car's initial value was 44000. C . The x-intercept is (0,20). It implies the car would have no more value 20 years since its purchase. D . The x-intercept is (20,0). It implies the car would have no more value 20 years since its purchase. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (20,0). It implies the car would have no more value 20 years since its purchase. " + "body": " A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The x-intercept is (0,24). It implies the car would have no more value 24 years since its purchase. B . The x-intercept is (24,0). It implies the car would have no more value 24 years since its purchase. C . The x-intercept is (0,52800). It implies the car's initial value was 52800. D . The x-intercept is (52800,0). It implies the car's initial value was 52800. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: A line s -intercept looks like . To find its intercept, we substitute in the equation with , and we have: The line s -intercept is . The correct solution is: The x-intercept is (24,0). It implies the car would have no more value 24 years since its purchase. " }, { "id": "section-standard-form-8-2-17-7", @@ -15271,7 +15190,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.48", "title": "", - "body": " A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (16,0). It implies the car would have no more value 16 years since its purchase. B . The y-intercept is (0,16). It implies the car would have no more value 16 years since its purchase. C . The y-intercept is (35200,0). It implies the car's initial value was 35200. D . The y-intercept is (0,35200). It implies the car's initial value was 35200. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: Since this line is in its slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,35200). It implies the car's initial value was 35200. " + "body": " A new car of a certain model costs . According to Blue Book, its value decreases by every year. Assume years since its purchase, the car s value is dollars. Use a linear equation to model the car s value. Find this line s -intercept, and interpret its meaning in this context. A . The y-intercept is (44000,0). It implies the car's initial value was 44000. B . The y-intercept is (0,20). It implies the car would have no more value 20 years since its purchase. C . The y-intercept is (20,0). It implies the car would have no more value 20 years since its purchase. D . The y-intercept is (0,44000). It implies the car's initial value was 44000. Assume years since its purchase, the car s value is dollars. A linear equation to model the car s value is: Since this line is in its slope-intercept mode, we can see its -intercept is . The correct solution is: The y-intercept is (0,44000). It implies the car's initial value was 44000. " }, { "id": "section-standard-form-8-3-2", @@ -15280,7 +15199,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "3.7.5.49", "title": "", - "body": " Fill in the variables and in with the numbers and You may only use each number once. For the steepest possible slope, must be , must be , and must be . For the shallowest possible slope, must be , must be , and must be . Let s put into slope-intercept form. Then, the equation becomes To make the steepest slope possible, we must maximize the fraction The possibilities are and The biggest of these is Thus, must equal 14, must equal 6, and must equal 7. To make the shallowest slope possible, we must minimize the fraction The smallest of the possible fractions is Thus, must equal 6, must equal 14, and must equal 7. " + "body": " Fill in the variables and in with the numbers and You may only use each number once. For the steepest possible slope, must be , must be , and must be . For the shallowest possible slope, must be , must be , and must be . Let s put into slope-intercept form. Then, the equation becomes To make the steepest slope possible, we must maximize the fraction The possibilities are and The biggest of these is Thus, must equal 9, must equal 6, and must equal 7. To make the shallowest slope possible, we must minimize the fraction The smallest of the possible fractions is Thus, must equal 6, must equal 9, and must equal 7. " }, { "id": "section-horizontal-vertical-parallel-and-perpendicular-lines", @@ -16452,6 +16371,519 @@ var ptx_lunr_docs = [ "title": "", "body": " Use whatever method you think best to plot . " }, +{ + "id": "review-graphing-lines", + "level": "1", + "url": "review-graphing-lines.html", + "type": "Section", + "number": "3.10", + "title": "Graphing Lines Chapter Review", + "body": " Graphing Lines Chapter Review Cartesian Coordinates In Section we covered the definition of the Cartesian Coordinate System and how to plot points using the - and -axes. On paper, sketch a Cartesian coordinate system with units, and then plot the following points: . A Cartesian grid with the four points plotted. a Cartesian grid with four points plotted; the point (3,2) is 3 units to the right of the origin and two units up; the point (-5,-1) is 5 units to the left of the origin and 1 unit down; the point (0,-3) is on the y-axis 3 units down; the point (4,0) is right 4 units along the x-axis. Graphing Equations In Section we covered how to plot solutions to equations to produce a graph of the equation. Graph the equation . Making a table for Set up the table Point Complete the table Point We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line. Graphing the Equation Use points from the table a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1) Connect the points in whatever pattern is apparent a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1);they are connected with a solid line with arrows on each end Exploring Two-Variable Data and Rate of Change In Section we covered how to find patterns in tables of data and how to calculate the rate of change between points in data. Write an equation in the form suggested by the pattern in the table. A table of linear data. We consider how the values change from one row to the next. From row to row, the -value increases by . Also, the -value decreases by from row to row. Since row-to-row change is always for and is always for , the rate of change from one row to another row is always the same: units of for every unit of . We know that the output for is . And our observation about the constant rate of change tells us that if we increase the input by units from , the output should decrease by , which is . So the output would be . So the equation is . Slope slope review In Section we covered the definition of slope and how to use slope triangles to calculate slope. There is also the slope formula which helps find the slope through any two points. Find the slope of the line in the following graph. The line with two points indicated. This is a grid with a line, passing the points (3,15) and (6,27). The line with a slope triangle drawn. This is a grid with a line, passing the points (3,15) and (6,27). A slope triangle is drawn, starting at (3,15), passing (6,15), and ends at (6,27). The label from (3,15) to (6,15) is \"6-3=3\"; the label from (6,15) to (6,27) is \"27-15=12\". We picked two points on the line, and then drew a slope triangle. Next, we will do: The line's slope is . Finding a Line's Slope by the Slope Formula Use the slope formula to find the slope of the line that passes through the points and . The line's slope is . Slope-Intercept Form In Section we covered the definition of slope intercept-form and both wrote equations in slope-intercept form and graphed lines given in slope-intercept form. Graph the line . Graphing First, plot the line's -intercept, . a Cartesian grid with the point (0,4) plotted The slope is . So we can try using a run of and a rise of or a run of and a rise of . the previous plot of the y-intercept with slope triangles drawn; they go right 2 and down 5 twice to the right; from the y-intercept they also go up 5 and left 2 once; the points plotted are (-2,9), (0,4), (2,-1), (4,-6) Arrowheads and labels are encouraged. the previous plot with a solid line drawn through all of the points; there are arrowheads on each end of the line Writing a Line's Equation in Slope-Intercept Form Based on Graph Given a line's graph, we can identify its -intercept, and then find its slope using a slope triangle. With a line's slope and -intercept, we can write its equation in the form of . Find the equation of the line in the graph. Graph of a line a coordinate graph of a line;three points on the line are (-3,12),(0,10) and (3,8) Identify the line's -intercept, . the previous graph showing the line crosses the y-axis at 10 or (0,10) Identify the line's slope using a slope triangle. Note that we can pick any two points on the line to create a slope triangle. We would get the same slope: the previous graph with slope triangles drawn moving right 3 units and down two units With the line's slope and -intercept , we can write the line's equation in slope-intercept form: . Point-Slope Form In Section we covered the definition of point-slope form and both wrote equations in point-slope form and graphed lines given in point-slope form. A line passes through and . Find this line's equation in point-slope form. . We will use the slope formula to find the slope first. After labeling those two points as , we have: Now the point-slope equation looks like . Next, we will use and substitute with and with , and we have: Standard Form In Section we covered the definition of standard form of a linear equation. We converted equations from standard form to slope-intercept form and vice versa. We also graphed lines from standard form by finding the intercepts of the line. Convert into slope-intercept form. Convert into standard form. The line's equation in slope-intercept form is . The line's equation in standard form is . To graph a line in standard form, we could first change it to slope-intercept form, and then graph the line by its -intercept and slope triangles. A second method is to graph the line by its -intercept and -intercept. Graph using its intercepts. And then use the intercepts to calculate the line's slope. We calculate the line's -intercept by substituting into the equation So the line's -intercept is . Similarly, we substitute into the equation to calculate the -intercept: So the line's -intercept is . With both intercepts' coordinates, we can graph the line: Graph of the graph of line 2x-3y=-6 with the x-intercept plotted at (-3,0) and the y-intercept at (0,2). Now that we have graphed the line we can read the slope. The rise is units and the run is units so the slope is . Horizontal, Vertical, Parallel, and Perpendicular Lines In Section we studied horizontal and vertical lines. We also covered the relationships between the slopes of parallel and perpendicular lines. Line 's equation is . Line is parallel to , and line also passes the point . Find an equation for line in point-slope form. Since parallel lines have the same slope, line 's slope is also . Since line also passes the point , we can write line 's equation in point-slope form: Two lines are perpendicular if and only if the product of their slopes is . Line 's equation is . Line is perpendicular to , and line also passes the point . Find an equation for line in slope-intercept form. Since line and are perpendicular, the product of their slopes is . Because line 's slope is given as , we can find line 's slope is . Since line also passes the point , we can write line 's equation in point-slope form: We can now convert this equation to slope-intercept form: Sketch the points , , , , , and on a Cartesian plane. a Cartesian graph with the points plotted;(8,2) is 8 units to the right and up 2;(5,5) is 5 to the right and up 5;(-3,0) is down 3 units on the y-axis;(0,14\/3)is 14\/3 or 4 2\/3 units up on the y-axis;(3,-2.5) is 3 units to the right and down 2.5;(-5,7) is left 5 units and up 7 Consider the equation Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer. We substitute in each ordered pair s - and -values into the equation , and see whether the resulting equation is true. In two cases, and , we have true equations: So and are solutions to the given equation. In the other two cases, and , we have false equations: So and are not solutions. Consider the equation Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer. We substitute in each ordered pair s - and -values into the equation , and see whether the resulting equation is true. In two cases, and , we have true equations: So and are solutions to the given equation. In the other two cases, and , we have false equations: So and are not solutions. Write an equation in the form suggested by the pattern in the table. The relationship between and in each row is not as clear here. Another popular approach for finding patterns: in each column, consider how the values change from one row to the next. From row to row, the -value increases by Also, the -value increases by from row to row. Since row-to-row change is always for and is always for the rate of change from one row to another row is always the same: units of for every unit of We know that the output for is And our observation about the constant rate of change tells us that if we increase the input by units from the output should increase by which is So the output would be So the equation is Write an equation in the form suggested by the pattern in the table. The relationship between and in each row is not as clear here. Another popular approach for finding patterns: in each column, consider how the values change from one row to the next. From row to row, the -value increases by Also, the -value decreases by from row to row. Since row-to-row change is always for and is always for the rate of change from one row to another row is always the same: units of for every unit of We know that the output for is And our observation about the constant rate of change tells us that if we increase the input by units from the output should increase by which is So the output would be So the equation is Find the slope of the line. graph of a line crossing the y-axis at -3; the line has an upward slant and also passes through the point (4,6) To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . graph of the line detailing a slope triangle from (0,-3) to (4,-3) to (4,6) This line s slope is . Find the slope of the line. graph of a line crossing the y-axis at 4; the line has an upward slant and also passes through the point (7,7) To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . graph of the line detailing a slope triangle from (0,4) to (7,4) to (7,7) This line s slope is . Find the slope of the line. graph of a line crossing the y-axis at -2; the line has an upward slant and also passes through the point (5,1) To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . graph of the line detailing a slope triangle from (0,-2) to (5,-2) to (5,1) This line s slope is . Below is a line s graph. The slope of this line is . To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . This line's slope is . Below is a line s graph. The slope of this line is . To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . This line's slope is . Below is a line s graph. The slope of this line is . To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . This line's slope is . A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: So the line s slope is . A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: So the line s slope is . A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This is a special line which is parallel to the -axis. So the line s slope is . A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This is a special line which is parallel to the -axis. So the line s slope is . A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: Since we cannot divide by , this line s slope does not exist. This is a special line which is parallel to the -axis; it is a vertical line. So the line s slope D oes N ot E xist. A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: Since we cannot divide by , this line s slope does not exist. This is a special line which is parallel to the -axis; it is a vertical line. So the line s slope D oes N ot E xist. A line s graph is given. What is this line s slope-intercept equation? A line's slope-intercept equation has the form , where is the slope and is the -coordinate of the -intercept. We first find the slope. To find the slope of a line from its graph, we identify two points, and then draw a slope triangle. It's wise to choose points with integer coordinates. For this problem, we choose and . Next, we draw a slope triangle and find the \"rise\" and \"run\". In this problem, the rise is and the run is . This line's slope is . It's clear in the graph that this line's -intercept is . So this line's slope-intercept equation is . A line s graph is given. What is this line s slope-intercept equation? A line's slope-intercept equation has the form , where is the slope and is the -coordinate of the -intercept. We first find the slope. To find the slope of a line from its graph, we identify two points, and then draw a slope triangle. It's wise to choose points with integer coordinates. For this problem, we choose and . Next, we draw a slope triangle and find the \"rise\" and \"run\". In this problem, the rise is and the run is . This line's slope is . It's clear in the graph that this line's -intercept is . So this line's slope-intercept equation is . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Five minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. This line s slope-intercept equation is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in slope-intercept form looks like , where is the slope, and is the -coordinate of the -intercept. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . The next step is to find the value of . There are at least two ways. One way is to substitute a known point into . We use the point . This line s slope-intercept equation is . Another way is to use the line s point-slope equation . Again, we use the point . This is the second way to find , and thus the line s slope-intercept equation is . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation . We have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Seven minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. This line s slope-intercept equation is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in slope-intercept form looks like , where is the slope, and is the -coordinate of the -intercept. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . The next step is to find the value of . There are at least two ways. One way is to substitute a known point into . We use the point . This line s slope-intercept equation is . Another way is to use the line s point-slope equation . Again, we use the point . This is the second way to find , and thus the line s slope-intercept equation is . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation . We have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . Fill out this table for the equation . The first row is an example. Points Values of and satisfying the equation Points Values of and satisfying the equation Fill out this table for the equation . The first row is an example. Points Values of and satisfying the equation Points Values of and satisfying the equation A line s graph is shown. Write an equation for the line. This line is horizontal, passing and . This implies its slope is . If a line's slope is , its equation has the form . By the graph, we can see that its -intercept is , so this line's equation is . A line s graph is shown. Write an equation for the line. This line is horizontal, passing and . This implies its slope is . If a line's slope is , its equation has the form . By the graph, we can see that its -intercept is , so this line's equation is . Line passes points and . Line passes points and . These two lines are . To find a line s slope, we can use the slope formula: For line , we have: This is a special line, . It is vertical with an undefined slope. For line , we have: This is a special line, . It is vertical with an undefined slope. Since both lines are vertical, these two lines are parallel. Line passes points and . Line passes points and . These two lines are . To find a line s slope, we can use the slope formula: For line , we have: This is a special line, . It is vertical with an undefined slope. For line , we have: This is a special line, . It is vertical with an undefined slope. Since both lines are vertical, these two lines are parallel. Line s equation is . Line is perpendicular to line and passes through the point . Find an equation for line in both point-slope form and slope-intercept form. An equation for in point-slope form is: . An equation for in slope-intercept form is: . When two lines are perpendicular, the product of their slopes is . It s given that line s slope is . So the slope of must be . Now we know line s slope, and a point on it: . We substitute these numbers into the generic formula for a point-slope equation for a line: and we have: The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and expand and simplify the right side. Line s equation is . Line is perpendicular to line and passes through the point . Find an equation for line in both point-slope form and slope-intercept form. An equation for in point-slope form is: . An equation for in slope-intercept form is: . When two lines are perpendicular, the product of their slopes is . It s given that line s slope is . So the slope of must be . Now we know line s slope, and a point on it: . We substitute these numbers into the generic formula for a point-slope equation for a line: and we have: The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and expand and simplify the right side. Graph the linear inequality . Graph the linear inequality . Graph the linear inequality . Graph the linear inequality . " +}, +{ + "id": "review-graphing-lines-2-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-2-3", + "type": "Example", + "number": "3.10.1", + "title": "", + "body": " On paper, sketch a Cartesian coordinate system with units, and then plot the following points: . A Cartesian grid with the four points plotted. a Cartesian grid with four points plotted; the point (3,2) is 3 units to the right of the origin and two units up; the point (-5,-1) is 5 units to the left of the origin and 1 unit down; the point (0,-3) is on the y-axis 3 units down; the point (4,0) is right 4 units along the x-axis. " +}, +{ + "id": "review-graphing-lines-3-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-3-3", + "type": "Example", + "number": "3.10.3", + "title": "", + "body": " Graph the equation . Making a table for Set up the table Point Complete the table Point We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line. Graphing the Equation Use points from the table a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1) Connect the points in whatever pattern is apparent a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1);they are connected with a solid line with arrows on each end " +}, +{ + "id": "review-graphing-lines-4-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-4-3", + "type": "Example", + "number": "3.10.6", + "title": "", + "body": " Write an equation in the form suggested by the pattern in the table. A table of linear data. We consider how the values change from one row to the next. From row to row, the -value increases by . Also, the -value decreases by from row to row. Since row-to-row change is always for and is always for , the rate of change from one row to another row is always the same: units of for every unit of . We know that the output for is . And our observation about the constant rate of change tells us that if we increase the input by units from , the output should decrease by , which is . So the output would be . So the equation is . " +}, +{ + "id": "review-graphing-lines-5-4", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-5-4", + "type": "Example", + "number": "3.10.8", + "title": "", + "body": " Find the slope of the line in the following graph. The line with two points indicated. This is a grid with a line, passing the points (3,15) and (6,27). The line with a slope triangle drawn. This is a grid with a line, passing the points (3,15) and (6,27). A slope triangle is drawn, starting at (3,15), passing (6,15), and ends at (6,27). The label from (3,15) to (6,15) is \"6-3=3\"; the label from (6,15) to (6,27) is \"27-15=12\". We picked two points on the line, and then drew a slope triangle. Next, we will do: The line's slope is . " +}, +{ + "id": "review-graphing-lines-5-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-5-5", + "type": "Example", + "number": "3.10.11", + "title": "Finding a Line’s Slope by the Slope Formula.", + "body": " Finding a Line's Slope by the Slope Formula Use the slope formula to find the slope of the line that passes through the points and . The line's slope is . " +}, +{ + "id": "review-graphing-lines-6-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-6-3", + "type": "Example", + "number": "3.10.12", + "title": "", + "body": " Graph the line . Graphing First, plot the line's -intercept, . a Cartesian grid with the point (0,4) plotted The slope is . So we can try using a run of and a rise of or a run of and a rise of . the previous plot of the y-intercept with slope triangles drawn; they go right 2 and down 5 twice to the right; from the y-intercept they also go up 5 and left 2 once; the points plotted are (-2,9), (0,4), (2,-1), (4,-6) Arrowheads and labels are encouraged. the previous plot with a solid line drawn through all of the points; there are arrowheads on each end of the line " +}, +{ + "id": "review-graphing-lines-6-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-6-5", + "type": "Example", + "number": "3.10.14", + "title": "", + "body": " Find the equation of the line in the graph. Graph of a line a coordinate graph of a line;three points on the line are (-3,12),(0,10) and (3,8) Identify the line's -intercept, . the previous graph showing the line crosses the y-axis at 10 or (0,10) Identify the line's slope using a slope triangle. Note that we can pick any two points on the line to create a slope triangle. We would get the same slope: the previous graph with slope triangles drawn moving right 3 units and down two units With the line's slope and -intercept , we can write the line's equation in slope-intercept form: . " +}, +{ + "id": "review-graphing-lines-7-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-7-3", + "type": "Example", + "number": "3.10.18", + "title": "", + "body": " A line passes through and . Find this line's equation in point-slope form. . We will use the slope formula to find the slope first. After labeling those two points as , we have: Now the point-slope equation looks like . Next, we will use and substitute with and with , and we have: " +}, +{ + "id": "review-graphing-lines-8-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-8-3", + "type": "Example", + "number": "3.10.19", + "title": "", + "body": " Convert into slope-intercept form. Convert into standard form. The line's equation in slope-intercept form is . The line's equation in standard form is . " +}, +{ + "id": "review-graphing-lines-8-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-8-5", + "type": "Example", + "number": "3.10.20", + "title": "", + "body": " Graph using its intercepts. And then use the intercepts to calculate the line's slope. We calculate the line's -intercept by substituting into the equation So the line's -intercept is . Similarly, we substitute into the equation to calculate the -intercept: So the line's -intercept is . With both intercepts' coordinates, we can graph the line: Graph of the graph of line 2x-3y=-6 with the x-intercept plotted at (-3,0) and the y-intercept at (0,2). Now that we have graphed the line we can read the slope. The rise is units and the run is units so the slope is . " +}, +{ + "id": "review-graphing-lines-9-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-9-3", + "type": "Example", + "number": "3.10.22", + "title": "", + "body": " Line 's equation is . Line is parallel to , and line also passes the point . Find an equation for line in point-slope form. Since parallel lines have the same slope, line 's slope is also . Since line also passes the point , we can write line 's equation in point-slope form: " +}, +{ + "id": "review-graphing-lines-9-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-9-5", + "type": "Example", + "number": "3.10.23", + "title": "", + "body": " Line 's equation is . Line is perpendicular to , and line also passes the point . Find an equation for line in slope-intercept form. Since line and are perpendicular, the product of their slopes is . Because line 's slope is given as , we can find line 's slope is . Since line also passes the point , we can write line 's equation in point-slope form: We can now convert this equation to slope-intercept form: " +}, +{ + "id": "review-graphing-lines-10-1-1", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-1-1", + "type": "Exercise", + "number": "3.10.9.1", + "title": "", + "body": " Sketch the points , , , , , and on a Cartesian plane. a Cartesian graph with the points plotted;(8,2) is 8 units to the right and up 2;(5,5) is 5 to the right and up 5;(-3,0) is down 3 units on the y-axis;(0,14\/3)is 14\/3 or 4 2\/3 units up on the y-axis;(3,-2.5) is 3 units to the right and down 2.5;(-5,7) is left 5 units and up 7 " +}, +{ + "id": "review-graphing-lines-10-1-2", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-1-2", + "type": "Exercise", + "number": "3.10.9.2", + "title": "", + "body": " " +}, +{ + "id": "review-graphing-lines-10-1-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-1-3", + "type": "Exercise", + "number": "3.10.9.3", + "title": "", + "body": " Consider the equation Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer. We substitute in each ordered pair s - and -values into the equation , and see whether the resulting equation is true. In two cases, and , we have true equations: So and are solutions to the given equation. In the other two cases, and , we have false equations: So and are not solutions. " +}, +{ + "id": "review-graphing-lines-10-1-4", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-1-4", + "type": "Exercise", + "number": "3.10.9.4", + "title": "", + "body": " Consider the equation Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer. We substitute in each ordered pair s - and -values into the equation , and see whether the resulting equation is true. In two cases, and , we have true equations: So and are solutions to the given equation. In the other two cases, and , we have false equations: So and are not solutions. " +}, +{ + "id": "review-graphing-lines-10-1-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-1-5", + "type": "Exercise", + "number": "3.10.9.5", + "title": "", + "body": " Write an equation in the form suggested by the pattern in the table. The relationship between and in each row is not as clear here. Another popular approach for finding patterns: in each column, consider how the values change from one row to the next. From row to row, the -value increases by Also, the -value increases by from row to row. Since row-to-row change is always for and is always for the rate of change from one row to another row is always the same: units of for every unit of We know that the output for is And our observation about the constant rate of change tells us that if we increase the input by units from the output should increase by which is So the output would be So the equation is " +}, +{ + "id": "review-graphing-lines-10-1-6", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-1-6", + "type": "Exercise", + "number": "3.10.9.6", + "title": "", + "body": " Write an equation in the form suggested by the pattern in the table. The relationship between and in each row is not as clear here. Another popular approach for finding patterns: in each column, consider how the values change from one row to the next. From row to row, the -value increases by Also, the -value decreases by from row to row. Since row-to-row change is always for and is always for the rate of change from one row to another row is always the same: units of for every unit of We know that the output for is And our observation about the constant rate of change tells us that if we increase the input by units from the output should increase by which is So the output would be So the equation is " +}, +{ + "id": "review-graphing-lines-10-2-1", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-2-1", + "type": "Exercise", + "number": "3.10.9.7", + "title": "", + "body": " Find the slope of the line. graph of a line crossing the y-axis at -3; the line has an upward slant and also passes through the point (4,6) To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . graph of the line detailing a slope triangle from (0,-3) to (4,-3) to (4,6) This line s slope is . " +}, +{ + "id": "review-graphing-lines-10-2-2", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-2-2", + "type": "Exercise", + "number": "3.10.9.8", + "title": "", + "body": " Find the slope of the line. graph of a line crossing the y-axis at 4; the line has an upward slant and also passes through the point (7,7) To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . graph of the line detailing a slope triangle from (0,4) to (7,4) to (7,7) This line s slope is . " +}, +{ + "id": "review-graphing-lines-10-2-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-2-3", + "type": "Exercise", + "number": "3.10.9.9", + "title": "", + "body": " Find the slope of the line. graph of a line crossing the y-axis at -2; the line has an upward slant and also passes through the point (5,1) To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . graph of the line detailing a slope triangle from (0,-2) to (5,-2) to (5,1) This line s slope is . " +}, +{ + "id": "review-graphing-lines-10-2-4", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-2-4", + "type": "Exercise", + "number": "3.10.9.10", + "title": "", + "body": " Below is a line s graph. The slope of this line is . To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . This line's slope is . " +}, +{ + "id": "review-graphing-lines-10-2-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-2-5", + "type": "Exercise", + "number": "3.10.9.11", + "title": "", + "body": " Below is a line s graph. The slope of this line is . To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . This line's slope is . " +}, +{ + "id": "review-graphing-lines-10-2-6", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-2-6", + "type": "Exercise", + "number": "3.10.9.12", + "title": "", + "body": " Below is a line s graph. The slope of this line is . To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose and . Next, we sketch a slope triangle and find the rise and run . In the sketch below, the rise is and the run is . This line's slope is . " +}, +{ + "id": "review-graphing-lines-10-3-1", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-1", + "type": "Exercise", + "number": "3.10.9.13", + "title": "", + "body": " A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: So the line s slope is . " +}, +{ + "id": "review-graphing-lines-10-3-2", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-2", + "type": "Exercise", + "number": "3.10.9.14", + "title": "", + "body": " A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: So the line s slope is . " +}, +{ + "id": "review-graphing-lines-10-3-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-3", + "type": "Exercise", + "number": "3.10.9.15", + "title": "", + "body": " A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This is a special line which is parallel to the -axis. So the line s slope is . " +}, +{ + "id": "review-graphing-lines-10-3-4", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-4", + "type": "Exercise", + "number": "3.10.9.16", + "title": "", + "body": " A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: This is a special line which is parallel to the -axis. So the line s slope is . " +}, +{ + "id": "review-graphing-lines-10-3-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-5", + "type": "Exercise", + "number": "3.10.9.17", + "title": "", + "body": " A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: Since we cannot divide by , this line s slope does not exist. This is a special line which is parallel to the -axis; it is a vertical line. So the line s slope D oes N ot E xist. " +}, +{ + "id": "review-graphing-lines-10-3-6", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-6", + "type": "Exercise", + "number": "3.10.9.18", + "title": "", + "body": " A line passes through the points and . Find this line s slope. To find a line s slope, we can use the slope formula: First, we mark which number corresponds to which variable in the formula: Now we substitute these numbers into the corresponding variables in the slope formula: Since we cannot divide by , this line s slope does not exist. This is a special line which is parallel to the -axis; it is a vertical line. So the line s slope D oes N ot E xist. " +}, +{ + "id": "review-graphing-lines-10-3-7", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-7", + "type": "Exercise", + "number": "3.10.9.19", + "title": "", + "body": " A line s graph is given. What is this line s slope-intercept equation? A line's slope-intercept equation has the form , where is the slope and is the -coordinate of the -intercept. We first find the slope. To find the slope of a line from its graph, we identify two points, and then draw a slope triangle. It's wise to choose points with integer coordinates. For this problem, we choose and . Next, we draw a slope triangle and find the \"rise\" and \"run\". In this problem, the rise is and the run is . This line's slope is . It's clear in the graph that this line's -intercept is . So this line's slope-intercept equation is . " +}, +{ + "id": "review-graphing-lines-10-3-8", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-8", + "type": "Exercise", + "number": "3.10.9.20", + "title": "", + "body": " A line s graph is given. What is this line s slope-intercept equation? A line's slope-intercept equation has the form , where is the slope and is the -coordinate of the -intercept. We first find the slope. To find the slope of a line from its graph, we identify two points, and then draw a slope triangle. It's wise to choose points with integer coordinates. For this problem, we choose and . Next, we draw a slope triangle and find the \"rise\" and \"run\". In this problem, the rise is and the run is . This line's slope is . It's clear in the graph that this line's -intercept is . So this line's slope-intercept equation is . " +}, +{ + "id": "review-graphing-lines-10-3-9", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-9", + "type": "Exercise", + "number": "3.10.9.21", + "title": "", + "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . " +}, +{ + "id": "review-graphing-lines-10-3-10", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-10", + "type": "Exercise", + "number": "3.10.9.22", + "title": "", + "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . " +}, +{ + "id": "review-graphing-lines-10-3-11", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-11", + "type": "Exercise", + "number": "3.10.9.23", + "title": "", + "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " +}, +{ + "id": "review-graphing-lines-10-3-12", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-3-12", + "type": "Exercise", + "number": "3.10.9.24", + "title": "", + "body": " A line passes through the points and . Find this line s equation in point-slope form. Using the point , this line s point-slope form equation is . Using the point , this line s point-slope form equation is . A line s equation in point-slope form looks like where is the slope of the line and is a point that the line passes through. We first need to find the line s slope. To find a line s slope, we can use the slope formula: We mark which number corresponds to which variable in the formula: Now we substitute these values into the corresponding variables in the slope formula: Now we have . The next step is to use a point that we know the line passes through. If we choose to use the point , we have: If we choose to use the point , we have: Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise. " +}, +{ + "id": "review-graphing-lines-10-4", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-4", + "type": "Exercise", + "number": "3.10.9.25", + "title": "", + "body": " Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Five minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. This line s slope-intercept equation is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in slope-intercept form looks like , where is the slope, and is the -coordinate of the -intercept. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . The next step is to find the value of . There are at least two ways. One way is to substitute a known point into . We use the point . This line s slope-intercept equation is . Another way is to use the line s point-slope equation . Again, we use the point . This is the second way to find , and thus the line s slope-intercept equation is . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation . We have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. " +}, +{ + "id": "review-graphing-lines-10-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-5", + "type": "Exercise", + "number": "3.10.9.26", + "title": "", + "body": " Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose grams. Seven minutes since the experiment started, the remaining gas had a mass of grams. Let be the number of minutes that have passed since the experiment started, and let be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time. This line s slope-intercept equation is . minutes after the experiment started, there would be grams of gas left. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone. A line s equation in slope-intercept form looks like , where is the slope, and is the -coordinate of the -intercept. Since the gas is leaking at a rate of grams per minute, we know that the slope of the linear model is . So now we have that . The next step is to find the value of . There are at least two ways. One way is to substitute a known point into . We use the point . This line s slope-intercept equation is . Another way is to use the line s point-slope equation . Again, we use the point . This is the second way to find , and thus the line s slope-intercept equation is . After minutes, we can find the mass of the remaining gas if we substitute in for in the equation . We have: So after minutes, there are grams of gas remaining. To find when the gas will be all gone, we need to find when there are grams of gas left. We can substitute in for in the equation , and we have: So after minutes, the gas will be all gone. " +}, +{ + "id": "review-graphing-lines-10-6", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-6", + "type": "Exercise", + "number": "3.10.9.27", + "title": "", + "body": " Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line " +}, +{ + "id": "review-graphing-lines-10-7", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-7", + "type": "Exercise", + "number": "3.10.9.28", + "title": "", + "body": " Find the -intercept and -intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row. -value -value Location (as an ordered pair) -intercept -intercept -intercept and -intercept of the line A line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: This line's -intercept is . Next, a line's -intercept is on the -axis, implying that its -value must be . To find a line's -intercept, we substitute in . In this problem we have: The line's -intercept is . The entries for the table are: -value -value Location -intercept -intercept -intercept and -intercept of the line " +}, +{ + "id": "review-graphing-lines-10-8-1", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-1", + "type": "Exercise", + "number": "3.10.9.29", + "title": "", + "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . " +}, +{ + "id": "review-graphing-lines-10-8-2", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-2", + "type": "Exercise", + "number": "3.10.9.30", + "title": "", + "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this form, is the line s slope, and is the coordinate on the -axis where the line intercepts the -axis. In this problem, the line s equation is given as . It would be helpful to algebraically rearrange this into slope-intercept form: . Now we can see the line s slope is , and its -intercept has coordinates . " +}, +{ + "id": "review-graphing-lines-10-8-3", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-3", + "type": "Exercise", + "number": "3.10.9.31", + "title": "", + "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . " +}, +{ + "id": "review-graphing-lines-10-8-4", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-4", + "type": "Exercise", + "number": "3.10.9.32", + "title": "", + "body": " Find the line s slope and -intercept. A line has equation . This line s slope is . This line s -intercept is . When an equation of a line is written in the form , it is said to be in slope-intercept form . In this problem, the line s equation is given in the standard form . If we algebraically rearrange the equation from standard form to slope-intercept form , it will be easy to see the slope and -intercept. Now we can see the line s slope is , and its -intercept has coordinates . " +}, +{ + "id": "review-graphing-lines-10-8-5", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-5", + "type": "Exercise", + "number": "3.10.9.33", + "title": "", + "body": " Fill out this table for the equation . The first row is an example. Points Values of and satisfying the equation Points Values of and satisfying the equation " +}, +{ + "id": "review-graphing-lines-10-8-6", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-6", + "type": "Exercise", + "number": "3.10.9.34", + "title": "", + "body": " Fill out this table for the equation . The first row is an example. Points Values of and satisfying the equation Points Values of and satisfying the equation " +}, +{ + "id": "review-graphing-lines-10-8-7", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-7", + "type": "Exercise", + "number": "3.10.9.35", + "title": "", + "body": " A line s graph is shown. Write an equation for the line. This line is horizontal, passing and . This implies its slope is . If a line's slope is , its equation has the form . By the graph, we can see that its -intercept is , so this line's equation is . " +}, +{ + "id": "review-graphing-lines-10-8-8", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-8", + "type": "Exercise", + "number": "3.10.9.36", + "title": "", + "body": " A line s graph is shown. Write an equation for the line. This line is horizontal, passing and . This implies its slope is . If a line's slope is , its equation has the form . By the graph, we can see that its -intercept is , so this line's equation is . " +}, +{ + "id": "review-graphing-lines-10-8-9", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-9", + "type": "Exercise", + "number": "3.10.9.37", + "title": "", + "body": " Line passes points and . Line passes points and . These two lines are . To find a line s slope, we can use the slope formula: For line , we have: This is a special line, . It is vertical with an undefined slope. For line , we have: This is a special line, . It is vertical with an undefined slope. Since both lines are vertical, these two lines are parallel. " +}, +{ + "id": "review-graphing-lines-10-8-10", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-10", + "type": "Exercise", + "number": "3.10.9.38", + "title": "", + "body": " Line passes points and . Line passes points and . These two lines are . To find a line s slope, we can use the slope formula: For line , we have: This is a special line, . It is vertical with an undefined slope. For line , we have: This is a special line, . It is vertical with an undefined slope. Since both lines are vertical, these two lines are parallel. " +}, +{ + "id": "review-graphing-lines-10-8-11", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-11", + "type": "Exercise", + "number": "3.10.9.39", + "title": "", + "body": " Line s equation is . Line is perpendicular to line and passes through the point . Find an equation for line in both point-slope form and slope-intercept form. An equation for in point-slope form is: . An equation for in slope-intercept form is: . When two lines are perpendicular, the product of their slopes is . It s given that line s slope is . So the slope of must be . Now we know line s slope, and a point on it: . We substitute these numbers into the generic formula for a point-slope equation for a line: and we have: The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and expand and simplify the right side. " +}, +{ + "id": "review-graphing-lines-10-8-12", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-12", + "type": "Exercise", + "number": "3.10.9.40", + "title": "", + "body": " Line s equation is . Line is perpendicular to line and passes through the point . Find an equation for line in both point-slope form and slope-intercept form. An equation for in point-slope form is: . An equation for in slope-intercept form is: . When two lines are perpendicular, the product of their slopes is . It s given that line s slope is . So the slope of must be . Now we know line s slope, and a point on it: . We substitute these numbers into the generic formula for a point-slope equation for a line: and we have: The slope-intercept form of a line equation looks like , where is the slope and is the -intercept. We can take the point-slope form and expand and simplify the right side. " +}, +{ + "id": "review-graphing-lines-10-8-13", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-13", + "type": "Exercise", + "number": "3.10.9.41", + "title": "", + "body": " Graph the linear inequality . " +}, +{ + "id": "review-graphing-lines-10-8-14", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-14", + "type": "Exercise", + "number": "3.10.9.42", + "title": "", + "body": " Graph the linear inequality . " +}, +{ + "id": "review-graphing-lines-10-8-15", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-15", + "type": "Exercise", + "number": "3.10.9.43", + "title": "", + "body": " Graph the linear inequality . " +}, +{ + "id": "review-graphing-lines-10-8-16", + "level": "2", + "url": "review-graphing-lines.html#review-graphing-lines-10-8-16", + "type": "Exercise", + "number": "3.10.9.44", + "title": "", + "body": " Graph the linear inequality . " +}, { "id": "section-solving-a-system-by-graphing", "level": "1", @@ -16459,7 +16891,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "4.1", "title": "Solving a System by Graphing", - "body": " Solving a System by Graphing PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In we learned a few ways we can graph one line in the plane. Now we will graph two lines simultaneoulsy and use what we see to identify the solution set to a system of two linear equations in two variables . Alternative Video Lesson Systems of Two Linear Equations in Two Variables There are times when two linear equations (each one with the same two variables) are relevant to some situation at the same time. When this happens, we have a system of two equations in two variables which maybe we can write this way: The large left brace indicates that this is a collection of two distinct equations, not one equation that was somehow algebraically manipulated into an equivalent equation. In the system above, , , , and are specific numbers, while and are the variables. The two line equations in the system above are both in slope-intercept form, but that is not required. Each line equation could be written in some other form and we would still call this a system of two equations in two variables. Let's explore an example of how these things can arise. Fabiana and David are running at constant speeds in parallel lanes on a track. David starts out ahead of Fabiana, but Fabiana is running faster. We want to determine when Fabiana will catch up with David. Suppose that Fabiana is running with a speed of . If she started out at position meters, then her position in meters after seconds is given by David is running more slowly, at only . But he had a head start, starting out at a position meters ahead of Fabiana. So David's position in meters after seconds is given by One of these equations represents Fabiana, and the other represents David. But imagine if you could find a point that worked as a solution to Fabiana's equation and also as a solution to David's equation. That would mean there is a moment in time ( seconds after the clock started) when Fabiana's position equals David's position. (Both runners would be meters from the starting point.) In other words, this would be the moment when Fabiana catches up to David. So we consider the two equations together: We have a system of two equations in two variables. And if we can find a common solution to both equations, we have figured out something meaningful. A solution to a system is an ordered pair that is a solution to both equations in the system. Of course the variables may be different letters they should be whatever two variables the equations are using. In , there is a solution: . You can verify that it works for both equations. Could you find that ordered pair yourself? In this section and the ones that follow, we will learn some techniques for finding it yourself. In this case, the solution tells us that it takes 8 for Fabiana to catch up with David. And when she does, they are 18 from the starting point. For future reference, we can have some formal defintions now. System of Linear Equations Systems of linear equations Linear systems of equations Systems of linear equations Solution(s) of system of linear equations A system of linear equations is any pairing of two (or more) linear equations. (But in this book we are only examining systems with two equations with two variables.) A solution to a system of linear equations is any point that is a solution to each of the equations in the system. The solution set to a system of linear equations is the collection of all solutions to the system. The solution set may be empty, may consist of one point only, or may have more. Solving a System of Equations by Graphing Graphing systems of linear equations to solve Systems of linear equations solved by graphing If the two equations in a system are both easy to graph, it might happen that we can find the solution to a system by graphing them both on the same axis system. Systems of linear equations applications Let's return to Fabiana and David from . The system of equations was And each of these two lines is straightforward to graph. Fabiana's line has vertical intercept at and slope . So we can plot it by starting at the origin and using slope triangles that move to the right, then up . David's line has vertical intercept at and slope . So we can plot it by starting at and using slope triangles that move to the right, then up . David and Fabiana's distances a coordinate plane with a line for Fabiana and a line for David; Fabiana's line has a y-intercept of (0,0) and David's line has a y-intercept of (0,2); the lines cross at the point (8,18) As we can see in , the two line equations cross at the point . Note that this means is a solution to Fabiana's equation and also to David's equation. So there is a moment in time ( 8 ) when both runners are at the same position ( 18 ). This means that is the solution to our system of linear equation. And it means that it takes Fabiana 8 to catch up to David. The lesson of is that we can find a solution to a system by graphing both lines and identifying where they cross. Determine the solution to the system of equations graphed in . Graph of a System of Equations a cartesian plot with two intersecting lines; one line has a y-intercept of -2 and a slope of -1\/4; the other line has a y-intercept of 5 and a slope of 2 The two lines intersect where and , so there is only one solution. It is the point . The solution set is . Determine the solution to the system of equations graphed below. The two lines intersect where and so the solution is the point Now let's solve a system where we need to make the graph ourselves. Solve the following system of equations by graphing: Notice that each of these equations is written in slope-intercept form. The first equation, , has slope and -intercept . The second equation, , has slope and -intercept . We can use this information to graph both lines. and . a Cartesian grid with two intersecting lines; one line has a y-intercept of 4 and a slope of 1\/2; the other line has a y-intercept of -5 and a slope of -1 It appears that the two lines intersect where and , so the solution to the system of equations would be the point . However we should be careful. Maybe the lines are poorly drawn, or maybe they cross at a point close to that is too close for us to see. We should check that actually works as a solution to each of the original equations, since we have those equations. This verifies that is the solution, and we write the solution set as . Solve the following system of equations by graphing. Both of these equations are written in slope-intercept form. The first equation, has slope and -intercept The second equation, has slope and -intercept We can use this information to graph both lines. a Cartesian grid with two intersecting lines; one line has a y-intercept of -5 and a slope of 2\/3; the other line has a y-intercept of 2 and a slope of -1\/2 It appears that the two lines intersect where and so the solution to the system of equations would be the point However we should check that actually works as a solution to each of the original equations. This verifies that is the solution. Solve the following system of equations by graphing. Note that the equations are in standard form. To review plotting a line equation that is in standard form, see Section 3.7. Both of these equations are written in standard form. The first equation, has -intercept at and -intercept at The second equation, has -intercept at and -intercept at We can use this information to graph both lines. a Cartesian grid with two intersecting lines; one line has a x-intercept of -12 and y-intercept of 4; the other line has x-intercept of -4 and y-intercept of -4 It appears that the two lines intersect where and so the solution to the system of equations would be the point However we should check that actually works as a solution to each of the original equations. This verifies that is the solution. Solve the following system of equations by graphing. Note that the equations are in point-slope form. To review plotting a line equation that is in point-slope form, see Section 3.6. Both of these equations are written in point-slope form. The first equation, passes through and has slope The second equation, passes through and has slope We can use this information to graph both lines. a Cartesian grid with two intersecting lines; one line passes through (2,1) with slope 3; the other line passes through (-1,-1) with slope -1\/2 It appears that the two lines intersect where and so the solution to the system of equations would be the point However we should check that actually works as a solution to each of the original equations. This verifies that is the solution. Systems of linear equations applications A college has a north campus and a south campus. The north campus has thousand students, but enrollment has been declining by thousand students per year. The newer south campus has only thousand students, but enrollment has been increasing by thousand students per year. If these trends continue, in how many years would the two campuses have the same number of students? Write a system of equations for this scenario, then solve it graphically to find the answer. Write two equations that form a system for this scenario. The north campus has initial population value and decreases with rate thousand students per year. So one equation is The south campus has initial population value and increases with rate thousand students per year. So the other equation is Plot the lines for the system of two equations. For plotting, it may be helpful to write the system with the decimals converted to fractions: So we see that we can use and as -intercepts. And then for the first line, use slope triangles where we move units to the right and then units down. While for the second line, use slope triangles where we move units to the right and then units up. a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8) Based on the graph, how long will it be until the two campuses have the same number of students? How many students will each campus have at that time? The lines cross at So after years, each campus will have thousand students. Special Systems of Equations Systems of linear equations with no solution Systems of linear equations with infinitely many solutions When we studied linear equations in one variable, there were two special cases discussed in detail in . One of these special cases was like with the equation , where there simply is no solution at all. The solution set is empty. The other special case is like with the equation , where there are infinitely many solutions. When solving systems of two linear equations in two variables, we have similar special cases to consider. Parallel Lines Consider the graphs of two lines with the same slope, and . Graphs of and a coordinate grid with two parallel lines; one line has a y-intercept of -4 and the other has a y-intercept of 1; they both have a slope of 2 For the system of equations what would the solution be? Since the two lines have the same slope, they are parallel lines and they never cross. This means that there is no solution to this system of equations. Because if there were a solution then it would be a point that is on the first line and also on the second line. So it would be a point where the two lines would cross. When there are no solutions to a system, we can simply say that. Or we can write that the solution set is the empty set, which we can write as or . When a system of two linear equations has no solution, we call the system inconsistent . The idea is that while it may possible for and to Inconsistent system have the relationship given by the first equation, and while it may possible for and to have the relationship given by the second equation, it's not possible for both of these relationships to happen at the same time: they are inconsistent. The Symbol is not Zero The symbol is a special symbol that represents an empty set, a set with no Empty set numbers in it. You could also write the empty set as braces with nothing in between them, Set empty like . But is a fancy alternative. This symbol is not the same thing as the number zero. The symbol represents a set with nothing in it. The symbol represents the number zero. If you have an empty carton of eggs, the carton itself would be , while is the count of how many eggs are in that carton. Coinciding Lines Consider the graphs of two lines with equations and . In other words, the system: To solve this system of equations, we want to graph each line. The first equation is in slope-intercept form and we graph it by starting at the -intercept and using the slope to make slope triangles. The second equation is in standard form. We find its -intercept is and its -intercept is , and use this to plot the line. The two lines are plotted together in . Now we can see these two lines are actually the same line. They coincide. Can we really solve this system? Finding a solution means finding a point that is a solution to each of the two lines in the system. But apparently any point on this one line we see is actually a solution to both of the original line equations. So we have an infinite number of solutions. All points that fall on that one line are in the solution set. Graphs of and a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-4) and a slope of 2 It may be enough for us to report that there are infinitely many solutions, not just one. But we could also be more specific and use set-builder notation. We want to write that any ordered pair that satisfies one (or the other) of the two original line equations is actually a solution to the system. We can write that the solution set is . When a system of two linear equations has infinitely many solutions, we call the system dependent . The two equation may not have looked identical, but it turned out that Dependent system they represented the same line. So in a sense, the two equations depend on one another (since actually they are equaivalent equations). In , what would have happened if we had decided to convert the second line equation into slope-intercept form? This is the literally the same as the first equation from that system. This is a different way to show that these two equations represent coinciding lines. Notice that for a system of equations with infinite solutions like , we didn't say that everything is a solution. It's only the points that are on that coinciding line that are solutions. It would be incorrect to say that the solution set is all real numbers or as all ordered pairs . Solve the following system of equations by graphing. If we graph these lines, we find they are the same line. a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0) So there are infinitely many solutions: all points on that common line. Solve the following system of equations by graphing. If we graph these lines, we find they are parallel and never cross. a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0) So there are no solutions to this system. We can summarize the possibilities for what a soltuion set to a system of two linear equations in two variables. Three Types of Systems of Two Linear Equations in Two Variables Crossing Lines If two linear equations make lines with different slopes, the system has one solution. The one solution is the place where the lines cross. Parallel Lines If two linear equations make lines with the same slope but different -intercepts, the system has no solution. Coinciding Lines If two linear equations make lines with the same slope and the same -intercept (in other words, they make the same line), the system has infinitely many solutions. This solution set consists of all ordered pairs on that line. What is the purpose of the one big left brace in a system of two equations? When you find a solution to a system of two linear equations in two variables, why should you check the solution? When you are checking a solution to a system of two linear equations in two variables, would it be good enough to only substitute the numbers into one of the original two equations? Why or why not? Suppose you have a system of two linear equations, and you know the system has exactly one solution. What can you say about the slopes of the two lines? Review Select the equations\/inequalities that are linear with one variable. None of the above None of the above Skills Practice Check a Possible Solution to a System Check if the given point is a solution to the given system of two linear equations. Is a solution? Is a solution? Is a solution? Is a solution? Is a solution? Is a solution? Identify Solution from Graph The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (-4, -3) a coordinate plot of two lines that cross at (-3, 4) a coordinate plot of two lines that cross at (-2, 3) a coordinate plot of two lines that cross at (-1, -4) a coordinate plot of two parallel lines a coordinate plot of two parallel lines See How Many Solutions Simply by looking closely at the linear equations, determine how many solutions the system will have. And state whether the system is dependeent, inconsistent, or neither. Solve a System Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-1, 8) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (2, 3) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (3, -4) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (5, 8) So the only solution is If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (7, 1) So there are no solutions. If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-8, -3) So there are no solutions. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-8, -5) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-9, -4) So the only solution is If we graph these lines, we find they are parallel and never cross. a coordinate plot of two parallel lines So the there are no solutions. If we graph these lines, we find they are parallel and never cross. a coordinate plot of two parallel lines So the there are no solutions. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-3, 8) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-4, (-2)) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, -1) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-3, -7) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-1, 5) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (1, -3) So the only solution is Applications Ezekiel is organizing an office lunch party. His budget for beverages is and he will try to spend all of it. He assume that people will attend, and there should be two beverages avaiable per person. Ezekiel has decided to order cans of a fancy beverage that cost each, and cans of a cheaper beverage that cost each. How many of each should he order? Write two equations that form a system for this scenario. Since there are people and there should be two drinks per person, one equation is Since the first type costs each, the second type costs each, and the total should be another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,141.75\/1.75) and (141.75\/2.25,0); the other line passes through (0,2*36.5) and (2*36.5,0); the lines cross at (28,45) Based on the graph, how many of the more expensive beverage should Ezekiel buy? How many of the less expensive beverage should Ezekiel buy? The lines cross at So Ezekiel should buy 28 of the more expensive drink, and 45 of the less expensive drink. At the local hardware store, Rhys bought a hammer and four boxes of nails. The total cost was The hammer costs more than a box of nails. How much does a hammer cost and how much does one box of nails cost? Write two equations that form a system for this scenario. Since the total cost of the hammer and four boxes of nails is one equation is Since the hammer costs more than a box of nails, another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,36\/4) and (36,0); the other line passes through (0,-16) and (16,0); the lines cross at (20,4) Based on the graph, how much does the hammer cost? How much does a box of nails cost? The lines cross at So a hammer costs and a box of nails costs $4. Donavan enters a grassy field from some dense trees. His dog is standing out in the field, away. As soon as they see each other, they start running toward each other. Donavan runs with speed and his dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. Donavan starts out at position running with speed So one equation is The dog starts running with speed but getting closer to Donavan. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,75) with slope 3; the other line passes through (0,0) with slope -12; the lines cross at (5,15) Based on the graph, how long will it be until they meet? How far will that be from the place where Donavan started? The lines cross at So after seconds, they will meet 15 feet away from where Donavan started. A cyclist riding at rides past a dog. A moment later, when the bicycle is away, the dog begins to chase the bicycle at a speed of Assuming the cyclist does not change course or speed, how long will it be until the dog catches up to the bicycle? Write two equations that form a system for this scenario. The bicycle starts out away and is traveling at So one equation is The dog starts running at time with speed So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,50) with slope 25; the other line passes through (0,0) with slope 30; the lines cross at (10,300) Based on the graph, how long will it be until the dog catches up with the bicycle? How far will that be from the place where the dog started running? The lines cross at So after seconds, the cyclist and the dog are each 300 feet away from where the dog started running. " + "body": " Solving a System by Graphing PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In we learned a few ways we can graph one line in the plane. Now we will graph two lines simultaneoulsy and use what we see to identify the solution set to a system of two linear equations in two variables . Alternative Video Lesson Systems of Two Linear Equations in Two Variables There are times when two linear equations (each one with the same two variables) are relevant to some situation at the same time. When this happens, we have a system of two equations in two variables which maybe we can write this way: The large left brace indicates that this is a collection of two distinct equations, not one equation that was somehow algebraically manipulated into an equivalent equation. In the system above, , , , and are specific numbers, while and are the variables. The two line equations in the system above are both in slope-intercept form, but that is not required. Each line equation could be written in some other form and we would still call this a system of two equations in two variables. Let's explore an example of how these things can arise. Fabiana and David are running at constant speeds in parallel lanes on a track. David starts out ahead of Fabiana, but Fabiana is running faster. We want to determine when Fabiana will catch up with David. Suppose that Fabiana is running with a speed of . If she started out at position meters, then her position in meters after seconds is given by David is running more slowly, at only . But he had a head start, starting out at a position meters ahead of Fabiana. So David's position in meters after seconds is given by One of these equations represents Fabiana, and the other represents David. But imagine if you could find a point that worked as a solution to Fabiana's equation and also as a solution to David's equation. That would mean there is a moment in time ( seconds after the clock started) when Fabiana's position equals David's position. (Both runners would be meters from the starting point.) In other words, this would be the moment when Fabiana catches up to David. So we consider the two equations together: We have a system of two equations in two variables. And if we can find a common solution to both equations, we have figured out something meaningful. A solution to a system is an ordered pair that is a solution to both equations in the system. Of course the variables may be different letters they should be whatever two variables the equations are using. In , there is a solution: . You can verify that it works for both equations. Could you find that ordered pair yourself? In this section and the ones that follow, we will learn some techniques for finding it yourself. In this case, the solution tells us that it takes 8 for Fabiana to catch up with David. And when she does, they are 18 from the starting point. For future reference, we can have some formal defintions now. System of Linear Equations Systems of linear equations Linear systems of equations Systems of linear equations Solution(s) of system of linear equations A system of linear equations is any pairing of two (or more) linear equations. (But in this book we are only examining systems with two equations with two variables.) A solution to a system of linear equations is any point that is a solution to each of the equations in the system. The solution set to a system of linear equations is the collection of all solutions to the system. The solution set may be empty, may consist of one point only, or may have more. Solving a System of Equations by Graphing Graphing systems of linear equations to solve Systems of linear equations solved by graphing If the two equations in a system are both easy to graph, it might happen that we can find the solution to a system by graphing them both on the same axis system. Systems of linear equations applications Let's return to Fabiana and David from . The system of equations was And each of these two lines is straightforward to graph. Fabiana's line has vertical intercept at and slope . So we can plot it by starting at the origin and using slope triangles that move to the right, then up . David's line has vertical intercept at and slope . So we can plot it by starting at and using slope triangles that move to the right, then up . David and Fabiana's distances a coordinate plane with a line for Fabiana and a line for David; Fabiana's line has a y-intercept of (0,0) and David's line has a y-intercept of (0,2); the lines cross at the point (8,18) As we can see in , the two line equations cross at the point . Note that this means is a solution to Fabiana's equation and also to David's equation. So there is a moment in time ( 8 ) when both runners are at the same position ( 18 ). This means that is the solution to our system of linear equation. And it means that it takes Fabiana 8 to catch up to David. The lesson of is that we can find a solution to a system by graphing both lines and identifying where they cross. Determine the solution to the system of equations graphed in . Graph of a System of Equations a cartesian plot with two intersecting lines; one line has a y-intercept of -2 and a slope of -1\/4; the other line has a y-intercept of 5 and a slope of 2 The two lines intersect where and , so there is only one solution. It is the point . The solution set is . Determine the solution to the system of equations graphed below. The two lines intersect where and so the solution is the point Now let's solve a system where we need to make the graph ourselves. Solve the following system of equations by graphing: Notice that each of these equations is written in slope-intercept form. The first equation, , has slope and -intercept . The second equation, , has slope and -intercept . We can use this information to graph both lines. and . a Cartesian grid with two intersecting lines; one line has a y-intercept of 4 and a slope of 1\/2; the other line has a y-intercept of -5 and a slope of -1 It appears that the two lines intersect where and , so the solution to the system of equations would be the point . However we should be careful. Maybe the lines are poorly drawn, or maybe they cross at a point close to that is too close for us to see. We should check that actually works as a solution to each of the original equations, since we have those equations. This verifies that is the solution, and we write the solution set as . Solve the following system of equations by graphing. Both of these equations are written in slope-intercept form. The first equation, has slope and -intercept The second equation, has slope and -intercept We can use this information to graph both lines. a Cartesian grid with two intersecting lines; one line has a y-intercept of -5 and a slope of 2\/3; the other line has a y-intercept of 2 and a slope of -1\/2 It appears that the two lines intersect where and so the solution to the system of equations would be the point However we should check that actually works as a solution to each of the original equations. This verifies that is the solution. Solve the following system of equations by graphing. Note that the equations are in standard form. To review plotting a line equation that is in standard form, see Section 3.7. Both of these equations are written in standard form. The first equation, has -intercept at and -intercept at The second equation, has -intercept at and -intercept at We can use this information to graph both lines. a Cartesian grid with two intersecting lines; one line has a x-intercept of -12 and y-intercept of 4; the other line has x-intercept of -4 and y-intercept of -4 It appears that the two lines intersect where and so the solution to the system of equations would be the point However we should check that actually works as a solution to each of the original equations. This verifies that is the solution. Solve the following system of equations by graphing. Note that the equations are in point-slope form. To review plotting a line equation that is in point-slope form, see Section 3.6. Both of these equations are written in point-slope form. The first equation, passes through and has slope The second equation, passes through and has slope We can use this information to graph both lines. a Cartesian grid with two intersecting lines; one line passes through (2,1) with slope 3; the other line passes through (-1,-1) with slope -1\/2 It appears that the two lines intersect where and so the solution to the system of equations would be the point However we should check that actually works as a solution to each of the original equations. This verifies that is the solution. Systems of linear equations applications A college has a north campus and a south campus. The north campus has thousand students, but enrollment has been declining by thousand students per year. The newer south campus has only thousand students, but enrollment has been increasing by thousand students per year. If these trends continue, in how many years would the two campuses have the same number of students? Write a system of equations for this scenario, then solve it graphically to find the answer. Write two equations that form a system for this scenario. The north campus has initial population value and decreases with rate thousand students per year. So one equation is The south campus has initial population value and increases with rate thousand students per year. So the other equation is Plot the lines for the system of two equations. For plotting, it may be helpful to write the system with the decimals converted to fractions: So we see that we can use and as -intercepts. And then for the first line, use slope triangles where we move units to the right and then units down. While for the second line, use slope triangles where we move units to the right and then units up. a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8) Based on the graph, how long will it be until the two campuses have the same number of students? How many students will each campus have at that time? The lines cross at So after years, each campus will have thousand students. Special Systems of Equations Systems of linear equations with no solution Systems of linear equations with infinitely many solutions When we studied linear equations in one variable, there were two special cases discussed in detail in . One of these special cases was like with the equation , where there simply is no solution at all. The solution set is empty. The other special case is like with the equation , where there are infinitely many solutions. When solving systems of two linear equations in two variables, we have similar special cases to consider. Parallel Lines Consider the graphs of two lines with the same slope, and . Graphs of and a coordinate grid with two parallel lines; one line has a y-intercept of -4 and the other has a y-intercept of 1; they both have a slope of 2 For the system of equations what would the solution be? Since the two lines have the same slope, they are parallel lines and they never cross. This means that there is no solution to this system of equations. Because if there were a solution then it would be a point that is on the first line and also on the second line. So it would be a point where the two lines would cross. When there are no solutions to a system, we can simply say that. Or we can write that the solution set is the empty set, which we can write as or . When a system of two linear equations has no solution, we call the system inconsistent . The idea is that while it may possible for and to Inconsistent system have the relationship given by the first equation, and while it may possible for and to have the relationship given by the second equation, it's not possible for both of these relationships to happen at the same time: they are inconsistent. The Symbol is not Zero The symbol is a special symbol that represents an empty set, a set with no Empty set numbers in it. You could also write the empty set as braces with nothing in between them, Set empty like . But is a fancy alternative. This symbol is not the same thing as the number zero. The symbol represents a set with nothing in it. The symbol represents the number zero. If you have an empty carton of eggs, the carton itself would be , while is the count of how many eggs are in that carton. Coinciding Lines Consider the graphs of two lines with equations and . In other words, the system: To solve this system of equations, we want to graph each line. The first equation is in slope-intercept form and we graph it by starting at the -intercept and using the slope to make slope triangles. The second equation is in standard form. We find its -intercept is and its -intercept is , and use this to plot the line. The two lines are plotted together in . Now we can see these two lines are actually the same line. They coincide. Can we really solve this system? Finding a solution means finding a point that is a solution to each of the two lines in the system. But apparently any point on this one line we see is actually a solution to both of the original line equations. So we have an infinite number of solutions. All points that fall on that one line are in the solution set. Graphs of and a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-4) and a slope of 2 It may be enough for us to report that there are infinitely many solutions, not just one. But we could also be more specific and use set-builder notation. We want to write that any ordered pair that satisfies one (or the other) of the two original line equations is actually a solution to the system. We can write that the solution set is . When a system of two linear equations has infinitely many solutions, we call the system dependent . The two equation may not have looked identical, but it turned out that Dependent system they represented the same line. So in a sense, the two equations depend on one another (since actually they are equaivalent equations). In , what would have happened if we had decided to convert the second line equation into slope-intercept form? This is the literally the same as the first equation from that system. This is a different way to show that these two equations represent coinciding lines. Notice that for a system of equations with infinite solutions like , we didn't say that everything is a solution. It's only the points that are on that coinciding line that are solutions. It would be incorrect to say that the solution set is all real numbers or as all ordered pairs . Solve the following system of equations by graphing. If we graph these lines, we find they are the same line. a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0) So there are infinitely many solutions: all points on that common line. Solve the following system of equations by graphing. If we graph these lines, we find they are parallel and never cross. a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0) So there are no solutions to this system. We can summarize the possibilities for what a soltuion set to a system of two linear equations in two variables. Three Types of Systems of Two Linear Equations in Two Variables Crossing Lines If two linear equations make lines with different slopes, the system has one solution. The one solution is the place where the lines cross. Parallel Lines If two linear equations make lines with the same slope but different -intercepts, the system has no solution. Coinciding Lines If two linear equations make lines with the same slope and the same -intercept (in other words, they make the same line), the system has infinitely many solutions. This solution set consists of all ordered pairs on that line. What is the purpose of the one big left brace in a system of two equations? When you find a solution to a system of two linear equations in two variables, why should you check the solution? When you are checking a solution to a system of two linear equations in two variables, would it be good enough to only substitute the numbers into one of the original two equations? Why or why not? Suppose you have a system of two linear equations, and you know the system has exactly one solution. What can you say about the slopes of the two lines? Review Select the equations\/inequalities that are linear with one variable. None of the above None of the above Skills Practice Check a Possible Solution to a System Check if the given point is a solution to the given system of two linear equations. Is a solution? Is a solution? Is a solution? Is a solution? Is a solution? Is a solution? Identify Solution from Graph The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (-2, 5) a coordinate plot of two lines that cross at (-1, 2) a coordinate plot of two lines that cross at (1, -3) a coordinate plot of two lines that cross at (2, 5) a coordinate plot of two parallel lines a coordinate plot of two parallel lines See How Many Solutions Simply by looking closely at the linear equations, determine how many solutions the system will have. And state whether the system is dependeent, inconsistent, or neither. Solve a System Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (3, 5) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (5, -2) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (7, -7) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (8, 3) So the only solution is If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-7, -1) So there are no solutions. If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-5, -8) So there are no solutions. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-4, -3) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (2, -6) So the only solution is If we graph these lines, we find they are parallel and never cross. a coordinate plot of two parallel lines So the there are no solutions. If we graph these lines, we find they are parallel and never cross. a coordinate plot of two parallel lines So the there are no solutions. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, 2) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-2, (-1)) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (, ) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-2, -5) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (1, 7) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (3, -1) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (5, -6) So the only solution is Applications Rhys is organizing an office lunch party. His budget for beverages is and he will try to spend all of it. He assume that people will attend, and there should be two beverages avaiable per person. Rhys has decided to order cans of a fancy beverage that cost each, and cans of a cheaper beverage that cost each. How many of each should he order? Write two equations that form a system for this scenario. Since there are people and there should be two drinks per person, one equation is Since the first type costs each, the second type costs each, and the total should be another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,42\/2) and (42\/3,0); the other line passes through (0,2*8.5) and (2*8.5,0); the lines cross at (8,9) Based on the graph, how many of the more expensive beverage should Rhys buy? How many of the less expensive beverage should Rhys buy? The lines cross at So Rhys should buy 8 of the more expensive drink, and 9 of the less expensive drink. At the local hardware store, Marshall bought a hammer and four boxes of nails. The total cost was The hammer costs more than a box of nails. How much does a hammer cost and how much does one box of nails cost? Write two equations that form a system for this scenario. Since the total cost of the hammer and four boxes of nails is one equation is Since the hammer costs more than a box of nails, another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,52\/4) and (52,0); the other line passes through (0,-37) and (37,0); the lines cross at (40,3) Based on the graph, how much does the hammer cost? How much does a box of nails cost? The lines cross at So a hammer costs and a box of nails costs $3. Aryana enters a grassy field from some dense trees. Her dog is standing out in the field, away. As soon as they see each other, they start running toward each other. Aryana runs with speed and her dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. Aryana starts out at position running with speed So one equation is The dog starts running with speed but getting closer to Aryana. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,55) with slope 3; the other line passes through (0,0) with slope -8; the lines cross at (5,15) Based on the graph, how long will it be until they meet? How far will that be from the place where Aryana started? The lines cross at So after seconds, they will meet 15 feet away from where Aryana started. A cyclist riding at rides past a dog. A moment later, when the bicycle is away, the dog begins to chase the bicycle at a speed of Assuming the cyclist does not change course or speed, how long will it be until the dog catches up to the bicycle? Write two equations that form a system for this scenario. The bicycle starts out away and is traveling at So one equation is The dog starts running at time with speed So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,40) with slope 26; the other line passes through (0,0) with slope 30; the lines cross at (10,300) Based on the graph, how long will it be until the dog catches up with the bicycle? How far will that be from the place where the dog started running? The lines cross at So after seconds, the cyclist and the dog are each 300 feet away from where the dog started running. " }, { "id": "section-solving-a-system-by-graphing-2", @@ -16792,7 +17224,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.9", "title": "", - "body": " a coordinate plot of two lines that cross at (-4, -3) " + "body": " a coordinate plot of two lines that cross at (-2, 5) " }, { "id": "see-solution-from-graph-slope-intercept-copy", @@ -16801,7 +17233,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.10", "title": "", - "body": " a coordinate plot of two lines that cross at (-3, 4) " + "body": " a coordinate plot of two lines that cross at (-1, 2) " }, { "id": "see-solution-from-graph-point-slope", @@ -16810,7 +17242,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.11", "title": "", - "body": " a coordinate plot of two lines that cross at (-2, 3) " + "body": " a coordinate plot of two lines that cross at (1, -3) " }, { "id": "see-solution-from-graph-point-slope-copy", @@ -16819,7 +17251,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.12", "title": "", - "body": " a coordinate plot of two lines that cross at (-1, -4) " + "body": " a coordinate plot of two lines that cross at (2, 5) " }, { "id": "see-solution-from-graph-point-slope-parallel", @@ -16900,7 +17332,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.21", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-1, 8) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (3, 5) So the only solution is " }, { "id": "solve-system-graphically-slope-intercept-copy", @@ -16909,7 +17341,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.22", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (2, 3) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (5, -2) So the only solution is " }, { "id": "solve-system-graphically-point-slope", @@ -16918,7 +17350,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.23", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (3, -4) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (7, -7) So the only solution is " }, { "id": "solve-system-graphically-point-slope-copy", @@ -16927,7 +17359,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.24", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (5, 8) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (8, 3) So the only solution is " }, { "id": "solve-system-graphically-point-slope-no-solution", @@ -16936,7 +17368,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.25", "title": "", - "body": " If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (7, 1) So there are no solutions. " + "body": " If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-7, -1) So there are no solutions. " }, { "id": "solve-system-graphically-point-slope-no-solution-copy", @@ -16945,7 +17377,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.26", "title": "", - "body": " If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-8, -3) So there are no solutions. " + "body": " If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-5, -8) So there are no solutions. " }, { "id": "solve-system-graphically-standard", @@ -16954,7 +17386,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.27", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-8, -5) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-4, -3) So the only solution is " }, { "id": "solve-system-graphically-standard-copy", @@ -16963,7 +17395,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.28", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-9, -4) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (2, -6) So the only solution is " }, { "id": "solve-system-graphically-standard-slope-intercept", @@ -17008,7 +17440,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.33", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-3, 8) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, 2) So the only solution is " }, { "id": "solve-system-graphically-standard-point-slope-copy", @@ -17017,7 +17449,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.34", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-4, (-2)) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-2, (-1)) So the only solution is " }, { "id": "solve-system-graphically-standard-point-slope-infinite-solutions", @@ -17044,7 +17476,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.37", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, -1) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-2, -5) So the only solution is " }, { "id": "solve-system-graphically-slope-intercept-point-slope-copy", @@ -17053,7 +17485,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.38", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-3, -7) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (1, 7) So the only solution is " }, { "id": "solve-system-graphically-slope-intercept-point-slope-infinite-solutions", @@ -17062,7 +17494,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.39", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-1, 5) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (3, -1) So the only solution is " }, { "id": "solve-system-graphically-slope-intercept-point-slope-infinite-solutions-copy", @@ -17071,7 +17503,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.40", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (1, -3) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (5, -6) So the only solution is " }, { "id": "solve-system-application-party-planning", @@ -17080,7 +17512,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.41", "title": "", - "body": " Ezekiel is organizing an office lunch party. His budget for beverages is and he will try to spend all of it. He assume that people will attend, and there should be two beverages avaiable per person. Ezekiel has decided to order cans of a fancy beverage that cost each, and cans of a cheaper beverage that cost each. How many of each should he order? Write two equations that form a system for this scenario. Since there are people and there should be two drinks per person, one equation is Since the first type costs each, the second type costs each, and the total should be another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,141.75\/1.75) and (141.75\/2.25,0); the other line passes through (0,2*36.5) and (2*36.5,0); the lines cross at (28,45) Based on the graph, how many of the more expensive beverage should Ezekiel buy? How many of the less expensive beverage should Ezekiel buy? The lines cross at So Ezekiel should buy 28 of the more expensive drink, and 45 of the less expensive drink. " + "body": " Rhys is organizing an office lunch party. His budget for beverages is and he will try to spend all of it. He assume that people will attend, and there should be two beverages avaiable per person. Rhys has decided to order cans of a fancy beverage that cost each, and cans of a cheaper beverage that cost each. How many of each should he order? Write two equations that form a system for this scenario. Since there are people and there should be two drinks per person, one equation is Since the first type costs each, the second type costs each, and the total should be another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,42\/2) and (42\/3,0); the other line passes through (0,2*8.5) and (2*8.5,0); the lines cross at (8,9) Based on the graph, how many of the more expensive beverage should Rhys buy? How many of the less expensive beverage should Rhys buy? The lines cross at So Rhys should buy 8 of the more expensive drink, and 9 of the less expensive drink. " }, { "id": "solve-system-application-hardware-store", @@ -17089,7 +17521,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.42", "title": "", - "body": " At the local hardware store, Rhys bought a hammer and four boxes of nails. The total cost was The hammer costs more than a box of nails. How much does a hammer cost and how much does one box of nails cost? Write two equations that form a system for this scenario. Since the total cost of the hammer and four boxes of nails is one equation is Since the hammer costs more than a box of nails, another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,36\/4) and (36,0); the other line passes through (0,-16) and (16,0); the lines cross at (20,4) Based on the graph, how much does the hammer cost? How much does a box of nails cost? The lines cross at So a hammer costs and a box of nails costs $4. " + "body": " At the local hardware store, Marshall bought a hammer and four boxes of nails. The total cost was The hammer costs more than a box of nails. How much does a hammer cost and how much does one box of nails cost? Write two equations that form a system for this scenario. Since the total cost of the hammer and four boxes of nails is one equation is Since the hammer costs more than a box of nails, another equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,52\/4) and (52,0); the other line passes through (0,-37) and (37,0); the lines cross at (40,3) Based on the graph, how much does the hammer cost? How much does a box of nails cost? The lines cross at So a hammer costs and a box of nails costs $3. " }, { "id": "solve-system-application-greet-dog", @@ -17098,7 +17530,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.43", "title": "", - "body": " Donavan enters a grassy field from some dense trees. His dog is standing out in the field, away. As soon as they see each other, they start running toward each other. Donavan runs with speed and his dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. Donavan starts out at position running with speed So one equation is The dog starts running with speed but getting closer to Donavan. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,75) with slope 3; the other line passes through (0,0) with slope -12; the lines cross at (5,15) Based on the graph, how long will it be until they meet? How far will that be from the place where Donavan started? The lines cross at So after seconds, they will meet 15 feet away from where Donavan started. " + "body": " Aryana enters a grassy field from some dense trees. Her dog is standing out in the field, away. As soon as they see each other, they start running toward each other. Aryana runs with speed and her dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. Aryana starts out at position running with speed So one equation is The dog starts running with speed but getting closer to Aryana. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,55) with slope 3; the other line passes through (0,0) with slope -8; the lines cross at (5,15) Based on the graph, how long will it be until they meet? How far will that be from the place where Aryana started? The lines cross at So after seconds, they will meet 15 feet away from where Aryana started. " }, { "id": "solve-system-application-dog-cyclist", @@ -17107,7 +17539,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.1.5.44", "title": "", - "body": " A cyclist riding at rides past a dog. A moment later, when the bicycle is away, the dog begins to chase the bicycle at a speed of Assuming the cyclist does not change course or speed, how long will it be until the dog catches up to the bicycle? Write two equations that form a system for this scenario. The bicycle starts out away and is traveling at So one equation is The dog starts running at time with speed So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,50) with slope 25; the other line passes through (0,0) with slope 30; the lines cross at (10,300) Based on the graph, how long will it be until the dog catches up with the bicycle? How far will that be from the place where the dog started running? The lines cross at So after seconds, the cyclist and the dog are each 300 feet away from where the dog started running. " + "body": " A cyclist riding at rides past a dog. A moment later, when the bicycle is away, the dog begins to chase the bicycle at a speed of Assuming the cyclist does not change course or speed, how long will it be until the dog catches up to the bicycle? Write two equations that form a system for this scenario. The bicycle starts out away and is traveling at So one equation is The dog starts running at time with speed So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,40) with slope 26; the other line passes through (0,0) with slope 30; the lines cross at (10,300) Based on the graph, how long will it be until the dog catches up with the bicycle? How far will that be from the place where the dog started running? The lines cross at So after seconds, the cyclist and the dog are each 300 feet away from where the dog started running. " }, { "id": "section-substitution", @@ -17116,7 +17548,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "4.2", "title": "Substitution", - "body": " Substitution PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In , we used graphing to solve a system of two linear equations. While graphing is insightful, it can be time-consuming. And what happens if the actual solution to the system does not have whole number coordinates will you be able to see exactly what are the coordinates of the crossing? And what if your hand-drawn graph has some small imperfection that makes it look like the lines are crossing at the wrong place? There are two main alternatives to graphing that can find the solution to a system of two linear equations. They use pencil-and-paper algebra instead of a picture. This section covers one of those alternative methods, and the next section covers the other. Alternative Video Lesson Substitution Systems of linear equations solved by substitution Substitution method Systems of linear equations applications The Interview Once upon a time, the New York Times published an article about the movie, The Interview . It included the following quote: The Interview generated roughly million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday. Sony did not say how much of that total represented digital rentals versus sales. The studio said there were about two million transactions overall. Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.) First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let be the number of rental transactions and let be the number of sales transactions. If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both and are measured in transactions . The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is: Without the units: The price of each rental was . That means the problem has given us a rate of to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by , and that would give us the total number of dollars generated from rentals. This is . Similarly, the price of each sale was , so the revenue from sales was . The total revenue was million, which we can represent with this equation: Without the units: Here is our system of equations, with the commas removed: To solve the system, we will use a method called substitution . The idea is to use one equation to isolate . Then substitute this for the that's in the other equation. This leaves you with one equation where the only variable is . And we can handle that directly. The first equation from the system is an easy one to isolate : This tells us that the expression is equal to , so we can substitute that in place of in the second equation: At this point, we know that . This tells us that out of the million transactions, roughly were from online sales. Recall that we isolated previously and found . In summary, there were roughly ssales and roughly rentals. In , we rounded the solution values because only whole numbers make sense in the context of the problem. It was especially acceptable to round because the original numbers we worked with were already rounded. In fact, it would be OK to round even more to something like and as long as we communicate clearly that we rounded. In other exercises where there is no context, it is not OK to round. Solutions should be communicated with their exact values. Solve the system of equations using substitution: To use substitution, we need to isolate one of the variables in one of our equations. Looking over both equations, it will be easiest to isolate in the first equation: Next, we substitute in for in the second equation, giving us a linear equation in only one variable . And this is an equation that we may solve using skills from . Now that we have the value for , we need to find the value for . We already isolated , and it's easiest to just use that equation. At this point we think the solution is the point or in other words: , . It's only human to make mistakes though, so we should check that the solution actually works. To check it, try using , in both of the original equations. We conclude then that this system of equations is true when and . The solution is the point and we write the solution set as . Use substitution to solve the system. Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do is add to each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve for the one variable that remains. Solve for the other variable. Sometimes it makes more sense to start the process using the second equation instead of the first. And sometimes it makes more sense to isolate the second variable instead of the first. Solve this system of equations using substitution: We need to isolate one of the variables in one of the equations. Looking over both equations, it will be easiest to isolate in the second equation because that is where the coefficient is smallest. We are going to have to divide by some coefficient; it might as well be the smallest one to keep the fraction arithmetic as simple as we can. Note that there are fractions once we've isolated . We should take care with the steps that follow to make sure that the fraction arithmetic is correct. Substitute in for in the first equation, and that leads to having a single linear equation with only one variable. We can solve this as we did in . Note the step in the middle where we clear denominators. Now that we have the value for , we need to find the value for . We already isolated , and it's easiest to just use that equation. To check the solution we think we've found, try using , in both of the original equations. We conclude then that this system of equations is true when and . The solution is the point and we write the solution set as . A system may start out with fractions among the coefficients. Just as we learned in , the algebra can go more smoothly if we clear the denominators before doing more work. Solve the system of equations using the substitution method. When a system of equations has fraction coefficients, it may be helpful to clear the denomintors . With each equation, multiply each side by the least common multiple of that equation's denominators. In the first equation, the least common multiple of the denominators is , so: In the second equation, the least common multiple of the denominators is , so: Now we have a system that is equivalent to the original system of equations, but there are no fraction coefficients: The second equation has already isolateed , so we will substitute in for in the first equation. And we have solved for . To find , we know , so we have: The solution is . (This should be checked in the original two equations though.) Real world applications can lead to a system where both equations are either in point-slope form or slope-intercept form. This is helpful, since it means one variable has already been isolated. Solve the system of equations using the substitution method. In the first equation, is already isolated. So we can substitute in for in the second equation: . Some people prefer to think of this as setting the two right sides equal to each other. That works too: if is equal to , and is also equal to , then these expressions are equal to each other. Either way: And then: And the solution is . For summary reference, here is the general procedure. Solving Systems of Equations by Substitution Systems of linear equations solved by substitution Substitution method To solve a system of two linear equations in two variables by substitution, If helpful, clear denominators from each equation using the least common multiple of each equation's denominators. Isolate one variable in one of the equations. Substitute that expression into the other equation. There should now only be one variable in that equation. Solve that equation for the one remaining variable. Substitute that value into the earlier result where the first variable was isolated, and solve for that variable. Verify your solution using the original two equations. Applications Systems of linear equations applications Two Different Interest Rates Notah made some large purchases with his two credit cards one month and took on a total of in debt from the two cards. He didn't make any payments the first month, so the two credit card debts each started accruing interest. That month, his Visa card charged interest and his Mastercard charged interest. Because of this, Notah's total debt grew by . How much money did Notah charge to each card? We start by clearly defining two variables for the two unknowns. Let be the amount charged to the Visa card (in dollars) and let be the amount charged to the Mastercard (in dollars). To set up the equations, notice that we are given two different total dollar amounts. One is the total debt Notah initially took on, . So we have: Or without units: The other total we were given is the total amount of interest, . The Visa had dollars charged to it and accrues interest. So is the dollar amount of interest that comes from using the Visa card. Similarly, is the dollar amount of interest from using the Mastercard. Together: Or without units: As a system, we write: To solve this system by substitution, notice that it will be easier to isolate one of the variables in the first equation. We'll isolate : Now we substitute in for in the second equation: And then we can determine the value of by using the earlier equation where we isolated : In summary, Notah charged to the Visa and to the Mastercard. We should check that these numbers work as solutions to our original system and that they make sense in context. (For instance, if one of these numbers were negative, or was something small like , they wouldn't make sense as credit card debt.) The next two examples are called mixture problems , because they involve mixing two quantities together to form a combination and we want to find out how much of each quantity to mix. Mixing Solutions with Two Different Concentrations Mixture problems LaVonda is a meticulous bartender and she needs to serve milliliters of Rob Roy, an alcoholic cocktail that is alcohol by volume. The main ingredients are scotch that is alcohol and vermouth that is alcohol. How many milliliters of each ingredient should she mix together to make the concentration she needs? The two unknowns are the quantities of each ingredient. Let be the amount of scotch (in ) and let be the amount of vermouth (in ). One quantity given to us in the problem is 600 . Since this is the total volume of the mixed drink, we must have: Or without units: To build the second equation, we have to think about the alcohol concentrations for the pure scotch, pure vermouth, and the mixed Rob Roy. It can be tricky to think about percentages like these correctly. One strategy is to focus on the amount of alcohol (in ). If we have milliliters of scotch that is alcohol, then is the actual amount of alcohol (in ) in that scotch. Similarly, is the amount of alcohol in the vermouth. And the final cocktail is 600 of liquid that is alcohol. So it has milliliters of alcohol. This is telling us: Or without units: So our system is: To solve this system, we'll isolate in the first equation: And then substitute in for in the second equation with : Now we determine using the equation where we had isolated : In summary, LaVonda needs to combine 400 of scotch with 200 of vermouth to create 600 of Rob Roy. Let's take a moment to consider estimation to ask whether the solution to is reasonable. LaVonda will mix scotch ( concentration) with vermouth ( concentration) and wants to end with a concentration. Is closer to or ? It's closer to so we should expect there to be more scotch than vermouth. This agrees with the solution we found. Mixing a Coffee Blend Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for per pound. They have some coffee beans from Columbia that sell for per pound and some coffee beans from Honduras that sell for per pound. How many pounds of each should they mix to make pounds of the blend? Write two equations that form a system for this scenario. We have pounds of Columbian coffee and pounds of Honduran coffee. Since there must be pounds total, one equation is The total cost of the Colombian coffee in the blend will be and the total cost of the Honduran coffee in the blend will be All together, the blend has a total cost of dollars. So another equation is Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do subtract from each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use. In summary, Desi needs to mix pounds of the Honduran coffee beans with pounds of the Columbian coffee beans to create this blend. Solving Special Systems of Equations with Substitution Remember the two special cases for a system of two linear equtions? We studied them in . If the two lines have the same slope, then they might be distinct lines that never meet, and then the system has no solutions. Or they might coincide as the same line, in which case there are infinitely many solutions represented by all the points on that line. In these cases, when we try to use substitution, interesting things happen. A System with No Solution Systems of linear equations with no solution Solve the system of equations using the substitution method: Since the first equation has already isolated , we will substitute in for in the second equation, and we have: Even though we were only intending to replace in the second equation, it turned out that all instances of disappear too. There are no variables at all in what remains. This will happen whenever the two lines from a system have the same slope. In this case, since is false no matter what values and might be, there can be no solution to the system. If we graphed the two lines from this system, we would see them as parallel and distinct . We can say that there are no solutions, or that the solution set is empty. The solution set is , or . For verification, let's re-write the second equation in slope-intercept form: So the system is equivalent to: Now it is easier to see that the two lines have the same slope but different -intercepts. The lines are parallel and distinct confirming that there should be no solution to the system. A System with Infinitely Many Solutions Systems of linear equations with infinitely many solutions Solve the system of equations using the substitution method: Since is already isolated in the first line, we will use that and substitute in for in the second equation: Once again, after the substitution we find ourselves with an equation where there are no variables. This time, it is an outright true equation; really does equal . What does this means for the original system of equations though? Apparently as long as equals , then both of the original equations are true. It's obvious that makes the first equation true, but our algebra above shows that also makes the second equation true. So there are infinitely many solutions. If we want to, we can write the solution set using set-builder notation: . Ultimately the two lines from the system were actually the same line. For verification, let's re-write the second equation in slope-intercept form: So the system is equivalent to: Now it is clear that the two equations represent the same line. So every point on that line is a solution, and there are infinitely many solutions. Give an example of a system of two equations in and where it would be nicer to solve the system using substitution than by graphing the two lines that the equations define. Explain why substitution would be nicer than graphing for your example system. What can be helpful if you have a system of two linear equations in two variables where there are fractions appearing in the equations? In an application problem, thinking about the can help you understand how to set up equations. Review Skills Practice Solve the system of equations using substitution. PTX:ERROR: WeBWorK problem webwork-1459 with seed 1459 is either empty or failed to compile Use -a to halt with full PG and returned content PTX:ERROR: WeBWorK problem webwork-substitution-fraction-coefficients-standard-slope-intercept with seed 1476 is either empty or failed to compile Use -a to halt with full PG and returned content Applications A rectangle s length is shorter than two times its width. The rectangle s perimeter is Find the rectangle s length and width. A school fund raising event sold a total of tickets and generated a total revenue of Each adult ticket cost and each child ticket cost How many adult tickets and how many child tickets were sold? One telecom company charges a monthly fee of and for each Gigabyte (GB) of data transmitted. A rival telecom company charges a monthly fee of and for each GB of data transmitted. How many GB of data would you have to use for the monthly bill to be the same for the two providers? And what would that monthly bill be? A local restaurant has two locations. At one location, the revenue this month is but it has been decreasing by per month. At the other location, the annual revenue this month is and it has been increasing by per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be? An algebra exam has questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth points, and short-answer questions, each worth points. How many questions are there of each type? Lizette invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Lizette earned a total of in interest. How much money did she invest in each account? Molly invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Molly s total fell from to How much money did she invest in each account? Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with married. Balsamburg had a population with married. After the merge, the new city has a total of residents, with married. How many residents did each town have before the merge? Xzavier poured some alcohol solution and some alcohol solution together into a beaker, and then the beaker had of alcohol solution. How much of each solution did Xzavier pour into the beaker? A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost per pound, and cashews cost per pound. The company is targeting a product that will cost them per pound worth of ingredients. How much of each type of nut should go into a bag? " + "body": " Substitution PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In , we used graphing to solve a system of two linear equations. While graphing is insightful, it can be time-consuming. And what happens if the actual solution to the system does not have whole number coordinates will you be able to see exactly what are the coordinates of the crossing? And what if your hand-drawn graph has some small imperfection that makes it look like the lines are crossing at the wrong place? There are two main alternatives to graphing that can find the solution to a system of two linear equations. They use pencil-and-paper algebra instead of a picture. This section covers one of those alternative methods, and the next section covers the other. Alternative Video Lesson Substitution The Interview Systems of linear equations solved by substitution Substitution method Systems of linear equations applications Once upon a time, the New York Times published an article about the movie, The Interview . It included the following quote: The Interview generated roughly million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday. Sony did not say how much of that total represented digital rentals versus sales. The studio said there were about two million transactions overall. Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.) First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let be the number of rental transactions and let be the number of sales transactions. If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both and are measured in transactions . The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is: Without the units: The price of each rental was . That means the problem has given us a rate of to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by , and that would give us the total number of dollars generated from rentals. This is . Similarly, the price of each sale was , so the revenue from sales was . The total revenue was million, which we can represent with this equation: Without the units: Here is our system of equations, with the commas removed: To solve the system, we will use a method called substitution . The idea is to use one equation to isolate . Then substitute this for the that's in the other equation. This leaves you with one equation where the only variable is . And we can handle that directly. The first equation from the system is an easy one to isolate : This tells us that the expression is equal to , so we can substitute that in place of in the second equation: At this point, we know that . This tells us that out of the million transactions, roughly were from online sales. Recall that we isolated previously and found . In summary, there were roughly ssales and roughly rentals. In , we rounded the solution values because only whole numbers make sense in the context of the problem. It was especially acceptable to round because the original numbers we worked with were already rounded. In fact, it would be OK to round even more to something like and as long as we communicate clearly that we rounded. In other exercises where there is no context, it is not OK to round. Solutions should be communicated with their exact values. Solve the system of equations using substitution: To use substitution, we need to isolate one of the variables in one of our equations. Looking over both equations, it will be easiest to isolate in the first equation: Next, we substitute in for in the second equation, giving us a linear equation in only one variable . And this is an equation that we may solve using skills from . Now that we have the value for , we need to find the value for . We already isolated , and it's easiest to just use that equation. At this point we think the solution is the point or in other words: , . It's only human to make mistakes though, so we should check that the solution actually works. To check it, try using , in both of the original equations. We conclude then that this system of equations is true when and . The solution is the point and we write the solution set as . Use substitution to solve the system. Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do is add to each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve for the one variable that remains. Solve for the other variable. Sometimes it makes more sense to start the process using the second equation instead of the first. And sometimes it makes more sense to isolate the second variable instead of the first. Solve this system of equations using substitution: We need to isolate one of the variables in one of the equations. Looking over both equations, it will be easiest to isolate in the second equation because that is where the coefficient is smallest. We are going to have to divide by some coefficient; it might as well be the smallest one to keep the fraction arithmetic as simple as we can. Note that there are fractions once we've isolated . We should take care with the steps that follow to make sure that the fraction arithmetic is correct. Substitute in for in the first equation, and that leads to having a single linear equation with only one variable. We can solve this as we did in . Note the step in the middle where we clear denominators. Now that we have the value for , we need to find the value for . We already isolated , and it's easiest to just use that equation. To check the solution we think we've found, try using , in both of the original equations. We conclude then that this system of equations is true when and . The solution is the point and we write the solution set as . A system may start out with fractions among the coefficients. Just as we learned in , the algebra can go more smoothly if we clear the denominators before doing more work. Solve the system of equations using the substitution method. When a system of equations has fraction coefficients, it may be helpful to clear the denomintors . With each equation, multiply each side by the least common multiple of that equation's denominators. In the first equation, the least common multiple of the denominators is , so: In the second equation, the least common multiple of the denominators is , so: Now we have a system that is equivalent to the original system of equations, but there are no fraction coefficients: The second equation has already isolateed , so we will substitute in for in the first equation. And we have solved for . To find , we know , so we have: The solution is . (This should be checked in the original two equations though.) Real world applications can lead to a system where both equations are either in point-slope form or slope-intercept form. This is helpful, since it means one variable has already been isolated. Solve the system of equations using the substitution method. In the first equation, is already isolated. So we can substitute in for in the second equation: . Some people prefer to think of this as setting the two right sides equal to each other. That works too: if is equal to , and is also equal to , then these expressions are equal to each other. Either way: And then: And the solution is . For summary reference, here is the general procedure. Solving Systems of Equations by Substitution Systems of linear equations solved by substitution Substitution method To solve a system of two linear equations in two variables by substitution, If helpful, clear denominators from each equation using the least common multiple of each equation's denominators. Isolate one variable in one of the equations. Substitute that expression into the other equation. There should now only be one variable in that equation. Solve that equation for the one remaining variable. Substitute that value into the earlier result where the first variable was isolated, and solve for that variable. Verify your solution using the original two equations. Applications Systems of linear equations applications Two Different Interest Rates Notah made some large purchases with his two credit cards one month and took on a total of in debt from the two cards. He didn't make any payments the first month, so the two credit card debts each started accruing interest. That month, his Visa card charged interest and his Mastercard charged interest. Because of this, Notah's total debt grew by . How much money did Notah charge to each card? We start by clearly defining two variables for the two unknowns. Let be the amount charged to the Visa card (in dollars) and let be the amount charged to the Mastercard (in dollars). To set up the equations, notice that we are given two different total dollar amounts. One is the total debt Notah initially took on, . So we have: Or without units: The other total we were given is the total amount of interest, . The Visa had dollars charged to it and accrues interest. So is the dollar amount of interest that comes from using the Visa card. Similarly, is the dollar amount of interest from using the Mastercard. Together: Or without units: As a system, we write: To solve this system by substitution, notice that it will be easier to isolate one of the variables in the first equation. We'll isolate : Now we substitute in for in the second equation: And then we can determine the value of by using the earlier equation where we isolated : In summary, Notah charged to the Visa and to the Mastercard. We should check that these numbers work as solutions to our original system and that they make sense in context. (For instance, if one of these numbers were negative, or was something small like , they wouldn't make sense as credit card debt.) The next two examples are called mixture problems , because they involve mixing two quantities together to form a combination and we want to find out how much of each quantity to mix. Mixing Solutions with Two Different Concentrations Mixture problems LaVonda is a meticulous bartender and she needs to serve milliliters of Rob Roy, an alcoholic cocktail that is alcohol by volume. The main ingredients are scotch that is alcohol and vermouth that is alcohol. How many milliliters of each ingredient should she mix together to make the concentration she needs? The two unknowns are the quantities of each ingredient. Let be the amount of scotch (in ) and let be the amount of vermouth (in ). One quantity given to us in the problem is 600 . Since this is the total volume of the mixed drink, we must have: Or without units: To build the second equation, we have to think about the alcohol concentrations for the pure scotch, pure vermouth, and the mixed Rob Roy. It can be tricky to think about percentages like these correctly. One strategy is to focus on the amount of alcohol (in ). If we have milliliters of scotch that is alcohol, then is the actual amount of alcohol (in ) in that scotch. Similarly, is the amount of alcohol in the vermouth. And the final cocktail is 600 of liquid that is alcohol. So it has milliliters of alcohol. This is telling us: Or without units: So our system is: To solve this system, we'll isolate in the first equation: And then substitute in for in the second equation with : Now we determine using the equation where we had isolated : In summary, LaVonda needs to combine 400 of scotch with 200 of vermouth to create 600 of Rob Roy. Let's take a moment to consider estimation to ask whether the solution to is reasonable. LaVonda will mix scotch ( concentration) with vermouth ( concentration) and wants to end with a concentration. Is closer to or ? It's closer to so we should expect there to be more scotch than vermouth. This agrees with the solution we found. Mixing a Coffee Blend Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for per pound. They have some coffee beans from Columbia that sell for per pound and some coffee beans from Honduras that sell for per pound. How many pounds of each should they mix to make pounds of the blend? Write two equations that form a system for this scenario. We have pounds of Columbian coffee and pounds of Honduran coffee. Since there must be pounds total, one equation is The total cost of the Colombian coffee in the blend will be and the total cost of the Honduran coffee in the blend will be All together, the blend has a total cost of dollars. So another equation is Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do subtract from each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use. In summary, Desi needs to mix pounds of the Honduran coffee beans with pounds of the Columbian coffee beans to create this blend. Solving Special Systems of Equations with Substitution Remember the two special cases for a system of two linear equtions? We studied them in . If the two lines have the same slope, then they might be distinct lines that never meet, and then the system has no solutions. Or they might coincide as the same line, in which case there are infinitely many solutions represented by all the points on that line. In these cases, when we try to use substitution, interesting things happen. A System with No Solution Systems of linear equations with no solution Solve the system of equations using the substitution method: Since the first equation has already isolated , we will substitute in for in the second equation, and we have: Even though we were only intending to replace in the second equation, it turned out that all instances of disappear too. There are no variables at all in what remains. This will happen whenever the two lines from a system have the same slope. In this case, since is false no matter what values and might be, there can be no solution to the system. If we graphed the two lines from this system, we would see them as parallel and distinct . We can say that there are no solutions, or that the solution set is empty. The solution set is , or . For verification, let's re-write the second equation in slope-intercept form: So the system is equivalent to: Now it is easier to see that the two lines have the same slope but different -intercepts. The lines are parallel and distinct confirming that there should be no solution to the system. A System with Infinitely Many Solutions Systems of linear equations with infinitely many solutions Solve the system of equations using the substitution method: Since is already isolated in the first line, we will use that and substitute in for in the second equation: Once again, after the substitution we find ourselves with an equation where there are no variables. This time, it is an outright true equation; really does equal . What does this means for the original system of equations though? Apparently as long as equals , then both of the original equations are true. It's obvious that makes the first equation true, but our algebra above shows that also makes the second equation true. So there are infinitely many solutions. If we want to, we can write the solution set using set-builder notation: . Ultimately the two lines from the system were actually the same line. For verification, let's re-write the second equation in slope-intercept form: So the system is equivalent to: Now it is clear that the two equations represent the same line. So every point on that line is a solution, and there are infinitely many solutions. Give an example of a system of two equations in and where it would be nicer to solve the system using substitution than by graphing the two lines that the equations define. Explain why substitution would be nicer than graphing for your example system. What can be helpful if you have a system of two linear equations in two variables where there are fractions appearing in the equations? In an application problem, thinking about the can help you understand how to set up equations. Skills Practice Solve the system of equations using substitution. Applications A rectangle s length is shorter than four times its width. The rectangle s perimeter is Find the rectangle s length and width. A school fund raising event sold a total of tickets and generated a total revenue of Each adult ticket cost and each child ticket cost How many adult tickets and how many child tickets were sold? One telecom company charges a monthly fee of and for each Gigabyte (GB) of data transmitted. A rival telecom company charges a monthly fee of and for each GB of data transmitted. How many GB of data would you have to use for the monthly bill to be the same for the two providers? And what would that monthly bill be? A local restaurant has two locations. At one location, the revenue this month is but it has been decreasing by per month. At the other location, the annual revenue this month is and it has been increasing by per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be? An algebra exam has questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth points, and short-answer questions, each worth points. How many questions are there of each type? Rocky invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Rocky earned a total of in interest. How much money did he invest in each account? Tyler invested a total of in two investments. Their savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Tyler s total fell from to How much money did they invest in each account? Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with Democrats. Balsamburg had a population with Democrats. After the merge, the new city has a total of residents, with Democrats. How many residents did each town have before the merge? Colin poured some alcohol solution and some alcohol solution together into a beaker, and then the beaker had of alcohol solution. How much of each solution did Colin pour into the beaker? A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost per pound, and cashews cost per pound. The company is targeting a product that will cost them per pound worth of ingredients. How much of each type of nut should go into a bag? " }, { "id": "section-substitution-2", @@ -17143,21 +17575,21 @@ var ptx_lunr_docs = [ "type": "Example", "number": "4.2.2", "title": "The Interview.", - "body": " Systems of linear equations applications The Interview Once upon a time, the New York Times published an article about the movie, The Interview . It included the following quote: The Interview generated roughly million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday. Sony did not say how much of that total represented digital rentals versus sales. The studio said there were about two million transactions overall. Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.) First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let be the number of rental transactions and let be the number of sales transactions. If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both and are measured in transactions . The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is: Without the units: The price of each rental was . That means the problem has given us a rate of to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by , and that would give us the total number of dollars generated from rentals. This is . Similarly, the price of each sale was , so the revenue from sales was . The total revenue was million, which we can represent with this equation: Without the units: Here is our system of equations, with the commas removed: To solve the system, we will use a method called substitution . The idea is to use one equation to isolate . Then substitute this for the that's in the other equation. This leaves you with one equation where the only variable is . And we can handle that directly. The first equation from the system is an easy one to isolate : This tells us that the expression is equal to , so we can substitute that in place of in the second equation: At this point, we know that . This tells us that out of the million transactions, roughly were from online sales. Recall that we isolated previously and found . In summary, there were roughly ssales and roughly rentals. " + "body": " The Interview Systems of linear equations solved by substitution Substitution method Systems of linear equations applications Once upon a time, the New York Times published an article about the movie, The Interview . It included the following quote: The Interview generated roughly million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday. Sony did not say how much of that total represented digital rentals versus sales. The studio said there were about two million transactions overall. Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.) First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let be the number of rental transactions and let be the number of sales transactions. If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both and are measured in transactions . The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is: Without the units: The price of each rental was . That means the problem has given us a rate of to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by , and that would give us the total number of dollars generated from rentals. This is . Similarly, the price of each sale was , so the revenue from sales was . The total revenue was million, which we can represent with this equation: Without the units: Here is our system of equations, with the commas removed: To solve the system, we will use a method called substitution . The idea is to use one equation to isolate . Then substitute this for the that's in the other equation. This leaves you with one equation where the only variable is . And we can handle that directly. The first equation from the system is an easy one to isolate : This tells us that the expression is equal to , so we can substitute that in place of in the second equation: At this point, we know that . This tells us that out of the million transactions, roughly were from online sales. Recall that we isolated previously and found . In summary, there were roughly ssales and roughly rentals. " }, { - "id": "section-substitution-4-5", + "id": "section-substitution-4-3", "level": "2", - "url": "section-substitution.html#section-substitution-4-5", + "url": "section-substitution.html#section-substitution-4-3", "type": "Remark", "number": "4.2.3", "title": "", "body": " In , we rounded the solution values because only whole numbers make sense in the context of the problem. It was especially acceptable to round because the original numbers we worked with were already rounded. In fact, it would be OK to round even more to something like and as long as we communicate clearly that we rounded. In other exercises where there is no context, it is not OK to round. Solutions should be communicated with their exact values. " }, { - "id": "section-substitution-4-6", + "id": "section-substitution-4-4", "level": "2", - "url": "section-substitution.html#section-substitution-4-6", + "url": "section-substitution.html#section-substitution-4-4", "type": "Example", "number": "4.2.4", "title": "", @@ -17173,27 +17605,27 @@ var ptx_lunr_docs = [ "body": " Use substitution to solve the system. Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do is add to each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve for the one variable that remains. Solve for the other variable. " }, { - "id": "section-substitution-4-9", + "id": "section-substitution-4-7", "level": "2", - "url": "section-substitution.html#section-substitution-4-9", + "url": "section-substitution.html#section-substitution-4-7", "type": "Example", "number": "4.2.6", "title": "", "body": " Solve this system of equations using substitution: We need to isolate one of the variables in one of the equations. Looking over both equations, it will be easiest to isolate in the second equation because that is where the coefficient is smallest. We are going to have to divide by some coefficient; it might as well be the smallest one to keep the fraction arithmetic as simple as we can. Note that there are fractions once we've isolated . We should take care with the steps that follow to make sure that the fraction arithmetic is correct. Substitute in for in the first equation, and that leads to having a single linear equation with only one variable. We can solve this as we did in . Note the step in the middle where we clear denominators. Now that we have the value for , we need to find the value for . We already isolated , and it's easiest to just use that equation. To check the solution we think we've found, try using , in both of the original equations. We conclude then that this system of equations is true when and . The solution is the point and we write the solution set as . " }, { - "id": "section-substitution-4-11", + "id": "section-substitution-4-9", "level": "2", - "url": "section-substitution.html#section-substitution-4-11", + "url": "section-substitution.html#section-substitution-4-9", "type": "Example", "number": "4.2.7", "title": "", "body": " Solve the system of equations using the substitution method. When a system of equations has fraction coefficients, it may be helpful to clear the denomintors . With each equation, multiply each side by the least common multiple of that equation's denominators. In the first equation, the least common multiple of the denominators is , so: In the second equation, the least common multiple of the denominators is , so: Now we have a system that is equivalent to the original system of equations, but there are no fraction coefficients: The second equation has already isolateed , so we will substitute in for in the first equation. And we have solved for . To find , we know , so we have: The solution is . (This should be checked in the original two equations though.) " }, { - "id": "section-substitution-4-13", + "id": "section-substitution-4-11", "level": "2", - "url": "section-substitution.html#section-substitution-4-13", + "url": "section-substitution.html#section-substitution-4-11", "type": "Example", "number": "4.2.8", "title": "", @@ -17566,7 +17998,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.30", "title": "", - "body": " PTX:ERROR: WeBWorK problem webwork-1459 with seed 1459 is either empty or failed to compile Use -a to halt with full PG and returned content " + "body": " " }, { "id": "substitution-fraction-solutions-integer-coefficients-point-slope-standard", @@ -17719,7 +18151,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.47", "title": "", - "body": " PTX:ERROR: WeBWorK problem webwork-substitution-fraction-coefficients-standard-slope-intercept with seed 1476 is either empty or failed to compile Use -a to halt with full PG and returned content " + "body": " " }, { "id": "substitution-fraction-coefficients-standard-slope-intercept-copy", @@ -17773,7 +18205,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.53", "title": "", - "body": " A rectangle s length is shorter than two times its width. The rectangle s perimeter is Find the rectangle s length and width. " + "body": " A rectangle s length is shorter than four times its width. The rectangle s perimeter is Find the rectangle s length and width. " }, { "id": "school-fundraiser", @@ -17818,7 +18250,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.58", "title": "", - "body": " Lizette invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Lizette earned a total of in interest. How much money did she invest in each account? " + "body": " Rocky invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Rocky earned a total of in interest. How much money did he invest in each account? " }, { "id": "mixed-investments-loss", @@ -17827,7 +18259,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.59", "title": "", - "body": " Molly invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Molly s total fell from to How much money did she invest in each account? " + "body": " Tyler invested a total of in two investments. Their savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Tyler s total fell from to How much money did they invest in each account? " }, { "id": "town-merger", @@ -17836,7 +18268,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.60", "title": "", - "body": " Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with married. Balsamburg had a population with married. After the merge, the new city has a total of residents, with married. How many residents did each town have before the merge? " + "body": " Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with Democrats. Balsamburg had a population with Democrats. After the merge, the new city has a total of residents, with Democrats. How many residents did each town have before the merge? " }, { "id": "alcohol-solution-mixture", @@ -17845,7 +18277,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.2.5.61", "title": "", - "body": " Xzavier poured some alcohol solution and some alcohol solution together into a beaker, and then the beaker had of alcohol solution. How much of each solution did Xzavier pour into the beaker? " + "body": " Colin poured some alcohol solution and some alcohol solution together into a beaker, and then the beaker had of alcohol solution. How much of each solution did Colin pour into the beaker? " }, { "id": "mixed-nuts", @@ -17863,7 +18295,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "4.3", "title": "Elimination", - "body": " Elimination PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In , we used graphing to solve a system of two linear equations. In , we learned the substitution technique. In this section, we learn another technique for solving a system of linear equations called elimination or the addition method . Alternative Video Lesson Solving Systems of Equations by Elimination Systems of linear equations solved by elimination Elimination method Addition method Elimination method Systems of linear equations applications Alicia has to send to her two grandchildren Bobbie and Cedric. Since Bobbie is entering college and has some textbook expenses, Alicia wants to send Bobbie more than she sends Cedric. How much money should she give to each grandchild? You may have a good way to answer this quickly, but we will use this example to demonstrate a new technique for solving a system of two linear equations. Let be the dollar amount that she sends to Bobbie, and be the dollar amount that she sends to Cedric. (Note that we begin the process of solving a word problem by clearly defining the variables, including their units.) Since the total she has to give is , we can write . And since she wants to give more to Bobbie, we can write . So we have the system of equations: We could solve this system using graphing or substitution, but here is a new method. If we add together the left sides from the two equations, it should equal the sum of the right sides : Note that the variable is eliminated. This happened because the and the perfectly cancel each other out when they are added. With only one variable left, it doesn't take much effort to finish solving for : . To finish solving this system of equations, we still need the value of . One easy way to find is to substitute our value for into one of the original equations: To check our work, substitute and into the original equations: This confirms that our solution is correct. In summary, Alicia should give to Bobbie and to Cedric. This method for solving a system of equations worked because and add to zero. Once the -terms were eliminated we were able to solve for . This method is called the elimination method or the addition method , because we add the corresponding sides from the two equations and that eliminates a variable. Most of the time, just adding the sides together does not eliminate one of the variables. We can still use this method, but it needs a little set up work first. Let's look at an example where we need to adjust one of the equations. Scaling One Equation Solve the system of equations using the elimination method. We want to see whether it will be easier to eliminate or . We see that the coefficients of in each equation are and , and the coefficients of are and . Because is a multiple of and the coefficients already have opposite signs, the variable will be easier to eliminate. To eliminate the terms, first multiply each side of the first equation by . Why? Because then the -terms will be and , so they can be added to eliminate . We now have an equivalent system of equations where the -terms can be eliminated: So we have . To solve for , one option is to substitute in for into either of the original equations. We use the first original equation, : We think the solution is and . We can check this using both of the original equations: So the solution to this system is and the solution set is . Use elimination to solve the system. What number could you multiply by each term in the second equation that would be helpful? If we multiply the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Here's an example where we have to scale both equations. Scaling Both Equations Solve the system of equations using the elimination method. Considering the coefficients of ( and ) and the coefficients of ( and ) we see that we cannot eliminate the or the variable by scaling a single equation. There's just no place where one coefficient divides the corresponding coefficient. But we can scale both equations. The -terms already have opposite signs, so we choose to eliminate . The least common multiple of and is , so we can target turning the terms into and . We can scale the first equation by and the second equation by so that the equations have terms and , which will cancel when added. To solve for , we'll replace with in : We'll check the system using and in each of the original equations: So the system's solution is and the solution set is . Try a similar exercise. Use elimination to solve the system. What numbers could you multiply each term in the first and second equations by that would be helpful? If we multiply the first equation through by and the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the first by and the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Systems of linear equations applications Meal Planning Javed is on a meal plan where his breakfast needs to have calories and grams of fat. A small avocado contains calories and grams of fat. He has bagels with calories and grams of fat. Write and solve a system of equations to determine how much bagel and how much avocado Javed could eat to meet his target calories and fat. To write this system of equations, we first need to define our variables. Let be the number of avocados consumed and let be the number of bagels consumed. (In the end, both and might be fractions.) For our first equation, we can count calories contributed from eating avocados and bagels: Or, without the units: . For a second equation, we can count the grams of fat: Or, without the units: . So the system of equations is: Since none of the coefficients are equal to , it may be easier to use the elimination method to solve this system than it would be to use substitution. Looking at the terms and , we can eliminate the variable if we multiply the second equation by to get the term : When we add the corresponding sides from the two equations together we have: Now we know that Javed should eat of a bagel (or one and one-quarter bagels). To determine how much avocados he should eat, we can substitute in for in either of our original equations. We think the solution is , . To check this, see if these numbers work as solutions in the original equations: In summary, Javed can eat of a bagel (so one and one-quarter bagel) and of an avocado in order to consume exactly calories and grams of fat. For summary reference, here is the general procedure. Solving Systems of Equations by Elimination Systems of linear equations solved by elimination To solve a system of equations by elimination, Algebraically manipulate both equations into standard form if they are not already that way. Scale one or both of the equations to force one of the variables to have equal but opposite coefficients in the two equations. Add the corresponding sides of the two equations together, which should have the effect that one variable is eliminated entirely. Solve the resulting equation for the one remaining variable. Substitute that value into either of the original equations to find the other variable. Verify your solution using the original two equations. Solving Special Systems of Equations with Elimination Remember the two special cases for a system of two linear equations? In , we learned how such systems have lines that are parallel when graphed. If you are using the substitution method, we learned in that after you make the substitution and simplify, you get an equation that is either outright true or outright false. With these systems, what happens when you use the elimination method? Let's see. A System with Infinitely Many Solutions Systems of linear equations with infinitely many solutions Solve the system of equations using the elimination method. To eliminate the -terms, we multiply each term in the first equation by , and we have: You might notice that the equations look very similar. Adding the respective sides of the equation, we have: Not just one, but both of the variables have been eliminated. This is not giving us any restrictions on what - and -values would solve this system. The two equations were essentially the same, after scaling them. So they represent the same line, and the solution set contains all pairs that lie on that line. We can write the solution set as . A System with No Solution Systems of linear equations with no solution Solve the system of equations using the elimination method. To eliminate the -terms, we will scale the first equation by and the second by : Adding the respective sides of the equation, we have: Again, not just one, but both of the variables have been eliminated. In this case, the statement is outright false no matter what and are. So the system has no solution. Substitution versus Elimination In every example so far from this section, both equations were in standard form, . And all of the coefficients were integers. If none of the coefficients are equal to then it is usually easier to use the elimination method, because otherwise you will probably have some fraction arithmetic to do in the middle of the substitution method. If there is a coefficient of , then it's a matter of preference. When one of the coefficients is , it's not so bad to isolate that variable and make a substitution. Systems of linear equations applications A college used to have a north campus with students and a south campus with students. The percentage of students at the north campus who self-identify as LGBTQ was three times the percentage at the south campus. After the merge, of students identify as LGBTQ. What percentage of students on each campus identified as LGBTQ before the merge? We will define as the percentage (as a decimal) of students at the north campus and as the percentage (as a decimal) of students at the south campus that identified as LGBTQ. Since the percentage of students at the north campus was three times the percentage at the south campus, we have one equation: For a second equation, we will count LGBTQ students at the various campuses. At the north campus, multiply the population, , by the percentage to get . This represents be the actual count of LGBTQ students at that campus. Similarly, the south campus had LGBTQ students, and the combined school has . When we combine the two campuses, we have: So we have a system of two equations: Because the first equation already has isolated, this is a good time to not use the elimination method. We can directly substitute in for in our second equation and solve for : And then we can determine using the first equation: So before the merge, of the north campus students self-identified as LGBTQ and of the south campus students self-identified as LGBTQ. If you need to solve a system, and one of the equations is not in standard form, substitution may be easier. But you also may find it easier to convert the equations into standard form and use elimination. Additionally, if the system's coefficients are fractions or decimals, it can help to take one step where you clear denominators and\/or decimal points and then decide which technique you would like to use. Solve the system of equations using the method of your choice. First, we can clear the denominators by using the least common multiple of the denominators in each equation, similarly to in . We have: We could convert the first equation into standard form by subtracting from both sides, and then use elimination. Alternatively, since the -variable in the first equation has coefficient , you might prefer the substitution method. Isolating gives us , and we could substitute that in to the second equation. It's really up to you. As an exercise, try it both ways. Hopefully you get the same solution either way, and you can reflect on which method feels more comfortable to you. This next example is an application exercise with decimal coefficients, and it demonstrates how you might clear decimals, similarly to clearing denominators. Systems of linear equations applications A penny is made from copper and zinc. A chemistry reference says copper has a density of 9 and zinc has a density of 7.1 . A penny's mass is 2.5 and its volume is 0.35 . How many each of copper and zinc go into one penny? Let be the volume of copper and be the volume of zinc in one penny, both measured in . Since the total volume is 0.35 , one equation is: Or without units: . For a second equation, we will examine the masses of copper and zinc. Since copper has a density of 9 and we are using to represent the volume of copper, the mass of copper in one penny is . Similarly, the mass of zinc is . Since the total mass is 2.5 , we have the equation: Or without units: . So we have a system of equations: We can eliminate the decimals by multiplying by the appropriate power of . If each term in the first equation is multiplied by , and each term in the second equation is multiplied by , there are no more decimal points. Now to set up elimination, scale each equation again to eliminate : Adding the corresponding sides from the two equations gives , from which we find . So there is about 0.342 of zinc in a penny. To solve for , we can use one of the original equations: Therefore there is about 0.342 of zinc and 0.008 of copper in a penny. If a variable in a system is already isolated in one of the equations, or has a coefficient of , consider using the substitution method. If both equations are in standard form or none of the coefficients are equal to , we suggest using the elimination method. Either way, if you have fraction or decimal coefficients, it may help to scale your equations so that only integer coefficients remain. What is another name for the elimination method ? To use the elimination method, usually the first step is to at least one equation. Describe a good situation to use the substitution method instead of the elimination method for solving a system of two linear equations in two variables. Review and Warmup Skills Practice Solve the system of equations using elimination. Applications An algebra exam has questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth points, and short-answer questions, each worth points. How many questions are there of each type? A school fund raising event sold a total of tickets and generated a total revenue of Each adult ticket cost and each child ticket cost How many adult tickets and how many child tickets were sold? Colin invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Colin earned a total of in interest. How much money did he invest in each account? Elias invested a total of in two investments. His savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Elias s total fell from to How much money did he invest in each account? Challenge Find the value of so that the system of equations has an infinite number of solutions. To have an infinite numbers of solutions, the two equations must represeent the same line coinciding. So the second equation comes from multiplying or dividing all the terms in the first equation by a certain factor. In this case, if you divide the first equation s terms by -5, the result is the second equation. So must equal " + "body": " Elimination PCC Course Content and Outcome Guide MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG MTH 60 CCOG In , we used graphing to solve a system of two linear equations. In , we learned the substitution technique. In this section, we learn another technique for solving a system of linear equations called elimination or the addition method . Alternative Video Lesson Solving Systems of Equations by Elimination Systems of linear equations solved by elimination Elimination method Addition method Elimination method Systems of linear equations applications Alicia has to send to her two grandchildren Bobbie and Cedric. Since Bobbie is entering college and has some textbook expenses, Alicia wants to send Bobbie more than she sends Cedric. How much money should she give to each grandchild? You may have a good way to answer this quickly, but we will use this example to demonstrate a new technique for solving a system of two linear equations. Let be the dollar amount that she sends to Bobbie, and be the dollar amount that she sends to Cedric. (Note that we begin the process of solving a word problem by clearly defining the variables, including their units.) Since the total she has to give is , we can write . And since she wants to give more to Bobbie, we can write . So we have the system of equations: We could solve this system using graphing or substitution, but here is a new method. If we add together the left sides from the two equations, it should equal the sum of the right sides : Note that the variable is eliminated. This happened because the and the perfectly cancel each other out when they are added. With only one variable left, it doesn't take much effort to finish solving for : . To finish solving this system of equations, we still need the value of . One easy way to find is to substitute our value for into one of the original equations: To check our work, substitute and into the original equations: This confirms that our solution is correct. In summary, Alicia should give to Bobbie and to Cedric. This method for solving a system of equations worked because and add to zero. Once the -terms were eliminated we were able to solve for . This method is called the elimination method or the addition method , because we add the corresponding sides from the two equations and that eliminates a variable. Most of the time, just adding the sides together does not eliminate one of the variables. We can still use this method, but it needs a little set up work first. Let's look at an example where we need to adjust one of the equations. Scaling One Equation Solve the system of equations using the elimination method. We want to see whether it will be easier to eliminate or . We see that the coefficients of in each equation are and , and the coefficients of are and . Because is a multiple of and the coefficients already have opposite signs, the variable will be easier to eliminate. To eliminate the terms, first multiply each side of the first equation by . Why? Because then the -terms will be and , so they can be added to eliminate . We now have an equivalent system of equations where the -terms can be eliminated: So we have . To solve for , one option is to substitute in for into either of the original equations. We use the first original equation, : We think the solution is and . We can check this using both of the original equations: So the solution to this system is and the solution set is . Use elimination to solve the system. What number could you multiply by each term in the second equation that would be helpful? If we multiply the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Here's an example where we have to scale both equations. Scaling Both Equations Solve the system of equations using the elimination method. Considering the coefficients of ( and ) and the coefficients of ( and ) we see that we cannot eliminate the or the variable by scaling a single equation. There's just no place where one coefficient divides the corresponding coefficient. But we can scale both equations. The -terms already have opposite signs, so we choose to eliminate . The least common multiple of and is , so we can target turning the terms into and . We can scale the first equation by and the second equation by so that the equations have terms and , which will cancel when added. To solve for , we'll replace with in : We'll check the system using and in each of the original equations: So the system's solution is and the solution set is . Try a similar exercise. Use elimination to solve the system. What numbers could you multiply each term in the first and second equations by that would be helpful? If we multiply the first equation through by and the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the first by and the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Meal Planning Systems of linear equations applications Javed is on a meal plan where his breakfast needs to have calories and grams of fat. A small avocado contains calories and grams of fat. He has bagels with calories and grams of fat. Write and solve a system of equations to determine how much bagel and how much avocado Javed could eat to meet his target calories and fat. To write this system of equations, we first need to define our variables. Let be the number of avocados consumed and let be the number of bagels consumed. (In the end, both and might be fractions.) For our first equation, we can count calories contributed from eating avocados and bagels: Or, without the units: . For a second equation, we can count the grams of fat: Or, without the units: . So the system of equations is: Since none of the coefficients are equal to , it may be easier to use the elimination method to solve this system than it would be to use substitution. Looking at the terms and , we can eliminate the variable if we multiply the second equation by to get the term : When we add the corresponding sides from the two equations together we have: Now we know that Javed should eat of a bagel (or one and one-quarter bagels). To determine how much avocados he should eat, we can substitute in for in either of our original equations. We think the solution is , . To check this, see if these numbers work as solutions in the original equations: In summary, Javed can eat of a bagel (so one and one-quarter bagel) and of an avocado in order to consume exactly calories and grams of fat. For summary reference, here is the general procedure. Solving Systems of Equations by Elimination Systems of linear equations solved by elimination To solve a system of equations by elimination, Algebraically manipulate both equations into standard form if they are not already that way. Scale one or both of the equations to force one of the variables to have equal but opposite coefficients in the two equations. Add the corresponding sides of the two equations together, which should have the effect that one variable is eliminated entirely. Solve the resulting equation for the one remaining variable. Substitute that value into either of the original equations to find the other variable. Verify your solution using the original two equations. Solving Special Systems of Equations with Elimination Remember the two special cases for a system of two linear equations? In , we learned how such systems have lines that are parallel when graphed. If you are using the substitution method, we learned in that after you make the substitution and simplify, you get an equation that is either outright true or outright false. With these systems, what happens when you use the elimination method? Let's see. A System with Infinitely Many Solutions Systems of linear equations with infinitely many solutions Solve the system of equations using the elimination method. To eliminate the -terms, we multiply each term in the first equation by , and we have: You might notice that the equations look very similar. Adding the respective sides of the equation, we have: Not just one, but both of the variables have been eliminated. This is not giving us any restrictions on what - and -values would solve this system. The two equations were essentially the same, after scaling them. So they represent the same line, and the solution set contains all pairs that lie on that line. We can write the solution set as . A System with No Solution Systems of linear equations with no solution Solve the system of equations using the elimination method. To eliminate the -terms, we will scale the first equation by and the second by : Adding the respective sides of the equation, we have: Again, not just one, but both of the variables have been eliminated. In this case, the statement is outright false no matter what and are. So the system has no solution. Substitution versus Elimination In every example so far from this section, both equations were in standard form, . And all of the coefficients were integers. If none of the coefficients are equal to then it is usually easier to use the elimination method, because otherwise you will probably have some fraction arithmetic to do in the middle of the substitution method. If there is a coefficient of , then it's a matter of preference. When one of the coefficients is , it's not so bad to isolate that variable and make a substitution. Systems of linear equations applications A college used to have a north campus with students and a south campus with students. The percentage of students at the north campus who self-identify as LGBTQ was three times the percentage at the south campus. After the merge, of students identify as LGBTQ. What percentage of students on each campus identified as LGBTQ before the merge? We will define as the percentage (as a decimal) of students at the north campus and as the percentage (as a decimal) of students at the south campus that identified as LGBTQ. Since the percentage of students at the north campus was three times the percentage at the south campus, we have one equation: For a second equation, we will count LGBTQ students at the various campuses. At the north campus, multiply the population, , by the percentage to get . This represents be the actual count of LGBTQ students at that campus. Similarly, the south campus had LGBTQ students, and the combined school has . When we combine the two campuses, we have: So we have a system of two equations: Because the first equation already has isolated, this is a good time to not use the elimination method. We can directly substitute in for in our second equation and solve for : And then we can determine using the first equation: So before the merge, of the north campus students self-identified as LGBTQ and of the south campus students self-identified as LGBTQ. If you need to solve a system, and one of the equations is not in standard form, substitution may be easier. But you also may find it easier to convert the equations into standard form and use elimination. Additionally, if the system's coefficients are fractions or decimals, it can help to take one step where you clear denominators and\/or decimal points and then decide which technique you would like to use. Solve the system of equations using the method of your choice. First, we can clear the denominators by using the least common multiple of the denominators in each equation, similarly to in . We have: We could convert the first equation into standard form by subtracting from both sides, and then use elimination. Alternatively, since the -variable in the first equation has coefficient , you might prefer the substitution method. Isolating gives us , and we could substitute that in to the second equation. It's really up to you. As an exercise, try it both ways. Hopefully you get the same solution either way, and you can reflect on which method feels more comfortable to you. This next example is an application exercise with decimal coefficients, and it demonstrates how you might clear decimals, similarly to clearing denominators. Systems of linear equations applications A penny is made from copper and zinc. A chemistry reference says copper has a density of 9 and zinc has a density of 7.1 . A penny's mass is 2.5 and its volume is 0.35 . How many each of copper and zinc go into one penny? Let be the volume of copper and be the volume of zinc in one penny, both measured in . Since the total volume is 0.35 , one equation is: Or without units: . For a second equation, we will examine the masses of copper and zinc. Since copper has a density of 9 and we are using to represent the volume of copper, the mass of copper in one penny is . Similarly, the mass of zinc is . Since the total mass is 2.5 , we have the equation: Or without units: . So we have a system of equations: We can eliminate the decimals by multiplying by the appropriate power of . If each term in the first equation is multiplied by , and each term in the second equation is multiplied by , there are no more decimal points. Now to set up elimination, scale each equation again to eliminate : Adding the corresponding sides from the two equations gives , from which we find . So there is about 0.342 of zinc in a penny. To solve for , we can use one of the original equations: Therefore there is about 0.342 of zinc and 0.008 of copper in a penny. If a variable in a system is already isolated in one of the equations, or has a coefficient of , consider using the substitution method. If both equations are in standard form or none of the coefficients are equal to , we suggest using the elimination method. Either way, if you have fraction or decimal coefficients, it may help to scale your equations so that only integer coefficients remain. What is another name for the elimination method ? To use the elimination method, usually the first step is to at least one equation. Describe a good situation to use the substitution method instead of the elimination method for solving a system of two linear equations in two variables. Skills Practice Solve the system of equations using elimination. Applications An algebra exam has questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth points, and short-answer questions, each worth points. How many questions are there of each type? A school fund raising event sold a total of tickets and generated a total revenue of Each adult ticket cost and each child ticket cost How many adult tickets and how many child tickets were sold? Garrison invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Garrison earned a total of in interest. How much money did he invest in each account? Jenny invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Jenny s total fell from to How much money did she invest in each account? Challenge Find the value of so that the system of equations has an infinite number of solutions. To have an infinite numbers of solutions, the two equations must represeent the same line coinciding. So the second equation comes from multiplying or dividing all the terms in the first equation by a certain factor. In this case, if you divide the first equation s terms by -4, the result is the second equation. So must equal " }, { "id": "section-elimination-2", @@ -17944,7 +18376,7 @@ var ptx_lunr_docs = [ "type": "Example", "number": "4.3.7", "title": "Meal Planning.", - "body": " Systems of linear equations applications Meal Planning Javed is on a meal plan where his breakfast needs to have calories and grams of fat. A small avocado contains calories and grams of fat. He has bagels with calories and grams of fat. Write and solve a system of equations to determine how much bagel and how much avocado Javed could eat to meet his target calories and fat. To write this system of equations, we first need to define our variables. Let be the number of avocados consumed and let be the number of bagels consumed. (In the end, both and might be fractions.) For our first equation, we can count calories contributed from eating avocados and bagels: Or, without the units: . For a second equation, we can count the grams of fat: Or, without the units: . So the system of equations is: Since none of the coefficients are equal to , it may be easier to use the elimination method to solve this system than it would be to use substitution. Looking at the terms and , we can eliminate the variable if we multiply the second equation by to get the term : When we add the corresponding sides from the two equations together we have: Now we know that Javed should eat of a bagel (or one and one-quarter bagels). To determine how much avocados he should eat, we can substitute in for in either of our original equations. We think the solution is , . To check this, see if these numbers work as solutions in the original equations: In summary, Javed can eat of a bagel (so one and one-quarter bagel) and of an avocado in order to consume exactly calories and grams of fat. " + "body": " Meal Planning Systems of linear equations applications Javed is on a meal plan where his breakfast needs to have calories and grams of fat. A small avocado contains calories and grams of fat. He has bagels with calories and grams of fat. Write and solve a system of equations to determine how much bagel and how much avocado Javed could eat to meet his target calories and fat. To write this system of equations, we first need to define our variables. Let be the number of avocados consumed and let be the number of bagels consumed. (In the end, both and might be fractions.) For our first equation, we can count calories contributed from eating avocados and bagels: Or, without the units: . For a second equation, we can count the grams of fat: Or, without the units: . So the system of equations is: Since none of the coefficients are equal to , it may be easier to use the elimination method to solve this system than it would be to use substitution. Looking at the terms and , we can eliminate the variable if we multiply the second equation by to get the term : When we add the corresponding sides from the two equations together we have: Now we know that Javed should eat of a bagel (or one and one-quarter bagels). To determine how much avocados he should eat, we can substitute in for in either of our original equations. We think the solution is , . To check this, see if these numbers work as solutions in the original equations: In summary, Javed can eat of a bagel (so one and one-quarter bagel) and of an avocado in order to consume exactly calories and grams of fat. " }, { "id": "algorithm-solving-systems-of-equations-by-elimination", @@ -18340,7 +18772,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.3.5.35", "title": "", - "body": " Colin invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Colin earned a total of in interest. How much money did he invest in each account? " + "body": " Garrison invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Garrison earned a total of in interest. How much money did he invest in each account? " }, { "id": "mixed-investments-loss-elimination", @@ -18349,7 +18781,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.3.5.36", "title": "", - "body": " Elias invested a total of in two investments. His savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Elias s total fell from to How much money did he invest in each account? " + "body": " Jenny invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Jenny s total fell from to How much money did she invest in each account? " }, { "id": "set-up-system-with-infinitely-many-solutions", @@ -18358,7 +18790,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.3.5.37", "title": "", - "body": " Find the value of so that the system of equations has an infinite number of solutions. To have an infinite numbers of solutions, the two equations must represeent the same line coinciding. So the second equation comes from multiplying or dividing all the terms in the first equation by a certain factor. In this case, if you divide the first equation s terms by -5, the result is the second equation. So must equal " + "body": " Find the value of so that the system of equations has an infinite number of solutions. To have an infinite numbers of solutions, the two equations must represeent the same line coinciding. So the second equation comes from multiplying or dividing all the terms in the first equation by a certain factor. In this case, if you divide the first equation s terms by -4, the result is the second equation. So must equal " }, { "id": "review-systems-of-linear-equations", @@ -18367,7 +18799,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "4.4", "title": "Systems of Two Linear Equations Chapter Review", - "body": " Systems of Two Linear Equations Chapter Review Solving a System by Graphing A system of two linear equations in two variables (or system for short) is a collection of two linear equations, each using the same two variables. A system is often presented like: but the two equations do not need to be in standard form like the two equations above. Systems of two linear equations in two variables arise in applications where, of course, there are two unknown quantities. But also there is some background information that lets you logically put together two ways that the variables must relate to each other. See and for some examples. A solution to a system is an ordered pair of numbers that can be substituted in to the two equations for the two variables, and make both equations true (not just one or the other). It turns out that a system of two linear equations in two variables will always have either exactly one solution, no solutions at all, or infinitely many solutions that all lie along one straight line. One method to find the solution(s) to a system is to graph the two lines. Most of the time, the lines cross sat one point, and this one point is the only solution to the system. If you graph precisely and accurately, then you can discern the coordinates of this crossing point and you know the solution to the system. Graphing might reveal that the two lines are parallel and never cross. Then you have an inconsistent system, and there is no solution. Or you might find that the two lines are actually the same line. This means you have a dependent system and there are infinitely many solutions (which are all the points along that common line). The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (1, 4) Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (2, 6) So the only solution is Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (9, 8) So the only solution is Substitution One alternative technique to solve a system of two linear equations in two variables is called substitution . With this technique, you choose one of the variables from one of the equations and isolate that variable. Then substitute that expression in for this variable in the other equation. Most of the time this leaves you with one equation in only one variable, so you can use skills from to solve for that one remaining variable. With one of the variables solved for, it's not much more work to solve for the other. If either of the equations can be simplified (for example by distributing, by combining like terms, or by scaling all terms to clear denominators) then you should do that as a first step. If there are like terms on opposite sides, you should use algebra to combine them together on one side. Then it is wise to identify which of the four instances of a variable (among the two equations) has the simplest coefficient, and choose that variable as the one to isolate. For example if any of the variables have or as their coefficient, then you can isolate that variable without using division, and without potentially bringing more fractions into the process. You may find that after you make a substitution and simplify, both of the variables disappear instead of just one. When this happens, you either had an inconsistent system (with no solution) or a dependent system (with infinitely many solutions). If the equation you now have is an outright false equation (like ) then you had an inconsistent system. If it is an outright true equation (like ) then you had a dependent system. Use substitution to solve the system. Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do is add to each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve for the one variable that remains. Solve for the other variable. Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for per pound. They have some coffee beans from Columbia that sell for per pound and some coffee beans from Honduras that sell for per pound. How many pounds of each should they mix to make pounds of the blend? Write two equations that form a system for this scenario. We have pounds of Columbian coffee and pounds of Honduran coffee. Since there must be pounds total, one equation is The total cost of the Colombian coffee in the blend will be and the total cost of the Honduran coffee in the blend will be All together, the blend has a total cost of dollars. So another equation is Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do subtract from each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use. In summary, Desi needs to mix pounds of the Honduran coffee beans with pounds of the Columbian coffee beans to create this blend. Solve the system of equations using substitution. Elimination Yet another option for solving a system of two equations in two variables is to use the elimination method, also known as the addition method. With this technique you cleverly scale one or both equations so that terms that go along with one of the variables have opposite coefficients. Then you can add the left sides together and the right sides together, and the resulting equation only has one variable. From there it is easy to solve for that one variable, and then use that solution to solve for the other variable. Just as with the substitution method, it is wise to begin by simplifying the equations if they can be simplified. And scaling the equations just to clear denominators or decimal points. Furthermore, since you will be adding terms from different equations, it is wise to covert each equation into standard form so that the corresponding terms are aligned. You may find that after you add to equations to eliminate a variable that both of the variables have been eliminated. When this happens, you either had an inconsistent system (with no solution) or a dependent system (with infinitely many solutions). If the equation you now have is an outright false equation (like ) then you had an inconsistent system. If it is an outright true equation (like ) then you had a dependent system. Use elimination to solve the system. What number could you multiply by each term in the second equation that would be helpful? If we multiply the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Use elimination to solve the system. What numbers could you multiply each term in the first and second equations by that would be helpful? If we multiply the first equation through by and the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the first by and the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Solve the system of equations using elimination. Review Exercises for : Check if the given point is a solution to the given system of two linear equations. Is a solution? The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (1, -2) Simply by looking closely at the linear equations, determine how many solutions the system will have. And state whether the system is dependeent, inconsistent, or neither. Solve a System Solve the given system of linear equations graphically. If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (2, -3) So there are no solutions. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, (-6)) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-7, 8) So the only solution is El enters a grassy field from some dense trees. Their dog is standing out in the field, away. As soon as they see each other, they start running toward each other. El runs with speed and their dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. El starts out at position running with speed So one equation is The dog starts running with speed but getting closer to El. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,110) with slope 3; the other line passes through (0,0) with slope -8; the lines cross at (10,30) Based on the graph, how long will it be until they meet? How far will that be from the place where El started? The lines cross at So after seconds, they will meet 30 feet away from where El started. : Solve the system of equations using substitution. A rectangle s length is shorter than three times its width. The rectangle s perimeter is Find the rectangle s length and width. A local restaurant has two locations. At one location, the revenue this month is but it has been decreasing by per month. At the other location, the annual revenue this month is and it has been increasing by per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be? Geoffrey invested a total of in two investments. His savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Geoffrey s total fell from to How much money did he invest in each account? A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost per pound, and cashews cost per pound. The company is targeting a product that will cost them per pound worth of ingredients. How much of each type of nut should go into a bag? : Solve the system of equations using elimination. Clayton invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Clayton earned a total of in interest. How much money did he invest in each account? " + "body": " Systems of Two Linear Equations Chapter Review Solving a System by Graphing A system of two linear equations in two variables (or system for short) is a collection of two linear equations, each using the same two variables. A system is often presented like: but the two equations do not need to be in standard form like the two equations above. Systems of two linear equations in two variables arise in applications where, of course, there are two unknown quantities. But also there is some background information that lets you logically put together two ways that the variables must relate to each other. See and for some examples. A solution to a system is an ordered pair of numbers that can be substituted in to the two equations for the two variables, and make both equations true (not just one or the other). It turns out that a system of two linear equations in two variables will always have either exactly one solution, no solutions at all, or infinitely many solutions that all lie along one straight line. One method to find the solution(s) to a system is to graph the two lines. Most of the time, the lines cross sat one point, and this one point is the only solution to the system. If you graph precisely and accurately, then you can discern the coordinates of this crossing point and you know the solution to the system. Graphing might reveal that the two lines are parallel and never cross. Then you have an inconsistent system, and there is no solution. Or you might find that the two lines are actually the same line. This means you have a dependent system and there are infinitely many solutions (which are all the points along that common line). The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (3, 1) Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (6, -4) So the only solution is Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, 8) So the only solution is Substitution One alternative technique to solve a system of two linear equations in two variables is called substitution . With this technique, you choose one of the variables from one of the equations and isolate that variable. Then substitute that expression in for this variable in the other equation. Most of the time this leaves you with one equation in only one variable, so you can use skills from to solve for that one remaining variable. With one of the variables solved for, it's not much more work to solve for the other. If either of the equations can be simplified (for example by distributing, by combining like terms, or by scaling all terms to clear denominators) then you should do that as a first step. If there are like terms on opposite sides, you should use algebra to combine them together on one side. Then it is wise to identify which of the four instances of a variable (among the two equations) has the simplest coefficient, and choose that variable as the one to isolate. For example if any of the variables have or as their coefficient, then you can isolate that variable without using division, and without potentially bringing more fractions into the process. You may find that after you make a substitution and simplify, both of the variables disappear instead of just one. When this happens, you either had an inconsistent system (with no solution) or a dependent system (with infinitely many solutions). If the equation you now have is an outright false equation (like ) then you had an inconsistent system. If it is an outright true equation (like ) then you had a dependent system. Use substitution to solve the system. Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do is add to each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve for the one variable that remains. Solve for the other variable. Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for per pound. They have some coffee beans from Columbia that sell for per pound and some coffee beans from Honduras that sell for per pound. How many pounds of each should they mix to make pounds of the blend? Write two equations that form a system for this scenario. We have pounds of Columbian coffee and pounds of Honduran coffee. Since there must be pounds total, one equation is The total cost of the Colombian coffee in the blend will be and the total cost of the Honduran coffee in the blend will be All together, the blend has a total cost of dollars. So another equation is Isolate one of the variables in one of the equations. We choose to isolate in the first equation. All we need to do subtract from each side, and then Substitute the isolated expression into the other equation. Making the substitution: Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use. In summary, Desi needs to mix pounds of the Honduran coffee beans with pounds of the Columbian coffee beans to create this blend. Solve the system of equations using substitution. Elimination Yet another option for solving a system of two equations in two variables is to use the elimination method, also known as the addition method. With this technique you cleverly scale one or both equations so that terms that go along with one of the variables have opposite coefficients. Then you can add the left sides together and the right sides together, and the resulting equation only has one variable. From there it is easy to solve for that one variable, and then use that solution to solve for the other variable. Just as with the substitution method, it is wise to begin by simplifying the equations if they can be simplified. And scaling the equations just to clear denominators or decimal points. Furthermore, since you will be adding terms from different equations, it is wise to covert each equation into standard form so that the corresponding terms are aligned. You may find that after you add to equations to eliminate a variable that both of the variables have been eliminated. When this happens, you either had an inconsistent system (with no solution) or a dependent system (with infinitely many solutions). If the equation you now have is an outright false equation (like ) then you had an inconsistent system. If it is an outright true equation (like ) then you had a dependent system. Use elimination to solve the system. What number could you multiply by each term in the second equation that would be helpful? If we multiply the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Use elimination to solve the system. What numbers could you multiply each term in the first and second equations by that would be helpful? If we multiply the first equation through by and the second equation through by then the -terms will be and and adding will eliminate After multiplying the terms in the first by and the terms in the second equation by and adding the two equations together, solve for Now solve for as well, by substituting in for in one of the original equations. Solve the system of equations using elimination. Review Exercises for : Check if the given point is a solution to the given system of two linear equations. Is a solution? The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (4, -4) Simply by looking closely at the linear equations, determine how many solutions the system will have. And state whether the system is dependeent, inconsistent, or neither. Solve a System Solve the given system of linear equations graphically. If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-8, 0) So there are no solutions. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-8, 3) So the only solution is If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-4, 5) So the only solution is Broderick enters a grassy field from some dense trees. His dog is standing out in the field, away. As soon as they see each other, they start running toward each other. Broderick runs with speed and his dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. Broderick starts out at position running with speed So one equation is The dog starts running with speed but getting closer to Broderick. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,65) with slope 3; the other line passes through (0,0) with slope -10; the lines cross at (5,15) Based on the graph, how long will it be until they meet? How far will that be from the place where Broderick started? The lines cross at So after seconds, they will meet 15 feet away from where Broderick started. : Solve the system of equations using substitution. A rectangle s length is shorter than four times its width. The rectangle s perimeter is Find the rectangle s length and width. A local restaurant has two locations. At one location, the revenue this month is but it has been decreasing by per month. At the other location, the annual revenue this month is and it has been increasing by per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be? Kiana invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Kiana s total fell from to How much money did she invest in each account? A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost per pound, and cashews cost per pound. The company is targeting a product that will cost them per pound worth of ingredients. How much of each type of nut should go into a bag? : Solve the system of equations using elimination. Freddy invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Freddy earned a total of in interest. How much money did he invest in each account? " }, { "id": "review-systems-of-linear-equations-2-2", @@ -18403,7 +18835,7 @@ var ptx_lunr_docs = [ "type": "Checkpoint", "number": "4.4.1", "title": "", - "body": " The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (1, 4) " + "body": " The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (3, 1) " }, { "id": "solve-system-graphically-point-slope-copy2", @@ -18412,7 +18844,7 @@ var ptx_lunr_docs = [ "type": "Checkpoint", "number": "4.4.2", "title": "", - "body": " Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (2, 6) So the only solution is " + "body": " Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (6, -4) So the only solution is " }, { "id": "solve-system-graphically-standard-copy2", @@ -18421,7 +18853,7 @@ var ptx_lunr_docs = [ "type": "Checkpoint", "number": "4.4.3", "title": "", - "body": " Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (9, 8) So the only solution is " + "body": " Solve the given system of linear equations graphically. If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, 8) So the only solution is " }, { "id": "review-systems-of-linear-equations-6-2", @@ -18511,7 +18943,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.2", "title": "", - "body": " The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (1, -2) " + "body": " The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system. a coordinate plot of two lines that cross at (4, -4) " }, { "id": "see-infinite-solutions-copy2", @@ -18529,7 +18961,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.4", "title": "", - "body": " If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (2, -3) So there are no solutions. " + "body": " If we graph these lines, we find they are parallel and do not cross. a coordinate plot of two lines that cross at (-8, 0) So there are no solutions. " }, { "id": "solve-system-graphically-standard-point-slope-copy2", @@ -18538,7 +18970,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.5", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-5, (-6)) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-8, 3) So the only solution is " }, { "id": "solve-system-graphically-slope-intercept-point-slope-infinite-solutions-copy2", @@ -18547,7 +18979,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.6", "title": "", - "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-7, 8) So the only solution is " + "body": " If we graph these lines, we find they cross. a coordinate plot of two lines that cross at (-4, 5) So the only solution is " }, { "id": "solve-system-application-greet-dog-copy", @@ -18556,7 +18988,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.7", "title": "", - "body": " El enters a grassy field from some dense trees. Their dog is standing out in the field, away. As soon as they see each other, they start running toward each other. El runs with speed and their dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. El starts out at position running with speed So one equation is The dog starts running with speed but getting closer to El. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,110) with slope 3; the other line passes through (0,0) with slope -8; the lines cross at (10,30) Based on the graph, how long will it be until they meet? How far will that be from the place where El started? The lines cross at So after seconds, they will meet 30 feet away from where El started. " + "body": " Broderick enters a grassy field from some dense trees. His dog is standing out in the field, away. As soon as they see each other, they start running toward each other. Broderick runs with speed and his dog runs with speed How long will it be until they meet? Write two equations that form a system for this scenario. Broderick starts out at position running with speed So one equation is The dog starts running with speed but getting closer to Broderick. So we ll use a negative rate of change. And the dog starts out away. So the other equation is Plot the lines for the system of two equations. Plotting the two slope-intercept equations: a Cartesian grid with two intersecting lines; one line passes through (0,65) with slope 3; the other line passes through (0,0) with slope -10; the lines cross at (5,15) Based on the graph, how long will it be until they meet? How far will that be from the place where Broderick started? The lines cross at So after seconds, they will meet 15 feet away from where Broderick started. " }, { "id": "substitution-integer-solutions-integer-coefficients-point-slope-copy2", @@ -18637,7 +19069,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.16", "title": "", - "body": " A rectangle s length is shorter than three times its width. The rectangle s perimeter is Find the rectangle s length and width. " + "body": " A rectangle s length is shorter than four times its width. The rectangle s perimeter is Find the rectangle s length and width. " }, { "id": "restaurant-location-revenues-copy", @@ -18655,7 +19087,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.18", "title": "", - "body": " Geoffrey invested a total of in two investments. His savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Geoffrey s total fell from to How much money did he invest in each account? " + "body": " Kiana invested a total of in two investments. Her savings account pays interest annually. A riskier stock investment lost at the end of the year. At the end of the year, Kiana s total fell from to How much money did she invest in each account? " }, { "id": "mixed-nuts-copy", @@ -18718,7 +19150,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "4.4.25", "title": "", - "body": " Clayton invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Clayton earned a total of in interest. How much money did he invest in each account? " + "body": " Freddy invested a total of in two investments. His savings account pays interest annually. A riskier stock investment earned at the end of the year. At the end of the year, Freddy earned a total of in interest. How much money did he invest in each account? " }, { "id": "section-arithmetic-with-negative-numbers", @@ -19420,7 +19852,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "A.2", "title": "Fractions and Fraction Arithmetic", - "body": " Fractions and Fraction Arithmetic The word fraction comes from the Latin word fractio , which means break into pieces. For thousands of years, cultures from all over the world have used fractions to understand parts of a whole. Alternative Video Lesson Visualizing Fractions Parts of a Whole One approach to understanding fractions is to think of them as parts of a whole. In Figure , we see whole divided into parts. Since parts are shaded, we have an illustration of the fraction . The denominator tells us how many parts to cut up the whole; since we have parts, they're called sevenths. The numerator tells us how many sevenths to consider. fraction numerator fraction denominator numerator denominator Representing as parts of a whole. a rectangle that is seven times as wide as it is tall; the entire rectangle is shaded; there is a 1 in the center of the rectangle; to its right, it is labeled 'one whole'; another rectangle of the same size is aligned directly below the first rectangle; it is subdivided equally into seven squares; the first three squares are shaded; there is a 1\/7 in the center of each of the first three squares; to its right, it is labeled 'three sevenths' A Fraction as Parts of a Whole To visualize the fraction you might cut a rectangle into equal parts, and then count up of them. You could cut a rectangle into equal pieces, and then of them would represent We can also locate fractions on number lines. When ticks are equally spread apart, as in Figure , each tick represents a fraction. Representing on a number line. a number line with a curved arrow emanating from 0 and landing to the right at 1\/7, then again from 1\/7 to 2\/7, and again from 2\/7 to 3\/7 A Fraction on a Number Line In the given number line, what fraction is marked? a number line with -1, 0, and 1 marked; there are evenly spaced ticks, with eight ticks between -1 and 0, eight between 0 and 1, and so on; there is a dot marked at the fifth tick to the right from 0 There are subdivisions between and and the mark is at the fifth subdivision. So the mark is of the way from to and therefore represents the fraction Division Fractions can also be understood through division. For example, we can view the fraction as divided into equal parts, as in Figure . Just one of those parts represents . Representing on a number line. a number line with a ruler superimposed over the segment from 0 to 3; the ruler is subdivided into seven pieces; an arrow points down from the first ruler tick to a point on the number line that is marked 3\/7 Seeing a Fraction as Division Arithmetic The fraction can be thought of as dividing the whole number into equal-sized parts. Since means the same as it can be thought of as dividing into equal parts. Equivalent Fractions It's common to have two fractions that represent the same amount. Consider and represented in various ways in Figures . and as equal parts of a whole one rectangle that is five times as wide as it is tall; the entire rectangle is shaded; there is a 1 in the center of the rectangle; a second rectangle that is also five times as wide as it is tall; it is subdivided equally into five squares; the first two squares are shaded; there is a 1\/5 in the center of each of the first two squares; a third rectangle that is also five times as wide as it is tall; it is subdivided equally into fifteen adjacent rectangles; the first six of these smaller rectangles are shaded; there is a 1\/15 in the center of each of the first six smaller rectangles and as equal on a number line a number line with a curved arrow emanating from 0 and landing to the right at 1\/5, then again from 1\/5 to 2\/5; there is a number line with a curved arrow emanating from 0 and landing to the right at 1\/15, then again from 1\/15 to 2\/15, then again from 2\/15 to 3\/15, then again from 3\/15 to 4\/15, then again from 4\/15 to 5\/15, and then again from 5\/15 to 6\/15 and as equal results from division a number line with a ruler superimposed over the segment from 0 to 2; the ruler is subdivided into five pieces; an arrow points down from the first ruler tick to a point on the number line that is marked 2\/5; there is a second number line with a ruler superimposed over the segment from 0 to 6; the ruler is subdivided into fifteen pieces; an arrow points down from the first ruler tick to a point on the number line that is marked 6\/15 Those two fractions, and are equal, as those figures demonstrate. In addition, both fractions are equal to as a decimal. If we must work with this number, the fraction that uses smaller numbers, , is preferable. Working with smaller numbers decreases the likelihood of making a human arithmetic error and it also increases the chances that you might make useful observations about the nature of that number. So if you are handed a fraction like , it is important to try to reduce it to lowest terms. reduced fraction fraction reduction The most important skill you can have to help you do this is to know the multiplication table well. If you know it well, you know that and , so you can break down the numerator and denominator that way. Both the numerator and denominator are divisible by , so they can be factored out and then as factors, cancel out. Reducing Fractions Reduce these fractions into lowest terms. With we have which reduces to With we have which reduces to With we have which reduces to Sometimes it is useful to do the opposite of reducing a fraction, and build up the fraction to use larger numbers. Building Up a Fraction Sayid scored on a recent exam. Build up this fraction so that the denominator is so that Sayid can understand what percent score he earned. To change the denominator from to it needs to be multiplied by So we calculate So the fraction is equivalent to (This means Sayid scored an ) Multiplying with Fractions Suppose a recipe calls for cup of milk, but we'd like to quadruple the recipe (make it four times as big). We'll need four times as much milk, and one way to measure this out is to fill a measuring cup to full, four times: four measuring cups, each filled two-thirds with liquid When you count up the shaded thirds, there are eight of them. So multiplying by the whole number , the result is . Mathematically: Multiplying a Fraction and a Whole Number When you multiply a whole number by a fraction, you may just multiply the whole number by the numerator and leave the denominator alone. In other words, as long as is not , then a whole number and a fraction multiply this way: We could also use multiplication to decrease amounts. Suppose we needed to cut the recipe down to just one fifth. Instead of four of the cup milk, we need one fifth of the cup milk. So instead of multiplying by , we multiply by . But how much is of cup? If we cut the measuring cup into five equal vertical strips along with the three equal horizontal strips, then in total there are subdivisions of the cup. Two of those sections represent of the cup. a measuring cup divided evenly into three horizontal sections, but also evenly into five vertical sections; in total, there are fifteen sections; the two that correspond to the lower two-thirds and leftmost one-fifth are shaded In the end, we have of a cup. The denominator came from multiplying and , the denominators of the fractions we had to multiply. The numerator came from multiplying and , the numerators of the fractions we had to multiply. Multiplication with Fractions multiplication of fractions As long as and are not , then fractions multiply this way: Fraction Multiplication Simplify these fraction products. Multiplying numerators gives and multiplying denominators gives The answer is Before we multiply fractions, note that reduces to and reduces to So we just have Multiplying numerators gives and multiplying denominators gives The result should be negative, so the answer is Before we multiply fractions, note that reduces to So we have Both the numerator of the first fraction and denominator of the second fraction are divisible by so it helps to reduce both fractions accordingly and get Both the denominator of the first fraction and numerator of the second fraction are divisible by so it helps to reduce both fractions accordingly and get Now we are just multiplying by so the result is Division with Fractions How does division with fractions work? Are we able to compute\/simplify each of these examples? We know that when we divide something by , this is the same as multiplying it by . Conversely, dividing a number or expression by is the same as multiplying by , or just . The more general property is that when we divide a number or expression by , this is equivalent to multiplying by the reciprocal . Division with Fractions As long as , and are not , then division with fractions works this way: With our examples from the beginning of this subsection: Fraction Division Simplify these fraction division expressions. Adding and Subtracting Fractions With whole numbers and integers, operations of addition and subtraction are relatively straightforward. The situation is almost as straightforward with fractions if the two fractions have the same denominator . Consider In the same way that tacos and tacos make tacos, we have: Adding\/Subtracting with Fractions Having the Same Denominator addition of fractions with the same denominator To add or subtract two fractions having the same denominator, keep that denominator, and add or subtract the numerators. If it's possible, useful, or required of you, simplify the result by reducing to lowest terms. Fraction Addition and Subtraction Add or subtract these fractions. Since the denominators are both we can add the numerators: The answer is Since the denominators are both we can subtract the numerators: The answer is but that reduces to Whenever we'd like to combine fractional amounts that don't represent the same number of parts of a whole (that is, when the denominators are different), finding sums and differences is more complicated. Quarters and Dimes Find the sum . Does this seem intimidating? Consider this: of a dollar is a quarter, and so of a dollar is cents. of a dollar is a dime, and so of a dollar is cents. So if you know what to look for, the expression is like adding cents and cents, which gives you cents. As a fraction of one dollar, that is . So we can report . (Although we should probably reduce that last fraction to .) This example was not something you can apply to other fraction addition situations, because the denominators here worked especially well with money amounts. But there is something we can learn here. The fraction was equivalent to , and the other fraction was equivalent to . These equivalent fractions have the same denominator and are therefore easy to add. What we saw happen was: This realization gives us a strategy for adding (or subtracting) fractions. Adding\/Subtracting Fractions with Different Denominators addition of fractions with different denominators To add (or subtract) generic fractions together, use their denominators to find a common denominator . fraction common denominator common denominator This means some whole number that is a whole multiple of both of the original denominators. Then rewrite the two fractions as equivalent fractions that use this common denominator. Write the result keeping that denominator and adding (or subtracting) the numerators. Reduce the fraction if that is useful or required. Let's add . The denominators are and , so the number would be a good common denominator. Using Some Flour A chef had cups of flour and needed to use cup to thicken a sauce. How much flour is left? We need to compute The denominators are and One common denominator is so we move to rewrite each fraction using as the denominator: The numerical result is but a pure number does not answer this question. The amount of flour remaining is cups . Mixed Numbers and Improper Fractions A simple recipe for bread contains only a few ingredients: tablespoons yeast tablespoons kosher salt cups unbleached, all-purpose flour (more for dusting) Each ingredient is listed as a mixed number fraction mixed number mixed number that quickly communicates how many whole amounts and how many parts are needed. It's useful for quickly communicating a practical amount of something you are cooking with, measuring on a ruler, purchasing at the grocery store, etc. But it causes trouble in an algebra class. The number means one and one half. So really, The trouble is that with , you have two numbers written right next to each other. Normally with two math expressions written right next to each other, they should be multiplied , not added . But with a mixed number, they should be added. Fortunately we just reviewed how to add fractions. If we need to do any arithmetic with a mixed number like , we can treat it as and simplify to get a nice fraction instead: . A fraction like is called an improper fraction fraction improper improper fraction because it's actually larger than . And a proper fraction would be something small that is only part of a whole instead of more than a whole. Review and Warmup Which letter is on the number line? Which letter is m on the number line? The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the second tick from , so it can be represented as , which reduces to . The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the third tick left from , so it can be represented as , which reduces to . The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the 4th tick from , so it can be represented as , which reduces to . The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the 5th tick to the left of , so it can be represented as . Reducing Fractions Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, goes into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this case, is the final answer. Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Notice that when the numerator s only prime factor, , is canceled, we have to leave a in the numerator. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, is the final answer. Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, again goes into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Notice that when all prime factors in the numerator are canceled, we have to add a 1 in the numerator. Reduce the fraction . Method 1: We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: , until the prime number is greater than or equal to . No prime number goes into both and , so the fraction is not reducible. Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Notice that no numbers can be canceled out, so the fraction is not reducible. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, divides into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, divides into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Note that if the denominator is , the result is simply an integer. Building Fractions Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Multiplying\/Dividing Fractions Multiply: To multiply two fractions, simply multiply both numerators and both denominators. We have: The answer to this question is . Multiply: To multiply two fractions, simply multiply both numerators and both denominators. We have: The answer to this question is . Multiply: To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it s easier to reduce fractions before multiplying: The answer to this question is . Note that we divided the first numerator and the second denominator by . Multiply: To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it s easier to reduce fractions before multiplying: The answer to this question is . Note that we divided the first numerator and the second denominator by . Multiply: To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a 1 as the denominator: Next, simply multiply both numerators and both denominators: The answer to this question is . Multiply: To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a 1 as the denominator: Next, simply multiply both numerators and both denominators: The answer to this question is . Multiply: To multiply two fractions, multiply both numerators and both denominators. Don t forget to reduce the fraction if possible. Remember: a positive number times a negative number is a negative number. We have: Multiply: To multiply two fractions, multiply both numerators and both denominators. Don t forget to reduce the fraction if possible. Remember: a positive number times a negative number is a negative number. We have: Multiply: To multiply an integer with a fraction, first we rewrite the integer as a fraction by using 1 as the denominator: Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers: Multiply: To multiply an integer with a fraction, first we rewrite the integer as a fraction by using 1 as the denominator: Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers: Multiply: When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Multiply: When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Multiply: When we multiply fractions, if there is an integer, we change the integer to a fraction by writing as its denominator. For example, . When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Multiply: When we multiply fractions, if there is an integer, we change the integer to a fraction by writing as its denominator. For example, . When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Divide: When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Divide: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Divide: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Multiply: In order to multiply two mixed numbers, like it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is: If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is . Multiply: In order to multiply two mixed numbers, like it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is: If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is . Adding\/Subtracting Fractions Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The result is: Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The result is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The whole process is: The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The whole process is: The answer is . Add: To add two fractions with different denominators, we first have to find the common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find the common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full solution is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full solution is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The answer is . Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer to this question is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer to this question is . Add: When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions: Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator . We will rewrite the first fraction: Finally, we add the numerators and keep the denominator unchanged. The whole process is: The answer to this question is . Add: When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions: Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator . We will rewrite the first fraction: Finally, we add the numerators and keep the denominator unchanged. The whole process is: The answer to this question is . Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Subtract: To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged. Subtract: To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged. Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the subtraction. The whole process is: The answer is . Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the subtraction. The whole process is: The answer is . Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . The first fraction already has the common denominator. Look at the second fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Finally, we can do the subtraction. The whole process is: Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . The first fraction already has the common denominator. Look at the second fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Finally, we can do the subtraction. The whole process is: Subtract: To add or subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the denominator of the second fraction to by: Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is: Subtract: To add or subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the denominator of the second fraction to by: Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is: Subtract: When we subtract a negative number, first we change those two negative signs to addition: Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. Subtract: When we subtract a negative number, first we change those two negative signs to addition: Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. Subtract: When doing arithmetic with fractions, it helps to rewrite any integers as fractions: Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator . We will change the first fraction to have the common denominator: Finally, we subtract the numerators and keep the denominator unchanged. The answer to this question is . Subtract: When doing arithmetic with fractions, it helps to rewrite any integers as fractions: Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator . We will change the first fraction to have the common denominator: Finally, we subtract the numerators and keep the denominator unchanged. The answer to this question is . Applications Kurt walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Kurt walk altogether? Kurt walked a total of of a mile. This is an addition problem. Kurt walked a total of miles altogether. Jerry walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Jerry walk altogether? Jerry walked a total of of a mile. This is an addition problem. Jerry walked a total of miles altogether. Samantha and Randi are sharing a pizza. Samantha ate of the pizza, and Randi ate of the pizza. How much of the pizza was eaten in total? They ate of the pizza. This is an addition problem. They ate of the pizza. A trail s total length is of a mile. It has two legs. The first leg is of a mile long. How long is the second leg? The second leg is of a mile in length. This is a subtraction problem. The second leg is of a mile in length. A trail s total length is of a mile. It has two legs. The first leg is of a mile long. How long is the second leg? The second leg is of a mile in length. This is a subtraction problem. The second leg is of a mile in length. Carl is participating in a running event. In the first hour, he completed of the total distance. After another hour, in total he had completed of the total distance. What fraction of the total distance did Carl complete during the second hour? Carl completed of the distance during the second hour. This is a subtraction problem. Carl completed of the distance during the second hour. The pie chart represents a school s student population. Together, white and black students make up of the school s population. By the pie chart, the school has white students, and black students. We will use addition to find the total portion of white and black students at this school. Together, white and black students make up of the school s population. Each page of a book is inches in height, and consists of a header (a top margin), a footer (a bottom margin), and the middle part (the body). The header is of an inch thick and the middle part is inches from top to bottom. What is the thickness of the footer? The footer is of an inch thick. From a page s total height, we subtract the thickness of the header and the height of the body to find the thickness of the footer: The footer is inches thick. Sherial and Wendy are sharing a pizza. Sherial ate of the pizza, and Wendy ate of the pizza. How much more pizza did Sherial eat than Wendy? Sherial ate more of the pizza than Wendy ate. To find the difference of two numbers, we use subtraction: Sherial ate more of the pizza than Wendy ate. Hayden and Maygen are sharing a pizza. Hayden ate of the pizza, and Maygen ate of the pizza. How much more pizza did Hayden eat than Maygen? Hayden ate more of the pizza than Maygen ate. To find the difference of two numbers, we use subtraction: Hayden ate more of the pizza than Maygen ate. A school had a fund-raising event. The revenue came from three resources: ticket sales, auction sales, and donations. Ticket sales account for of the total revenue; auction sales account for of the total revenue. What fraction of the revenue came from donations? of the revenue came from donations. The revenue consists of parts. The total revenue is like a big ; ticket sales account for ; auction sales account for . To find the missing piece (from donations), we use subtraction. Note that a common denominator of , and is . So of the revenue came from donations. A few years back, a car was purchased for . Today it is worth of its original value. What is the car s current value? The car s current value is . When we use the word of in situations like of , we can translate of into the multiplication symbol. In this problem, to find of , we do: The car is worth now. A few years back, a car was purchased for . Today it is worth of its original value. What is the car s current value? The car s current value is . When we use the word of in situations like of , we can translate of into the multiplication symbol. In this problem, to find of , we do: The car is worth now. The pie chart represents a school s student population. more of the school is white students than black students. By the pie chart, the school has white students, and black students. We will use subtraction to find the difference: A town has residents in total, of which are white\/Caucasian Americans. How many white\/Caucasian Americans reside in this town? There are white\/Caucasian Americans residing in this town. When we use the word of in expressions like of , we can translate of into the multiplication symbol. In this problem, to find of , we compute: There are white\/Caucasian Americans residing in this town. A company received a grant, and decided to spend of this grant in research and development next year. Out of the money set aside for research and development, will be used to buy new equipment. What fraction of the grant will be used to buy new equipment? of the grant will be used to buy new equipment. For this problem, we are trying to find of . The word of implies multiplication: So of the grant will be used to buy new equipment. A food bank just received kilograms of emergency food. Each family in need is to receive kilograms of food. How many families can be served with the kilograms of food? families can be served with the kilograms of food. We can address this problem as repeatedly taking away kilograms from kilograms, which implies a division problem: families can be served with the kilograms of food. A construction team maintains a -mile-long sewage pipe. Each day, the team can cover of a mile. How many days will it take the team to complete the maintenance of the entire sewage pipe? It will take the team days to complete maintaining the entire sewage pipe. In this problem, the team maintains miles of sewage pipe every day, until they complete all miles. This is like repeatedly taking away miles from miles, a division problem. It will take the team days to complete maintaining the entire sewage pipe. A child is stacking up tiles. Each tile s height is of a centimeter. How many layers of tiles are needed to reach centimeters in total height? To reach the total height of centimeters, layers of tiles are needed. In this problem, each layer of tile is centimeters tall. The child will keep stacking up layers until the total height reaches centimeters. This is like asking how many centimeters are there in centimeters, a division problem. To reach the total height of centimeters, layers of tiles are needed. A restaurant made cups of pudding for a festival. Customers at the festival will be served of a cup of pudding per serving. How many customers can the restaurant serve at the festival with the cups of pudding? The restaurant can serve customers at the festival with the cups of pudding. There are a total of cups of pudding, with each serving being cup. To find how many servings can be served, we need to find how many cups are there in cups, a division problem: The restaurant can serve customers at the festival with the cups of pudding. Shortcut Since each serving has cup, we know each cup has servings. So cups has servings. A piece of lumber in your garage is inches long. A second is inches long. If you lay them end to end, what will the total length be? The total length will be inches. To find the total length in inches, we add the two lengths together. So the total length is inches. A piece of lumber in your garage is inches long. A second is inches long. If you lay them end to end, what will the total length be? The total length will be inches. To find the total length in inches, we add the two lengths together. So the total length is inches. Each page of a book consists of a header, a footer and the middle part. The header is inches in height; the footer is inches in height; and the middle part is inches in height. What is the total height of each page in this book? Use mixed number in your answer if needed. Each page in this book is inches in height. We add up the height of the header, middle part and footer to find the total height of a page: Each page in this book is inches in height. To pave the road on Ellis Street, the crew used tons of cement on the first day, and used tons on the second day. How many tons of cement were used in all? tons of cement were used in all. This problem is obviously an addition problem. To add mixed number, we break each mixed number into an integer and a fraction, and then add up integers and fractions separately. tons of cement were used in all. When driving on a high way, noticed a sign saying exit to Johnstown is miles away, while exit to Jerrystown is miles away. How far is Johnstown from Jerrystown? Johnstown and Jerrystown are miles apart. To find the distance between Johnstown and Jerrystown, we need to find the difference between their distance. This implies subtraction. We could treat subtraction as adding a negative : From here, there are two methods to continue. Method 1 We can split from the integer, and do : Note that we changed to . Method2 We could simply change the whole number into a fraction without splitting : ran more miles than . A cake recipe needs cups of flour. Using this recipe, to bake cakes, how many cups of flour are needed? To bake cakes, cups of flour are needed. Each cake needs cups of flour. To find how many cups of flour are needed to bake cakes, we use multiplication: To bake cakes, cups of flour are needed. Sketching Fractions Sketch a number line showing each fraction. (Be sure to carefully indicate the correct number of equal parts of the whole.) Sketch a number line showing each fraction. (Be sure to carefully indicate the correct number of equal parts of the whole.) Sketch a picture of the product , using a number line or rectangles. Sketch a picture of the sum , using a number line or rectangles. Challenge Given that simplify Since the fractions have the same denominator, we can just add numerators and keep that denominator. Given that simplify Since the fractions don t have the same denominator, we have to make them have like denominators. Multiply the first fraction by Given that simplify Since the fractions don t have the same denominator, we have to make them have like denominators. Multiply the first fraction by " + "body": " Fractions and Fraction Arithmetic The word fraction comes from the Latin word fractio , which means break into pieces. For thousands of years, cultures from all over the world have used fractions to understand parts of a whole. Alternative Video Lesson Visualizing Fractions Parts of a Whole One approach to understanding fractions is to think of them as parts of a whole. In Figure , we see whole divided into parts. Since parts are shaded, we have an illustration of the fraction . The denominator tells us how many parts to cut up the whole; since we have parts, they're called sevenths. The numerator tells us how many sevenths to consider. fraction numerator fraction denominator numerator denominator Representing as parts of a whole. a rectangle that is seven times as wide as it is tall; the entire rectangle is shaded; there is a 1 in the center of the rectangle; to its right, it is labeled 'one whole'; another rectangle of the same size is aligned directly below the first rectangle; it is subdivided equally into seven squares; the first three squares are shaded; there is a 1\/7 in the center of each of the first three squares; to its right, it is labeled 'three sevenths' A Fraction as Parts of a Whole To visualize the fraction you might cut a rectangle into equal parts, and then count up of them. You could cut a rectangle into equal pieces, and then of them would represent We can also locate fractions on number lines. When ticks are equally spread apart, as in Figure , each tick represents a fraction. Representing on a number line. a number line with a curved arrow emanating from 0 and landing to the right at 1\/7, then again from 1\/7 to 2\/7, and again from 2\/7 to 3\/7 A Fraction on a Number Line In the given number line, what fraction is marked? a number line with -1, 0, and 1 marked; there are evenly spaced ticks, with eight ticks between -1 and 0, eight between 0 and 1, and so on; there is a dot marked at the fifth tick to the right from 0 There are subdivisions between and and the mark is at the fifth subdivision. So the mark is of the way from to and therefore represents the fraction Division Fractions can also be understood through division. For example, we can view the fraction as divided into equal parts, as in Figure . Just one of those parts represents . Representing on a number line. a number line with a ruler superimposed over the segment from 0 to 3; the ruler is subdivided into seven pieces; an arrow points down from the first ruler tick to a point on the number line that is marked 3\/7 Seeing a Fraction as Division Arithmetic The fraction can be thought of as dividing the whole number into equal-sized parts. Since means the same as it can be thought of as dividing into equal parts. Equivalent Fractions It's common to have two fractions that represent the same amount. Consider and represented in various ways in Figures . and as equal parts of a whole one rectangle that is five times as wide as it is tall; the entire rectangle is shaded; there is a 1 in the center of the rectangle; a second rectangle that is also five times as wide as it is tall; it is subdivided equally into five squares; the first two squares are shaded; there is a 1\/5 in the center of each of the first two squares; a third rectangle that is also five times as wide as it is tall; it is subdivided equally into fifteen adjacent rectangles; the first six of these smaller rectangles are shaded; there is a 1\/15 in the center of each of the first six smaller rectangles and as equal on a number line a number line with a curved arrow emanating from 0 and landing to the right at 1\/5, then again from 1\/5 to 2\/5; there is a number line with a curved arrow emanating from 0 and landing to the right at 1\/15, then again from 1\/15 to 2\/15, then again from 2\/15 to 3\/15, then again from 3\/15 to 4\/15, then again from 4\/15 to 5\/15, and then again from 5\/15 to 6\/15 and as equal results from division a number line with a ruler superimposed over the segment from 0 to 2; the ruler is subdivided into five pieces; an arrow points down from the first ruler tick to a point on the number line that is marked 2\/5; there is a second number line with a ruler superimposed over the segment from 0 to 6; the ruler is subdivided into fifteen pieces; an arrow points down from the first ruler tick to a point on the number line that is marked 6\/15 Those two fractions, and are equal, as those figures demonstrate. In addition, both fractions are equal to as a decimal. If we must work with this number, the fraction that uses smaller numbers, , is preferable. Working with smaller numbers decreases the likelihood of making a human arithmetic error and it also increases the chances that you might make useful observations about the nature of that number. So if you are handed a fraction like , it is important to try to reduce it to lowest terms. reduced fraction fraction reduction The most important skill you can have to help you do this is to know the multiplication table well. If you know it well, you know that and , so you can break down the numerator and denominator that way. Both the numerator and denominator are divisible by , so they can be factored out and then as factors, cancel out. Reducing Fractions Reduce these fractions into lowest terms. With we have which reduces to With we have which reduces to With we have which reduces to Sometimes it is useful to do the opposite of reducing a fraction, and build up the fraction to use larger numbers. Building Up a Fraction Sayid scored on a recent exam. Build up this fraction so that the denominator is so that Sayid can understand what percent score he earned. To change the denominator from to it needs to be multiplied by So we calculate So the fraction is equivalent to (This means Sayid scored an ) Multiplying with Fractions Suppose a recipe calls for cup of milk, but we'd like to quadruple the recipe (make it four times as big). We'll need four times as much milk, and one way to measure this out is to fill a measuring cup to full, four times: four measuring cups, each filled two-thirds with liquid When you count up the shaded thirds, there are eight of them. So multiplying by the whole number , the result is . Mathematically: Multiplying a Fraction and a Whole Number When you multiply a whole number by a fraction, you may just multiply the whole number by the numerator and leave the denominator alone. In other words, as long as is not , then a whole number and a fraction multiply this way: We could also use multiplication to decrease amounts. Suppose we needed to cut the recipe down to just one fifth. Instead of four of the cup milk, we need one fifth of the cup milk. So instead of multiplying by , we multiply by . But how much is of cup? If we cut the measuring cup into five equal vertical strips along with the three equal horizontal strips, then in total there are subdivisions of the cup. Two of those sections represent of the cup. a measuring cup divided evenly into three horizontal sections, but also evenly into five vertical sections; in total, there are fifteen sections; the two that correspond to the lower two-thirds and leftmost one-fifth are shaded In the end, we have of a cup. The denominator came from multiplying and , the denominators of the fractions we had to multiply. The numerator came from multiplying and , the numerators of the fractions we had to multiply. Multiplication with Fractions multiplication of fractions As long as and are not , then fractions multiply this way: Fraction Multiplication Simplify these fraction products. Multiplying numerators gives and multiplying denominators gives The answer is Before we multiply fractions, note that reduces to and reduces to So we just have Multiplying numerators gives and multiplying denominators gives The result should be negative, so the answer is Before we multiply fractions, note that reduces to So we have Both the numerator of the first fraction and denominator of the second fraction are divisible by so it helps to reduce both fractions accordingly and get Both the denominator of the first fraction and numerator of the second fraction are divisible by so it helps to reduce both fractions accordingly and get Now we are just multiplying by so the result is Division with Fractions How does division with fractions work? Are we able to compute\/simplify each of these examples? We know that when we divide something by , this is the same as multiplying it by . Conversely, dividing a number or expression by is the same as multiplying by , or just . The more general property is that when we divide a number or expression by , this is equivalent to multiplying by the reciprocal . Division with Fractions As long as , and are not , then division with fractions works this way: With our examples from the beginning of this subsection: Fraction Division Simplify these fraction division expressions. Adding and Subtracting Fractions With whole numbers and integers, operations of addition and subtraction are relatively straightforward. The situation is almost as straightforward with fractions if the two fractions have the same denominator . Consider In the same way that tacos and tacos make tacos, we have: Adding\/Subtracting with Fractions Having the Same Denominator addition of fractions with the same denominator To add or subtract two fractions having the same denominator, keep that denominator, and add or subtract the numerators. If it's possible, useful, or required of you, simplify the result by reducing to lowest terms. Fraction Addition and Subtraction Add or subtract these fractions. Since the denominators are both we can add the numerators: The answer is Since the denominators are both we can subtract the numerators: The answer is but that reduces to Whenever we'd like to combine fractional amounts that don't represent the same number of parts of a whole (that is, when the denominators are different), finding sums and differences is more complicated. Quarters and Dimes Find the sum . Does this seem intimidating? Consider this: of a dollar is a quarter, and so of a dollar is cents. of a dollar is a dime, and so of a dollar is cents. So if you know what to look for, the expression is like adding cents and cents, which gives you cents. As a fraction of one dollar, that is . So we can report . (Although we should probably reduce that last fraction to .) This example was not something you can apply to other fraction addition situations, because the denominators here worked especially well with money amounts. But there is something we can learn here. The fraction was equivalent to , and the other fraction was equivalent to . These equivalent fractions have the same denominator and are therefore easy to add. What we saw happen was: This realization gives us a strategy for adding (or subtracting) fractions. Adding\/Subtracting Fractions with Different Denominators addition of fractions with different denominators To add (or subtract) generic fractions together, use their denominators to find a common denominator . fraction common denominator common denominator This means some whole number that is a whole multiple of both of the original denominators. Then rewrite the two fractions as equivalent fractions that use this common denominator. Write the result keeping that denominator and adding (or subtracting) the numerators. Reduce the fraction if that is useful or required. Let's add . The denominators are and , so the number would be a good common denominator. Using Some Flour A chef had cups of flour and needed to use cup to thicken a sauce. How much flour is left? We need to compute The denominators are and One common denominator is so we move to rewrite each fraction using as the denominator: The numerical result is but a pure number does not answer this question. The amount of flour remaining is cups . Mixed Numbers and Improper Fractions A simple recipe for bread contains only a few ingredients: tablespoons yeast tablespoons kosher salt cups unbleached, all-purpose flour (more for dusting) Each ingredient is listed as a mixed number fraction mixed number mixed number that quickly communicates how many whole amounts and how many parts are needed. It's useful for quickly communicating a practical amount of something you are cooking with, measuring on a ruler, purchasing at the grocery store, etc. But it causes trouble in an algebra class. The number means one and one half. So really, The trouble is that with , you have two numbers written right next to each other. Normally with two math expressions written right next to each other, they should be multiplied , not added . But with a mixed number, they should be added. Fortunately we just reviewed how to add fractions. If we need to do any arithmetic with a mixed number like , we can treat it as and simplify to get a nice fraction instead: . A fraction like is called an improper fraction fraction improper improper fraction because it's actually larger than . And a proper fraction would be something small that is only part of a whole instead of more than a whole. Review and Warmup Which letter is on the number line? Which letter is m on the number line? The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the third tick from , so it can be represented as , which reduces to . The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the second tick left from , so it can be represented as , which reduces to . The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the third tick from , so it can be represented as , which reduces to . The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the 5th tick to the left of , so it can be represented as . Reducing Fractions Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, goes into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this case, is the final answer. Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Notice that when the numerator s only prime factor, , is canceled, we have to leave a in the numerator. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, is the final answer. Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, again goes into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Reduce the fraction . Method 1: We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: , until the prime number is greater than or equal to . No prime number goes into both and , so the fraction is not reducible. Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Notice that no numbers can be canceled out, so the fraction is not reducible. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, divides into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, divides into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Note that if the denominator is , the result is simply an integer. Building Fractions Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Find an equivalent fraction to with denominator . From to , we multiplied by . For a fraction, when we multiply the same number in both the numerator and denominator, the fraction s value doesn t change. We have: Multiplying\/Dividing Fractions Multiply: To multiply two fractions, simply multiply both numerators and both denominators. We have: The answer to this question is . Multiply: To multiply two fractions, simply multiply both numerators and both denominators. We have: The answer to this question is . Multiply: To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it s easier to reduce fractions before multiplying: The answer to this question is . Note that we divided the first numerator and the second denominator by . Multiply: To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it s easier to reduce fractions before multiplying: The answer to this question is . Note that we divided the first numerator and the second denominator by . Multiply: To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a 1 as the denominator: Next, simply multiply both numerators and both denominators: The answer to this question is . Multiply: To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a 1 as the denominator: Next, simply multiply both numerators and both denominators: The answer to this question is . Multiply: To multiply two fractions, multiply both numerators and both denominators. Don t forget to reduce the fraction if possible. Remember: a positive number times a negative number is a negative number. We have: Multiply: To multiply two fractions, multiply both numerators and both denominators. Don t forget to reduce the fraction if possible. Remember: a positive number times a negative number is a negative number. We have: Multiply: To multiply an integer with a fraction, first we rewrite the integer as a fraction by using 1 as the denominator: Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers: Multiply: To multiply an integer with a fraction, first we rewrite the integer as a fraction by using 1 as the denominator: Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers: Multiply: When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Multiply: When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Multiply: When we multiply fractions, if there is an integer, we change the integer to a fraction by writing as its denominator. For example, . When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Multiply: When we multiply fractions, if there is an integer, we change the integer to a fraction by writing as its denominator. For example, . When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers. The full computation is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When we do division with fractions, as with we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have Then we do fraction multiplication as usual. The full process is: Divide: When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Divide: When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Divide: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Divide: When we do fraction multiplication\/division, we need to rewrite any integer as a fraction. In this problem, we will do this with: Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have Next, we do the fraction multiplication. The full process is: Multiply: In order to multiply two mixed numbers, like it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is: If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is . Multiply: In order to multiply two mixed numbers, like it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is: If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is . Adding\/Subtracting Fractions Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The result is: Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The result is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The whole process is: The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The whole process is: The answer is . Add: To add two fractions with different denominators, we first have to find the common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find the common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full solution is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full solution is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The answer is . Add: To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged. The answer is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the addition. The full process is: Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer to this question is . Add: To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. The full process is: The answer to this question is . Add: When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions: Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator . We will rewrite the first fraction: Finally, we add the numerators and keep the denominator unchanged. The whole process is: The answer to this question is . Add: When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions: Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator . We will rewrite the first fraction: Finally, we add the numerators and keep the denominator unchanged. The whole process is: The answer to this question is . Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Add: To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all denominators divide into the multiple. The largest denominator is . We list its first few multiples: Note that all denominators, each divide into . The full computation is: Don t forget to reduce the fraction by dividing into both the numerator and the denominator. Subtract: To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged. Subtract: To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged. Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the subtraction. The whole process is: The answer is . Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: The second fraction already has the common denominator. Finally, we can do the subtraction. The whole process is: The answer is . Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . The first fraction already has the common denominator. Look at the second fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Finally, we can do the subtraction. The whole process is: Subtract: To subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . The first fraction already has the common denominator. Look at the second fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Finally, we can do the subtraction. The whole process is: Subtract: To add or subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the denominator of the second fraction to by: Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is: Subtract: To add or subtract two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the denominator of the second fraction to by: Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is: Subtract: When we subtract a negative number, first we change those two negative signs to addition: Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. Subtract: When we subtract a negative number, first we change those two negative signs to addition: Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of and is . Look at the first fraction . To change the denominator to , we multiply at both the top and bottom, and we have: Similarly, we change the second fraction s denominator to : Now we can add those two fractions. Subtract: When doing arithmetic with fractions, it helps to rewrite any integers as fractions: Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator . We will change the first fraction to have the common denominator: Finally, we subtract the numerators and keep the denominator unchanged. The answer to this question is . Subtract: When doing arithmetic with fractions, it helps to rewrite any integers as fractions: Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator . We will change the first fraction to have the common denominator: Finally, we subtract the numerators and keep the denominator unchanged. The answer to this question is . Applications Thanh walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Thanh walk altogether? Thanh walked a total of of a mile. This is an addition problem. Thanh walked a total of miles altogether. Blake walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Blake walk altogether? Blake walked a total of of a mile. This is an addition problem. Blake walked a total of miles altogether. Diane and Peter are sharing a pizza. Diane ate of the pizza, and Peter ate of the pizza. How much of the pizza was eaten in total? They ate of the pizza. This is an addition problem. They ate of the pizza. A trail s total length is of a mile. It has two legs. The first leg is of a mile long. How long is the second leg? The second leg is of a mile in length. This is a subtraction problem. The second leg is of a mile in length. A trail s total length is of a mile. It has two legs. The first leg is of a mile long. How long is the second leg? The second leg is of a mile in length. This is a subtraction problem. The second leg is of a mile in length. Huynh is participating in a running event. In the first hour, he completed of the total distance. After another hour, in total he had completed of the total distance. What fraction of the total distance did Huynh complete during the second hour? Huynh completed of the distance during the second hour. This is a subtraction problem. Huynh completed of the distance during the second hour. The pie chart represents a school s student population. Together, white and black students make up of the school s population. By the pie chart, the school has white students, and black students. We will use addition to find the total portion of white and black students at this school. Together, white and black students make up of the school s population. Each page of a book is inches in height, and consists of a header (a top margin), a footer (a bottom margin), and the middle part (the body). The header is of an inch thick and the middle part is inches from top to bottom. What is the thickness of the footer? The footer is of an inch thick. From a page s total height, we subtract the thickness of the header and the height of the body to find the thickness of the footer: The footer is inches thick. Penelope and Michele are sharing a pizza. Penelope ate of the pizza, and Michele ate of the pizza. How much more pizza did Penelope eat than Michele? Penelope ate more of the pizza than Michele ate. To find the difference of two numbers, we use subtraction: Penelope ate more of the pizza than Michele ate. Samantha and Anthony are sharing a pizza. Samantha ate of the pizza, and Anthony ate of the pizza. How much more pizza did Samantha eat than Anthony? Samantha ate more of the pizza than Anthony ate. To find the difference of two numbers, we use subtraction: Samantha ate more of the pizza than Anthony ate. A school had a fund-raising event. The revenue came from three resources: ticket sales, auction sales, and donations. Ticket sales account for of the total revenue; auction sales account for of the total revenue. What fraction of the revenue came from donations? of the revenue came from donations. The revenue consists of parts. The total revenue is like a big ; ticket sales account for ; auction sales account for . To find the missing piece (from donations), we use subtraction. Note that a common denominator of , and is . So of the revenue came from donations. A few years back, a car was purchased for . Today it is worth of its original value. What is the car s current value? The car s current value is . When we use the word of in situations like of , we can translate of into the multiplication symbol. In this problem, to find of , we do: The car is worth now. A few years back, a car was purchased for . Today it is worth of its original value. What is the car s current value? The car s current value is . When we use the word of in situations like of , we can translate of into the multiplication symbol. In this problem, to find of , we do: The car is worth now. The pie chart represents a school s student population. more of the school is white students than black students. By the pie chart, the school has white students, and black students. We will use subtraction to find the difference: A town has residents in total, of which are Asian Americans. How many Asian Americans reside in this town? There are Asian Americans residing in this town. When we use the word of in expressions like of , we can translate of into the multiplication symbol. In this problem, to find of , we compute: There are Asian Americans residing in this town. A company received a grant, and decided to spend of this grant in research and development next year. Out of the money set aside for research and development, will be used to buy new equipment. What fraction of the grant will be used to buy new equipment? of the grant will be used to buy new equipment. For this problem, we are trying to find of . The word of implies multiplication: So of the grant will be used to buy new equipment. A food bank just received kilograms of emergency food. Each family in need is to receive kilograms of food. How many families can be served with the kilograms of food? families can be served with the kilograms of food. We can address this problem as repeatedly taking away kilograms from kilograms, which implies a division problem: families can be served with the kilograms of food. A construction team maintains a -mile-long sewage pipe. Each day, the team can cover of a mile. How many days will it take the team to complete the maintenance of the entire sewage pipe? It will take the team days to complete maintaining the entire sewage pipe. In this problem, the team maintains miles of sewage pipe every day, until they complete all miles. This is like repeatedly taking away miles from miles, a division problem. It will take the team days to complete maintaining the entire sewage pipe. A child is stacking up tiles. Each tile s height is of a centimeter. How many layers of tiles are needed to reach centimeters in total height? To reach the total height of centimeters, layers of tiles are needed. In this problem, each layer of tile is centimeters tall. The child will keep stacking up layers until the total height reaches centimeters. This is like asking how many centimeters are there in centimeters, a division problem. To reach the total height of centimeters, layers of tiles are needed. A restaurant made cups of pudding for a festival. Customers at the festival will be served of a cup of pudding per serving. How many customers can the restaurant serve at the festival with the cups of pudding? The restaurant can serve customers at the festival with the cups of pudding. There are a total of cups of pudding, with each serving being cup. To find how many servings can be served, we need to find how many cups are there in cups, a division problem: The restaurant can serve customers at the festival with the cups of pudding. Shortcut Since each serving has cup, we know each cup has servings. So cups has servings. A piece of lumber in your garage is inches long. A second is inches long. If you lay them end to end, what will the total length be? The total length will be inches. To find the total length in inches, we add the two lengths together. So the total length is inches. A piece of lumber in your garage is inches long. A second is inches long. If you lay them end to end, what will the total length be? The total length will be inches. To find the total length in inches, we add the two lengths together. So the total length is inches. Each page of a book consists of a header, a footer and the middle part. The header is inches in height; the footer is inches in height; and the middle part is inches in height. What is the total height of each page in this book? Use mixed number in your answer if needed. Each page in this book is inches in height. We add up the height of the header, middle part and footer to find the total height of a page: Each page in this book is inches in height. To pave the road on Ellis Street, the crew used tons of cement on the first day, and used tons on the second day. How many tons of cement were used in all? tons of cement were used in all. This problem is obviously an addition problem. To add mixed number, we break each mixed number into an integer and a fraction, and then add up integers and fractions separately. tons of cement were used in all. When driving on a high way, noticed a sign saying exit to Johnstown is miles away, while exit to Jerrystown is miles away. How far is Johnstown from Jerrystown? Johnstown and Jerrystown are miles apart. To find the distance between Johnstown and Jerrystown, we need to find the difference between their distance. This implies subtraction. We could treat subtraction as adding a negative : From here, there are two methods to continue. Method 1 We can split from the integer, and do : Note that we changed to . Method2 We could simply change the whole number into a fraction without splitting : ran more miles than . A cake recipe needs cups of flour. Using this recipe, to bake cakes, how many cups of flour are needed? To bake cakes, cups of flour are needed. Each cake needs cups of flour. To find how many cups of flour are needed to bake cakes, we use multiplication: To bake cakes, cups of flour are needed. Sketching Fractions Sketch a number line showing each fraction. (Be sure to carefully indicate the correct number of equal parts of the whole.) Sketch a number line showing each fraction. (Be sure to carefully indicate the correct number of equal parts of the whole.) Sketch a picture of the product , using a number line or rectangles. Sketch a picture of the sum , using a number line or rectangles. Challenge Given that simplify Since the fractions have the same denominator, we can just add numerators and keep that denominator. Given that simplify Since the fractions don t have the same denominator, we have to make them have like denominators. Multiply the first fraction by Given that simplify Since the fractions don t have the same denominator, we have to make them have like denominators. Multiply the first fraction by " }, { "id": "section-fractions-and-fraction-arithmetic-2-2", @@ -19726,7 +20158,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.3", "title": "", - "body": " The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the second tick from , so it can be represented as , which reduces to . " + "body": " The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the third tick from , so it can be represented as , which reduces to . " }, { "id": "section-fractions-and-fraction-arithmetic-9-1-5", @@ -19735,7 +20167,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.4", "title": "", - "body": " The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the third tick left from , so it can be represented as , which reduces to . " + "body": " The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the second tick left from , so it can be represented as , which reduces to . " }, { "id": "section-fractions-and-fraction-arithmetic-9-1-6", @@ -19744,7 +20176,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.5", "title": "", - "body": " The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the 4th tick from , so it can be represented as , which reduces to . " + "body": " The dot in the graph can be represented by what fraction? On the number line, the segment between and is cut into pieces. The blue dot is marked at the third tick from , so it can be represented as , which reduces to . " }, { "id": "section-fractions-and-fraction-arithmetic-9-1-7", @@ -19780,7 +20212,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.9", "title": "", - "body": " Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, again goes into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. Notice that when all prime factors in the numerator are canceled, we have to add a 1 in the numerator. " + "body": " Reduce the fraction . There are at least two methods to reduce the fraction . Method 1:We find a number that divides into both the numerator , and the denominator . Check the first few prime numbers one by one: In this case, divides into both the numerator and denominator of . We divide into both numbers, and we have: Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, again goes into both the numerator and denominator for . We divide into both numbers, and we have: Finally, we have: Method 2: A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator. " }, { "id": "section-fractions-and-fraction-arithmetic-9-2-5", @@ -20392,7 +20824,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.77", "title": "", - "body": " Kurt walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Kurt walk altogether? Kurt walked a total of of a mile. This is an addition problem. Kurt walked a total of miles altogether. " + "body": " Thanh walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Thanh walk altogether? Thanh walked a total of of a mile. This is an addition problem. Thanh walked a total of miles altogether. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-3", @@ -20401,7 +20833,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.78", "title": "", - "body": " Jerry walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Jerry walk altogether? Jerry walked a total of of a mile. This is an addition problem. Jerry walked a total of miles altogether. " + "body": " Blake walked of a mile in the morning, and then walked of a mile in the afternoon. How far did Blake walk altogether? Blake walked a total of of a mile. This is an addition problem. Blake walked a total of miles altogether. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-4", @@ -20410,7 +20842,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.79", "title": "", - "body": " Samantha and Randi are sharing a pizza. Samantha ate of the pizza, and Randi ate of the pizza. How much of the pizza was eaten in total? They ate of the pizza. This is an addition problem. They ate of the pizza. " + "body": " Diane and Peter are sharing a pizza. Diane ate of the pizza, and Peter ate of the pizza. How much of the pizza was eaten in total? They ate of the pizza. This is an addition problem. They ate of the pizza. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-5", @@ -20437,7 +20869,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.82", "title": "", - "body": " Carl is participating in a running event. In the first hour, he completed of the total distance. After another hour, in total he had completed of the total distance. What fraction of the total distance did Carl complete during the second hour? Carl completed of the distance during the second hour. This is a subtraction problem. Carl completed of the distance during the second hour. " + "body": " Huynh is participating in a running event. In the first hour, he completed of the total distance. After another hour, in total he had completed of the total distance. What fraction of the total distance did Huynh complete during the second hour? Huynh completed of the distance during the second hour. This is a subtraction problem. Huynh completed of the distance during the second hour. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-8", @@ -20464,7 +20896,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.85", "title": "", - "body": " Sherial and Wendy are sharing a pizza. Sherial ate of the pizza, and Wendy ate of the pizza. How much more pizza did Sherial eat than Wendy? Sherial ate more of the pizza than Wendy ate. To find the difference of two numbers, we use subtraction: Sherial ate more of the pizza than Wendy ate. " + "body": " Penelope and Michele are sharing a pizza. Penelope ate of the pizza, and Michele ate of the pizza. How much more pizza did Penelope eat than Michele? Penelope ate more of the pizza than Michele ate. To find the difference of two numbers, we use subtraction: Penelope ate more of the pizza than Michele ate. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-11", @@ -20473,7 +20905,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.86", "title": "", - "body": " Hayden and Maygen are sharing a pizza. Hayden ate of the pizza, and Maygen ate of the pizza. How much more pizza did Hayden eat than Maygen? Hayden ate more of the pizza than Maygen ate. To find the difference of two numbers, we use subtraction: Hayden ate more of the pizza than Maygen ate. " + "body": " Samantha and Anthony are sharing a pizza. Samantha ate of the pizza, and Anthony ate of the pizza. How much more pizza did Samantha eat than Anthony? Samantha ate more of the pizza than Anthony ate. To find the difference of two numbers, we use subtraction: Samantha ate more of the pizza than Anthony ate. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-12", @@ -20518,7 +20950,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.2.7.91", "title": "", - "body": " A town has residents in total, of which are white\/Caucasian Americans. How many white\/Caucasian Americans reside in this town? There are white\/Caucasian Americans residing in this town. When we use the word of in expressions like of , we can translate of into the multiplication symbol. In this problem, to find of , we compute: There are white\/Caucasian Americans residing in this town. " + "body": " A town has residents in total, of which are Asian Americans. How many Asian Americans reside in this town? There are Asian Americans residing in this town. When we use the word of in expressions like of , we can translate of into the multiplication symbol. In this problem, to find of , we compute: There are Asian Americans residing in this town. " }, { "id": "section-fractions-and-fraction-arithmetic-9-6-17", @@ -20689,7 +21121,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "A.3", "title": "Absolute Value and Square Root", - "body": " Absolute Value and Square Root In this section, we will learn the basics of absolute value and square root . These are actions you can do to a given number, often changing the number into something else. Alternative Video Lesson Introduction to Absolute Value absolute value formal definition absolute value The absolute value of a number is the distance between that number and on a number line. For the absolute value of , we write . Let's look at and , the absolute value of and the absolute value of . and a number line marking the values -2 and 2 as being 2 units from the value 0 Since the distance between and on the number line is units, the absolute value of is . We write . Since the distance between and on the number line is also units, the absolute value of is also . We write . Absolute Value absolute value Taking the absolute value of a number results in whatever the positive version of that number is. This is because the real meaning of absolute value is its distance from zero. Calculating Absolute Value Try calculating some absolute values. is units away from on a number line, so Another way to think about this is that the positive version of is is units away from on a number line, so Another way to think about this is that the positive version of is is units away from on a number line, so Another way to think about this is that the positive version of is Absolute Value Does Not Exactly Make Everything Positive Students may see an expression like and incorrectly think it is OK to make everything positive and write . This is incorrect since works out to be , not , as we are actually taking the absolute value of (the equivalent number inside the absolute value). Square Root Facts If you have learned your basic multiplication table, you know: Multiplication table with squares The numbers along the diagonal are special; they are known as perfect squares . perfect squares And for working with square roots, it will be helpful if you can memorize these first few perfect square numbers. Taking a square root is the opposite action of squaring a number. For example, when you square , the result is . So when you take the square root of , the result is . Just knowing that comes about as lets us realize that is the square root of . This is why memorizing the perfect squares from the multiplication table can be so helpful. The notation we use for taking a square root is the radical , radical . For example, the square root of is denoted . And now we know enough to be able to write . Tossing in a few extra special square roots, it's advisable to memorize the following: Calculating Square Roots with a Calculator Most square roots are actually numbers with decimal places that go on forever. Take as an example: Since is between and , then must be somewhere between and . There are no whole numbers between and , so must be some number with decimal places. If the decimal places eventually stopped, then squaring it would give you another number with decimal places that stop further out. But squaring it gives you with no decimal places. So the only possibility is that is a decimal between and that goes on forever. With a calculator, we can see: Actually the decimal will not terminate, and that is why we used the symbol instead of an equals sign. To get we rounded down slightly from the true value of . With a calculator, we can check that , a little shy of . Square Roots of Fractions We can calculate the square root of some fractions by hand, such as . The idea is the same: can you think of a number that you would square to get ? Being familiar with fraction multiplication, we know that and so . Square Roots of Fractions Try calculating some absolute values. Since and then Since and then Since and then Square Root of Negative Numbers Imaginary Numbers Mathematicians imagined a new type of number, neither positive nor negative, that would square to a negative result. But that is beyond the scope of this chapter. Can we find the square root of a negative number, such as ? That would mean that there is some number out there that multiplies by itself to make . Would be positive or negative? Either way, once you square it (multiply it by itself) the result would be positive. So it couldn't possibly square to . So there is no square root of or of any negative number for that matter. If you are confronted with an expression like , or any other square root of a negative number, you can state that there is no real square root or that the result does not exist (as a real number). Review and Warmup Evaluate the expressions. Evaluate the expressions. Absolute Value Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. Solution: Solution: Solution: Evaluate the following. Solution: Solution: Solution: Evaluate the following. The absolute value of measures how many units are between and on the number line, so . The absolute value of measures how many units are between and on the number line, so . The absolute value symbols cannot affect what are outside them, so we have . Similar to Part c, we have . Evaluate the following. The absolute value of measures how many units are between and on the number line, so . The absolute value of measures how many units are between and on the number line, so . The absolute value symbols cannot affect what are outside them, so we have . Similar to Part c, we have . Evaluate the following. Solution: Solution: Solution: Solution: Solution: Evaluate the following. Solution: Solution: Solution: Solution: Solution: Which of the following are square numbers? There may be more than one correct answer. You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are ABD. Which of the following are square numbers? There may be more than one correct answer. You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are DEF. Square Roots Evaluate the following. = = = You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are: Evaluate the following. = = = You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are: Evaluate the following. = = Square root is the opposite operation of square. Since , we have: . The answers are: Note that square root of a negative number does not exist. This is because any number squared is always positive or . Evaluate the following. = = Square root is the opposite operation of square. Since , we have: . The answers are: Note that square root of a negative number does not exist. This is because any number squared is always positive or . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Use a calculator to approximate with a decimal. A calcultaor shows that . Evaluate the following. Use a calculator to approximate with a decimal. A calcultaor shows that . Evaluate the following. . Evaluate the following. . Evaluate the following. . Evaluate the following. . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . Evaluate the following. . Evaluate the following. Evaluate the following. Evaluate the following. = . Because the denominator in this problem is a perfect square, we can easily evaluate the square root: Evaluate the following. = . Because the denominator in this problem is a perfect square, we can easily evaluate the square root: " + "body": " Absolute Value and Square Root In this section, we will learn the basics of absolute value and square root . These are actions you can do to a given number, often changing the number into something else. Alternative Video Lesson Introduction to Absolute Value absolute value formal definition absolute value The absolute value of a number is the distance between that number and on a number line. For the absolute value of , we write . Let's look at and , the absolute value of and the absolute value of . and a number line marking the values -2 and 2 as being 2 units from the value 0 Since the distance between and on the number line is units, the absolute value of is . We write . Since the distance between and on the number line is also units, the absolute value of is also . We write . Absolute Value absolute value Taking the absolute value of a number results in whatever the positive version of that number is. This is because the real meaning of absolute value is its distance from zero. Calculating Absolute Value Try calculating some absolute values. is units away from on a number line, so Another way to think about this is that the positive version of is is units away from on a number line, so Another way to think about this is that the positive version of is is units away from on a number line, so Another way to think about this is that the positive version of is Absolute Value Does Not Exactly Make Everything Positive Students may see an expression like and incorrectly think it is OK to make everything positive and write . This is incorrect since works out to be , not , as we are actually taking the absolute value of (the equivalent number inside the absolute value). Square Root Facts If you have learned your basic multiplication table, you know: Multiplication table with squares The numbers along the diagonal are special; they are known as perfect squares . perfect squares And for working with square roots, it will be helpful if you can memorize these first few perfect square numbers. Taking a square root is the opposite action of squaring a number. For example, when you square , the result is . So when you take the square root of , the result is . Just knowing that comes about as lets us realize that is the square root of . This is why memorizing the perfect squares from the multiplication table can be so helpful. The notation we use for taking a square root is the radical , radical . For example, the square root of is denoted . And now we know enough to be able to write . Tossing in a few extra special square roots, it's advisable to memorize the following: Calculating Square Roots with a Calculator Most square roots are actually numbers with decimal places that go on forever. Take as an example: Since is between and , then must be somewhere between and . There are no whole numbers between and , so must be some number with decimal places. If the decimal places eventually stopped, then squaring it would give you another number with decimal places that stop further out. But squaring it gives you with no decimal places. So the only possibility is that is a decimal between and that goes on forever. With a calculator, we can see: Actually the decimal will not terminate, and that is why we used the symbol instead of an equals sign. To get we rounded down slightly from the true value of . With a calculator, we can check that , a little shy of . Square Roots of Fractions We can calculate the square root of some fractions by hand, such as . The idea is the same: can you think of a number that you would square to get ? Being familiar with fraction multiplication, we know that and so . Square Roots of Fractions Try calculating some absolute values. Since and then Since and then Since and then Square Root of Negative Numbers Imaginary Numbers Mathematicians imagined a new type of number, neither positive nor negative, that would square to a negative result. But that is beyond the scope of this chapter. Can we find the square root of a negative number, such as ? That would mean that there is some number out there that multiplies by itself to make . Would be positive or negative? Either way, once you square it (multiply it by itself) the result would be positive. So it couldn't possibly square to . So there is no square root of or of any negative number for that matter. If you are confronted with an expression like , or any other square root of a negative number, you can state that there is no real square root or that the result does not exist (as a real number). Review and Warmup Evaluate the expressions. Evaluate the expressions. Absolute Value Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. The absolute value of is the distance between and on the number line. In this case, since is units away from 0, the answer is . Evaluate the following. Solution: Solution: Solution: Evaluate the following. Solution: Solution: Solution: Evaluate the following. The absolute value of measures how many units are between and on the number line, so . The absolute value of measures how many units are between and on the number line, so . The absolute value symbols cannot affect what are outside them, so we have . Similar to Part c, we have . Evaluate the following. The absolute value of measures how many units are between and on the number line, so . The absolute value of measures how many units are between and on the number line, so . The absolute value symbols cannot affect what are outside them, so we have . Similar to Part c, we have . Evaluate the following. Solution: Solution: Solution: Solution: Solution: Evaluate the following. Solution: Solution: Solution: Solution: Solution: Which of the following are square numbers? There may be more than one correct answer. You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are DEF. Which of the following are square numbers? There may be more than one correct answer. You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are DEF. Square Roots Evaluate the following. = = = You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are: Evaluate the following. = = = You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are: Evaluate the following. = = Square root is the opposite operation of square. Since , we have: . The answers are: Note that square root of a negative number does not exist. This is because any number squared is always positive or . Evaluate the following. = = Square root is the opposite operation of square. Since , we have: . The answers are: Note that square root of a negative number does not exist. This is because any number squared is always positive or . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Do not use a calculator. = = = First, recognize that is a square number. Since , we have . Next, since , we have . Finally, since , we have . Evaluate the following. Use a calculator to approximate with a decimal. A calcultaor shows that . Evaluate the following. Use a calculator to approximate with a decimal. A calcultaor shows that . Evaluate the following. . Evaluate the following. . Evaluate the following. . Evaluate the following. . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . The square root of a negative number is not a real number . Evaluate the following. . Evaluate the following. . Evaluate the following. Evaluate the following. Evaluate the following. = . Because the denominator in this problem is a perfect square, we can easily evaluate the square root: Evaluate the following. = . Because the denominator in this problem is a perfect square, we can easily evaluate the square root: " }, { "id": "section-absolute-value-and-square-root-2-1", @@ -20923,7 +21355,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.3.6.15", "title": "", - "body": " Which of the following are square numbers? There may be more than one correct answer. You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are ABD. " + "body": " Which of the following are square numbers? There may be more than one correct answer. You can quickly find the first few square numbers on scratch paper: It would be very helpful if you can memorize the above 12 square numbers. So the correct answers are DEF. " }, { "id": "section-absolute-value-and-square-root-8-4", @@ -21175,7 +21607,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "A.4", "title": "Percentages", - "body": " Percentages Percent-related problems arise in everyday life. This section reviews some basic calculations that can be made with percentages. Alternative Video Lesson Converting Percents, Decimals, and English In many situations when translating from English to math, the word of translates as multiplication. Also the word is (and many similar words related to to be ) translates to an equals sign. For example: Here is another example, this time involving a percentage. We know that is of , so we can say: Translate each statement involving percents below into an equation. Define any variables used. (Solving these equations is an exercise). How much is of ? is what percent of ? is of how much money? Each question can be translated from English into a math equation by reading it slowly and looking for the right signals. The word is means about the same thing as the equals sign. How much is a question phrase, and we can let be the unknown amount (in dollars). The word of translates to multiplication, as discussed earlier. So we have: Let be the unknown value. We have: With this setup, is going to be a decimal value ( ) that you would translate into a percentage ( ). Let be the unknown amount (in dollars). We have: Solve each equation from Example . How much is of is . is what percent of is . is of how much money? is . Setting up and Solving Percent Equations An important skill for solving percent-related problems is to boil down a complicated word problem into a simple form like is of . Let's look at some further examples. In Fall 2016, Portland Community College had enrolled students. According to Figure , how many Black students were enrolled at PCC in Fall 2016? Racial breakdown of PCC students in Fall 2016 a pie chart that indicates white students are 68%; Hispanic students are 11%; Asian students are 8%; Black students are 6%; and students of other ethnicities make up 7% After reading this word problem and the chart, we can translate the problem into what is of ? Let be the number of Black students enrolled at PCC in Fall 2016. We can set up and solve the equation: There was not much solving to do, since the variable we wanted to isolate was already isolated. As of Fall 2016, Portland Community College had Black students. Note: this is not likely to be perfectly accurate, because the numbers we started with ( enrolled students and ) appear to be rounded. The bar graph in Figure displays how many students are in each class at a local high school. According to the bar graph, what percentage of the school's student population is freshman? Number of students at a high school by class a bar graph indicating there are 134 freshmen, 103 sophomores, 96 juniors, and 86 seniors The school's total number of students is: With that calculated, we can translate the main question: What percentage of the school's student population is freshman? into: What percent of is ? Using to represent the unknown quantity, we write and solve the equation: Approximately of the school's student population is freshman. When solving equations that do not have context we state the solution set. However, when solving an equation or inequality that arises in an application problem (such as the context of the high school in Example ), it makes more sense to summarize our result with a sentence, using the context of the application. This allows us to communicate the full result, including appropriate units. Carlos just received his monthly paycheck. His gross pay (the amount before taxes and related things are deducted) was , and his total tax and other deductions was . The rest was deposited directly into his checking account. What percent of his gross pay went into his checking account? Train yourself to read the word problem and not try to pick out numbers to substitute into formulas. You may find it helps to read the problem over to yourself three or more times before you attempt to solve it. There are three dollar amounts to discuss in this problem, and many students fall into a trap of using the wrong values in the wrong places. There is the gross pay, the amount that was deducted, and the amount that was deposited. Only two of these have been explicitly written down. We need to use subtraction to find the dollar amount that was deposited: Now, we can translate the main question: What percent of his gross pay went into his checking account? into: What percent of is ? Using to represent the unknown quantity, we write and solve the equation: Approximately of his gross pay went into his checking account. Alexis sells cars for a living, and earns of the dealership s sales profit as commission. In a certain month, she plans to earn in commissions. How much total sales profit does she need to bring in for the dealership? Alexis needs to bring in in sales profit. Be careful that you do not calculate of That might be what a student would do who doesn t thoroughly read the question. If you have ever trained yourself to quickly find numbers in word problems and substitute them into formulas, you must unlearn this. The issue is that is not the dealership s sales profit, and if you mistakenly multiply then makes no sense as an answer to this question. How could Alexis bring in only of sales profit, and earn in commission? We can translate the problem into is of what? Letting be the sales profit for the dealership (in dollars), we can write and solve the equation: To earn in commission, Alexis needs to bring in approximately of sales profit for the dealership. According to e-Literate , the average cost of a new college textbook has been increasing. Find the percentage of increase from 2009 to 2013. Average New Textbook Price from 2009 to 2013 a plot over time indicating average textbook price was $62.00 in 2009; $65.11 in 2010; $68.87 in 2011; $72.11 in 2012; and $79.00 in 2013 The actual amount of increase from 2009 to 2013 was , dollars. We need to answer the question is what percent of ? Note that we are comparing the to , not to . In these situations where one amount is the earlier amount, the earlier original amount is the one that represents . Let represent the percent of increase. We can set up and solve the equation: From 2009 to 2013, the average cost of a new textbook increased by approximately . Last month, a full tank of gas for a car you drive cost you You hear on the news that gas prices have risen by By how much, in dollars, has the cost of a full tank gone up? A full tank of gas now costs more than it did last month. Let represent the amount of increase. We can set up and solve the equation: A full tank now costs more than it did last month. Enrollment at your neighborhood's elementary school two years ago was children. After a increase last year and a decrease this year, what's the new enrollment? It is tempting to think that increasing by and then decreasing by would bring the enrollment right back to where it started. But the decrease applies to the enrollment after it had already increased. So that decrease is going to translate to more students lost than were gained. Using as corresponding to the enrollment from two years ago, the enrollment last year was of that. But then using as corresponding to the enrollment from last year, the enrollment this year was of that. So we can set up and solve the equation We would round and report that enrollment is now students. (The percentage rise and fall of were probably rounded in the first place, which is why we did not end up with a whole number.) Review and Warmup Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal point to the left twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal point to the left twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Basic Percentage Calculation of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence , is the percentage , is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence , is the percentage , is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is about of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is approximately of . Method 2 We will use the percentage formula to solve this problem. This translation from English to math should help you remember the percentage formula. Assume is (as a percent) of . We have: is approximately of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is approximately of . Answer with a percent. is about of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is approximately of . Method 2 We will use the percentage formula to solve this problem. This translation from English to math should help you remember the percentage formula. Assume is (as a percent) of . We have: is approximately of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is approximately of . Applications A town has registered residents. Among them, were Democrats, were Republicans. The rest were Independents. How many registered Independents live in this town? There are registered Independent residents in this town. It s given there are Democrats, Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents: So there are Independents. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are registered Independent residents in this town. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are registered Independent residents in this town. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are registered Independent residents in this town. A town has registered residents. Among them, were Democrats, were Republicans. The rest were Independents. How many registered Independents live in this town? There are registered Independent residents in this town. It s given there are Democrats, Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents: So there are Independents. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are registered Independent residents in this town. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are registered Independent residents in this town. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are registered Independent residents in this town. Sharell is paying a dinner bill of . Sharell plans to pay in tips. How much tip will Sharell pay? Sharell will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Sharell will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Sharell will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Sharell will pay in tip. Lindsay is paying a dinner bill of . Lindsay plans to pay in tips. How much tip will Lindsay pay? Lindsay will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Lindsay will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Lindsay will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Lindsay will pay in tip. Emiliano is paying a dinner bill of . Emiliano plans to pay in tips. How much in total (including bill and tip) will Emiliano pay? Emiliano will pay in total (including bill and tip). First, we need to find how much tip Emiliano paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Emiliano will pay in tip. In total, he will pay . Emiliano will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Emiliano will pay in tip. In total, he will pay . Emiliano will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Emiliano will pay in tip. In total, he will pay . Emiliano will pay in total (including bill and tip). Alyson is paying a dinner bill of . Alyson plans to pay in tips. How much in total (including bill and tip) will Alyson pay? Alyson will pay in total (including bill and tip). First, we need to find how much tip Alyson paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Alyson will pay in tip. In total, she will pay . Alyson will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Alyson will pay in tip. In total, she will pay . Alyson will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Alyson will pay in tip. In total, she will pay . Alyson will pay in total (including bill and tip). A watch s wholesale price was . The retailer marked up the price by . What s the watch s new price (markup price)? The watch s markup price is . First, we need to find the amount of increase in price. It s given that the watch s price was marked up by of its original price, . The problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: The amount of price increase was , so the new price is . So the watch s markup price is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: The amount of price increase was , so the new price is . So the watch s markup price is . Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: The amount of price increase was , so the new price is . So the watch s markup price is . A watch s wholesale price was . The retailer marked up the price by . What s the watch s new price (markup price)? The watch s markup price is . First, we need to find the amount of increase in price. It s given that the watch s price was marked up by of its original price, . The problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: The amount of price increase was , so the new price is . So the watch s markup price is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: The amount of price increase was , so the new price is . So the watch s markup price is . Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: The amount of price increase was , so the new price is . So the watch s markup price is . In the past few seasons basketball games, Connor attempted free throws, and made of them. What percent of free throws did Connor make? Connor made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Connor made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Connor made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Connor made of free throws in the past few seasons. In the past few seasons basketball games, Stephen attempted free throws, and made of them. What percent of free throws did Stephen make? Stephen made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Stephen made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Stephen made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Stephen made of free throws in the past few seasons. A painting is on sale at . Its original price was . What percentage is this off its original price? The painting was off its original price. The price changed from to , implying the markdown was . Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The painting was off its original price. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: The painting was off its original price. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: The painting was off its original price. A painting is on sale at . Its original price was . What percentage is this off its original price? The painting was off its original price. The price changed from to , implying the markdown was . Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The painting was off its original price. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: The painting was off its original price. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: The painting was off its original price. The pie chart represents a collector s collection of signatures from various artists. If the collector has a total of signatures, there are signatures by Sting. We can use subtraction to find the percent of Sting signatures: So there are Sting signatures. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are signatures by Sting in this collection. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are signatures by Sting in this collection. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are signatures by Sting in this collection. The pie chart represents a collector s collection of signatures from various artists. If the collector has a total of signatures, there are signatures by Sting. We can use subtraction to find the percent of Sting signatures: So there are Sting signatures. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are signatures by Sting in this collection. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are signatures by Sting in this collection. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are signatures by Sting in this collection. In the last election, of a county s residents, or people, turned out to vote. How many residents live in this county? This county has residents. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: This county has residents. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: This county has residents. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: This county has residents. In the last election, of a county s residents, or people, turned out to vote. How many residents live in this county? This county has residents. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: This county has residents. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: This county has residents. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: This county has residents. grams of pure alcohol was used to produce a bottle of alcohol solution. What is the weight of the solution in grams? The alcohol solution weighs . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The alcohol solution weighs . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The alcohol solution weighs . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The alcohol solution weighs . grams of pure alcohol was used to produce a bottle of alcohol solution. What is the weight of the solution in grams? The alcohol solution weighs . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The alcohol solution weighs . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The alcohol solution weighs . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The alcohol solution weighs . Hayden paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . Carmen paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . Selena sells cars for a living. Each month, she earns of base pay, plus a certain percentage of commission from her sales. One month, Selena made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Selena earn? Selena earned in commission. Selena s pay is made up of base pay and commission. In that month, Selena earned a total of , with of base pay. This implies Selena earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Selena earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Selena earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Selena earned in commission. Thanh sells cars for a living. Each month, he earns of base pay, plus a certain percentage of commission from his sales. One month, Thanh made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Thanh earn? Thanh earned in commission. Thanh s pay is made up of base pay and commission. In that month, Thanh earned a total of , with of base pay. This implies Thanh earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Thanh earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Thanh earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Thanh earned in commission. The following is a nutrition fact label from a certain macaroni and cheese box. The highlighted row means each serving of macaroni and cheese in this box contains of fat, which is of an average person s daily intake of fat. What s the recommended daily intake of fat for an average person? The recommended daily intake of fat for an average person is . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The recommended daily intake of fat for an average person is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The recommended daily intake of fat for an average person is . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The recommended daily intake of fat for an average person is . The following is a nutrition fact label from a certain macaroni and cheese box. The highlighted row means each serving of macaroni and cheese in this box contains of fat, which is of an average person s daily intake of fat. What s the recommended daily intake of fat for an average person? The recommended daily intake of fat for an average person is . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The recommended daily intake of fat for an average person is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The recommended daily intake of fat for an average person is . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The recommended daily intake of fat for an average person is . A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey. What percent of students ride Bus #1? Approximately of students ride Bus #1. First, we find the total number of students who participated in this survey: Now, this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Approximately of students ride Bus #1. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Approximately of students ride Bus #1. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Approximately of students ride Bus #1. A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey. What percent of students ride Bus #1? Approximately of students ride Bus #1. First, we find the total number of students who participated in this survey: Now, this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Approximately of students ride Bus #1. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Approximately of students ride Bus #1. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Approximately of students ride Bus #1. Scot earned of interest from a mutual fund, which was of his total investment. How much money did Scot invest into this mutual fund? Scot invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Scot invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Scot invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Scot invested in this mutual fund. Sharell earned of interest from a mutual fund, which was of his total investment. How much money did Sharell invest into this mutual fund? Sharell invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Sharell invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Sharell invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Sharell invested in this mutual fund. A town has registered residents. Among them, there are Democrats and Republicans. The rest are Independents. What percentage of registered voters in this town are Independents? In this town, of all registered voters are Independents. Out of registered voters, there are Democrats and Republicans. This implies there are Independents. Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: In this town, of all registered voters are Independents. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: In this town, of all registered voters are Independents. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: In this town, of all registered voters are Independents. A town has registered residents. Among them, there are Democrats and Republicans. The rest are Independents. What percentage of registered voters in this town are Independents? In this town, of all registered voters are Independents. Out of registered voters, there are Democrats and Republicans. This implies there are Independents. Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: In this town, of all registered voters are Independents. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: In this town, of all registered voters are Independents. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: In this town, of all registered voters are Independents. Percent Increase\/Decrease The population of cats in a shelter decreased from to . What is the percentage decrease of the shelter s cat population? The percentage decrease is . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Next, since we started with cats, we need to ask: is what percent of ? Assume is of , so out of corresponds to out of . Method 1 We will use proportion to solve this problem. The percentage decrease of the shelter s cat population is . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let be the unknown percentage, and let the decrease be (percent) of . That means: The percentage decrease of the shelter s cat population is . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the shelter s cat population is . The population of cats in a shelter decreased from to . What is the percentage decrease of the shelter s cat population? The percentage decrease is . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Next, since we started with cats, we need to ask: is what percent of ? Assume is of , so out of corresponds to out of . Method 1 We will use proportion to solve this problem. The percentage decrease of the shelter s cat population is . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let be the unknown percentage, and let the decrease be (percent) of . That means: The percentage decrease of the shelter s cat population is . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the shelter s cat population is . The population of cats in a shelter increased from to . What is the percentage increase of the shelter s cat population? The percentage increase is approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of increase is , which is . Next, since we started with cats, we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage increase of the shelter s cat population is approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the increase be (as a percent) of . That means: The percentage increase of the shelter s cat population is approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage increase is approximately . The percentage increase of the shelter s cat population is approximately . The population of cats in a shelter increased from to . What is the percentage increase of the shelter s cat population? The percentage increase is approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of increase is , which is . Next, since we started with cats, we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage increase of the shelter s cat population is approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the increase be (as a percent) of . That means: The percentage increase of the shelter s cat population is approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage increase is approximately . The percentage increase of the shelter s cat population is approximately . Last year, a small town s population was . This year, the population decreased to . What is the percentage decrease? The percentage decrease of the town s population was approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Since the city s population was initially , we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage decrease of the town s population was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the decrease be (as a percent) of . That means: The percentage decrease of the town s population was approximately . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the town s population was approximately . Last year, a small town s population was . This year, the population decreased to . What is the percentage decrease? The percentage decrease of the town s population was approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Since the city s population was initially , we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage decrease of the town s population was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the decrease be (as a percent) of . That means: The percentage decrease of the town s population was approximately . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the town s population was approximately . Your salary used to be per year. You had to take a pay cut. After the cut, your salary was per year. Then, you earned a raise. After the raise, your salary was per year. Question 1 Your original salary was per year. The amount of cut was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of increase: After the cut, your salary became dollars per year. Question 2 Next, the amount of pay raise was of . Notice that it s incorrect to find of , because the annual salary has changed from to . We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was dollars. We have: Or we can use the percentage formula to find the amount of pay cut: After the raise, your salary became per year. After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut). Your salary used to be per year. You had to take a pay cut. After the cut, your salary was per year. Then, you earned a raise. After the raise, your salary was per year. Question 1 Your original salary was per year. The amount of cut was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of increase: After the cut, your salary became dollars per year. Question 2 Next, the amount of pay raise was of . Notice that it s incorrect to find of , because the annual salary has changed from to . We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was dollars. We have: Or we can use the percentage formula to find the amount of pay cut: After the raise, your salary became per year. After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut). A house was bought two years ago at the price of . Each year, the house s value decreased by . What s the house s value this year? The house s value this year is . The house s value two years ago was . After one year, the house s value decreased by . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After one year, the house s value became . After another year, the house s value decreased by another . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After two years, the house s value became . The house s value this year is . A house was bought two years ago at the price of . Each year, the house s value decreased by . What s the house s value this year? The house s value this year is . The house s value two years ago was . After one year, the house s value decreased by . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After one year, the house s value became . After another year, the house s value decreased by another . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After two years, the house s value became . The house s value this year is . This line graph shows a certain stock's price change over a few days. From 11\/1 to 11\/5, what is the stock price s percentage change? From 11\/1 to 11\/5, the stock price s percentage change was approximately . To calculate the percentage increase\/decrease, first we find the amount of decrease\/decrease, and then we find this increase\/decrease is what percent of the original value. In this problem, the stock s price was on 11\/1, and the price changed to on 11\/5. Their difference was , so the amount of change was . Next, we need to find: is what percent of the original value, . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the change be (as a percent) of . That means: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage change was approximately . From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . This line graph shows a certain stock's price change over a few days. From 11\/1 to 11\/5, what is the stock price s percentage change? From 11\/1 to 11\/5, the stock price s percentage change was approximately . To calculate the percentage increase\/decrease, first we find the amount of decrease\/decrease, and then we find this increase\/decrease is what percent of the original value. In this problem, the stock s price was on 11\/1, and the price changed to on 11\/5. Their difference was , so the amount of change was . Next, we need to find: is what percent of the original value, . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the change be (as a percent) of . That means: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage change was approximately . From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . " + "body": " Percentages Percent-related problems arise in everyday life. This section reviews some basic calculations that can be made with percentages. Alternative Video Lesson Converting Percents, Decimals, and English In many situations when translating from English to math, the word of translates as multiplication. Also the word is (and many similar words related to to be ) translates to an equals sign. For example: Here is another example, this time involving a percentage. We know that is of , so we can say: Translate each statement involving percents below into an equation. Define any variables used. (Solving these equations is an exercise). How much is of ? is what percent of ? is of how much money? Each question can be translated from English into a math equation by reading it slowly and looking for the right signals. The word is means about the same thing as the equals sign. How much is a question phrase, and we can let be the unknown amount (in dollars). The word of translates to multiplication, as discussed earlier. So we have: Let be the unknown value. We have: With this setup, is going to be a decimal value ( ) that you would translate into a percentage ( ). Let be the unknown amount (in dollars). We have: Solve each equation from Example . How much is of is . is what percent of is . is of how much money? is . Setting up and Solving Percent Equations An important skill for solving percent-related problems is to boil down a complicated word problem into a simple form like is of . Let's look at some further examples. In Fall 2016, Portland Community College had enrolled students. According to Figure , how many Black students were enrolled at PCC in Fall 2016? Racial breakdown of PCC students in Fall 2016 a pie chart that indicates white students are 68%; Hispanic students are 11%; Asian students are 8%; Black students are 6%; and students of other ethnicities make up 7% After reading this word problem and the chart, we can translate the problem into what is of ? Let be the number of Black students enrolled at PCC in Fall 2016. We can set up and solve the equation: There was not much solving to do, since the variable we wanted to isolate was already isolated. As of Fall 2016, Portland Community College had Black students. Note: this is not likely to be perfectly accurate, because the numbers we started with ( enrolled students and ) appear to be rounded. The bar graph in Figure displays how many students are in each class at a local high school. According to the bar graph, what percentage of the school's student population is freshman? Number of students at a high school by class a bar graph indicating there are 134 freshmen, 103 sophomores, 96 juniors, and 86 seniors The school's total number of students is: With that calculated, we can translate the main question: What percentage of the school's student population is freshman? into: What percent of is ? Using to represent the unknown quantity, we write and solve the equation: Approximately of the school's student population is freshman. When solving equations that do not have context we state the solution set. However, when solving an equation or inequality that arises in an application problem (such as the context of the high school in Example ), it makes more sense to summarize our result with a sentence, using the context of the application. This allows us to communicate the full result, including appropriate units. Carlos just received his monthly paycheck. His gross pay (the amount before taxes and related things are deducted) was , and his total tax and other deductions was . The rest was deposited directly into his checking account. What percent of his gross pay went into his checking account? Train yourself to read the word problem and not try to pick out numbers to substitute into formulas. You may find it helps to read the problem over to yourself three or more times before you attempt to solve it. There are three dollar amounts to discuss in this problem, and many students fall into a trap of using the wrong values in the wrong places. There is the gross pay, the amount that was deducted, and the amount that was deposited. Only two of these have been explicitly written down. We need to use subtraction to find the dollar amount that was deposited: Now, we can translate the main question: What percent of his gross pay went into his checking account? into: What percent of is ? Using to represent the unknown quantity, we write and solve the equation: Approximately of his gross pay went into his checking account. Alexis sells cars for a living, and earns of the dealership s sales profit as commission. In a certain month, she plans to earn in commissions. How much total sales profit does she need to bring in for the dealership? Alexis needs to bring in in sales profit. Be careful that you do not calculate of That might be what a student would do who doesn t thoroughly read the question. If you have ever trained yourself to quickly find numbers in word problems and substitute them into formulas, you must unlearn this. The issue is that is not the dealership s sales profit, and if you mistakenly multiply then makes no sense as an answer to this question. How could Alexis bring in only of sales profit, and earn in commission? We can translate the problem into is of what? Letting be the sales profit for the dealership (in dollars), we can write and solve the equation: To earn in commission, Alexis needs to bring in approximately of sales profit for the dealership. According to e-Literate , the average cost of a new college textbook has been increasing. Find the percentage of increase from 2009 to 2013. Average New Textbook Price from 2009 to 2013 a plot over time indicating average textbook price was $62.00 in 2009; $65.11 in 2010; $68.87 in 2011; $72.11 in 2012; and $79.00 in 2013 The actual amount of increase from 2009 to 2013 was , dollars. We need to answer the question is what percent of ? Note that we are comparing the to , not to . In these situations where one amount is the earlier amount, the earlier original amount is the one that represents . Let represent the percent of increase. We can set up and solve the equation: From 2009 to 2013, the average cost of a new textbook increased by approximately . Last month, a full tank of gas for a car you drive cost you You hear on the news that gas prices have risen by By how much, in dollars, has the cost of a full tank gone up? A full tank of gas now costs more than it did last month. Let represent the amount of increase. We can set up and solve the equation: A full tank now costs more than it did last month. Enrollment at your neighborhood's elementary school two years ago was children. After a increase last year and a decrease this year, what's the new enrollment? It is tempting to think that increasing by and then decreasing by would bring the enrollment right back to where it started. But the decrease applies to the enrollment after it had already increased. So that decrease is going to translate to more students lost than were gained. Using as corresponding to the enrollment from two years ago, the enrollment last year was of that. But then using as corresponding to the enrollment from last year, the enrollment this year was of that. So we can set up and solve the equation We would round and report that enrollment is now students. (The percentage rise and fall of were probably rounded in the first place, which is why we did not end up with a whole number.) Review and Warmup Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal point to the left twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal point to the left twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following decimals as percentages. To change a decimal to a percentage, we move the decimal point to the right twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Write the following percentages as decimals. To change a percentage to a decimal, we move the decimal to the left twice. Basic Percentage Calculation of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: of is . Method 3 In the sentence of is what, what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence , is the percentage , is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence , is the percentage , is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . of is . Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: of is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: of is . Method 3 In the sentence of what is , is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: of is . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is of . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: is of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is of . Answer with a percent. is about of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is approximately of . Method 2 We will use the percentage formula to solve this problem. This translation from English to math should help you remember the percentage formula. Assume is (as a percent) of . We have: is approximately of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is approximately of . Answer with a percent. is about of . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: is approximately of . Method 2 We will use the percentage formula to solve this problem. This translation from English to math should help you remember the percentage formula. Assume is (as a percent) of . We have: is approximately of . Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: is approximately of . Applications A town has registered residents. Among them, were Democrats, were Republicans. The rest were Independents. How many registered Independents live in this town? There are registered Independent residents in this town. It s given there are Democrats, Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents: So there are Independents. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are registered Independent residents in this town. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are registered Independent residents in this town. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are registered Independent residents in this town. A town has registered residents. Among them, were Democrats, were Republicans. The rest were Independents. How many registered Independents live in this town? There are registered Independent residents in this town. It s given there are Democrats, Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents: So there are Independents. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are registered Independent residents in this town. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are registered Independent residents in this town. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are registered Independent residents in this town. Eric is paying a dinner bill of . Eric plans to pay in tips. How much tip will Eric pay? Eric will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Eric will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Eric will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Eric will pay in tip. Amber is paying a dinner bill of . Amber plans to pay in tips. How much tip will Amber pay? Amber will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Amber will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Amber will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Amber will pay in tip. Phil is paying a dinner bill of . Phil plans to pay in tips. How much in total (including bill and tip) will Phil pay? Phil will pay in total (including bill and tip). First, we need to find how much tip Phil paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Phil will pay in tip. In total, he will pay . Phil will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Phil will pay in tip. In total, he will pay . Phil will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Phil will pay in tip. In total, he will pay . Phil will pay in total (including bill and tip). Kim is paying a dinner bill of . Kim plans to pay in tips. How much in total (including bill and tip) will Kim pay? Kim will pay in total (including bill and tip). First, we need to find how much tip Kim paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Kim will pay in tip. In total, she will pay . Kim will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Kim will pay in tip. In total, she will pay . Kim will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Kim will pay in tip. In total, she will pay . Kim will pay in total (including bill and tip). A watch s wholesale price was . The retailer marked up the price by . What s the watch s new price (markup price)? The watch s markup price is . First, we need to find the amount of increase in price. It s given that the watch s price was marked up by of its original price, . The problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: The amount of price increase was , so the new price is . So the watch s markup price is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: The amount of price increase was , so the new price is . So the watch s markup price is . Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: The amount of price increase was , so the new price is . So the watch s markup price is . A watch s wholesale price was . The retailer marked up the price by . What s the watch s new price (markup price)? The watch s markup price is . First, we need to find the amount of increase in price. It s given that the watch s price was marked up by of its original price, . The problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: The amount of price increase was , so the new price is . So the watch s markup price is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: The amount of price increase was , so the new price is . So the watch s markup price is . Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: The amount of price increase was , so the new price is . So the watch s markup price is . In the past few seasons basketball games, Michele attempted free throws, and made of them. What percent of free throws did Michele make? Michele made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Michele made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Michele made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Michele made of free throws in the past few seasons. In the past few seasons basketball games, Grant attempted free throws, and made of them. What percent of free throws did Grant make? Grant made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Grant made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Grant made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Grant made of free throws in the past few seasons. A painting is on sale at . Its original price was . What percentage is this off its original price? The painting was off its original price. The price changed from to , implying the markdown was . Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The painting was off its original price. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: The painting was off its original price. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: The painting was off its original price. A painting is on sale at . Its original price was . What percentage is this off its original price? The painting was off its original price. The price changed from to , implying the markdown was . Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The painting was off its original price. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: The painting was off its original price. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: The painting was off its original price. The pie chart represents a collector s collection of signatures from various artists. If the collector has a total of signatures, there are signatures by Sting. We can use subtraction to find the percent of Sting signatures: So there are Sting signatures. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are signatures by Sting in this collection. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are signatures by Sting in this collection. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are signatures by Sting in this collection. The pie chart represents a collector s collection of signatures from various artists. If the collector has a total of signatures, there are signatures by Sting. We can use subtraction to find the percent of Sting signatures: So there are Sting signatures. Now this problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: There are signatures by Sting in this collection. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: There are signatures by Sting in this collection. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: There are signatures by Sting in this collection. In the last election, of a county s residents, or people, turned out to vote. How many residents live in this county? This county has residents. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: This county has residents. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: This county has residents. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: This county has residents. In the last election, of a county s residents, or people, turned out to vote. How many residents live in this county? This county has residents. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: This county has residents. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: This county has residents. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: This county has residents. grams of pure alcohol was used to produce a bottle of alcohol solution. What is the weight of the solution in grams? The alcohol solution weighs . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The alcohol solution weighs . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The alcohol solution weighs . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The alcohol solution weighs . grams of pure alcohol was used to produce a bottle of alcohol solution. What is the weight of the solution in grams? The alcohol solution weighs . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The alcohol solution weighs . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The alcohol solution weighs . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The alcohol solution weighs . Shane paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . Kenji paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . Diane sells cars for a living. Each month, she earns of base pay, plus a certain percentage of commission from her sales. One month, Diane made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Diane earn? Diane earned in commission. Diane s pay is made up of base pay and commission. In that month, Diane earned a total of , with of base pay. This implies Diane earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Diane earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Diane earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Diane earned in commission. Emiliano sells cars for a living. Each month, he earns of base pay, plus a certain percentage of commission from his sales. One month, Emiliano made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Emiliano earn? Emiliano earned in commission. Emiliano s pay is made up of base pay and commission. In that month, Emiliano earned a total of , with of base pay. This implies Emiliano earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Emiliano earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Emiliano earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Emiliano earned in commission. The following is a nutrition fact label from a certain macaroni and cheese box. The highlighted row means each serving of macaroni and cheese in this box contains of fat, which is of an average person s daily intake of fat. What s the recommended daily intake of fat for an average person? The recommended daily intake of fat for an average person is . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The recommended daily intake of fat for an average person is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The recommended daily intake of fat for an average person is . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The recommended daily intake of fat for an average person is . The following is a nutrition fact label from a certain macaroni and cheese box. The highlighted row means each serving of macaroni and cheese in this box contains of fat, which is of an average person s daily intake of fat. What s the recommended daily intake of fat for an average person? The recommended daily intake of fat for an average person is . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The recommended daily intake of fat for an average person is . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The recommended daily intake of fat for an average person is . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The recommended daily intake of fat for an average person is . A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey. What percent of students ride Bus #1? Approximately of students ride Bus #1. First, we find the total number of students who participated in this survey: Now, this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Approximately of students ride Bus #1. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Approximately of students ride Bus #1. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Approximately of students ride Bus #1. A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey. What percent of students ride Bus #1? Approximately of students ride Bus #1. First, we find the total number of students who participated in this survey: Now, this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Approximately of students ride Bus #1. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Approximately of students ride Bus #1. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Approximately of students ride Bus #1. Irene earned of interest from a mutual fund, which was of his total investment. How much money did Irene invest into this mutual fund? Irene invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Irene invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Irene invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Irene invested in this mutual fund. Farshad earned of interest from a mutual fund, which was of his total investment. How much money did Farshad invest into this mutual fund? Farshad invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Farshad invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Farshad invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Farshad invested in this mutual fund. A town has registered residents. Among them, there are Democrats and Republicans. The rest are Independents. What percentage of registered voters in this town are Independents? In this town, of all registered voters are Independents. Out of registered voters, there are Democrats and Republicans. This implies there are Independents. Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: In this town, of all registered voters are Independents. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: In this town, of all registered voters are Independents. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: In this town, of all registered voters are Independents. A town has registered residents. Among them, there are Democrats and Republicans. The rest are Independents. What percentage of registered voters in this town are Independents? In this town, of all registered voters are Independents. Out of registered voters, there are Democrats and Republicans. This implies there are Independents. Now this problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: In this town, of all registered voters are Independents. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: In this town, of all registered voters are Independents. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: In this town, of all registered voters are Independents. Percent Increase\/Decrease The population of cats in a shelter decreased from to . What is the percentage decrease of the shelter s cat population? The percentage decrease is . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Next, since we started with cats, we need to ask: is what percent of ? Assume is of , so out of corresponds to out of . Method 1 We will use proportion to solve this problem. The percentage decrease of the shelter s cat population is . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let be the unknown percentage, and let the decrease be (percent) of . That means: The percentage decrease of the shelter s cat population is . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the shelter s cat population is . The population of cats in a shelter decreased from to . What is the percentage decrease of the shelter s cat population? The percentage decrease is . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Next, since we started with cats, we need to ask: is what percent of ? Assume is of , so out of corresponds to out of . Method 1 We will use proportion to solve this problem. The percentage decrease of the shelter s cat population is . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let be the unknown percentage, and let the decrease be (percent) of . That means: The percentage decrease of the shelter s cat population is . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the shelter s cat population is . The population of cats in a shelter increased from to . What is the percentage increase of the shelter s cat population? The percentage increase is approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of increase is , which is . Next, since we started with cats, we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage increase of the shelter s cat population is approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the increase be (as a percent) of . That means: The percentage increase of the shelter s cat population is approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage increase is approximately . The percentage increase of the shelter s cat population is approximately . The population of cats in a shelter increased from to . What is the percentage increase of the shelter s cat population? The percentage increase is approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of increase is , which is . Next, since we started with cats, we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage increase of the shelter s cat population is approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the increase be (as a percent) of . That means: The percentage increase of the shelter s cat population is approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage increase is approximately . The percentage increase of the shelter s cat population is approximately . Last year, a small town s population was . This year, the population decreased to . What is the percentage decrease? The percentage decrease of the town s population was approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Since the city s population was initially , we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage decrease of the town s population was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the decrease be (as a percent) of . That means: The percentage decrease of the town s population was approximately . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the town s population was approximately . Last year, a small town s population was . This year, the population decreased to . What is the percentage decrease? The percentage decrease of the town s population was approximately . To calculate the percentage increase\/decrease, first we find the amount of increase\/decrease by doing a simple subtraction calculation, and then we find the percentage increase\/decrease. In this problem, the amount of decrease is . Since the city s population was initially , we need to ask: is what percent of ? Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The percentage decrease of the town s population was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the decrease be (as a percent) of . That means: The percentage decrease of the town s population was approximately . Method 3 We first divide the new number by the original number : So the new number is of the original number, implying the percentage decrease is . The percentage decrease of the town s population was approximately . Your salary used to be per year. You had to take a pay cut. After the cut, your salary was per year. Then, you earned a raise. After the raise, your salary was per year. Question 1 Your original salary was per year. The amount of cut was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of increase: After the cut, your salary became dollars per year. Question 2 Next, the amount of pay raise was of . Notice that it s incorrect to find of , because the annual salary has changed from to . We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was dollars. We have: Or we can use the percentage formula to find the amount of pay cut: After the raise, your salary became per year. After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut). Your salary used to be per year. You had to take a pay cut. After the cut, your salary was per year. Then, you earned a raise. After the raise, your salary was per year. Question 1 Your original salary was per year. The amount of cut was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of increase: After the cut, your salary became dollars per year. Question 2 Next, the amount of pay raise was of . Notice that it s incorrect to find of , because the annual salary has changed from to . We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was dollars. We have: Or we can use the percentage formula to find the amount of pay cut: After the raise, your salary became per year. After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut). A house was bought two years ago at the price of . Each year, the house s value decreased by . What s the house s value this year? The house s value this year is . The house s value two years ago was . After one year, the house s value decreased by . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After one year, the house s value became . After another year, the house s value decreased by another . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After two years, the house s value became . The house s value this year is . A house was bought two years ago at the price of . Each year, the house s value decreased by . What s the house s value this year? The house s value this year is . The house s value two years ago was . After one year, the house s value decreased by . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After one year, the house s value became . After another year, the house s value decreased by another . The amount of decrease was of . There are two methods to find this amount. We can use proportion. Let dollars be the amount of decrease, then we have: Or we can use the percentage formula to find the amount of decrease: After two years, the house s value became . The house s value this year is . This line graph shows a certain stock's price change over a few days. From 11\/1 to 11\/5, what is the stock price s percentage change? From 11\/1 to 11\/5, the stock price s percentage change was approximately . To calculate the percentage increase\/decrease, first we find the amount of decrease\/decrease, and then we find this increase\/decrease is what percent of the original value. In this problem, the stock s price was on 11\/1, and the price changed to on 11\/5. Their difference was , so the amount of change was . Next, we need to find: is what percent of the original value, . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the change be (as a percent) of . That means: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage change was approximately . From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . This line graph shows a certain stock's price change over a few days. From 11\/1 to 11\/5, what is the stock price s percentage change? From 11\/1 to 11\/5, the stock price s percentage change was approximately . To calculate the percentage increase\/decrease, first we find the amount of decrease\/decrease, and then we find this increase\/decrease is what percent of the original value. In this problem, the stock s price was on 11\/1, and the price changed to on 11\/5. Their difference was , so the amount of change was . Next, we need to find: is what percent of the original value, . Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Let the change be (as a percent) of . That means: From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . Method 3 We first divide the new number by the original number : So the new number is approximately of the original number, implying the percentage change was approximately . From 11\/1 to 11\/5, the percentage change in the stock s value was approximately . " }, { "id": "section-percentages-2-2", @@ -21571,7 +22003,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.33", "title": "", - "body": " Sharell is paying a dinner bill of . Sharell plans to pay in tips. How much tip will Sharell pay? Sharell will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Sharell will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Sharell will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Sharell will pay in tip. " + "body": " Eric is paying a dinner bill of . Eric plans to pay in tips. How much tip will Eric pay? Eric will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Eric will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Eric will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Eric will pay in tip. " }, { "id": "section-percentages-5-5-5", @@ -21580,7 +22012,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.34", "title": "", - "body": " Lindsay is paying a dinner bill of . Lindsay plans to pay in tips. How much tip will Lindsay pay? Lindsay will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Lindsay will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Lindsay will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Lindsay will pay in tip. " + "body": " Amber is paying a dinner bill of . Amber plans to pay in tips. How much tip will Amber pay? Amber will pay in tip. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Amber will pay in tip. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Amber will pay in tip. Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Amber will pay in tip. " }, { "id": "section-percentages-5-5-6", @@ -21589,7 +22021,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.35", "title": "", - "body": " Emiliano is paying a dinner bill of . Emiliano plans to pay in tips. How much in total (including bill and tip) will Emiliano pay? Emiliano will pay in total (including bill and tip). First, we need to find how much tip Emiliano paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Emiliano will pay in tip. In total, he will pay . Emiliano will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Emiliano will pay in tip. In total, he will pay . Emiliano will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Emiliano will pay in tip. In total, he will pay . Emiliano will pay in total (including bill and tip). " + "body": " Phil is paying a dinner bill of . Phil plans to pay in tips. How much in total (including bill and tip) will Phil pay? Phil will pay in total (including bill and tip). First, we need to find how much tip Phil paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Phil will pay in tip. In total, he will pay . Phil will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Phil will pay in tip. In total, he will pay . Phil will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Phil will pay in tip. In total, he will pay . Phil will pay in total (including bill and tip). " }, { "id": "section-percentages-5-5-7", @@ -21598,7 +22030,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.36", "title": "", - "body": " Alyson is paying a dinner bill of . Alyson plans to pay in tips. How much in total (including bill and tip) will Alyson pay? Alyson will pay in total (including bill and tip). First, we need to find how much tip Alyson paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Alyson will pay in tip. In total, she will pay . Alyson will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Alyson will pay in tip. In total, she will pay . Alyson will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Alyson will pay in tip. In total, she will pay . Alyson will pay in total (including bill and tip). " + "body": " Kim is paying a dinner bill of . Kim plans to pay in tips. How much in total (including bill and tip) will Kim pay? Kim will pay in total (including bill and tip). First, we need to find how much tip Kim paid. This problem can be boiled down to this question: What is of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume of is , so out of corresponds to out of . We will write and solve the proportion: Kim will pay in tip. In total, she will pay . Kim will pay in total (including bill and tip). Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: What is of ? Assume is of . We have: Kim will pay in tip. In total, she will pay . Kim will pay in total (including bill and tip). Method 3 In the sentence What is of , what is the percentage , is the rate , is the base (following the word of ). By the formula , we do a multiplication to solve the problem: Kim will pay in tip. In total, she will pay . Kim will pay in total (including bill and tip). " }, { "id": "section-percentages-5-5-8", @@ -21625,7 +22057,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.39", "title": "", - "body": " In the past few seasons basketball games, Connor attempted free throws, and made of them. What percent of free throws did Connor make? Connor made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Connor made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Connor made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Connor made of free throws in the past few seasons. " + "body": " In the past few seasons basketball games, Michele attempted free throws, and made of them. What percent of free throws did Michele make? Michele made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Michele made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Michele made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Michele made of free throws in the past few seasons. " }, { "id": "section-percentages-5-5-11", @@ -21634,7 +22066,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.40", "title": "", - "body": " In the past few seasons basketball games, Stephen attempted free throws, and made of them. What percent of free throws did Stephen make? Stephen made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Stephen made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Stephen made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Stephen made of free throws in the past few seasons. " + "body": " In the past few seasons basketball games, Grant attempted free throws, and made of them. What percent of free throws did Grant make? Grant made of free throws in the past few seasons. This problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Grant made of free throws in the past few seasons. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Grant made of free throws in the past few seasons. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Grant made of free throws in the past few seasons. " }, { "id": "section-percentages-5-5-12", @@ -21715,7 +22147,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.49", "title": "", - "body": " Hayden paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . " + "body": " Shane paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . " }, { "id": "section-percentages-5-5-21", @@ -21724,7 +22156,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.50", "title": "", - "body": " Carmen paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . " + "body": " Kenji paid a dinner and left , or , in tips. How much was the original bill (without counting the tip)? The original bill (not including the tip) was . This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: The original bill (not including the tip) was . Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: The original bill (not including the tip) was . Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: The original bill (not including the tip) was . " }, { "id": "section-percentages-5-5-22", @@ -21733,7 +22165,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.51", "title": "", - "body": " Selena sells cars for a living. Each month, she earns of base pay, plus a certain percentage of commission from her sales. One month, Selena made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Selena earn? Selena earned in commission. Selena s pay is made up of base pay and commission. In that month, Selena earned a total of , with of base pay. This implies Selena earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Selena earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Selena earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Selena earned in commission. " + "body": " Diane sells cars for a living. Each month, she earns of base pay, plus a certain percentage of commission from her sales. One month, Diane made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Diane earn? Diane earned in commission. Diane s pay is made up of base pay and commission. In that month, Diane earned a total of , with of base pay. This implies Diane earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Diane earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Diane earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Diane earned in commission. " }, { "id": "section-percentages-5-5-23", @@ -21742,7 +22174,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.52", "title": "", - "body": " Thanh sells cars for a living. Each month, he earns of base pay, plus a certain percentage of commission from his sales. One month, Thanh made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Thanh earn? Thanh earned in commission. Thanh s pay is made up of base pay and commission. In that month, Thanh earned a total of , with of base pay. This implies Thanh earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Thanh earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Thanh earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Thanh earned in commission. " + "body": " Emiliano sells cars for a living. Each month, he earns of base pay, plus a certain percentage of commission from his sales. One month, Emiliano made in sales, and earned a total of in that month (including base pay and commission). What percent commission did Emiliano earn? Emiliano earned in commission. Emiliano s pay is made up of base pay and commission. In that month, Emiliano earned a total of , with of base pay. This implies Emiliano earned in commission, out of in sales. Now the problem can be boiled down to this question: is what percent of ? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Emiliano earned in commission. Method 2 We will use the percentage formula to solve this problem. This translation from English to may should help you remember the percentage formula. Assume is (as a percent) of . We have: Emiliano earned in commission. Method 3 In the sentence is what percent of , is the percentage , what percent is the rate , is the base (following the word of ). By the formula , we do a division to solve the problem: Emiliano earned in commission. " }, { "id": "section-percentages-5-5-24", @@ -21787,7 +22219,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.57", "title": "", - "body": " Scot earned of interest from a mutual fund, which was of his total investment. How much money did Scot invest into this mutual fund? Scot invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Scot invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Scot invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Scot invested in this mutual fund. " + "body": " Irene earned of interest from a mutual fund, which was of his total investment. How much money did Irene invest into this mutual fund? Irene invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Irene invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Irene invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Irene invested in this mutual fund. " }, { "id": "section-percentages-5-5-29", @@ -21796,7 +22228,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.4.3.58", "title": "", - "body": " Sharell earned of interest from a mutual fund, which was of his total investment. How much money did Sharell invest into this mutual fund? Sharell invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Sharell invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Sharell invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Sharell invested in this mutual fund. " + "body": " Farshad earned of interest from a mutual fund, which was of his total investment. How much money did Farshad invest into this mutual fund? Farshad invested in this mutual fund. This problem can be boiled down to this question: is of what? We will show multiple methods to solve this problem. Method 1 We will use proportion to solve this problem. Assume is of , so out of corresponds to out of . We will write and solve the proportion: Farshad invested in this mutual fund. Method 2 We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula. The question is: is of what? Assume is of . We have: Farshad invested in this mutual fund. Method 3 In the sentence is of what, is the percentage , is the rate , what is the base (following the word of ). By the formula , we do a division to solve the problem: Farshad invested in this mutual fund. " }, { "id": "section-percentages-5-5-30", @@ -21961,9 +22393,9 @@ var ptx_lunr_docs = [ "body": " 3 is doubled, then squared: a square whose width is labeled as 2*3 yd = 6 yd wide and whose height is also 6 yd; the square's area is labeled as being area 36 yd^2 " }, { - "id": "section-order-of-operations-3-9", + "id": "order-of-operations-checkpoint-aminusgroupgroupbplusctimesd", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-3-9", + "url": "section-order-of-operations.html#order-of-operations-checkpoint-aminusgroupgroupbplusctimesd", "type": "Checkpoint", "number": "A.5.4", "title": "", @@ -22006,27 +22438,27 @@ var ptx_lunr_docs = [ "body": " There are several different ways to write multiplication. We can use the symbols , , and to mean multiplication. We can also write two things right next to each other with no symbol in between them to mean multiplication. That is what is happening in Item , where the is written right next to the with no symbol in between. multiplication explicit multiplication implicit Using a symbol for multiplication is called explicit multiplication and not writing any symbol at all is called implicit multiplication . For this textbook, explicit and implicit multiplication have the same priority in the order of operations. However there are some conventions out in the real world where implicit multiplication has a higher priority in the order of operations than explicit multiplication. You may have seen memes with expressions like that play on how the real world has more than one convention for the order of operations. " }, { - "id": "section-order-of-operations-4-9", + "id": "order-of-operations-checkpoint-stepbystep", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-4-9", + "url": "section-order-of-operations.html#order-of-operations-checkpoint-stepbystep", "type": "Checkpoint", "number": "A.5.8", "title": "Practice with order of operations.", "body": "Practice with order of operations Simplify this expression one step at a time, using the order of operations. " }, { - "id": "section-order-of-operations-4-10", + "id": "order-of-operations-checkpoint-adivgroupgroupbdivcplusdpluse", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-4-10", + "url": "section-order-of-operations.html#order-of-operations-checkpoint-adivgroupgroupbdivcplusdpluse", "type": "Checkpoint", "number": "A.5.9", "title": "", "body": " Simplify With the expression we ll simplify what s inside the parentheses according to the order of operations, and then take divided by that expression as our last step: " }, { - "id": "section-order-of-operations-4-11", + "id": "order-of-operations-checkpoint-groupaminusgroupbsquareddivgroupcminusdcubed", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-4-11", + "url": "section-order-of-operations.html#order-of-operations-checkpoint-groupaminusgroupbsquareddivgroupcminusdcubed", "type": "Checkpoint", "number": "A.5.10", "title": "", @@ -22042,18 +22474,18 @@ var ptx_lunr_docs = [ "body": " Use the order of operations to simplify the following expressions. . The absolute value bars group the so we must do that subtraction first. Then we take the absolute value and continue: It would be a mistake to subtract first, because that is multiplied by the . So subtracting would violate the order of operations. . The radical is grouping , which must be simplified first. Then we take the square root and continue: . We recognize that the fraction bar is creating two groups. We should simplify the numerator and denominator separately according to the order of operations, and proceed from there: " }, { - "id": "section-order-of-operations-5-6", + "id": "order-of-operations-checkpoint-impliedgrouping", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-5-6", + "url": "section-order-of-operations.html#order-of-operations-checkpoint-impliedgrouping", "type": "Checkpoint", "number": "A.5.12", "title": "Implied Grouping.", "body": "Implied Grouping Use the order of operations to evaluate We start by identifying the innermost, highest priority operations: " }, { - "id": "section-order-of-operations-5-7", + "id": "order-of-operations-checkpoint-impliedgrouping2", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-5-7", + "url": "section-order-of-operations.html#order-of-operations-checkpoint-impliedgrouping2", "type": "Checkpoint", "number": "A.5.13", "title": "", @@ -22078,549 +22510,549 @@ var ptx_lunr_docs = [ "body": "Negating and Raising to Powers Compute the following. In each part, the first expression asks you to exponentiate and then negate the result. The second expression has a negative number raised to a power. and and and and and and " }, { - "id": "section-order-of-operations-7-1-2-2", + "id": "order-of-operations-aminusbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-2", + "url": "section-order-of-operations.html#order-of-operations-aminusbplusc", "type": "Exercise", "number": "A.5.5.1", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-3", + "id": "order-of-operations-aplusbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-3", + "url": "section-order-of-operations.html#order-of-operations-aplusbtimesc", "type": "Exercise", "number": "A.5.5.2", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-4", + "id": "order-of-operations-aminusbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-4", + "url": "section-order-of-operations.html#order-of-operations-aminusbtimesc", "type": "Exercise", "number": "A.5.5.3", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-5", + "id": "order-of-operations-aplusbc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-5", + "url": "section-order-of-operations.html#order-of-operations-aplusbc", "type": "Exercise", "number": "A.5.5.4", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-6", + "id": "order-of-operations-aminusbc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-6", + "url": "section-order-of-operations.html#order-of-operations-aminusbc", "type": "Exercise", "number": "A.5.5.5", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-7", + "id": "order-of-operations-aplusbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-7", + "url": "section-order-of-operations.html#order-of-operations-aplusbdivc", "type": "Exercise", "number": "A.5.5.6", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-8", + "id": "order-of-operations-aminusbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-8", + "url": "section-order-of-operations.html#order-of-operations-aminusbdivc", "type": "Exercise", "number": "A.5.5.7", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-9", + "id": "order-of-operations-aplusbslashc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-9", + "url": "section-order-of-operations.html#order-of-operations-aplusbslashc", "type": "Exercise", "number": "A.5.5.8", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-10", + "id": "order-of-operations-aminusbslashc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-10", + "url": "section-order-of-operations.html#order-of-operations-aminusbslashc", "type": "Exercise", "number": "A.5.5.9", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-11", + "id": "order-of-operations-adivbslashc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-11", + "url": "section-order-of-operations.html#order-of-operations-adivbslashc", "type": "Exercise", "number": "A.5.5.10", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-12", + "id": "order-of-operations-aplusbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-12", + "url": "section-order-of-operations.html#order-of-operations-aplusbtoc", "type": "Exercise", "number": "A.5.5.11", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-13", + "id": "order-of-operations-aminusbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-13", + "url": "section-order-of-operations.html#order-of-operations-aminusbtoc", "type": "Exercise", "number": "A.5.5.12", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-14", + "id": "order-of-operations-atimesbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-14", + "url": "section-order-of-operations.html#order-of-operations-atimesbtoc", "type": "Exercise", "number": "A.5.5.13", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-15", + "id": "order-of-operations-adivbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-15", + "url": "section-order-of-operations.html#order-of-operations-adivbtoc", "type": "Exercise", "number": "A.5.5.14", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-16", + "id": "order-of-operations-aslashbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-16", + "url": "section-order-of-operations.html#order-of-operations-aslashbtoc", "type": "Exercise", "number": "A.5.5.15", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-2-17", + "id": "order-of-operations-atimesbcaretc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-2-17", + "url": "section-order-of-operations.html#order-of-operations-atimesbcaretc", "type": "Exercise", "number": "A.5.5.16", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-2", + "id": "order-of-operations-aminusgroupbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-2", + "url": "section-order-of-operations.html#order-of-operations-aminusgroupbplusc", "type": "Exercise", "number": "A.5.5.17", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-3", + "id": "order-of-operations-aminusgroupbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-3", + "url": "section-order-of-operations.html#order-of-operations-aminusgroupbminusc", "type": "Exercise", "number": "A.5.5.18", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-4", + "id": "order-of-operations-atimesgroupbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-4", + "url": "section-order-of-operations.html#order-of-operations-atimesgroupbplusc", "type": "Exercise", "number": "A.5.5.19", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-5", + "id": "order-of-operations-groupaplusbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-5", + "url": "section-order-of-operations.html#order-of-operations-groupaplusbtimesc", "type": "Exercise", "number": "A.5.5.20", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-6", + "id": "order-of-operations-atimesgroupbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-6", + "url": "section-order-of-operations.html#order-of-operations-atimesgroupbminusc", "type": "Exercise", "number": "A.5.5.21", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-7", + "id": "order-of-operations-groupaminusbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-7", + "url": "section-order-of-operations.html#order-of-operations-groupaminusbtimesc", "type": "Exercise", "number": "A.5.5.22", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-8", + "id": "order-of-operations-agroupbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-8", + "url": "section-order-of-operations.html#order-of-operations-agroupbplusc", "type": "Exercise", "number": "A.5.5.23", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-9", + "id": "order-of-operations-groupaplusbc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-9", + "url": "section-order-of-operations.html#order-of-operations-groupaplusbc", "type": "Exercise", "number": "A.5.5.24", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-10", + "id": "order-of-operations-agroupbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-10", + "url": "section-order-of-operations.html#order-of-operations-agroupbminusc", "type": "Exercise", "number": "A.5.5.25", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-11", + "id": "order-of-operations-groupaminusbc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-11", + "url": "section-order-of-operations.html#order-of-operations-groupaminusbc", "type": "Exercise", "number": "A.5.5.26", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-12", + "id": "order-of-operations-adivgroupbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-12", + "url": "section-order-of-operations.html#order-of-operations-adivgroupbplusc", "type": "Exercise", "number": "A.5.5.27", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-13", + "id": "order-of-operations-groupaplusbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-13", + "url": "section-order-of-operations.html#order-of-operations-groupaplusbdivc", "type": "Exercise", "number": "A.5.5.28", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-14", + "id": "order-of-operations-adivgroupbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-14", + "url": "section-order-of-operations.html#order-of-operations-adivgroupbminusc", "type": "Exercise", "number": "A.5.5.29", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-15", + "id": "order-of-operations-groupaminusbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-15", + "url": "section-order-of-operations.html#order-of-operations-groupaminusbdivc", "type": "Exercise", "number": "A.5.5.30", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-16", + "id": "order-of-operations-adivgroupbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-16", + "url": "section-order-of-operations.html#order-of-operations-adivgroupbtimesc", "type": "Exercise", "number": "A.5.5.31", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-17", + "id": "order-of-operations-adivgroupbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-17", + "url": "section-order-of-operations.html#order-of-operations-adivgroupbdivc", "type": "Exercise", "number": "A.5.5.32", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-18", + "id": "order-of-operations-groupaplusbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-18", + "url": "section-order-of-operations.html#order-of-operations-groupaplusbtoc", "type": "Exercise", "number": "A.5.5.33", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-19", + "id": "order-of-operations-groupaminusbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-19", + "url": "section-order-of-operations.html#order-of-operations-groupaminusbtoc", "type": "Exercise", "number": "A.5.5.34", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-20", + "id": "order-of-operations-groupatimesbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-20", + "url": "section-order-of-operations.html#order-of-operations-groupatimesbtoc", "type": "Exercise", "number": "A.5.5.35", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-21", + "id": "order-of-operations-groupadivbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-21", + "url": "section-order-of-operations.html#order-of-operations-groupadivbtoc", "type": "Exercise", "number": "A.5.5.36", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-22", + "id": "order-of-operations-neggroupaplusb", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-22", + "url": "section-order-of-operations.html#order-of-operations-neggroupaplusb", "type": "Exercise", "number": "A.5.5.37", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-23", + "id": "order-of-operations-neggroupaminusb", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-23", + "url": "section-order-of-operations.html#order-of-operations-neggroupaminusb", "type": "Exercise", "number": "A.5.5.38", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-24", + "id": "order-of-operations-neggroupatimeesb", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-24", + "url": "section-order-of-operations.html#order-of-operations-neggroupatimeesb", "type": "Exercise", "number": "A.5.5.39", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-25", + "id": "order-of-operations-neggroupadivb", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-25", + "url": "section-order-of-operations.html#order-of-operations-neggroupadivb", "type": "Exercise", "number": "A.5.5.40", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-26", + "id": "order-of-operations-neggroupatob", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-26", + "url": "section-order-of-operations.html#order-of-operations-neggroupatob", "type": "Exercise", "number": "A.5.5.41", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-3-27", + "id": "order-of-operations-groupnegatob", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-3-27", + "url": "section-order-of-operations.html#order-of-operations-groupnegatob", "type": "Exercise", "number": "A.5.5.42", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-2", + "id": "order-of-operations-aminusabsbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-2", + "url": "section-order-of-operations.html#order-of-operations-aminusabsbminusc", "type": "Exercise", "number": "A.5.5.43", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-3", + "id": "order-of-operations-atimesabsbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-3", + "url": "section-order-of-operations.html#order-of-operations-atimesabsbminusc", "type": "Exercise", "number": "A.5.5.44", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-4", + "id": "order-of-operations-absaminusbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-4", + "url": "section-order-of-operations.html#order-of-operations-absaminusbtimesc", "type": "Exercise", "number": "A.5.5.45", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-5", + "id": "order-of-operations-aabsbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-5", + "url": "section-order-of-operations.html#order-of-operations-aabsbminusc", "type": "Exercise", "number": "A.5.5.46", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-6", + "id": "order-of-operations-absaminusbc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-6", + "url": "section-order-of-operations.html#order-of-operations-absaminusbc", "type": "Exercise", "number": "A.5.5.47", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-7", + "id": "order-of-operations-adivabsbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-7", + "url": "section-order-of-operations.html#order-of-operations-adivabsbminusc", "type": "Exercise", "number": "A.5.5.48", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-8", + "id": "order-of-operations-absaminusbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-8", + "url": "section-order-of-operations.html#order-of-operations-absaminusbdivc", "type": "Exercise", "number": "A.5.5.49", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-9", + "id": "order-of-operations-absaminusbtoc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-9", + "url": "section-order-of-operations.html#order-of-operations-absaminusbtoc", "type": "Exercise", "number": "A.5.5.50", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-10", + "id": "order-of-operations-negabsaplusb", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-10", + "url": "section-order-of-operations.html#order-of-operations-negabsaplusb", "type": "Exercise", "number": "A.5.5.51", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-11", + "id": "order-of-operations-negabsaminusb", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-11", + "url": "section-order-of-operations.html#order-of-operations-negabsaminusb", "type": "Exercise", "number": "A.5.5.52", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-12", + "id": "order-of-operations-sqrtaplusbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-12", + "url": "section-order-of-operations.html#order-of-operations-sqrtaplusbplusc", "type": "Exercise", "number": "A.5.5.53", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-13", + "id": "order-of-operations-aplussqrtbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-13", + "url": "section-order-of-operations.html#order-of-operations-aplussqrtbplusc", "type": "Exercise", "number": "A.5.5.54", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-14", + "id": "order-of-operations-sqrtaminusbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-14", + "url": "section-order-of-operations.html#order-of-operations-sqrtaminusbminusc", "type": "Exercise", "number": "A.5.5.55", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-15", + "id": "order-of-operations-aminussqrtbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-15", + "url": "section-order-of-operations.html#order-of-operations-aminussqrtbminusc", "type": "Exercise", "number": "A.5.5.56", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-16", + "id": "order-of-operations-asqrtbplusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-16", + "url": "section-order-of-operations.html#order-of-operations-asqrtbplusc", "type": "Exercise", "number": "A.5.5.57", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-17", + "id": "order-of-operations-asqrtbminusc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-17", + "url": "section-order-of-operations.html#order-of-operations-asqrtbminusc", "type": "Exercise", "number": "A.5.5.58", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-18", + "id": "order-of-operations-asqrtbtimesc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-18", + "url": "section-order-of-operations.html#order-of-operations-asqrtbtimesc", "type": "Exercise", "number": "A.5.5.59", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-19", + "id": "order-of-operations-asqrtbdivc", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-19", + "url": "section-order-of-operations.html#order-of-operations-asqrtbdivc", "type": "Exercise", "number": "A.5.5.60", "title": "", "body": " " }, { - "id": "section-order-of-operations-7-1-4-20", + "id": "order-of-operations-fractiongrouping", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-20", + "url": "section-order-of-operations.html#order-of-operations-fractiongrouping", "type": "Exercise", "number": "A.5.5.61", "title": "", @@ -22654,9 +23086,9 @@ var ptx_lunr_docs = [ "body": " " }, { - "id": "section-order-of-operations-7-1-4-24", + "id": "order-of-operations-exponentgrouping", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-1-4-24", + "url": "section-order-of-operations.html#order-of-operations-exponentgrouping", "type": "Exercise", "number": "A.5.5.65", "title": "", @@ -22690,9 +23122,9 @@ var ptx_lunr_docs = [ "body": " " }, { - "id": "section-order-of-operations-7-2-2", + "id": "order-of-operations-challenge", "level": "2", - "url": "section-order-of-operations.html#section-order-of-operations-7-2-2", + "url": "section-order-of-operations.html#order-of-operations-challenge", "type": "Exercise", "number": "A.5.5.69", "title": "", @@ -22705,7 +23137,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "A.6", "title": "Set Notation and Types of Numbers", - "body": " Set Notation and Types of Numbers When we talk about how many or how much of something we have, it often makes sense to use different types of numbers. For example, if we are counting dogs in a shelter, the possibilities are only . (It would be difficult to have of a dog.) On the other hand if you were weighing a dog in pounds, it doesn't make sense to only allow yourself to work with whole numbers. The dog might weigh something like pounds. These examples highlight how certain kinds of numbers are appropriate for certain situations. We'll classify various types of numbers in this section. Alternative Video Lesson Set Notation What is the mathematical difference between these three lists? To a mathematician, the last one, is an ordered triple. What matters is not merely the three numbers, but also the order in which they come. The ordered triple is not the same as ; they have the same numbers in them, but the order has changed. For some context, February has days; then March has days; then April has days. The order of the three numbers is meaningful in that context. With curly braces and , a mathematician sees a collection of numbers and does not particularly care in which order they are written. Such a collection is called a set . set All that matters is that these numbers are part of a collection. They've been written in some particular order because that's necessary to write them down. But you might as well have put the three numbers in a bag and shaken up the bag. For some context, maybe your favorite three NBA players have jersey numbers , , and , and you like them all equally well. It doesn't really matter what order you use to list them. So we can say: What about just writing ? This list of three numbers is ambiguous. Without the curly braces or parentheses, it's unclear to a reader if the order is important. Set notation set notation is the use of curly braces to surround a list\/collection of numbers, and we will use set notation frequently in this section. Set Notation Practice using (and not using) set notation. According to Google, the three most common error codes from visiting a web site are 403 , 404 , and 500 . Without knowing which error code is most common, express this set mathematically. Since we only have to describe a collection of three numbers and their order doesn t matter, we can write {403,404,500}. Error code 500 is the most common. Error code 403 is the least common of these three. And that leaves 404 in the middle. Express the error codes in a mathematical way that appreciates how frequently they happen, from most often to least often. Now we must describe the same three numbers and we want readers to know that the order we are writing the numbers matters. We can write (500,404,403). Different Number Sets types of numbers In the introduction, we mentioned how different sets of numbers are appropriate for different situations. Here are the basic sets of numbers that are used in basic algebra. Natural Numbers natural numbers When we count, we begin: and continue on in that pattern. These numbers are known as natural numbers . Whole Numbers whole numbers If we include zero, then we have the set of whole numbers . has no standard symbol, but some options are , , and . Integers integers If we include the negatives of whole numbers, then we have the set of integers . . A is used because one word in German for numbers is Zahlen. Rational Numbers rational numbers A rational number is any number that can be written as a fraction of integers, where the denominator is nonzero. Alternatively, a rational number is any number that can be written with a decimal that terminates or that repeats. A is used because fractions are q uotients of integers. Irrational Numbers irrational numbers Any number that cannot be written as a fraction of integers belongs to the set of irrational numbers . Another way to say this is that any number whose decimal places goes on forever without repeating is an irrational number . Some examples include , , There is no standard symbol for the set of irrational numbers. Real Numbers real numbers Any number that can be marked somewhere on a number line is a real number . Real numbers might be the only numbers you are familiar with. For a number to not be real, you have to start considering things called complex numbers , which are not our concern right now. The set of real numbers can be denoted with for short. Types of Numbers a disc represents all real numbers; the disc is separated into two areas: one for irrational numbers with examples like pi, e, sqrt(15), and 1.010010001..., and the other for rational numbers with examples like 3\/17, 1.25, and 4.3 repeating; within the rational area, there is a disc representing integers, with -42 as an example; within the integer area there is a region representing whole numbers with 0 as an example; within the whole numbers area there is a region representing natural numbers, with 23 as an example Rational Numbers in Other Forms Any number that can be written as a ratio of integers is rational, even if it's not written that way at first. For example, these numbers might not look rational to you at first glance: , , , and . But they are all rational, because they can respectively be written as , , , and . Determine If Numbers Are This Type or That Type Determine which numbers from the set are natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers. All of these numbers are real numbers, because all of these numbers can be positioned on the real number line. Each real number is either rational or irrational, and not both. , , , and are rational because we can see directly that their decimal expressions terminate. is also rational, because its decimal expression repeats. is rational because it is a ratio of integers. And last but not least, is rational, because that's the same thing as . This leaves only and as irrational numbers. Their decimal expressions go on forever without entering a repetitive cycle. Only , , , and (which is really ) are integers. Of these, only , , and are whole numbers, because whole numbers excludes the negative integers. Of these, only and are natural numbers, because the natural numbers exclude . Give an example of a whole number that is not an integer. Give an example of an integer that is not a whole number. Give an example of a rational number that is not an integer. Give an example of a irrational number. Give an example of a irrational number that is also an integer. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question. Any negative integer, like , is not a whole number, but is an integer. Any terminating decimal, like , is a rational number, but is not an integer. is the easiest number to remember as an irrational number. Another constant worth knowing is . Finally, the square root of most integers are irrational, like and . All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers. In the introduction, we mentioned that the different types of numbers are appropriate in different situation. Which number set do you think is most appropriate in each of the following situations? The number of people in a math class that play the ukulele. This number is best considered as a . The number of people who play the ukulele could be so the whole numbers are the appropriate set. The hypotenuse s length in a given right triangle. This number is best considered as a . A hypotenuse s length could be (which is irrational), or any other positive number. So the real numbers are the appropriate set. The proportion of people in a math class that have a cat. This number is best considered as a . This proportion will be a ratio of integers, as both the total number of people in the class and the number of people who have a cat are integers. So the rational numbers are the appropriate set. The number of people in the room with you who have the same birthday as you. This number is best considered as a . We know that the number of people must be a counting number, and since you are in the room with yourself, there is at least one person in that room with your birthday. So the natural numbers are the appropriate set. The total revenue (in dollars) generated for ticket sales at a Timbers soccer game. This number is best considered as a . The total revenue will be some number of dollars and cents, such as which is a terminating decimal and thus a rational number. So the rational numbers are the appropriate set. Converting Repeating Decimals to Fractions We have learned that a terminating decimal number is a rational number. It's easy to convert a terminating decimal number into a fraction of integers: you just need to multiply and divide by one of the numbers in the set . For example, when we say the number out loud, we say one hundred and twenty-three thousandths. While that's a lot to say, it makes it obvious that this number can be written as a ratio: . Similarly, , demonstrating how any terminating decimal can be written as a fraction. Repeating decimals can also be written as a fraction. To understand how, use a calculator to find the decimal for, say, and You will find that . The pattern is that dividing a number by a number from with the same number of digits will create a repeating decimal that starts as . and then repeats the numerator. We can use this observation to reverse engineer some fractions from repeating decimals. Write the rational number as a fraction. The three -digit number repeats after the decimal. So we will make use of the three -digit denominator And we have Write the rational number as a fraction. The two -digit number repeats after the decimal. So we will make use of the two -digit denominator And we have But this fraction can be reduced to Converting a repeating decimal to a fraction is not always quite this straightforward. There are complications if the number takes a few digits before it begins repeating. For your interest, here is one example on how to do that. Can we convert the repeating decimal to a fraction? The trick is to separate its terminating part from its repeating part, like this: . Now note that the terminating part is , and the repeating part is almost like our earlier examples, except it has an extra right after the decimal. So we have: . With what we learned in the earlier examples and basic fraction arithmetic, we can continue: Check that this is right by entering into a calculator and seeing if it returns the decimal we started with, . Review and Warmup Write the decimal number as a fraction. = is read as eighty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = is read as ninety-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.95. is read as ninety-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.65. is read as sixty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = is read as 322 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = is read as 496 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the fraction as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. Write the fraction as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. Write the mixed number as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number. Write the mixed number as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number. Set Notation There are two numbers that you can square to get Express this collection of two numbers using set notation. There are four positive, even, one-digit numbers. Express this collection of four numbers using set notation. There are six two-digit perfect square numbers. Express this collection of six numbers using set notation. There is a set of three small positive integers where you can square all three numbers, then add the results, and get Express this collection of three numbers using set notation. Types of Numbers Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are CF. Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are EH. Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are BCEH. Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are CDFH. Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are ACDEGH. Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are ABCDEG. Which of the following are irrational numbers? There may be more than one correct answer. To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . The irrational numbers here are are GH. Which of the following are irrational numbers? There may be more than one correct answer. To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . The irrational numbers here are are AE. Which of the following are real numbers? There may be more than one correct answer. All of these numbers are real numbers, since they can all be located on a number line. Which of the following are real numbers? There may be more than one correct answer. All of these numbers are real numbers, since they can all be located on a number line. Determine the validity of each statement by selecting True or False. The number is an integer The number is an integer, but not a whole number The number is irrational The number is irrational The number is an integer, but not a whole number Determine the validity of each statement by selecting True or False. The number is rational, but not a natural number The number is a real number, but not a rational number The number is an integer that is also a natural number The number is rational, but not an integer The number is rational, but not a whole number In each situation, which number set do you think is most appropriate? The number of dogs a student has owned throughout their lifetime. This number is best considered as a . The difference between the projected annual expenditures and the actual annual expenditures for a given company. This number is best considered as a . The length around swimming pool in the shape of a half circle with radius This number is best considered as a . The proportion of students at a college who own a car. This number is best considered as a . The width of a sheet of paper, in inches. This number is best considered as a . The number of people eating in a non-empty restaurant. This number is best considered as a . Give an example of a whole number that is not an integer. Give an example of an integer that is not a whole number. Give an example of a rational number that is not an integer. Give an example of a irrational number. Give an example of a irrational number that is also an integer. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question. Any negative integer, like , is not a whole number, but is an integer. Any terminating decimal, like , is a rational number, but is not an integer. is the easiest number to remember as an irrational number. Another constant worth knowing is . Finally, the square root of most integers are irrational, like and . All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers. Writing Decimals as Fractions Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Challenge Imagine making up a number with the following pattern. After the decimal point, write the natural numbers 1, 2, 3, 4, 5, etc. The decimal digits will extend forever with this pattern: Is the number a rational number or an irrational number? The number is irrational. The decimal will never end and it will never repeat. " + "body": " Set Notation and Types of Numbers When we talk about how many or how much of something we have, it often makes sense to use different types of numbers. For example, if we are counting dogs in a shelter, the possibilities are only . (It would be difficult to have of a dog.) On the other hand if you were weighing a dog in pounds, it doesn't make sense to only allow yourself to work with whole numbers. The dog might weigh something like pounds. These examples highlight how certain kinds of numbers are appropriate for certain situations. We'll classify various types of numbers in this section. Alternative Video Lesson Set Notation What is the mathematical difference between these three lists? To a mathematician, the last one, is an ordered triple. What matters is not merely the three numbers, but also the order in which they come. The ordered triple is not the same as ; they have the same numbers in them, but the order has changed. For some context, February has days; then March has days; then April has days. The order of the three numbers is meaningful in that context. With curly braces and , a mathematician sees a collection of numbers and does not particularly care in which order they are written. Such a collection is called a set . set All that matters is that these numbers are part of a collection. They've been written in some particular order because that's necessary to write them down. But you might as well have put the three numbers in a bag and shaken up the bag. For some context, maybe your favorite three NBA players have jersey numbers , , and , and you like them all equally well. It doesn't really matter what order you use to list them. So we can say: What about just writing ? This list of three numbers is ambiguous. Without the curly braces or parentheses, it's unclear to a reader if the order is important. Set notation set notation is the use of curly braces to surround a list\/collection of numbers, and we will use set notation frequently in this section. Set Notation Practice using (and not using) set notation. According to Google, the three most common error codes from visiting a web site are 403 , 404 , and 500 . Without knowing which error code is most common, express this set mathematically. Since we only have to describe a collection of three numbers and their order doesn t matter, we can write {403,404,500}. Error code 500 is the most common. Error code 403 is the least common of these three. And that leaves 404 in the middle. Express the error codes in a mathematical way that appreciates how frequently they happen, from most often to least often. Now we must describe the same three numbers and we want readers to know that the order we are writing the numbers matters. We can write (500,404,403). Different Number Sets types of numbers In the introduction, we mentioned how different sets of numbers are appropriate for different situations. Here are the basic sets of numbers that are used in basic algebra. Natural Numbers natural numbers When we count, we begin: and continue on in that pattern. These numbers are known as natural numbers . Whole Numbers whole numbers If we include zero, then we have the set of whole numbers . has no standard symbol, but some options are , , and . Integers integers If we include the negatives of whole numbers, then we have the set of integers . . A is used because one word in German for numbers is Zahlen. Rational Numbers rational numbers A rational number is any number that can be written as a fraction of integers, where the denominator is nonzero. Alternatively, a rational number is any number that can be written with a decimal that terminates or that repeats. A is used because fractions are q uotients of integers. Irrational Numbers irrational numbers Any number that cannot be written as a fraction of integers belongs to the set of irrational numbers . Another way to say this is that any number whose decimal places goes on forever without repeating is an irrational number . Some examples include , , There is no standard symbol for the set of irrational numbers. Real Numbers real numbers Any number that can be marked somewhere on a number line is a real number . Real numbers might be the only numbers you are familiar with. For a number to not be real, you have to start considering things called complex numbers , which are not our concern right now. The set of real numbers can be denoted with for short. Types of Numbers a disc represents all real numbers; the disc is separated into two areas: one for irrational numbers with examples like pi, e, sqrt(15), and 1.010010001..., and the other for rational numbers with examples like 3\/17, 1.25, and 4.3 repeating; within the rational area, there is a disc representing integers, with -42 as an example; within the integer area there is a region representing whole numbers with 0 as an example; within the whole numbers area there is a region representing natural numbers, with 23 as an example Rational Numbers in Other Forms Any number that can be written as a ratio of integers is rational, even if it's not written that way at first. For example, these numbers might not look rational to you at first glance: , , , and . But they are all rational, because they can respectively be written as , , , and . Determine If Numbers Are This Type or That Type Determine which numbers from the set are natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers. All of these numbers are real numbers, because all of these numbers can be positioned on the real number line. Each real number is either rational or irrational, and not both. , , , and are rational because we can see directly that their decimal expressions terminate. is also rational, because its decimal expression repeats. is rational because it is a ratio of integers. And last but not least, is rational, because that's the same thing as . This leaves only and as irrational numbers. Their decimal expressions go on forever without entering a repetitive cycle. Only , , , and (which is really ) are integers. Of these, only , , and are whole numbers, because whole numbers excludes the negative integers. Of these, only and are natural numbers, because the natural numbers exclude . Give an example of a whole number that is not an integer. Give an example of an integer that is not a whole number. Give an example of a rational number that is not an integer. Give an example of a irrational number. Give an example of a irrational number that is also an integer. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question. Any negative integer, like , is not a whole number, but is an integer. Any terminating decimal, like , is a rational number, but is not an integer. is the easiest number to remember as an irrational number. Another constant worth knowing is . Finally, the square root of most integers are irrational, like and . All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers. In the introduction, we mentioned that the different types of numbers are appropriate in different situation. Which number set do you think is most appropriate in each of the following situations? The number of people in a math class that play the ukulele. This number is best considered as a . The number of people who play the ukulele could be so the whole numbers are the appropriate set. The hypotenuse s length in a given right triangle. This number is best considered as a . A hypotenuse s length could be (which is irrational), or any other positive number. So the real numbers are the appropriate set. The proportion of people in a math class that have a cat. This number is best considered as a . This proportion will be a ratio of integers, as both the total number of people in the class and the number of people who have a cat are integers. So the rational numbers are the appropriate set. The number of people in the room with you who have the same birthday as you. This number is best considered as a . We know that the number of people must be a counting number, and since you are in the room with yourself, there is at least one person in that room with your birthday. So the natural numbers are the appropriate set. The total revenue (in dollars) generated for ticket sales at a Timbers soccer game. This number is best considered as a . The total revenue will be some number of dollars and cents, such as which is a terminating decimal and thus a rational number. So the rational numbers are the appropriate set. Converting Repeating Decimals to Fractions We have learned that a terminating decimal number is a rational number. It's easy to convert a terminating decimal number into a fraction of integers: you just need to multiply and divide by one of the numbers in the set . For example, when we say the number out loud, we say one hundred and twenty-three thousandths. While that's a lot to say, it makes it obvious that this number can be written as a ratio: . Similarly, , demonstrating how any terminating decimal can be written as a fraction. Repeating decimals can also be written as a fraction. To understand how, use a calculator to find the decimal for, say, and You will find that . The pattern is that dividing a number by a number from with the same number of digits will create a repeating decimal that starts as . and then repeats the numerator. We can use this observation to reverse engineer some fractions from repeating decimals. Write the rational number as a fraction. The three -digit number repeats after the decimal. So we will make use of the three -digit denominator And we have Write the rational number as a fraction. The two -digit number repeats after the decimal. So we will make use of the two -digit denominator And we have But this fraction can be reduced to Converting a repeating decimal to a fraction is not always quite this straightforward. There are complications if the number takes a few digits before it begins repeating. For your interest, here is one example on how to do that. Can we convert the repeating decimal to a fraction? The trick is to separate its terminating part from its repeating part, like this: . Now note that the terminating part is , and the repeating part is almost like our earlier examples, except it has an extra right after the decimal. So we have: . With what we learned in the earlier examples and basic fraction arithmetic, we can continue: Check that this is right by entering into a calculator and seeing if it returns the decimal we started with, . Review and Warmup Write the decimal number as a fraction. = is read as fifteen hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = is read as twenty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.75. is read as seventy-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.45. is read as forty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = is read as 502 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the decimal number as a fraction. = is read as 664 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. Write the fraction as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. Write the fraction as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. Write the mixed number as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number. Write the mixed number as a decimal number. = = To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number. Set Notation There are two numbers that you can square to get Express this collection of two numbers using set notation. There are four positive, even, one-digit numbers. Express this collection of four numbers using set notation. There are six two-digit perfect square numbers. Express this collection of six numbers using set notation. There is a set of three small positive integers where you can square all three numbers, then add the results, and get Express this collection of three numbers using set notation. Types of Numbers Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are AG. Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are CF. Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are ACDH. Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are ADEF. Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are ABCDEG. Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are BCDEFH. Which of the following are irrational numbers? There may be more than one correct answer. To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . The irrational numbers here are are DF. Which of the following are irrational numbers? There may be more than one correct answer. To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . The irrational numbers here are are AE. Which of the following are real numbers? There may be more than one correct answer. All of these numbers are real numbers, since they can all be located on a number line. Which of the following are real numbers? There may be more than one correct answer. All of these numbers are real numbers, since they can all be located on a number line. Determine the validity of each statement by selecting True or False. The number is irrational The number is an integer The number is an integer, but not a whole number The number is irrational The number is irrational Determine the validity of each statement by selecting True or False. The number is rational, but not a whole number The number is a real number, but not a rational number The number is a real number, but not an irrational number The number is a real number, but not a rational number The number is rational In each situation, which number set do you think is most appropriate? The number of dogs a student has owned throughout their lifetime. This number is best considered as a . The difference between the projected annual expenditures and the actual annual expenditures for a given company. This number is best considered as a . The length around swimming pool in the shape of a half circle with radius This number is best considered as a . The proportion of students at a college who own a car. This number is best considered as a . The width of a sheet of paper, in inches. This number is best considered as a . The number of people eating in a non-empty restaurant. This number is best considered as a . Give an example of a whole number that is not an integer. Give an example of an integer that is not a whole number. Give an example of a rational number that is not an integer. Give an example of a irrational number. Give an example of a irrational number that is also an integer. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question. Any negative integer, like , is not a whole number, but is an integer. Any terminating decimal, like , is a rational number, but is not an integer. is the easiest number to remember as an irrational number. Another constant worth knowing is . Finally, the square root of most integers are irrational, like and . All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers. Writing Decimals as Fractions Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Write the rational number as a fraction. Challenge Imagine making up a number with the following pattern. After the decimal point, write the natural numbers 1, 2, 3, 4, 5, etc. The decimal digits will extend forever with this pattern: Is the number a rational number or an irrational number? The number is irrational. The decimal will never end and it will never repeat. " }, { "id": "section-set-notation-and-types-of-numbers-2-2", @@ -22822,7 +23254,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.1", "title": "", - "body": " Write the decimal number as a fraction. = is read as eighty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " + "body": " Write the decimal number as a fraction. = is read as fifteen hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " }, { "id": "section-set-notation-and-types-of-numbers-6-1-3", @@ -22831,7 +23263,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.2", "title": "", - "body": " Write the decimal number as a fraction. = is read as ninety-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " + "body": " Write the decimal number as a fraction. = is read as twenty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " }, { "id": "section-set-notation-and-types-of-numbers-6-1-4", @@ -22840,7 +23272,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.3", "title": "", - "body": " Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.95. is read as ninety-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " + "body": " Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.75. is read as seventy-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " }, { "id": "section-set-notation-and-types-of-numbers-6-1-5", @@ -22849,7 +23281,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.4", "title": "", - "body": " Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.65. is read as sixty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " + "body": " Write the decimal number as a fraction. = The integer part of the decimal will not change. We will only change the decimal part--0.45. is read as forty-five hundredth, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " }, { "id": "section-set-notation-and-types-of-numbers-6-1-6", @@ -22858,7 +23290,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.5", "title": "", - "body": " Write the decimal number as a fraction. = is read as 322 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " + "body": " Write the decimal number as a fraction. = is read as 502 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " }, { "id": "section-set-notation-and-types-of-numbers-6-1-7", @@ -22867,7 +23299,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.6", "title": "", - "body": " Write the decimal number as a fraction. = is read as 496 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " + "body": " Write the decimal number as a fraction. = is read as 664 thousandths, so it can be written as , which can be reduced to by dividing in both the numerator and denominator. " }, { "id": "section-set-notation-and-types-of-numbers-6-2-1", @@ -22948,7 +23380,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.15", "title": "", - "body": " Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are CF. " + "body": " Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are AG. " }, { "id": "section-set-notation-and-types-of-numbers-6-5-3", @@ -22957,7 +23389,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.16", "title": "", - "body": " Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are EH. " + "body": " Which of the following are whole numbers? There may be more than one correct answer. To be a whole number, you have to be one of the numbers . Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example is a whole number, since . So the correct answers are CF. " }, { "id": "section-set-notation-and-types-of-numbers-6-6-1", @@ -22966,7 +23398,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.17", "title": "", - "body": " Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are BCEH. " + "body": " Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are ACDH. " }, { "id": "section-set-notation-and-types-of-numbers-6-6-2", @@ -22975,7 +23407,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.18", "title": "", - "body": " Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are CDFH. " + "body": " Which of the following are integers? There may be more than one correct answer. To be an integer, you have to be one of the numbers . Sometimes integers are explicitly written this way, and sometimes they are hidden. For example is an integer, since . So the correct answers are ADEF. " }, { "id": "section-set-notation-and-types-of-numbers-6-7-1", @@ -22984,7 +23416,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.19", "title": "", - "body": " Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are ACDEGH. " + "body": " Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are ABCDEG. " }, { "id": "section-set-notation-and-types-of-numbers-6-7-2", @@ -22993,7 +23425,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.20", "title": "", - "body": " Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are ABCDEG. " + "body": " Which of the following are rational numbers? There may be more than one correct answer. To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . So the correct answers are BCDEFH. " }, { "id": "section-set-notation-and-types-of-numbers-6-8-1", @@ -23002,7 +23434,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.21", "title": "", - "body": " Which of the following are irrational numbers? There may be more than one correct answer. To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . The irrational numbers here are are GH. " + "body": " Which of the following are irrational numbers? There may be more than one correct answer. To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example is a rational number, since . The irrational numbers here are are DF. " }, { "id": "section-set-notation-and-types-of-numbers-6-8-2", @@ -23038,7 +23470,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.25", "title": "", - "body": " Determine the validity of each statement by selecting True or False. The number is an integer The number is an integer, but not a whole number The number is irrational The number is irrational The number is an integer, but not a whole number " + "body": " Determine the validity of each statement by selecting True or False. The number is irrational The number is an integer The number is an integer, but not a whole number The number is irrational The number is irrational " }, { "id": "section-set-notation-and-types-of-numbers-6-10-2", @@ -23047,7 +23479,7 @@ var ptx_lunr_docs = [ "type": "Exercise", "number": "A.6.4.26", "title": "", - "body": " Determine the validity of each statement by selecting True or False. The number is rational, but not a natural number The number is a real number, but not a rational number The number is an integer that is also a natural number The number is rational, but not an integer The number is rational, but not a whole number " + "body": " Determine the validity of each statement by selecting True or False. The number is rational, but not a whole number The number is a real number, but not a rational number The number is a real number, but not an irrational number The number is a real number, but not a rational number The number is rational " }, { "id": "section-set-notation-and-types-of-numbers-6-11", diff --git a/orcca.html b/orcca.html index fa9aa3850..1042a45cc 100644 --- a/orcca.html +++ b/orcca.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/part-linear-equations-and-lines.html b/part-linear-equations-and-lines.html index 9df557893..a7b471adc 100644 --- a/part-linear-equations-and-lines.html +++ b/part-linear-equations-and-lines.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/review-graphing-lines.html b/review-graphing-lines.html index 9d770444e..e2f54a079 100644 --- a/review-graphing-lines.html +++ b/review-graphing-lines.html @@ -77,7 +77,7 @@ "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "https://pretextbook.org/js/lib" + "pretext": "_static/pretext/js/lib" } }, "startup": { @@ -89,9 +89,8 @@ )} } }; - - - + @@ -99,16 +98,16 @@ - - - - - - - - - - + + + + + + + + + + @@ -131,15 +130,15 @@ eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.1.4'; +eBookConfig.runestone_version = '7.2.18'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - - + + Skip to main content
    @@ -147,7 +146,7 @@

    Open Resources for Community College Algebra

    Faculty

    -
    • @@ -317,53 +316,53 @@

    Search Results:

  • 2.1 Solving Multistep Linear Equations
  • 2.2 Solving Multistep Linear Inequalities
  • 2.3 Equations and Inequalities with Fractions
  • 2.4 Special Solution Sets
  • 2.5 Isolating a Linear Variable
  • 2.6 Linear Equations and Inequalities Chapter Review
    - +
    • @@ -373,101 +372,101 @@

    Search Results:

  • 3.1 Cartesian Coordinates
  • 3.2 Graphing Equations
  • 3.3 Exploring Two-Variable Data and Rate of Change
  • 3.4 Slope
  • 3.5 Slope-Intercept Form
  • 3.6 Point-Slope Form
  • 3.7 Standard Form
  • 3.8 Horizontal, Vertical, Parallel, and Perpendicular Lines
  • 3.9 Summary of Graphing Lines
  • 3.10 Graphing Lines Chapter Review
    • @@ -478,36 +477,36 @@

    Search Results:

  • 4.1 Solving a System by Graphing
  • 4.2 Substitution
  • 4.3 Elimination
  • 4.4 Systems of Two Linear Equations Chapter Review
    - +
    • @@ -522,63 +521,63 @@

    Search Results:

  • A.1 Arithmetic with Negative Numbers
  • A.2 Fractions and Fraction Arithmetic
  • A.3 Absolute Value and Square Root
  • A.4 Percentages
  • A.5 Order of Operations
  • A.6 Set Notation and Types of Numbers
    • @@ -588,39 +587,36 @@

    Search Results:

    C PCC Course Content and Outcome Guides
    -
  • D Answers to Exercises
    • -
  • Index
    • +
  • D Answers to Exercises
    • +
  • Index

    • Section 3.10 Graphing Lines Chapter Review

    -

    +

    Subsection 3.10.1 Cartesian Coordinates

    -
    In Section 1 we covered the definition of the Cartesian Coordinate System and how to plot points using the \(x\)- and \(y\)-axes.
    -

    +
    In Section 1 we covered the definition of the Cartesian Coordinate System and how to plot points using the \(x\)- and \(y\)-axes.
    +

    Example 3.10.1.

    -
    On paper, sketch a Cartesian coordinate system with units, and then plot the following points: \((3,2),(-5,-1),(0,-3),(4,0)\text{.}\) +
    On paper, sketch a Cartesian coordinate system with units, and then plot the following points: \((3,2),(-5,-1),(0,-3),(4,0)\text{.}\)
    -
    -Explanation.
    a Cartesian grid with four points plotted; the point (3,2) is 3 units to the right of the origin and two units up; the point (-5,-1) is 5 units to the left of the origin and 1 unit down; the point (0,-3) is on the y-axis 3 units down; the point (4,0) is right 4 units along the x-axis.
    -
    Figure 3.10.2. A Cartesian grid with the four points plotted.
    -

    +
    Explanation.
    a Cartesian grid with four points plotted; the point (3,2) is 3 units to the right of the origin and two units up; the point (-5,-1) is 5 units to the left of the origin and 1 unit down; the point (0,-3) is on the y-axis 3 units down; the point (4,0) is right 4 units along the x-axis.
    +
    Figure 3.10.2. A Cartesian grid with the four points plotted.

    Subsection 3.10.2 Graphing Equations

    -
    In Section 2 we covered how to plot solutions to equations to produce a graph of the equation.
    -

    +
    In Section 2 we covered how to plot solutions to equations to produce a graph of the equation.
    +

    Example 3.10.3.

    -
    Graph the equation \(y=-2x+5\text{.}\) +
    Graph the equation \(y=-2x+5\text{.}\)
    -
    -Explanation.
    -
    -
    +
    Explanation.
    +
    +
    @@ -653,7 +649,7 @@

    Search Results:

    \(x\) \(y=-2x+5\)
    (a) Set up the table
    -
    +
    @@ -687,22 +683,21 @@

    Search Results:

    \(x\) \(y=-2x+5\)
    (b) Complete the table
    -
    Figure 3.10.4. Making a table for \(y=-2x+5\)
    We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line.
    -
    a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1)
    (a) Use points from the table
    -
    a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1);they are connected with a solid line with arrows on each end
    (b) Connect the points in whatever pattern is apparent
    +
    Figure 3.10.4. Making a table for \(y=-2x+5\)
    We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line.
    +
    a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1)
    (a) Use points from the table
    +
    a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1);they are connected with a solid line with arrows on each end
    (b) Connect the points in whatever pattern is apparent
    Figure 3.10.5. Graphing the Equation \(y=-2x+5\)
    -
    -

    +

    Subsection 3.10.3 Exploring Two-Variable Data and Rate of Change

    -
    In Section 3 we covered how to find patterns in tables of data and how to calculate the rate of change between points in data.
    -

    +
    In Section 3 we covered how to find patterns in tables of data and how to calculate the rate of change between points in data.
    +

    Example 3.10.6.

    -
    Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
    -
    +
    Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
    +
    @@ -726,10 +721,9 @@

    Search Results:

    \(x\) \(y\)
    Figure 3.10.7. A table of linear data.
    -
    -Explanation.
    +
    Explanation.
    -
    We consider how the values change from one row to the next. From row to row, the \(x\)-value increases by \(1\text{.}\) Also, the \(y\)-value decreases by \(2\) from row to row.
    +
    We consider how the values change from one row to the next. From row to row, the \(x\)-value increases by \(1\text{.}\) Also, the \(y\)-value decreases by \(2\) from row to row.
    @@ -762,23 +756,21 @@

    Search Results:

    \(\leftarrow{}-2\)
    -
    Since row-to-row change is always \(1\) for \(x\) and is always \(-2\) for \(y\text{,}\) the rate of change from one row to another row is always the same: \(-2\) units of \(y\) for every \(1\) unit of \(x\text{.}\) -
    We know that the output for \(x = 0\) is \(y = -4\text{.}\) And our observation about the constant rate of change tells us that if we increase the input by \(x\) units from \(0\text{,}\) the output should decrease by \(\overbrace{(-2)+(-2)+\cdots+(-2)}^{x\text{ times}}\text{,}\) which is \(-2x\text{.}\) So the output would be \(-4-2x\text{.}\) -
    So the equation is \(y=-2x-4\text{.}\) +
    Since row-to-row change is always \(1\) for \(x\) and is always \(-2\) for \(y\text{,}\) the rate of change from one row to another row is always the same: \(-2\) units of \(y\) for every \(1\) unit of \(x\text{.}\) +
    We know that the output for \(x = 0\) is \(y = -4\text{.}\) And our observation about the constant rate of change tells us that if we increase the input by \(x\) units from \(0\text{,}\) the output should decrease by \(\overbrace{(-2)+(-2)+\cdots+(-2)}^{x\text{ times}}\text{,}\) which is \(-2x\text{.}\) So the output would be \(-4-2x\text{.}\) +
    So the equation is \(y=-2x-4\text{.}\)
    -
    -

    +

    Subsection 3.10.4 Slope

    -
    In Section 4 we covered the definition of slope and how to use slope triangles to calculate slope. There is also the slope formula which helps find the slope through any two points.
    -

    +
    In Section 4 we covered the definition of slope and how to use slope triangles to calculate slope. There is also the slope formula which helps find the slope through any two points.
    +

    Example 3.10.8.

    -
    Find the slope of the line in the following graph.
    This is a grid with a line, passing the points (3,15) and (6,27).
    -
    Figure 3.10.9. The line with two points indicated.
    -Explanation.
    -
    This is a grid with a line, passing the points (3,15) and (6,27). A slope triangle is drawn, starting at (3,15), passing (6,15), and ends at (6,27). The label from (3,15) to (6,15) is "6-3=3"; the label from (6,15) to (6,27) is "27-15=12".
    Figure 3.10.10. The line with a slope triangle drawn.
    -
    +
    Find the slope of the line in the following graph.
    This is a grid with a line, passing the points (3,15) and (6,27).
    +
    Figure 3.10.9. The line with two points indicated.
    Explanation.
    +
    This is a grid with a line, passing the points (3,15) and (6,27). A slope triangle is drawn, starting at (3,15), passing (6,15), and ends at (6,27). The label from (3,15) to (6,15) is "6-3=3"; the label from (6,15) to (6,27) is "27-15=12".
    Figure 3.10.10. The line with a slope triangle drawn.
    +
    We picked two points on the line, and then drew a slope triangle. Next, we will do:
    \begin{equation*} @@ -788,15 +780,13 @@

    Search Results:

    The line’s slope is \(4\text{.}\)
    -
    -

    +

    Example 3.10.11. Finding a Line’s Slope by the Slope Formula.

    -
    Use the slope formula to find the slope of the line that passes through the points \((-5,25)\) and \((4,-2)\text{.}\) +
    Use the slope formula to find the slope of the line that passes through the points \((-5,25)\) and \((4,-2)\text{.}\)
    -
    -Explanation.
    -
    +
    Explanation.
    +
    \begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-2-(25)}{4-(-5)}\\ @@ -806,52 +796,46 @@

    Search Results:

    The line’s slope is \(-3\text{.}\)
    -
    -

    +

    Subsection 3.10.5 Slope-Intercept Form

    -
    In Section 5 we covered the definition of slope intercept-form and both wrote equations in slope-intercept form and graphed lines given in slope-intercept form.
    -

    +
    In Section 5 we covered the definition of slope intercept-form and both wrote equations in slope-intercept form and graphed lines given in slope-intercept form.
    +

    Example 3.10.12.

    -
    Graph the line \(y=-\frac{5}{2}x+4\text{.}\) +
    Graph the line \(y=-\frac{5}{2}x+4\text{.}\)
    -
    -Explanation.
    -
    a Cartesian grid with the point (0,4) plotted
    (a) First, plot the line’s \(y\)-intercept, \((0,4)\text{.}\)
    -
    the previous plot of the y-intercept with slope triangles drawn; they go right 2 and down 5 twice to the right; from the y-intercept they also go up 5 and left 2 once; the points plotted are (-2,9), (0,4), (2,-1), (4,-6)
    (b) The slope is \(-\frac{5}{2}=\frac{-5}{2}=\frac{5}{-2}\text{.}\) So we can try using a “run” of \(2\) and a “rise” of \(-5\) or a “run” of \(-2\) and a “rise” of \(5\text{.}\)
    -
    the previous plot with a solid line drawn through all of the points; there are arrowheads on each end of the line
    (c) Arrowheads and labels are encouraged.
    +
    Explanation.
    +
    a Cartesian grid with the point (0,4) plotted
    (a) First, plot the line’s \(y\)-intercept, \((0,4)\text{.}\)
    +
    the previous plot of the y-intercept with slope triangles drawn; they go right 2 and down 5 twice to the right; from the y-intercept they also go up 5 and left 2 once; the points plotted are (-2,9), (0,4), (2,-1), (4,-6)
    (b) The slope is \(-\frac{5}{2}=\frac{-5}{2}=\frac{5}{-2}\text{.}\) So we can try using a “run” of \(2\) and a “rise” of \(-5\) or a “run” of \(-2\) and a “rise” of \(5\text{.}\)
    +
    the previous plot with a solid line drawn through all of the points; there are arrowheads on each end of the line
    (c) Arrowheads and labels are encouraged.
    -
    Figure 3.10.13. Graphing \(y=-\frac{5}{2}x+4\)
    -

    Writing a Line’s Equation in Slope-Intercept Form Based on Graph.

    Given a line’s graph, we can identify its \(y\)-intercept, and then find its slope using a slope triangle. With a line’s slope and \(y\)-intercept, we can write its equation in the form of \(y=mx+b\text{.}\) -

    +
    Figure 3.10.13. Graphing \(y=-\frac{5}{2}x+4\)

    Writing a Line’s Equation in Slope-Intercept Form Based on Graph.

    Given a line’s graph, we can identify its \(y\)-intercept, and then find its slope using a slope triangle. With a line’s slope and \(y\)-intercept, we can write its equation in the form of \(y=mx+b\text{.}\) +

    Example 3.10.14.

    -
    Find the equation of the line in the graph.
    -
    a coordinate graph of a line;three points on the line are (-3,12),(0,10) and (3,8)
    Figure 3.10.15. Graph of a line
    +
    Find the equation of the line in the graph.
    +
    a coordinate graph of a line;three points on the line are (-3,12),(0,10) and (3,8)
    Figure 3.10.15. Graph of a line
    -
    -Explanation.
    +
    Explanation.
    -
    the previous graph showing the line crosses the y-axis at 10 or (0,10)
    Figure 3.10.16. Identify the line’s \(y\)-intercept, \(10\text{.}\)
    -
    the previous graph with slope triangles drawn moving right 3 units and down two units
    Figure 3.10.17. Identify the line’s slope using a slope triangle. Note that we can pick any two points on the line to create a slope triangle. We would get the same slope: \(-\frac{2}{3}\)
    -
    With the line’s slope \(-\frac{2}{3}\) and \(y\)-intercept \(10\text{,}\) we can write the line’s equation in slope-intercept form: \(y=-\frac{2}{3}x+10\text{.}\) +
    the previous graph showing the line crosses the y-axis at 10 or (0,10)
    Figure 3.10.16. Identify the line’s \(y\)-intercept, \(10\text{.}\)
    +
    the previous graph with slope triangles drawn moving right 3 units and down two units
    Figure 3.10.17. Identify the line’s slope using a slope triangle. Note that we can pick any two points on the line to create a slope triangle. We would get the same slope: \(-\frac{2}{3}\)
    +
    With the line’s slope \(-\frac{2}{3}\) and \(y\)-intercept \(10\text{,}\) we can write the line’s equation in slope-intercept form: \(y=-\frac{2}{3}x+10\text{.}\)
    -
    -

    +

    Subsection 3.10.6 Point-Slope Form

    -
    In Section 6 we covered the definition of point-slope form and both wrote equations in point-slope form and graphed lines given in point-slope form.
    -

    +
    In Section 6 we covered the definition of point-slope form and both wrote equations in point-slope form and graphed lines given in point-slope form.
    +

    Example 3.10.18.

    -
    A line passes through \((-6,0)\) and \((9,-10)\text{.}\) Find this line’s equation in point-slope form. .
    -
    -Explanation.
    -
    -
    We will use the slope formula to find the slope first. After labeling those two points as \((\overset{x_1}{-6},\overset{y_1}{0}) \text{ and } (\overset{x_2}{9},\overset{y_2}{-10})\text{,}\) we have:
    -
    +
    A line passes through \((-6,0)\) and \((9,-10)\text{.}\) Find this line’s equation in point-slope form. .
    +
    Explanation.
    +
    +
    We will use the slope formula to find the slope first. After labeling those two points as \((\overset{x_1}{-6},\overset{y_1}{0}) \text{ and } (\overset{x_2}{9},\overset{y_2}{-10})\text{,}\) we have:
    +
    \begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-10-0}{9-(-6)}\\ @@ -859,31 +843,29 @@

    Search Results:

    \amp=-\frac{2}{3} \end{align*}
    -
    +
    Now the point-slope equation looks like \(y=-\frac{2}{3}(x-x_0)+y_0\text{.}\) Next, we will use \((9,-10)\) and substitute \(x_0\) with \(9\) and \(y_0\) with \(-10\text{,}\) and we have:
    -
    +
    \begin{align*} y\amp=-\frac{2}{3}(x-x_0)+y_0\\ y\amp=-\frac{2}{3}(x-9)+(-10)\\ y\amp=-\frac{2}{3}(x-9)-10 \end{align*}
    -
    -

    +

    Subsection 3.10.7 Standard Form

    -
    In Section 7 we covered the definition of standard form of a linear equation. We converted equations from standard form to slope-intercept form and vice versa. We also graphed lines from standard form by finding the intercepts of the line.
    -

    +
    In Section 7 we covered the definition of standard form of a linear equation. We converted equations from standard form to slope-intercept form and vice versa. We also graphed lines from standard form by finding the intercepts of the line.
    +

    Example 3.10.19.

    -
      -
    1. Convert \(2x+3y=6\) into slope-intercept form.
      2. -
    2. Convert \(y=-\frac{4}{7}x-3\) into standard form.
      3. +
        +
      1. Convert \(2x+3y=6\) into slope-intercept form.
        2. +
      2. Convert \(y=-\frac{4}{7}x-3\) into standard form.
      -
      -Explanation.
        -
      1. -
        +
        Explanation.
          +
        1. +
          \begin{align*} 2x+3y\amp=6\\ 2x+3y\subtractright{2x}\amp=6\subtractright{2x}\\ @@ -896,8 +878,8 @@

          Search Results:

          The line’s equation in slope-intercept form is \(y=-\frac{2}{3}x+2\text{.}\)
          2. -
        2. -
          +
        3. +
          \begin{align*} y\amp=-\frac{4}{7}x-3\\ \multiplyleft{7}y\amp=\multiplyleft{7}(-\frac{4}{7}x-3)\\ @@ -910,17 +892,15 @@

          Search Results:

          The line’s equation in standard form is \(4x+7y=-21\text{.}\)
          4. -
        -
    To graph a line in standard form, we could first change it to slope-intercept form, and then graph the line by its \(y\)-intercept and slope triangles. A second method is to graph the line by its \(x\)-intercept and \(y\)-intercept.
    -

    +

    To graph a line in standard form, we could first change it to slope-intercept form, and then graph the line by its \(y\)-intercept and slope triangles. A second method is to graph the line by its \(x\)-intercept and \(y\)-intercept.
    +

    Example 3.10.20.

    -
    Graph \(2x-3y=-6\) using its intercepts. And then use the intercepts to calculate the line’s slope.
    -
    -Explanation.
    -
    +
    Graph \(2x-3y=-6\) using its intercepts. And then use the intercepts to calculate the line’s slope.
    +
    Explanation.
    +
    We calculate the line’s \(x\)-intercept by substituting \(y=0\) into the equation
    -
    +
    \begin{align*} 2x-3y\amp=-6\\ 2x-3(\substitute{0})\amp=-6\\ @@ -930,9 +910,9 @@

    Search Results:

    So the line’s \(x\)-intercept is \((-3,0)\text{.}\)
    -
    +
    Similarly, we substitute \(x=0\) into the equation to calculate the \(y\)-intercept:
    -
    +
    \begin{align*} 2x-3y\amp=-6\\ 2(\substitute{0})-3y\amp=-6\\ @@ -942,41 +922,37 @@

    Search Results:

    So the line’s \(y\)-intercept is \((0,2)\text{.}\)
    -
    With both intercepts’ coordinates, we can graph the line:
    the graph of line 2x-3y=-6 with the x-intercept plotted at (-3,0) and the y-intercept at (0,2).
    -
    Figure 3.10.21. Graph of \(2x-3y=-6\)
    Now that we have graphed the line we can read the slope. The rise is \(2\) units and the run is \(3\) units so the slope is \(\frac{2}{3}\text{.}\) +
    With both intercepts’ coordinates, we can graph the line:
    the graph of line 2x-3y=-6 with the x-intercept plotted at (-3,0) and the y-intercept at (0,2).
    +
    Figure 3.10.21. Graph of \(2x-3y=-6\)
    Now that we have graphed the line we can read the slope. The rise is \(2\) units and the run is \(3\) units so the slope is \(\frac{2}{3}\text{.}\)
    -
    -

    +

    Subsection 3.10.8 Horizontal, Vertical, Parallel, and Perpendicular Lines

    -
    In Section 8 we studied horizontal and vertical lines. We also covered the relationships between the slopes of parallel and perpendicular lines.
    -

    +
    In Section 8 we studied horizontal and vertical lines. We also covered the relationships between the slopes of parallel and perpendicular lines.
    +

    Example 3.10.22.

    -
    Line \(m\)’s equation is \(y=-2x+20\text{.}\) Line \(n\) is parallel to \(m\text{,}\) and line \(n\) also passes the point \((4,-3)\text{.}\) Find an equation for line \(n\) in point-slope form.
    -
    -Explanation.
    +
    Line \(m\)’s equation is \(y=-2x+20\text{.}\) Line \(n\) is parallel to \(m\text{,}\) and line \(n\) also passes the point \((4,-3)\text{.}\) Find an equation for line \(n\) in point-slope form.
    +
    Explanation.
    Since parallel lines have the same slope, line \(n\)’s slope is also \(-2\text{.}\) Since line \(n\) also passes the point \((4,-3)\text{,}\) we can write line \(n\)’s equation in point-slope form:
    -
    +
    \begin{align*} y\amp=m(x-x_1)+y_1\\ y\amp=-2(x-4)+(-3)\\ y\amp=-2(x-4)-3 \end{align*}
    -
    -
    Two lines are perpendicular if and only if the product of their slopes is \(-1\text{.}\) +

    Two lines are perpendicular if and only if the product of their slopes is \(-1\text{.}\)
    -

    +

    Example 3.10.23.

    -
    Line \(m\)’s equation is \(y=-2x+20\text{.}\) Line \(n\) is perpendicular to \(m\text{,}\) and line \(q\) also passes the point \((4,-3)\text{.}\) Find an equation for line \(q\) in slope-intercept form.
    -
    -Explanation.
    -
    Since line \(m\) and \(q\) are perpendicular, the product of their slopes is \(-1\text{.}\) Because line \(m\)’s slope is given as \(-2\text{,}\) we can find line \(q\)’s slope is \(\frac{1}{2}\text{.}\) -
    +
    Line \(m\)’s equation is \(y=-2x+20\text{.}\) Line \(n\) is perpendicular to \(m\text{,}\) and line \(q\) also passes the point \((4,-3)\text{.}\) Find an equation for line \(q\) in slope-intercept form.
    +
    Explanation.
    +
    Since line \(m\) and \(q\) are perpendicular, the product of their slopes is \(-1\text{.}\) Because line \(m\)’s slope is given as \(-2\text{,}\) we can find line \(q\)’s slope is \(\frac{1}{2}\text{.}\) +
    Since line \(q\) also passes the point \((4,-3)\text{,}\) we can write line \(q\)’s equation in point-slope form:
    -
    +
    \begin{align*} y\amp=m(x-x_1)+y_1\\ y\amp=\frac{1}{2}(x-4)+(-3)\\ @@ -984,7 +960,7 @@

    Search Results:

    \end{align*}
    We can now convert this equation to slope-intercept form:
    -
    +
    \begin{align*} y\amp=\frac{1}{2}(x-4)-3\\ y\amp=\frac{1}{2}x-2-3\\ @@ -992,76 +968,76 @@

    Search Results:

    \end{align*}
    -
    -

    +

    Exercises 3.10.9 Exercises

    -
    +

    Exercise Group.

    -
    1.
    -
    Sketch the points \((8,2)\text{,}\) \((5,5)\text{,}\) \((-3,0)\text{,}\) \(\left(0,-\frac{14}{3}\right)\text{,}\) \((3,-2.5)\text{,}\) and \((-5,7)\) on a Cartesian plane.
    -
    -Answer.
    a Cartesian graph with the points plotted;(8,2) is 8 units to the right and up 2;(5,5) is 5 to the right and up 5;(-3,0) is down 3 units on the y-axis;(0,14/3)is 14/3 or 4 2/3 units up on the y-axis;(3,-2.5) is 3 units to the right and down 2.5;(-5,7) is left 5 units and up 7
    -
    2.
    -
    -
    +
    1.
    +
    Sketch the points \((8,2)\text{,}\) \((5,5)\text{,}\) \((-3,0)\text{,}\) \(\left(0,-\frac{14}{3}\right)\text{,}\) \((3,-2.5)\text{,}\) and \((-5,7)\) on a Cartesian plane.
    +
    Answer.
    a Cartesian graph with the points plotted;(8,2) is 8 units to the right and up 2;(5,5) is 5 to the right and up 5;(-3,0) is down 3 units on the y-axis;(0,14/3)is 14/3 or 4 2/3 units up on the y-axis;(3,-2.5) is 3 units to the right and down 2.5;(-5,7) is left 5 units and up 7
    2.
    +
    +
    -
    Locate each point in the graph:
    Write each point’s position as an ordered pair, like \((1,2)\text{.}\) -
    - - - - - - - - -
    -\(A=\) - -\(B=\) -
    -\(C=\) - -\(D=\) -
    +
    +\(A={}\) \(B={}\) \(C={}\) \(D={}\) +
    -Answer 1.
    \(\left(-2,2\right)\)
    -Answer 2.
    \(\left(0,-6\right)\)
    -Answer 3.
    \(\left(0,7\right)\)
    -Answer 4.
    \(\left(5,0\right)\)
    +
    Answer 1.
    \(\left(-2,2\right)\)
    Answer 2.
    \(\left(3,0\right)\)
    Answer 3.
    \(\left(0,-3\right)\)
    Answer 4.
    \(\left(4,-4\right)\)
    -
    3.
    -
    -
    +
    3.
    +
    +
    -
    Consider the equation
    \(y=-\frac{3}{4} x-2\)
    Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
      -
    • \(\displaystyle (-8,4)\)
      • -
    • \(\displaystyle (12,-7)\)
      • -
    • \(\displaystyle (-20,18)\)
      • -
    • \(\displaystyle (0,-2)\)
      • +
      Consider the equation
      \(y=-\frac{3}{4} x-5\)
      Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
        +
      • \(\displaystyle (0,-5)\)
        • +
      • \(\displaystyle (-4,3)\)
        • +
      • \(\displaystyle (16,-12)\)
        • +
      • \(\displaystyle (-12,4)\)
      -
    -
    4.
    -
    -
    +
    Explanation.
    +
    We substitute in each ordered pair’s \(x\)- and \(y\)-values into the equation \(y=-\frac{3}{4} x-5\text{,}\) and see whether the resulting equation is true.
    In two cases, \((-12,4)\) and \((0,-5)\text{,}\) we have true equations:
    \(\begin{aligned} +4 \amp \stackrel{?}{=}-\frac{3}{4}(-12)-5 \amp -5 \amp \stackrel{?}{=}-\frac{3}{4}(0)-5\\ +4 \amp \stackrel{?}{=}9-5 \amp -5 \amp \stackrel{?}{=}0-5\\ +4 \amp \stackrel{?}{=}4 \amp -5 \amp \stackrel{?}{=}-5\\ +\end{aligned}\)
    So \((-12,4)\) and \((0,-5)\) are solutions to the given equation.
    In the other two cases, \((16,-12)\) and \((-4,3)\text{,}\) we have false equations:
    \(\begin{aligned} +3 \amp \stackrel{?}{=}-\frac{3}{4}(-4)-5 \amp -12 \amp \stackrel{?}{=}-\frac{3}{4}(16)-5\\ +3 \amp \stackrel{?}{=}3-5 \amp -12 \amp \stackrel{?}{=}-12-5\\ +3 \amp \stackrel{?}{=}-2 \amp -12 \amp \stackrel{?}{=}-17\\ +\end{aligned}\)
    So \((16,-12)\) and \((-4,3)\) are not solutions.
    +
    +
    +
    4.
    +
    +
    -
    Consider the equation
    \(y=-\frac{5}{6} x-4\)
    Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
      -
    • \(\displaystyle (-30,21)\)
      • -
    • \(\displaystyle (12,-10)\)
      • -
    • \(\displaystyle (-24,21)\)
      • -
    • \(\displaystyle (0,-4)\)
      • +
      Consider the equation
      \(y=-\frac{5}{6} x-5\)
      Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
        +
      • \(\displaystyle (30,-28)\)
        • +
      • \(\displaystyle (-24,15)\)
        • +
      • \(\displaystyle (0,-5)\)
        • +
      • \(\displaystyle (-18,12)\)
      -
    -
    5.
    -
    -
    +
    Explanation.
    +
    We substitute in each ordered pair’s \(x\)- and \(y\)-values into the equation \(y=-\frac{5}{6} x-5\text{,}\) and see whether the resulting equation is true.
    In two cases, \((-24,15)\) and \((0,-5)\text{,}\) we have true equations:
    \(\begin{aligned} +15 \amp \stackrel{?}{=}-\frac{5}{6}(-24)-5 \amp -5 \amp \stackrel{?}{=}-\frac{5}{6}(0)-5\\ +15 \amp \stackrel{?}{=}20-5 \amp -5 \amp \stackrel{?}{=}0-5\\ +15 \amp \stackrel{?}{=}15 \amp -5 \amp \stackrel{?}{=}-5\\ +\end{aligned}\)
    So \((-24,15)\) and \((0,-5)\) are solutions to the given equation.
    In the other two cases, \((30,-28)\) and \((-18,12)\text{,}\) we have false equations:
    \(\begin{aligned} +12 \amp \stackrel{?}{=}-\frac{5}{6}(-18)-5 \amp -28 \amp \stackrel{?}{=}-\frac{5}{6}(30)-5\\ +12 \amp \stackrel{?}{=}15-5 \amp -28 \amp \stackrel{?}{=}-25-5\\ +12 \amp \stackrel{?}{=}10 \amp -28 \amp \stackrel{?}{=}-30\\ +\end{aligned}\)
    So \((30,-28)\) and \((-18,12)\) are not solutions.
    +
    +
    +
    5.
    +
    +
    -
    Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
    +
    Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
    - + @@ -1082,16 +1058,52 @@

    Exercise Group.

    \(x\)
    \(x\)
    		 \(y\)
    -Answer.
    \(y = 2x-4\)
    +
    Answer.
    \(y = 2x-4\)
    Explanation.
    +
    The relationship between \(x\) and \(y\) in each row is not as clear here. Another popular approach for finding patterns: in each column, consider how the values change from one row to the next. From row to row, the \(x\)-value increases by \(1\text{.}\) Also, the \(y\)-value increases by \(2\) from row to row.
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    \(x\)\(y\)
    \(0\)\({-4}\)
    \({}+1\rightarrow\)\(1\)\({-2}\)\(\leftarrow{}+2\)
    \({}+1\rightarrow\)\(2\)\({0}\)\(\leftarrow{}+2\)
    \({}+1\rightarrow\)\(3\)\({2}\)\(\leftarrow{}+2\)
    Since row-to-row change is always \(1\) for \(x\) and is always \(2\) for \(y\text{,}\) the rate of change from one row to another row is always the same: \(2\) units of \(y\) for every \(1\) unit of \(x\text{.}\) +
    We know that the output for \(x = 0\) is \(y = -4\text{.}\) And our observation about the constant rate of change tells us that if we increase the input by \(x\) units from \(0\text{,}\) the output should increase by \(\overbrace{(2)+(2)+\cdots+(2)}^{x\text{ times}}\text{,}\) which is \(2x\text{.}\) So the output would be \({-4+2x}\text{.}\) +
    So the equation is \({y = 2x-4}\text{.}\) +
    +
    -
    6.
    -
    -
    +
    6.
    +
    +
    -
    Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
    +
    Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
    - + @@ -1112,237 +1124,555 @@

    Exercise Group.

    \(x\)
    \(x\)
    		 \(y\)
    -Answer.
    \(y = 3-5x\)
    +
    Answer.
    \(y = 3-5x\)
    Explanation.
    +
    The relationship between \(x\) and \(y\) in each row is not as clear here. Another popular approach for finding patterns: in each column, consider how the values change from one row to the next. From row to row, the \(x\)-value increases by \(1\text{.}\) Also, the \(y\)-value decreases by \(5\) from row to row.
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    \(x\)\(y\)
    \(0\)\({3}\)
    \({}+1\rightarrow\)\(1\)\({-2}\)\(\leftarrow{}-5\)
    \({}+1\rightarrow\)\(2\)\({-7}\)\(\leftarrow{}-5\)
    \({}+1\rightarrow\)\(3\)\({-12}\)\(\leftarrow{}-5\)
    Since row-to-row change is always \(1\) for \(x\) and is always \(-5\) for \(y\text{,}\) the rate of change from one row to another row is always the same: \(-5\) units of \(y\) for every \(1\) unit of \(x\text{.}\) +
    We know that the output for \(x = 0\) is \(y = 3\text{.}\) And our observation about the constant rate of change tells us that if we increase the input by \(x\) units from \(0\text{,}\) the output should increase by \(\overbrace{(-5)+(-5)+\cdots+(-5)}^{x\text{ times}}\text{,}\) which is \(-5x\text{.}\) So the output would be \({3-5x}\text{.}\) +
    So the equation is \({y = 3-5x}\text{.}\) +
    +
    -
    +

    Exercise Group.

    -
    7.
    -
    -
    +
    7.
    +
    +
    -
    Find the slope of the line.
    graph of a line crossing the y-axis at 4; the line has an upward slant and also passes through the point (8,13)
    +
    Find the slope of the line.
    graph of a line crossing the y-axis at -3; the line has an upward slant and also passes through the point (4,6)
    -Answer.
    \({\frac{9}{8}}\)
    +
    Answer.
    \({\frac{9}{4}}\)
    Explanation.
    +
    To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,-3)\) and \((4,6)\text{.}\) +
    Next, we sketch a slope triangle and find the rise and run. In the sketch below, the rise is \(9\) and the run is \(4\text{.}\) +
    graph of the line detailing a slope triangle from (0,-3) to (4,-3) to (4,6)
    +\begin{equation*} +\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{9}{4} +\end{equation*} +
    This line’s slope is \({{\frac{9}{4}}}\text{.}\)
    +
    -
    8.
    -
    -
    +
    +
    8.
    +
    +
    -
    Find the slope of the line.
    graph of a line crossing the y-axis at -1; the line has an upward slant and also passes through the point (5,2)
    +
    Find the slope of the line.
    graph of a line crossing the y-axis at 4; the line has an upward slant and also passes through the point (7,7)
    -Answer.
    \({\frac{3}{5}}\)
    +
    Answer.
    \({\frac{3}{7}}\)
    Explanation.
    +
    To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,4)\) and \((7,7)\text{.}\) +
    Next, we sketch a slope triangle and find the rise and run. In the sketch below, the rise is \(3\) and the run is \(7\text{.}\) +
    graph of the line detailing a slope triangle from (0,4) to (7,4) to (7,7)
    +\begin{equation*} +\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{3}{7} +\end{equation*} +
    This line’s slope is \({{\frac{3}{7}}}\text{.}\) +
    +
    -
    9.
    -
    -
    +
    9.
    +
    +
    -
    Find the slope of the line.
    graph of a line crossing the y-axis at 3; the line has an upward slant and also passes through the point (3,7)
    +
    Find the slope of the line.
    graph of a line crossing the y-axis at -2; the line has an upward slant and also passes through the point (5,1)
    -Answer.
    \({\frac{4}{3}}\)
    +
    Answer.
    \({\frac{3}{5}}\)
    Explanation.
    +
    To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,-2)\) and \((5,1)\text{.}\) +
    Next, we sketch a slope triangle and find the rise and run. In the sketch below, the rise is \(3\) and the run is \(5\text{.}\) +
    graph of the line detailing a slope triangle from (0,-2) to (5,-2) to (5,1)
    +\begin{equation*} +\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{3}{5} +\end{equation*} +
    This line’s slope is \({{\frac{3}{5}}}\text{.}\)
    +
    -
    10.
    -
    -
    +
    +
    10.
    +
    +
    -
    Below is a line’s graph.
    The slope of this line is .
    +
    Below is a line’s graph.
    The slope of this line is .
    -Answer.
    \(-{\frac{3}{7}}\)
    +
    Answer.
    \(-{\frac{3}{7}}\)
    Explanation.
    +
    To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,2)\) and \((7,-1)\text{.}\) +
    Next, we sketch a slope triangle and find the rise and run. In the sketch below, the rise is \(-3\) and the run is \(7\text{.}\) +
    +\begin{equation*} +\begin{aligned}\text{slope}\amp =\frac{\text{rise}}{\text{run}}\\ +\amp =\frac{-3}{7}\\ +\end{aligned} +\end{equation*} +
    This line’s slope is \(-{\frac{3}{7}}\text{.}\)
    +
    -
    11.
    -
    -
    +
    +
    11.
    +
    +
    -
    Below is a line’s graph.
    The slope of this line is .
    +
    Below is a line’s graph.
    The slope of this line is .
    -Answer.
    \(-{\frac{4}{7}}\)
    +
    Answer.
    \(-{\frac{4}{3}}\)
    Explanation.
    +
    To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,-1)\) and \((3,-5)\text{.}\) +
    Next, we sketch a slope triangle and find the rise and run. In the sketch below, the rise is \(-4\) and the run is \(3\text{.}\) +
    +\begin{equation*} +\begin{aligned}\text{slope}\amp =\frac{\text{rise}}{\text{run}}\\ +\amp =\frac{-4}{3}\\ +\end{aligned} +\end{equation*} +
    This line’s slope is \(-{\frac{4}{3}}\text{.}\) +
    +
    -
    12.
    -
    -
    +
    12.
    +
    +
    -
    Below is a line’s graph.
    The slope of this line is .
    +
    Below is a line’s graph.
    The slope of this line is .
    -Answer.
    \(-{\frac{5}{2}}\)
    +
    Answer.
    \(-{\frac{5}{4}}\)
    Explanation.
    +
    To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,1)\) and \((4,-4)\text{.}\) +
    Next, we sketch a slope triangle and find the rise and run. In the sketch below, the rise is \(-5\) and the run is \(4\text{.}\) +
    +\begin{equation*} +\begin{aligned}\text{slope}\amp =\frac{\text{rise}}{\text{run}}\\ +\amp =\frac{-5}{4}\\ +\end{aligned} +\end{equation*} +
    This line’s slope is \(-{\frac{5}{4}}\text{.}\) +
    +
    -
    +

    Exercise Group.

    -
    13.
    -
    -
    +
    13.
    +
    +
    -
    A line passes through the points \((-6,{7})\) and \((12,{-14})\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((-8,{-3})\) and \((24,{-31})\text{.}\) Find this line’s slope.
    -Answer.
    \(-{\frac{7}{6}}\)
    +
    Answer.
    \(-{\frac{7}{8}}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    +\begin{equation*} +\text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} +\end{equation*} +
    First, we mark which number corresponds to which variable in the formula:
    +\begin{equation*} +(-8,{-3}) \longrightarrow (x_{1},y_{1}) +\end{equation*} +
    +\begin{equation*} +(24,{-31}) \longrightarrow (x_{2},y_{2}) +\end{equation*} +
    Now we substitute these numbers into the corresponding variables in the slope formula:
    +\begin{equation*} +\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{{-31}-((-3))}{24-(-8)}\\ +\amp =\frac{{-28}}{32}\\ +\amp =-\frac{7}{8} \end{aligned} +\end{equation*} +
    So the line’s slope is \(\displaystyle{-\frac{7}{8}}\text{.}\) +
    +
    -
    14.
    -
    -
    +
    14.
    +
    +
    -
    A line passes through the points \((-8,{23})\) and \((8,{-5})\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((-12,{12})\) and \((18,{-23})\text{.}\) Find this line’s slope.
    -Answer.
    \(-{\frac{7}{4}}\)
    +
    Answer.
    \(-{\frac{7}{6}}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    +\begin{equation*} +\text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} +\end{equation*} +
    First, we mark which number corresponds to which variable in the formula:
    +\begin{equation*} +(-12,{12}) \longrightarrow (x_{1},y_{1}) +\end{equation*} +
    +\begin{equation*} +(18,{-23}) \longrightarrow (x_{2},y_{2}) +\end{equation*} +
    Now we substitute these numbers into the corresponding variables in the slope formula:
    +\begin{equation*} +\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{{-23}-{12}}{18-(-12)}\\ +\amp =\frac{{-35}}{30}\\ +\amp =-\frac{7}{6} \end{aligned} +\end{equation*} +
    So the line’s slope is \(\displaystyle{-\frac{7}{6}}\text{.}\) +
    +
    -
    15.
    -
    -
    +
    15.
    +
    +
    -
    A line passes through the points \((5,7)\) and \((-4,7)\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((2,6)\) and \((-1,6)\text{.}\) Find this line’s slope.
    -Answer.
    \(0\)
    +
    Answer.
    \(0\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (2,6) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-1,6) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{6-6}{-1-2}\\ +\amp =\frac{0}{-3}\\ +\amp ={0} \end{aligned}}\)
    This is a special line which is parallel to the \(x\)-axis.
    So the line’s slope is \({0}\text{.}\)
    +
    -
    16.
    -
    -
    +
    +
    16.
    +
    +
    -
    A line passes through the points \((4,9)\) and \((-2,9)\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((5,9)\) and \((-4,9)\text{.}\) Find this line’s slope.
    -Answer.
    \(0\)
    +
    Answer.
    \(0\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (5,9) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-4,9) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{9-9}{-4-5}\\ +\amp =\frac{0}{-9}\\ +\amp ={0} \end{aligned}}\)
    This is a special line which is parallel to the \(x\)-axis.
    So the line’s slope is \({0}\text{.}\) +
    +
    -
    17.
    -
    -
    +
    17.
    +
    +
    -
    A line passes through the points \((-10,-4)\) and \((-10,1)\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((-10,-3)\) and \((-10,4)\text{.}\) Find this line’s slope.
    -Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    +
    Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-10,-3) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-10,4) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{4-(-3)}{-10 + 10}\\ +\amp =\frac{7}{0} \end{aligned}}\)
    Since we cannot divide by \(0\text{,}\) this line’s slope does not exist. This is a special line which is parallel to the \(y\)-axis; it is a vertical line.
    So the line’s slope Does Not Exist.
    +
    -
    18.
    -
    -
    +
    18.
    +
    +
    -
    A line passes through the points \((-7,-1)\) and \((-7,3)\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((-8,-5)\) and \((-8,1)\text{.}\) Find this line’s slope.
    -Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    +
    Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-8,-5) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-8,1) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{1-(-5)}{-8 + 8}\\ +\amp =\frac{6}{0} \end{aligned}}\)
    Since we cannot divide by \(0\text{,}\) this line’s slope does not exist. This is a special line which is parallel to the \(y\)-axis; it is a vertical line.
    So the line’s slope Does Not Exist.
    +
    -
    19.
    -
    -
    +
    19.
    +
    +
    -
    A line’s graph is given. What is this line’s slope-intercept equation?
    +
    A line’s graph is given. What is this line’s slope-intercept equation?
    -Answer.
    \(y=-\frac{3}{5}x+1\)
    +
    Answer.
    \(y=-\frac{3}{8}x\)
    Explanation.
    +
    A line’s slope-intercept equation has the form \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-coordinate of the \(y\)-intercept. We first find the slope.
    To find the slope of a line from its graph, we identify two points, and then draw a slope triangle. It’s wise to choose points with integer coordinates. For this problem, we choose \((0,0)\) and \((8,-3)\text{.}\) +
    Next, we draw a slope triangle and find the "rise" and "run". In this problem, the rise is \(-3\) and the run is \(8\text{.}\) +
    +\begin{equation*} +\begin{aligned}\text{slope}\amp =\frac{\text{rise}}{\text{run}}\\\amp =\frac{-3}{8}\\\amp =-{\frac{3}{8}}\end{aligned} +\end{equation*} +
    This line’s slope is \(-{\frac{3}{8}}\text{.}\) +
    It’s clear in the graph that this line’s \(y\)-intercept is \((0,0)\text{.}\) +
    So this line’s slope-intercept equation is \(y= -{\frac{3}{8}} x+ 0\text{.}\)
    +
    -
    20.
    -
    -
    +
    +
    20.
    +
    +
    -
    A line’s graph is given. What is this line’s slope-intercept equation?
    +
    A line’s graph is given. What is this line’s slope-intercept equation?
    -Answer.
    \(y=-\frac{4}{3}x-1\)
    +
    Answer.
    \(y=-\frac{4}{5}x+4\)
    Explanation.
    +
    A line’s slope-intercept equation has the form \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-coordinate of the \(y\)-intercept. We first find the slope.
    To find the slope of a line from its graph, we identify two points, and then draw a slope triangle. It’s wise to choose points with integer coordinates. For this problem, we choose \((0,4)\) and \((5,0)\text{.}\) +
    Next, we draw a slope triangle and find the "rise" and "run". In this problem, the rise is \(-4\) and the run is \(5\text{.}\) +
    +\begin{equation*} +\begin{aligned}\text{slope}\amp =\frac{\text{rise}}{\text{run}}\\\amp =\frac{-4}{5}\\\amp =-{\frac{4}{5}}\end{aligned} +\end{equation*} +
    This line’s slope is \(-{\frac{4}{5}}\text{.}\) +
    It’s clear in the graph that this line’s \(y\)-intercept is \((0,4)\text{.}\) +
    So this line’s slope-intercept equation is \(y= -{\frac{4}{5}} x+ 4\text{.}\) +
    +
    -
    21.
    -
    -
    +
    21.
    +
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 8x-5y=20 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 2x-5y=-10 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -Answer 1.
    \({\frac{8}{5}}\)
    -Answer 2.
    \(\left(0,-4\right)\)
    -
    -
    -
    22.
    -
    -
    +
    Answer 1.
    \({\frac{2}{5}}\)
    Answer 2.
    \(\left(0,2\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 2x-5y=-10 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +2x-5y \amp = -10 \\ +2x-5y\mathbf{{}-2x} \amp = -10\mathbf{{}-2x} \\ +-5y \amp = -2x - 10 \\ +\frac{-5y}{-5} \amp = \frac{-2x}{-5} + \frac{-10}{-5} \\ +y \amp = \frac{2}{5}x + 2 +\end{aligned} +}\)
    Now we can see the line’s slope is \(\displaystyle{ \frac{2}{5} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,2\right)}\text{.}\) +
    +
    +
    +
    +
    22.
    +
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 5x-6y=-18 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 7x-6y=18 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -Answer 1.
    \({\frac{5}{6}}\)
    -Answer 2.
    \(\left(0,3\right)\)
    -
    -
    -
    23.
    -
    -
    +
    Answer 1.
    \({\frac{7}{6}}\)
    Answer 2.
    \(\left(0,-3\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 7x-6y=18 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +7x-6y \amp = 18 \\ +7x-6y\mathbf{{}-7x} \amp = 18\mathbf{{}-7x} \\ +-6y \amp = -7x+18 \\ +\frac{-6y}{-6} \amp = \frac{-7x}{-6} + \frac{18}{-6} \\ +y \amp = \frac{7}{6}x - 3 +\end{aligned} +}\)
    Now we can see the line’s slope is \(\displaystyle{ \frac{7}{6} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,-3\right)}\text{.}\) +
    +
    +
    + +
    23.
    +
    +
    -
    A line passes through the points \((-9,{-18})\) and \((-3,{-4})\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((-9,{-18})\text{,}\) this line’s point-slope form equation is .
    Using the point \((-3,{-4})\text{,}\) this line’s point-slope form equation is .
    +
    A line passes through the points \((-15,{-28})\) and \((5,{0})\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((-15,{-28})\text{,}\) this line’s point-slope form equation is .
    Using the point \((5,{0})\text{,}\) this line’s point-slope form equation is .
    -Answer 1.
    \(y = \frac{7}{3}\mathopen{}\left(x+9\right)+-18\)
    -Answer 2.
    \(y = \frac{7}{3}\mathopen{}\left(x+3\right)+-4\)
    -
    -
    -
    24.
    -
    -
    +
    Answer 1.
    \(y = \frac{7}{5}\mathopen{}\left(x+15\right)+-28\)
    Answer 2.
    \(y = \frac{7}{5}\mathopen{}\left(x-5\right)\)
    Explanation.
    +
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_0\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-15,{-28}) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (5,{0}) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{{0}-((-28))}{5-(-15)}\\\amp =\frac{{28}}{20}\\\amp ={{\frac{7}{5}}}\end{aligned} }\)
    Now we have \(y={{\frac{7}{5}}}(x-x_{0})+y_0\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((-15,{-28})\text{,}\) we have:
    \(\begin{aligned} +y \amp = m(x-x_{0})+y_0 \\ +{y} \amp = {{\frac{7}{5}}}(x + 15)+{-28} +\end{aligned}\)
    If we choose to use the point \((5,{0})\text{,}\) we have:
    \(\begin{aligned} +y \amp = m(x-x_{0})+y_0 \\ +{y} \amp = {{\frac{7}{5}}}(x-5)+{0} +\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    +
    +
    + +
    24.
    +
    +
    -
    A line passes through the points \((15,{15})\) and \((0,{-9})\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((15,{15})\text{,}\) this line’s point-slope form equation is .
    Using the point \((0,{-9})\text{,}\) this line’s point-slope form equation is .
    +
    A line passes through the points \((7,{17})\) and \((-7,{1})\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((7,{17})\text{,}\) this line’s point-slope form equation is .
    Using the point \((-7,{1})\text{,}\) this line’s point-slope form equation is .
    -Answer 1.
    \(y = \frac{8}{5}\mathopen{}\left(x-15\right)+15\)
    -Answer 2.
    \(y = \frac{8}{5}\mathopen{}\left(x-0\right)+-9\)
    +
    Answer 1.
    \(y = \frac{8}{7}\mathopen{}\left(x-7\right)+17\)
    Answer 2.
    \(y = \frac{8}{7}\mathopen{}\left(x+7\right)+1\)
    Explanation.
    +
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_0\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (7,{17}) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-7,{1}) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{{1}-{17}}{-7-7}\\\amp =\frac{{-16}}{-14}\\\amp ={{\frac{8}{7}}}\end{aligned} }\)
    Now we have \(y={{\frac{8}{7}}}(x-x_{0})+y_0\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((7,{17})\text{,}\) we have:
    \(\begin{aligned} +y \amp = m(x-x_{0})+y_0 \\ +{y} \amp = {{\frac{8}{7}}}(x-7)+{17} +\end{aligned}\)
    If we choose to use the point \((-7,{1})\text{,}\) we have:
    \(\begin{aligned} +y \amp = m(x-x_{0})+y_0 \\ +{y} \amp = {{\frac{8}{7}}}(x + 7)+{1} +\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    +
    -

    25.

    -
    -
    +

    25.

    +
    +
    -
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(5.6\) grams. Seven minutes since the experiment started, the remaining gas had a mass of \(240.8\) grams.
    +
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(7.6\) grams. Five minutes since the experiment started, the remaining gas had a mass of \(334.4\) grams.
    Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
      -
    1. This line’s slope-intercept equation is .
      2. -
    2. -\(34\) minutes after the experiment started, there would be grams of gas left.
      3. -
    3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
      4. +
    4. This line’s slope-intercept equation is .
      5. +
    5. +\(37\) minutes after the experiment started, there would be grams of gas left.
      6. +
    6. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
    -Answer 1.
    \(y = -5.6x+280\)
    -Answer 2.
    \(89.6\)
    -Answer 3.
    \(50\)
    +
    Answer 1.
    \(y = -7.6x+372.4\)
    Answer 2.
    \(91.2\)
    Answer 3.
    \(49\)
    Explanation.
      +
    1. +
      A line’s equation in slope-intercept form looks like \(y=mx+b\text{,}\) where \(m\) is the slope, and \(y\) is the \(y\)-coordinate of the \(y\)-intercept. Since the gas is leaking at a rate of \(7.6\) grams per minute, we know that the slope of the linear model is \(-7.6\text{.}\) +
      +
      So now we have that \(y=-7.6x+b\text{.}\) The next step is to find the value of \(b\text{.}\) There are at least two ways.
      +
      One way is to substitute a known point into \(y=-7.6x+b\text{.}\) We use the point \((5,334.4)\text{.}\) +
      +
      \(\begin{aligned} +y \amp = -7.6x + b \\ +334.4 \amp = -7.6 \cdot 5 + b \\ +334.4 \amp = -38 + b \\ +334.4\mathbf{{} + 38} \amp = -38 + b\mathbf{{} + 38} \\ +372.4 \amp = b\\ +b \amp = 372.4 +\end{aligned}\)
      +
      This line’s slope-intercept equation is \(y=-7.6x+372.4\text{.}\) +
      +
      Another way is to use the line’s point-slope equation \(y-y_{1}=m(x-x_{1})\text{.}\) Again, we use the point \((5,334.4)\text{.}\) +
      +
      \(\begin{aligned} +y-y_{1} \amp = m(x-x_{1}) \\ +y-334.4 \amp = -7.6(x-5) \\ +y-334.4 \amp = -7.6x - 7.6 \cdot (-5) \\ +y-334.4 \amp = -7.6x+38 \\ +y-334.4\mathbf{{}+334.4} \amp = -7.6x+38\mathbf{{}+334.4} \\ +y \amp = -7.6x+372.4 +\end{aligned}\)
      +
      This is the second way to find \(b\text{,}\) and thus the line’s slope-intercept equation is \(y=-7.6x+372.4\text{.}\) +
      +
      2. +
    2. +
      After \(37\) minutes, we can find the mass of the remaining gas if we substitute \(37\) in for \(x\) in the equation \(y=-7.6x+372.4\text{.}\) We have:
      +
      \(\displaystyle{\begin{aligned} +y \amp = -7.6x+372.4 \\ +y \amp = -7.6 \cdot 37 +372.4 \\ +y \amp = -281.2+372.4 \\ +y \amp = 91.2 +\end{aligned} +}\)
      +
      So after \(37\) minutes, there are \(91.2\) grams of gas remaining.
      +
      3. +
    3. +
      To find when the gas will be all gone, we need to find when there are \(0\) grams of gas left. We can substitute \(0\) in for \(y\) in the equation \(y=-7.6x+372.4\text{,}\) and we have:
      +
      \(\displaystyle{\begin{aligned} +y \amp = -7.6x+372.4 \\ +0 \amp = -7.6x +372.4 \\ +0\mathbf{{}-372.4} \amp = -7.6x +372.4\mathbf{{}-372.4}\\ +-372.4 \amp = -7.6x \\ +\frac{-372.4}{-7.6} \amp = \frac{-7.6x}{-7.6} \\ +49 \amp = x +\end{aligned} +}\)
      +
      So after \(49\) minutes, the gas will be all gone.
      +
      4. +
    -

    26.

    -
    -
    +

    26.

    +
    +
    -
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(2.5\) grams. Ten minutes since the experiment started, the remaining gas had a mass of \(75\) grams.
    +
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(4.5\) grams. Seven minutes since the experiment started, the remaining gas had a mass of \(148.5\) grams.
    Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
      -
    1. This line’s slope-intercept equation is .
      2. -
    2. -\(33\) minutes after the experiment started, there would be grams of gas left.
      3. -
    3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
      4. +
    4. This line’s slope-intercept equation is .
      5. +
    5. +\(37\) minutes after the experiment started, there would be grams of gas left.
      6. +
    6. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
    -Answer 1.
    \(y = -2.5x+100\)
    -Answer 2.
    \(17.5\)
    -Answer 3.
    \(40\)
    +
    Answer 1.
    \(y = -4.5x+180\)
    Answer 2.
    \(13.5\)
    Answer 3.
    \(40\)
    Explanation.
      +
    1. +
      A line’s equation in slope-intercept form looks like \(y=mx+b\text{,}\) where \(m\) is the slope, and \(y\) is the \(y\)-coordinate of the \(y\)-intercept. Since the gas is leaking at a rate of \(4.5\) grams per minute, we know that the slope of the linear model is \(-4.5\text{.}\) +
      +
      So now we have that \(y=-4.5x+b\text{.}\) The next step is to find the value of \(b\text{.}\) There are at least two ways.
      +
      One way is to substitute a known point into \(y=-4.5x+b\text{.}\) We use the point \((7,148.5)\text{.}\) +
      +
      \(\begin{aligned} +y \amp = -4.5x + b \\ +148.5 \amp = -4.5 \cdot 7 + b \\ +148.5 \amp = -31.5 + b \\ +148.5\mathbf{{} + 31.5} \amp = -31.5 + b\mathbf{{} + 31.5} \\ +180 \amp = b\\ +b \amp = 180 +\end{aligned}\)
      +
      This line’s slope-intercept equation is \(y=-4.5x+180\text{.}\) +
      +
      Another way is to use the line’s point-slope equation \(y-y_{1}=m(x-x_{1})\text{.}\) Again, we use the point \((7,148.5)\text{.}\) +
      +
      \(\begin{aligned} +y-y_{1} \amp = m(x-x_{1}) \\ +y-148.5 \amp = -4.5(x-7) \\ +y-148.5 \amp = -4.5x - 4.5 \cdot (-7) \\ +y-148.5 \amp = -4.5x+31.5 \\ +y-148.5\mathbf{{}+148.5} \amp = -4.5x+31.5\mathbf{{}+148.5} \\ +y \amp = -4.5x+180 +\end{aligned}\)
      +
      This is the second way to find \(b\text{,}\) and thus the line’s slope-intercept equation is \(y=-4.5x+180\text{.}\) +
      +
      2. +
    2. +
      After \(37\) minutes, we can find the mass of the remaining gas if we substitute \(37\) in for \(x\) in the equation \(y=-4.5x+180\text{.}\) We have:
      +
      \(\displaystyle{\begin{aligned} +y \amp = -4.5x+180 \\ +y \amp = -4.5 \cdot 37 +180 \\ +y \amp = -166.5+180 \\ +y \amp = 13.5 +\end{aligned} +}\)
      +
      So after \(37\) minutes, there are \(13.5\) grams of gas remaining.
      +
      3. +
    3. +
      To find when the gas will be all gone, we need to find when there are \(0\) grams of gas left. We can substitute \(0\) in for \(y\) in the equation \(y=-4.5x+180\text{,}\) and we have:
      +
      \(\displaystyle{\begin{aligned} +y \amp = -4.5x+180 \\ +0 \amp = -4.5x +180 \\ +0\mathbf{{}-180} \amp = -4.5x +180\mathbf{{}-180}\\ +-180 \amp = -4.5x \\ +\frac{-180}{-4.5} \amp = \frac{-4.5x}{-4.5} \\ +40 \amp = x +\end{aligned} +}\)
      +
      So after \(40\) minutes, the gas will be all gone.
      +
      4. +
    -

    27.

    -
    -
    +

    27.

    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -2 x + 7 y = -14 +2 x + 3 y = -18 \end{equation*}
    @@ -1368,24 +1698,68 @@

    Exercise Group.

    -\(x\)-intercept and \(y\)-intercept of the line \(2 x+7 y=-14\) +\(x\)-intercept and \(y\)-intercept of the line \(2 x+3 y=-18\)
    -Answer 1.
    \(0\)
    -Answer 2.
    \(-2\)
    -Answer 3.
    \(\left(0,-2\right)\)
    -Answer 4.
    \(-7\)
    -Answer 5.
    \(0\)
    -Answer 6.
    \(\left(-7,0\right)\)
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(-6\)
    Answer 3.
    \(\left(0,-6\right)\)
    Answer 4.
    \(-9\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-9,0\right)\)
    Explanation.
    +
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    +\begin{equation*} +\begin{aligned} +2 x+3 y \amp = -18\\ +2 (0)+3 y \amp = -18\\ +3 y \amp = -18 \\ +\frac{3 y}{3} \amp = \frac{-18}{3}\\ +y \amp = -6 +\end{aligned} +\end{equation*} +
    This line’s \(y\)-intercept is \((0,-6)\text{.}\) +
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    +\begin{equation*} +\begin{aligned} +2 x+3 y \amp = -18\\ +2 x+3 (0) \amp = -18\\ +2 x \amp = -18\\ +\frac{2 x}{2} \amp = \frac{-18}{2}\\ +x \amp = -9 +\end{aligned} +\end{equation*} +
    The line’s \(x\)-intercept is \((-9,0)\text{.}\) +
    The entries for the table are:
    + + + + + + + + + + + + + + + + + + +
    +\(x\)-value +\(y\)-valueLocation
    +\(y\)-intercept\(0\)\(-6\)\((0,-6)\)
    +\(x\)-intercept\(-9\)\(0\)\((-9,0)\)
    +\(x\)-intercept and \(y\)-intercept of the line \(2 x+3 y=-18\) +
    +
    -

    28.

    -
    -
    +

    28.

    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -4 x + 5 y = -60 +2 x + 7 y = -14 \end{equation*}
    @@ -1411,78 +1785,161 @@

    Exercise Group.

    -\(x\)-intercept and \(y\)-intercept of the line \(4 x+5 y=-60\) +\(x\)-intercept and \(y\)-intercept of the line \(2 x+7 y=-14\)
    -Answer 1.
    \(0\)
    -Answer 2.
    \(-12\)
    -Answer 3.
    \(\left(0,-12\right)\)
    -Answer 4.
    \(-15\)
    -Answer 5.
    \(0\)
    -Answer 6.
    \(\left(-15,0\right)\)
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(-2\)
    Answer 3.
    \(\left(0,-2\right)\)
    Answer 4.
    \(-7\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-7,0\right)\)
    Explanation.
    +
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    +\begin{equation*} +\begin{aligned} +2 x+7 y \amp = -14\\ +2 (0)+7 y \amp = -14\\ +7 y \amp = -14 \\ +\frac{7 y}{7} \amp = \frac{-14}{7}\\ +y \amp = -2 +\end{aligned} +\end{equation*} +
    This line’s \(y\)-intercept is \((0,-2)\text{.}\) +
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    +\begin{equation*} +\begin{aligned} +2 x+7 y \amp = -14\\ +2 x+7 (0) \amp = -14\\ +2 x \amp = -14\\ +\frac{2 x}{2} \amp = \frac{-14}{2}\\ +x \amp = -7 +\end{aligned} +\end{equation*} +
    The line’s \(x\)-intercept is \((-7,0)\text{.}\) +
    The entries for the table are:
    + + + + + + + + + + + + + + + + + + +
    +\(x\)-value +\(y\)-valueLocation
    +\(y\)-intercept\(0\)\(-2\)\((0,-2)\)
    +\(x\)-intercept\(-7\)\(0\)\((-7,0)\)
    +\(x\)-intercept and \(y\)-intercept of the line \(2 x+7 y=-14\) +
    +
    -
    +

    Exercise Group.

    -
    29.
    -
    -
    +
    29.
    +
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -{5}x+y= -8 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -{5}x+y= -3 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -Answer 1.
    \(5\)
    -Answer 2.
    \(\left(0,-8\right)\)
    +
    Answer 1.
    \(5\)
    Answer 2.
    \(\left(0,-3\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.
    In this problem, the line’s equation is given as \(\displaystyle{ -{5}x+y= -3 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\) +
    \(\displaystyle{\begin{aligned} +-{5}x+y \amp = -3 \\ +-{5}x+y\mathbf{{}+{5}x} \amp = -3\mathbf{{}+{5}x} \\ +y \amp = {5}x - 3 +\end{aligned} +}\)
    Now we can see the line’s slope is \({5}\text{,}\) and its \(y\)-intercept has coordinates \((0,-3)\text{.}\) +
    +
    -
    30.
    -
    -
    +
    30.
    +
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -x-y= 6 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -x-y= -1 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -Answer 1.
    \(-1\)
    -Answer 2.
    \(\left(0,-6\right)\)
    -
    -
    -
    31.
    -
    -
    +
    Answer 1.
    \(-1\)
    Answer 2.
    \(\left(0,-\left(-1\right)\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.
    In this problem, the line’s equation is given as \(\displaystyle{ -x-y= -1 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\) +
    \(\displaystyle{\begin{aligned} +-x-y \amp = -1 \\ +-x-y\mathbf{{}+x} \amp = -1\mathbf{{}+x} \\ +-y \amp = x - 1 \\ +(-1) \cdot (-y) \amp = (-1) \cdot (x - 1) \\ +y \amp = (-1) \cdot x+(-1) \cdot -1 \\ +y \amp = -x + 1 \\ +\end{aligned} +}\)
    Now we can see the line’s slope is \({-1}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,1\right)}\text{.}\) +
    +
    +
    +
    +
    31.
    +
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 12x+9y=4 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 12x+15y=4 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -Answer 1.
    \(-{\frac{4}{3}}\)
    -Answer 2.
    \(\left(0,\frac{4}{9}\right)\)
    -
    -
    -
    32.
    -
    -
    +
    Answer 1.
    \(-{\frac{4}{5}}\)
    Answer 2.
    \(\left(0,\frac{4}{15}\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 12x+15y=4 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +12x+15y \amp = 4 \\ +12x+15y\mathbf{{}-12x} \amp = 4\mathbf{{}-12x} \\ +15y \amp = -12x +4 \\ +\frac{15y}{15} \amp = \frac{-12x}{15} + \frac{4}{15}\\ +y \amp = -\frac{12}{15}x + \frac{4}{15}\\ +y \amp = -\frac{4}{5}x + \frac{4}{15} +\end{aligned} +}\)
    Now we can see the line’s slope is \(-\frac{4}{5}\text{,}\) and its \(y\)-intercept has coordinates \(\left(0,\frac{4}{15}\right)\text{.}\) +
    +
    +
    +
    +
    32.
    +
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 15x+6y=1 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 20x+12y=5 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -Answer 1.
    \(-{\frac{5}{2}}\)
    -Answer 2.
    \(\left(0,\frac{1}{6}\right)\)
    -
    -
    -
    33.
    -
    -
    +
    Answer 1.
    \(-{\frac{5}{3}}\)
    Answer 2.
    \(\left(0,\frac{5}{12}\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 20x+12y=5 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +20x+12y \amp = 5 \\ +20x+12y\mathbf{{}-20x} \amp = 5\mathbf{{}-20x} \\ +12y \amp = -20x +5 \\ +\frac{12y}{12} \amp = \frac{-20x}{12} + \frac{5}{12}\\ +y \amp = -\frac{20}{12}x + \frac{5}{12}\\ +y \amp = -\frac{5}{3}x + \frac{5}{12} +\end{aligned} +}\)
    Now we can see the line’s slope is \(-\frac{5}{3}\text{,}\) and its \(y\)-intercept has coordinates \(\left(0,\frac{5}{12}\right)\text{.}\) +
    +
    +
    + +
    33.
    +
    +
    -
    Fill out this table for the equation \(x=-2\text{.}\) The first row is an example.
    +
    Fill out this table for the equation \(x=-3\text{.}\) The first row is an example.
    - - + + @@ -1509,26 +1966,51 @@

    Exercise Group.

    -
    \(x\) \(y\) Points
    \(-2\) \(-3\)\(\left(-2,-3\right)\)\(-3\)\(\left(-3,-3\right)\)
    \(2\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-2\) +
    Values of \(x\) and \(y\) satisfying the equation \(x=-3\)
    -Answer 1.
    \(-2\)
    -Answer 2.
    \(\left(-2,-2\right)\)
    -Answer 3.
    \(-2\)
    -Answer 4.
    \(\left(-2,-1\right)\)
    -Answer 5.
    \(-2\)
    -Answer 6.
    \(\left(-2,0\right)\)
    -Answer 7.
    \(-2\)
    -Answer 8.
    \(\left(-2,1\right)\)
    -Answer 9.
    \(-2\)
    -Answer 10.
    \(\left(-2,2\right)\)
    -
    -
    -
    34.
    -
    -
    +
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(-3,-2\right)\)
    Answer 3.
    \(-3\)
    Answer 4.
    \(\left(-3,-1\right)\)
    Answer 5.
    \(-3\)
    Answer 6.
    \(\left(-3,0\right)\)
    Answer 7.
    \(-3\)
    Answer 8.
    \(\left(-3,1\right)\)
    Answer 9.
    \(-3\)
    Answer 10.
    \(\left(-3,2\right)\)
    Explanation.
    +
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    \(x\)\(y\)Points
    \(-3\)\(-2\)\(\left(-3,-2\right)\)
    \(-3\)\(-1\)\(\left(-3,-1\right)\)
    \(-3\)\(0\)\(\left(-3,0\right)\)
    \(-3\)\(1\)\(\left(-3,1\right)\)
    \(-3\)\(2\)\(\left(-3,2\right)\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-3\) +
    +
    +
    + +
    34.
    +
    +
    -
    Fill out this table for the equation \(x=-1\text{.}\) The first row is an example.
    +
    Fill out this table for the equation \(x=-1\text{.}\) The first row is an example.
    @@ -1567,119 +2049,158 @@

    Exercise Group.

    \(x\) \(y\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-1\)
    -Answer 1.
    \(-1\)
    -Answer 2.
    \(\left(-1,-2\right)\)
    -Answer 3.
    \(-1\)
    -Answer 4.
    \(\left(-1,-1\right)\)
    -Answer 5.
    \(-1\)
    -Answer 6.
    \(\left(-1,0\right)\)
    -Answer 7.
    \(-1\)
    -Answer 8.
    \(\left(-1,1\right)\)
    -Answer 9.
    \(-1\)
    -Answer 10.
    \(\left(-1,2\right)\)
    -
    -
    -
    35.
    -
    -
    -
    -
    A line’s graph is shown. Write an equation for the line.
    -
    -Answer.
    \(y = -4\)
    +
    Answer 1.
    \(-1\)
    Answer 2.
    \(\left(-1,-2\right)\)
    Answer 3.
    \(-1\)
    Answer 4.
    \(\left(-1,-1\right)\)
    Answer 5.
    \(-1\)
    Answer 6.
    \(\left(-1,0\right)\)
    Answer 7.
    \(-1\)
    Answer 8.
    \(\left(-1,1\right)\)
    Answer 9.
    \(-1\)
    Answer 10.
    \(\left(-1,2\right)\)
    Explanation.
    +
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    \(x\)\(y\)Points
    \(-1\)\(-2\)\(\left(-1,-2\right)\)
    \(-1\)\(-1\)\(\left(-1,-1\right)\)
    \(-1\)\(0\)\(\left(-1,0\right)\)
    \(-1\)\(1\)\(\left(-1,1\right)\)
    \(-1\)\(2\)\(\left(-1,2\right)\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-1\) +
    +
    -
    36.
    -
    -
    +
    35.
    +
    +
    -
    A line’s graph is shown. Write an equation for the line.
    +
    A line’s graph is shown. Write an equation for the line.
    -Answer.
    \(y = -3\)
    +
    Answer.
    \(y = -4\)
    Explanation.
    +
    This line is horizontal, passing \((0,-4)\) and \((1,-4)\text{.}\) This implies its slope is \(0\text{.}\) +
    If a line’s slope is \(0\text{,}\) its equation has the form \(y=b\text{.}\) By the graph, we can see that its \(y\)-intercept is \(-4\text{,}\) so this line’s equation is \(y=-4\text{.}\)
    +
    -
    37.
    -
    -
    -
    -
    Line \(m\) passes points \((-5,-3)\) and \((-5,0)\text{.}\) -
    Line \(n\) passes points \((-7,-6)\) and \((-7,1)\text{.}\) -
    -
    These two lines are
    -
      -
    • parallel
      • -
    • perpendicular
      • -
    • neither parallel nor perpendicular
      • -
    -
    .
    +
    36.
    +
    +
    +
    +
    A line’s graph is shown. Write an equation for the line.
    -Answer.
    \(\text{parallel}\)
    +
    Answer.
    \(y = -3\)
    Explanation.
    +
    This line is horizontal, passing \((0,-3)\) and \((1,-3)\text{.}\) This implies its slope is \(0\text{.}\) +
    If a line’s slope is \(0\text{,}\) its equation has the form \(y=b\text{.}\) By the graph, we can see that its \(y\)-intercept is \(-3\text{,}\) so this line’s equation is \(y=-3\text{.}\)
    +
    -
    38.
    -
    -
    -
    -
    Line \(m\) passes points \((-3,10)\) and \((-3,9)\text{.}\) -
    Line \(n\) passes points \((8,-6)\) and \((8,-9)\text{.}\) -
    -
    These two lines are
    -
      -
    • parallel
      • -
    • perpendicular
      • -
    • neither parallel nor perpendicular
      • -
    -
    .
    +
    37.
    +
    +
    +
    +
    Line \(m\) passes points \((-6,1)\) and \((-6,-10)\text{.}\) +
    Line \(n\) passes points \((0,-2)\) and \((0,8)\text{.}\) +
    These two lines are .
    -Answer.
    \(\text{parallel}\)
    -
    -
    -
    39.
    -
    -
    +
    Answer.
    \(\text{parallel}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-10-1}{-6-(-6)}\\ +\amp =\frac{-11}{0}\\ +\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=-6\text{.}\) It is vertical with an undefined slope.
    For line \(n\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{8-(-2)}{0-0}\\ +\amp =\frac{10}{0}\\ +\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=0\text{.}\) It is vertical with an undefined slope.
    Since both lines are vertical, these two lines are parallel.
    +
    +
    + +
    38.
    +
    +
    -
    Line \(k\)’s equation is \(y={-{\frac{5}{7}}x+1}\text{.}\) -
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-2,{-{\frac{29}{5}}})\text{.}\) -
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    +
    Line \(m\) passes points \((-3,-6)\) and \((-3,-2)\text{.}\) +
    Line \(n\) passes points \((9,0)\) and \((9,8)\text{.}\) +
    These two lines are .
    -Answer 1.
    \(y = {\frac{7}{5}}\mathopen{}\left(x+2\right)+-{\frac{29}{5}}\)
    -Answer 2.
    \(y = {\frac{7}{5}}x-3\)
    -
    -
    -
    40.
    -
    -
    +
    Answer.
    \(\text{parallel}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-2-(-6)}{-3-(-3)}\\ +\amp =\frac{4}{0}\\ +\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=-3\text{.}\) It is vertical with an undefined slope.
    For line \(n\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{8-0}{9-9}\\ +\amp =\frac{8}{0}\\ +\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=9\text{.}\) It is vertical with an undefined slope.
    Since both lines are vertical, these two lines are parallel.
    +
    +
    + +
    39.
    +
    +
    +
    +
    Line \(k\)’s equation is \(y={{\frac{5}{8}}x+2}\text{.}\) +
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-3,{{\frac{19}{5}}})\text{.}\) +
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    +
    +
    Answer 1.
    \(y = -{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{19}{5}}\)
    Answer 2.
    \(y = -{\frac{8}{5}}x-1\)
    Explanation.
    +
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) It’s given that line \(k\)’s slope is \({{\frac{5}{8}}}\text{.}\) So the slope of \(\ell\) must be \({-{\frac{8}{5}}}\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-3,{{\frac{19}{5}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{19}{5}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {-{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{19}{5}}}\\ +y \amp = {-{\frac{8}{5}}}x - {-{\frac{8}{5}}}(-3) +{{\frac{19}{5}}}\\ +y \amp = {-{\frac{8}{5}}x-1} +\end{aligned} +}\)
    +
    +
    +
    +
    40.
    +
    +
    -
    Line \(k\)’s equation is \(y={-{\frac{6}{7}}x-4}\text{.}\) -
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-1,{-{\frac{37}{6}}})\text{.}\) -
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    +
    Line \(k\)’s equation is \(y={{\frac{6}{5}}x-4}\text{.}\) +
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-2,{{\frac{20}{3}}})\text{.}\) +
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -Answer 1.
    \(y = {\frac{7}{6}}\mathopen{}\left(x+1\right)+-{\frac{37}{6}}\)
    -Answer 2.
    \(y = {\frac{7}{6}}x-5\)
    +
    Answer 1.
    \(y = -{\frac{5}{6}}\mathopen{}\left(x+2\right)+{\frac{20}{3}}\)
    Answer 2.
    \(y = -{\frac{5}{6}}x+5\)
    Explanation.
    +
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) It’s given that line \(k\)’s slope is \({{\frac{6}{5}}}\text{.}\) So the slope of \(\ell\) must be \({-{\frac{5}{6}}}\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-2,{{\frac{20}{3}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{5}{6}}\mathopen{}\left(x+2\right)+{\frac{20}{3}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {-{\frac{5}{6}}\mathopen{}\left(x+2\right)+{\frac{20}{3}}}\\ +y \amp = {-{\frac{5}{6}}}x - {-{\frac{5}{6}}}(-2) +{{\frac{20}{3}}}\\ +y \amp = {-{\frac{5}{6}}x+5} +\end{aligned} +}\)
    +
    -
    41.
    -
    Graph the linear inequality \(y\gt \frac{4}{3}x+1\text{.}\) +
    41.
    +
    Graph the linear inequality \(y\gt \frac{4}{3}x+1\text{.}\)
    -
    -Answer.
    -
    42.
    -
    Graph the linear inequality \(y\leq -\frac{1}{2}x-3\text{.}\) +
    Answer.
    42.
    +
    Graph the linear inequality \(y\leq -\frac{1}{2}x-3\text{.}\)
    -
    -Answer.
    -
    43.
    -
    Graph the linear inequality \(y\geq 3\text{.}\) +
    Answer.
    43.
    +
    Graph the linear inequality \(y\geq 3\text{.}\)
    -
    -Answer.
    -
    44.
    -
    Graph the linear inequality \(3x+2y\lt -6\text{.}\) +
    Answer.
    44.
    +
    Graph the linear inequality \(3x+2y\lt -6\text{.}\)
    -
    -Answer.
    -
    +
    Answer.
    diff --git a/review-linear-equations-and-inequalities.html b/review-linear-equations-and-inequalities.html index f75d5cfc2..239a1497f 100644 --- a/review-linear-equations-and-inequalities.html +++ b/review-linear-equations-and-inequalities.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/review-systems-of-linear-equations.html b/review-systems-of-linear-equations.html index a5534b80c..0749cdb5d 100644 --- a/review-systems-of-linear-equations.html +++ b/review-systems-of-linear-equations.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -603,24 +617,24 @@

    Search Results:

    Checkpoint 4.4.1.
    The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system.
    -
    -
    -
    a coordinate plot of two lines that cross at (1, 4)
    +
    +
    +
    a coordinate plot of two lines that cross at (3, 1)

    Checkpoint 4.4.2.

    Solve the given system of linear equations graphically.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-{\frac{8}{7}}\mathopen{}\left(x-9\right)-2}\\ -y \amp= {{\frac{1}{4}}\mathopen{}\left(x+6\right)+4} +y \amp= {-{\frac{3}{10}}\mathopen{}\left(x+4\right)-1}\\ +y \amp= {-{\frac{4}{13}}\mathopen{}\left(x+7\right)} \end{alignedat} \right.\)
    -
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (2, 6)
    So the only solution is \((2,6)\text{.}\) +
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (6, -4)
    So the only solution is \((6,-4)\text{.}\)
    @@ -628,17 +642,17 @@

    Search Results:

    Checkpoint 4.4.3.
    Solve the given system of linear equations graphically.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp {-4x+3y = -12}\\ -\amp {2x-3y = -6} +\amp {4x+5y = 20}\\ +\amp {x+y = 3} \end{alignedat} \right.\)
    -
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (9, 8)
    So the only solution is \((9,8)\text{.}\) +
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-5, 8)
    So the only solution is \((-5,8)\text{.}\)
    @@ -648,11 +662,11 @@

    Search Results:

    You may find that after you make a substitution and simplify, both of the variables disappear instead of just one. When this happens, you either had an inconsistent system (with no solution) or a dependent system (with infinitely many solutions). If the equation you now have is an outright false equation (like \(0=1\)) then you had an inconsistent system. If it is an outright true equation (like \(2=2\)) then you had a dependent system.

    Checkpoint 4.4.4.

    -
    -
    +
    +
    -
    -
    Use substitution to solve the system.
    +
    +
    Use substitution to solve the system.
    \begin{equation*} \begin{aligned} \left\{ @@ -664,15 +678,15 @@

    Search Results:

    \end{aligned} \end{equation*}
    -

    (a)

    -
    Isolate one of the variables in one of the equations.
    -
    Explanation.
    We choose to isolate \(x\) in the first equation. All we need to do is add \(3y\) to each side, and then \(x=3y+1\text{.}\) -

    (b)

    -
    Substitute the isolated expression into the other equation.
    -
    Explanation.
    Making the substitution: \(2\left(3y+1\right)+y=-3\text{.}\) -

    (c)

    -
    Solve for the one variable that remains.
    -
    Explanation.
    +

    (a)

    +
    Isolate one of the variables in one of the equations.
    +
    Explanation.
    We choose to isolate \(x\) in the first equation. All we need to do is add \(3y\) to each side, and then \(x=3y+1\text{.}\) +

    (b)

    +
    Substitute the isolated expression into the other equation.
    +
    Explanation.
    Making the substitution: \(2\left(3y+1\right)+y=-3\text{.}\) +

    (c)

    +
    Solve for the one variable that remains.
    +
    Explanation.
    \begin{equation*} \begin{aligned} 2\left(3y+1\right)+y \amp= -3\\ @@ -682,9 +696,9 @@

    Search Results:

    y\amp=-\frac{5}{7} \end{aligned} \end{equation*} -

    (d)

    -
    Solve for the other variable.
    -
    Explanation.
    +

    (d)

    +
    Solve for the other variable.
    +
    Explanation.
    \begin{equation*} \begin{aligned} x \amp= 3y+1\\ @@ -698,25 +712,25 @@

    Search Results:

    Checkpoint 4.4.5.

    -
    -
    +
    +
    -
    Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for \(\$12.50\) per pound. They have some coffee beans from Columbia that sell for \(\$9.00\) per pound and some coffee beans from Honduras that sell for \(\$14.00\) per pound. How many pounds of each should they mix to make \(30\) pounds of the blend?

    (a)

    -
    Write two equations that form a system for this scenario.
    -
    Explanation.
    -
    We have \(C\) pounds of Columbian coffee and \(H\) pounds of Honduran coffee. Since there must be \(30\) pounds total, one equation is \(C+H=30\text{.}\) -
    The total cost of the Colombian coffee in the blend will be \(9C\text{,}\) and the total cost of the Honduran coffee in the blend will be \(14H\text{.}\) All together, the blend has a total cost of \(12.50\cdot30=375\) dollars. So another equation is \(9C+14H=375\text{.}\) +
    Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for \(\$12.50\) per pound. They have some coffee beans from Columbia that sell for \(\$9.00\) per pound and some coffee beans from Honduras that sell for \(\$14.00\) per pound. How many pounds of each should they mix to make \(30\) pounds of the blend?

    (a)

    +
    Write two equations that form a system for this scenario.
    +
    Explanation.
    +
    We have \(C\) pounds of Columbian coffee and \(H\) pounds of Honduran coffee. Since there must be \(30\) pounds total, one equation is \(C+H=30\text{.}\) +
    The total cost of the Colombian coffee in the blend will be \(9C\text{,}\) and the total cost of the Honduran coffee in the blend will be \(14H\text{.}\) All together, the blend has a total cost of \(12.50\cdot30=375\) dollars. So another equation is \(9C+14H=375\text{.}\)
    -

    (b)

    -
    Isolate one of the variables in one of the equations.
    -
    Explanation.
    We choose to isolate \(C\) in the first equation. All we need to do subtract \(H\) from each side, and then \(C = 30 - H\text{.}\) -

    (c)

    -
    Substitute the isolated expression into the other equation.
    -
    Explanation.
    Making the substitution: \(9(30 - H) + 14H = 375\text{.}\) -

    (d)

    -
    Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use.
    -
    Explanation.
    -
    +

    (b)

    +
    Isolate one of the variables in one of the equations.
    +
    Explanation.
    We choose to isolate \(C\) in the first equation. All we need to do subtract \(H\) from each side, and then \(C = 30 - H\text{.}\) +

    (c)

    +
    Substitute the isolated expression into the other equation.
    +
    Explanation.
    Making the substitution: \(9(30 - H) + 14H = 375\text{.}\) +

    (d)

    +
    Solve this equation in one variable and then solve for the other variable using the isolation from a previous step. Finally, report how much of each type of coffee Desi should use.
    +
    Explanation.
    +
    \begin{equation*} \begin{aligned} 9(30 - H) + 14H \amp= 375\\ @@ -726,26 +740,26 @@

    Search Results:

    H\amp=21 \end{aligned} \end{equation*} -
    +
    \begin{equation*} \begin{aligned} C \amp= 30 - H\\ C \amp= 30 - 21 = 9 \end{aligned} \end{equation*} -
    In summary, Desi needs to mix \(21\) pounds of the Honduran coffee beans with \(9\) pounds of the Columbian coffee beans to create this blend.
    +
    In summary, Desi needs to mix \(21\) pounds of the Honduran coffee beans with \(9\) pounds of the Columbian coffee beans to create this blend.

    Checkpoint 4.4.6.

    Solve the system of equations using substitution.
    -
    -
    -
    \(\left\{ +
    +
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-5d+1f = 1} \\ -\amp {f = 5\mathopen{}\left(d-5\right)+32} +\amp {5n+1T = -9} \\ +\amp {T = -5\mathopen{}\left(n-\left(-1\right)\right)+3} \end{alignedat} \right.\)

    Elimination.

    @@ -754,11 +768,11 @@

    Search Results:

    You may find that after you add to equations to eliminate a variable that both of the variables have been eliminated. When this happens, you either had an inconsistent system (with no solution) or a dependent system (with infinitely many solutions). If the equation you now have is an outright false equation (like \(0=1\)) then you had an inconsistent system. If it is an outright true equation (like \(2=2\)) then you had a dependent system.

    Checkpoint 4.4.7.

    -
    -
    +
    +
    -
    -
    Use elimination to solve the system.
    +
    +
    Use elimination to solve the system.
    \begin{equation*} \begin{aligned} \left\{ @@ -770,13 +784,13 @@

    Search Results:

    \end{aligned} \end{equation*}
    -

    (a)

    -
    What number could you multiply by each term in the second equation that would be helpful?
    -
    Explanation.
    If we multiply the second equation through by \(-3\) then the \(y\)-terms will be \(6y\) and \(-6y\text{,}\) and adding will eliminate \(y\text{.}\) -

    (b)

    -
    After multiplying the terms in the second equation by \(-3\) and adding the two equations together, solve for \(x\text{.}\) +

    (a)

    +
    What number could you multiply by each term in the second equation that would be helpful?
    +
    Explanation.
    If we multiply the second equation through by \(-3\) then the \(y\)-terms will be \(6y\) and \(-6y\text{,}\) and adding will eliminate \(y\text{.}\) +

    (b)

    +
    After multiplying the terms in the second equation by \(-3\) and adding the two equations together, solve for \(x\text{.}\)
    -
    Explanation.
    +
    Explanation.
    \begin{equation*} \begin{aligned} \amp\left\{\begin{alignedat}{4}5x\amp+{}\amp 6y \amp={}\amp -7 \\\multiplyleft{-3}(4x\amp+{}\amp 2y) \amp={}\amp\multiplyleft{-3}(-1)\end{alignedat}\right.\\ @@ -786,9 +800,9 @@

    Search Results:

    \amp\phantom{\Bigg\{}\begin{alignedat}{4}{-7}x\amp\phantom{{}+{}}\amp\phantom{6y}\amp=-4\\\phantom{-12x}\amp\amp x\amp=\frac{4}{7}\end{alignedat} \end{aligned} \end{equation*} -

    (c)

    -
    Now solve for \(y\) as well, by substituting \(\frac{4}{7}\) in for \(x\) in one of the original equations.
    -
    Explanation.
    +

    (c)

    +
    Now solve for \(y\) as well, by substituting \(\frac{4}{7}\) in for \(x\) in one of the original equations.
    +
    Explanation.
    \begin{equation*} \begin{aligned} 4x + 2y \amp= -1\\ @@ -805,11 +819,11 @@

    Search Results:

    Checkpoint 4.4.8.

    -
    -
    +
    +
    -
    -
    Use elimination to solve the system.
    +
    +
    Use elimination to solve the system.
    \begin{equation*} \begin{aligned} \left\{ @@ -821,13 +835,13 @@

    Search Results:

    \end{aligned} \end{equation*}
    -

    (a)

    -
    What numbers could you multiply each term in the first and second equations by that would be helpful?
    -
    Explanation.
    If we multiply the first equation through by \(5\) and the second equation through by \(-3\) then the \(x\)-terms will be \(15x\) and \(-15x\text{,}\) and adding will eliminate \(x\text{.}\) -

    (b)

    -
    After multiplying the terms in the first by \(5\) and the terms in the second equation by \(-3\text{,}\) and adding the two equations together, solve for \(y\text{.}\) +

    (a)

    +
    What numbers could you multiply each term in the first and second equations by that would be helpful?
    +
    Explanation.
    If we multiply the first equation through by \(5\) and the second equation through by \(-3\) then the \(x\)-terms will be \(15x\) and \(-15x\text{,}\) and adding will eliminate \(x\text{.}\) +

    (b)

    +
    After multiplying the terms in the first by \(5\) and the terms in the second equation by \(-3\text{,}\) and adding the two equations together, solve for \(y\text{.}\)
    -
    Explanation.
    +
    Explanation.
    \begin{equation*} \begin{aligned} \amp\left\{\begin{alignedat}{4}\multiplyleft{5}(3x\amp+{}\amp 4y) \amp={}\amp \multiplyleft{5}(-26) \\\multiplyleft{-3}(5x\amp+{}\amp 5y) \amp={}\amp\multiplyleft{-3}(-40)\end{alignedat}\right.\\ @@ -836,9 +850,9 @@

    Search Results:

    \amp\phantom{\Bigg\{}\begin{alignedat}{4}\phantom{-15x-{}}\amp\amp\phantom{1}5y\amp=-10\\\amp\amp y\amp =-2\end{alignedat} \end{aligned} \end{equation*} -

    (c)

    -
    Now solve for \(x\) as well, by substituting \(-2\) in for \(y\) in one of the original equations.
    -
    Explanation.
    +

    (c)

    +
    Now solve for \(x\) as well, by substituting \(-2\) in for \(y\) in one of the original equations.
    +
    Explanation.
    \begin{equation*} \begin{aligned} 3x + 4y \amp= -26\\ @@ -854,12 +868,12 @@

    Search Results:

    Checkpoint 4.4.9.
    Solve the system of equations using elimination.
    -
    -
    -
    \(\left\{ +
    +
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-15u+22c = 18+25c} \\ -\amp {-20u+3c = -15u+22} +\amp {-2F+\left(-9\right)R = 74+7R} \\ +\amp {3F+\left(-4\right)R = -2F+\left(-9\right)} \end{alignedat} \right.\)

    @@ -868,38 +882,38 @@

    Search Results:

    Section 1: Solving a System by Graphing

    1.
    Check if the given point is a solution to the given system of two linear equations.
    -
    -
    +
    +
    -
    Is \({\left(0,4\right)}\) a solution?
    \(\left\{ +
    Is \({\left(4,0\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{1}{3}}\mathopen{}\left(x-6\right)+6}\\ -y \amp= {2\mathopen{}\left(x+2\right)} +y \amp= {2\mathopen{}\left(x-2\right)-4}\\ +y \amp= {-{\frac{1}{4}}\mathopen{}\left(x-8\right)-1} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{Yes, it is a solution.}\)
    +
    Answer.
    \(\text{Yes, it is a solution.}\)
    2.
    The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system.
    -
    -
    +
    +
    -
    a coordinate plot of two lines that cross at (1, -2)
    -
    Answer.
    \(\left(1,-2\right)\)
    +
    a coordinate plot of two lines that cross at (4, -4)
    +
    Answer.
    \(\left(4,-4\right)\)
    3.
    Simply by looking closely at the linear equations, determine how many solutions the system will have. And state whether the system is dependeent, inconsistent, or neither.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{1}{3}}x+5}\\ -y \amp= {{\frac{1}{3}}\mathopen{}\left(x+9\right)+2} +y \amp= {-{\frac{11}{5}}x+9}\\ +y \amp= {-{\frac{11}{5}}\mathopen{}\left(x-5\right)-2} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\text{Infinitely many}\)
    Answer 2.
    \(\text{Dependent}\)
    +
    Answer 1.
    \(\text{Infinitely many}\)
    Answer 2.
    \(\text{Dependent}\)
    @@ -907,51 +921,51 @@
    Solve a System.
    Solve the given system of linear equations graphically.
    4.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-{\frac{1}{6}}\mathopen{}\left(x+4\right)-2}\\ -y \amp= {-{\frac{1}{6}}\mathopen{}\left(x+6\right)+1} +y \amp= {-{\frac{4}{15}}\mathopen{}\left(x-7\right)-4}\\ +y \amp= {-{\frac{4}{15}}\mathopen{}\left(x+7\right)+8} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-4,-2\right), \left(2,-3\right)\right\}, \left\{\text{line}, \text{solid}, \left(-6,1\right), \left(0,0\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    -
    If we graph these lines, we find they are parallel and do not cross.
    a coordinate plot of two lines that cross at (2, -3)
    So there are no solutions.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(7,-4\right), \left(22,-8\right)\right\}, \left\{\text{line}, \text{solid}, \left(-7,8\right), \left(8,4\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    If we graph these lines, we find they are parallel and do not cross.
    a coordinate plot of two lines that cross at (-8, 0)
    So there are no solutions.
    5.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp y={{\frac{11}{3}}\mathopen{}\left(x+2\right)+5}\\ -\amp {-3x+5y = -15} +\amp y={-{\frac{4}{7}}\mathopen{}\left(x-6\right)-5}\\ +\amp {-3x-4y = 12} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(5,0\right), \left(0,-3\right)\right\}, \left\{\text{line}, \text{solid}, \left(-2,5\right), \left(-5,-6\right)\right\}\)
    Answer 2.
    \(\left(-5,-6\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-5, (-6))
    So the only solution is \((-5,-6)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-4,0\right), \left(0,-3\right)\right\}, \left\{\text{line}, \text{solid}, \left(6,-5\right), \left(-8,3\right)\right\}\)
    Answer 2.
    \(\left(-8,3\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-8, 3)
    So the only solution is \((-8,3)\text{.}\)
    6.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-{\frac{2}{7}}x+6}\\ -y \amp= {-{\frac{2}{7}}\mathopen{}\left(x+7\right)+8} +y \amp= {-{\frac{1}{2}}x+3}\\ +y \amp= {-{\frac{1}{2}}\mathopen{}\left(x+4\right)+5} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,6\right), \left(7,4\right)\right\}\)
    Answer 2.
    \(\left(-7,8\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-7, 8)
    So the only solution is \((-7,8)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,3\right), \left(2,2\right)\right\}\)
    Answer 2.
    \(\left(-4,5\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-4, 5)
    So the only solution is \((-4,5)\text{.}\)
    @@ -960,27 +974,27 @@
    Solve a System.
    7.
    -
    -
    +
    +
    -
    El enters a grassy field from some dense trees. Their dog is standing out in the field, \({110\ {\rm ft}}\) away. As soon as they see each other, they start running toward each other. El runs with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) and their dog runs with speed \({8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) How long will it be until they meet?
    (a)
    -
    Write two equations that form a system for this scenario.
    +
    Broderick enters a grassy field from some dense trees. His dog is standing out in the field, \({65\ {\rm ft}}\) away. As soon as they see each other, they start running toward each other. Broderick runs with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) and his dog runs with speed \({10\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) How long will it be until they meet?
    (a)
    +
    Write two equations that form a system for this scenario.
    -
    Answer 1.
    \(d = 3t\)
    Answer 2.
    \(d = 110+\left(-8\right)t\)
    Explanation.
    -
    El starts out at position \(d=0\text{,}\) running with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) So one equation is \({d = 3t}\text{.}\) -
    The dog starts running with speed \({8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) but getting closer to El. So we’ll use a negative rate of change. And the dog starts out \({110\ {\rm ft}}\) away. So the other equation is \({d = 110+\left(-8\right)t}\text{.}\) +
    Answer 1.
    \(d = 3t\)
    Answer 2.
    \(d = 65+\left(-10\right)t\)
    Explanation.
    +
    Broderick starts out at position \(d=0\text{,}\) running with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) So one equation is \({d = 3t}\text{.}\) +
    The dog starts running with speed \({10\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) but getting closer to Broderick. So we’ll use a negative rate of change. And the dog starts out \({65\ {\rm ft}}\) away. So the other equation is \({d = 65+\left(-10\right)t}\text{.}\)
    -
    (b)
    -
    Plot the lines for the system of two equations.
    +
    (b)
    +
    Plot the lines for the system of two equations.
    -
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,110\right), \left(1,102\right)\right\}\)
    Explanation.
    -
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,110) with slope 3; the other line passes through (0,0) with slope -8; the lines cross at (10,30)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,65\right), \left(1,55\right)\right\}\)
    Explanation.
    +
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,65) with slope 3; the other line passes through (0,0) with slope -10; the lines cross at (5,15)
    -
    (c)
    -
    Based on the graph, how long will it be until they meet?
    How far will that be from the place where El started?
    +
    (c)
    +
    Based on the graph, how long will it be until they meet?
    How far will that be from the place where Broderick started?
    -
    Answer 1.
    \(10\ {\rm s}\)
    Answer 2.
    \(30\ {\rm m}\)
    Explanation.
    The lines cross at \((10,30)\text{.}\) So after \(10\) seconds, they will meet 30 feet away from where El started.
    +
    Answer 1.
    \(5\ {\rm s}\)
    Answer 2.
    \(15\ {\rm m}\)
    Explanation.
    The lines cross at \((5,15)\text{.}\) So after \(5\) seconds, they will meet 15 feet away from where Broderick started.

    Section 2: Substitution

    @@ -989,138 +1003,138 @@
    Exercise Group.
    Solve the system of equations using substitution.
    8.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {n = 3\mathopen{}\left(p-5\right)+5} \\ -\amp {n = -5\mathopen{}\left(p-4\right)-6} +\amp {d = -3\mathopen{}\left(z+2\right)+16} \\ +\amp {d = 5\mathopen{}\left(z+1\right)-3} \end{alignedat} \right.\)
    -
    Answer.
    \(p = 3, n = -1\)
    +
    Answer.
    \(z = 1, d = 7\)
    9.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {W = 3\mathopen{}\left(u+5\right)-1} \\ -\amp {W = -u-10} +\amp {K = -3E+24} \\ +\amp {K = -\left(E-4\right)+6} \end{alignedat} \right.\)
    -
    Answer.
    \(u = -6, W = -4\)
    +
    Answer.
    \(E = 7, K = 3\)
    10.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {5A+\left(-4\right)E = -23} \\ -\amp {E = 5\mathopen{}\left(A-\left(-4\right)\right)+\left(-18\right)} +\amp {-3K+1s = 5} \\ +\amp {s = -\left(K-2\right)+\left(-1\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(A = 1, E = 7\)
    +
    Answer.
    \(K = -2, s = -1\)
    11.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {6F+2l = 46} \\ -\amp {l = -3\mathopen{}\left(F-\left(-3\right)\right)+32} +\amp {8Q+\left(-2\right)b = 52} \\ +\amp {b = 4\mathopen{}\left(Q-1\right)+\left(-22\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    12.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-2K+1T = 4} \\ -\amp {T = 8K+9} +\amp {-9X+1H = 1} \\ +\amp {H = 4X+3} \end{alignedat} \right.\)
    -
    Answer.
    \(K = -{\frac{5}{6}}, T = {\frac{7}{3}}\)
    +
    Answer.
    \(X = {\frac{2}{5}}, H = {\frac{23}{5}}\)
    13.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {A = 4.6R+2.7} \\ -\amp {A = -6.1R-1.5} +\amp {r = 2.1c-5.3} \\ +\amp {r = 1.5c-1.7} \end{alignedat} \right.\)
    -
    Answer.
    \(R = 0.392523, A = 4.50561\)
    +
    Answer.
    \(c = -6, r = -17.9\)
    14.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {j = {\frac{2}{3}}\mathopen{}\left(X+3\right) - {\frac{4}{3}}} \\ -\amp {j = {\frac{4}{5}}\mathopen{}\left(X+3\right) - {\frac{11}{15}}} +\amp {Z = {\frac{1}{2}}\mathopen{}\left(h-4\right)+{\frac{5}{2}}} \\ +\amp {Z = {\frac{2}{3}}\mathopen{}\left(h-5\right)+{\frac{29}{6}}} \end{alignedat} \right.\)
    -
    Answer.
    \(X = -{\frac{15}{2}}, j = -{\frac{13}{3}}\)
    +
    Answer.
    \(h = -6, Z = -{\frac{5}{2}}\)
    15.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {R = 0.111111d+0.5} \\ -\amp {0.166667d+\left(-1\right)R = -0.75} +\amp {G = 9n+\left(-2\right)} \\ +\amp {0.5n+\left(-1\right)G = -1.25} \end{alignedat} \right.\)
    -
    Answer.
    \(d = -{\frac{9}{2}}, R = {\frac{21}{2}}\)
    +
    Answer.
    \(n = {\frac{13}{34}}, G = -{\frac{13}{17}}\)
    16.
    -
    -
    +
    +
    -
    A rectangle’s length is \({7\ {\rm ft}}\) shorter than three times its width. The rectangle’s perimeter is \({103\ {\rm ft}}\text{.}\) Find the rectangle’s length and width.
    +
    A rectangle’s length is \({6\ {\rm ft}}\) shorter than four times its width. The rectangle’s perimeter is \({189\ {\rm ft}}\text{.}\) Find the rectangle’s length and width.
    -
    Answer 1.
    \(L = 3W-7\)
    Answer 2.
    \(2L+2W = 103\)
    Answer 3.
    \(36.875\ {\rm ft}\)
    Answer 4.
    \(14.625\ {\rm ft}\)
    +
    Answer 1.
    \(L = 4W-6\)
    Answer 2.
    \(2L+2W = 189\)
    Answer 3.
    \(74.4\ {\rm ft}\)
    Answer 4.
    \(20.1\ {\rm ft}\)
    17.
    -
    -
    +
    +
    -
    A local restaurant has two locations. At one location, the revenue this month is \({\$86{,}000}\) but it has been decreasing by \({\$3{,}500}\) per month. At the other location, the annual revenue this month is \({\$43{,}000}\) and it has been increasing by \({\$2{,}500}\) per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be?
    +
    A local restaurant has two locations. At one location, the revenue this month is \({\$90{,}000}\) but it has been decreasing by \({\$4{,}000}\) per month. At the other location, the annual revenue this month is \({\$59{,}000}\) and it has been increasing by \({\$2{,}000}\) per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be?
    -
    Answer 1.
    \(R = 86000-3500t\)
    Answer 2.
    \(R = 43000+2500t\)
    Answer 3.
    \(7.16667\ {\rm months}\)
    Answer 4.
    \(\$0\)
    +
    Answer 1.
    \(R = 90000-4000t\)
    Answer 2.
    \(R = 59000+2000t\)
    Answer 3.
    \(5.16667\ {\rm months}\)
    Answer 4.
    \(\$0\)
    18.
    -
    -
    +
    +
    -
    Geoffrey invested a total of \({\$5{,}100}\) in two investments. His savings account pays \({2.5\%}\) interest annually. A riskier stock investment lost \({6.5\%}\) at the end of the year. At the end of the year, Geoffrey’s total fell from \({\$5{,}100}\) to \({\$4{,}840.50}\text{.}\) How much money did he invest in each account?
    +
    Kiana invested a total of \({\$3{,}500}\) in two investments. Her savings account pays \({2\%}\) interest annually. A riskier stock investment lost \({6\%}\) at the end of the year. At the end of the year, Kiana’s total fell from \({\$3{,}500}\) to \({\$3{,}338}\text{.}\) How much money did she invest in each account?
    -
    Answer 1.
    \(x+y = 5100\)
    Answer 2.
    \(0.025x-0.065y = -259.5\)
    Answer 3.
    \(\$1{,}800\)
    Answer 4.
    \(\$3{,}300\)
    +
    Answer 1.
    \(x+y = 3500\)
    Answer 2.
    \(0.02x-0.06y = -162\)
    Answer 3.
    \(\$1{,}200\)
    Answer 4.
    \(\$2{,}300\)
    19.
    -
    -
    +
    +
    -
    A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost \({\$1.70}\) per pound, and cashews cost \({\$5.78}\) per pound. The company is targeting a product that will cost them \({\$2.60}\) per pound worth of ingredients. How much of each type of nut should go into a bag?
    +
    A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost \({\$1.74}\) per pound, and cashews cost \({\$5.70}\) per pound. The company is targeting a product that will cost them \({\$2.50}\) per pound worth of ingredients. How much of each type of nut should go into a bag?
    -
    Answer 1.
    \(P+C = 1\)
    Answer 2.
    \(1.7P+5.78C = 2.6\)
    Answer 3.
    \(0.779412\ {\rm lb}\)
    Answer 4.
    \(0.220588\ {\rm lb}\)
    +
    Answer 1.
    \(P+C = 1\)
    Answer 2.
    \(1.74P+5.7C = 2.5\)
    Answer 3.
    \(0.808081\ {\rm lb}\)
    Answer 4.
    \(0.191919\ {\rm lb}\)

    Section 3: Elimination

    @@ -1129,75 +1143,75 @@
    Exercise Group.
    Solve the system of equations using elimination.
    20.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {15F+\left(-5\right)f = 70} \\ -\amp {5F+4f = -5} +\amp {-5Q+\left(-12\right)V = -41} \\ +\amp {4Q+\left(-3\right)V = -5} \end{alignedat} \right.\)
    -
    Answer.
    \(F = 3, f = -5\)
    +
    Answer.
    \(Q = 1, V = 3\)
    21.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {1K+\left(-4\right)M = -30} \\ -\amp {1K+\left(-2\right)M = -18} +\amp {-1W+\left(-1\right)B = -4} \\ +\amp {-1W+2B = -13} \end{alignedat} \right.\)
    -
    Answer.
    \(K = -6, M = 6\)
    +
    Answer.
    \(W = 7, B = -3\)
    22.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {1R+\left(-1\right)u = 3.8} \\ -\amp {1R+\left(-1.8\right)u = -3.8} +\amp {-0.7c+\left(-1\right)k = 0.6} \\ +\amp {0.1c+0.6k = -7.4} \end{alignedat} \right.\)
    -
    Answer.
    \(R = 13.3, u = 9.5\)
    +
    Answer.
    \(c = 22, k = -16\)
    23.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.833333X+\left(-1.66667\right)d = -0.8} \\ -\amp {0.75X+\left(-1.5\right)d = 0.166667} +\amp {2.5h+1.33333R = -1.66667} \\ +\amp {0.5h+0.266667R = 0.666667} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{no solutions}\)
    +
    Answer.
    \(\text{no solutions}\)
    24.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-4d+\left(-6\right)K = 3+\left(-5\right)K} \\ -\amp {4d+2K = -4d+\left(-6\right)} +\amp {-6n+\left(-36\right)A = -9+\left(-37\right)A} \\ +\amp {-30n+4A = -6n+\left(-36\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    25.
    -
    -
    +
    +
    -
    Clayton invested a total of \({\$4{,}900}\) in two investments. His savings account pays \({1.5\%}\) interest annually. A riskier stock investment earned \({6.5\%}\) at the end of the year. At the end of the year, Clayton earned a total of \({\$228.50}\) in interest. How much money did he invest in each account?
    +
    Freddy invested a total of \({\$3{,}300}\) in two investments. His savings account pays \({1\%}\) interest annually. A riskier stock investment earned \({6\%}\) at the end of the year. At the end of the year, Freddy earned a total of \({\$138}\) in interest. How much money did he invest in each account?
    -
    Answer 1.
    \(x+y = 4900\)
    Answer 2.
    \(0.015x+0.065y = 228.5\)
    Answer 3.
    \(\$1{,}800\)
    Answer 4.
    \(\$3{,}100\)
    +
    Answer 1.
    \(x+y = 3300\)
    Answer 2.
    \(0.01x+0.06y = 138\)
    Answer 3.
    \(\$1{,}200\)
    Answer 4.
    \(\$2{,}100\)
    diff --git a/review-variables-expressions-and-equations.html b/review-variables-expressions-and-equations.html index b1de546dc..678e7d44a 100644 --- a/review-variables-expressions-and-equations.html +++ b/review-variables-expressions-and-equations.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-absolute-value-and-square-root.html b/section-absolute-value-and-square-root.html index 5ef9b4160..3d7986744 100644 --- a/section-absolute-value-and-square-root.html +++ b/section-absolute-value-and-square-root.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -602,26 +616,26 @@

    Search Results:

    Checkpoint A.3.5. Calculating Absolute Value.
    Try calculating some absolute values.
    -
    -
    +
    +
    -

    (a)

    -
    +

    (a)

    +
    \(\abs{57}=\)
    -
    Explanation.
    +
    Explanation.
    \(57\) is \(57\) units away from \(0\) on a number line, so \(\abs{57}=57\text{.}\) Another way to think about this is that the “positive version” of \(57\) is \(57\text{.}\) -

    (b)

    -
    +

    (b)

    +
    \(\abs{-43}=\)
    -
    Explanation.
    +
    Explanation.
    \(-43\) is \(43\) units away from \(0\) on a number line, so \(\abs{-43}=43\text{.}\) Another way to think about this is that the “positive version” of \(-43\) is \(43\text{.}\) -

    (c)

    -
    +

    (c)

    +
    \(\abs{\frac{2}{-5}}=\)
    -
    Explanation.
    +
    Explanation.
    \(\frac{2}{-5}\) is \(\frac{2}{5}\) units away from \(0\) on a number line, so \(\abs{\frac{2}{-5}}=\frac{2}{5}\text{.}\) Another way to think about this is that the “positive version” of \(\frac{2}{-5}\) is \(\frac{2}{5}\text{.}\)
    • @@ -812,24 +826,24 @@

    Search Results:

    Checkpoint A.3.8. Square Roots of Fractions.
    Try calculating some absolute values.
    -
    -
    +
    +
    -

    (a)

    -
    +

    (a)

    +
    \(\sqrt{\dfrac{1}{25}}=\)
    -
    Explanation.
    Since \(\sqrt{1}=1\) and \(\sqrt{25}=5\text{,}\) then \(\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\text{.}\) -

    (b)

    -
    +
    Explanation.
    Since \(\sqrt{1}=1\) and \(\sqrt{25}=5\text{,}\) then \(\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\text{.}\) +

    (b)

    +
    \(\sqrt{\dfrac{4}{9}}=\)
    -
    Explanation.
    Since \(\sqrt{4}=2\) and \(\sqrt{9}=3\text{,}\) then \(\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}\text{.}\) -

    (c)

    -
    +
    Explanation.
    Since \(\sqrt{4}=2\) and \(\sqrt{9}=3\text{,}\) then \(\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}\text{.}\) +

    (c)

    +
    \(\sqrt{\dfrac{81}{121}}=\)
    -
    Explanation.
    Since \(\sqrt{81}=9\) and \(\sqrt{121}=11\text{,}\) then \(\sqrt{\dfrac{81}{121}}=\dfrac{9}{11}\text{.}\) +
    Explanation.
    Since \(\sqrt{81}=9\) and \(\sqrt{121}=11\text{,}\) then \(\sqrt{\dfrac{81}{121}}=\dfrac{9}{11}\text{.}\)

    @@ -844,54 +858,54 @@

    Search Results:

    Review and Warmup.

    1.

    -
    -
    +
    +
    -
    +
    Evaluate the expressions.
      -
    1. \(\displaystyle 1^2\)
      2. -
    2. \(\displaystyle 3^2\)
      3. -
    3. \(\displaystyle 5^2\)
      4. -
    4. \(\displaystyle 7^2\)
      5. -
    5. \(\displaystyle 9^2\)
      6. -
    6. \(\displaystyle 11^2\)
      7. +
    7. \(\displaystyle 1^2\)
      8. +
    8. \(\displaystyle 3^2\)
      9. +
    9. \(\displaystyle 5^2\)
      10. +
    10. \(\displaystyle 7^2\)
      11. +
    11. \(\displaystyle 9^2\)
      12. +
    12. \(\displaystyle 11^2\)
    -
    Answer 1.
    \(1\)
    Answer 2.
    \(9\)
    Answer 3.
    \(25\)
    Answer 4.
    \(49\)
    Answer 5.
    \(81\)
    Answer 6.
    \(121\)
    Explanation.
      -
    1. \(\displaystyle 1\)
      2. -
    2. \(\displaystyle 9\)
      3. -
    3. \(\displaystyle 25\)
      4. -
    4. \(\displaystyle 49\)
      5. -
    5. \(\displaystyle 81\)
      6. -
    6. \(\displaystyle 121\)
      7. +
      Answer 1.
      \(1\)
      Answer 2.
      \(9\)
      Answer 3.
      \(25\)
      Answer 4.
      \(49\)
      Answer 5.
      \(81\)
      Answer 6.
      \(121\)
      Explanation.
        +
      1. \(\displaystyle 1\)
        2. +
      2. \(\displaystyle 9\)
        3. +
      3. \(\displaystyle 25\)
        4. +
      4. \(\displaystyle 49\)
        5. +
      5. \(\displaystyle 81\)
        6. +
      6. \(\displaystyle 121\)

    2.

    -
    -
    +
    +
    -
    +
    Evaluate the expressions.
      -
    1. \(\displaystyle 2^2\)
      2. -
    2. \(\displaystyle 4^2\)
      3. -
    3. \(\displaystyle 6^2\)
      4. -
    4. \(\displaystyle 8^2\)
      5. -
    5. \(\displaystyle 10^2\)
      6. -
    6. \(\displaystyle 12^2\)
      7. +
    7. \(\displaystyle 2^2\)
      8. +
    8. \(\displaystyle 4^2\)
      9. +
    9. \(\displaystyle 6^2\)
      10. +
    10. \(\displaystyle 8^2\)
      11. +
    11. \(\displaystyle 10^2\)
      12. +
    12. \(\displaystyle 12^2\)
    -
    Answer 1.
    \(4\)
    Answer 2.
    \(16\)
    Answer 3.
    \(36\)
    Answer 4.
    \(64\)
    Answer 5.
    \(100\)
    Answer 6.
    \(144\)
    Explanation.
      -
    1. \(\displaystyle 4\)
      2. -
    2. \(\displaystyle 16\)
      3. -
    3. \(\displaystyle 36\)
      4. -
    4. \(\displaystyle 64\)
      5. -
    5. \(\displaystyle 100\)
      6. -
    6. \(\displaystyle 144\)
      7. +
      Answer 1.
      \(4\)
      Answer 2.
      \(16\)
      Answer 3.
      \(36\)
      Answer 4.
      \(64\)
      Answer 5.
      \(100\)
      Answer 6.
      \(144\)
      Explanation.
        +
      1. \(\displaystyle 4\)
        2. +
      2. \(\displaystyle 16\)
        3. +
      3. \(\displaystyle 36\)
        4. +
      4. \(\displaystyle 64\)
        5. +
      5. \(\displaystyle 100\)
        6. +
      6. \(\displaystyle 144\)
    @@ -902,271 +916,271 @@

    Review and Warmup.

    Absolute Value.

    3.

    -
    -
    +
    +
    -
    Evaluate the following.
    -\(\displaystyle{ |{-6}|= }\) +
    Evaluate the following.
    +\(\displaystyle{ |{-2}|= }\)
    -
    Answer.
    \(6\)
    Explanation.
    The absolute value of \(-6\) is the distance between \(-6\) and \(0\) on the number line. In this case, since \(-6\) is \(6\) units away from 0, the answer is \(6\text{.}\) +
    Answer.
    \(2\)
    Explanation.
    The absolute value of \(-2\) is the distance between \(-2\) and \(0\) on the number line. In this case, since \(-2\) is \(2\) units away from 0, the answer is \(2\text{.}\)

    4.

    -
    -
    +
    +
    -
    Evaluate the following.
    -\(\displaystyle{ |{-4}|= }\) +
    Evaluate the following.
    +\(\displaystyle{ |{1}|= }\)
    -
    Answer.
    \(4\)
    Explanation.
    The absolute value of \(-4\) is the distance between \(-4\) and \(0\) on the number line. In this case, since \(-4\) is \(4\) units away from 0, the answer is \(4\text{.}\) +
    Answer.
    \(1\)
    Explanation.
    The absolute value of \(1\) is the distance between \(1\) and \(0\) on the number line. In this case, since \(1\) is \(1\) units away from 0, the answer is \(1\text{.}\)

    5.

    -
    -
    +
    +
    -
    Evaluate the following.
    -\(\displaystyle{ \left\lvert-74.74\right\rvert= }\) +
    Evaluate the following.
    +\(\displaystyle{ \left\lvert53.63\right\rvert= }\)
    -
    Answer.
    \(74.74\)
    Explanation.
    The absolute value of \(-74.74\) is the distance between \(-74.74\) and \(0\) on the number line. In this case, since \(-74.74\) is \(74.74\) units away from 0, the answer is \(74.74\text{.}\) +
    Answer.
    \(53.63\)
    Explanation.
    The absolute value of \(53.63\) is the distance between \(53.63\) and \(0\) on the number line. In this case, since \(53.63\) is \(53.63\) units away from 0, the answer is \(53.63\text{.}\)

    6.

    -
    -
    +
    +
    -
    Evaluate the following.
    +
    Evaluate the following.
    \(\displaystyle{ \left\lvert-15.76\right\rvert= }\)
    -
    Answer.
    \(15.76\)
    Explanation.
    The absolute value of \(-15.76\) is the distance between \(-15.76\) and \(0\) on the number line. In this case, since \(-15.76\) is \(15.76\) units away from 0, the answer is \(15.76\text{.}\) +
    Answer.
    \(15.76\)
    Explanation.
    The absolute value of \(-15.76\) is the distance between \(-15.76\) and \(0\) on the number line. In this case, since \(-15.76\) is \(15.76\) units away from 0, the answer is \(15.76\text{.}\)

    7.

    -
    -
    +
    +
    -
    Evaluate the following.
    -\(\displaystyle{ \left\lvert{-{\frac{16}{45}}}\right\rvert= }\) +
    Evaluate the following.
    +\(\displaystyle{ \left\lvert{{\frac{73}{69}}}\right\rvert= }\)
    -
    Answer.
    \({\frac{16}{45}}\)
    Explanation.
    The absolute value of \({-{\frac{16}{45}}}\) is the distance between \({-{\frac{16}{45}}}\) and \(0\) on the number line. In this case, since \({-{\frac{16}{45}}}\) is \({{\frac{16}{45}}}\) units away from 0, the answer is \({{\frac{16}{45}}}\text{.}\) +
    Answer.
    \({\frac{73}{69}}\)
    Explanation.
    The absolute value of \({{\frac{73}{69}}}\) is the distance between \({{\frac{73}{69}}}\) and \(0\) on the number line. In this case, since \({{\frac{73}{69}}}\) is \({{\frac{73}{69}}}\) units away from 0, the answer is \({{\frac{73}{69}}}\text{.}\)

    8.

    -
    -
    +
    +
    -
    Evaluate the following.
    -\(\displaystyle{ \left\lvert{{\frac{54}{41}}}\right\rvert= }\) +
    Evaluate the following.
    +\(\displaystyle{ \left\lvert{-{\frac{13}{9}}}\right\rvert= }\)
    -
    Answer.
    \({\frac{54}{41}}\)
    Explanation.
    The absolute value of \({{\frac{54}{41}}}\) is the distance between \({{\frac{54}{41}}}\) and \(0\) on the number line. In this case, since \({{\frac{54}{41}}}\) is \({{\frac{54}{41}}}\) units away from 0, the answer is \({{\frac{54}{41}}}\text{.}\) +
    Answer.
    \({\frac{13}{9}}\)
    Explanation.
    The absolute value of \({-{\frac{13}{9}}}\) is the distance between \({-{\frac{13}{9}}}\) and \(0\) on the number line. In this case, since \({-{\frac{13}{9}}}\) is \({{\frac{13}{9}}}\) units away from 0, the answer is \({{\frac{13}{9}}}\text{.}\)

    9.

    -
    -
    +
    +
    -
    Evaluate the following.
      -
    1. -\(\displaystyle{ - \lvert 2-9 \rvert = }\) +
      Evaluate the following.
        +
      1. +\(\displaystyle{ - \lvert 1-8 \rvert = }\)
        2. -
      2. -\(\displaystyle{ \lvert -2-9 \rvert = }\) +
      3. +\(\displaystyle{ \lvert -1-8 \rvert = }\)
        4. -
      4. -\(\displaystyle{ -2 \lvert 9-2 \rvert = }\) +
      5. +\(\displaystyle{ -4 \lvert 8-1 \rvert = }\)
      -
      Answer 1.
      \(-7\)
      Answer 2.
      \(11\)
      Answer 3.
      \(-14\)
      Explanation.
        -
      1. -
        Solution:
        -
        \(\begin{aligned}[t] -- \lvert 2-9 \rvert \amp = - \lvert -7 \rvert \\ +
        Answer 1.
        \(-7\)
        Answer 2.
        \(9\)
        Answer 3.
        \(-28\)
        Explanation.
          +
        1. +
          Solution:
          +
          \(\begin{aligned}[t] +- \lvert 1-8 \rvert \amp = - \lvert -7 \rvert \\ \amp = {-7} \end{aligned}\)
          2. -
        2. -
          Solution:
          -
          \(\begin{aligned}[t] -\lvert -2-9 \rvert \amp = \lvert -11 \rvert \\ -\amp = {11} +
        3. +
          Solution:
          +
          \(\begin{aligned}[t] +\lvert -1-8 \rvert \amp = \lvert -9 \rvert \\ +\amp = {9} \end{aligned}\)
          4. -
        4. -
          Solution:
          -
          \(\begin{aligned}[t] --2 \lvert 9-2 \rvert \amp = -2 \cdot \lvert 7 \rvert \\ -\amp = -2 \cdot 7 \\ -\amp = {-14} +
        5. +
          Solution:
          +
          \(\begin{aligned}[t] +-4 \lvert 8-1 \rvert \amp = -4 \cdot \lvert 7 \rvert \\ +\amp = -4 \cdot 7 \\ +\amp = {-28} \end{aligned}\)

    10.

    -
    -
    +
    +
    -
    Evaluate the following.
      -
    1. -\(\displaystyle{ - \lvert 1-6 \rvert = }\) +
      Evaluate the following.
        +
      1. +\(\displaystyle{ - \lvert 4-6 \rvert = }\)
        2. -
      2. -\(\displaystyle{ \lvert -1-6 \rvert = }\) +
      3. +\(\displaystyle{ \lvert -4-6 \rvert = }\)
        4. -
      4. -\(\displaystyle{ -2 \lvert 6-1 \rvert = }\) +
      5. +\(\displaystyle{ -4 \lvert 6-4 \rvert = }\)
      -
      Answer 1.
      \(-5\)
      Answer 2.
      \(7\)
      Answer 3.
      \(-10\)
      Explanation.
        -
      1. -
        Solution:
        -
        \(\begin{aligned}[t] -- \lvert 1-6 \rvert \amp = - \lvert -5 \rvert \\ -\amp = {-5} +
        Answer 1.
        \(-2\)
        Answer 2.
        \(10\)
        Answer 3.
        \(-8\)
        Explanation.
          +
        1. +
          Solution:
          +
          \(\begin{aligned}[t] +- \lvert 4-6 \rvert \amp = - \lvert -2 \rvert \\ +\amp = {-2} \end{aligned}\)
          2. -
        2. -
          Solution:
          -
          \(\begin{aligned}[t] -\lvert -1-6 \rvert \amp = \lvert -7 \rvert \\ -\amp = {7} +
        3. +
          Solution:
          +
          \(\begin{aligned}[t] +\lvert -4-6 \rvert \amp = \lvert -10 \rvert \\ +\amp = {10} \end{aligned}\)
          4. -
        4. -
          Solution:
          -
          \(\begin{aligned}[t] --2 \lvert 6-1 \rvert \amp = -2 \cdot \lvert 5 \rvert \\ -\amp = -2 \cdot 5 \\ -\amp = {-10} +
        5. +
          Solution:
          +
          \(\begin{aligned}[t] +-4 \lvert 6-4 \rvert \amp = -4 \cdot \lvert 2 \rvert \\ +\amp = -4 \cdot 2 \\ +\amp = {-8} \end{aligned}\)

    11.

    -
    -
    +
    +
    -
    Evaluate the following.
      -
    1. -\(\displaystyle{ |{2}|= }\) +
      Evaluate the following.
        +
      1. +\(\displaystyle{ |{4}|= }\)
        2. -
      2. -\(\displaystyle{ |{-2}|= }\) +
      3. +\(\displaystyle{ |{-4}|= }\)
        4. -
      4. -\(\displaystyle{ -|{2}|= }\) +
      5. +\(\displaystyle{ -|{4}|= }\)
        6. -
      6. -\(\displaystyle{ -|{-2}|= }\) +
      7. +\(\displaystyle{ -|{-4}|= }\)
      -
      Answer 1.
      \(2\)
      Answer 2.
      \(2\)
      Answer 3.
      \(-2\)
      Answer 4.
      \(-2\)
      Explanation.
        -
      1. The absolute value of \(2\) measures how many units are between \(2\) and \(0\) on the number line, so \(|{2}|=2\text{.}\) +
        Answer 1.
        \(4\)
        Answer 2.
        \(4\)
        Answer 3.
        \(-4\)
        Answer 4.
        \(-4\)
        Explanation.
          +
        1. The absolute value of \(4\) measures how many units are between \(4\) and \(0\) on the number line, so \(|{4}|=4\text{.}\)
          2. -
        2. The absolute value of \(-2\) measures how many units are between \(-2\) and \(0\) on the number line, so \(|{-2}|=2\text{.}\) +
        3. The absolute value of \(-4\) measures how many units are between \(-4\) and \(0\) on the number line, so \(|{-4}|=4\text{.}\)
          4. -
        4. The absolute value symbols cannot affect what are outside them, so we have \(-|{2}|=-2\text{.}\) +
        5. The absolute value symbols cannot affect what are outside them, so we have \(-|{4}|=-4\text{.}\)
          6. -
        6. Similar to Part c, we have \(-|{-2}|=-2\text{.}\) +
        7. Similar to Part c, we have \(-|{-4}|=-4\text{.}\)

    12.

    -
    -
    +
    +
    -
    Evaluate the following.
      -
    1. -\(\displaystyle{ |{3}|= }\) +
      Evaluate the following.
        +
      1. +\(\displaystyle{ |{5}|= }\)
        2. -
      2. -\(\displaystyle{ |{-3}|= }\) +
      3. +\(\displaystyle{ |{-5}|= }\)
        4. -
      4. -\(\displaystyle{ -|{3}|= }\) +
      5. +\(\displaystyle{ -|{5}|= }\)
        6. -
      6. -\(\displaystyle{ -|{-3}|= }\) +
      7. +\(\displaystyle{ -|{-5}|= }\)
      -
      Answer 1.
      \(3\)
      Answer 2.
      \(3\)
      Answer 3.
      \(-3\)
      Answer 4.
      \(-3\)
      Explanation.
        -
      1. The absolute value of \(3\) measures how many units are between \(3\) and \(0\) on the number line, so \(|{3}|=3\text{.}\) +
        Answer 1.
        \(5\)
        Answer 2.
        \(5\)
        Answer 3.
        \(-5\)
        Answer 4.
        \(-5\)
        Explanation.
          +
        1. The absolute value of \(5\) measures how many units are between \(5\) and \(0\) on the number line, so \(|{5}|=5\text{.}\)
          2. -
        2. The absolute value of \(-3\) measures how many units are between \(-3\) and \(0\) on the number line, so \(|{-3}|=3\text{.}\) +
        3. The absolute value of \(-5\) measures how many units are between \(-5\) and \(0\) on the number line, so \(|{-5}|=5\text{.}\)
          4. -
        4. The absolute value symbols cannot affect what are outside them, so we have \(-|{3}|=-3\text{.}\) +
        5. The absolute value symbols cannot affect what are outside them, so we have \(-|{5}|=-5\text{.}\)
          6. -
        6. Similar to Part c, we have \(-|{-3}|=-3\text{.}\) +
        7. Similar to Part c, we have \(-|{-5}|=-5\text{.}\)

    13.

    -
    -
    +
    +
    -
    Evaluate the following.
      -
    1. -\(\displaystyle{ \left\lvert 4 \right\rvert = }\) +
      Evaluate the following.
        +
      1. +\(\displaystyle{ \left\lvert 6 \right\rvert = }\)
        2. -
      2. -\(\displaystyle{ \left\lvert -1 \right\rvert = }\) +
      3. +\(\displaystyle{ \left\lvert -3 \right\rvert = }\)
        4. -
      4. +
      5. \(\displaystyle{ \left\lvert 0 \right\rvert = }\)
        6. -
      6. -\(\displaystyle{ \left\lvert {14+\left(-2\right)} \right\rvert = }\) +
      7. +\(\displaystyle{ \left\lvert {12+\left(-7\right)} \right\rvert = }\)
        8. -
      8. -\(\displaystyle{ \left\lvert {-10-\left(-3\right)} \right\rvert = }\) +
      9. +\(\displaystyle{ \left\lvert {-7-\left(-3\right)} \right\rvert = }\)
      -
      Answer 1.
      \(4\)
      Answer 2.
      \(1\)
      Answer 3.
      \(0\)
      Answer 4.
      \(12\)
      Answer 5.
      \(7\)
      Explanation.
        -
      1. -
        Solution:
        -
        \(\displaystyle{ \left| 4 \right| = {4} }\)
        +
        Answer 1.
        \(6\)
        Answer 2.
        \(3\)
        Answer 3.
        \(0\)
        Answer 4.
        \(5\)
        Answer 5.
        \(4\)
        Explanation.
          +
        1. +
          Solution:
          +
          \(\displaystyle{ \left| 6 \right| = {6} }\)
          2. -
        2. -
          Solution:
          -
          \(\displaystyle{\left| -1 \right| = {1} }\)
          +
        3. +
          Solution:
          +
          \(\displaystyle{\left| -3 \right| = {3} }\)
          4. -
        4. -
          Solution:
          -
          \(\displaystyle{\left| 0 \right| = {0} }\)
          +
        5. +
          Solution:
          +
          \(\displaystyle{\left| 0 \right| = {0} }\)
          6. -
        6. -
          Solution:
          -
          \(\displaystyle{ \begin{aligned}[t] -\left| {14+\left(-2\right)} \right| \amp = \left| 12 \right| \\ -\amp = {12} +
        7. +
          Solution:
          +
          \(\displaystyle{ \begin{aligned}[t] +\left| {12+\left(-7\right)} \right| \amp = \left| 5 \right| \\ +\amp = {5} \end{aligned} }\)
          8. -
        8. -
          Solution:
          -
          \(\displaystyle{ \begin{aligned}[t] -\lvert {-10-\left(-3\right)} \rvert \amp = \lvert -7 \rvert \\ -\amp = {7} +
        9. +
          Solution:
          +
          \(\displaystyle{ \begin{aligned}[t] +\lvert {-7-\left(-3\right)} \rvert \amp = \lvert -4 \rvert \\ +\amp = {4} \end{aligned} }\)
          10. @@ -1174,53 +1188,53 @@

          Absolute Value.

    14.

    -
    -
    +
    +
    -
    Evaluate the following.
      -
    1. -\(\displaystyle{ \left\lvert 5 \right\rvert = }\) +
      Evaluate the following.
        +
      1. +\(\displaystyle{ \left\lvert 7 \right\rvert = }\)
        2. -
      2. -\(\displaystyle{ \left\lvert -4 \right\rvert = }\) +
      3. +\(\displaystyle{ \left\lvert -6 \right\rvert = }\)
        4. -
      4. +
      5. \(\displaystyle{ \left\lvert 0 \right\rvert = }\)
        6. -
      6. -\(\displaystyle{ \left\lvert {19+\left(-5\right)} \right\rvert = }\) +
      7. +\(\displaystyle{ \left\lvert {17+\left(-10\right)} \right\rvert = }\)
        8. -
      8. -\(\displaystyle{ \left\lvert {-10-\left(-1\right)} \right\rvert = }\) +
      9. +\(\displaystyle{ \left\lvert {-7-\left(-1\right)} \right\rvert = }\)
      -
      Answer 1.
      \(5\)
      Answer 2.
      \(4\)
      Answer 3.
      \(0\)
      Answer 4.
      \(14\)
      Answer 5.
      \(9\)
      Explanation.
        -
      1. -
        Solution:
        -
        \(\displaystyle{ \left| 5 \right| = {5} }\)
        +
        Answer 1.
        \(7\)
        Answer 2.
        \(6\)
        Answer 3.
        \(0\)
        Answer 4.
        \(7\)
        Answer 5.
        \(6\)
        Explanation.
          +
        1. +
          Solution:
          +
          \(\displaystyle{ \left| 7 \right| = {7} }\)
          2. -
        2. -
          Solution:
          -
          \(\displaystyle{\left| -4 \right| = {4} }\)
          +
        3. +
          Solution:
          +
          \(\displaystyle{\left| -6 \right| = {6} }\)
          4. -
        4. -
          Solution:
          -
          \(\displaystyle{\left| 0 \right| = {0} }\)
          +
        5. +
          Solution:
          +
          \(\displaystyle{\left| 0 \right| = {0} }\)
          6. -
        6. -
          Solution:
          -
          \(\displaystyle{ \begin{aligned}[t] -\left| {19+\left(-5\right)} \right| \amp = \left| 14 \right| \\ -\amp = {14} +
        7. +
          Solution:
          +
          \(\displaystyle{ \begin{aligned}[t] +\left| {17+\left(-10\right)} \right| \amp = \left| 7 \right| \\ +\amp = {7} \end{aligned} }\)
          8. -
        8. -
          Solution:
          -
          \(\displaystyle{ \begin{aligned}[t] -\lvert {-10-\left(-1\right)} \rvert \amp = \lvert -9 \rvert \\ -\amp = {9} +
        9. +
          Solution:
          +
          \(\displaystyle{ \begin{aligned}[t] +\lvert {-7-\left(-1\right)} \rvert \amp = \lvert -6 \rvert \\ +\amp = {6} \end{aligned} }\)
          10. @@ -1231,19 +1245,19 @@

          Absolute Value.

        15.

        -
        -
        +
        +
        -
        Which of the following are square numbers? There may be more than one correct answer.
          -
        • \(\displaystyle 49\)
          • -
        • \(\displaystyle 9\)
          • -
        • \(\displaystyle 143\)
          • -
        • \(\displaystyle 16\)
          • -
        • \(\displaystyle 61\)
          • -
        • \(\displaystyle 46\)
          • +
          Which of the following are square numbers? There may be more than one correct answer.
            +
          • \(\displaystyle 73\)
            • +
          • \(\displaystyle 113\)
            • +
          • \(\displaystyle 46\)
            • +
          • \(\displaystyle 4\)
            • +
          • \(\displaystyle 1\)
            • +
          • \(\displaystyle 100\)
          -
          Explanation.
          -
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] +
          Explanation.
          +
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] 1^{2} \amp = 1 \\ 2^{2} \amp = 4 \\ 3^{2} \amp = 9 \\ @@ -1257,23 +1271,23 @@

          Absolute Value.

          11^{2} \amp = 121 \\ 12^{2} \amp = 144 \\ \end{aligned} -}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are ABD.
          +}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are DEF.

        16.

        -
        -
        +
        +
        -
        Which of the following are square numbers? There may be more than one correct answer.
          -
        • \(\displaystyle 132\)
          • -
        • \(\displaystyle 141\)
          • -
        • \(\displaystyle 7\)
          • -
        • \(\displaystyle 49\)
          • -
        • \(\displaystyle 64\)
          • -
        • \(\displaystyle 144\)
          • +
          Which of the following are square numbers? There may be more than one correct answer.
            +
          • \(\displaystyle 74\)
            • +
          • \(\displaystyle 71\)
            • +
          • \(\displaystyle 116\)
            • +
          • \(\displaystyle 64\)
            • +
          • \(\displaystyle 121\)
            • +
          • \(\displaystyle 36\)
          -
          Explanation.
          -
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] +
          Explanation.
          +
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] 1^{2} \amp = 1 \\ 2^{2} \amp = 4 \\ 3^{2} \amp = 9 \\ @@ -1287,30 +1301,30 @@

          Absolute Value.

          11^{2} \amp = 121 \\ 12^{2} \amp = 144 \\ \end{aligned} -}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are DEF.
          +}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are DEF.

        Square Roots.

        17.

        -
        -
        +
        +
        -
        Evaluate the following.
          -
        1. -\(\displaystyle{ \sqrt{100} }\) = +
          Evaluate the following.
            +
          1. +\(\displaystyle{ \sqrt{144} }\) =
            2. -
          2. -\(\displaystyle{ \sqrt{49} }\) = +
          3. +\(\displaystyle{ \sqrt{25} }\) =
            4. -
          4. -\(\displaystyle{ \sqrt{1} }\) = +
          5. +\(\displaystyle{ \sqrt{121} }\) =
          -
          Answer 1.
          \(10\)
          Answer 2.
          \(7\)
          Answer 3.
          \(1\)
          Explanation.
          -
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] +
          Answer 1.
          \(12\)
          Answer 2.
          \(5\)
          Answer 3.
          \(11\)
          Explanation.
          +
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] 1^{2} \amp = 1 \\ 2^{2} \amp = 4 \\ 3^{2} \amp = 9 \\ @@ -1324,32 +1338,32 @@

          Square Roots.

          11^{2} \amp = 121 \\ 12^{2} \amp = 144 \\ \end{aligned} -}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are:
            -
          1. \(\displaystyle \displaystyle{\sqrt{100} = 10}\)
            2. -
          2. \(\displaystyle \displaystyle{\sqrt{49} = 7}\)
            3. -
          3. \(\displaystyle \displaystyle{\sqrt{1} = 1}\)
            4. +}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are:
            +
          1. \(\displaystyle \displaystyle{\sqrt{144} = 12}\)
            2. +
          2. \(\displaystyle \displaystyle{\sqrt{25} = 5}\)
            3. +
          3. \(\displaystyle \displaystyle{\sqrt{121} = 11}\)

        18.

        -
        -
        +
        +
        -
        Evaluate the following.
          -
        1. -\(\displaystyle{ \sqrt{121} }\) = +
          Evaluate the following.
            +
          1. +\(\displaystyle{ \sqrt{1} }\) =
            2. -
          2. -\(\displaystyle{ \sqrt{9} }\) = +
          3. +\(\displaystyle{ \sqrt{4} }\) =
            4. -
          4. -\(\displaystyle{ \sqrt{49} }\) = +
          5. +\(\displaystyle{ \sqrt{36} }\) =
          -
          Answer 1.
          \(11\)
          Answer 2.
          \(3\)
          Answer 3.
          \(7\)
          Explanation.
          -
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] +
          Answer 1.
          \(1\)
          Answer 2.
          \(2\)
          Answer 3.
          \(6\)
          Explanation.
          +
          You can quickly find the first few square numbers on scratch paper:
          \(\displaystyle{\begin{aligned}[t] 1^{2} \amp = 1 \\ 2^{2} \amp = 4 \\ 3^{2} \amp = 9 \\ @@ -1363,56 +1377,56 @@

          Square Roots.

          11^{2} \amp = 121 \\ 12^{2} \amp = 144 \\ \end{aligned} -}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are:
            -
          1. \(\displaystyle \displaystyle{\sqrt{121} = 11}\)
            2. -
          2. \(\displaystyle \displaystyle{\sqrt{9} = 3}\)
            3. -
          3. \(\displaystyle \displaystyle{\sqrt{49} = 7}\)
            4. +}\)
          It would be very helpful if you can memorize the above 12 square numbers.
          So the correct answers are:
            +
          1. \(\displaystyle \displaystyle{\sqrt{1} = 1}\)
            2. +
          2. \(\displaystyle \displaystyle{\sqrt{4} = 2}\)
            3. +
          3. \(\displaystyle \displaystyle{\sqrt{36} = 6}\)

        19.

        -
        -
        +
        +
        -
        Evaluate the following.
          -
        1. -\(\displaystyle{ \sqrt{{{\frac{144}{121}}}} }\) = +
          Evaluate the following.
            +
          1. +\(\displaystyle{ \sqrt{{{\frac{9}{100}}}} }\) =
            2. -
          2. -\(\displaystyle{ \sqrt{{-{\frac{9}{100}}}} }\) = +
          3. +\(\displaystyle{ \sqrt{{-{\frac{64}{25}}}} }\) =
          -
          Answer 1.
          \({\frac{12}{11}}\)
          Answer 2.
          \(\text{not a real number}\)
          Explanation.
          -
          “Square root” is the opposite operation of “square.”
          Since \(\left(\frac{2}{3}\right)^{2} = \left(\frac{2}{3}\right)\left(\frac{2}{3}\right) = \frac{4}{9}\text{,}\) we have: \(\sqrt{\frac{4}{9}} = \frac{2}{3}\text{.}\) -
          The answers are:
            -
          1. \(\displaystyle \displaystyle{\sqrt{{{\frac{144}{121}}}} = {{\frac{12}{11}}} }\)
            2. -
          2. \(\displaystyle \displaystyle{\sqrt{{-{\frac{9}{100}}}} \text{ is not a real number} }\)
            3. -
          Note that square root of a negative number does not exist. This is because any number squared is always positive or \(0\text{.}\) +
          Answer 1.
          \({\frac{3}{10}}\)
          Answer 2.
          \(\text{not a real number}\)
          Explanation.
          +
          “Square root” is the opposite operation of “square.”
          Since \(\left(\frac{2}{3}\right)^{2} = \left(\frac{2}{3}\right)\left(\frac{2}{3}\right) = \frac{4}{9}\text{,}\) we have: \(\sqrt{\frac{4}{9}} = \frac{2}{3}\text{.}\) +
          The answers are:
            +
          1. \(\displaystyle \displaystyle{\sqrt{{{\frac{9}{100}}}} = {{\frac{3}{10}}} }\)
            2. +
          2. \(\displaystyle \displaystyle{\sqrt{{-{\frac{64}{25}}}} \text{ is not a real number} }\)
            3. +
          Note that square root of a negative number does not exist. This is because any number squared is always positive or \(0\text{.}\)

        20.

        -
        -
        +
        +
        -
        Evaluate the following.
          -
        1. -\(\displaystyle{ \sqrt{{{\frac{4}{49}}}} }\) = +
          Evaluate the following.
            +
          1. +\(\displaystyle{ \sqrt{{{\frac{16}{25}}}} }\) =
            2. -
          2. -\(\displaystyle{ \sqrt{{-{\frac{25}{144}}}} }\) = +
          3. +\(\displaystyle{ \sqrt{{-{\frac{4}{49}}}} }\) =
          -
          Answer 1.
          \({\frac{2}{7}}\)
          Answer 2.
          \(\text{not a real number}\)
          Explanation.
          -
          “Square root” is the opposite operation of “square.”
          Since \(\left(\frac{2}{3}\right)^{2} = \left(\frac{2}{3}\right)\left(\frac{2}{3}\right) = \frac{4}{9}\text{,}\) we have: \(\sqrt{\frac{4}{9}} = \frac{2}{3}\text{.}\) -
          The answers are:
            -
          1. \(\displaystyle \displaystyle{\sqrt{{{\frac{4}{49}}}} = {{\frac{2}{7}}} }\)
            2. -
          2. \(\displaystyle \displaystyle{\sqrt{{-{\frac{25}{144}}}} \text{ is not a real number} }\)
            3. -
          Note that square root of a negative number does not exist. This is because any number squared is always positive or \(0\text{.}\) +
          Answer 1.
          \({\frac{4}{5}}\)
          Answer 2.
          \(\text{not a real number}\)
          Explanation.
          +
          “Square root” is the opposite operation of “square.”
          Since \(\left(\frac{2}{3}\right)^{2} = \left(\frac{2}{3}\right)\left(\frac{2}{3}\right) = \frac{4}{9}\text{,}\) we have: \(\sqrt{\frac{4}{9}} = \frac{2}{3}\text{.}\) +
          The answers are:
            +
          1. \(\displaystyle \displaystyle{\sqrt{{{\frac{16}{25}}}} = {{\frac{4}{5}}} }\)
            2. +
          2. \(\displaystyle \displaystyle{\sqrt{{-{\frac{4}{49}}}} \text{ is not a real number} }\)
            3. +
          Note that square root of a negative number does not exist. This is because any number squared is always positive or \(0\text{.}\)
          @@ -1424,145 +1438,145 @@

          Square Roots.

          Exercise Group.

          21.

          -
          -
          +
          +
          -
          Evaluate the following.
          Do not use a calculator.
            -
          1. -\(\displaystyle{ \sqrt{16} }\) = +
            Evaluate the following.
            Do not use a calculator.
              +
            1. +\(\displaystyle{ \sqrt{36} }\) =
              2. -
            2. -\(\displaystyle{ \sqrt{0.16} }\) = +
            3. +\(\displaystyle{ \sqrt{0.36} }\) =
              4. -
            4. -\(\displaystyle{ \sqrt{1600} }\) = +
            5. +\(\displaystyle{ \sqrt{3600} }\) =
            -
            Answer 1.
            \(4\)
            Answer 2.
            \(0.4\)
            Answer 3.
            \(40\)
            Explanation.
            -
            First, recognize that \(16\) is a square number. Since \(4\cdot4=16\text{,}\) we have \(\sqrt{16}=4\text{.}\) -
            Next, since \(0.4\cdot0.4=0.16\text{,}\) we have \(\sqrt{0.16}=0.4\text{.}\) -
            Finally, since \(40\cdot40=1600\text{,}\) we have \(\sqrt{1600}=40\text{.}\) +
            Answer 1.
            \(6\)
            Answer 2.
            \(0.6\)
            Answer 3.
            \(60\)
            Explanation.
            +
            First, recognize that \(36\) is a square number. Since \(6\cdot6=36\text{,}\) we have \(\sqrt{36}=6\text{.}\) +
            Next, since \(0.6\cdot0.6=0.36\text{,}\) we have \(\sqrt{0.36}=0.6\text{.}\) +
            Finally, since \(60\cdot60=3600\text{,}\) we have \(\sqrt{3600}=60\text{.}\)

          22.

          -
          -
          +
          +
          -
          Evaluate the following.
          Do not use a calculator.
            -
          1. -\(\displaystyle{ \sqrt{25} }\) = +
            Evaluate the following.
            Do not use a calculator.
              +
            1. +\(\displaystyle{ \sqrt{49} }\) =
              2. -
            2. -\(\displaystyle{ \sqrt{0.25} }\) = +
            3. +\(\displaystyle{ \sqrt{0.49} }\) =
              4. -
            4. -\(\displaystyle{ \sqrt{2500} }\) = +
            5. +\(\displaystyle{ \sqrt{4900} }\) =
            -
            Answer 1.
            \(5\)
            Answer 2.
            \(0.5\)
            Answer 3.
            \(50\)
            Explanation.
            -
            First, recognize that \(25\) is a square number. Since \(5\cdot5=25\text{,}\) we have \(\sqrt{25}=5\text{.}\) -
            Next, since \(0.5\cdot0.5=0.25\text{,}\) we have \(\sqrt{0.25}=0.5\text{.}\) -
            Finally, since \(50\cdot50=2500\text{,}\) we have \(\sqrt{2500}=50\text{.}\) +
            Answer 1.
            \(7\)
            Answer 2.
            \(0.7\)
            Answer 3.
            \(70\)
            Explanation.
            +
            First, recognize that \(49\) is a square number. Since \(7\cdot7=49\text{,}\) we have \(\sqrt{49}=7\text{.}\) +
            Next, since \(0.7\cdot0.7=0.49\text{,}\) we have \(\sqrt{0.49}=0.7\text{.}\) +
            Finally, since \(70\cdot70=4900\text{,}\) we have \(\sqrt{4900}=70\text{.}\)

          23.

          -
          -
          +
          +
          -
          Evaluate the following.
          Do not use a calculator.
            -
          1. -\(\displaystyle{ \sqrt{36} }\) = +
            Evaluate the following.
            Do not use a calculator.
              +
            1. +\(\displaystyle{ \sqrt{64} }\) =
              2. -
            2. -\(\displaystyle{ \sqrt{3600} }\) = +
            3. +\(\displaystyle{ \sqrt{6400} }\) =
              4. -
            4. -\(\displaystyle{ \sqrt{360000} }\) = +
            5. +\(\displaystyle{ \sqrt{640000} }\) =
            -
            Answer 1.
            \(6\)
            Answer 2.
            \(60\)
            Answer 3.
            \(600\)
            Explanation.
            -
            First, recognize that \(36\) is a square number. Since \(6\cdot6=36\text{,}\) we have \(\sqrt{36}=6\text{.}\) -
            Next, since \(60\cdot60=3600\text{,}\) we have \(\sqrt{3600}=60\text{.}\) -
            Finally, since \(600\cdot600=360000\text{,}\) we have \(\sqrt{360000}=600\text{.}\) +
            Answer 1.
            \(8\)
            Answer 2.
            \(80\)
            Answer 3.
            \(800\)
            Explanation.
            +
            First, recognize that \(64\) is a square number. Since \(8\cdot8=64\text{,}\) we have \(\sqrt{64}=8\text{.}\) +
            Next, since \(80\cdot80=6400\text{,}\) we have \(\sqrt{6400}=80\text{.}\) +
            Finally, since \(800\cdot800=640000\text{,}\) we have \(\sqrt{640000}=800\text{.}\)

          24.

          -
          -
          +
          +
          -
          Evaluate the following.
          Do not use a calculator.
            -
          1. -\(\displaystyle{ \sqrt{49} }\) = +
            Evaluate the following.
            Do not use a calculator.
              +
            1. +\(\displaystyle{ \sqrt{81} }\) =
              2. -
            2. -\(\displaystyle{ \sqrt{4900} }\) = +
            3. +\(\displaystyle{ \sqrt{8100} }\) =
              4. -
            4. -\(\displaystyle{ \sqrt{490000} }\) = +
            5. +\(\displaystyle{ \sqrt{810000} }\) =
            -
            Answer 1.
            \(7\)
            Answer 2.
            \(70\)
            Answer 3.
            \(700\)
            Explanation.
            -
            First, recognize that \(49\) is a square number. Since \(7\cdot7=49\text{,}\) we have \(\sqrt{49}=7\text{.}\) -
            Next, since \(70\cdot70=4900\text{,}\) we have \(\sqrt{4900}=70\text{.}\) -
            Finally, since \(700\cdot700=490000\text{,}\) we have \(\sqrt{490000}=700\text{.}\) +
            Answer 1.
            \(9\)
            Answer 2.
            \(90\)
            Answer 3.
            \(900\)
            Explanation.
            +
            First, recognize that \(81\) is a square number. Since \(9\cdot9=81\text{,}\) we have \(\sqrt{81}=9\text{.}\) +
            Next, since \(90\cdot90=8100\text{,}\) we have \(\sqrt{8100}=90\text{.}\) +
            Finally, since \(900\cdot900=810000\text{,}\) we have \(\sqrt{810000}=900\text{.}\)

          25.

          -
          -
          +
          +
          -
          Evaluate the following.
          Do not use a calculator.
            -
          1. -\(\displaystyle{ \sqrt{64} }\) = +
            Evaluate the following.
            Do not use a calculator.
              +
            1. +\(\displaystyle{ \sqrt{121} }\) =
              2. -
            2. -\(\displaystyle{ \sqrt{0.64} }\) = +
            3. +\(\displaystyle{ \sqrt{1.21} }\) =
              4. -
            4. -\(\displaystyle{ \sqrt{0.0064} }\) = +
            5. +\(\displaystyle{ \sqrt{0.0121} }\) =
            -
            Answer 1.
            \(8\)
            Answer 2.
            \(0.8\)
            Answer 3.
            \(0.08\)
            Explanation.
            -
            First, recognize that \(64\) is a square number. Since \(8\cdot8=64\text{,}\) we have \(\sqrt{64}=8\text{.}\) -
            Next, since \(0.8\cdot0.8=0.64\text{,}\) we have \(\sqrt{0.64}=0.8\text{.}\) -
            Finally, since \(0.08\cdot0.08=0.0064\text{,}\) we have \(\sqrt{0.0064}=0.08\text{.}\) +
            Answer 1.
            \(11\)
            Answer 2.
            \(1.1\)
            Answer 3.
            \(0.11\)
            Explanation.
            +
            First, recognize that \(121\) is a square number. Since \(11\cdot11=121\text{,}\) we have \(\sqrt{121}=11\text{.}\) +
            Next, since \(1.1\cdot1.1=1.21\text{,}\) we have \(\sqrt{1.21}=1.1\text{.}\) +
            Finally, since \(0.11\cdot0.11=0.0121\text{,}\) we have \(\sqrt{0.0121}=0.11\text{.}\)

          26.

          -
          -
          +
          +
          -
          Evaluate the following.
          Do not use a calculator.
            -
          1. -\(\displaystyle{ \sqrt{100} }\) = +
            Evaluate the following.
            Do not use a calculator.
              +
            1. +\(\displaystyle{ \sqrt{144} }\) =
              2. -
            2. -\(\displaystyle{ \sqrt{1} }\) = +
            3. +\(\displaystyle{ \sqrt{1.44} }\) =
              4. -
            4. -\(\displaystyle{ \sqrt{0.01} }\) = +
            5. +\(\displaystyle{ \sqrt{0.0144} }\) =
            -
            Answer 1.
            \(10\)
            Answer 2.
            \(1\)
            Answer 3.
            \(0.1\)
            Explanation.
            -
            First, recognize that \(100\) is a square number. Since \(10\cdot10=100\text{,}\) we have \(\sqrt{100}=10\text{.}\) -
            Next, since \(1\cdot1=1\text{,}\) we have \(\sqrt{1}=1\text{.}\) -
            Finally, since \(0.1\cdot0.1=0.01\text{,}\) we have \(\sqrt{0.01}=0.1\text{.}\) +
            Answer 1.
            \(12\)
            Answer 2.
            \(1.2\)
            Answer 3.
            \(0.12\)
            Explanation.
            +
            First, recognize that \(144\) is a square number. Since \(12\cdot12=144\text{,}\) we have \(\sqrt{144}=12\text{.}\) +
            Next, since \(1.2\cdot1.2=1.44\text{,}\) we have \(\sqrt{1.44}=1.2\text{.}\) +
            Finally, since \(0.12\cdot0.12=0.0144\text{,}\) we have \(\sqrt{0.0144}=0.12\text{.}\)
            @@ -1574,26 +1588,26 @@

            Exercise Group.

            Exercise Group.

            27.

            -
            -
            +
            +
            -
            Evaluate the following.
            Use a calculator to approximate with a decimal.
            -\(\sqrt{99}\approx\) +
            Evaluate the following.
            Use a calculator to approximate with a decimal.
            +\(\sqrt{13}\approx\)
            -
            Answer.
            \(9.94987\)
            Explanation.
            A calcultaor shows that \(\sqrt{99}\approx{9.94987}\text{.}\) +
            Answer.
            \(3.60555\)
            Explanation.
            A calcultaor shows that \(\sqrt{13}\approx{3.60555}\text{.}\)

            28.

            -
            -
            +
            +
            -
            Evaluate the following.
            Use a calculator to approximate with a decimal.
            -\(\sqrt{103}\approx\) +
            Evaluate the following.
            Use a calculator to approximate with a decimal.
            +\(\sqrt{27}\approx\)
            -
            Answer.
            \(10.1489\)
            Explanation.
            A calcultaor shows that \(\sqrt{103}\approx{10.1489}\text{.}\) +
            Answer.
            \(5.19615\)
            Explanation.
            A calcultaor shows that \(\sqrt{27}\approx{5.19615}\text{.}\)
            @@ -1604,171 +1618,171 @@

            Exercise Group.

            Exercise Group.

            29.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{\sqrt{{{\frac{4}{25}}}}={}}\).
            +
            Evaluate the following.
            +\(\displaystyle{\sqrt{{{\frac{16}{121}}}}={}}\).
            -
            Answer.
            \(\frac{2}{5}\)
            Explanation.
            \(\displaystyle{\begin{aligned}[t] -\sqrt{\frac{4}{25}} \amp = \frac{\sqrt{4}}{\sqrt{25}} \\ -\amp = \frac{2}{5} +
            Answer.
            \(\frac{4}{11}\)
            Explanation.
            \(\displaystyle{\begin{aligned}[t] +\sqrt{\frac{16}{121}} \amp = \frac{\sqrt{16}}{\sqrt{121}} \\ +\amp = \frac{4}{11} \end{aligned} }\)

            30.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{\sqrt{{{\frac{9}{49}}}}={}}\).
            +
            Evaluate the following.
            +\(\displaystyle{\sqrt{{{\frac{25}{121}}}}={}}\).
            -
            Answer.
            \(\frac{3}{7}\)
            Explanation.
            \(\displaystyle{\begin{aligned}[t] -\sqrt{\frac{9}{49}} \amp = \frac{\sqrt{9}}{\sqrt{49}} \\ -\amp = \frac{3}{7} +
            Answer.
            \(\frac{5}{11}\)
            Explanation.
            \(\displaystyle{\begin{aligned}[t] +\sqrt{\frac{25}{121}} \amp = \frac{\sqrt{25}}{\sqrt{121}} \\ +\amp = \frac{5}{11} \end{aligned} }\)

            31.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(-\sqrt{25}={}\).
            +
            Evaluate the following.
            +\(-\sqrt{49}={}\).
            -
            Answer.
            \(-5\)
            Explanation.
            \(\displaystyle{ -\sqrt{25}=-5 }\)
            +
            Answer.
            \(-7\)
            Explanation.
            \(\displaystyle{ -\sqrt{49}=-7 }\)

            32.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(-\sqrt{36}={}\).
            +
            Evaluate the following.
            +\(-\sqrt{64}={}\).
            -
            Answer.
            \(-6\)
            Explanation.
            \(\displaystyle{ -\sqrt{36}=-6 }\)
            +
            Answer.
            \(-8\)
            Explanation.
            \(\displaystyle{ -\sqrt{64}=-8 }\)

            33.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\sqrt{-49}=\).
            +
            Evaluate the following.
            +\(\sqrt{-100}=\).
            -
            Answer.
            \(\text{not a real number}\)
            Explanation.
            The square root of a negative number is not a real number.
            +
            Answer.
            \(\text{not a real number}\)
            Explanation.
            The square root of a negative number is not a real number.

            34.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\sqrt{-81}=\).
            +
            Evaluate the following.
            +\(\sqrt{-121}=\).
            -
            Answer.
            \(\text{not a real number}\)
            Explanation.
            The square root of a negative number is not a real number.
            +
            Answer.
            \(\text{not a real number}\)
            Explanation.
            The square root of a negative number is not a real number.

            35.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{\sqrt{-{{\frac{81}{100}}}}={}}\).
            +
            Evaluate the following.
            +\(\displaystyle{\sqrt{-{{\frac{121}{144}}}}={}}\).
            -
            Answer.
            \(\text{not a real number}\hbox{ or }0.9i\)
            Explanation.
            The square root of a negative number is not a real number.
            +
            Answer.
            \(\text{not a real number}\hbox{ or }0.916667i\)
            Explanation.
            The square root of a negative number is not a real number.

            36.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{\sqrt{-{{\frac{100}{121}}}}={}}\).
            +
            Evaluate the following.
            +\(\displaystyle{\sqrt{-{{\frac{1}{100}}}}={}}\).
            -
            Answer.
            \(\text{not a real number}\hbox{ or }0.909091i\)
            Explanation.
            The square root of a negative number is not a real number.
            +
            Answer.
            \(\text{not a real number}\hbox{ or }0.1i\)
            Explanation.
            The square root of a negative number is not a real number.

            37.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{-\sqrt{{{\frac{121}{144}}}}={}}\).
            +
            Evaluate the following.
            +\(\displaystyle{-\sqrt{{{\frac{9}{16}}}}={}}\).
            -
            Answer.
            \(\frac{-11}{12}\)
            Explanation.
            \(\displaystyle{ +
            Answer.
            \(\frac{-3}{4}\)
            Explanation.
            \(\displaystyle{ \begin{aligned}[t] --\sqrt{\frac{121}{144}} \amp = - \frac{\sqrt{121}}{\sqrt{144}} \\ -\amp = - \frac{11}{12} +-\sqrt{\frac{9}{16}} \amp = - \frac{\sqrt{9}}{\sqrt{16}} \\ +\amp = - \frac{3}{4} \end{aligned} }\)

            38.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{-\sqrt{{{\frac{1}{25}}}}={}}\).
            +
            Evaluate the following.
            +\(\displaystyle{-\sqrt{{{\frac{16}{81}}}}={}}\).
            -
            Answer.
            \(\frac{-1}{5}\)
            Explanation.
            \(\displaystyle{ +
            Answer.
            \(\frac{-4}{9}\)
            Explanation.
            \(\displaystyle{ \begin{aligned}[t] --\sqrt{\frac{1}{25}} \amp = - \frac{\sqrt{1}}{\sqrt{25}} \\ -\amp = - \frac{1}{5} +-\sqrt{\frac{16}{81}} \amp = - \frac{\sqrt{16}}{\sqrt{81}} \\ +\amp = - \frac{4}{9} \end{aligned} }\)

            39.

            -
            -
            +
            +
            -
            Evaluate the following.
              -
            1. -\(\displaystyle{\sqrt{25}-\sqrt{16}=}\) +
              Evaluate the following.
                +
              1. +\(\displaystyle{\sqrt{100}-\sqrt{36}=}\)
                2. -
              2. -\(\displaystyle{\sqrt{25-16}=}\) +
              3. +\(\displaystyle{\sqrt{100-36}=}\)
              -
              Answer 1.
              \(1\)
              Answer 2.
              \(3\)
              Explanation.
                -
              1. \(\displaystyle \displaystyle{\begin{aligned}[t] -\sqrt{25}-\sqrt{16} \amp = 5 - 4 \\ -\amp = {1} +
                Answer 1.
                \(4\)
                Answer 2.
                \(8\)
                Explanation.
                  +
                1. \(\displaystyle \displaystyle{\begin{aligned}[t] +\sqrt{100}-\sqrt{36} \amp = 10 - 6 \\ +\amp = {4} \end{aligned} }\)
                  2. -
                2. \(\displaystyle \displaystyle{ +
                3. \(\displaystyle \displaystyle{ \begin{aligned}[t] -\sqrt{25-16} \amp = \sqrt{9} \\ -\amp = {3} +\sqrt{100-36} \amp = \sqrt{64} \\ +\amp = {8} \end{aligned} }\)

            40.

            -
            -
            +
            +
            -
            Evaluate the following.
              -
            1. +
              Evaluate the following.
                +
              1. \(\displaystyle{\sqrt{25}-\sqrt{9}=}\)
                2. -
              2. +
              3. \(\displaystyle{\sqrt{25-9}=}\)
              -
              Answer 1.
              \(2\)
              Answer 2.
              \(4\)
              Explanation.
                -
              1. \(\displaystyle \displaystyle{\begin{aligned}[t] +
                Answer 1.
                \(2\)
                Answer 2.
                \(4\)
                Explanation.
                  +
                1. \(\displaystyle \displaystyle{\begin{aligned}[t] \sqrt{25}-\sqrt{9} \amp = 5 - 3 \\ \amp = {2} \end{aligned} }\)
                  2. -
                2. \(\displaystyle \displaystyle{ +
                3. \(\displaystyle \displaystyle{ \begin{aligned}[t] \sqrt{25-9} \amp = \sqrt{16} \\ \amp = {4} @@ -1778,26 +1792,26 @@

                  Exercise Group.

            41.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{\frac{5}{\sqrt{{16}}}}\) = .
            +
            Evaluate the following.
            +\(\displaystyle{\frac{7}{\sqrt{{4}}}}\) = .
            -
            Answer.
            \(\frac{5}{4}\)
            Explanation.
            -
            Because the denominator in this problem is a perfect square, we can easily evaluate the square root:
            \(\displaystyle{ \frac{5}{\sqrt{{16}}} = \frac{5}{4} }\)
            +
            Answer.
            \(\frac{7}{2}\)
            Explanation.
            +
            Because the denominator in this problem is a perfect square, we can easily evaluate the square root:
            \(\displaystyle{ \frac{7}{\sqrt{{4}}} = \frac{7}{2} }\)

            42.

            -
            -
            +
            +
            -
            Evaluate the following.
            -\(\displaystyle{\frac{5}{\sqrt{{81}}}}\) = .
            +
            Evaluate the following.
            +\(\displaystyle{\frac{7}{\sqrt{{64}}}}\) = .
            -
            Answer.
            \(\frac{5}{9}\)
            Explanation.
            -
            Because the denominator in this problem is a perfect square, we can easily evaluate the square root:
            \(\displaystyle{ \frac{5}{\sqrt{{81}}} = \frac{5}{9} }\)
            +
            Answer.
            \(\frac{7}{8}\)
            Explanation.
            +
            Because the denominator in this problem is a perfect square, we can easily evaluate the square root:
            \(\displaystyle{ \frac{7}{\sqrt{{64}}} = \frac{7}{8} }\)
            diff --git a/section-algebraic-properties-and-simplifying-expressions.html b/section-algebraic-properties-and-simplifying-expressions.html index 2fbdc51ef..1cab88341 100644 --- a/section-algebraic-properties-and-simplifying-expressions.html +++ b/section-algebraic-properties-and-simplifying-expressions.html @@ -420,7 +420,7 @@

            Search Results:

            @@ -455,6 +455,20 @@

            Search Results:

          2. 3.9.5 Exercises
            3. +
          3. +
            3.10 Graphing Lines Chapter Review
            + +
          4. diff --git a/section-arithmetic-with-negative-numbers.html b/section-arithmetic-with-negative-numbers.html index 3cde60829..266b3998b 100644 --- a/section-arithmetic-with-negative-numbers.html +++ b/section-arithmetic-with-negative-numbers.html @@ -420,7 +420,7 @@

            Search Results:

            @@ -455,6 +455,20 @@

            Search Results:

          5. 3.9.5 Exercises
            6. +
          6. +
            3.10 Graphing Lines Chapter Review
            + +
          7. @@ -607,21 +621,21 @@

            Search Results:

          Checkpoint A.1.4. Identify “Minus” Signs.

          -
          -
          +
          +
          -
          In each expression, how many negative signs and subtraction signs are there?

          (a)

          -
          +
          In each expression, how many negative signs and subtraction signs are there?

          (a)

          +
          \(1-9\) has negative signs and subtraction signs.
          -
          Explanation.
          -\(1-9\) has zero negative signs and one subtraction sign.

          (b)

          -
          +
          Explanation.
          +\(1-9\) has zero negative signs and one subtraction sign.

          (b)

          +
          \(-12+(-50)\) has negative signs and subtraction signs.
          -
          Explanation.
          -\(-12+(-50)\) has two negative signs and zero subtraction signs.

          (c)

          -
          +
          Explanation.
          +\(-12+(-50)\) has two negative signs and zero subtraction signs.

          (c)

          +
          \(\dfrac{-13-(-15)-17}{23-4}\) has negative signs and subtraction signs.
          -
          Explanation.
          +
          Explanation.
          \(\dfrac{-13-(-15)-17}{23-4}\) has two negative signs and three subtraction signs.

    @@ -687,28 +701,28 @@

    Search Results:

    Checkpoint A.1.9. Addition with Negative Numbers.

    -
    -
    +
    +
    -
    Take a moment to practice adding when at least one negative number is involved. The expectation is that readers can make these calculations here without a calculator.

    (a)

    -
    Add \(-1+9\text{.}\) +
    Take a moment to practice adding when at least one negative number is involved. The expectation is that readers can make these calculations here without a calculator.

    (a)

    +
    Add \(-1+9\text{.}\)
    -
    Explanation.
    The two numbers have opposite sign, so we can think to subtract \(9-1=8\text{.}\) Of the two numbers we added, the positive is larger, so we stick with postive \(8\) as the answer.

    (b)

    -
    Add \(-12+(-98)\text{.}\) +
    Explanation.
    The two numbers have opposite sign, so we can think to subtract \(9-1=8\text{.}\) Of the two numbers we added, the positive is larger, so we stick with postive \(8\) as the answer.

    (b)

    +
    Add \(-12+(-98)\text{.}\)
    -
    Explanation.
    The two numbers are both negative, so we can add \(12+98=110\text{,}\) and take the negative of that as the answer: \(-110\text{.}\) -

    (c)

    -
    Add \(100+(-123)\text{.}\) +
    Explanation.
    The two numbers are both negative, so we can add \(12+98=110\text{,}\) and take the negative of that as the answer: \(-110\text{.}\) +

    (c)

    +
    Add \(100+(-123)\text{.}\)
    -
    Explanation.
    The two numbers have opposite sign, so we can think to subtract \(123-100=23\text{.}\) Of the two numbers we added, the negative is larger, so we take the negative of \(23\) as the answer. That is, the answer is \(-23\text{.}\) -

    (d)

    -
    Find the sum \(-2.1+(-2.1)\text{.}\) +
    Explanation.
    The two numbers have opposite sign, so we can think to subtract \(123-100=23\text{.}\) Of the two numbers we added, the negative is larger, so we take the negative of \(23\) as the answer. That is, the answer is \(-23\text{.}\) +

    (d)

    +
    Find the sum \(-2.1+(-2.1)\text{.}\)
    -
    Explanation.
    The two numbers are both negative, so we can add \(2.1+2.1=4.2\text{,}\) and take the negative of that as the answer: \(-4.2\text{.}\) -

    (e)

    -
    Find the sum \(-34.67+81.53\text{.}\) +
    Explanation.
    The two numbers are both negative, so we can add \(2.1+2.1=4.2\text{,}\) and take the negative of that as the answer: \(-4.2\text{.}\) +

    (e)

    +
    Find the sum \(-34.67+81.53\text{.}\)
    -
    Explanation.
    The two numbers have opposite sign, so we can think to subtract \(81.53-34.67=46.86\text{.}\) Of the two numbers we added, the positive is larger, so we stick with postive \(46.86\) as the answer.
    +
    Explanation.
    The two numbers have opposite sign, so we can think to subtract \(81.53-34.67=46.86\text{.}\) Of the two numbers we added, the positive is larger, so we stick with postive \(46.86\) as the answer.

    Subsection A.1.3 Subtracting @@ -760,29 +774,29 @@

    Search Results:

    Checkpoint A.1.10. Subtraction with Negative Numbers.

    -
    -
    +
    +
    -
    Take a moment to practice subtracting when at least one negative number is involved. The expectation is that readers can make these calculations here without a calculator.

    (a)

    -
    Subtract \(-1\) from \(9\text{.}\) +
    Take a moment to practice subtracting when at least one negative number is involved. The expectation is that readers can make these calculations here without a calculator.

    (a)

    +
    Subtract \(-1\) from \(9\text{.}\)
    -
    Explanation.
    After writing this as \(9-(-1)\text{,}\) we can rewrite it as \(9+1\) and get \(10\text{.}\) -

    (b)

    -
    Subtract \(32-50\text{.}\) +
    Explanation.
    After writing this as \(9-(-1)\text{,}\) we can rewrite it as \(9+1\) and get \(10\text{.}\) +

    (b)

    +
    Subtract \(32-50\text{.}\)
    -
    Explanation.
    Subtrcting in the oppsite order with the larger number first, \(50-32=18\text{.}\) But since we were asked to subtract the larger number from the smaller number, the answer is \(-18\text{.}\) -

    (c)

    -
    Subtract \(108-(-108)\text{.}\) +
    Explanation.
    Subtrcting in the oppsite order with the larger number first, \(50-32=18\text{.}\) But since we were asked to subtract the larger number from the smaller number, the answer is \(-18\text{.}\) +

    (c)

    +
    Subtract \(108-(-108)\text{.}\)
    -
    Explanation.
    After writing this as \(108-(-108)\text{,}\) we can rewrite it as \(108+108\) and get \(216\text{.}\) -

    (d)

    -
    Find the difference \(-5.9-(-3.1)\text{.}\) +
    Explanation.
    After writing this as \(108-(-108)\text{,}\) we can rewrite it as \(108+108\) and get \(216\text{.}\) +

    (d)

    +
    Find the difference \(-5.9-(-3.1)\text{.}\)
    -
    Explanation.
    After writing this as \(-5.9-(-3.1)\text{,}\) we can rewrite it as \(-5.9+3.1\text{.}\) Now it is the sum of two numbers of opposite sign, so we can subtract \(5.9-3.1\) to get \(2.8\text{.}\) But we were adding numbers where the negative number was larger, so the final answer should be \(-2.8\text{.}\) -

    (e)

    -
    Find the difference \(-12.04-17.2\text{.}\) +
    Explanation.
    After writing this as \(-5.9-(-3.1)\text{,}\) we can rewrite it as \(-5.9+3.1\text{.}\) Now it is the sum of two numbers of opposite sign, so we can subtract \(5.9-3.1\) to get \(2.8\text{.}\) But we were adding numbers where the negative number was larger, so the final answer should be \(-2.8\text{.}\) +

    (e)

    +
    Find the difference \(-12.04-17.2\text{.}\)
    -
    Explanation.
    Since we are subtracting a positive number from a negative number, the result should be an even more negative number. We can add \(12.04+17.2\) to get \(29.24\text{,}\) but our final answer should be the opposite, \(-29.24\text{.}\) +
    Explanation.
    Since we are subtracting a positive number from a negative number, the result should be an even more negative number. We can add \(12.04+17.2\) to get \(29.24\text{,}\) but our final answer should be the opposite, \(-29.24\text{.}\)

    @@ -804,25 +818,25 @@

    Search Results:

    Checkpoint A.1.14. Multiplication with Negative Numbers.

    -
    -
    +
    +
    -
    Here are some practice exercises with multiplication and signed numbers. The expectation is that readers can make these calculations here without a calculator.

    (a)

    -
    Multiply \(-13\cdot2\text{.}\) +
    Here are some practice exercises with multiplication and signed numbers. The expectation is that readers can make these calculations here without a calculator.

    (a)

    +
    Multiply \(-13\cdot2\text{.}\)
    -
    Explanation.
    Since \(13\cdot2=26\text{,}\) and we are multiplying numbers of opposite signs, the answer is negative: \(-26\text{.}\) -

    (b)

    -
    Find the product of \(30\) and \(-50\text{.}\) +
    Explanation.
    Since \(13\cdot2=26\text{,}\) and we are multiplying numbers of opposite signs, the answer is negative: \(-26\text{.}\) +

    (b)

    +
    Find the product of \(30\) and \(-50\text{.}\)
    -
    Explanation.
    Since \(30\cdot50=1500\text{,}\) and we are multiplying numbers of opposite signs, the answer is negative: \(-1500\text{.}\) -

    (c)

    -
    Compute \(-12(-7)\text{.}\) +
    Explanation.
    Since \(30\cdot50=1500\text{,}\) and we are multiplying numbers of opposite signs, the answer is negative: \(-1500\text{.}\) +

    (c)

    +
    Compute \(-12(-7)\text{.}\)
    -
    Explanation.
    Since \(12\cdot7=84\text{,}\) and we are multiplying numbers of the same sign, the answer is positive: \(84\text{.}\) -

    (d)

    -
    Find the product \(-285(0)\text{.}\) +
    Explanation.
    Since \(12\cdot7=84\text{,}\) and we are multiplying numbers of the same sign, the answer is positive: \(84\text{.}\) +

    (d)

    +
    Find the product \(-285(0)\text{.}\)
    -
    Explanation.
    Any number multiplied by \(0\) is \(0\text{.}\) +
    Explanation.
    Any number multiplied by \(0\) is \(0\text{.}\)

    @@ -861,25 +875,25 @@

    Search Results:

    Checkpoint A.1.16. Exponents with Negative Bases.

    -
    -
    +
    +
    -
    Here is some practice with natural exponents on negative bases. The expectation is that readers can make these calculations here without a calculator.

    (a)

    -
    Compute \((-8)^2\text{.}\) +
    Here is some practice with natural exponents on negative bases. The expectation is that readers can make these calculations here without a calculator.

    (a)

    +
    Compute \((-8)^2\text{.}\)
    -
    Explanation.
    Since \(8^2\) is \(64\) and we are raising a negative number to an even power, the answer is positive: \(64\text{.}\) -

    (b)

    -
    Calculate the power \((-1)^{203}\text{.}\) +
    Explanation.
    Since \(8^2\) is \(64\) and we are raising a negative number to an even power, the answer is positive: \(64\text{.}\) +

    (b)

    +
    Calculate the power \((-1)^{203}\text{.}\)
    -
    Explanation.
    Since \(1^{203}\) is \(1\) and we are raising a negative number to an odd power, the answer is negative: \(-1\text{.}\) -

    (c)

    -
    Find \((-3)^3\text{.}\) +
    Explanation.
    Since \(1^{203}\) is \(1\) and we are raising a negative number to an odd power, the answer is negative: \(-1\text{.}\) +

    (c)

    +
    Find \((-3)^3\text{.}\)
    -
    Explanation.
    Since \(3^{3}\) is \(27\) and we are raising a negative number to an odd power, the answer is negative: \(-27\text{.}\) -

    (d)

    -
    Calculate \(-5^2\text{.}\) +
    Explanation.
    Since \(3^{3}\) is \(27\) and we are raising a negative number to an odd power, the answer is negative: \(-27\text{.}\) +

    (d)

    +
    Calculate \(-5^2\text{.}\)
    -
    Explanation.
    Careful: here we are raising positive \(5\) to the second power to get \(25\) and then negating the result: \(-25\text{.}\) Since we don’t see “\((-5)^2\text{,}\)” the answer is not positive \(25\text{.}\) +
    Explanation.
    Careful: here we are raising positive \(5\) to the second power to get \(25\) and then negating the result: \(-25\text{.}\) Since we don’t see “\((-5)^2\text{,}\)” the answer is not positive \(25\text{.}\)

    @@ -921,173 +935,173 @@

    Search Results:

    Exercise Group.

    1.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle -8+(-3)\)
      2. -
    2. \(\displaystyle -6+(-7)\)
      3. -
    3. \(\displaystyle -2+(-9)\)
      4. +
    4. \(\displaystyle -10+(-3)\)
      5. +
    5. \(\displaystyle -7+(-4)\)
      6. +
    6. \(\displaystyle -1+(-9)\)
    -
    Answer 1.
    \(-11\)
    Answer 2.
    \(-13\)
    Answer 3.
    \(-11\)
    Explanation.
    -
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-8+(-3)\text{.}\) -
    First, find \(-8\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(3\) units. We will reach \({-11}\) on the number line, which is the answer.
    So, \(-8+(-3)={-11}\text{.}\) -
    Similarly, \(-6+(-7)={-13}\text{,}\) and \(-2+(-9)={-11}\text{.}\) -
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-8+(-3)\text{.}\) -
    The first number is \(-8\text{.}\) Since it’s negative, it’s like you lost \(8\) dollars while gambling at the casino this morning.
    The second number is \(-3\text{.}\) Since it’s negative, it’s like you lost \(3\) dollars again while gambling in the casino this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(8\) dollars and then lost \(3\) dollars, it makes sense that you lost a total of \({11}\) dollars. So the final answer is: \(-8+(-3)={-11}\text{.}\) +
    Answer 1.
    \(-13\)
    Answer 2.
    \(-11\)
    Answer 3.
    \(-10\)
    Explanation.
    +
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-10+(-3)\text{.}\) +
    First, find \(-10\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(3\) units. We will reach \({-13}\) on the number line, which is the answer.
    So, \(-10+(-3)={-13}\text{.}\) +
    Similarly, \(-7+(-4)={-11}\text{,}\) and \(-1+(-9)={-10}\text{.}\) +
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-10+(-3)\text{.}\) +
    The first number is \(-10\text{.}\) Since it’s negative, it’s like you lost \(10\) dollars while gambling at the casino this morning.
    The second number is \(-3\text{.}\) Since it’s negative, it’s like you lost \(3\) dollars again while gambling in the casino this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(10\) dollars and then lost \(3\) dollars, it makes sense that you lost a total of \({13}\) dollars. So the final answer is: \(-10+(-3)={-13}\text{.}\)

    2.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle -8+(-1)\)
      2. -
    2. \(\displaystyle -5+(-3)\)
      3. -
    3. \(\displaystyle -2+(-7)\)
      4. +
    4. \(\displaystyle -10+(-1)\)
      5. +
    5. \(\displaystyle -5+(-6)\)
      6. +
    6. \(\displaystyle -1+(-7)\)
    -
    Answer 1.
    \(-9\)
    Answer 2.
    \(-8\)
    Answer 3.
    \(-9\)
    Explanation.
    -
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-8+(-1)\text{.}\) -
    First, find \(-8\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(1\) units. We will reach \({-9}\) on the number line, which is the answer.
    So, \(-8+(-1)={-9}\text{.}\) -
    Similarly, \(-5+(-3)={-8}\text{,}\) and \(-2+(-7)={-9}\text{.}\) -
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-8+(-1)\text{.}\) -
    The first number is \(-8\text{.}\) Since it’s negative, it’s like you lost \(8\) dollars while gambling at the casino this morning.
    The second number is \(-1\text{.}\) Since it’s negative, it’s like you lost \(1\) dollars again while gambling in the casino this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(8\) dollars and then lost \(1\) dollars, it makes sense that you lost a total of \({9}\) dollars. So the final answer is: \(-8+(-1)={-9}\text{.}\) +
    Answer 1.
    \(-11\)
    Answer 2.
    \(-11\)
    Answer 3.
    \(-8\)
    Explanation.
    +
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-10+(-1)\text{.}\) +
    First, find \(-10\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(1\) units. We will reach \({-11}\) on the number line, which is the answer.
    So, \(-10+(-1)={-11}\text{.}\) +
    Similarly, \(-5+(-6)={-11}\text{,}\) and \(-1+(-7)={-8}\text{.}\) +
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-10+(-1)\text{.}\) +
    The first number is \(-10\text{.}\) Since it’s negative, it’s like you lost \(10\) dollars while gambling at the casino this morning.
    The second number is \(-1\text{.}\) Since it’s negative, it’s like you lost \(1\) dollars again while gambling in the casino this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(10\) dollars and then lost \(1\) dollars, it makes sense that you lost a total of \({11}\) dollars. So the final answer is: \(-10+(-1)={-11}\text{.}\)

    3.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle 1+(-8)\)
      2. -
    2. \(\displaystyle 5+(-2)\)
      3. -
    3. \(\displaystyle 5+(-5)\)
      4. +
    4. \(\displaystyle 2+(-9)\)
      5. +
    5. \(\displaystyle 10+(-4)\)
      6. +
    6. \(\displaystyle 10+(-10)\)
    -
    Answer 1.
    \(-7\)
    Answer 2.
    \(3\)
    Answer 3.
    \(0\)
    Explanation.
    -
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(1+(-8)\text{.}\) -
    First, find \(1\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(8\) units. We will reach \({-7}\) on the number line, which is the answer.
    So, \(1+(-8)={-7}\text{.}\) -
    Similarly, \(5+(-2)={3}\text{,}\) and \(5+(-5)={0}\text{.}\) -
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(1+(-8)\text{.}\) -
    The first number is \(1\text{.}\) Since it’s positive, it’s like you won \(1\) dollars while gambling at the casino this morning.
    The second number is \(-8\text{.}\) Since it’s negative, it’s like you lost \(8\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you won \(1\) dollars and then lost \(8\) dollars, it makes sense that you lost the difference of \(8\) and \(1\) dollars, which is \({7}\) dollars. So the final answer is: \(1+(-8)={-7}\text{.}\) +
    Answer 1.
    \(-7\)
    Answer 2.
    \(6\)
    Answer 3.
    \(0\)
    Explanation.
    +
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(2+(-9)\text{.}\) +
    First, find \(2\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(9\) units. We will reach \({-7}\) on the number line, which is the answer.
    So, \(2+(-9)={-7}\text{.}\) +
    Similarly, \(10+(-4)={6}\text{,}\) and \(10+(-10)={0}\text{.}\) +
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(2+(-9)\text{.}\) +
    The first number is \(2\text{.}\) Since it’s positive, it’s like you won \(2\) dollars while gambling at the casino this morning.
    The second number is \(-9\text{.}\) Since it’s negative, it’s like you lost \(9\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you won \(2\) dollars and then lost \(9\) dollars, it makes sense that you lost the difference of \(9\) and \(2\) dollars, which is \({7}\) dollars. So the final answer is: \(2+(-9)={-7}\text{.}\)

    4.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle 1+(-9)\)
      2. -
    2. \(\displaystyle 8+(-3)\)
      3. -
    3. \(\displaystyle 4+(-4)\)
      4. +
    4. \(\displaystyle 2+(-10)\)
      5. +
    5. \(\displaystyle 7+(-1)\)
      6. +
    6. \(\displaystyle 9+(-9)\)
    -
    Answer 1.
    \(-8\)
    Answer 2.
    \(5\)
    Answer 3.
    \(0\)
    Explanation.
    -
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(1+(-9)\text{.}\) -
    First, find \(1\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(9\) units. We will reach \({-8}\) on the number line, which is the answer.
    So, \(1+(-9)={-8}\text{.}\) -
    Similarly, \(8+(-3)={5}\text{,}\) and \(4+(-4)={0}\text{.}\) -
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(1+(-9)\text{.}\) -
    The first number is \(1\text{.}\) Since it’s positive, it’s like you won \(1\) dollars while gambling at the casino this morning.
    The second number is \(-9\text{.}\) Since it’s negative, it’s like you lost \(9\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you won \(1\) dollars and then lost \(9\) dollars, it makes sense that you lost the difference of \(9\) and \(1\) dollars, which is \({8}\) dollars. So the final answer is: \(1+(-9)={-8}\text{.}\) +
    Answer 1.
    \(-8\)
    Answer 2.
    \(6\)
    Answer 3.
    \(0\)
    Explanation.
    +
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(2+(-10)\text{.}\) +
    First, find \(2\) on the number line. Next, since we are adding a negative number, we move left, in the negative direction, by \(10\) units. We will reach \({-8}\) on the number line, which is the answer.
    So, \(2+(-10)={-8}\text{.}\) +
    Similarly, \(7+(-1)={6}\text{,}\) and \(9+(-9)={0}\text{.}\) +
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(2+(-10)\text{.}\) +
    The first number is \(2\text{.}\) Since it’s positive, it’s like you won \(2\) dollars while gambling at the casino this morning.
    The second number is \(-10\text{.}\) Since it’s negative, it’s like you lost \(10\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you won \(2\) dollars and then lost \(10\) dollars, it makes sense that you lost the difference of \(10\) and \(2\) dollars, which is \({8}\) dollars. So the final answer is: \(2+(-10)={-8}\text{.}\)

    5.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle -6+2\)
      2. -
    2. \(\displaystyle -4+10\)
      3. -
    3. \(\displaystyle -7+7\)
      4. +
    4. \(\displaystyle -7+3\)
      5. +
    5. \(\displaystyle -3+9\)
      6. +
    6. \(\displaystyle -2+2\)
    -
    Answer 1.
    \(-4\)
    Answer 2.
    \(6\)
    Answer 3.
    \(0\)
    Explanation.
    -
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-6+2\text{.}\) -
    First, find \(-6\) on the number line. Next, since we are adding a positive number, we move right, in the positive direction, by \(2\) units. We will reach \({-4}\) on the number line, which is the answer.
    So, \(-6+(2)={-4}\text{.}\) -
    Similarly, \(-4+(10)={6}\text{,}\) and \(-7+(7)={0}\text{.}\) -
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-6+2\text{.}\) -
    The first number is \(-6\text{.}\) Since it’s negative, it’s like you lost \(6\) dollars while gambling at the casino this morning.
    The second number is \(2\text{.}\) Since it’s positive, it’s like you won \(2\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you lost \(6\) dollars and then won \(2\) dollars, it makes sense that you lost the difference of \(6\) and \(2\) dollars, which is \({4}\) dollars. So the final answer is: \(-6+2={-4}\text{.}\) +
    Answer 1.
    \(-4\)
    Answer 2.
    \(6\)
    Answer 3.
    \(0\)
    Explanation.
    +
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-7+3\text{.}\) +
    First, find \(-7\) on the number line. Next, since we are adding a positive number, we move right, in the positive direction, by \(3\) units. We will reach \({-4}\) on the number line, which is the answer.
    So, \(-7+(3)={-4}\text{.}\) +
    Similarly, \(-3+(9)={6}\text{,}\) and \(-2+(2)={0}\text{.}\) +
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-7+3\text{.}\) +
    The first number is \(-7\text{.}\) Since it’s negative, it’s like you lost \(7\) dollars while gambling at the casino this morning.
    The second number is \(3\text{.}\) Since it’s positive, it’s like you won \(3\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you lost \(7\) dollars and then won \(3\) dollars, it makes sense that you lost the difference of \(7\) and \(3\) dollars, which is \({4}\) dollars. So the final answer is: \(-7+3={-4}\text{.}\)

    6.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle -8+2\)
      2. -
    2. \(\displaystyle -2+7\)
      3. -
    3. \(\displaystyle -7+7\)
      4. +
    4. \(\displaystyle -9+3\)
      5. +
    5. \(\displaystyle -4+6\)
      6. +
    6. \(\displaystyle -2+2\)
    -
    Answer 1.
    \(-6\)
    Answer 2.
    \(5\)
    Answer 3.
    \(0\)
    Explanation.
    -
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-8+2\text{.}\) -
    First, find \(-8\) on the number line. Next, since we are adding a positive number, we move right, in the positive direction, by \(2\) units. We will reach \({-6}\) on the number line, which is the answer.
    So, \(-8+(2)={-6}\text{.}\) -
    Similarly, \(-2+(7)={5}\text{,}\) and \(-7+(7)={0}\text{.}\) -
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-8+2\text{.}\) -
    The first number is \(-8\text{.}\) Since it’s negative, it’s like you lost \(8\) dollars while gambling at the casino this morning.
    The second number is \(2\text{.}\) Since it’s positive, it’s like you won \(2\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you lost \(8\) dollars and then won \(2\) dollars, it makes sense that you lost the difference of \(8\) and \(2\) dollars, which is \({6}\) dollars. So the final answer is: \(-8+2={-6}\text{.}\) +
    Answer 1.
    \(-6\)
    Answer 2.
    \(2\)
    Answer 3.
    \(0\)
    Explanation.
    +
    Here are two different explanations of how two negative numbers can be added together.
    METHOD 1
    Use a number line. Let’s find \(-9+3\text{.}\) +
    First, find \(-9\) on the number line. Next, since we are adding a positive number, we move right, in the positive direction, by \(3\) units. We will reach \({-6}\) on the number line, which is the answer.
    So, \(-9+(3)={-6}\text{.}\) +
    Similarly, \(-4+(6)={2}\text{,}\) and \(-2+(2)={0}\text{.}\) +
    METHOD 2
    A second method asks you to think in terms of money. Let’s find \(-9+3\text{.}\) +
    The first number is \(-9\text{.}\) Since it’s negative, it’s like you lost \(9\) dollars while gambling at the casino this morning.
    The second number is \(3\text{.}\) Since it’s positive, it’s like you won \(3\) dollars while gambling at the casino this afternoon.
    Since you lost more money than you won, overall you have lost, implying the answer is negative.
    Since you lost \(9\) dollars and then won \(3\) dollars, it makes sense that you lost the difference of \(9\) and \(3\) dollars, which is \({6}\) dollars. So the final answer is: \(-9+3={-6}\text{.}\)

    7.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle -55+(-83)\)
      2. -
    2. \(\displaystyle -29+53\)
      3. -
    3. \(\displaystyle 42+(-62)\)
      4. +
    4. \(\displaystyle -37+(-12)\)
      5. +
    5. \(\displaystyle -43+95\)
      6. +
    6. \(\displaystyle 88+(-71)\)
    -
    Answer 1.
    \(-138\)
    Answer 2.
    \(24\)
    Answer 3.
    \(-20\)
    +
    Answer 1.
    \(-49\)
    Answer 2.
    \(52\)
    Answer 3.
    \(17\)

    8.

    -
    -
    +
    +
    -
    +
    Add the following.
      -
    1. \(\displaystyle -45+(-24)\)
      2. -
    2. \(\displaystyle -82+28\)
      3. -
    3. \(\displaystyle 41+(-17)\)
      4. +
    4. \(\displaystyle -27+(-43)\)
      5. +
    5. \(\displaystyle -96+70\)
      6. +
    6. \(\displaystyle 87+(-27)\)
    -
    Answer 1.
    \(-69\)
    Answer 2.
    \(-54\)
    Answer 3.
    \(24\)
    +
    Answer 1.
    \(-70\)
    Answer 2.
    \(-26\)
    Answer 3.
    \(60\)
    @@ -1097,127 +1111,127 @@

    Exercise Group.

    Exercise Group.

    9.

    -
    -
    +
    +
    -
    +
    Subtract the following.
      -
    1. \(\displaystyle 4-8\)
      2. -
    2. \(\displaystyle 8-4\)
      3. -
    3. \(\displaystyle 3-15\)
      4. +
    4. \(\displaystyle 5-7\)
      5. +
    5. \(\displaystyle 7-2\)
      6. +
    6. \(\displaystyle 6-14\)
    -
    Answer 1.
    \(-4\)
    Answer 2.
    \(4\)
    Answer 3.
    \(-12\)
    Explanation.
    -
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(4-8\text{.}\) First, we rewrite it as
    \(4+(-8)\)
    Find \(4\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(8\) units. We will reach \({-4}\) on the number line, which is the answer.
    So \(4-8={-4}\text{.}\) -
    Similarly, \(8-4={4}\) and \(3-15={-12}\text{.}\) -
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(4-8\text{.}\) First, we rewrite it as
    \(4+(-8)\text{.}\)
    The first number is \(4\text{.}\) Since it’s positive, it’s like you won \(4\) dollars at the casino this morning.
    The second number is \(-8\text{.}\) Since it’s negative, it’s like you lost \(8\) dollars in the casino later this afternoon.
    Since you lost more money than you won, you ended up losing money overall, implying the answer is negative.
    Since you won \(4\) dollars and then lost \(8\) dollars, it makes sense that you lost the difference of \(8\) and \(4\) dollars, which is \({4}\) dollars. So the final answer is: \(4-8={-4}\text{.}\) +
    Answer 1.
    \(-2\)
    Answer 2.
    \(5\)
    Answer 3.
    \(-8\)
    Explanation.
    +
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(5-7\text{.}\) First, we rewrite it as
    \(5+(-7)\)
    Find \(5\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(7\) units. We will reach \({-2}\) on the number line, which is the answer.
    So \(5-7={-2}\text{.}\) +
    Similarly, \(7-2={5}\) and \(6-14={-8}\text{.}\) +
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(5-7\text{.}\) First, we rewrite it as
    \(5+(-7)\text{.}\)
    The first number is \(5\text{.}\) Since it’s positive, it’s like you won \(5\) dollars at the casino this morning.
    The second number is \(-7\text{.}\) Since it’s negative, it’s like you lost \(7\) dollars in the casino later this afternoon.
    Since you lost more money than you won, you ended up losing money overall, implying the answer is negative.
    Since you won \(5\) dollars and then lost \(7\) dollars, it makes sense that you lost the difference of \(7\) and \(5\) dollars, which is \({2}\) dollars. So the final answer is: \(5-7={-2}\text{.}\)

    10.

    -
    -
    +
    +
    -
    +
    Subtract the following.
      -
    1. \(\displaystyle 5-6\)
      2. -
    2. \(\displaystyle 5-3\)
      3. -
    3. \(\displaystyle 3-20\)
      4. +
    4. \(\displaystyle 1-10\)
      5. +
    5. \(\displaystyle 10-1\)
      6. +
    6. \(\displaystyle 6-19\)
    -
    Answer 1.
    \(-1\)
    Answer 2.
    \(2\)
    Answer 3.
    \(-17\)
    Explanation.
    -
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(5-6\text{.}\) First, we rewrite it as
    \(5+(-6)\)
    Find \(5\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(6\) units. We will reach \({-1}\) on the number line, which is the answer.
    So \(5-6={-1}\text{.}\) -
    Similarly, \(5-3={2}\) and \(3-20={-17}\text{.}\) -
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(5-6\text{.}\) First, we rewrite it as
    \(5+(-6)\text{.}\)
    The first number is \(5\text{.}\) Since it’s positive, it’s like you won \(5\) dollars at the casino this morning.
    The second number is \(-6\text{.}\) Since it’s negative, it’s like you lost \(6\) dollars in the casino later this afternoon.
    Since you lost more money than you won, you ended up losing money overall, implying the answer is negative.
    Since you won \(5\) dollars and then lost \(6\) dollars, it makes sense that you lost the difference of \(6\) and \(5\) dollars, which is \({1}\) dollars. So the final answer is: \(5-6={-1}\text{.}\) +
    Answer 1.
    \(-9\)
    Answer 2.
    \(9\)
    Answer 3.
    \(-13\)
    Explanation.
    +
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(1-10\text{.}\) First, we rewrite it as
    \(1+(-10)\)
    Find \(1\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(10\) units. We will reach \({-9}\) on the number line, which is the answer.
    So \(1-10={-9}\text{.}\) +
    Similarly, \(10-1={9}\) and \(6-19={-13}\text{.}\) +
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(1-10\text{.}\) First, we rewrite it as
    \(1+(-10)\text{.}\)
    The first number is \(1\text{.}\) Since it’s positive, it’s like you won \(1\) dollars at the casino this morning.
    The second number is \(-10\text{.}\) Since it’s negative, it’s like you lost \(10\) dollars in the casino later this afternoon.
    Since you lost more money than you won, you ended up losing money overall, implying the answer is negative.
    Since you won \(1\) dollars and then lost \(10\) dollars, it makes sense that you lost the difference of \(10\) and \(1\) dollars, which is \({9}\) dollars. So the final answer is: \(1-10={-9}\text{.}\)

    11.

    -
    -
    +
    +
    -
    +
    Subtract the following.
      -
    1. \(\displaystyle -1-5\)
      2. -
    2. \(\displaystyle -8-2\)
      3. -
    3. \(\displaystyle -8-8\)
      4. +
    4. \(\displaystyle -5-3\)
      5. +
    5. \(\displaystyle -9-5\)
      6. +
    6. \(\displaystyle -4-4\)
    -
    Answer 1.
    \(-6\)
    Answer 2.
    \(-10\)
    Answer 3.
    \(-16\)
    Explanation.
    -
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-1-5\text{.}\) First, we rewrite it as
    \(-1+(-5)\)
    Find \(-1\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(5\) units. We will reach \({-6}\) on the number line, which is the answer.
    So \(-1-5={-6}\text{.}\) -
    Similarly, \(-8-2={-10}\) and \(-8-8={-16}\text{.}\) -
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-1-5\text{.}\) First, we rewrite it as
    \(-1+(-5)\text{.}\)
    The first number is \(-1\text{.}\) Since it’s negative, it’s like you lost \(1\) dollars at the casino this morning.
    The second number is \(-5\text{.}\) Since it’s also negative, it’s like you lost \(5\) dollars more in the casino later this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(1\) dollars and then lost \(5\) dollars, it makes sense that you lost the total of \(1\) and \(5\) dollars, which is \({6}\) dollars. So the final answer is: \(-1-5={-6}\text{.}\) +
    Answer 1.
    \(-8\)
    Answer 2.
    \(-14\)
    Answer 3.
    \(-8\)
    Explanation.
    +
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-5-3\text{.}\) First, we rewrite it as
    \(-5+(-3)\)
    Find \(-5\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(3\) units. We will reach \({-8}\) on the number line, which is the answer.
    So \(-5-3={-8}\text{.}\) +
    Similarly, \(-9-5={-14}\) and \(-4-4={-8}\text{.}\) +
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-5-3\text{.}\) First, we rewrite it as
    \(-5+(-3)\text{.}\)
    The first number is \(-5\text{.}\) Since it’s negative, it’s like you lost \(5\) dollars at the casino this morning.
    The second number is \(-3\text{.}\) Since it’s also negative, it’s like you lost \(3\) dollars more in the casino later this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(5\) dollars and then lost \(3\) dollars, it makes sense that you lost the total of \(5\) and \(3\) dollars, which is \({8}\) dollars. So the final answer is: \(-5-3={-8}\text{.}\)

    12.

    -
    -
    +
    +
    -
    +
    Subtract the following.
      -
    1. \(\displaystyle -5-3\)
      2. -
    2. \(\displaystyle -6-1\)
      3. -
    3. \(\displaystyle -8-8\)
      4. +
    4. \(\displaystyle -4-2\)
      5. +
    5. \(\displaystyle -7-3\)
      6. +
    6. \(\displaystyle -4-4\)
    -
    Answer 1.
    \(-8\)
    Answer 2.
    \(-7\)
    Answer 3.
    \(-16\)
    Explanation.
    -
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-5-3\text{.}\) First, we rewrite it as
    \(-5+(-3)\)
    Find \(-5\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(3\) units. We will reach \({-8}\) on the number line, which is the answer.
    So \(-5-3={-8}\text{.}\) -
    Similarly, \(-6-1={-7}\) and \(-8-8={-16}\text{.}\) -
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-5-3\text{.}\) First, we rewrite it as
    \(-5+(-3)\text{.}\)
    The first number is \(-5\text{.}\) Since it’s negative, it’s like you lost \(5\) dollars at the casino this morning.
    The second number is \(-3\text{.}\) Since it’s also negative, it’s like you lost \(3\) dollars more in the casino later this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(5\) dollars and then lost \(3\) dollars, it makes sense that you lost the total of \(5\) and \(3\) dollars, which is \({8}\) dollars. So the final answer is: \(-5-3={-8}\text{.}\) +
    Answer 1.
    \(-6\)
    Answer 2.
    \(-10\)
    Answer 3.
    \(-8\)
    Explanation.
    +
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-4-2\text{.}\) First, we rewrite it as
    \(-4+(-2)\)
    Find \(-4\) on a number line. Since we are adding a negative number, we move left, in the negative direction, by \(2\) units. We will reach \({-6}\) on the number line, which is the answer.
    So \(-4-2={-6}\text{.}\) +
    Similarly, \(-7-3={-10}\) and \(-4-4={-8}\text{.}\) +
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-4-2\text{.}\) First, we rewrite it as
    \(-4+(-2)\text{.}\)
    The first number is \(-4\text{.}\) Since it’s negative, it’s like you lost \(4\) dollars at the casino this morning.
    The second number is \(-2\text{.}\) Since it’s also negative, it’s like you lost \(2\) dollars more in the casino later this afternoon.
    Since you lost twice, you ended up losing a lot, implying the answer is negative.
    Since you lost \(4\) dollars and then lost \(2\) dollars, it makes sense that you lost the total of \(4\) and \(2\) dollars, which is \({6}\) dollars. So the final answer is: \(-4-2={-6}\text{.}\)

    13.

    -
    -
    +
    +
    -
    +
    Subtract the following.
      -
    1. \(\displaystyle -1-(-6)\)
      2. -
    2. \(\displaystyle -6-(-2)\)
      3. -
    3. \(\displaystyle -3-(-3)\)
      4. +
    4. \(\displaystyle -2-(-8)\)
      5. +
    5. \(\displaystyle -5-(-1)\)
      6. +
    6. \(\displaystyle -5-(-5)\)
    -
    Answer 1.
    \(5\)
    Answer 2.
    \(-4\)
    Answer 3.
    \(0\)
    Explanation.
    -
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-1-(-6)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-1+6\)
    Find \(-1\) on a number line. Since we are adding a positive number we move right, in the positive direction, by \(6\) units. We will reach \({5}\) on the number line, which is the answer.
    So \(-1-(-6)={5}\text{.}\) -
    Similarly, \(-6-(-2)={-4}\) and \(-3-(-3)={0}\text{.}\) -
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-1-(-6)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-1+6\text{.}\)
    The first number is \(-1\text{.}\) Since it’s negative, it’s like you lost \(1\) dollars at the casino this morning.
    The second number is \(6\text{.}\) Since it’s positive, it’s like you won \(6\) dollars more in the casino later this afternoon.
    Since you won more than you lost, you have a net gain overall, implying the answer is positve.
    Since you lost \(1\) dollars and then won \(6\) dollars, it makes sense that you won the difference between \(6\) and \(1\) dollars, which is \({5}\) dollars. So the final answer is: \(-1-(-6)={5}\text{.}\) This is because we can rewrite the subtraction of a negative number in the previous expression as \(-1+(+6)={5}\text{.}\) +
    Answer 1.
    \(6\)
    Answer 2.
    \(-4\)
    Answer 3.
    \(0\)
    Explanation.
    +
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-2-(-8)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-2+8\)
    Find \(-2\) on a number line. Since we are adding a positive number we move right, in the positive direction, by \(8\) units. We will reach \({6}\) on the number line, which is the answer.
    So \(-2-(-8)={6}\text{.}\) +
    Similarly, \(-5-(-1)={-4}\) and \(-5-(-5)={0}\text{.}\) +
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-2-(-8)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-2+8\text{.}\)
    The first number is \(-2\text{.}\) Since it’s negative, it’s like you lost \(2\) dollars at the casino this morning.
    The second number is \(8\text{.}\) Since it’s positive, it’s like you won \(8\) dollars more in the casino later this afternoon.
    Since you won more than you lost, you have a net gain overall, implying the answer is positve.
    Since you lost \(2\) dollars and then won \(8\) dollars, it makes sense that you won the difference between \(8\) and \(2\) dollars, which is \({6}\) dollars. So the final answer is: \(-2-(-8)={6}\text{.}\) This is because we can rewrite the subtraction of a negative number in the previous expression as \(-2+(+8)={6}\text{.}\)

    14.

    -
    -
    +
    +
    -
    +
    Subtract the following.
      -
    1. \(\displaystyle -2-(-8)\)
      2. -
    2. \(\displaystyle -9-(-2)\)
      3. -
    3. \(\displaystyle -3-(-3)\)
      4. +
    4. \(\displaystyle -3-(-7)\)
      5. +
    5. \(\displaystyle -8-(-1)\)
      6. +
    6. \(\displaystyle -5-(-5)\)
    -
    Answer 1.
    \(6\)
    Answer 2.
    \(-7\)
    Answer 3.
    \(0\)
    Explanation.
    -
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-2-(-8)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-2+8\)
    Find \(-2\) on a number line. Since we are adding a positive number we move right, in the positive direction, by \(8\) units. We will reach \({6}\) on the number line, which is the answer.
    So \(-2-(-8)={6}\text{.}\) -
    Similarly, \(-9-(-2)={-7}\) and \(-3-(-3)={0}\text{.}\) -
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-2-(-8)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-2+8\text{.}\)
    The first number is \(-2\text{.}\) Since it’s negative, it’s like you lost \(2\) dollars at the casino this morning.
    The second number is \(8\text{.}\) Since it’s positive, it’s like you won \(8\) dollars more in the casino later this afternoon.
    Since you won more than you lost, you have a net gain overall, implying the answer is positve.
    Since you lost \(2\) dollars and then won \(8\) dollars, it makes sense that you won the difference between \(8\) and \(2\) dollars, which is \({6}\) dollars. So the final answer is: \(-2-(-8)={6}\text{.}\) This is because we can rewrite the subtraction of a negative number in the previous expression as \(-2+(+8)={6}\text{.}\) +
    Answer 1.
    \(4\)
    Answer 2.
    \(-7\)
    Answer 3.
    \(0\)
    Explanation.
    +
    It helps to understand that the subtraction sign means that we are adding the opposite. For example, \(3-2=3+(-2)\text{.}\) For many students, it’s easier to change “minus a number” into “adding the opposite number”. This way of looking at subtraction will help us understand more complicated topics later, like subtracting a negative number.
    METHOD 1
    One way is to use a number line. Let’s do this for \(-3-(-7)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-3+7\)
    Find \(-3\) on a number line. Since we are adding a positive number we move right, in the positive direction, by \(7\) units. We will reach \({4}\) on the number line, which is the answer.
    So \(-3-(-7)={4}\text{.}\) +
    Similarly, \(-8-(-1)={-7}\) and \(-5-(-5)={0}\text{.}\) +
    METHOD 2
    A second method asks you to think in context; for example when money is involved or the temperature changes. Let’s do this for \(-3-(-7)\text{.}\) First, we rewrite the subtraction as addition of the opposite number:
    \(-3+7\text{.}\)
    The first number is \(-3\text{.}\) Since it’s negative, it’s like you lost \(3\) dollars at the casino this morning.
    The second number is \(7\text{.}\) Since it’s positive, it’s like you won \(7\) dollars more in the casino later this afternoon.
    Since you won more than you lost, you have a net gain overall, implying the answer is positve.
    Since you lost \(3\) dollars and then won \(7\) dollars, it makes sense that you won the difference between \(7\) and \(3\) dollars, which is \({4}\) dollars. So the final answer is: \(-3-(-7)={4}\text{.}\) This is because we can rewrite the subtraction of a negative number in the previous expression as \(-3+(+7)={4}\text{.}\)
    @@ -1229,105 +1243,105 @@

    Exercise Group.

    Exercise Group.

    15.

    -
    -
    +
    +
    -
    +
    Perform the given addition and subtraction.
      -
    1. \(\displaystyle {-16-5+\left(-9\right)}\)
      2. -
    2. \(\displaystyle {3-\left(-18\right)+\left(-16\right)}\)
      3. +
    3. \(\displaystyle {-14-3+\left(-1\right)}\)
      4. +
    4. \(\displaystyle {8-\left(-12\right)+\left(-17\right)}\)
    -
    Answer 1.
    \(-30\)
    Answer 2.
    \(5\)
    Explanation.
    -
    a.
    \(\begin{aligned} -{-16-5+\left(-9\right)} \amp =-21+(-9)\\ -\amp =-21 - 9\\ -\amp =-30 -\end{aligned}\)
    b.
    \(\begin{aligned} -{3-\left(-18\right)+\left(-16\right)} \amp =3+18+(-16) \\ -\amp =21+(-16) \\ -\amp =21 - 16 \\ -\amp =5 +
    Answer 1.
    \(-18\)
    Answer 2.
    \(3\)
    Explanation.
    +
    a.
    \(\begin{aligned} +{-14-3+\left(-1\right)} \amp =-17+(-1)\\ +\amp =-17 - 1\\ +\amp =-18 +\end{aligned}\)
    b.
    \(\begin{aligned} +{8-\left(-12\right)+\left(-17\right)} \amp =8+12+(-17) \\ +\amp =20+(-17) \\ +\amp =20 - 17 \\ +\amp =3 \end{aligned}\)

    16.

    -
    -
    +
    +
    -
    +
    Perform the given addition and subtraction.
      -
    1. \(\displaystyle {-15-1+\left(-5\right)}\)
      2. -
    2. \(\displaystyle {10-\left(-18\right)+\left(-11\right)}\)
      3. +
    3. \(\displaystyle {-13-9+\left(-7\right)}\)
      4. +
    4. \(\displaystyle {5-\left(-12\right)+\left(-12\right)}\)
    -
    Answer 1.
    \(-21\)
    Answer 2.
    \(17\)
    Explanation.
    -
    a.
    \(\begin{aligned} -{-15-1+\left(-5\right)} \amp =-16+(-5)\\ -\amp =-16 - 5\\ -\amp =-21 -\end{aligned}\)
    b.
    \(\begin{aligned} -{10-\left(-18\right)+\left(-11\right)} \amp =10+18+(-11) \\ -\amp =28+(-11) \\ -\amp =28 - 11 \\ -\amp =17 +
    Answer 1.
    \(-29\)
    Answer 2.
    \(5\)
    Explanation.
    +
    a.
    \(\begin{aligned} +{-13-9+\left(-7\right)} \amp =-22+(-7)\\ +\amp =-22 - 7\\ +\amp =-29 +\end{aligned}\)
    b.
    \(\begin{aligned} +{5-\left(-12\right)+\left(-12\right)} \amp =5+12+(-12) \\ +\amp =17+(-12) \\ +\amp =17 - 12 \\ +\amp =5 \end{aligned}\)

    17.

    -
    -
    +
    +
    -
    +
    Perform the given addition and subtraction.
      -
    1. \(\displaystyle {-14-8+\left(-1\right)}\)
      2. -
    2. \(\displaystyle {7-\left(-18\right)+\left(-16\right)}\)
      3. +
    3. \(\displaystyle {-12-6+\left(-2\right)}\)
      4. +
    4. \(\displaystyle {2-\left(-12\right)+\left(-17\right)}\)
    -
    Answer 1.
    \(-23\)
    Answer 2.
    \(9\)
    Explanation.
    -
    a.
    \(\begin{aligned} -{-14-8+\left(-1\right)} \amp =-22+(-1)\\ -\amp =-22 - 1\\ -\amp =-23 -\end{aligned}\)
    b.
    \(\begin{aligned} -{7-\left(-18\right)+\left(-16\right)} \amp =7+18+(-16) \\ -\amp =25+(-16) \\ -\amp =25 - 16 \\ -\amp =9 +
    Answer 1.
    \(-20\)
    Answer 2.
    \(-3\)
    Explanation.
    +
    a.
    \(\begin{aligned} +{-12-6+\left(-2\right)} \amp =-18+(-2)\\ +\amp =-18 - 2\\ +\amp =-20 +\end{aligned}\)
    b.
    \(\begin{aligned} +{2-\left(-12\right)+\left(-17\right)} \amp =2+12+(-17) \\ +\amp =14+(-17) \\ +\amp =14 - 17 \\ +\amp =-3 \end{aligned}\)

    18.

    -
    -
    +
    +
    -
    +
    Perform the given addition and subtraction.
      -
    1. \(\displaystyle {-13-4+\left(-7\right)}\)
      2. -
    2. \(\displaystyle {5-\left(-18\right)+\left(-11\right)}\)
      3. +
    3. \(\displaystyle {-10-2+\left(-8\right)}\)
      4. +
    4. \(\displaystyle {9-\left(-13\right)+\left(-12\right)}\)
    -
    Answer 1.
    \(-24\)
    Answer 2.
    \(12\)
    Explanation.
    -
    a.
    \(\begin{aligned} -{-13-4+\left(-7\right)} \amp =-17+(-7)\\ -\amp =-17 - 7\\ -\amp =-24 -\end{aligned}\)
    b.
    \(\begin{aligned} -{5-\left(-18\right)+\left(-11\right)} \amp =5+18+(-11) \\ -\amp =23+(-11) \\ -\amp =23 - 11 \\ -\amp =12 +
    Answer 1.
    \(-20\)
    Answer 2.
    \(10\)
    Explanation.
    +
    a.
    \(\begin{aligned} +{-10-2+\left(-8\right)} \amp =-12+(-8)\\ +\amp =-12 - 8\\ +\amp =-20 +\end{aligned}\)
    b.
    \(\begin{aligned} +{9-\left(-13\right)+\left(-12\right)} \amp =9+13+(-12) \\ +\amp =22+(-12) \\ +\amp =22 - 12 \\ +\amp =10 \end{aligned}\)
    @@ -1339,153 +1353,153 @@

    Exercise Group.

    Exercise Group.

    19.

    -
    -
    +
    +
    -
    +
    Multiply the following.
      -
    1. \(\displaystyle (-8)\cdot(-3)\)
      2. -
    2. \(\displaystyle (-4)\cdot2\)
      3. -
    3. \(\displaystyle 5\cdot(-5)\)
      4. -
    4. \(\displaystyle (-5)\cdot0\)
      5. +
    5. \(\displaystyle (-10)\cdot(-1)\)
      6. +
    6. \(\displaystyle (-5)\cdot5\)
      7. +
    7. \(\displaystyle 8\cdot(-5)\)
      8. +
    8. \(\displaystyle (-2)\cdot0\)
    -
    Answer 1.
    \(24\)
    Answer 2.
    \(-8\)
    Answer 3.
    \(-25\)
    Answer 4.
    \(0\)
    Explanation.
    -
    The rules for multiplying positive and negative numbers are:
      -
    • \(\displaystyle{ \text{positive} \cdot \text{positive} = \text{positive} }\text{,}\)
      • -
    • \(\displaystyle{ \text{positive} \cdot \text{negative} = \text{negative} }\text{,}\)
      • -
    • \(\displaystyle{ \text{negative} \cdot \text{positive} = \text{negative} }\text{,}\)
      • -
    • \(\displaystyle{ \text{negative} \cdot \text{negative} = \text{positive} }\text{.}\)
      • -
    The solutions are:
      -
    1. \(\displaystyle \displaystyle{ (-8)\cdot(-3)={24}, }\)
      2. -
    2. \(\displaystyle \displaystyle{ (-4)\cdot2={-8}, }\)
      3. -
    3. \(\displaystyle \displaystyle{ 5\cdot(-5)={-25}, }\)
      4. -
    4. \(\displaystyle \displaystyle{ (-5)\cdot0={0}. }\)
      5. +
      Answer 1.
      \(10\)
      Answer 2.
      \(-25\)
      Answer 3.
      \(-40\)
      Answer 4.
      \(0\)
      Explanation.
      +
      The rules for multiplying positive and negative numbers are:
        +
      • \(\displaystyle{ \text{positive} \cdot \text{positive} = \text{positive} }\text{,}\)
        • +
      • \(\displaystyle{ \text{positive} \cdot \text{negative} = \text{negative} }\text{,}\)
        • +
      • \(\displaystyle{ \text{negative} \cdot \text{positive} = \text{negative} }\text{,}\)
        • +
      • \(\displaystyle{ \text{negative} \cdot \text{negative} = \text{positive} }\text{.}\)
        • +
      The solutions are:
        +
      1. \(\displaystyle \displaystyle{ (-10)\cdot(-1)={10}, }\)
        2. +
      2. \(\displaystyle \displaystyle{ (-5)\cdot5={-25}, }\)
        3. +
      3. \(\displaystyle \displaystyle{ 8\cdot(-5)={-40}, }\)
        4. +
      4. \(\displaystyle \displaystyle{ (-2)\cdot0={0}. }\)

    20.

    -
    -
    +
    +
    -
    +
    Multiply the following.
      -
    1. \(\displaystyle (-8)\cdot(-1)\)
      2. -
    2. \(\displaystyle (-7)\cdot7\)
      3. -
    3. \(\displaystyle 5\cdot(-1)\)
      4. -
    4. \(\displaystyle (-4)\cdot0\)
      5. +
    5. \(\displaystyle (-10)\cdot(-2)\)
      6. +
    6. \(\displaystyle (-7)\cdot4\)
      7. +
    7. \(\displaystyle 8\cdot(-2)\)
      8. +
    8. \(\displaystyle (-10)\cdot0\)
    -
    Answer 1.
    \(8\)
    Answer 2.
    \(-49\)
    Answer 3.
    \(-5\)
    Answer 4.
    \(0\)
    Explanation.
    -
    The rules for multiplying positive and negative numbers are:
      -
    • \(\displaystyle{ \text{positive} \cdot \text{positive} = \text{positive} }\text{,}\)
      • -
    • \(\displaystyle{ \text{positive} \cdot \text{negative} = \text{negative} }\text{,}\)
      • -
    • \(\displaystyle{ \text{negative} \cdot \text{positive} = \text{negative} }\text{,}\)
      • -
    • \(\displaystyle{ \text{negative} \cdot \text{negative} = \text{positive} }\text{.}\)
      • -
    The solutions are:
      -
    1. \(\displaystyle \displaystyle{ (-8)\cdot(-1)={8}, }\)
      2. -
    2. \(\displaystyle \displaystyle{ (-7)\cdot7={-49}, }\)
      3. -
    3. \(\displaystyle \displaystyle{ 5\cdot(-1)={-5}, }\)
      4. -
    4. \(\displaystyle \displaystyle{ (-4)\cdot0={0}. }\)
      5. +
      Answer 1.
      \(20\)
      Answer 2.
      \(-28\)
      Answer 3.
      \(-16\)
      Answer 4.
      \(0\)
      Explanation.
      +
      The rules for multiplying positive and negative numbers are:
        +
      • \(\displaystyle{ \text{positive} \cdot \text{positive} = \text{positive} }\text{,}\)
        • +
      • \(\displaystyle{ \text{positive} \cdot \text{negative} = \text{negative} }\text{,}\)
        • +
      • \(\displaystyle{ \text{negative} \cdot \text{positive} = \text{negative} }\text{,}\)
        • +
      • \(\displaystyle{ \text{negative} \cdot \text{negative} = \text{positive} }\text{.}\)
        • +
      The solutions are:
        +
      1. \(\displaystyle \displaystyle{ (-10)\cdot(-2)={20}, }\)
        2. +
      2. \(\displaystyle \displaystyle{ (-7)\cdot4={-28}, }\)
        3. +
      3. \(\displaystyle \displaystyle{ 8\cdot(-2)={-16}, }\)
        4. +
      4. \(\displaystyle \displaystyle{ (-10)\cdot0={0}. }\)

    21.

    -
    -
    +
    +
    -
    +
    Multiply the following.
      -
    1. \(\displaystyle (-3)\cdot(-6)\cdot(-3)\)
      2. -
    2. \(\displaystyle 5\cdot(-9)\cdot(-4)\)
      3. -
    3. \(\displaystyle (-84)\cdot(-67)\cdot0\)
      4. +
    4. \(\displaystyle (-3)\cdot(-6)\cdot(-4)\)
      5. +
    5. \(\displaystyle 2\cdot(-7)\cdot(-4)\)
      6. +
    6. \(\displaystyle (-97)\cdot(-54)\cdot0\)
    -
    Answer 1.
    \(-54\)
    Answer 2.
    \(180\)
    Answer 3.
    \(0\)
    Explanation.
    -
    a.
    \(\begin{aligned}[t] -(-3)\cdot(-6)\cdot(-3) \amp = 18\cdot(-3) \\ -\amp = {-54} -\end{aligned}\)
    b.
    \(\begin{aligned}[t] -5\cdot(-9)\cdot(-4) \amp = -45\cdot(-4) \\ -\amp = {180} -\end{aligned}\)
    c.
    \(-84\cdot(-67)\cdot0 =0\)
    Note that in the last case there is no need to waste time multiplying the first two numbers because any number multiplied by \(0\) equals \(0\text{.}\) +
    Answer 1.
    \(-72\)
    Answer 2.
    \(56\)
    Answer 3.
    \(0\)
    Explanation.
    +
    a.
    \(\begin{aligned}[t] +(-3)\cdot(-6)\cdot(-4) \amp = 18\cdot(-4) \\ +\amp = {-72} +\end{aligned}\)
    b.
    \(\begin{aligned}[t] +2\cdot(-7)\cdot(-4) \amp = -14\cdot(-4) \\ +\amp = {56} +\end{aligned}\)
    c.
    \(-97\cdot(-54)\cdot0 =0\)
    Note that in the last case there is no need to waste time multiplying the first two numbers because any number multiplied by \(0\) equals \(0\text{.}\)

    22.

    -
    -
    +
    +
    -
    +
    Multiply the following.
      -
    1. \(\displaystyle (-3)\cdot(-4)\cdot(-5)\)
      2. -
    2. \(\displaystyle 3\cdot(-9)\cdot(-1)\)
      3. -
    3. \(\displaystyle (-82)\cdot(-55)\cdot0\)
      4. +
    4. \(\displaystyle (-2)\cdot(-4)\cdot(-2)\)
      5. +
    5. \(\displaystyle 6\cdot(-7)\cdot(-2)\)
      6. +
    6. \(\displaystyle (-95)\cdot(-72)\cdot0\)
    -
    Answer 1.
    \(-60\)
    Answer 2.
    \(27\)
    Answer 3.
    \(0\)
    Explanation.
    -
    a.
    \(\begin{aligned}[t] -(-3)\cdot(-4)\cdot(-5) \amp = 12\cdot(-5) \\ -\amp = {-60} -\end{aligned}\)
    b.
    \(\begin{aligned}[t] -3\cdot(-9)\cdot(-1) \amp = -27\cdot(-1) \\ -\amp = {27} -\end{aligned}\)
    c.
    \(-82\cdot(-55)\cdot0 =0\)
    Note that in the last case there is no need to waste time multiplying the first two numbers because any number multiplied by \(0\) equals \(0\text{.}\) +
    Answer 1.
    \(-16\)
    Answer 2.
    \(84\)
    Answer 3.
    \(0\)
    Explanation.
    +
    a.
    \(\begin{aligned}[t] +(-2)\cdot(-4)\cdot(-2) \amp = 8\cdot(-2) \\ +\amp = {-16} +\end{aligned}\)
    b.
    \(\begin{aligned}[t] +6\cdot(-7)\cdot(-2) \amp = -42\cdot(-2) \\ +\amp = {84} +\end{aligned}\)
    c.
    \(-95\cdot(-72)\cdot0 =0\)
    Note that in the last case there is no need to waste time multiplying the first two numbers because any number multiplied by \(0\) equals \(0\text{.}\)

    23.

    -
    -
    +
    +
    -
    Multiply the following.
      -
    1. \(\displaystyle \displaystyle{ (-1)(-2)(-2)(-1) }\)
      2. -
    2. \(\displaystyle \displaystyle{ (-2)(-3)(-1)(-2) }\)
      3. +
      Multiply the following.
        +
      1. \(\displaystyle \displaystyle{ (-2)(-2)(-1)(-2) }\)
        2. +
      2. \(\displaystyle \displaystyle{ (1)(-2)(3)(-2) }\)
      -
      Answer 1.
      \(4\)
      Answer 2.
      \(12\)
      Explanation.
      -
      In multiplication, each pair of negative signs cancel. If all negative signs are canceled, the product is positive; if there is one negative sign left, the product is negative.
      a.
      \(\begin{aligned}[t] -\amp \phantom{{}=}(-1)(-2)(-2)(-1) \\ -\amp = (1)(2)(2)(1) \\ -\amp = {4} -\end{aligned}\)
      b.
      \(\begin{aligned}[t] -\amp \phantom{{}=}(-2)(-3)(-1)(-2) \\ -\amp = (2)(3)(1)(2) \\ +
      Answer 1.
      \(8\)
      Answer 2.
      \(12\)
      Explanation.
      +
      In multiplication, each pair of negative signs cancel. If all negative signs are canceled, the product is positive; if there is one negative sign left, the product is negative.
      a.
      \(\begin{aligned}[t] +\amp \phantom{{}=}(-2)(-2)(-1)(-2) \\ +\amp = (2)(2)(1)(2) \\ +\amp = {8} +\end{aligned}\)
      b.
      \(\begin{aligned}[t] +\amp \phantom{{}=}(1)(-2)(3)(-2) \\ +\amp = (1)(2)(3)(2) \\ \amp = {12} \end{aligned}\)

    24.

    -
    -
    +
    +
    -
    Multiply the following.
      -
    1. \(\displaystyle \displaystyle{ (-2)(-1)(-3)(-3) }\)
      2. -
    2. \(\displaystyle \displaystyle{ (-3)(-1)(2)(-3) }\)
      3. +
      Multiply the following.
        +
      1. \(\displaystyle \displaystyle{ (-2)(-1)(-3)(-1) }\)
        2. +
      2. \(\displaystyle \displaystyle{ (3)(-2)(-1)(-1) }\)
      -
      Answer 1.
      \(18\)
      Answer 2.
      \(-18\)
      Explanation.
      -
      In multiplication, each pair of negative signs cancel. If all negative signs are canceled, the product is positive; if there is one negative sign left, the product is negative.
      a.
      \(\begin{aligned}[t] -\amp \phantom{{}=}(-2)(-1)(-3)(-3) \\ -\amp = (2)(1)(3)(3) \\ -\amp = {18} -\end{aligned}\)
      b.
      \(\begin{aligned}[t] -\amp \phantom{{}=}(-3)(-1)(2)(-3) \\ -\amp = -(3)(1)(2)(3) \\ -\amp = {-18} +
      Answer 1.
      \(6\)
      Answer 2.
      \(-6\)
      Explanation.
      +
      In multiplication, each pair of negative signs cancel. If all negative signs are canceled, the product is positive; if there is one negative sign left, the product is negative.
      a.
      \(\begin{aligned}[t] +\amp \phantom{{}=}(-2)(-1)(-3)(-1) \\ +\amp = (2)(1)(3)(1) \\ +\amp = {6} +\end{aligned}\)
      b.
      \(\begin{aligned}[t] +\amp \phantom{{}=}(3)(-2)(-1)(-1) \\ +\amp = -(3)(2)(1)(1) \\ +\amp = {-6} \end{aligned}\)
      @@ -1497,144 +1511,144 @@

      Exercise Group.

      Exercise Group.

      25.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle \frac{-60}{-6}\)
        2. -
      2. \(\displaystyle \frac{36}{-6}\)
        3. -
      3. \(\displaystyle \frac{-25}{5}\)
        4. +
      4. \(\displaystyle \frac{-32}{-4}\)
        5. +
      5. \(\displaystyle \frac{70}{-7}\)
        6. +
      6. \(\displaystyle \frac{-32}{8}\)
      -
      Answer 1.
      \(10\)
      Answer 2.
      \(-6\)
      Answer 3.
      \(-5\)
      Explanation.
      -
      The rules for dividing positive numbers are the same as those for multiplication:
      \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
      \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
      The solutions are:
        -
      1. \(\displaystyle \displaystyle{ \frac{-60}{-6}={10}, }\)
        2. -
      2. \(\displaystyle \displaystyle{ \frac{36}{-6}={-6}, }\)
        3. -
      3. \(\displaystyle \displaystyle{ \frac{-25}{5}={-5}. }\)
        4. +
        Answer 1.
        \(8\)
        Answer 2.
        \(-10\)
        Answer 3.
        \(-4\)
        Explanation.
        +
        The rules for dividing positive numbers are the same as those for multiplication:
        \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
        \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
        The solutions are:
          +
        1. \(\displaystyle \displaystyle{ \frac{-32}{-4}={8}, }\)
          2. +
        2. \(\displaystyle \displaystyle{ \frac{70}{-7}={-10}, }\)
          3. +
        3. \(\displaystyle \displaystyle{ \frac{-32}{8}={-4}. }\)

      26.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle \frac{-35}{-5}\)
        2. -
      2. \(\displaystyle \frac{15}{-5}\)
        3. -
      3. \(\displaystyle \frac{-45}{5}\)
        4. +
      4. \(\displaystyle \frac{-15}{-3}\)
        5. +
      5. \(\displaystyle \frac{40}{-5}\)
        6. +
      6. \(\displaystyle \frac{-64}{8}\)
      -
      Answer 1.
      \(7\)
      Answer 2.
      \(-3\)
      Answer 3.
      \(-9\)
      Explanation.
      -
      The rules for dividing positive numbers are the same as those for multiplication:
      \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
      \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
      The solutions are:
        -
      1. \(\displaystyle \displaystyle{ \frac{-35}{-5}={7}, }\)
        2. -
      2. \(\displaystyle \displaystyle{ \frac{15}{-5}={-3}, }\)
        3. -
      3. \(\displaystyle \displaystyle{ \frac{-45}{5}={-9}. }\)
        4. +
        Answer 1.
        \(5\)
        Answer 2.
        \(-8\)
        Answer 3.
        \(-8\)
        Explanation.
        +
        The rules for dividing positive numbers are the same as those for multiplication:
        \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
        \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
        The solutions are:
          +
        1. \(\displaystyle \displaystyle{ \frac{-15}{-3}={5}, }\)
          2. +
        2. \(\displaystyle \displaystyle{ \frac{40}{-5}={-8}, }\)
          3. +
        3. \(\displaystyle \displaystyle{ \frac{-64}{8}={-8}. }\)

      27.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle \frac{-4}{-1}\)
        2. -
      2. \(\displaystyle \frac{5}{-1}\)
        3. -
      3. \(\displaystyle \frac{110}{-110}\)
        4. -
      4. \(\displaystyle \frac{-10}{-10}\)
        5. -
      5. \(\displaystyle \frac{3}{0}\)
        6. -
      6. \(\displaystyle \frac{0}{-7}\)
        7. +
      7. \(\displaystyle \frac{-2}{-1}\)
        8. +
      8. \(\displaystyle \frac{4}{-1}\)
        9. +
      9. \(\displaystyle \frac{200}{-200}\)
        10. +
      10. \(\displaystyle \frac{-16}{-16}\)
        11. +
      11. \(\displaystyle \frac{13}{0}\)
        12. +
      12. \(\displaystyle \frac{0}{-8}\)
      -
      Answer 1.
      \(4\)
      Answer 2.
      \(-5\)
      Answer 3.
      \(-1\)
      Answer 4.
      \(1\)
      Answer 5.
      \(\text{undefined}\)
      Answer 6.
      \(0\)
      Explanation.
      -
      The rules for determining the sign of the result when dividing positive and negative numbers are the same as those for multiplication:
      \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
      \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
      In addition, \(0\) divided by any number (except \(0\)) is \(0\text{;}\) any number divided by \(0\) is undefined.
      The solutions are:
        -
      1. \(\displaystyle \displaystyle{ \frac{-4}{-1}={4}, }\)
        2. -
      2. \(\displaystyle \displaystyle{ \frac{5}{-1}={-5}, }\)
        3. -
      3. \(\displaystyle \displaystyle{ \frac{110}{-110}={-1}, }\)
        4. -
      4. \(\displaystyle \displaystyle{ \frac{-10}{-10}={1}, }\)
        5. -
      5. -\(\displaystyle{ \frac{3}{0} }\) is undefined
        6. -
      6. \(\displaystyle \displaystyle{ \frac{0}{-7}={0}. }\)
        7. +
        Answer 1.
        \(2\)
        Answer 2.
        \(-4\)
        Answer 3.
        \(-1\)
        Answer 4.
        \(1\)
        Answer 5.
        \(\text{undefined}\)
        Answer 6.
        \(0\)
        Explanation.
        +
        The rules for determining the sign of the result when dividing positive and negative numbers are the same as those for multiplication:
        \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
        \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
        In addition, \(0\) divided by any number (except \(0\)) is \(0\text{;}\) any number divided by \(0\) is undefined.
        The solutions are:
          +
        1. \(\displaystyle \displaystyle{ \frac{-2}{-1}={2}, }\)
          2. +
        2. \(\displaystyle \displaystyle{ \frac{4}{-1}={-4}, }\)
          3. +
        3. \(\displaystyle \displaystyle{ \frac{200}{-200}={-1}, }\)
          4. +
        4. \(\displaystyle \displaystyle{ \frac{-16}{-16}={1}, }\)
          5. +
        5. +\(\displaystyle{ \frac{13}{0} }\) is undefined
          6. +
        6. \(\displaystyle \displaystyle{ \frac{0}{-8}={0}. }\)

      28.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle \frac{-3}{-1}\)
        2. -
      2. \(\displaystyle \frac{10}{-1}\)
        3. -
      3. \(\displaystyle \frac{160}{-160}\)
        4. -
      4. \(\displaystyle \frac{-13}{-13}\)
        5. -
      5. \(\displaystyle \frac{3}{0}\)
        6. -
      6. \(\displaystyle \frac{0}{-3}\)
        7. +
      7. \(\displaystyle \frac{-10}{-1}\)
        8. +
      8. \(\displaystyle \frac{8}{-1}\)
        9. +
      9. \(\displaystyle \frac{140}{-140}\)
        10. +
      10. \(\displaystyle \frac{-19}{-19}\)
        11. +
      11. \(\displaystyle \frac{13}{0}\)
        12. +
      12. \(\displaystyle \frac{0}{-4}\)
      -
      Answer 1.
      \(3\)
      Answer 2.
      \(-10\)
      Answer 3.
      \(-1\)
      Answer 4.
      \(1\)
      Answer 5.
      \(\text{undefined}\)
      Answer 6.
      \(0\)
      Explanation.
      -
      The rules for determining the sign of the result when dividing positive and negative numbers are the same as those for multiplication:
      \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
      \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
      \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
      In addition, \(0\) divided by any number (except \(0\)) is \(0\text{;}\) any number divided by \(0\) is undefined.
      The solutions are:
        -
      1. \(\displaystyle \displaystyle{ \frac{-3}{-1}={3}, }\)
        2. -
      2. \(\displaystyle \displaystyle{ \frac{10}{-1}={-10}, }\)
        3. -
      3. \(\displaystyle \displaystyle{ \frac{160}{-160}={-1}, }\)
        4. -
      4. \(\displaystyle \displaystyle{ \frac{-13}{-13}={1}, }\)
        5. -
      5. -\(\displaystyle{ \frac{3}{0} }\) is undefined
        6. -
      6. \(\displaystyle \displaystyle{ \frac{0}{-3}={0}. }\)
        7. +
        Answer 1.
        \(10\)
        Answer 2.
        \(-8\)
        Answer 3.
        \(-1\)
        Answer 4.
        \(1\)
        Answer 5.
        \(\text{undefined}\)
        Answer 6.
        \(0\)
        Explanation.
        +
        The rules for determining the sign of the result when dividing positive and negative numbers are the same as those for multiplication:
        \(\displaystyle{ \text{positive} \div \text{positive} = \text{positive} }\text{,}\)
        \(\displaystyle{ \text{positive} \div \text{negative} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{positive} = \text{negative} }\text{,}\)
        \(\displaystyle{ \text{negative} \div \text{negative} = \text{positive} }\text{.}\)
        In addition, \(0\) divided by any number (except \(0\)) is \(0\text{;}\) any number divided by \(0\) is undefined.
        The solutions are:
          +
        1. \(\displaystyle \displaystyle{ \frac{-10}{-1}={10}, }\)
          2. +
        2. \(\displaystyle \displaystyle{ \frac{8}{-1}={-8}, }\)
          3. +
        3. \(\displaystyle \displaystyle{ \frac{140}{-140}={-1}, }\)
          4. +
        4. \(\displaystyle \displaystyle{ \frac{-19}{-19}={1}, }\)
          5. +
        5. +\(\displaystyle{ \frac{13}{0} }\) is undefined
          6. +
        6. \(\displaystyle \displaystyle{ \frac{0}{-4}={0}. }\)

      29.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle (-9)^{2}\)
        2. -
      2. \(\displaystyle -4^{2}\)
        3. +
      3. \(\displaystyle (-9)^{2}\)
        4. +
      4. \(\displaystyle -4^{2}\)
      -
      Answer 1.
      \(81\)
      Answer 2.
      \(-16\)
      Explanation.
      -
      Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
      The solutions are:
        -
      1. -
        \(\begin{aligned}[t] +
        Answer 1.
        \(81\)
        Answer 2.
        \(-16\)
        Explanation.
        +
        Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
        The solutions are:
          +
        1. +
          \(\begin{aligned}[t] (-9)^{2}\amp =(-9)\cdot(-9)\\ \amp ={81} \end{aligned}\)
          -
          Note that we cannot simply evaluate \(9^{2}\text{.}\) Instead, we have to apply the square to the whole quantity inside the parentheses- the result is a positive answer.
          +
          Note that we cannot simply evaluate \(9^{2}\text{.}\) Instead, we have to apply the square to the whole quantity inside the parentheses- the result is a positive answer.
          2. -
        2. -
          \(\begin{aligned}[t] +
        3. +
          \(\begin{aligned}[t] -4^{2}\amp =-1\cdot4^2\\ \amp =-1\cdot4\cdot4\\ \amp = -1\cdot 16\\ \amp ={-16} \end{aligned}\)
          -
          Note that the negative sign is like “negative one multiplied by....”. For example:
          -
            -
          • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
            • -
          • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
            • -
          • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            • +
            Note that the negative sign is like “negative one multiplied by....”. For example:
            +
              +
            • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
              • +
            • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
              • +
            • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            -
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\) +
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\)
            -
            So, we can change \(-4^{2}\) into \(-1\cdot4^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(4^2\) before multiplying by \(-1\text{.}\) +
            So, we can change \(-4^{2}\) into \(-1\cdot4^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(4^2\) before multiplying by \(-1\text{.}\)
        @@ -1642,42 +1656,42 @@

        Exercise Group.

      30.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle (-7)^{2}\)
        2. -
      2. \(\displaystyle -6^{2}\)
        3. +
      3. \(\displaystyle (-7)^{2}\)
        4. +
      4. \(\displaystyle -6^{2}\)
      -
      Answer 1.
      \(49\)
      Answer 2.
      \(-36\)
      Explanation.
      -
      Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
      The solutions are:
        -
      1. -
        \(\begin{aligned}[t] +
        Answer 1.
        \(49\)
        Answer 2.
        \(-36\)
        Explanation.
        +
        Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
        The solutions are:
          +
        1. +
          \(\begin{aligned}[t] (-7)^{2}\amp =(-7)\cdot(-7)\\ \amp ={49} \end{aligned}\)
          -
          Note that we cannot simply evaluate \(7^{2}\text{.}\) Instead, we have to apply the square to the whole quantity inside the parentheses- the result is a positive answer.
          +
          Note that we cannot simply evaluate \(7^{2}\text{.}\) Instead, we have to apply the square to the whole quantity inside the parentheses- the result is a positive answer.
          2. -
        2. -
          \(\begin{aligned}[t] +
        3. +
          \(\begin{aligned}[t] -6^{2}\amp =-1\cdot6^2\\ \amp =-1\cdot6\cdot6\\ \amp = -1\cdot 36\\ \amp ={-36} \end{aligned}\)
          -
          Note that the negative sign is like “negative one multiplied by....”. For example:
          -
            -
          • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
            • -
          • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
            • -
          • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            • +
            Note that the negative sign is like “negative one multiplied by....”. For example:
            +
              +
            • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
              • +
            • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
              • +
            • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            -
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\) +
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\)
            -
            So, we can change \(-6^{2}\) into \(-1\cdot6^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(6^2\) before multiplying by \(-1\text{.}\) +
            So, we can change \(-6^{2}\) into \(-1\cdot6^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(6^2\) before multiplying by \(-1\text{.}\)
        @@ -1685,42 +1699,42 @@

        Exercise Group.

      31.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle (-3)^{3}\)
        2. -
      2. \(\displaystyle -1^{3}\)
        3. +
      3. \(\displaystyle (-3)^{3}\)
        4. +
      4. \(\displaystyle -1^{3}\)
      -
      Answer 1.
      \(-27\)
      Answer 2.
      \(-1\)
      Explanation.
      -
      Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
      The quantities evaluate to:
        -
      1. -
        \(\begin{aligned}[t] +
        Answer 1.
        \(-27\)
        Answer 2.
        \(-1\)
        Explanation.
        +
        Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
        The quantities evaluate to:
          +
        1. +
          \(\begin{aligned}[t] (-3)^{3}\amp =(-3)\cdot(-3)\cdot(-3)\\ \amp ={-27} \end{aligned}\)
          -
          When we multiply three negative numbers, the result is negative. Two negative signs cancelled, but one is left, making the result negative.
          +
          When we multiply three negative numbers, the result is negative. Two negative signs cancelled, but one is left, making the result negative.
          2. -
        2. -
          \(\begin{aligned}[t] +
        3. +
          \(\begin{aligned}[t] -1^{3}\amp =-1\cdot1^3\\ \amp =-1\cdot1\cdot1\cdot1\\ \amp = -1\cdot 1\\ \amp ={-1} \end{aligned}\)
          -
          Note that the negative sign is like “negative one multiplied by....”. For example:
          -
            -
          • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
            • -
          • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
            • -
          • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            • +
            Note that the negative sign is like “negative one multiplied by....”. For example:
            +
              +
            • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
              • +
            • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
              • +
            • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            -
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\) +
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\)
            -
            So, we can change \(-1^{3}\) into \(-1\cdot1^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(1^2\) before multiplying by \(-1\text{.}\) +
            So, we can change \(-1^{3}\) into \(-1\cdot1^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(1^2\) before multiplying by \(-1\text{.}\)
        @@ -1728,42 +1742,42 @@

        Exercise Group.

      32.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle (-2)^{3}\)
        2. -
      2. \(\displaystyle -4^{3}\)
        3. +
      3. \(\displaystyle (-2)^{3}\)
        4. +
      4. \(\displaystyle -4^{3}\)
      -
      Answer 1.
      \(-8\)
      Answer 2.
      \(-64\)
      Explanation.
      -
      Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
      The quantities evaluate to:
        -
      1. -
        \(\begin{aligned}[t] +
        Answer 1.
        \(-8\)
        Answer 2.
        \(-64\)
        Explanation.
        +
        Remember to use order of operations here: parentheses take priority over exponents, and exponents take priority over multiplication.
        The quantities evaluate to:
          +
        1. +
          \(\begin{aligned}[t] (-2)^{3}\amp =(-2)\cdot(-2)\cdot(-2)\\ \amp ={-8} \end{aligned}\)
          -
          When we multiply three negative numbers, the result is negative. Two negative signs cancelled, but one is left, making the result negative.
          +
          When we multiply three negative numbers, the result is negative. Two negative signs cancelled, but one is left, making the result negative.
          2. -
        2. -
          \(\begin{aligned}[t] +
        3. +
          \(\begin{aligned}[t] -4^{3}\amp =-1\cdot4^3\\ \amp =-1\cdot4\cdot4\cdot4\\ \amp = -1\cdot 64\\ \amp ={-64} \end{aligned}\)
          -
          Note that the negative sign is like “negative one multiplied by....”. For example:
          -
            -
          • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
            • -
          • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
            • -
          • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            • +
            Note that the negative sign is like “negative one multiplied by....”. For example:
            +
              +
            • \(\displaystyle \displaystyle{-2=-1\cdot2}\)
              • +
            • \(\displaystyle \displaystyle{-3=-1\cdot3}\)
              • +
            • \(\displaystyle \displaystyle{-4=-1\cdot4}\)
            -
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\) +
            You can see the pattern. This demonstrates that the negative sign can be viewed as multiplication by \(-1\text{.}\)
            -
            So, we can change \(-4^{3}\) into \(-1\cdot4^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(4^2\) before multiplying by \(-1\text{.}\) +
            So, we can change \(-4^{3}\) into \(-1\cdot4^{2}\text{.}\) Since exponents take priority over multiplication, we evaluate \(4^2\) before multiplying by \(-1\text{.}\)
        @@ -1771,34 +1785,34 @@

        Exercise Group.

      33.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle 4^{2}\)
        2. -
      2. \(\displaystyle 2^{3}\)
        3. -
      3. \(\displaystyle (-4)^{2}\)
        4. -
      4. \(\displaystyle (-3)^{3}\)
        5. +
      5. \(\displaystyle 4^{2}\)
        6. +
      6. \(\displaystyle 2^{3}\)
        7. +
      7. \(\displaystyle (-4)^{2}\)
        8. +
      8. \(\displaystyle (-3)^{3}\)
      -
      Answer 1.
      \(16\)
      Answer 2.
      \(8\)
      Answer 3.
      \(16\)
      Answer 4.
      \(-27\)
      Explanation.
      -
      The solutions are:
        -
      1. \(\displaystyle \begin{aligned}[t] +
        Answer 1.
        \(16\)
        Answer 2.
        \(8\)
        Answer 3.
        \(16\)
        Answer 4.
        \(-27\)
        Explanation.
        +
        The solutions are:
          +
        1. \(\displaystyle \begin{aligned}[t] 4^{2}\amp =4\cdot4\\ \amp ={16} \end{aligned}\)
          2. -
        2. \(\displaystyle \begin{aligned}[t] +
        3. \(\displaystyle \begin{aligned}[t] 2^{3}\amp =2\cdot2\cdot2\\ \amp ={8} \end{aligned}\)
          4. -
        4. \(\displaystyle \begin{aligned}[t] +
        5. \(\displaystyle \begin{aligned}[t] (-4)^{2}\amp =(-4)\cdot(-4)\\ \amp ={16} \end{aligned}\)
          6. -
        6. \(\displaystyle \begin{aligned}[t] +
        7. \(\displaystyle \begin{aligned}[t] (-3)^{3}\amp =(-3)\cdot(-3)\cdot(-3)\\ \amp ={-27} \end{aligned}\)
          8. @@ -1807,34 +1821,34 @@

          Exercise Group.

      34.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle 5^{2}\)
        2. -
      2. \(\displaystyle 4^{3}\)
        3. -
      3. \(\displaystyle (-3)^{2}\)
        4. -
      4. \(\displaystyle (-5)^{3}\)
        5. +
      5. \(\displaystyle 5^{2}\)
        6. +
      6. \(\displaystyle 4^{3}\)
        7. +
      7. \(\displaystyle (-3)^{2}\)
        8. +
      8. \(\displaystyle (-5)^{3}\)
      -
      Answer 1.
      \(25\)
      Answer 2.
      \(64\)
      Answer 3.
      \(9\)
      Answer 4.
      \(-125\)
      Explanation.
      -
      The solutions are:
        -
      1. \(\displaystyle \begin{aligned}[t] +
        Answer 1.
        \(25\)
        Answer 2.
        \(64\)
        Answer 3.
        \(9\)
        Answer 4.
        \(-125\)
        Explanation.
        +
        The solutions are:
          +
        1. \(\displaystyle \begin{aligned}[t] 5^{2}\amp =5\cdot5\\ \amp ={25} \end{aligned}\)
          2. -
        2. \(\displaystyle \begin{aligned}[t] +
        3. \(\displaystyle \begin{aligned}[t] 4^{3}\amp =4\cdot4\cdot4\\ \amp ={64} \end{aligned}\)
          4. -
        4. \(\displaystyle \begin{aligned}[t] +
        5. \(\displaystyle \begin{aligned}[t] (-3)^{2}\amp =(-3)\cdot(-3)\\ \amp ={9} \end{aligned}\)
          6. -
        6. \(\displaystyle \begin{aligned}[t] +
        7. \(\displaystyle \begin{aligned}[t] (-5)^{3}\amp =(-5)\cdot(-5)\cdot(-5)\\ \amp ={-125} \end{aligned}\)
          8. @@ -1843,53 +1857,53 @@

          Exercise Group.

      35.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle 1^{10}\)
        2. -
      2. \(\displaystyle (-1)^{11}\)
        3. -
      3. \(\displaystyle (-1)^{12}\)
        4. -
      4. \(\displaystyle 0^{20}\)
        5. +
      5. \(\displaystyle 1^{10}\)
        6. +
      6. \(\displaystyle (-1)^{11}\)
        7. +
      7. \(\displaystyle (-1)^{12}\)
        8. +
      8. \(\displaystyle 0^{20}\)
      -
      Answer 1.
      \(1\)
      Answer 2.
      \(-1\)
      Answer 3.
      \(1\)
      Answer 4.
      \(0\)
      Explanation.
      -
      The solutions are:
        -
      1. \(\displaystyle \displaystyle{ 1^{10}=1, }\)
        2. -
      2. \(\displaystyle \displaystyle{ (-1)^{11}=-1, }\)
        3. -
      3. \(\displaystyle \displaystyle{ (-1)^{12}=1, }\)
        4. -
      4. \(\displaystyle \displaystyle{ 0^{20}=0. }\)
        5. -
      When two negative numbers are multiplied together they make a positive number; each pair of negative signs “cancel each other out”.
      When an odd number of negative numbers are multiplied together, each pair of negative signs “cancel out” but there is still one negative sign left. That’s why \((-1)^{\text{odd number}}=-1\text{.}\) -
      Similarly, when an even number of negative numbers are multiplied together each pair of negative signs “cancel out” and there are no negative signs left. That’s why \((-1)^{\text{even number}}=1\text{.}\) +
      Answer 1.
      \(1\)
      Answer 2.
      \(-1\)
      Answer 3.
      \(1\)
      Answer 4.
      \(0\)
      Explanation.
      +
      The solutions are:
        +
      1. \(\displaystyle \displaystyle{ 1^{10}=1, }\)
        2. +
      2. \(\displaystyle \displaystyle{ (-1)^{11}=-1, }\)
        3. +
      3. \(\displaystyle \displaystyle{ (-1)^{12}=1, }\)
        4. +
      4. \(\displaystyle \displaystyle{ 0^{20}=0. }\)
        5. +
      When two negative numbers are multiplied together they make a positive number; each pair of negative signs “cancel each other out”.
      When an odd number of negative numbers are multiplied together, each pair of negative signs “cancel out” but there is still one negative sign left. That’s why \((-1)^{\text{odd number}}=-1\text{.}\) +
      Similarly, when an even number of negative numbers are multiplied together each pair of negative signs “cancel out” and there are no negative signs left. That’s why \((-1)^{\text{even number}}=1\text{.}\)

      36.

      -
      -
      +
      +
      -
      +
      Evaluate the following.
        -
      1. \(\displaystyle 1^{5}\)
        2. -
      2. \(\displaystyle (-1)^{13}\)
        3. -
      3. \(\displaystyle (-1)^{18}\)
        4. -
      4. \(\displaystyle 0^{18}\)
        5. +
      5. \(\displaystyle 1^{5}\)
        6. +
      6. \(\displaystyle (-1)^{13}\)
        7. +
      7. \(\displaystyle (-1)^{18}\)
        8. +
      8. \(\displaystyle 0^{18}\)
      -
      Answer 1.
      \(1\)
      Answer 2.
      \(-1\)
      Answer 3.
      \(1\)
      Answer 4.
      \(0\)
      Explanation.
      -
      The solutions are:
        -
      1. \(\displaystyle \displaystyle{ 1^{5}=1, }\)
        2. -
      2. \(\displaystyle \displaystyle{ (-1)^{13}=-1, }\)
        3. -
      3. \(\displaystyle \displaystyle{ (-1)^{18}=1, }\)
        4. -
      4. \(\displaystyle \displaystyle{ 0^{18}=0. }\)
        5. -
      When two negative numbers are multiplied together they make a positive number; each pair of negative signs “cancel each other out”.
      When an odd number of negative numbers are multiplied together, each pair of negative signs “cancel out” but there is still one negative sign left. That’s why \((-1)^{\text{odd number}}=-1\text{.}\) -
      Similarly, when an even number of negative numbers are multiplied together each pair of negative signs “cancel out” and there are no negative signs left. That’s why \((-1)^{\text{even number}}=1\text{.}\) +
      Answer 1.
      \(1\)
      Answer 2.
      \(-1\)
      Answer 3.
      \(1\)
      Answer 4.
      \(0\)
      Explanation.
      +
      The solutions are:
        +
      1. \(\displaystyle \displaystyle{ 1^{5}=1, }\)
        2. +
      2. \(\displaystyle \displaystyle{ (-1)^{13}=-1, }\)
        3. +
      3. \(\displaystyle \displaystyle{ (-1)^{18}=1, }\)
        4. +
      4. \(\displaystyle \displaystyle{ 0^{18}=0. }\)
        5. +
      When two negative numbers are multiplied together they make a positive number; each pair of negative signs “cancel each other out”.
      When an odd number of negative numbers are multiplied together, each pair of negative signs “cancel out” but there is still one negative sign left. That’s why \((-1)^{\text{odd number}}=-1\text{.}\) +
      Similarly, when an even number of negative numbers are multiplied together each pair of negative signs “cancel out” and there are no negative signs left. That’s why \((-1)^{\text{even number}}=1\text{.}\)
      @@ -1901,200 +1915,200 @@

      Exercise Group.

      Exercise Group.

      37.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ -8.73 + (-29.4) }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ -1.52 + (-64.3) }\)
      -
      Answer.
      \(-38.13\)
      Explanation.
      -
      When we add up two negative numbers, the result will be negative.
      We need to find the sum of their absolute value. To add two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{+2}8.73 \\ -\amp \underline{+2 9.4\phantom{ad}} \\ -\amp \phantom{+}38.13 +
      Answer.
      \(-65.82\)
      Explanation.
      +
      When we add up two negative numbers, the result will be negative.
      We need to find the sum of their absolute value. To add two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{+6}1.52 \\ +\amp \underline{+6 4.3\phantom{ad}} \\ +\amp \phantom{+}65.82 \end{aligned} -}\)
      Since the result is negative, we have \(-8.73 + (-29.4) = -38.13\text{.}\) +}\)
      Since the result is negative, we have \(-1.52 + (-64.3) = -65.82\text{.}\)

      38.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ -9.47 + (-89.8) }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ -2.26 + (-34.7) }\)
      -
      Answer.
      \(-99.27\)
      Explanation.
      -
      When we add up two negative numbers, the result will be negative.
      We need to find the sum of their absolute value. To add two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{+8}9.47 \\ -\amp \underline{+8 9.8\phantom{ad}} \\ -\amp \phantom{+}99.27 +
      Answer.
      \(-36.96\)
      Explanation.
      +
      When we add up two negative numbers, the result will be negative.
      We need to find the sum of their absolute value. To add two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{+3}2.26 \\ +\amp \underline{+3 4.7\phantom{ad}} \\ +\amp \phantom{+}36.96 \end{aligned} -}\)
      Since the result is negative, we have \(-9.47 + (-89.8) = -99.27\text{.}\) +}\)
      Since the result is negative, we have \(-2.26 + (-34.7) = -36.96\text{.}\)

      39.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ 5.1 - 1.69 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ 6.8 - 5.04 }\)
      -
      Answer.
      \(3.41\)
      Explanation.
      -
      To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}5.10 \\ -\amp \underline{-1.69\phantom{ad}} \\ -\amp \phantom{-}3.41 +
      Answer.
      \(1.76\)
      Explanation.
      +
      To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}6.80 \\ +\amp \underline{-5.04\phantom{ad}} \\ +\amp \phantom{-}1.76 \end{aligned} -}\)
      Note that we added an extra \(0\) to the end of \(5.1\) before doing subtraction.
      +}\)
      Note that we added an extra \(0\) to the end of \(6.8\) before doing subtraction.

      40.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ 5.7 - 3.39 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ 6.5 - 2.74 }\)
      -
      Answer.
      \(2.31\)
      Explanation.
      -
      To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}5.70 \\ -\amp \underline{-3.39\phantom{ad}} \\ -\amp \phantom{-}2.31 +
      Answer.
      \(3.76\)
      Explanation.
      +
      To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}6.50 \\ +\amp \underline{-2.74\phantom{ad}} \\ +\amp \phantom{-}3.76 \end{aligned} -}\)
      Note that we added an extra \(0\) to the end of \(5.7\) before doing subtraction.
      +}\)
      Note that we added an extra \(0\) to the end of \(6.5\) before doing subtraction.

      41.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ - 5.09 + 6.4 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ - 5.54 + 7.2 }\)
      -
      Answer.
      \(1.31\)
      Explanation.
      -
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}6.40 \\ -\amp \underline{-5.09\phantom{ad}} \\ -\amp \phantom{-}1.31 +
      Answer.
      \(1.66\)
      Explanation.
      +
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}7.20 \\ +\amp \underline{-5.54\phantom{ad}} \\ +\amp \phantom{-}1.66 \end{aligned} -}\)
      Note that we added an extra \(0\) to the end of \(6.4\) before doing subtraction.
      Finally, since the absolute value of \(- 5.09\) is smaller than the absolute value of \(6.4\text{,}\) the result is positive.
      \(- 5.09 + 6.4 = 1.31\)
      +}\)
      Note that we added an extra \(0\) to the end of \(7.2\) before doing subtraction.
      Finally, since the absolute value of \(- 5.54\) is smaller than the absolute value of \(7.2\text{,}\) the result is positive.
      \(- 5.54 + 7.2 = 1.66\)

      42.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ - 2.79 + 6.1 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ - 2.24 + 7.8 }\)
      -
      Answer.
      \(3.31\)
      Explanation.
      -
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}6.10 \\ -\amp \underline{-2.79\phantom{ad}} \\ -\amp \phantom{-}3.31 +
      Answer.
      \(5.56\)
      Explanation.
      +
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}7.80 \\ +\amp \underline{-2.24\phantom{ad}} \\ +\amp \phantom{-}5.56 \end{aligned} -}\)
      Note that we added an extra \(0\) to the end of \(6.1\) before doing subtraction.
      Finally, since the absolute value of \(- 2.79\) is smaller than the absolute value of \(6.1\text{,}\) the result is positive.
      \(- 2.79 + 6.1 = 3.31\)
      +}\)
      Note that we added an extra \(0\) to the end of \(7.8\) before doing subtraction.
      Finally, since the absolute value of \(- 2.24\) is smaller than the absolute value of \(7.8\text{,}\) the result is positive.
      \(- 2.24 + 7.8 = 5.56\)

      43.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ - 5.48 - (-7.7) }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ - 5.94 - (-8.5) }\)
      -
      Answer.
      \(2.22\)
      Explanation.
      -
      When we see \(- 5.48 - (-7.7)\text{,}\) we first change those two negative symbol to a positive symbol. Now the problem becomes an addition problem:
      \(\displaystyle{ - 5.48 - (-7.7) = - 5.48 +7.7 }\)
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}7.70 \\ -\amp \underline{-5.48\phantom{ad}} \\ -\amp \phantom{-}2.22 +
      Answer.
      \(2.56\)
      Explanation.
      +
      When we see \(- 5.94 - (-8.5)\text{,}\) we first change those two negative symbol to a positive symbol. Now the problem becomes an addition problem:
      \(\displaystyle{ - 5.94 - (-8.5) = - 5.94 +8.5 }\)
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}8.50 \\ +\amp \underline{-5.94\phantom{ad}} \\ +\amp \phantom{-}2.56 \end{aligned} -}\)
      Note that we added an extra \(0\) to the end of \(7.7\) before doing subtraction.
      Finally, since the absolute value of \(- 5.48\) is smaller than the absolute value of \(7.7\text{,}\) the result is positive.
      \(- 5.48 + 7.7 = 2.22\)
      +}\)
      Note that we added an extra \(0\) to the end of \(8.5\) before doing subtraction.
      Finally, since the absolute value of \(- 5.94\) is smaller than the absolute value of \(8.5\text{,}\) the result is positive.
      \(- 5.94 + 8.5 = 2.56\)

      44.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ - 2.28 - (-7.4) }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ - 1.64 - (-9.2) }\)
      -
      Answer.
      \(5.12\)
      Explanation.
      -
      When we see \(- 2.28 - (-7.4)\text{,}\) we first change those two negative symbol to a positive symbol. Now the problem becomes an addition problem:
      \(\displaystyle{ - 2.28 - (-7.4) = - 2.28 +7.4 }\)
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}7.40 \\ -\amp \underline{-2.28\phantom{ad}} \\ -\amp \phantom{-}5.12 +
      Answer.
      \(7.56\)
      Explanation.
      +
      When we see \(- 1.64 - (-9.2)\text{,}\) we first change those two negative symbol to a positive symbol. Now the problem becomes an addition problem:
      \(\displaystyle{ - 1.64 - (-9.2) = - 1.64 +9.2 }\)
      When we add a negative number with a positive number, we need to find the difference of their absolute values. To subtract two decimals, make sure to line up the decimal points:
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}9.20 \\ +\amp \underline{-1.64\phantom{ad}} \\ +\amp \phantom{-}7.56 \end{aligned} -}\)
      Note that we added an extra \(0\) to the end of \(7.4\) before doing subtraction.
      Finally, since the absolute value of \(- 2.28\) is smaller than the absolute value of \(7.4\text{,}\) the result is positive.
      \(- 2.28 + 7.4 = 5.12\)
      +}\)
      Note that we added an extra \(0\) to the end of \(9.2\) before doing subtraction.
      Finally, since the absolute value of \(- 1.64\) is smaller than the absolute value of \(9.2\text{,}\) the result is positive.
      \(- 1.64 + 9.2 = 7.56\)

      45.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ 69 - 6.98 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ 87 - 5.44 }\)
      -
      Answer.
      \(62.02\)
      Explanation.
      -
      When doing decimal subtractions, make sure to line up the decimal points. For the integer \(69\text{,}\) the decimal point is at the end of the number, like in \(69.00\) -
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}69.00 \\ -\amp \underline{-\phantom{6}6.98\phantom{ad}} \\ -\amp \phantom{-}62.02 +
      Answer.
      \(81.56\)
      Explanation.
      +
      When doing decimal subtractions, make sure to line up the decimal points. For the integer \(87\text{,}\) the decimal point is at the end of the number, like in \(87.00\) +
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}87.00 \\ +\amp \underline{-\phantom{8}5.44\phantom{ad}} \\ +\amp \phantom{-}81.56 \end{aligned} }\)

      46.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ 76 - 1.68 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ 14 - 9.14 }\)
      -
      Answer.
      \(74.32\)
      Explanation.
      -
      When doing decimal subtractions, make sure to line up the decimal points. For the integer \(76\text{,}\) the decimal point is at the end of the number, like in \(76.00\) -
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}76.00 \\ -\amp \underline{-\phantom{7}1.68\phantom{ad}} \\ -\amp \phantom{-}74.32 +
      Answer.
      \(4.86\)
      Explanation.
      +
      When doing decimal subtractions, make sure to line up the decimal points. For the integer \(14\text{,}\) the decimal point is at the end of the number, like in \(14.00\) +
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}14.00 \\ +\amp \underline{-\phantom{1}9.14\phantom{ad}} \\ +\amp \phantom{-}4.86 \end{aligned} }\)

      47.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ -82 + 5.38 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ -20 + 3.84 }\)
      -
      Answer.
      \(-76.62\)
      Explanation.
      -
      When we add a negative number with a positive number, we need to find the difference of their absolute values. When doing decimal subtractions, make sure to line up the decimal points. For the integer \(-82\text{,}\) the decimal point is at the end of the number, like in \(-82.00\) -
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}82.00 \\ -\amp \underline{-\phantom{8}5.38\phantom{ad}} \\ -\amp \phantom{-}76.62 +
      Answer.
      \(-16.16\)
      Explanation.
      +
      When we add a negative number with a positive number, we need to find the difference of their absolute values. When doing decimal subtractions, make sure to line up the decimal points. For the integer \(-20\text{,}\) the decimal point is at the end of the number, like in \(-20.00\) +
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}20.00 \\ +\amp \underline{-\phantom{2}3.84\phantom{ad}} \\ +\amp \phantom{-}16.16 \end{aligned} -}\)
      Next, since the absolute value of \(-82\) is bigger than that of \(5.38\text{,}\) the result is negative.
      \(-82 + 5.38 = -76.62\)
      +}\)
      Next, since the absolute value of \(-20\) is bigger than that of \(3.84\text{,}\) the result is negative.
      \(-20 + 3.84 = -16.16\)

      48.

      -
      -
      +
      +
      -
      Simplify without using a calculator.
      \(\displaystyle{ -19 + 8.18 }\)
      +
      Simplify without using a calculator.
      \(\displaystyle{ -27 + 7.53 }\)
      -
      Answer.
      \(-10.82\)
      Explanation.
      -
      When we add a negative number with a positive number, we need to find the difference of their absolute values. When doing decimal subtractions, make sure to line up the decimal points. For the integer \(-19\text{,}\) the decimal point is at the end of the number, like in \(-19.00\) -
      \(\displaystyle{\begin{aligned}[t] -\amp \phantom{-}19.00 \\ -\amp \underline{-\phantom{1}8.18\phantom{ad}} \\ -\amp \phantom{-}10.82 +
      Answer.
      \(-19.47\)
      Explanation.
      +
      When we add a negative number with a positive number, we need to find the difference of their absolute values. When doing decimal subtractions, make sure to line up the decimal points. For the integer \(-27\text{,}\) the decimal point is at the end of the number, like in \(-27.00\) +
      \(\displaystyle{\begin{aligned}[t] +\amp \phantom{-}27.00 \\ +\amp \underline{-\phantom{2}7.53\phantom{ad}} \\ +\amp \phantom{-}19.47 \end{aligned} -}\)
      Next, since the absolute value of \(-19\) is bigger than that of \(8.18\text{,}\) the result is negative.
      \(-19 + 8.18 = -10.82\)
      +}\)
      Next, since the absolute value of \(-27\) is bigger than that of \(7.53\text{,}\) the result is negative.
      \(-27 + 7.53 = -19.47\)
      @@ -2105,102 +2119,102 @@

      Exercise Group.

      Exercise Group.

      49.

      -
      -
      +
      +
      -
      It’s given that \(26\cdot38=988\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ 0.026(-3.8) = }\)
      +
      It’s given that \(44\cdot23=1012\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ 4.4(-2.3) = }\)
      -
      Answer.
      \(-0.0988\)
      Explanation.
      -
      There is no operation symbol between \(0.026\) and \(-3.8\text{,}\) which implies multiplication.
      Next, since “positive times negative is negative,” we know the answer must be negative. Now we can focus on calculating \(0.026 \cdot 3.8\text{,}\) and in the end add back the negative symbol.
      It’s given that \(26\cdot38=988\text{.}\) We need to calculate \(0.026 \cdot 3.8\text{.}\) -
      Notice that \(0.026=\frac{26}{1000}\) and \(3.8=\frac{38}{10}\text{,}\) so we can calculate \(0.026 \cdot 3.8\) in this way:
      \(\displaystyle{ +
      Answer.
      \(-10.12\)
      Explanation.
      +
      There is no operation symbol between \(4.4\) and \(-2.3\text{,}\) which implies multiplication.
      Next, since “positive times negative is negative,” we know the answer must be negative. Now we can focus on calculating \(4.4 \cdot 2.3\text{,}\) and in the end add back the negative symbol.
      It’s given that \(44\cdot23=1012\text{.}\) We need to calculate \(4.4 \cdot 2.3\text{.}\) +
      Notice that \(4.4=\frac{44}{10}\) and \(2.3=\frac{23}{10}\text{,}\) so we can calculate \(4.4 \cdot 2.3\) in this way:
      \(\displaystyle{ \begin{aligned}[t] -\amp \phantom{{}=} 0.026 \cdot 3.8 \\ -\amp = \frac{26}{1000} \cdot \frac{38}{10} \\ -\amp = \frac{26\cdot38}{1000\cdot10} \\ -\amp = \frac{988}{10000} \\ -\amp = 0.0988 +\amp \phantom{{}=} 4.4 \cdot 2.3 \\ +\amp = \frac{44}{10} \cdot \frac{23}{10} \\ +\amp = \frac{44\cdot23}{10\cdot10} \\ +\amp = \frac{1012}{100} \\ +\amp = 10.12 \end{aligned} -}\)
      Once we understand the math, we can use this shortcut.
        -
      1. To change \(26\) to \(0.026\text{,}\) we moved the decimal point to the left by \(3\) place(s).
        2. -
      2. To change \(38\) to \(3.8\text{,}\) we moved the decimal point to the left by \(1\) place(s).
        3. -
      3. We move the decimal point of \(988\) to the left by \(3+1=4\) places to get \(0.026\cdot3.8=0.0988\text{.}\) +}\)
        Once we understand the math, we can use this shortcut.
          +
        1. To change \(44\) to \(4.4\text{,}\) we moved the decimal point to the left by \(1\) place(s).
          2. +
        2. To change \(23\) to \(2.3\text{,}\) we moved the decimal point to the left by \(1\) place(s).
          3. +
        3. We move the decimal point of \(1012\) to the left by \(1+1=2\) places to get \(4.4\cdot2.3=10.12\text{.}\)
          4. -
        Again, don’t forget to add back the negative symbol.
        \(0.026(-3.8)=-0.0988\)
        +
      Again, don’t forget to add back the negative symbol.
      \(4.4(-2.3)=-10.12\)

      50.

      -
      -
      +
      +
      -
      It’s given that \(33\cdot75=2475\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ 0.033(-0.075) = }\)
      +
      It’s given that \(51\cdot61=3111\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ 5.1(-0.061) = }\)
      -
      Answer.
      \(-0.002475\)
      Explanation.
      -
      There is no operation symbol between \(0.033\) and \(-0.075\text{,}\) which implies multiplication.
      Next, since “positive times negative is negative,” we know the answer must be negative. Now we can focus on calculating \(0.033 \cdot 0.075\text{,}\) and in the end add back the negative symbol.
      It’s given that \(33\cdot75=2475\text{.}\) We need to calculate \(0.033 \cdot 0.075\text{.}\) -
      Notice that \(0.033=\frac{33}{1000}\) and \(0.075=\frac{75}{1000}\text{,}\) so we can calculate \(0.033 \cdot 0.075\) in this way:
      \(\displaystyle{ +
      Answer.
      \(-0.3111\)
      Explanation.
      +
      There is no operation symbol between \(5.1\) and \(-0.061\text{,}\) which implies multiplication.
      Next, since “positive times negative is negative,” we know the answer must be negative. Now we can focus on calculating \(5.1 \cdot 0.061\text{,}\) and in the end add back the negative symbol.
      It’s given that \(51\cdot61=3111\text{.}\) We need to calculate \(5.1 \cdot 0.061\text{.}\) +
      Notice that \(5.1=\frac{51}{10}\) and \(0.061=\frac{61}{1000}\text{,}\) so we can calculate \(5.1 \cdot 0.061\) in this way:
      \(\displaystyle{ \begin{aligned}[t] -\amp \phantom{{}=} 0.033 \cdot 0.075 \\ -\amp = \frac{33}{1000} \cdot \frac{75}{1000} \\ -\amp = \frac{33\cdot75}{1000\cdot1000} \\ -\amp = \frac{2475}{1000000} \\ -\amp = 0.002475 +\amp \phantom{{}=} 5.1 \cdot 0.061 \\ +\amp = \frac{51}{10} \cdot \frac{61}{1000} \\ +\amp = \frac{51\cdot61}{10\cdot1000} \\ +\amp = \frac{3111}{10000} \\ +\amp = 0.3111 \end{aligned} -}\)
      Once we understand the math, we can use this shortcut.
        -
      1. To change \(33\) to \(0.033\text{,}\) we moved the decimal point to the left by \(3\) place(s).
        2. -
      2. To change \(75\) to \(0.075\text{,}\) we moved the decimal point to the left by \(3\) place(s).
        3. -
      3. We move the decimal point of \(2475\) to the left by \(3+3=6\) places to get \(0.033\cdot0.075=0.002475\text{.}\) +}\)
        Once we understand the math, we can use this shortcut.
          +
        1. To change \(51\) to \(5.1\text{,}\) we moved the decimal point to the left by \(1\) place(s).
          2. +
        2. To change \(61\) to \(0.061\text{,}\) we moved the decimal point to the left by \(3\) place(s).
          3. +
        3. We move the decimal point of \(3111\) to the left by \(1+3=4\) places to get \(5.1\cdot0.061=0.3111\text{.}\)
          4. -
        Again, don’t forget to add back the negative symbol.
        \(0.033(-0.075)=-0.002475\)
        +
      Again, don’t forget to add back the negative symbol.
      \(5.1(-0.061)=-0.3111\)

      51.

      -
      -
      +
      +
      -
      It’s given that \(49\cdot23=1127\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ (-0.049)(-2.3) }\)
      +
      It’s given that \(67\cdot97=6499\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ (-6.7)(-9.7) }\)
      -
      Answer.
      \(0.1127\)
      Explanation.
      -
      There is no operation symbols between \(0.049\) and \(2.3\text{,}\) which implies multiplication.
      Next, since “negative times negative is positive,” the problem becomes:
      \(\displaystyle{ (-0.049)(-2.3) = 0.049 \cdot 2.3 }\)
      We know the answer must be positive.
      It’s given that \(49\cdot23=1127\text{.}\) We need to calculate \(0.049 \cdot 2.3\text{.}\) -
      Notice that \(0.049=\frac{49}{1000}\) and \(2.3=\frac{23}{10}\text{,}\) so we can calculate \(0.049 \cdot 2.3\) in this way:
      \(\displaystyle{ +
      Answer.
      \(64.99\)
      Explanation.
      +
      There is no operation symbols between \(6.7\) and \(9.7\text{,}\) which implies multiplication.
      Next, since “negative times negative is positive,” the problem becomes:
      \(\displaystyle{ (-6.7)(-9.7) = 6.7 \cdot 9.7 }\)
      We know the answer must be positive.
      It’s given that \(67\cdot97=6499\text{.}\) We need to calculate \(6.7 \cdot 9.7\text{.}\) +
      Notice that \(6.7=\frac{67}{10}\) and \(9.7=\frac{97}{10}\text{,}\) so we can calculate \(6.7 \cdot 9.7\) in this way:
      \(\displaystyle{ \begin{aligned}[t] -\amp \phantom{{}=} 0.049 \cdot 2.3 \\ -\amp = \frac{49}{1000} \cdot \frac{23}{10} \\ -\amp = \frac{49\cdot23}{1000\cdot10} \\ -\amp = \frac{1127}{10000} \\ -\amp = 0.1127 +\amp \phantom{{}=} 6.7 \cdot 9.7 \\ +\amp = \frac{67}{10} \cdot \frac{97}{10} \\ +\amp = \frac{67\cdot97}{10\cdot10} \\ +\amp = \frac{6499}{100} \\ +\amp = 64.99 \end{aligned} -}\)
      Once we understand the math, we can use this shortcut.
        -
      1. To change \(49\) to \(0.049\text{,}\) we moved the decimal point to the left by \(3\) place(s).
        2. -
      2. To change \(23\) to \(2.3\text{,}\) we moved the decimal point to the left by \(1\) place(s).
        3. -
      3. We move the decimal point of \(1127\) to the left by \(3+1=4\) places to get \(0.049\cdot2.3=0.1127\text{.}\) +}\)
        Once we understand the math, we can use this shortcut.
          +
        1. To change \(67\) to \(6.7\text{,}\) we moved the decimal point to the left by \(1\) place(s).
          2. +
        2. To change \(97\) to \(9.7\text{,}\) we moved the decimal point to the left by \(1\) place(s).
          3. +
        3. We move the decimal point of \(6499\) to the left by \(1+1=2\) places to get \(6.7\cdot9.7=64.99\text{.}\)
          4. -
        \(\displaystyle{ (-0.049)(-2.3) = 0.1127 }\)
        +
      \(\displaystyle{ (-6.7)(-9.7) = 64.99 }\)

      52.

      -
      -
      +
      +
      -
      It’s given that \(55\cdot51=2805\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ (-0.055)(-0.051) }\)
      +
      It’s given that \(74\cdot45=3330\text{.}\) Use this fact to calculate the following without using a calculator.
      \(\displaystyle{ (-7.4)(-0.045) }\)
      -
      Answer.
      \(0.002805\)
      Explanation.
      -
      There is no operation symbols between \(0.055\) and \(0.051\text{,}\) which implies multiplication.
      Next, since “negative times negative is positive,” the problem becomes:
      \(\displaystyle{ (-0.055)(-0.051) = 0.055 \cdot 0.051 }\)
      We know the answer must be positive.
      It’s given that \(55\cdot51=2805\text{.}\) We need to calculate \(0.055 \cdot 0.051\text{.}\) -
      Notice that \(0.055=\frac{55}{1000}\) and \(0.051=\frac{51}{1000}\text{,}\) so we can calculate \(0.055 \cdot 0.051\) in this way:
      \(\displaystyle{ +
      Answer.
      \(0.333\)
      Explanation.
      +
      There is no operation symbols between \(7.4\) and \(0.045\text{,}\) which implies multiplication.
      Next, since “negative times negative is positive,” the problem becomes:
      \(\displaystyle{ (-7.4)(-0.045) = 7.4 \cdot 0.045 }\)
      We know the answer must be positive.
      It’s given that \(74\cdot45=3330\text{.}\) We need to calculate \(7.4 \cdot 0.045\text{.}\) +
      Notice that \(7.4=\frac{74}{10}\) and \(0.045=\frac{45}{1000}\text{,}\) so we can calculate \(7.4 \cdot 0.045\) in this way:
      \(\displaystyle{ \begin{aligned}[t] -\amp \phantom{{}=} 0.055 \cdot 0.051 \\ -\amp = \frac{55}{1000} \cdot \frac{51}{1000} \\ -\amp = \frac{55\cdot51}{1000\cdot1000} \\ -\amp = \frac{2805}{1000000} \\ -\amp = 0.002805 +\amp \phantom{{}=} 7.4 \cdot 0.045 \\ +\amp = \frac{74}{10} \cdot \frac{45}{1000} \\ +\amp = \frac{74\cdot45}{10\cdot1000} \\ +\amp = \frac{3330}{10000} \\ +\amp = 0.333 \end{aligned} -}\)
      Once we understand the math, we can use this shortcut.
        -
      1. To change \(55\) to \(0.055\text{,}\) we moved the decimal point to the left by \(3\) place(s).
        2. -
      2. To change \(51\) to \(0.051\text{,}\) we moved the decimal point to the left by \(3\) place(s).
        3. -
      3. We move the decimal point of \(2805\) to the left by \(3+3=6\) places to get \(0.055\cdot0.051=0.002805\text{.}\) +}\)
        Once we understand the math, we can use this shortcut.
          +
        1. To change \(74\) to \(7.4\text{,}\) we moved the decimal point to the left by \(1\) place(s).
          2. +
        2. To change \(45\) to \(0.045\text{,}\) we moved the decimal point to the left by \(3\) place(s).
          3. +
        3. We move the decimal point of \(3330\) to the left by \(1+3=4\) places to get \(7.4\cdot0.045=0.333\text{.}\)
          4. -
        \(\displaystyle{ (-0.055)(-0.051) = 0.002805 }\)
        +
      \(\displaystyle{ (-7.4)(-0.045) = 0.333 }\)
      @@ -2211,72 +2225,72 @@

      Exercise Group.

      Applications.

      53.

      -
      -
      +
      +
      -
      Consider the following situation in which you borrow money from your cousin:
        -
      • On June 1st, you borrowed \(1300\) dollars from your cousin.
        • -
      • On July 1st, you borrowed \(340\) more dollars from your cousin.
        • -
      • On August 1st, you paid back \(700\) dollars to your cousin.
        • -
      • On September 1st, you borrowed another \(950\) dollars from your cousin.
        • -
      How much money do you owe your cousin now?
      +
      Consider the following situation in which you borrow money from your cousin:
        +
      • On June 1st, you borrowed \(1400\) dollars from your cousin.
        • +
      • On July 1st, you borrowed \(500\) more dollars from your cousin.
        • +
      • On August 1st, you paid back \(670\) dollars to your cousin.
        • +
      • On September 1st, you borrowed another \(840\) dollars from your cousin.
        • +
      How much money do you owe your cousin now?
      -
      Answer.
      \(\$1{,}890\)
      Explanation.
      -
      Borrowing money is like adding a negative number. Paying money back is like adding a positive number. To model this situation, we do the following calculation:
      \(\begin{aligned} --1300+(-340)+700+(-950) \amp = -1640+700+(-950) \\ -\amp = -940 +(-950) \\ -\amp = -1890 -\end{aligned}\)
      Now you owe your cousin \({\$1{,}890}\text{.}\) +
      Answer.
      \(\$2{,}070\)
      Explanation.
      +
      Borrowing money is like adding a negative number. Paying money back is like adding a positive number. To model this situation, we do the following calculation:
      \(\begin{aligned} +-1400+(-500)+670+(-840) \amp = -1900+670+(-840) \\ +\amp = -1230 +(-840) \\ +\amp = -2070 +\end{aligned}\)
      Now you owe your cousin \({\$2{,}070}\text{.}\)

      54.

      -
      -
      +
      +
      -
      Consider the following scenario in which you study your bank account.
        -
      • On Jan. 1, you had a balance of \(-380\) dollars in your bank account.
        • -
      • On Jan. 2, your bank charged \(50\) dollar overdraft fee.
        • -
      • On Jan. 3, you deposited \(880\) dollars.
        • -
      • On Jan. 10, you withdrew \(630\) dollars.
        • -
      What is your balance on Jan. 11?
      +
      Consider the following scenario in which you study your bank account.
        +
      • On Jan. 1, you had a balance of \(-470\) dollars in your bank account.
        • +
      • On Jan. 2, your bank charged \(45\) dollar overdraft fee.
        • +
      • On Jan. 3, you deposited \(840\) dollars.
        • +
      • On Jan. 10, you withdrew \(780\) dollars.
        • +
      What is your balance on Jan. 11?
      -
      Answer.
      \(-\$180\)
      Explanation.
      -
      Withdrawing money and being charge a fee is like adding a negative number to your bank balance. Depositing money is like adding a positive number to your bank balance. To model this situation, we do the following calculation:
      \(\begin{aligned} --380+(-50)+880+(-630)\amp = -430+880+(-630) \\ -\amp = 450 +(-630) \\ -\amp = -180 -\end{aligned}\)
      Your balance is \({-\$180}\) on Jan. 11.
      +
      Answer.
      \(-\$455\)
      Explanation.
      +
      Withdrawing money and being charge a fee is like adding a negative number to your bank balance. Depositing money is like adding a positive number to your bank balance. To model this situation, we do the following calculation:
      \(\begin{aligned} +-470+(-45)+840+(-780)\amp = -515+840+(-780) \\ +\amp = 325 +(-780) \\ +\amp = -455 +\end{aligned}\)
      Your balance is \({-\$455}\) on Jan. 11.

      55.

      -
      -
      +
      +
      -
      A mountain is \(1100\) feet above sea level. A trench is \(360\) feet below sea level. What is the difference in elevation between the mountain top and the bottom of the trench?
      +
      A mountain is \(1100\) feet above sea level. A trench is \(360\) feet below sea level. What is the difference in elevation between the mountain top and the bottom of the trench?
      -
      Answer.
      \(1460\ {\rm ft}\)
      Explanation.
      -
      The height of the mountain is simply \(1100\) feet because it is above sea level.
      The bottom of the trench is \(-360\) feet in height. Note that it is negative because it is below sea level.
      To find the difference in height, we use subtraction:
      \(\begin{aligned} +
      Answer.
      \(1460\ {\rm ft}\)
      Explanation.
      +
      The height of the mountain is simply \(1100\) feet because it is above sea level.
      The bottom of the trench is \(-360\) feet in height. Note that it is negative because it is below sea level.
      To find the difference in height, we use subtraction:
      \(\begin{aligned} 1100-(-360) \amp = 1100+360\\ \amp = 1460 -\end{aligned}\)
      The difference in elevation from the mountain top to the bottom of the trench is \({1460\ {\rm ft}}\text{.}\) +\end{aligned}\)
      The difference in elevation from the mountain top to the bottom of the trench is \({1460\ {\rm ft}}\text{.}\)

      56.

      -
      -
      +
      +
      -
      A mountain is \(1200\) feet above sea level. A trench is \(420\) feet below sea level. What is the difference in elevation between the mountain top and the bottom of the trench?
      +
      A mountain is \(1200\) feet above sea level. A trench is \(420\) feet below sea level. What is the difference in elevation between the mountain top and the bottom of the trench?
      -
      Answer.
      \(1620\ {\rm ft}\)
      Explanation.
      -
      The height of the mountain is simply \(1200\) feet because it is above sea level.
      The bottom of the trench is \(-420\) feet in height. Note that it is negative because it is below sea level.
      To find the difference in height, we use subtraction:
      \(\begin{aligned} +
      Answer.
      \(1620\ {\rm ft}\)
      Explanation.
      +
      The height of the mountain is simply \(1200\) feet because it is above sea level.
      The bottom of the trench is \(-420\) feet in height. Note that it is negative because it is below sea level.
      To find the difference in height, we use subtraction:
      \(\begin{aligned} 1200-(-420) \amp = 1200+420\\ \amp = 1620 -\end{aligned}\)
      The difference in elevation from the mountain top to the bottom of the trench is \({1620\ {\rm ft}}\text{.}\) +\end{aligned}\)
      The difference in elevation from the mountain top to the bottom of the trench is \({1620\ {\rm ft}}\text{.}\)
      @@ -2287,26 +2301,26 @@

      Applications.

      Challenge.

      57.

      -
      -
      +
      +
      -
      Select the correct word to make each statement true.
      (a)
      -
      A positive number minus a positive number is negative.
      +
      Select the correct word to make each statement true.
      (a)
      +
      A positive number minus a positive number is negative.
      -
      Answer.
      \(\text{sometimes}\)
      Explanation.
      A positive number minus a positive number is sometimes negative. For example, consider \(6 - 3 = 3\) and consider \(6 - 8 = -2\text{.}\) +
      Answer.
      \(\text{sometimes}\)
      Explanation.
      A positive number minus a positive number is sometimes negative. For example, consider \(6 - 3 = 3\) and consider \(6 - 8 = -2\text{.}\)
      -
      (b)
      -
      A negative number plus a negative number is negative.
      +
      (b)
      +
      A negative number plus a negative number is negative.
      -
      Answer.
      \(\text{always}\)
      Explanation.
      A negative number plus a negative number is always negative.
      -
      (c)
      -
      A positive number minus a negative number is positive.
      +
      Answer.
      \(\text{always}\)
      Explanation.
      A negative number plus a negative number is always negative.
      +
      (c)
      +
      A positive number minus a negative number is positive.
      -
      Answer.
      \(\text{always}\)
      Explanation.
      A positive number minus a negative number is always positive.
      -
      (d)
      -
      A negative number multiplied by a negative number is negative.
      +
      Answer.
      \(\text{always}\)
      Explanation.
      A positive number minus a negative number is always positive.
      +
      (d)
      +
      A negative number multiplied by a negative number is negative.
      -
      Answer.
      \(\text{never}\)
      Explanation.
      A negative number multiplied by a negative number is never negative. A negative number multiplied by a negative number is always positive.
      +
      Answer.
      \(\text{never}\)
      Explanation.
      A negative number multiplied by a negative number is never negative. A negative number multiplied by a negative number is always positive.
      diff --git a/section-cartesian-coordinates.html b/section-cartesian-coordinates.html index b4ed21eac..a2f6228f9 100644 --- a/section-cartesian-coordinates.html +++ b/section-cartesian-coordinates.html @@ -420,7 +420,7 @@

      Search Results:

      @@ -455,6 +455,20 @@

      Search Results:

    3. 3.9.5 Exercises
      4. +
    4. +
      3.10 Graphing Lines Chapter Review
      + +
    5. diff --git a/section-combining-like-terms.html b/section-combining-like-terms.html index b25037c03..d42bf4127 100644 --- a/section-combining-like-terms.html +++ b/section-combining-like-terms.html @@ -420,7 +420,7 @@

      Search Results:

      @@ -455,6 +455,20 @@

      Search Results:

    6. 3.9.5 Exercises
      7. +
    7. +
      3.10 Graphing Lines Chapter Review
      + +
    8. diff --git a/section-comparison-symbols-and-notation-for-intervals.html b/section-comparison-symbols-and-notation-for-intervals.html index afce98d80..63be2bde9 100644 --- a/section-comparison-symbols-and-notation-for-intervals.html +++ b/section-comparison-symbols-and-notation-for-intervals.html @@ -420,7 +420,7 @@

      Search Results:

      @@ -455,6 +455,20 @@

      Search Results:

    9. 3.9.5 Exercises
      10. +
    10. +
      3.10 Graphing Lines Chapter Review
      + +
    11. @@ -628,7 +642,7 @@

      Search Results:

      Use the \(\gt\) symbol to arrange the following numbers in order from greatest to least.
      (a)
      \({-7.6}\quad{6}\quad{-6}\quad{9.5}\quad{8}\)
      Explanation.
      -
      We can order these numbers by placing these numbers on a number line.
      And so we see the answer is \(AnswerEvaluator=HASH(0x561336f98588)\text{.}\) +
      We can order these numbers by placing these numbers on a number line.
      And so we see the answer is \(AnswerEvaluator=HASH(0x561336ec9a68)\text{.}\)
      (b)
      @@ -638,7 +652,7 @@

      Search Results:

      Explanation.
      We can order these numbers by placing these numbers on a number line. Knowing or computing their decimals helps with this: \(\pi\approx3.141\ldots\) and \(\frac{10}{3}\approx3.333\ldots\text{.}\) -
      And so we see the answer is \(AnswerEvaluator=HASH(0x561336fa0030)\text{.}\) +
      And so we see the answer is \(AnswerEvaluator=HASH(0x561336ecced8)\text{.}\)
      12. diff --git a/section-elimination.html b/section-elimination.html index 8aaaab16c..190e0d867 100644 --- a/section-elimination.html +++ b/section-elimination.html @@ -420,7 +420,7 @@

      Search Results:

      @@ -455,6 +455,20 @@

      Search Results:

    12. 3.9.5 Exercises
      13. +
    13. +
      3.10 Graphing Lines Chapter Review
      + +
    14. @@ -706,7 +720,7 @@

      Search Results:

    Checkpoint 4.3.4.

    -
    +
    @@ -819,7 +833,7 @@

    Search Results:

    Checkpoint 4.3.6.
    Try a similar exercise.
    -
    +
    @@ -1257,447 +1271,447 @@

    Search Results:

    Describe a good situation to use the substitution method instead of the elimination method for solving a system of two linear equations in two variables.

    Exercises 4.3.5 Exercises

    -

    Review and Warmup

    Skills Practice

    -
    +

    Skills Practice

    +
    Exercise Group.
    -
    Solve the system of equations using elimination.
    +
    Solve the system of equations using elimination.
    1.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-1v+\left(-3\right)J = 13} \\ -\amp {1v+\left(-1\right)J = 7} +\amp {5F+2y = 1} \\ +\amp {-1F+\left(-2\right)y = -5} \end{alignedat} \right.\)
    -
    Answer.
    \(v = 2, J = -5\)
    +
    Answer.
    \(F = -1, y = 3\)
    2.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-1B+\left(-4\right)s = -17} \\ -\amp {-4B+4s = 52} +\amp {2L+3h = 6} \\ +\amp {-5L+\left(-3\right)h = -24} \end{alignedat} \right.\)
    -
    Answer.
    \(B = -7, s = 6\)
    +
    Answer.
    \(L = 6, h = -2\)
    3.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {1G+\left(-1\right)a = -4} \\ -\amp {2G+\left(-3\right)a = -11} +\amp {5R+\left(-1\right)N = -14} \\ +\amp {-1R+1N = -2} \end{alignedat} \right.\)
    -
    Answer.
    \(G = -1, a = 3\)
    +
    Answer.
    \(R = -4, N = -6\)
    4.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-2L+1G = -15} \\ -\amp {-1L+2G = -12} +\amp {4Y+4w = 32} \\ +\amp {-5Y+1w = -10} \end{alignedat} \right.\)
    -
    Answer.
    \(L = 6, G = -3\)
    +
    Answer.
    \(Y = 3, w = 5\)
    5.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-2S+\left(-25\right)q = 158} \\ -\amp {2S+\left(-5\right)q = 22} +\amp {-2d+5f = 22} \\ +\amp {-1d+\left(-1\right)f = 4} \end{alignedat} \right.\)
    -
    Answer.
    \(S = -4, q = -6\)
    +
    Answer.
    \(d = -6, f = 2\)
    6.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-2Y+2X = 4} \\ -\amp {-4Y+6X = 18} +\amp {4j+\left(-5\right)M = 24} \\ +\amp {12j+4M = -4} \end{alignedat} \right.\)
    -
    Answer.
    \(Y = 3, X = 5\)
    +
    Answer.
    \(j = 1, M = -4\)
    7.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-3e+\left(-3\right)F = 15} \\ -\amp {4e+2F = -22} +\amp {4p+\left(-4\right)v = 56} \\ +\amp {5p+3v = 14} \end{alignedat} \right.\)
    -
    Answer.
    \(e = -6, F = 1\)
    +
    Answer.
    \(p = 7, v = -7\)
    8.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-3j+5p = -23} \\ -\amp {2j+3p = -10} +\amp {-5u+3c = 22} \\ +\amp {2u+\left(-4\right)c = -20} \end{alignedat} \right.\)
    -
    Answer.
    \(j = 1, p = -4\)
    +
    Answer.
    \(u = -2, c = 4\)
    9.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {9q+1W = -5} \\ -\amp {18q+2W = 7} +\amp {6A+\left(-1\right)K = 7} \\ +\amp {-24A+4K = -3} \end{alignedat} \right.\)
    Answer.
    \(\text{no solutions}\)
    10.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-1v+5D = -5} \\ -\amp {4v+\left(-20\right)D = -7} +\amp {-1F+\left(-7\right)s = 7} \\ +\amp {-2F+\left(-14\right)s = 5} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{no solutions}\)
    +
    Answer.
    \(\text{no solutions}\)
    11.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-3A+\left(-4\right)k = -4} \\ -\amp {-3A+1k = -14} +\amp {-1L+1a = 4} \\ +\amp {-1L+5a = 28} \end{alignedat} \right.\)
    -
    Answer.
    \(A = 4, k = -2\)
    +
    Answer.
    \(L = 2, a = 6\)
    12.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-3G+\left(-3\right)T = 33} \\ -\amp {-4G+\left(-3\right)T = 38} +\amp {3R+2H = -17} \\ +\amp {3R+\left(-3\right)H = -27} \end{alignedat} \right.\)
    -
    Answer.
    \(G = -5, T = -6\)
    +
    Answer.
    \(R = -7, H = 2\)
    13.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {3L+\left(-4\right)A = -8} \\ -\amp {-1L+4A = -1} +\amp {-1X+1q = 3} \\ +\amp {-3X+1q = 4} \end{alignedat} \right.\)
    -
    Answer.
    \(L = -{\frac{9}{2}}, A = -{\frac{11}{8}}\)
    +
    Answer.
    \(X = -{\frac{1}{2}}, q = {\frac{5}{2}}\)
    14.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {5S+1i = 2} \\ -\amp {-1S+\left(-2\right)i = -6} +\amp {-1d+\left(-1\right)Y = 8} \\ +\amp {-1d+1Y = 1} \end{alignedat} \right.\)
    -
    Answer.
    \(S = -{\frac{2}{9}}, i = {\frac{28}{9}}\)
    +
    Answer.
    \(d = -{\frac{9}{2}}, Y = -{\frac{7}{2}}\)
    15.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {2Y+\left(-5\right)Q = -1} \\ -\amp {-4Y+6Q = -8} +\amp {-8i+\left(-9\right)G = -7} \\ +\amp {7i+9G = 9} \end{alignedat} \right.\)
    -
    Answer.
    \(Y = {\frac{23}{4}}, Q = {\frac{5}{2}}\)
    +
    Answer.
    \(i = -2, G = {\frac{23}{9}}\)
    16.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {4e+\left(-4\right)z = 6} \\ -\amp {-2e+3z = -7} +\amp {p+\left(-4\right)n = 3} \\ +\amp {-2p+5n = 2} \end{alignedat} \right.\)
    -
    Answer.
    \(e = -{\frac{5}{2}}, z = -4\)
    +
    Answer.
    \(p = -{\frac{23}{8}}, n = -{\frac{3}{4}}\)
    17.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {0.6j+\left(-1\right)g = -2} \\ -\amp {0.4j+\left(-0.6\right)g = 6} +\amp {-0.3u+\left(-1\right)W = -5.2} \\ +\amp {1.5u+\left(-2.4\right)W = 2.5} \end{alignedat} \right.\)
    -
    Answer.
    \(j = 180, g = 110\)
    +
    Answer.
    \(u = 6.74775, W = 3.17568\)
    18.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-1p+\left(-1.6\right)P = -3.8} \\ -\amp {-0.4p+1.2P = 6.8} +\amp {-1A+0.3E = -7} \\ +\amp {-1.5A+0.3E = 3.2} \end{alignedat} \right.\)
    -
    Answer.
    \(p = -3.43478, P = 4.52174\)
    +
    Answer.
    \(A = -20.4, E = -91.3333\)
    19.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {0.333333v+0.333333x = 2.5} \\ -\amp {0.2v+1x = -0.166667} +\amp {0.75F+\left(-0.666667\right)l = 0.333333} \\ +\amp {1.5F+1l = -2.5} \end{alignedat} \right.\)
    -
    Answer.
    \(v = {\frac{115}{12}}, x = -{\frac{5}{2}}\)
    +
    Answer.
    \(F = -{\frac{16}{21}}, l = -{\frac{12}{7}}\)
    20.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.75A+0.5e = 0.333333} \\ -\amp {1.25A+\left(-1\right)e = -0.666667} +\amp {0.666667K+0.666667T = -0.6} \\ +\amp {1K+\left(-0.5\right)T = 1.5} \end{alignedat} \right.\)
    -
    Answer.
    \(A = 0, e = -{\frac{2}{11}}\)
    +
    Answer.
    \(K = {\frac{7}{10}}, T = {\frac{3}{20}}\)
    21.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {0.833333G+\left(-0.4\right)M = -0.666667} \\ -\amp {0.25G+\left(-0.12\right)M = -0.2} +\amp {2.5R+\left(-0.2\right)A = -0.8} \\ +\amp {1.33333R+\left(-0.106667\right)A = -0.426667} \end{alignedat} \right.\)
    Answer.
    \(\text{infinitely many solutions}\)
    22.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.666667L+\left(-1.66667\right)u = -0.2} \\ -\amp {0.25L+\left(-0.625\right)u = -0.075} +\amp {0.2X+0.8j = 0.833333} \\ +\amp {1.33333X+5.33333j = 5.55556} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    23.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {2.5S+\left(-0.2\right)c = -0.666667} \\ -\amp {0.25S+\left(-0.02\right)c = 0.333333} +\amp {0.6d+\left(-0.6\right)R = -0.8} \\ +\amp {1.33333d+\left(-1.33333\right)R = -0.6} \end{alignedat} \right.\)
    Answer.
    \(\text{no solutions}\)
    24.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.2Y+1.33333J = -0.2} \\ -\amp {0.333333Y+2.22222J = 0.166667} +\amp {0.5i+0.166667z = 0.833333} \\ +\amp {1.33333i+0.444444z = 0.666667} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{no solutions}\)
    +
    Answer.
    \(\text{no solutions}\)
    25.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {4t+\left(-4\right)d = -4} \\ -\amp {2d+\left(-4\right)t = 0} +\amp {3h+2p = -23} \\ +\amp {-2p+1h = -5} \end{alignedat} \right.\)
    -
    Answer.
    \(d = 2, t = 1\)
    +
    Answer.
    \(p = -1, h = -7\)
    26.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-1a+4j = -25} \\ -\amp {-4j+2a = 22} +\amp {5P+2u = 27} \\ +\amp {5u+\left(-5\right)P = 15} \end{alignedat} \right.\)
    -
    Answer.
    \(j = -7, a = -3\)
    +
    Answer.
    \(u = 6, P = 3\)
    27.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-5p = 12-1H} \\ -\amp {5H = -1p+34} +\amp {1z = -8-5w} \\ +\amp {1w = 1z+2} \end{alignedat} \right.\)
    -
    Answer.
    \(p = -1, H = 7\)
    +
    Answer.
    \(z = -3, w = -1\)
    28.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-5v = -33-\left(-1\right)q} \\ -\amp {-2q = 4v+\left(-30\right)} +\amp {-1F = 22-\left(-5\right)f} \\ +\amp {4f = -5F+\left(-5\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(v = 6, q = 3\)
    +
    Answer.
    \(F = 3, f = -5\)
    29.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-5A+\left(-20\right)Y = 17+\left(-23\right)Y} \\ -\amp {-3A+12Y = -5A+\left(-20\right)} +\amp {1K+\left(-42\right)M = -30+\left(-38\right)M} \\ +\amp {6K+\left(-2\right)M = 1K+\left(-42\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(A = -4, Y = -1\)
    +
    Answer.
    \(K = -6, M = 6\)
    30.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-8G+8E = -44+4E} \\ -\amp {-12G+\left(-4\right)E = -8G+8} +\amp {1R+\left(-22\right)u = -8+\left(-19\right)u} \\ +\amp {6R+\left(-9\right)u = 1R+\left(-22\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(G = 3, E = -5\)
    +
    Answer.
    \(R = 1, u = 3\)
    31.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-8L+\left(-18\right)n = -9+\left(-19\right)n} \\ -\amp {-24L+2n = -8L+\left(-18\right)} +\amp {-1X+\left(-8\right)d = 2+\left(-4\right)d} \\ +\amp {3X+16d = -1X+\left(-8\right)} \end{alignedat} \right.\)
    Answer.
    \(\text{infinitely many solutions}\)
    32.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-1R+36W = -9+32W} \\ -\amp {3R+\left(-16\right)W = -1R+36} +\amp {-4d+\left(-6\right)K = 3+\left(-5\right)K} \\ +\amp {4d+2K = -4d+\left(-6\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    -

    Applications

    +

    Applications

    33.
    -
    +
    -
    An algebra exam has \(16\) questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth \(4\) points, and short-answer questions, each worth \(8\) points. How many questions are there of each type?
    +
    An algebra exam has \(21\) questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth \(4\) points, and short-answer questions, each worth \(6\) points. How many questions are there of each type?
    -
    Answer 1.
    \(M+S = 16\)
    Answer 2.
    \(4M+8S = 100\)
    Answer 3.
    \(7\)
    Answer 4.
    \(9\)
    +
    Answer 1.
    \(M+S = 21\)
    Answer 2.
    \(4M+6S = 100\)
    Answer 3.
    \(13\)
    Answer 4.
    \(8\)
    34.
    -
    +
    -
    A school fund raising event sold a total of \(124\) tickets and generated a total revenue of \({\$476.50}\text{.}\) Each adult ticket cost \({\$5}\text{,}\) and each child ticket cost \({\$1.50}\text{.}\) How many adult tickets and how many child tickets were sold?
    +
    A school fund raising event sold a total of \(126\) tickets and generated a total revenue of \({\$571.50}\text{.}\) Each adult ticket cost \({\$6.50}\text{,}\) and each child ticket cost \({\$1}\text{.}\) How many adult tickets and how many child tickets were sold?
    -
    Answer 1.
    \(A+C = 124\)
    Answer 2.
    \(5A+1.5C = 476.5\)
    Answer 3.
    \(83\)
    Answer 4.
    \(41\)
    +
    Answer 1.
    \(A+C = 126\)
    Answer 2.
    \(6.5A+1C = 571.5\)
    Answer 3.
    \(81\)
    Answer 4.
    \(45\)
    35.
    -
    +
    -
    Colin invested a total of \({\$5{,}800}\) in two investments. His savings account pays \({2.5\%}\) interest annually. A riskier stock investment earned \({6.5\%}\) at the end of the year. At the end of the year, Colin earned a total of \({\$305}\) in interest. How much money did he invest in each account?
    +
    Garrison invested a total of \({\$4{,}200}\) in two investments. His savings account pays \({2\%}\) interest annually. A riskier stock investment earned \({6\%}\) at the end of the year. At the end of the year, Garrison earned a total of \({\$204}\) in interest. How much money did he invest in each account?
    -
    Answer 1.
    \(x+y = 5800\)
    Answer 2.
    \(0.025x+0.065y = 305\)
    Answer 3.
    \(\$1{,}800\)
    Answer 4.
    \(\$4{,}000\)
    +
    Answer 1.
    \(x+y = 4200\)
    Answer 2.
    \(0.02x+0.06y = 204\)
    Answer 3.
    \(\$1{,}200\)
    Answer 4.
    \(\$3{,}000\)
    36.
    -
    +
    -
    Elias invested a total of \({\$5{,}500}\) in two investments. His savings account pays \({2\%}\) interest annually. A riskier stock investment lost \({4\%}\) at the end of the year. At the end of the year, Elias’s total fell from \({\$5{,}500}\) to \({\$5{,}310}\text{.}\) How much money did he invest in each account?
    +
    Jenny invested a total of \({\$5{,}000}\) in two investments. Her savings account pays \({1.5\%}\) interest annually. A riskier stock investment lost \({3.5\%}\) at the end of the year. At the end of the year, Jenny’s total fell from \({\$5{,}000}\) to \({\$4{,}865}\text{.}\) How much money did she invest in each account?
    -
    Answer 1.
    \(x+y = 5500\)
    Answer 2.
    \(0.02x-0.04y = -190\)
    Answer 3.
    \(\$1{,}500\)
    Answer 4.
    \(\$4{,}000\)
    +
    Answer 1.
    \(x+y = 5000\)
    Answer 2.
    \(0.015x-0.035y = -135\)
    Answer 3.
    \(\$2{,}000\)
    Answer 4.
    \(\$3{,}000\)
    -

    Challenge

    +

    Challenge

    37.
    -
    -
    +
    +
    -
    Find the value of \(b\) so that the system of equations has an infinite number of solutions.
    +
    Find the value of \(b\) so that the system of equations has an infinite number of solutions.
    \begin{equation*} -\left\{\begin{aligned} {-20x+35y} \amp = -5\\ -{4x-by} \amp = 1 \end{aligned}\right. +\left\{\begin{aligned} {-16x+12y} \amp = -8\\ +{4x-by} \amp = 2 \end{aligned}\right. \end{equation*}
    -
    Answer.
    \(7\)
    Explanation.
    To have an infinite numbers of solutions, the two equations must represeent the same line coinciding. So the second equation comes from multiplying or dividing all the terms in the first equation by a certain factor. In this case, if you divide the first equation’s terms by -5, the result is the second equation. So \(b\) must equal \({7}\text{.}\) +
    Answer.
    \(3\)
    Explanation.
    To have an infinite numbers of solutions, the two equations must represeent the same line coinciding. So the second equation comes from multiplying or dividing all the terms in the first equation by a certain factor. In this case, if you divide the first equation’s terms by -4, the result is the second equation. So \(b\) must equal \({3}\text{.}\)
    diff --git a/section-equations-and-inequalities-with-fractions.html b/section-equations-and-inequalities-with-fractions.html index ba20fb7e3..6b61e6aea 100644 --- a/section-equations-and-inequalities-with-fractions.html +++ b/section-equations-and-inequalities-with-fractions.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-equations-inequalities-and-solutions.html b/section-equations-inequalities-and-solutions.html index c6b12002f..fe07db162 100644 --- a/section-equations-inequalities-and-solutions.html +++ b/section-equations-inequalities-and-solutions.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-exploring-two-variable-data-and-rate-of-change.html b/section-exploring-two-variable-data-and-rate-of-change.html index 0a29dd3e2..a5fa146b1 100644 --- a/section-exploring-two-variable-data-and-rate-of-change.html +++ b/section-exploring-two-variable-data-and-rate-of-change.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-fractions-and-fraction-arithmetic.html b/section-fractions-and-fraction-arithmetic.html index 1af336518..c890ff6f2 100644 --- a/section-fractions-and-fraction-arithmetic.html +++ b/section-fractions-and-fraction-arithmetic.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -591,11 +605,11 @@

    Search Results:

    • Checkpoint A.2.3. A Fraction as Parts of a Whole.

    -
    -
    +
    +
    -
    To visualize the fraction \(\frac{14}{35}\text{,}\) you might cut a rectangle into equal parts, and then count up of them.
    -
    Explanation.
    You could cut a rectangle into \(35\) equal pieces, and then \(14\) of them would represent \(\frac{14}{35}\text{.}\) +
    To visualize the fraction \(\frac{14}{35}\text{,}\) you might cut a rectangle into equal parts, and then count up of them.
    +
    Explanation.
    You could cut a rectangle into \(35\) equal pieces, and then \(14\) of them would represent \(\frac{14}{35}\text{.}\)
    @@ -604,11 +618,11 @@

    Search Results:

    Checkpoint A.2.5. A Fraction on a Number Line.

    -
    -
    +
    +
    -
    In the given number line, what fraction is marked?
    a number line with -1, 0, and 1 marked; there are evenly spaced ticks, with eight ticks between -1 and 0, eight between 0 and 1, and so on; there is a dot marked at the fifth tick to the right from 0
    -
    Explanation.
    There are \(8\) subdivisions between \(0\) and \(1\text{,}\) and the mark is at the fifth subdivision. So the mark is \(\frac{5}{8}\) of the way from \(0\) to \(1\) and therefore represents the fraction \(\frac{5}{8}\text{.}\) +
    In the given number line, what fraction is marked?
    a number line with -1, 0, and 1 marked; there are evenly spaced ticks, with eight ticks between -1 and 0, eight between 0 and 1, and so on; there is a dot marked at the fifth tick to the right from 0
    +
    Explanation.
    There are \(8\) subdivisions between \(0\) and \(1\text{,}\) and the mark is at the fifth subdivision. So the mark is \(\frac{5}{8}\) of the way from \(0\) to \(1\) and therefore represents the fraction \(\frac{5}{8}\text{.}\)

    Division.

    Fractions can also be understood through division.
    @@ -618,11 +632,11 @@

    Search Results:

    Checkpoint A.2.7. Seeing a Fraction as Division Arithmetic.

    -
    -
    +
    +
    -
    The fraction \(\frac{21}{40}\) can be thought of as dividing the whole number into equal-sized parts.
    -
    Explanation.
    Since \(\frac{21}{40}\) means the same as \(21\div40\text{,}\) it can be thought of as dividing \(21\) into \(40\) equal parts.
    +
    The fraction \(\frac{21}{40}\) can be thought of as dividing the whole number into equal-sized parts.
    +
    Explanation.
    Since \(\frac{21}{40}\) means the same as \(21\div40\text{,}\) it can be thought of as dividing \(21\) into \(40\) equal parts.

    Subsection A.2.2 Equivalent Fractions @@ -648,43 +662,43 @@

    Search Results:

    Checkpoint A.2.11. Reducing Fractions.

    -
    -
    +
    +
    -
    Reduce these fractions into lowest terms.

    (a)

    -
    +
    Reduce these fractions into lowest terms.

    (a)

    +
    \(\dfrac{14}{42}=\)
    -
    Explanation.
    With \(\frac{14}{42}\text{,}\) we have \(\frac{2\cdot7}{2\cdot3\cdot7}\text{,}\) which reduces to \(\frac{1}{3}\text{.}\) -

    (b)

    -
    +
    Explanation.
    With \(\frac{14}{42}\text{,}\) we have \(\frac{2\cdot7}{2\cdot3\cdot7}\text{,}\) which reduces to \(\frac{1}{3}\text{.}\) +

    (b)

    +
    \(\dfrac{8}{30}=\)
    -
    Explanation.
    With \(\frac{8}{30}\text{,}\) we have \(\frac{2\cdot2\cdot2}{2\cdot3\cdot5}\text{,}\) which reduces to \(\frac{4}{15}\text{.}\) -

    (c)

    -
    +
    Explanation.
    With \(\frac{8}{30}\text{,}\) we have \(\frac{2\cdot2\cdot2}{2\cdot3\cdot5}\text{,}\) which reduces to \(\frac{4}{15}\text{.}\) +

    (c)

    +
    \(\dfrac{70}{90}=\)
    -
    Explanation.
    With \(\frac{70}{90}\text{,}\) we have \(\frac{7\cdot10}{9\cdot10}\text{,}\) which reduces to \(\frac{7}{9}\text{.}\) +
    Explanation.
    With \(\frac{70}{90}\text{,}\) we have \(\frac{7\cdot10}{9\cdot10}\text{,}\) which reduces to \(\frac{7}{9}\text{.}\)
    Sometimes it is useful to do the opposite of reducing a fraction, and build up the fraction to use larger numbers.

    Checkpoint A.2.12. Building Up a Fraction.

    -
    -
    +
    +
    -
    Sayid scored \(\frac{21}{25}\) on a recent exam. Build up this fraction so that the denominator is \(100\text{,}\) so that Sayid can understand what percent score he earned.
    -
    Explanation.
    -
    To change the denominator from \(25\) to \(100\text{,}\) it needs to be multiplied by \(4\text{.}\) So we calculate
    +
    Sayid scored \(\frac{21}{25}\) on a recent exam. Build up this fraction so that the denominator is \(100\text{,}\) so that Sayid can understand what percent score he earned.
    +
    Explanation.
    +
    To change the denominator from \(25\) to \(100\text{,}\) it needs to be multiplied by \(4\text{.}\) So we calculate
    \begin{equation*} \begin{aligned} \frac{21}{25}\amp=\frac{21\cdot4}{25\cdot4}\\ \amp=\frac{84}{100} \end{aligned} \end{equation*} -
    So the fraction \(\frac{21}{25}\) is equivalent to \(\frac{84}{100}\text{.}\) (This means Sayid scored an \(84\%\text{.}\))
    +
    So the fraction \(\frac{21}{25}\) is equivalent to \(\frac{84}{100}\text{.}\) (This means Sayid scored an \(84\%\text{.}\))

    @@ -738,29 +752,29 @@

    Search Results:

    Checkpoint A.2.17. Fraction Multiplication.

    -
    -
    +
    +
    -
    Simplify these fraction products.

    (a)

    -
    +
    Simplify these fraction products.

    (a)

    +
    \(\dfrac{1}{3}\cdot\dfrac{10}{7}=\)
    -
    Explanation.
    Multiplying numerators gives \(10\text{,}\) and multiplying denominators gives \(21\text{.}\) The answer is \(\frac{10}{21}\text{.}\) -

    (b)

    -
    +
    Explanation.
    Multiplying numerators gives \(10\text{,}\) and multiplying denominators gives \(21\text{.}\) The answer is \(\frac{10}{21}\text{.}\) +

    (b)

    +
    \(\dfrac{12}{3}\cdot\dfrac{15}{3}=\)
    -
    Explanation.
    Before we multiply fractions, note that \(\frac{12}{3}\) reduces to \(4\text{,}\) and \(\frac{15}{3}\) reduces to \(5\text{.}\) So we just have \(4\cdot5=20\text{.}\) -

    (c)

    -
    +
    Explanation.
    Before we multiply fractions, note that \(\frac{12}{3}\) reduces to \(4\text{,}\) and \(\frac{15}{3}\) reduces to \(5\text{.}\) So we just have \(4\cdot5=20\text{.}\) +

    (c)

    +
    \(-\dfrac{14}{5}\cdot\dfrac{2}{3}=\)
    -
    Explanation.
    Multiplying numerators gives \(28\text{,}\) and multiplying denominators gives \(15\text{.}\) The result should be negative, so the answer is \(-\frac{28}{15}\text{.}\) -

    (d)

    -
    +
    Explanation.
    Multiplying numerators gives \(28\text{,}\) and multiplying denominators gives \(15\text{.}\) The result should be negative, so the answer is \(-\frac{28}{15}\text{.}\) +

    (d)

    +
    \(\dfrac{70}{27}\cdot\dfrac{12}{-20}=\)
    -
    Explanation.
    Before we multiply fractions, note that \(\frac{12}{-20}\) reduces to \(\frac{-3}{5}\text{.}\) So we have \(\frac{70}{27}\cdot\frac{-3}{5}\text{.}\) Both the numerator of the first fraction and denominator of the second fraction are divisible by \(5\text{,}\) so it helps to reduce both fractions accordingly and get \(\frac{14}{27}\cdot\frac{-3}{1}\text{.}\) Both the denominator of the first fraction and numerator of the second fraction are divisible by \(3\text{,}\) so it helps to reduce both fractions accordingly and get \(\frac{14}{9}\cdot\frac{-1}{1}\text{.}\) Now we are just multiplying \(\frac{14}{9}\) by \(-1\text{,}\) so the result is \(\frac{-14}{9}\text{.}\) +
    Explanation.
    Before we multiply fractions, note that \(\frac{12}{-20}\) reduces to \(\frac{-3}{5}\text{.}\) So we have \(\frac{70}{27}\cdot\frac{-3}{5}\text{.}\) Both the numerator of the first fraction and denominator of the second fraction are divisible by \(5\text{,}\) so it helps to reduce both fractions accordingly and get \(\frac{14}{27}\cdot\frac{-3}{1}\text{.}\) Both the denominator of the first fraction and numerator of the second fraction are divisible by \(3\text{,}\) so it helps to reduce both fractions accordingly and get \(\frac{14}{9}\cdot\frac{-1}{1}\text{.}\) Now we are just multiplying \(\frac{14}{9}\) by \(-1\text{,}\) so the result is \(\frac{-14}{9}\text{.}\)

    @@ -821,38 +835,38 @@

    Search Results:

    Checkpoint A.2.20. Fraction Division.

    -
    -
    +
    +
    -
    Simplify these fraction division expressions.

    (a)

    -
    +
    Simplify these fraction division expressions.

    (a)

    +
    \(\dfrac{1}{3}\div\dfrac{10}{7}=\)
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] \frac{1}{3}\div\frac{10}{7}\amp=\frac{1}{3}\cdot\frac{7}{10}\\ \amp=\frac{7}{30}\\ \ -\end{aligned}\)

    (b)

    -
    +\end{aligned}\)

    (b)

    +
    \(\dfrac{12}{5}\div5=\)
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] \frac{12}{5}\div5\amp=\frac{12}{5}\cdot\frac{1}{5}\\ \amp=\frac{12}{25}\\ \ -\end{aligned}\)

    (c)

    -
    +\end{aligned}\)

    (c)

    +
    \(-14\div\dfrac{3}{2}=\)
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] -14\div\frac{3}{2}\amp=-14\cdot\frac{2}{3}\\ \amp=-\frac{14}{1}\cdot\frac{2}{3}\\ \amp=-\frac{28}{3} -\end{aligned}\)

    (d)

    -
    +\end{aligned}\)

    (d)

    +
    \(\dfrac{70}{9}\div\dfrac{11}{-20}=\)
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] \frac{70}{9}\div\frac{11}{-20}\amp=-\frac{70}{9}\cdot\frac{20}{11}\\ \amp=-\frac{1400}{99}\\ \ @@ -893,19 +907,19 @@

    Search Results:

    Checkpoint A.2.22. Fraction Addition and Subtraction.

    -
    -
    +
    +
    -
    Add or subtract these fractions.

    (a)

    -
    +
    Add or subtract these fractions.

    (a)

    +
    \(\frac{1}{3}+\frac{10}{3}=\)
    -
    Explanation.
    Since the denominators are both \(3\text{,}\) we can add the numerators: \(1+10=11\text{.}\) The answer is \(\frac{11}{3}\text{.}\) -

    (b)

    -
    +
    Explanation.
    Since the denominators are both \(3\text{,}\) we can add the numerators: \(1+10=11\text{.}\) The answer is \(\frac{11}{3}\text{.}\) +

    (b)

    +
    \(\frac{13}{6}-\frac{5}{6}=\)
    -
    Explanation.
    Since the denominators are both \(6\text{,}\) we can subtract the numerators: \(13-5=8\text{.}\) The answer is \(\frac{8}{6}\text{,}\) but that reduces to \(\frac{4}{3}\text{.}\) +
    Explanation.
    Since the denominators are both \(6\text{,}\) we can subtract the numerators: \(13-5=8\text{.}\) The answer is \(\frac{8}{6}\text{,}\) but that reduces to \(\frac{4}{3}\text{.}\)
    Whenever we’d like to combine fractional amounts that don’t represent the same number of parts of a whole (that is, when the denominators are different), finding sums and differences is more complicated.
    @@ -955,13 +969,13 @@

    Search Results:

    Checkpoint A.2.26. Using Some Flour.

    -
    -
    +
    +
    -
    A chef had \(\frac{2}{3}\) cups of flour and needed to use \(\frac{1}{8}\) cup to thicken a sauce. How much flour is left? +
    A chef had \(\frac{2}{3}\) cups of flour and needed to use \(\frac{1}{8}\) cup to thicken a sauce. How much flour is left?
    -
    Explanation.
    -
    We need to compute \(\frac{2}{3} - \frac{1}{8}\text{.}\) The denominators are \(3\) and \(8\text{.}\) One common denominator is \(24\text{,}\) so we move to rewrite each fraction using \(24\) as the denominator:
    +
    Explanation.
    +
    We need to compute \(\frac{2}{3} - \frac{1}{8}\text{.}\) The denominators are \(3\) and \(8\text{.}\) One common denominator is \(24\text{,}\) so we move to rewrite each fraction using \(24\) as the denominator:
    \begin{equation*} \begin{aligned} \frac{2}{3} - \frac{1}{8}\amp=\frac{2\cdot8}{3\cdot8} - \frac{1\cdot3}{8\cdot3}\\ @@ -969,7 +983,7 @@

    Search Results:

    \amp=\frac{13}{24} \end{aligned} \end{equation*} -
    The numerical result is \(\frac{13}{24}\text{,}\) but a pure number does not answer this question. The amount of flour remaining is \(\frac{13}{24}\) cups.
    +
    The numerical result is \(\frac{13}{24}\text{,}\) but a pure number does not answer this question. The amount of flour remaining is \(\frac{13}{24}\) cups.

    @@ -1016,56 +1030,56 @@

    Search Results:

    Review and Warmup.

    1.

    -
    -
    +
    +
    -
    Which letter is \({-{\frac{33}{4}}}\) on the number line?
    -
    Answer.
    \(5\)
    +
    Which letter is \({-{\frac{25}{4}}}\) on the number line?
    +
    Answer.
    \(5\)

    2.

    -
    -
    +
    +
    -
    Which letter is \({{\frac{19}{4}}}\)m on the number line?
    -
    Answer.
    \(\text{D}\)
    +
    Which letter is \({{\frac{23}{4}}}\)m on the number line?
    +
    Answer.
    \(\text{D}\)

    3.

    -
    -
    +
    +
    -
    The dot in the graph can be represented by what fraction?
    +
    The dot in the graph can be represented by what fraction?
    -
    Answer.
    \({\frac{1}{2}}\)
    Explanation.
    On the number line, the segment between \(0\) and \(1\) is cut into \(4\) pieces. The blue dot is marked at the second tick from \(0\text{,}\) so it can be represented as \(\frac{2}{4}\text{,}\) which reduces to \({{\frac{1}{2}}}\text{.}\) +
    Answer.
    \({\frac{1}{2}}\)
    Explanation.
    On the number line, the segment between \(0\) and \(1\) is cut into \(6\) pieces. The blue dot is marked at the third tick from \(0\text{,}\) so it can be represented as \(\frac{3}{6}\text{,}\) which reduces to \({{\frac{1}{2}}}\text{.}\)

    4.

    -
    -
    +
    +
    -
    The dot in the graph can be represented by what fraction?
    +
    The dot in the graph can be represented by what fraction?
    -
    Answer.
    \(-{\frac{1}{2}}\)
    Explanation.
    On the number line, the segment between \(-1\) and \(0\) is cut into \(6\) pieces. The blue dot is marked at the third tick left from \(0\text{,}\) so it can be represented as \(-{\frac{3}{6}}\text{,}\) which reduces to \({-{\frac{1}{2}}}\text{.}\) +
    Answer.
    \(-{\frac{1}{4}}\)
    Explanation.
    On the number line, the segment between \(-1\) and \(0\) is cut into \(8\) pieces. The blue dot is marked at the second tick left from \(0\text{,}\) so it can be represented as \(-{\frac{2}{8}}\text{,}\) which reduces to \({-{\frac{1}{4}}}\text{.}\)

    5.

    -
    -
    +
    +
    -
    The dot in the graph can be represented by what fraction?
    +
    The dot in the graph can be represented by what fraction?
    -
    Answer.
    \(6 {\textstyle\frac{2}{3}}\)
    Explanation.
    On the number line, the segment between \(6\) and \(7\) is cut into \(6\) pieces. The blue dot is marked at the 4th tick from \(6\text{,}\) so it can be represented as \(6{\frac{4}{6}}\text{,}\) which reduces to \({6 {\textstyle\frac{2}{3}}}\text{.}\) +
    Answer.
    \(9 {\textstyle\frac{1}{2}}\)
    Explanation.
    On the number line, the segment between \(9\) and \(10\) is cut into \(6\) pieces. The blue dot is marked at the third tick from \(9\text{,}\) so it can be represented as \(9{\frac{3}{6}}\text{,}\) which reduces to \({9 {\textstyle\frac{1}{2}}}\text{.}\)

    6.

    -
    -
    +
    +
    -
    The dot in the graph can be represented by what fraction?
    +
    The dot in the graph can be represented by what fraction?
    -
    Answer.
    \(-5 {\textstyle\frac{1}{2}}\)
    Explanation.
    On the number line, the segment between \(-6\) and \(-5\) is cut into \(10\) pieces. The blue dot is marked at the 5th tick to the left of \(-5\text{,}\) so it can be represented as \(-5{\frac{5}{10}}={-5 {\textstyle\frac{1}{2}}}\text{.}\) +
    Answer.
    \(-3 {\textstyle\frac{1}{2}}\)
    Explanation.
    On the number line, the segment between \(-4\) and \(-3\) is cut into \(10\) pieces. The blue dot is marked at the 5th tick to the left of \(-3\text{,}\) so it can be represented as \(-3{\frac{5}{10}}={-3 {\textstyle\frac{1}{2}}}\text{.}\)
    @@ -1076,124 +1090,122 @@

    Review and Warmup.

    Reducing Fractions.

    7.

    -
    -
    +
    +
    -
    Reduce the fraction \(\displaystyle{ \frac{3}{27} }\text{.}\) +
    Reduce the fraction \(\displaystyle{ \frac{2}{8} }\text{.}\)
    -
    Answer.
    \({\frac{1}{9}}\)
    Explanation.
    -
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{3}{27} }\text{.}\) -
    Method 1:We find a number that divides into both the numerator \(3\text{,}\) and the denominator \(27\text{.}\) -
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) -
    In this case, \(3\) goes into both the numerator and denominator of \(\displaystyle{ \frac{3}{27} }\text{.}\) We divide \(3\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned} -\frac{3}{27} \amp = \frac{3 \div 3}{27\div 3}\\ -\amp = \frac{1}{9}\end{aligned} }\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers.
    In this case, \(\displaystyle{\frac{1}{9}}\) is the final answer.
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] -\frac{3}{27} \amp = \frac{1 \cdot 3}{3 \cdot 3 \cdot 3} \\ -\amp =\frac{1}{3 \cdot 3}\\ -\amp = \frac{1}{9} +
    Answer.
    \({\frac{1}{4}}\)
    Explanation.
    +
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{2}{8} }\text{.}\) +
    Method 1:We find a number that divides into both the numerator \(2\text{,}\) and the denominator \(8\text{.}\) +
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) +
    In this case, \(2\) goes into both the numerator and denominator of \(\displaystyle{ \frac{2}{8} }\text{.}\) We divide \(2\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned} +\frac{2}{8} \amp = \frac{2 \div 2}{8\div 2}\\ +\amp = \frac{1}{4}\end{aligned} }\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers.
    In this case, \(\displaystyle{\frac{1}{4}}\) is the final answer.
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] +\frac{2}{8} \amp = \frac{1 \cdot 2}{2 \cdot 2 \cdot 2} \\ +\amp =\frac{1}{2 \cdot 2}\\ +\amp = \frac{1}{4} \end{aligned} -}\)
    Notice that when the numerator’s only prime factor, \(3\text{,}\) is canceled, we have to leave a \(1\) in the numerator.
    +}\)
    Notice that when the numerator’s only prime factor, \(2\text{,}\) is canceled, we have to leave a \(1\) in the numerator.

    8.

    -
    -
    +
    +
    -
    Reduce the fraction \(\displaystyle{ \frac{35}{63} }\text{.}\) +
    Reduce the fraction \(\displaystyle{ \frac{10}{45} }\text{.}\)
    -
    Answer.
    \({\frac{5}{9}}\)
    Explanation.
    -
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{35}{63} }\text{.}\) -
    Method 1:We find a number that divides into both the numerator \(35\text{,}\) and the denominator \(63\text{.}\) -
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) -
    In this case, \(7\) divides into both the numerator and denominator of \(\displaystyle{ \frac{35}{63} }\text{.}\) We divide \(7\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{35}{63} \amp = \frac{35 \div 7}{63\div 7}\\ \amp = \frac{5}{9} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers.
    In this case, \(\displaystyle{\frac{5}{9}}\) is the final answer.
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] -\frac{35}{63} \amp = \frac{5 \cdot 7}{3 \cdot 3 \cdot 7} \\ -\amp =\frac{5}{3 \cdot 3}\\ -\amp = \frac{5}{9} +
    Answer.
    \({\frac{2}{9}}\)
    Explanation.
    +
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{10}{45} }\text{.}\) +
    Method 1:We find a number that divides into both the numerator \(10\text{,}\) and the denominator \(45\text{.}\) +
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) +
    In this case, \(5\) divides into both the numerator and denominator of \(\displaystyle{ \frac{10}{45} }\text{.}\) We divide \(5\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{10}{45} \amp = \frac{10 \div 5}{45\div 5}\\ \amp = \frac{2}{9} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers.
    In this case, \(\displaystyle{\frac{2}{9}}\) is the final answer.
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] +\frac{10}{45} \amp = \frac{2 \cdot 5}{3 \cdot 3 \cdot 5} \\ +\amp =\frac{2}{3 \cdot 3}\\ +\amp = \frac{2}{9} \end{aligned} }\)

    9.

    -
    -
    +
    +
    -
    Reduce the fraction \(\displaystyle{ \frac{10}{80} }\text{.}\) +
    Reduce the fraction \(\displaystyle{ \frac{28}{98} }\text{.}\)
    -
    Answer.
    \({\frac{1}{8}}\)
    Explanation.
    -
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{10}{80} }\text{.}\) -
    Method 1:We find a number that divides into both the numerator \(10\text{,}\) and the denominator \(80\text{.}\) -
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) -
    In this case, \(2\) divides into both the numerator and denominator of \(\displaystyle{ \frac{10}{80} }\text{.}\) We divide \(2\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{10}{80} \amp = \frac{10 \div 2}{80\div 2}\\ \amp = \frac{5}{40} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, \(5\) again goes into both the numerator and denominator for \(\displaystyle{ \frac{5}{40} }\text{.}\) We divide \(5\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{5}{40} \amp = \frac{5 \div 5}{40\div 5}\\ \amp = \frac{1}{8} \end{aligned}}\)
    Finally, we have: \(\displaystyle{ \frac{10}{80} = \frac{1}{8} }\) -
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] -\frac{10}{80} \amp = \frac{1 \cdot 2 \cdot 5}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 5} \\ -\amp =\frac{1}{2 \cdot 2 \cdot 2}\\ -\amp = \frac{1}{8} +
    Answer.
    \({\frac{2}{7}}\)
    Explanation.
    +
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{28}{98} }\text{.}\) +
    Method 1:We find a number that divides into both the numerator \(28\text{,}\) and the denominator \(98\text{.}\) +
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) +
    In this case, \(2\) divides into both the numerator and denominator of \(\displaystyle{ \frac{28}{98} }\text{.}\) We divide \(2\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{28}{98} \amp = \frac{28 \div 2}{98\div 2}\\ \amp = \frac{14}{49} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers go into both numbers. In this case, \(7\) again goes into both the numerator and denominator for \(\displaystyle{ \frac{14}{49} }\text{.}\) We divide \(7\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{14}{49} \amp = \frac{14 \div 7}{49\div 7}\\ \amp = \frac{2}{7} \end{aligned}}\)
    Finally, we have: \(\displaystyle{ \frac{28}{98} = \frac{2}{7} }\) +
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] +\frac{28}{98} \amp = \frac{2 \cdot 2 \cdot 7}{2 \cdot 7 \cdot 7} \\ +\amp = \frac{2}{7} \end{aligned} -}\)
    Notice that when all prime factors in the numerator are canceled, we have to add a 1 in the numerator.
    +}\)

    10.

    -
    -
    +
    +
    -
    Reduce the fraction \(\displaystyle{ {{\frac{22}{45}}} }\text{.}\) +
    Reduce the fraction \(\displaystyle{ {{\frac{28}{33}}} }\text{.}\)
    -
    Answer.
    \({\frac{22}{45}}\)
    Explanation.
    -
    Method 1:
    We find a number that divides into both the numerator \(22\text{,}\) and the denominator \(45\text{.}\) -
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\text{,}\) until the prime number is greater than or equal to \(22\text{.}\) No prime number goes into both \(22\) and \(45\text{,}\) so the fraction \({{\frac{22}{45}}}\) is not reducible.
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] -\frac{22}{45} \amp = \frac{2 \cdot 11}{3 \cdot 3 \cdot 5} +
    Answer.
    \({\frac{28}{33}}\)
    Explanation.
    +
    Method 1:
    We find a number that divides into both the numerator \(28\text{,}\) and the denominator \(33\text{.}\) +
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\text{,}\) until the prime number is greater than or equal to \(28\text{.}\) No prime number goes into both \(28\) and \(33\text{,}\) so the fraction \({{\frac{28}{33}}}\) is not reducible.
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] +\frac{28}{33} \amp = \frac{2 \cdot 2 \cdot 7}{3 \cdot 11} \end{aligned} -}\)
    Notice that no numbers can be canceled out, so the fraction \({{\frac{22}{45}}}\) is not reducible.
    +}\)
    Notice that no numbers can be canceled out, so the fraction \({{\frac{28}{33}}}\) is not reducible.

    11.

    -
    -
    +
    +
    -
    Reduce the fraction \(\displaystyle{ \frac{126}{98} }\text{.}\) +
    Reduce the fraction \(\displaystyle{ \frac{100}{70} }\text{.}\)
    -
    Answer.
    \({\frac{9}{7}}\)
    Explanation.
    -
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{126}{98} }\text{.}\) -
    Method 1:We find a number that divides into both the numerator \(126\text{,}\) and the denominator \(98\text{.}\) -
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) -
    In this case, \(2\) divides into both the numerator and denominator of \(\displaystyle{ \frac{126}{98} }\text{.}\) We divide \(2\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{126}{98} \amp = \frac{126 \div 2}{98\div 2}\\ \amp = \frac{63}{49} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, \(7\) divides into both the numerator and denominator for \(\displaystyle{ \frac{63}{49} }\text{.}\) We divide \(7\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{63}{49} \amp = \frac{63 \div 7}{49\div 7}\\\amp = \frac{9}{7} \end{aligned}}\)
    Finally, we have: \(\displaystyle{ \frac{126}{98} = \frac{9}{7} }\) -
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] -\frac{126}{98} \amp = \frac{2 \cdot 3 \cdot 3 \cdot 7}{2 \cdot 7 \cdot 7} \\ -\amp =\frac{3 \cdot 3}{7}\\ -\amp = \frac{9}{7} +
    Answer.
    \({\frac{10}{7}}\)
    Explanation.
    +
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{100}{70} }\text{.}\) +
    Method 1:We find a number that divides into both the numerator \(100\text{,}\) and the denominator \(70\text{.}\) +
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) +
    In this case, \(2\) divides into both the numerator and denominator of \(\displaystyle{ \frac{100}{70} }\text{.}\) We divide \(2\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{100}{70} \amp = \frac{100 \div 2}{70\div 2}\\ \amp = \frac{50}{35} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, \(5\) divides into both the numerator and denominator for \(\displaystyle{ \frac{50}{35} }\text{.}\) We divide \(5\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{50}{35} \amp = \frac{50 \div 5}{35\div 5}\\\amp = \frac{10}{7} \end{aligned}}\)
    Finally, we have: \(\displaystyle{ \frac{100}{70} = \frac{10}{7} }\) +
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] +\frac{100}{70} \amp = \frac{2 \cdot 2 \cdot 5 \cdot 5}{2 \cdot 5 \cdot 7} \\ +\amp =\frac{2 \cdot 5}{7}\\ +\amp = \frac{10}{7} \end{aligned} }\)

    12.

    -
    -
    +
    +
    -
    Reduce the fraction \(\displaystyle{ \frac{90}{15} }\text{.}\) +
    Reduce the fraction \(\displaystyle{ \frac{42}{6} }\text{.}\)
    -
    Answer.
    \(6\)
    Explanation.
    -
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{90}{15} }\text{.}\) -
    Method 1:We find a number that divides into both the numerator \(90\text{,}\) and the denominator \(15\text{.}\) -
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) -
    In this case, \(3\) divides into both the numerator and denominator of \(\displaystyle{ \frac{90}{15} }\text{.}\) We divide \(3\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{90}{15} \amp = \frac{90 \div 3}{15\div 3}\\ \amp = \frac{30}{5} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, \(5\) divides into both the numerator and denominator for \(\displaystyle{ \frac{30}{5} }\text{.}\) We divide \(5\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{30}{5} \amp = \frac{30 \div 5}{5\div 5}\\\amp = \frac{6}{1}\\\amp = 6\end{aligned} }\)
    Finally, we have: \(\displaystyle{ \frac{90}{15} = 6 }\) -
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] -\frac{90}{15} \amp = \frac{2 \cdot 3 \cdot 3 \cdot 5}{1 \cdot 3 \cdot 5} \\ -\amp =\frac{2 \cdot 3}{1}\\ -\amp = \frac{6}{1} \\ -\amp = 6 +
    Answer.
    \(7\)
    Explanation.
    +
    There are at least two methods to reduce the fraction \(\displaystyle{ \frac{42}{6} }\text{.}\) +
    Method 1:We find a number that divides into both the numerator \(42\text{,}\) and the denominator \(6\text{.}\) +
    Check the first few prime numbers one by one: \(2, 3, 5, 7, \ldots\) +
    In this case, \(2\) divides into both the numerator and denominator of \(\displaystyle{ \frac{42}{6} }\text{.}\) We divide \(2\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{42}{6} \amp = \frac{42 \div 2}{6\div 2}\\ \amp = \frac{21}{3} \end{aligned}}\)
    Next, check again whether any prime number divides into both the numerator and denominator. We need to keep trying until no prime numbers divide into both numbers. In this problem, \(3\) divides into both the numerator and denominator for \(\displaystyle{ \frac{21}{3} }\text{.}\) We divide \(3\) into both numbers, and we have:
    \(\displaystyle{ \begin{aligned}\frac{21}{3} \amp = \frac{21 \div 3}{3\div 3}\\\amp = \frac{7}{1}\\\amp = 7\end{aligned} }\)
    Finally, we have: \(\displaystyle{ \frac{42}{6} = 7 }\) +
    Method 2:
    A second method to reduce fraction is to prime factor both the numerator and the denominator, and then cancel out factors in pairs: one from the numerator and one from the denominator.
    \(\displaystyle{\begin{aligned}[t] +\frac{42}{6} \amp = \frac{2 \cdot 3 \cdot 7}{1 \cdot 2 \cdot 3} \\ +\amp = \frac{7}{1} \\ +\amp = 7 \end{aligned} -}\)
    Note that if the denominator is \(1\text{,}\) the result is simply an integer.
    +}\)
    Note that if the denominator is \(1\text{,}\) the result is simply an integer.
    @@ -1204,54 +1216,54 @@

    Reducing Fractions.

    Building Fractions.

    13.

    -
    -
    +
    +
    -
    Find an equivalent fraction to \(\frac{1}{5}\) with denominator \(25\text{.}\) +
    Find an equivalent fraction to \(\frac{6}{7}\) with denominator \(28\text{.}\)
    -
    Answer.
    \(\frac{5}{25}\)
    Explanation.
    -
    From \(5\) to \(25\text{,}\) we multiplied by \(25\div5=5\text{.}\) -
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{1}{5}=\frac{1\cdot5}{5\cdot5} = \frac{5}{25} }\)
    +
    Answer.
    \(\frac{24}{28}\)
    Explanation.
    +
    From \(7\) to \(28\text{,}\) we multiplied by \(28\div7=4\text{.}\) +
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{6}{7}=\frac{6\cdot4}{7\cdot4} = \frac{24}{28} }\)

    14.

    -
    -
    +
    +
    -
    Find an equivalent fraction to \(\frac{5}{6}\) with denominator \(12\text{.}\) +
    Find an equivalent fraction to \(\frac{5}{8}\) with denominator \(16\text{.}\)
    -
    Answer.
    \(\frac{10}{12}\)
    Explanation.
    -
    From \(6\) to \(12\text{,}\) we multiplied by \(12\div6=2\text{.}\) -
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{5}{6}=\frac{5\cdot2}{6\cdot2} = \frac{10}{12} }\)
    +
    Answer.
    \(\frac{10}{16}\)
    Explanation.
    +
    From \(8\) to \(16\text{,}\) we multiplied by \(16\div8=2\text{.}\) +
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{5}{8}=\frac{5\cdot2}{8\cdot2} = \frac{10}{16} }\)

    15.

    -
    -
    +
    +
    -
    Find an equivalent fraction to \(\frac{9}{14}\) with denominator \(70\text{.}\) +
    Find an equivalent fraction to \(\frac{5}{18}\) with denominator \(54\text{.}\)
    -
    Answer.
    \(\frac{45}{70}\)
    Explanation.
    -
    From \(9\) to \(45\text{,}\) we multiplied by \(45\div9=5\text{.}\) -
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{9}{14}=\frac{9\cdot5}{14\cdot5} = \frac{45}{70} }\)
    +
    Answer.
    \(\frac{15}{54}\)
    Explanation.
    +
    From \(5\) to \(15\text{,}\) we multiplied by \(15\div5=3\text{.}\) +
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{5}{18}=\frac{5\cdot3}{18\cdot3} = \frac{15}{54} }\)

    16.

    -
    -
    +
    +
    -
    Find an equivalent fraction to \(\frac{1}{16}\) with denominator \(64\text{.}\) +
    Find an equivalent fraction to \(\frac{1}{2}\) with denominator \(10\text{.}\)
    -
    Answer.
    \(\frac{4}{64}\)
    Explanation.
    -
    From \(1\) to \(4\text{,}\) we multiplied by \(4\div1=4\text{.}\) -
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{1}{16}=\frac{1\cdot4}{16\cdot4} = \frac{4}{64} }\)
    +
    Answer.
    \(\frac{5}{10}\)
    Explanation.
    +
    From \(1\) to \(5\text{,}\) we multiplied by \(5\div1=5\text{.}\) +
    For a fraction, when we multiply the same number in both the numerator and denominator, the fraction’s value doesn’t change. We have:
    \(\displaystyle{ \frac{1}{2}=\frac{1\cdot5}{2\cdot5} = \frac{5}{10} }\)
    @@ -1262,455 +1274,455 @@

    Building Fractions.

    Multiplying/Dividing Fractions.

    17.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ \frac{2}{7} \cdot \frac{2}{9} }\) +
    Multiply: \(\displaystyle{ \frac{2}{7} \cdot \frac{4}{9} }\)
    -
    Answer.
    \({\frac{4}{63}}\)
    Explanation.
    -
    To multiply two fractions, simply multiply both numerators and both denominators. We have:
    \(\displaystyle{ +
    Answer.
    \({\frac{8}{63}}\)
    Explanation.
    +
    To multiply two fractions, simply multiply both numerators and both denominators. We have:
    \(\displaystyle{ \begin{aligned}[t] -\frac{2}{7} \cdot \frac{2}{9} -\amp = \frac{2 \cdot 2}{7 \cdot 9} \\ -\amp = \frac{4}{63} +\frac{2}{7} \cdot \frac{4}{9} +\amp = \frac{2 \cdot 4}{7 \cdot 9} \\ +\amp = \frac{8}{63} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{\frac{4}{63}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{8}{63}}\text{.}\)

    18.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ \frac{1}{8} \cdot \frac{5}{6} }\) +
    Multiply: \(\displaystyle{ \frac{2}{7} \cdot \frac{4}{9} }\)
    -
    Answer.
    \({\frac{5}{48}}\)
    Explanation.
    -
    To multiply two fractions, simply multiply both numerators and both denominators. We have:
    \(\displaystyle{ +
    Answer.
    \({\frac{8}{63}}\)
    Explanation.
    +
    To multiply two fractions, simply multiply both numerators and both denominators. We have:
    \(\displaystyle{ \begin{aligned}[t] -\frac{1}{8} \cdot \frac{5}{6} -\amp = \frac{1 \cdot 5}{8 \cdot 6} \\ -\amp = \frac{5}{48} +\frac{2}{7} \cdot \frac{4}{9} +\amp = \frac{2 \cdot 4}{7 \cdot 9} \\ +\amp = \frac{8}{63} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{\frac{5}{48}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{8}{63}}\text{.}\)

    19.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ \frac{3}{11} \cdot \frac{4}{3} }\) +
    Multiply: \(\displaystyle{ \frac{6}{11} \cdot \frac{3}{14} }\)
    -
    Answer.
    \({\frac{4}{11}}\)
    Explanation.
    -
    To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it’s easier to reduce fractions before multiplying:
    \(\displaystyle{ +
    Answer.
    \({\frac{9}{77}}\)
    Explanation.
    +
    To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it’s easier to reduce fractions before multiplying:
    \(\displaystyle{ \begin{aligned}[t] -\frac{3}{11} \cdot \frac{4}{3} -\amp = \frac{3 \cdot 4}{11 \cdot 3} \\ -\amp = \frac{1 \cdot 4}{11 \cdot 1} \\ -\amp = \frac{4}{11} +\frac{6}{11} \cdot \frac{3}{14} +\amp = \frac{6 \cdot 3}{11 \cdot 14} \\ +\amp = \frac{3 \cdot 3}{11 \cdot 7} \\ +\amp = \frac{9}{77} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{\frac{4}{11}}\text{.}\) -
    Note that we divided the first numerator and the second denominator by \(3\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{9}{77}}\text{.}\) +
    Note that we divided the first numerator and the second denominator by \(2\text{.}\)

    20.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ \frac{5}{3} \cdot \frac{11}{10} }\) +
    Multiply: \(\displaystyle{ \frac{8}{11} \cdot \frac{15}{2} }\)
    -
    Answer.
    \({\frac{11}{6}}\)
    Explanation.
    -
    To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it’s easier to reduce fractions before multiplying:
    \(\displaystyle{ +
    Answer.
    \({\frac{60}{11}}\)
    Explanation.
    +
    To multiply two fractions, usually we simply multiply both numerators and both denominators. However, it’s easier to reduce fractions before multiplying:
    \(\displaystyle{ \begin{aligned}[t] -\frac{5}{3} \cdot \frac{11}{10} -\amp = \frac{5 \cdot 11}{3 \cdot 10} \\ -\amp = \frac{1 \cdot 11}{3 \cdot 2} \\ -\amp = \frac{11}{6} +\frac{8}{11} \cdot \frac{15}{2} +\amp = \frac{8 \cdot 15}{11 \cdot 2} \\ +\amp = \frac{4 \cdot 15}{11 \cdot 1} \\ +\amp = \frac{60}{11} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{\frac{11}{6}}\text{.}\) -
    Note that we divided the first numerator and the second denominator by \(5\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{60}{11}}\text{.}\) +
    Note that we divided the first numerator and the second denominator by \(2\text{.}\)

    21.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{4\cdot \frac{2}{5} }\) +
    Multiply: \(\displaystyle{8\cdot \frac{3}{5} }\)
    -
    Answer.
    \({\frac{8}{5}}\)
    Explanation.
    -
    To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a “1” as the denominator:
    \(\displaystyle{4 = \frac{4}{1} }\)
    Next, simply multiply both numerators and both denominators:
    \(\displaystyle{\begin{aligned}[t] -4\cdot \frac{2}{5} -\amp = \frac{4}{1} \cdot \frac{2}{5} \\ -\amp = \frac{4\cdot2}{1 \cdot 5} \\ -\amp = \frac{8}{5} +
    Answer.
    \({\frac{24}{5}}\)
    Explanation.
    +
    To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a “1” as the denominator:
    \(\displaystyle{8 = \frac{8}{1} }\)
    Next, simply multiply both numerators and both denominators:
    \(\displaystyle{\begin{aligned}[t] +8\cdot \frac{3}{5} +\amp = \frac{8}{1} \cdot \frac{3}{5} \\ +\amp = \frac{8\cdot3}{1 \cdot 5} \\ +\amp = \frac{24}{5} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{\frac{8}{5}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{24}{5}}\text{.}\)

    22.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{7\cdot \frac{3}{4} }\) +
    Multiply: \(\displaystyle{5\cdot \frac{4}{9} }\)
    -
    Answer.
    \({\frac{21}{4}}\)
    Explanation.
    -
    To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a “1” as the denominator:
    \(\displaystyle{7 = \frac{7}{1} }\)
    Next, simply multiply both numerators and both denominators:
    \(\displaystyle{\begin{aligned}[t] -7\cdot \frac{3}{4} -\amp = \frac{7}{1} \cdot \frac{3}{4} \\ -\amp = \frac{7\cdot3}{1 \cdot 4} \\ -\amp = \frac{21}{4} +
    Answer.
    \({\frac{20}{9}}\)
    Explanation.
    +
    To multiply an integer with a fraction, first we change the integer into a fraction by rewriting it with a “1” as the denominator:
    \(\displaystyle{5 = \frac{5}{1} }\)
    Next, simply multiply both numerators and both denominators:
    \(\displaystyle{\begin{aligned}[t] +5\cdot \frac{4}{9} +\amp = \frac{5}{1} \cdot \frac{4}{9} \\ +\amp = \frac{5\cdot4}{1 \cdot 9} \\ +\amp = \frac{20}{9} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{\frac{21}{4}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{20}{9}}\text{.}\)

    23.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{-\frac{7}{13} \cdot \frac{2}{7}}\) +
    Multiply: \(\displaystyle{-\frac{6}{11} \cdot \frac{7}{15}}\)
    -
    Answer.
    \(-{\frac{2}{13}}\)
    Explanation.
    -
    To multiply two fractions, multiply both numerators and both denominators. Don’t forget to reduce the fraction if possible.
    Remember: a positive number times a negative number is a negative number.
    We have:
    \(\displaystyle{\begin{aligned} --\frac{7}{13} \cdot \frac{2}{7} -\amp =-\frac{1\cdot7}{13} \cdot \frac{2}{1\cdot7} \\ -\amp =-\frac{1}{13} \cdot \frac{2}{1}\\ -\amp = -\frac{2}{13} +
    Answer.
    \(-{\frac{14}{55}}\)
    Explanation.
    +
    To multiply two fractions, multiply both numerators and both denominators. Don’t forget to reduce the fraction if possible.
    Remember: a positive number times a negative number is a negative number.
    We have:
    \(\displaystyle{\begin{aligned} +-\frac{6}{11} \cdot \frac{7}{15} +\amp =-\frac{2\cdot3}{11} \cdot \frac{7}{5\cdot3} \\ +\amp =-\frac{2}{11} \cdot \frac{7}{5}\\ +\amp = -\frac{14}{55} \end{aligned}}\)

    24.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{-\frac{9}{11} \cdot \frac{13}{15}}\) +
    Multiply: \(\displaystyle{-\frac{15}{13} \cdot \frac{13}{20}}\)
    -
    Answer.
    \(-{\frac{39}{55}}\)
    Explanation.
    -
    To multiply two fractions, multiply both numerators and both denominators. Don’t forget to reduce the fraction if possible.
    Remember: a positive number times a negative number is a negative number.
    We have:
    \(\displaystyle{\begin{aligned} --\frac{9}{11} \cdot \frac{13}{15} -\amp =-\frac{3\cdot3}{11} \cdot \frac{13}{5\cdot3} \\ -\amp =-\frac{3}{11} \cdot \frac{13}{5}\\ -\amp = -\frac{39}{55} +
    Answer.
    \(-{\frac{3}{4}}\)
    Explanation.
    +
    To multiply two fractions, multiply both numerators and both denominators. Don’t forget to reduce the fraction if possible.
    Remember: a positive number times a negative number is a negative number.
    We have:
    \(\displaystyle{\begin{aligned} +-\frac{15}{13} \cdot \frac{13}{20} +\amp =-\frac{3\cdot5}{13} \cdot \frac{13}{4\cdot5} \\ +\amp =-\frac{3}{13} \cdot \frac{13}{4}\\ +\amp = -\frac{3}{4} \end{aligned}}\)

    25.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{9\cdot\left( -{\frac{8}{9}} \right)}\) +
    Multiply: \(\displaystyle{36\cdot\left( -{\frac{1}{9}} \right)}\)
    -
    Answer.
    \(-8\)
    Explanation.
    -
    To multiply an integer with a fraction, first we rewrite the integer as a fraction by using “1” as the denominator:
    \(\displaystyle{ 9 = \frac{9}{1} }\)
    Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers:
    \(\displaystyle{\begin{aligned} -9\cdot\left( -{\frac{8}{9}} \right) -\amp =\frac{9}{1} \cdot \left(-{\frac{8}{9}}\right) \amp \hbox{We can divide both the first numerator and the second denominator by 9.}\\ -\amp =\frac{1}{1} \cdot \left(-{\frac{8}{1}}\right)\\ -\amp ={-8} +
    Answer.
    \(-4\)
    Explanation.
    +
    To multiply an integer with a fraction, first we rewrite the integer as a fraction by using “1” as the denominator:
    \(\displaystyle{ 36 = \frac{36}{1} }\)
    Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers:
    \(\displaystyle{\begin{aligned} +36\cdot\left( -{\frac{1}{9}} \right) +\amp =\frac{36}{1} \cdot \left(-{\frac{1}{9}}\right) \amp \hbox{We can divide both the first numerator and the second denominator by 9.}\\ +\amp =\frac{4}{1} \cdot \left(-{\frac{1}{1}}\right)\\ +\amp ={-4} \end{aligned}}\)

    26.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{14\cdot\left( -{\frac{9}{7}} \right)}\) +
    Multiply: \(\displaystyle{5\cdot\left( -{\frac{2}{5}} \right)}\)
    -
    Answer.
    \(-18\)
    Explanation.
    -
    To multiply an integer with a fraction, first we rewrite the integer as a fraction by using “1” as the denominator:
    \(\displaystyle{ 14 = \frac{14}{1} }\)
    Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers:
    \(\displaystyle{\begin{aligned} -14\cdot\left( -{\frac{9}{7}} \right) -\amp =\frac{14}{1} \cdot \left(-{\frac{9}{7}}\right) \amp \hbox{We can divide both the first numerator and the second denominator by 7.}\\ -\amp =\frac{2}{1} \cdot \left(-{\frac{9}{1}}\right)\\ -\amp ={-18} +
    Answer.
    \(-2\)
    Explanation.
    +
    To multiply an integer with a fraction, first we rewrite the integer as a fraction by using “1” as the denominator:
    \(\displaystyle{ 5 = \frac{5}{1} }\)
    Next, do the fraction multiplication. If you reduce numbers before multiplying across, you can avoid dealing with big numbers:
    \(\displaystyle{\begin{aligned} +5\cdot\left( -{\frac{2}{5}} \right) +\amp =\frac{5}{1} \cdot \left(-{\frac{2}{5}}\right) \amp \hbox{We can divide both the first numerator and the second denominator by 5.}\\ +\amp =\frac{1}{1} \cdot \left(-{\frac{2}{1}}\right)\\ +\amp ={-2} \end{aligned}}\)

    27.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ {{\frac{3}{4}}} \cdot {{\frac{35}{9}}} \cdot {{\frac{2}{49}}} }\) +
    Multiply: \(\displaystyle{ {{\frac{21}{4}}} \cdot {{\frac{5}{9}}} \cdot {{\frac{2}{25}}} }\)
    -
    Answer.
    \({\frac{5}{42}}\)
    Explanation.
    -
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} -{{\frac{3}{4}}} \cdot {{\frac{35}{9}}} \cdot {{\frac{2}{49}}} -\amp = \frac{1\cdot3}{2\cdot2} \cdot \frac{5\cdot7}{3\cdot3} \cdot \frac{1\cdot2}{7\cdot7} \\ -\amp = \frac{1}{2} \cdot \frac{5}{3} \cdot \frac{1}{7} \\ -\amp = {{\frac{5}{42}}} \\ +
    Answer.
    \({\frac{7}{30}}\)
    Explanation.
    +
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} +{{\frac{21}{4}}} \cdot {{\frac{5}{9}}} \cdot {{\frac{2}{25}}} +\amp = \frac{7\cdot3}{2\cdot2} \cdot \frac{1\cdot5}{3\cdot3} \cdot \frac{1\cdot2}{5\cdot5} \\ +\amp = \frac{7}{2} \cdot \frac{1}{3} \cdot \frac{1}{5} \\ +\amp = {{\frac{7}{30}}} \\ \end{aligned} }\)

    28.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ {{\frac{7}{4}}} \cdot {{\frac{15}{49}}} \cdot {{\frac{2}{9}}} }\) +
    Multiply: \(\displaystyle{ {{\frac{7}{9}}} \cdot {{\frac{5}{49}}} \cdot {{\frac{6}{25}}} }\)
    -
    Answer.
    \({\frac{5}{42}}\)
    Explanation.
    -
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} -{{\frac{7}{4}}} \cdot {{\frac{15}{49}}} \cdot {{\frac{2}{9}}} -\amp = \frac{1\cdot7}{2\cdot2} \cdot \frac{5\cdot3}{7\cdot7} \cdot \frac{1\cdot2}{3\cdot3} \\ -\amp = \frac{1}{2} \cdot \frac{5}{7} \cdot \frac{1}{3} \\ -\amp = {{\frac{5}{42}}} \\ +
    Answer.
    \({\frac{2}{105}}\)
    Explanation.
    +
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} +{{\frac{7}{9}}} \cdot {{\frac{5}{49}}} \cdot {{\frac{6}{25}}} +\amp = \frac{1\cdot7}{3\cdot3} \cdot \frac{1\cdot5}{7\cdot7} \cdot \frac{2\cdot3}{5\cdot5} \\ +\amp = \frac{1}{3} \cdot \frac{1}{7} \cdot \frac{2}{5} \\ +\amp = {{\frac{2}{105}}} \\ \end{aligned} }\)

    29.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ {{\frac{35}{2}}} \cdot {{\frac{1}{25}}} \cdot {6} }\) +
    Multiply: \(\displaystyle{ {{\frac{5}{3}}} \cdot {{\frac{7}{25}}} \cdot {6} }\)
    -
    Answer.
    \({\frac{21}{5}}\)
    Explanation.
    -
    When we multiply fractions, if there is an integer, we change the integer to a fraction by writing \(1\) as its denominator. For example, \(6 = \frac{6}{1}\text{.}\) -
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} -{{\frac{35}{2}}} \cdot {{\frac{1}{25}}} \cdot {6} -\amp = {{\frac{35}{2}}} \cdot {{\frac{1}{25}}} \cdot \frac{6}{1} \\ -\amp = \frac{7\cdot5}{1 \cdot2} \cdot \frac{1}{5\cdot5} \cdot \frac{2\cdot3}{1} \\ -\amp = \frac{7}{1} \cdot \frac{1}{5} \cdot \frac{3}{1} \\ -\amp = {{\frac{21}{5}}} \\ +
    Answer.
    \({\frac{14}{5}}\)
    Explanation.
    +
    When we multiply fractions, if there is an integer, we change the integer to a fraction by writing \(1\) as its denominator. For example, \(6 = \frac{6}{1}\text{.}\) +
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} +{{\frac{5}{3}}} \cdot {{\frac{7}{25}}} \cdot {6} +\amp = {{\frac{5}{3}}} \cdot {{\frac{7}{25}}} \cdot \frac{6}{1} \\ +\amp = \frac{1\cdot5}{1 \cdot3} \cdot \frac{7}{5\cdot5} \cdot \frac{3\cdot2}{1} \\ +\amp = \frac{1}{1} \cdot \frac{7}{5} \cdot \frac{2}{1} \\ +\amp = {{\frac{14}{5}}} \\ \end{aligned} }\)

    30.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{ {{\frac{2}{3}}} \cdot {{\frac{5}{4}}} \cdot {21} }\) +
    Multiply: \(\displaystyle{ {{\frac{6}{5}}} \cdot {{\frac{1}{4}}} \cdot {35} }\)
    -
    Answer.
    \({\frac{35}{2}}\)
    Explanation.
    -
    When we multiply fractions, if there is an integer, we change the integer to a fraction by writing \(1\) as its denominator. For example, \(21 = \frac{21}{1}\text{.}\) -
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} -{{\frac{2}{3}}} \cdot {{\frac{5}{4}}} \cdot {21} -\amp = {{\frac{2}{3}}} \cdot {{\frac{5}{4}}} \cdot \frac{21}{1} \\ -\amp = \frac{1\cdot2}{1 \cdot3} \cdot \frac{5}{2\cdot2} \cdot \frac{3\cdot7}{1} \\ -\amp = \frac{1}{1} \cdot \frac{5}{2} \cdot \frac{7}{1} \\ -\amp = {{\frac{35}{2}}} \\ +
    Answer.
    \({\frac{21}{2}}\)
    Explanation.
    +
    When we multiply fractions, if there is an integer, we change the integer to a fraction by writing \(1\) as its denominator. For example, \(35 = \frac{35}{1}\text{.}\) +
    When multiplying with fractions, reduce numerators and denominators if possible. This way, we can avoid dealing with big numbers.
    The full computation is:
    \(\displaystyle{\begin{aligned} +{{\frac{6}{5}}} \cdot {{\frac{1}{4}}} \cdot {35} +\amp = {{\frac{6}{5}}} \cdot {{\frac{1}{4}}} \cdot \frac{35}{1} \\ +\amp = \frac{3\cdot2}{1 \cdot5} \cdot \frac{1}{2\cdot2} \cdot \frac{5\cdot7}{1} \\ +\amp = \frac{3}{1} \cdot \frac{1}{2} \cdot \frac{7}{1} \\ +\amp = {{\frac{21}{2}}} \\ \end{aligned} }\)

    31.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{ \frac{3}{10} \div \frac{7}{3} }\) +
    Divide: \(\displaystyle{ \frac{4}{9} \div \frac{3}{2} }\)
    -
    Answer.
    \({\frac{9}{70}}\)
    Explanation.
    -
    When we do division with fractions, as with
    \(\displaystyle{\frac{3}{10} \div \frac{7}{3} }\)
    we first change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{3}{10} \cdot \frac{3}{7} }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] -\frac{3}{10} \div \frac{7}{3} -\amp = \frac{3}{10} \cdot \frac{3}{7} \\ -\amp = \frac{3 \cdot 3}{10 \cdot 7} \\ -\amp = \frac{9}{70} +
    Answer.
    \({\frac{8}{27}}\)
    Explanation.
    +
    When we do division with fractions, as with
    \(\displaystyle{\frac{4}{9} \div \frac{3}{2} }\)
    we first change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{4}{9} \cdot \frac{2}{3} }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] +\frac{4}{9} \div \frac{3}{2} +\amp = \frac{4}{9} \cdot \frac{2}{3} \\ +\amp = \frac{4 \cdot 2}{9 \cdot 3} \\ +\amp = \frac{8}{27} \end{aligned} }\)

    32.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{ \frac{4}{9} \div \frac{7}{5} }\) +
    Divide: \(\displaystyle{ \frac{4}{7} \div \frac{9}{4} }\)
    -
    Answer.
    \({\frac{20}{63}}\)
    Explanation.
    -
    When we do division with fractions, as with
    \(\displaystyle{\frac{4}{9} \div \frac{7}{5} }\)
    we first change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{4}{9} \cdot \frac{5}{7} }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] -\frac{4}{9} \div \frac{7}{5} -\amp = \frac{4}{9} \cdot \frac{5}{7} \\ -\amp = \frac{4 \cdot 5}{9 \cdot 7} \\ -\amp = \frac{20}{63} +
    Answer.
    \({\frac{16}{63}}\)
    Explanation.
    +
    When we do division with fractions, as with
    \(\displaystyle{\frac{4}{7} \div \frac{9}{4} }\)
    we first change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{4}{7} \cdot \frac{4}{9} }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] +\frac{4}{7} \div \frac{9}{4} +\amp = \frac{4}{7} \cdot \frac{4}{9} \\ +\amp = \frac{4 \cdot 4}{7 \cdot 9} \\ +\amp = \frac{16}{63} \end{aligned} }\)

    33.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{ \frac{1}{20} \div \left(-\frac{5}{8}\right) }\) +
    Divide: \(\displaystyle{ \frac{1}{25} \div \left(-\frac{7}{10}\right) }\)
    -
    Answer.
    \(-{\frac{2}{25}}\)
    Explanation.
    -
    When we do division with fractions, as with
    \(\displaystyle{\frac{1}{20} \div \left(-\frac{5}{8}\right) }\)
    we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have
    \(\displaystyle{\frac{1}{20} \cdot \left(-\frac{8}{5}\right) }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] -\frac{1}{20} \div \left(-\frac{5}{8}\right) -\amp = \frac{1}{20} \cdot \left(-\frac{8}{5}\right) \amp \hbox{We can divide the first denominator and the second numerator by 4.}\\ -\amp = \frac{1}{5 }\cdot\left(-\frac{2}{5}\right) \\ -\amp = -\frac{2}{25} +
    Answer.
    \(-{\frac{2}{35}}\)
    Explanation.
    +
    When we do division with fractions, as with
    \(\displaystyle{\frac{1}{25} \div \left(-\frac{7}{10}\right) }\)
    we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have
    \(\displaystyle{\frac{1}{25} \cdot \left(-\frac{10}{7}\right) }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] +\frac{1}{25} \div \left(-\frac{7}{10}\right) +\amp = \frac{1}{25} \cdot \left(-\frac{10}{7}\right) \amp \hbox{We can divide the first denominator and the second numerator by 5.}\\ +\amp = \frac{1}{5 }\cdot\left(-\frac{2}{7}\right) \\ +\amp = -\frac{2}{35} \end{aligned} }\)

    34.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{ \frac{7}{20} \div \left(-\frac{3}{16}\right) }\) +
    Divide: \(\displaystyle{ \frac{5}{4} \div \left(-\frac{7}{6}\right) }\)
    -
    Answer.
    \(-{\frac{28}{15}}\)
    Explanation.
    -
    When we do division with fractions, as with
    \(\displaystyle{\frac{7}{20} \div \left(-\frac{3}{16}\right) }\)
    we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have
    \(\displaystyle{\frac{7}{20} \cdot \left(-\frac{16}{3}\right) }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] -\frac{7}{20} \div \left(-\frac{3}{16}\right) -\amp = \frac{7}{20} \cdot \left(-\frac{16}{3}\right) \amp \hbox{We can divide the first denominator and the second numerator by 4.}\\ -\amp = \frac{7}{5 }\cdot\left(-\frac{4}{3}\right) \\ -\amp = -\frac{28}{15} +
    Answer.
    \(-{\frac{15}{14}}\)
    Explanation.
    +
    When we do division with fractions, as with
    \(\displaystyle{\frac{5}{4} \div \left(-\frac{7}{6}\right) }\)
    we first change division to multiplication, and at the same time, flip the second fraction. In this case, we have
    \(\displaystyle{\frac{5}{4} \cdot \left(-\frac{6}{7}\right) }\)
    Then we do fraction multiplication as usual. The full process is:
    \(\displaystyle{\begin{aligned}[t] +\frac{5}{4} \div \left(-\frac{7}{6}\right) +\amp = \frac{5}{4} \cdot \left(-\frac{6}{7}\right) \amp \hbox{We can divide the first denominator and the second numerator by 2.}\\ +\amp = \frac{5}{2 }\cdot\left(-\frac{3}{7}\right) \\ +\amp = -\frac{15}{14} \end{aligned} }\)

    35.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{-\frac{25}{6} \div (-15) }\) +
    Divide: \(\displaystyle{-\frac{5}{4} \div (-25) }\)
    -
    Answer.
    \({\frac{5}{18}}\)
    Explanation.
    -
    When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs:
    \(\displaystyle{ -\frac{25}{6} \div (-15) = \frac{25}{6} \div 15 }\)
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{15 = \frac{15}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{ \frac{25}{6} \cdot \frac{1}{15} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} -\left(-\frac{25}{6}\right) \div (-15) -\amp = \frac{25}{6} \div 15 \\ -\amp = \frac{25}{6} \div \frac{15}{1} \\ -\amp = \frac{25}{6} \cdot \frac{1}{15} \amp \hbox{We can simplify by dividing 25 and 15 by 5, making the numbers smaller.}\\ -\amp = \frac{5}{6}\cdot \frac{1}{3} \\ -\amp = \frac{5 \cdot 1}{6 \cdot 3} \\ -\amp = \frac{5}{18} +
    Answer.
    \({\frac{1}{20}}\)
    Explanation.
    +
    When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs:
    \(\displaystyle{ -\frac{5}{4} \div (-25) = \frac{5}{4} \div 25 }\)
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{25 = \frac{25}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{ \frac{5}{4} \cdot \frac{1}{25} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} +\left(-\frac{5}{4}\right) \div (-25) +\amp = \frac{5}{4} \div 25 \\ +\amp = \frac{5}{4} \div \frac{25}{1} \\ +\amp = \frac{5}{4} \cdot \frac{1}{25} \amp \hbox{We can simplify by dividing 5 and 25 by 5, making the numbers smaller.}\\ +\amp = \frac{1}{4}\cdot \frac{1}{5} \\ +\amp = \frac{1 \cdot 1}{4 \cdot 5} \\ +\amp = \frac{1}{20} \end{aligned} }\)

    36.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{-\frac{2}{3} \div (-6) }\) +
    Divide: \(\displaystyle{-\frac{10}{7} \div (-15) }\)
    -
    Answer.
    \({\frac{1}{9}}\)
    Explanation.
    -
    When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs:
    \(\displaystyle{ -\frac{2}{3} \div (-6) = \frac{2}{3} \div 6 }\)
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{6 = \frac{6}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{ \frac{2}{3} \cdot \frac{1}{6} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} -\left(-\frac{2}{3}\right) \div (-6) -\amp = \frac{2}{3} \div 6 \\ -\amp = \frac{2}{3} \div \frac{6}{1} \\ -\amp = \frac{2}{3} \cdot \frac{1}{6} \amp \hbox{We can simplify by dividing 2 and 6 by 2, making the numbers smaller.}\\ -\amp = \frac{1}{3}\cdot \frac{1}{3} \\ -\amp = \frac{1 \cdot 1}{3 \cdot 3} \\ -\amp = \frac{1}{9} +
    Answer.
    \({\frac{2}{21}}\)
    Explanation.
    +
    When a negative number is divided by a negative number, the result is positive. We can simply ignore those two negative signs:
    \(\displaystyle{ -\frac{10}{7} \div (-15) = \frac{10}{7} \div 15 }\)
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{15 = \frac{15}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{ \frac{10}{7} \cdot \frac{1}{15} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} +\left(-\frac{10}{7}\right) \div (-15) +\amp = \frac{10}{7} \div 15 \\ +\amp = \frac{10}{7} \div \frac{15}{1} \\ +\amp = \frac{10}{7} \cdot \frac{1}{15} \amp \hbox{We can simplify by dividing 10 and 15 by 5, making the numbers smaller.}\\ +\amp = \frac{2}{7}\cdot \frac{1}{3} \\ +\amp = \frac{2 \cdot 1}{7 \cdot 3} \\ +\amp = \frac{2}{21} \end{aligned} }\)

    37.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{45 \div \frac{9}{2} }\) +
    Divide: \(\displaystyle{28 \div \frac{7}{3} }\)
    -
    Answer.
    \(10\)
    Explanation.
    -
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{45 = \frac{45}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{45}{1} \cdot \frac{2}{9} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} -45\div \frac{9}{2} -\amp = \frac{45}{1} \div \frac{9}{2} \\ -\amp = \frac{45}{1} \cdot \frac{2}{9} \amp \hbox{We can divide 45 and 9 by 9 and made those two values smaller.}\\ -\amp = \frac{5}{1}\cdot\frac{2}{1} \\ -\amp = \frac{5\cdot2}{1 \cdot 1} \\ -\amp = 10 +
    Answer.
    \(12\)
    Explanation.
    +
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{28 = \frac{28}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{28}{1} \cdot \frac{3}{7} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} +28\div \frac{7}{3} +\amp = \frac{28}{1} \div \frac{7}{3} \\ +\amp = \frac{28}{1} \cdot \frac{3}{7} \amp \hbox{We can divide 28 and 7 by 7 and made those two values smaller.}\\ +\amp = \frac{4}{1}\cdot\frac{3}{1} \\ +\amp = \frac{4\cdot3}{1 \cdot 1} \\ +\amp = 12 \end{aligned} }\)

    38.

    -
    -
    +
    +
    -
    Divide: \(\displaystyle{7 \div \frac{7}{2} }\) +
    Divide: \(\displaystyle{4 \div \frac{4}{3} }\)
    -
    Answer.
    \(2\)
    Explanation.
    -
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{7 = \frac{7}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{7}{1} \cdot \frac{2}{7} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} -7\div \frac{7}{2} -\amp = \frac{7}{1} \div \frac{7}{2} \\ -\amp = \frac{7}{1} \cdot \frac{2}{7} \amp \hbox{We can divide 7 and 7 by 7 and made those two values smaller.}\\ -\amp = \frac{1}{1}\cdot\frac{2}{1} \\ -\amp = \frac{1\cdot2}{1 \cdot 1} \\ -\amp = 2 +
    Answer.
    \(3\)
    Explanation.
    +
    When we do fraction multiplication/division, we need to rewrite any integer as a fraction. In this problem, we will do this with:
    \(\displaystyle{4 = \frac{4}{1} }\)
    Next, we change division to multiplication, and at the same time flip the second fraction. In this case, we have
    \(\displaystyle{\frac{4}{1} \cdot \frac{3}{4} }\)
    Next, we do the fraction multiplication. The full process is:
    \(\displaystyle{\begin{aligned} +4\div \frac{4}{3} +\amp = \frac{4}{1} \div \frac{4}{3} \\ +\amp = \frac{4}{1} \cdot \frac{3}{4} \amp \hbox{We can divide 4 and 4 by 4 and made those two values smaller.}\\ +\amp = \frac{1}{1}\cdot\frac{3}{1} \\ +\amp = \frac{1\cdot3}{1 \cdot 1} \\ +\amp = 3 \end{aligned} }\)

    39.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{{1 {\textstyle\frac{7}{8}}} \cdot {2 {\textstyle\frac{2}{9}}} }\) +
    Multiply: \(\displaystyle{{2 {\textstyle\frac{11}{12}}} \cdot {1 {\textstyle\frac{7}{25}}} }\)
    -
    Answer.
    \(4 {\textstyle\frac{1}{6}}\)
    Explanation.
    -
    In order to multiply two mixed numbers, like
    \(\displaystyle{ {1 {\textstyle\frac{7}{8}}} \cdot {2 {\textstyle\frac{2}{9}}} }\)
    it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes
    \(\displaystyle{\frac{15}{8} \cdot \frac{20}{9}}\)
    Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is:
    \(\displaystyle{\begin{aligned}[t] -{1 {\textstyle\frac{7}{8}}} \cdot {2 {\textstyle\frac{2}{9}}} -\amp = \frac{15}{8} \cdot \frac{20}{9} \\ -\amp = \frac{5}{2} \cdot \frac{5}{3} \\ -\amp = \frac{25}{6} +
    Answer.
    \(3 {\textstyle\frac{11}{15}}\)
    Explanation.
    +
    In order to multiply two mixed numbers, like
    \(\displaystyle{ {2 {\textstyle\frac{11}{12}}} \cdot {1 {\textstyle\frac{7}{25}}} }\)
    it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes
    \(\displaystyle{\frac{35}{12} \cdot \frac{32}{25}}\)
    Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is:
    \(\displaystyle{\begin{aligned}[t] +{2 {\textstyle\frac{11}{12}}} \cdot {1 {\textstyle\frac{7}{25}}} +\amp = \frac{35}{12} \cdot \frac{32}{25} \\ +\amp = \frac{7}{3} \cdot \frac{8}{5} \\ +\amp = \frac{56}{15} \end{aligned} -}\)
    If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is \({4 {\textstyle\frac{1}{6}}}\text{.}\) +}\)
    If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is \({3 {\textstyle\frac{11}{15}}}\text{.}\)

    40.

    -
    -
    +
    +
    -
    Multiply: \(\displaystyle{{3 {\textstyle\frac{1}{9}}} \cdot {1 {\textstyle\frac{1}{8}}} }\) +
    Multiply: \(\displaystyle{{3 {\textstyle\frac{1}{9}}} \cdot {2 {\textstyle\frac{5}{8}}} }\)
    -
    Answer.
    \(3 {\textstyle\frac{1}{2}}\)
    Explanation.
    -
    In order to multiply two mixed numbers, like
    \(\displaystyle{ {3 {\textstyle\frac{1}{9}}} \cdot {1 {\textstyle\frac{1}{8}}} }\)
    it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes
    \(\displaystyle{\frac{28}{9} \cdot \frac{9}{8}}\)
    Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is:
    \(\displaystyle{\begin{aligned}[t] -{3 {\textstyle\frac{1}{9}}} \cdot {1 {\textstyle\frac{1}{8}}} -\amp = \frac{28}{9} \cdot \frac{9}{8} \\ -\amp = \frac{7}{1} \cdot \frac{1}{2} \\ -\amp = \frac{7}{2} +
    Answer.
    \(8 {\textstyle\frac{1}{6}}\)
    Explanation.
    +
    In order to multiply two mixed numbers, like
    \(\displaystyle{ {3 {\textstyle\frac{1}{9}}} \cdot {2 {\textstyle\frac{5}{8}}} }\)
    it helps to first change both mixed numbers to improper fractions. In this case, the problem becomes
    \(\displaystyle{\frac{28}{9} \cdot \frac{21}{8}}\)
    Then we do fraction multiplication as usual. To avoid dealing with big numbers, we reduce fractions before doing multiplications. The full solution is:
    \(\displaystyle{\begin{aligned}[t] +{3 {\textstyle\frac{1}{9}}} \cdot {2 {\textstyle\frac{5}{8}}} +\amp = \frac{28}{9} \cdot \frac{21}{8} \\ +\amp = \frac{7}{3} \cdot \frac{7}{2} \\ +\amp = \frac{49}{6} \end{aligned} -}\)
    If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is \({3 {\textstyle\frac{1}{2}}}\text{.}\) +}\)
    If we like, we may convert this answer to a mixed number since the original factors were mixed numbers. The solution is \({8 {\textstyle\frac{1}{6}}}\text{.}\)
    @@ -1722,174 +1734,174 @@

    Multiplying/Dividing Fractions.Adding/Subtracting Fractions.

    41.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{1}{16} + \frac{5}{16}}\) +
    Add: \(\displaystyle{\frac{1}{27} + \frac{5}{27}}\)
    -
    Answer.
    \({\frac{3}{8}}\)
    Explanation.
    -
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    The result is:
    \(\displaystyle{\begin{aligned} -\frac{1}{16} + \frac{5}{16} -\amp = \frac{1 + 5}{16} \\ -\amp = \frac{6}{16} \amp \hbox{We can reduce this by dividing 2 into both the numerator and the denominator.}\\ -\amp = \frac{3}{8} \\ +
    Answer.
    \({\frac{2}{9}}\)
    Explanation.
    +
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    The result is:
    \(\displaystyle{\begin{aligned} +\frac{1}{27} + \frac{5}{27} +\amp = \frac{1 + 5}{27} \\ +\amp = \frac{6}{27} \amp \hbox{We can reduce this by dividing 3 into both the numerator and the denominator.}\\ +\amp = \frac{2}{9} \\ \end{aligned} }\)

    42.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{2}{27} + \frac{1}{27}}\) +
    Add: \(\displaystyle{\frac{3}{40} + \frac{9}{40}}\)
    -
    Answer.
    \({\frac{1}{9}}\)
    Explanation.
    -
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    The result is:
    \(\displaystyle{\begin{aligned} -\frac{2}{27} + \frac{1}{27} -\amp = \frac{2 + 1}{27} \\ -\amp = \frac{3}{27} \amp \hbox{We can reduce this by dividing 3 into both the numerator and the denominator.}\\ -\amp = \frac{1}{9} \\ +
    Answer.
    \({\frac{3}{10}}\)
    Explanation.
    +
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    The result is:
    \(\displaystyle{\begin{aligned} +\frac{3}{40} + \frac{9}{40} +\amp = \frac{3 + 9}{40} \\ +\amp = \frac{12}{40} \amp \hbox{We can reduce this by dividing 4 into both the numerator and the denominator.}\\ +\amp = \frac{3}{10} \\ \end{aligned} }\)

    43.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{4}{9} + \frac{1}{27}}\) +
    Add: \(\displaystyle{\frac{1}{8} + \frac{11}{16}}\)
    -
    Answer.
    \({\frac{13}{27}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(9\) and \(27\) is \(27\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{4}{9}}\text{.}\) To change the denominator to \(27\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} \frac{4}{9} \amp = \frac{4 \cdot 3}{9 \cdot 3}\\ \amp = \frac{12}{27} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{4}{9} + \frac{1}{27} -\amp = \frac{12}{27} + \frac{1}{27} \\ -\amp = \frac{12 + 1}{27} \\ -\amp = \frac{13}{27} +
    Answer.
    \({\frac{13}{16}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(8\) and \(16\) is \(16\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{1}{8}}\text{.}\) To change the denominator to \(16\text{,}\) we multiply \(2\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} \frac{1}{8} \amp = \frac{1 \cdot 2}{8 \cdot 2}\\ \amp = \frac{2}{16} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The whole process is:
    \(\displaystyle{\begin{aligned} +\frac{1}{8} + \frac{11}{16} +\amp = \frac{2}{16} + \frac{11}{16} \\ +\amp = \frac{2 + 11}{16} \\ +\amp = \frac{13}{16} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{13}{27}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{13}{16}}\text{.}\)

    44.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{5}{7} + \frac{32}{35}}\) +
    Add: \(\displaystyle{\frac{1}{6} + \frac{13}{24}}\)
    -
    Answer.
    \({\frac{57}{35}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(35\) is \(35\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{5}{7}}\text{.}\) To change the denominator to \(35\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} \frac{5}{7} \amp = \frac{5 \cdot 5}{7 \cdot 5}\\ \amp = \frac{25}{35} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{5}{7} + \frac{32}{35} -\amp = \frac{25}{35} + \frac{32}{35} \\ -\amp = \frac{25 + 32}{35} \\ -\amp = \frac{57}{35} +
    Answer.
    \({\frac{17}{24}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(24\) is \(24\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{1}{6}}\text{.}\) To change the denominator to \(24\text{,}\) we multiply \(4\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} \frac{1}{6} \amp = \frac{1 \cdot 4}{6 \cdot 4}\\ \amp = \frac{4}{24} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The whole process is:
    \(\displaystyle{\begin{aligned} +\frac{1}{6} + \frac{13}{24} +\amp = \frac{4}{24} + \frac{13}{24} \\ +\amp = \frac{4 + 13}{24} \\ +\amp = \frac{17}{24} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{57}{35}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{17}{24}}\text{.}\)

    45.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{3}{8} + \frac{17}{24}}\) +
    Add: \(\displaystyle{\frac{2}{9} + \frac{1}{36}}\)
    -
    Answer.
    \({\frac{13}{12}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find the common denominator.
    In this case, the least common multiple of \(8\) and \(24\) is \(24\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{3}{8}}\text{.}\) To change the denominator to \(24\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{3}{8} \amp = \frac{3 \cdot 3}{8 \cdot 3}\\ \amp = \frac{9}{24}\end{aligned} }\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} -\frac{3}{8} + \frac{17}{24} -\amp = \frac{9}{24} + \frac{17}{24} \\ -\amp = \frac{9 + 17}{24} \\ -\amp = \frac{26}{24} \amp \hbox{We can reduce this by dividing 2 into both the numerator and the denominator.}\\ -\amp = \frac{13}{12} +
    Answer.
    \({\frac{1}{4}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find the common denominator.
    In this case, the least common multiple of \(9\) and \(36\) is \(36\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{2}{9}}\text{.}\) To change the denominator to \(36\text{,}\) we multiply \(4\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{2}{9} \amp = \frac{2 \cdot 4}{9 \cdot 4}\\ \amp = \frac{8}{36}\end{aligned} }\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} +\frac{2}{9} + \frac{1}{36} +\amp = \frac{8}{36} + \frac{1}{36} \\ +\amp = \frac{8 + 1}{36} \\ +\amp = \frac{9}{36} \amp \hbox{We can reduce this by dividing 9 into both the numerator and the denominator.}\\ +\amp = \frac{1}{4} \end{aligned} }\)

    46.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{1}{9} + \frac{35}{36}}\) +
    Add: \(\displaystyle{\frac{1}{9} + \frac{17}{36}}\)
    -
    Answer.
    \({\frac{13}{12}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find the common denominator.
    In this case, the least common multiple of \(9\) and \(36\) is \(36\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{1}{9}}\text{.}\) To change the denominator to \(36\text{,}\) we multiply \(4\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{9} \amp = \frac{1 \cdot 4}{9 \cdot 4}\\ \amp = \frac{4}{36}\end{aligned} }\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} -\frac{1}{9} + \frac{35}{36} -\amp = \frac{4}{36} + \frac{35}{36} \\ -\amp = \frac{4 + 35}{36} \\ -\amp = \frac{39}{36} \amp \hbox{We can reduce this by dividing 3 into both the numerator and the denominator.}\\ -\amp = \frac{13}{12} +
    Answer.
    \({\frac{7}{12}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find the common denominator.
    In this case, the least common multiple of \(9\) and \(36\) is \(36\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{1}{9}}\text{.}\) To change the denominator to \(36\text{,}\) we multiply \(4\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{9} \amp = \frac{1 \cdot 4}{9 \cdot 4}\\ \amp = \frac{4}{36}\end{aligned} }\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} +\frac{1}{9} + \frac{17}{36} +\amp = \frac{4}{36} + \frac{17}{36} \\ +\amp = \frac{4 + 17}{36} \\ +\amp = \frac{21}{36} \amp \hbox{We can reduce this by dividing 3 into both the numerator and the denominator.}\\ +\amp = \frac{7}{12} \end{aligned} }\)

    47.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{3}{10} + \frac{2}{5}}\) +
    Add: \(\displaystyle{\frac{1}{7} + \frac{2}{9}}\)
    -
    Answer.
    \({\frac{7}{10}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(5\) is \(10\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{3}{10}}\text{.}\) To change the denominator to \(10\text{,}\) we multiply \(1\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{3}{10} \amp = \frac{3 \cdot 1}{10 \cdot 1}\\ \amp = \frac{3}{10}\end{aligned} }\)
    Similarly, we change the second fraction’s denominator to \(10\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{2}{5} \amp = \frac{2 \cdot 2}{5 \cdot 2}\\ \amp = \frac{4}{10} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} -\frac{3}{10} + \frac{2}{5} -\amp = \frac{3}{10} + \frac{4}{10} \\ -\amp = \frac{3 + 4}{10} \\ -\amp = \frac{7}{10} +
    Answer.
    \({\frac{23}{63}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(9\) is \(63\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{1}{7}}\text{.}\) To change the denominator to \(63\text{,}\) we multiply \(9\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{7} \amp = \frac{1 \cdot 9}{7 \cdot 9}\\ \amp = \frac{9}{63}\end{aligned} }\)
    Similarly, we change the second fraction’s denominator to \(63\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{2}{9} \amp = \frac{2 \cdot 7}{9 \cdot 7}\\ \amp = \frac{14}{63} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} +\frac{1}{7} + \frac{2}{9} +\amp = \frac{9}{63} + \frac{14}{63} \\ +\amp = \frac{9 + 14}{63} \\ +\amp = \frac{23}{63} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{7}{10}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{23}{63}}\text{.}\)

    48.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{3}{7} + \frac{4}{9}}\) +
    Add: \(\displaystyle{\frac{1}{8} + \frac{1}{6}}\)
    -
    Answer.
    \({\frac{55}{63}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(9\) is \(63\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{3}{7}}\text{.}\) To change the denominator to \(63\text{,}\) we multiply \(9\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{3}{7} \amp = \frac{3 \cdot 9}{7 \cdot 9}\\ \amp = \frac{27}{63}\end{aligned} }\)
    Similarly, we change the second fraction’s denominator to \(63\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{4}{9} \amp = \frac{4 \cdot 7}{9 \cdot 7}\\ \amp = \frac{28}{63} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} -\frac{3}{7} + \frac{4}{9} -\amp = \frac{27}{63} + \frac{28}{63} \\ -\amp = \frac{27 + 28}{63} \\ -\amp = \frac{55}{63} +
    Answer.
    \({\frac{7}{24}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(8\) and \(6\) is \(24\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{1}{8}}\text{.}\) To change the denominator to \(24\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{8} \amp = \frac{1 \cdot 3}{8 \cdot 3}\\ \amp = \frac{3}{24}\end{aligned} }\)
    Similarly, we change the second fraction’s denominator to \(24\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{1}{6} \amp = \frac{1 \cdot 4}{6 \cdot 4}\\ \amp = \frac{4}{24} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} +\frac{1}{8} + \frac{1}{6} +\amp = \frac{3}{24} + \frac{4}{24} \\ +\amp = \frac{3 + 4}{24} \\ +\amp = \frac{7}{24} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{55}{63}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{7}{24}}\text{.}\)

    49.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{1}{6} + \frac{1}{10}}\) +
    Add: \(\displaystyle{\frac{1}{6} + \frac{1}{10}}\)
    -
    Answer.
    \({\frac{4}{15}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{1}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{6} \amp = \frac{1 \cdot 5}{6 \cdot 5}\\ \amp = \frac{5}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{1}{10} \amp = \frac{1 \cdot 3}{10 \cdot 3}\\ \amp = \frac{3}{30} \end{aligned}}\)
    Now we can add those two fractions. The full solution is:
    \(\displaystyle{\begin{aligned} +
    Answer.
    \({\frac{4}{15}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{1}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{6} \amp = \frac{1 \cdot 5}{6 \cdot 5}\\ \amp = \frac{5}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{1}{10} \amp = \frac{1 \cdot 3}{10 \cdot 3}\\ \amp = \frac{3}{30} \end{aligned}}\)
    Now we can add those two fractions. The full solution is:
    \(\displaystyle{\begin{aligned} \frac{1}{6} + \frac{1}{10} \amp = \frac{5}{30} + \frac{3}{30} \\ \amp = \frac{5 + 3}{30} \\ @@ -1902,16 +1914,16 @@

    Adding/Subtracting Fractions.

    50.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{3}{10} + \frac{1}{6}}\) +
    Add: \(\displaystyle{\frac{3}{10} + \frac{1}{6}}\)
    -
    Answer.
    \({\frac{7}{15}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{3}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{3}{10} \amp = \frac{3 \cdot 3}{10 \cdot 3}\\ \amp = \frac{9}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{1}{6} \amp = \frac{1 \cdot 5}{6 \cdot 5}\\ \amp = \frac{5}{30} \end{aligned}}\)
    Now we can add those two fractions. The full solution is:
    \(\displaystyle{\begin{aligned} +
    Answer.
    \({\frac{7}{15}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{3}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{3}{10} \amp = \frac{3 \cdot 3}{10 \cdot 3}\\ \amp = \frac{9}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{1}{6} \amp = \frac{1 \cdot 5}{6 \cdot 5}\\ \amp = \frac{5}{30} \end{aligned}}\)
    Now we can add those two fractions. The full solution is:
    \(\displaystyle{\begin{aligned} \frac{3}{10} + \frac{1}{6} \amp = \frac{9}{30} + \frac{5}{30} \\ \amp = \frac{9 + 5}{30} \\ @@ -1924,16 +1936,16 @@

    Adding/Subtracting Fractions.

    51.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{5}{6} + \frac{9}{10}}\) +
    Add: \(\displaystyle{\frac{5}{6} + \frac{9}{10}}\)
    -
    Answer.
    \({\frac{26}{15}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{5}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{5}{6} \amp = \frac{5 \cdot 5}{6 \cdot 5}\\ \amp = \frac{25}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{9}{10} \amp = \frac{9 \cdot 3}{10 \cdot 3}\\ \amp = \frac{27}{30} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} +
    Answer.
    \({\frac{26}{15}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{5}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{5}{6} \amp = \frac{5 \cdot 5}{6 \cdot 5}\\ \amp = \frac{25}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{9}{10} \amp = \frac{9 \cdot 3}{10 \cdot 3}\\ \amp = \frac{27}{30} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} \frac{5}{6} + \frac{9}{10} \amp = \frac{25}{30} + \frac{27}{30} \\ \amp = \frac{25 + 27}{30} \\ @@ -1946,16 +1958,16 @@

    Adding/Subtracting Fractions.

    52.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{\frac{4}{5} + \frac{7}{10}}\) +
    Add: \(\displaystyle{\frac{4}{5} + \frac{7}{10}}\)
    -
    Answer.
    \({\frac{3}{2}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(5\) and \(10\) is \(10\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{4}{5}}\text{.}\) To change the denominator to \(10\text{,}\) we multiply \(2\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{4}{5} \amp = \frac{4 \cdot 2}{5 \cdot 2}\\ \amp = \frac{8}{10} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(10\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{7}{10} \amp = \frac{7 \cdot 1}{10 \cdot 1}\\ \amp = \frac{7}{10} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} +
    Answer.
    \({\frac{3}{2}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(5\) and \(10\) is \(10\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{4}{5}}\text{.}\) To change the denominator to \(10\text{,}\) we multiply \(2\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{4}{5} \amp = \frac{4 \cdot 2}{5 \cdot 2}\\ \amp = \frac{8}{10} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(10\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{7}{10} \amp = \frac{7 \cdot 1}{10 \cdot 1}\\ \amp = \frac{7}{10} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} \frac{4}{5} + \frac{7}{10} \amp = \frac{8}{10} + \frac{7}{10} \\ \amp = \frac{8 + 7}{10} \\ @@ -1968,395 +1980,396 @@

    Adding/Subtracting Fractions.

    53.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{-\frac{2}{13} + \frac{7}{13}}\) +
    Add: \(\displaystyle{-\frac{1}{5} + \frac{2}{5}}\)
    -
    Answer.
    \({\frac{5}{13}}\)
    Explanation.
    -
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned}[t] --\frac{2}{13} + \frac{7}{13} -\amp = \frac{-2 + 7}{13} \\ -\amp = \frac{5}{13} +
    Answer.
    \({\frac{1}{5}}\)
    Explanation.
    +
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned}[t] +-\frac{1}{5} + \frac{2}{5} +\amp = \frac{-1 + 2}{5} \\ +\amp = \frac{1}{5} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{5}{13}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{1}{5}}\text{.}\)

    54.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{-\frac{2}{5} + \frac{4}{5}}\) +
    Add: \(\displaystyle{-\frac{2}{5} + \frac{4}{5}}\)
    -
    Answer.
    \({\frac{2}{5}}\)
    Explanation.
    -
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned}[t] +
    Answer.
    \({\frac{2}{5}}\)
    Explanation.
    +
    To add two fractions with the same denominator, we simply add up the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned}[t] -\frac{2}{5} + \frac{4}{5} \amp = \frac{-2 + 4}{5} \\ \amp = \frac{2}{5} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{2}{5}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{2}{5}}\text{.}\)

    55.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{-\frac{1}{9} + \frac{5}{36}}\) +
    Add: \(\displaystyle{-\frac{2}{7} + \frac{5}{14}}\)
    -
    Answer.
    \({\frac{1}{36}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of \(9\) and \(36\) is \(36\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{1}{9}}\text{.}\) To change the denominator to \(36\text{,}\) we multiply \(4\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{1}{9} \amp = -\frac{1 \cdot 4}{9 \cdot 4}\\ \amp = -\frac{4}{36} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} --\frac{1}{9} + \frac{5}{36} -\amp = -\frac{4}{36} + \frac{5}{36} \\ -\amp = \frac{-4 + 5}{36} \\ -\amp = \frac{1}{36} +
    Answer.
    \({\frac{1}{14}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of \(7\) and \(14\) is \(14\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{2}{7}}\text{.}\) To change the denominator to \(14\text{,}\) we multiply \(2\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{2}{7} \amp = -\frac{2 \cdot 2}{7 \cdot 2}\\ \amp = -\frac{4}{14} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} +-\frac{2}{7} + \frac{5}{14} +\amp = -\frac{4}{14} + \frac{5}{14} \\ +\amp = \frac{-4 + 5}{14} \\ +\amp = \frac{1}{14} \end{aligned} }\)

    56.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{-\frac{2}{7} + \frac{27}{35}}\) +
    Add: \(\displaystyle{-\frac{3}{8} + \frac{3}{16}}\)
    -
    Answer.
    \({\frac{17}{35}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of \(7\) and \(35\) is \(35\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{2}{7}}\text{.}\) To change the denominator to \(35\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{2}{7} \amp = -\frac{2 \cdot 5}{7 \cdot 5}\\ \amp = -\frac{10}{35} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} --\frac{2}{7} + \frac{27}{35} -\amp = -\frac{10}{35} + \frac{27}{35} \\ -\amp = \frac{-10 + 27}{35} \\ -\amp = \frac{17}{35} +
    Answer.
    \(-{\frac{3}{16}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator. In this case, the least common multiple of \(8\) and \(16\) is \(16\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{3}{8}}\text{.}\) To change the denominator to \(16\text{,}\) we multiply \(2\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{3}{8} \amp = -\frac{3 \cdot 2}{8 \cdot 2}\\ \amp = -\frac{6}{16} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the addition. The full process is:
    \(\displaystyle{\begin{aligned} +-\frac{3}{8} + \frac{3}{16} +\amp = -\frac{6}{16} + \frac{3}{16} \\ +\amp = \frac{-6 + 3}{16} \\ +\amp = \frac{-3}{16} \\ +\amp = -\frac{3}{16} \end{aligned} }\)

    57.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{-\frac{6}{7} + \frac{2}{5}}\) +
    Add: \(\displaystyle{-\frac{5}{8} + \frac{9}{10}}\)
    -
    Answer.
    \(-{\frac{16}{35}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(5\) is \(35\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{6}{7}}\text{.}\) To change the denominator to \(35\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} -\frac{6}{7} \amp = -\frac{6 \cdot 5}{7 \cdot 5}\\ \amp = -\frac{30}{35} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(35\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{2}{5} \amp = \frac{2 \cdot 7}{5 \cdot 7}\\ \amp = \frac{14}{35} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} --\frac{6}{7} + \frac{2}{5} -\amp = -\frac{30}{35} + \frac{14}{35} \\ -\amp = \frac{-30 + 14}{35} \\ -\amp = -\frac{16}{35} +
    Answer.
    \({\frac{11}{40}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(8\) and \(10\) is \(40\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{5}{8}}\text{.}\) To change the denominator to \(40\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} -\frac{5}{8} \amp = -\frac{5 \cdot 5}{8 \cdot 5}\\ \amp = -\frac{25}{40} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(40\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{9}{10} \amp = \frac{9 \cdot 4}{10 \cdot 4}\\ \amp = \frac{36}{40} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} +-\frac{5}{8} + \frac{9}{10} +\amp = -\frac{25}{40} + \frac{36}{40} \\ +\amp = \frac{-25 + 36}{40} \\ +\amp = \frac{11}{40} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{-\frac{16}{35}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{11}{40}}\text{.}\)

    58.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{-\frac{4}{7} + \frac{1}{8}}\) +
    Add: \(\displaystyle{-\frac{3}{8} + \frac{4}{7}}\)
    -
    Answer.
    \(-{\frac{25}{56}}\)
    Explanation.
    -
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(8\) is \(56\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{4}{7}}\text{.}\) To change the denominator to \(56\text{,}\) we multiply \(8\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} -\frac{4}{7} \amp = -\frac{4 \cdot 8}{7 \cdot 8}\\ \amp = -\frac{32}{56} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(56\text{:}\) -
    \(\displaystyle{ \begin{aligned}\frac{1}{8} \amp = \frac{1 \cdot 7}{8 \cdot 7}\\ \amp = \frac{7}{56} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} --\frac{4}{7} + \frac{1}{8} -\amp = -\frac{32}{56} + \frac{7}{56} \\ -\amp = \frac{-32 + 7}{56} \\ -\amp = -\frac{25}{56} +
    Answer.
    \({\frac{11}{56}}\)
    Explanation.
    +
    To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(8\) and \(7\) is \(56\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{3}{8}}\text{.}\) To change the denominator to \(56\text{,}\) we multiply \(7\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned} -\frac{3}{8} \amp = -\frac{3 \cdot 7}{8 \cdot 7}\\ \amp = -\frac{21}{56} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(56\text{:}\) +
    \(\displaystyle{ \begin{aligned}\frac{4}{7} \amp = \frac{4 \cdot 8}{7 \cdot 8}\\ \amp = \frac{32}{56} \end{aligned}}\)
    Now we can add those two fractions. The full process is:
    \(\displaystyle{\begin{aligned} +-\frac{3}{8} + \frac{4}{7} +\amp = -\frac{21}{56} + \frac{32}{56} \\ +\amp = \frac{-21 + 32}{56} \\ +\amp = \frac{11}{56} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{-\frac{25}{56}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{\frac{11}{56}}\text{.}\)

    59.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{ 1 + \frac{2}{3}}\) +
    Add: \(\displaystyle{ 3 + \frac{3}{10}}\)
    -
    Answer.
    \({\frac{5}{3}}\)
    Explanation.
    -
    When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions:
    \(\displaystyle{ 1 = \frac{1}{1} }\)
    Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator \(3\text{.}\) We will rewrite the first fraction:
    \(\displaystyle{ \begin{aligned}\frac{1}{1} \amp = \frac{1 \cdot 3}{1 \cdot 3}\\ \amp = \frac{3}{3} \end{aligned}}\)
    Finally, we add the numerators and keep the denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} -1 + \frac{2}{3} -\amp = \frac{3}{3} + \frac{2}{3} \\ -\amp = \frac{3 + 2}{3} \\ -\amp = \frac{5}{3} +
    Answer.
    \({\frac{33}{10}}\)
    Explanation.
    +
    When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions:
    \(\displaystyle{ 3 = \frac{3}{1} }\)
    Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator \(10\text{.}\) We will rewrite the first fraction:
    \(\displaystyle{ \begin{aligned}\frac{3}{1} \amp = \frac{3 \cdot 10}{1 \cdot 10}\\ \amp = \frac{30}{10} \end{aligned}}\)
    Finally, we add the numerators and keep the denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} +3 + \frac{3}{10} +\amp = \frac{30}{10} + \frac{3}{10} \\ +\amp = \frac{30 + 3}{10} \\ +\amp = \frac{33}{10} \end{aligned} -}\)
    The answer to this question is \({{\frac{5}{3}}}\text{.}\) +}\)
    The answer to this question is \({{\frac{33}{10}}}\text{.}\)

    60.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{ 2 + \frac{4}{9}}\) +
    Add: \(\displaystyle{ 4 + \frac{2}{7}}\)
    -
    Answer.
    \({\frac{22}{9}}\)
    Explanation.
    -
    When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions:
    \(\displaystyle{ 2 = \frac{2}{1} }\)
    Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator \(9\text{.}\) We will rewrite the first fraction:
    \(\displaystyle{ \begin{aligned}\frac{2}{1} \amp = \frac{2 \cdot 9}{1 \cdot 9}\\ \amp = \frac{18}{9} \end{aligned}}\)
    Finally, we add the numerators and keep the denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} -2 + \frac{4}{9} -\amp = \frac{18}{9} + \frac{4}{9} \\ -\amp = \frac{18 + 4}{9} \\ -\amp = \frac{22}{9} +
    Answer.
    \({\frac{30}{7}}\)
    Explanation.
    +
    When doing arithmetic with fractions, it is helpful to rewrite any integers as fractions:
    \(\displaystyle{ 4 = \frac{4}{1} }\)
    Next, to add two fractions, we need to find a common denominator. In this case, it is simply the second denominator \(7\text{.}\) We will rewrite the first fraction:
    \(\displaystyle{ \begin{aligned}\frac{4}{1} \amp = \frac{4 \cdot 7}{1 \cdot 7}\\ \amp = \frac{28}{7} \end{aligned}}\)
    Finally, we add the numerators and keep the denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} +4 + \frac{2}{7} +\amp = \frac{28}{7} + \frac{2}{7} \\ +\amp = \frac{28 + 2}{7} \\ +\amp = \frac{30}{7} \end{aligned} -}\)
    The answer to this question is \({{\frac{22}{9}}}\text{.}\) +}\)
    The answer to this question is \({{\frac{30}{7}}}\text{.}\)

    61.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{ {{\frac{1}{9}}} + {{\frac{2}{5}}} + {{\frac{1}{10}}} }\) +
    Add: \(\displaystyle{ {{\frac{2}{5}}} + {{\frac{1}{10}}} + {{\frac{1}{8}}} }\)
    -
    Answer.
    \({\frac{11}{18}}\)
    Explanation.
    -
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(10\text{.}\) We list its first few multiples:
    \(\displaystyle{ 10, 20, 30, 40, 50, 60, 70, 80, 90 }\)
    Note that all denominators, \(9, 5 \text{ and } 10\) each divide into \(90\text{.}\) -
    The full computation is:
    \(\displaystyle{\begin{aligned} -\frac{1}{9} + \frac{2}{5} + \frac{1}{10} -\amp = \frac{1 \cdot 10}{9 \cdot 10} + \frac{2 \cdot 18}{5 \cdot 18} + \frac{1 \cdot 9}{10 \cdot 9} \\ -\amp = \frac{10}{90} + \frac{36}{90} + \frac{9}{90} \\ -\amp = \frac{10+36+9}{90} \\ -\amp = \frac{55}{90} \\ -\amp = {{\frac{11}{18}}} +
    Answer.
    \({\frac{5}{8}}\)
    Explanation.
    +
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(10\text{.}\) We list its first few multiples:
    \(\displaystyle{ 10, 20, 30, 40 }\)
    Note that all denominators, \(5, 10 \text{ and } 8\) each divide into \(40\text{.}\) +
    The full computation is:
    \(\displaystyle{\begin{aligned} +\frac{2}{5} + \frac{1}{10} + \frac{1}{8} +\amp = \frac{2 \cdot 8}{5 \cdot 8} + \frac{1 \cdot 4}{10 \cdot 4} + \frac{1 \cdot 5}{8 \cdot 5} \\ +\amp = \frac{16}{40} + \frac{4}{40} + \frac{5}{40} \\ +\amp = \frac{16+4+5}{40} \\ +\amp = \frac{25}{40} \\ +\amp = {{\frac{5}{8}}} \end{aligned} -}\)
    Don’t forget to reduce the fraction \(\frac{55}{90}\) by dividing \(5\) into both the numerator and the denominator.
    +}\)
    Don’t forget to reduce the fraction \(\frac{25}{40}\) by dividing \(5\) into both the numerator and the denominator.

    62.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{ {{\frac{1}{3}}} + {{\frac{3}{10}}} + {{\frac{1}{6}}} }\) +
    Add: \(\displaystyle{ {{\frac{1}{3}}} + {{\frac{3}{8}}} + {{\frac{1}{6}}} }\)
    -
    Answer.
    \({\frac{4}{5}}\)
    Explanation.
    -
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(10\text{.}\) We list its first few multiples:
    \(\displaystyle{ 10, 20, 30 }\)
    Note that all denominators, \(3, 10 \text{ and } 6\) each divide into \(30\text{.}\) -
    The full computation is:
    \(\displaystyle{\begin{aligned} -\frac{1}{3} + \frac{3}{10} + \frac{1}{6} -\amp = \frac{1 \cdot 10}{3 \cdot 10} + \frac{3 \cdot 3}{10 \cdot 3} + \frac{1 \cdot 5}{6 \cdot 5} \\ -\amp = \frac{10}{30} + \frac{9}{30} + \frac{5}{30} \\ -\amp = \frac{10+9+5}{30} \\ -\amp = \frac{24}{30} \\ -\amp = {{\frac{4}{5}}} +
    Answer.
    \({\frac{7}{8}}\)
    Explanation.
    +
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(8\text{.}\) We list its first few multiples:
    \(\displaystyle{ 8, 16, 24 }\)
    Note that all denominators, \(3, 8 \text{ and } 6\) each divide into \(24\text{.}\) +
    The full computation is:
    \(\displaystyle{\begin{aligned} +\frac{1}{3} + \frac{3}{8} + \frac{1}{6} +\amp = \frac{1 \cdot 8}{3 \cdot 8} + \frac{3 \cdot 3}{8 \cdot 3} + \frac{1 \cdot 4}{6 \cdot 4} \\ +\amp = \frac{8}{24} + \frac{9}{24} + \frac{4}{24} \\ +\amp = \frac{8+9+4}{24} \\ +\amp = \frac{21}{24} \\ +\amp = {{\frac{7}{8}}} \end{aligned} -}\)
    Don’t forget to reduce the fraction \(\frac{24}{30}\) by dividing \(6\) into both the numerator and the denominator.
    +}\)
    Don’t forget to reduce the fraction \(\frac{21}{24}\) by dividing \(3\) into both the numerator and the denominator.

    63.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{ {{\frac{5}{6}}} + {{\frac{4}{9}}} + {{\frac{1}{10}}} }\) +
    Add: \(\displaystyle{ {{\frac{1}{6}}} + {{\frac{1}{5}}} + {{\frac{7}{10}}} }\)
    -
    Answer.
    \(1 {\textstyle\frac{17}{45}}\)
    Explanation.
    -
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(10\text{.}\) We list its first few multiples:
    \(\displaystyle{ 10, 20, 30, 40, 50, 60, 70, 80, 90 }\)
    Note that all denominators, \(6, 9 \text{ and } 10\) each divide into \(90\text{.}\) -
    The full computation is:
    \(\displaystyle{\begin{aligned} -\frac{5}{6} + \frac{4}{9} + \frac{1}{10} -\amp = \frac{5 \cdot 15}{6 \cdot 15} + \frac{4 \cdot 10}{9 \cdot 10} + \frac{1 \cdot 9}{10 \cdot 9} \\ -\amp = \frac{75}{90} + \frac{40}{90} + \frac{9}{90} \\ -\amp = \frac{75+40+9}{90} \\ -\amp = \frac{124}{90} \\ -\amp = \frac{62}{45} \\ -\amp = {1 {\textstyle\frac{17}{45}}} +
    Answer.
    \(1 {\textstyle\frac{1}{15}}\)
    Explanation.
    +
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(10\text{.}\) We list its first few multiples:
    \(\displaystyle{ 10, 20, 30 }\)
    Note that all denominators, \(6, 5 \text{ and } 10\) each divide into \(30\text{.}\) +
    The full computation is:
    \(\displaystyle{\begin{aligned} +\frac{1}{6} + \frac{1}{5} + \frac{7}{10} +\amp = \frac{1 \cdot 5}{6 \cdot 5} + \frac{1 \cdot 6}{5 \cdot 6} + \frac{7 \cdot 3}{10 \cdot 3} \\ +\amp = \frac{5}{30} + \frac{6}{30} + \frac{21}{30} \\ +\amp = \frac{5+6+21}{30} \\ +\amp = \frac{32}{30} \\ +\amp = \frac{16}{15} \\ +\amp = {1 {\textstyle\frac{1}{15}}} \end{aligned} -}\)
    Don’t forget to reduce the fraction \(\frac{124}{90}\) by dividing \(2\) into both the numerator and the denominator.
    +}\)
    Don’t forget to reduce the fraction \(\frac{32}{30}\) by dividing \(2\) into both the numerator and the denominator.

    64.

    -
    -
    +
    +
    -
    Add: \(\displaystyle{ {{\frac{2}{3}}} + {{\frac{5}{6}}} + {{\frac{3}{4}}} }\) +
    Add: \(\displaystyle{ {{\frac{1}{9}}} + {{\frac{3}{10}}} + {{\frac{5}{6}}} }\)
    -
    Answer.
    \(2 {\textstyle\frac{1}{4}}\)
    Explanation.
    -
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(6\text{.}\) We list its first few multiples:
    \(\displaystyle{ 6, 12 }\)
    Note that all denominators, \(3, 6 \text{ and } 4\) each divide into \(12\text{.}\) -
    The full computation is:
    \(\displaystyle{\begin{aligned} -\frac{2}{3} + \frac{5}{6} + \frac{3}{4} -\amp = \frac{2 \cdot 4}{3 \cdot 4} + \frac{5 \cdot 2}{6 \cdot 2} + \frac{3 \cdot 3}{4 \cdot 3} \\ -\amp = \frac{8}{12} + \frac{10}{12} + \frac{9}{12} \\ -\amp = \frac{8+10+9}{12} \\ -\amp = \frac{27}{12} \\ -\amp = \frac{9}{4} \\ -\amp = {2 {\textstyle\frac{1}{4}}} +
    Answer.
    \(1 {\textstyle\frac{11}{45}}\)
    Explanation.
    +
    To add fractions with different denominators, we need to find the common denominator first. One way is to list the multiples of the largest denominator, until all \(3\) denominators divide into the multiple.
    The largest denominator is \(10\text{.}\) We list its first few multiples:
    \(\displaystyle{ 10, 20, 30, 40, 50, 60, 70, 80, 90 }\)
    Note that all denominators, \(9, 10 \text{ and } 6\) each divide into \(90\text{.}\) +
    The full computation is:
    \(\displaystyle{\begin{aligned} +\frac{1}{9} + \frac{3}{10} + \frac{5}{6} +\amp = \frac{1 \cdot 10}{9 \cdot 10} + \frac{3 \cdot 9}{10 \cdot 9} + \frac{5 \cdot 15}{6 \cdot 15} \\ +\amp = \frac{10}{90} + \frac{27}{90} + \frac{75}{90} \\ +\amp = \frac{10+27+75}{90} \\ +\amp = \frac{112}{90} \\ +\amp = \frac{56}{45} \\ +\amp = {1 {\textstyle\frac{11}{45}}} \end{aligned} -}\)
    Don’t forget to reduce the fraction \(\frac{27}{12}\) by dividing \(3\) into both the numerator and the denominator.
    +}\)
    Don’t forget to reduce the fraction \(\frac{112}{90}\) by dividing \(2\) into both the numerator and the denominator.

    65.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{\frac{5}{21} - \frac{2}{21}}\) +
    Subtract: \(\displaystyle{\frac{15}{16} - \frac{5}{16}}\)
    -
    Answer.
    \({\frac{1}{7}}\)
    Explanation.
    -
    To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} -\frac{5}{21} - \frac{2}{21} -\amp = \frac{5 - 2}{21} \\ -\amp = \frac{3}{21} \amp \hbox{We can reduce this by dividing 3 into both the numerator and the denominator. +
    Answer.
    \({\frac{5}{8}}\)
    Explanation.
    +
    To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} +\frac{15}{16} - \frac{5}{16} +\amp = \frac{15 - 5}{16} \\ +\amp = \frac{10}{16} \amp \hbox{We can reduce this by dividing 2 into both the numerator and the denominator. }\\ -\amp = \frac{1}{7} \\ +\amp = \frac{5}{8} \\ \end{aligned} }\)

    66.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{\frac{33}{28} - \frac{17}{28}}\) +
    Subtract: \(\displaystyle{\frac{19}{32} - \frac{7}{32}}\)
    -
    Answer.
    \({\frac{4}{7}}\)
    Explanation.
    -
    To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} -\frac{33}{28} - \frac{17}{28} -\amp = \frac{33 - 17}{28} \\ -\amp = \frac{16}{28} \amp \hbox{We can reduce this by dividing 4 into both the numerator and the denominator. +
    Answer.
    \({\frac{3}{8}}\)
    Explanation.
    +
    To subtract two fractions with the same denominator, we simply subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} +\frac{19}{32} - \frac{7}{32} +\amp = \frac{19 - 7}{32} \\ +\amp = \frac{12}{32} \amp \hbox{We can reduce this by dividing 4 into both the numerator and the denominator. }\\ -\amp = \frac{4}{7} \\ +\amp = \frac{3}{8} \\ \end{aligned} }\)

    67.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{\frac{4}{7} - \frac{32}{35}}\) +
    Subtract: \(\displaystyle{\frac{4}{7} - \frac{19}{42}}\)
    -
    Answer.
    \(-{\frac{12}{35}}\)
    Explanation.
    -
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(35\) is \(35\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{4}{7}}\text{.}\) To change the denominator to \(35\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{4}{7} \amp = \frac{4 \cdot 5}{7 \cdot 5}\\ \amp = \frac{20}{35} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{4}{7} - \frac{32}{35} -\amp = \frac{20}{35} - \frac{32}{35} \\ -\amp = \frac{20 - 32}{35} \\ -\amp = -\frac{12}{35} +
    Answer.
    \({\frac{5}{42}}\)
    Explanation.
    +
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(7\) and \(42\) is \(42\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{4}{7}}\text{.}\) To change the denominator to \(42\text{,}\) we multiply \(6\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{4}{7} \amp = \frac{4 \cdot 6}{7 \cdot 6}\\ \amp = \frac{24}{42} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} +\frac{4}{7} - \frac{19}{42} +\amp = \frac{24}{42} - \frac{19}{42} \\ +\amp = \frac{24 - 19}{42} \\ +\amp = \frac{5}{42} \end{aligned} -}\)
    The answer is \(\displaystyle{-\frac{12}{35}}\text{.}\) +}\)
    The answer is \(\displaystyle{\frac{5}{42}}\text{.}\)

    68.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{\frac{2}{9} - \frac{2}{27}}\) +
    Subtract: \(\displaystyle{\frac{4}{9} - \frac{13}{27}}\)
    -
    Answer.
    \({\frac{4}{27}}\)
    Explanation.
    -
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(9\) and \(27\) is \(27\text{.}\) -
    Look at the first fraction \(\displaystyle{\frac{2}{9}}\text{.}\) To change the denominator to \(27\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{2}{9} \amp = \frac{2 \cdot 3}{9 \cdot 3}\\ \amp = \frac{6}{27} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{2}{9} - \frac{2}{27} -\amp = \frac{6}{27} - \frac{2}{27} \\ -\amp = \frac{6 - 2}{27} \\ -\amp = \frac{4}{27} +
    Answer.
    \(-{\frac{1}{27}}\)
    Explanation.
    +
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(9\) and \(27\) is \(27\text{.}\) +
    Look at the first fraction \(\displaystyle{\frac{4}{9}}\text{.}\) To change the denominator to \(27\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{4}{9} \amp = \frac{4 \cdot 3}{9 \cdot 3}\\ \amp = \frac{12}{27} \end{aligned}}\)
    The second fraction already has the common denominator.
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} +\frac{4}{9} - \frac{13}{27} +\amp = \frac{12}{27} - \frac{13}{27} \\ +\amp = \frac{12 - 13}{27} \\ +\amp = -\frac{1}{27} \end{aligned} -}\)
    The answer is \(\displaystyle{\frac{4}{27}}\text{.}\) +}\)
    The answer is \(\displaystyle{-\frac{1}{27}}\text{.}\)

    69.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{\frac{17}{18} - \frac{4}{9}}\) +
    Subtract: \(\displaystyle{\frac{11}{30} - \frac{1}{10}}\)
    -
    Answer.
    \({\frac{1}{2}}\)
    Explanation.
    -
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(18\) and \(9\) is \(18\text{.}\) -
    The first fraction already has the common denominator.
    Look at the second fraction \(\displaystyle{\frac{4}{9}}\text{.}\) To change the denominator to \(18\text{,}\) we multiply \(2\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{4}{9} \amp = \frac{4 \cdot 2}{9 \cdot 2}\\ \amp = \frac{8}{18} \end{aligned}}\)
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{17}{18} - \frac{4}{9} -\amp = \frac{17}{18} - \frac{8}{18} \\ -\amp = \frac{17 - 8}{18} \\ -\amp = \frac{9}{18} \amp \hbox{We can reduce this by dividing 9 into both the numerator and the denominator. +
    Answer.
    \({\frac{4}{15}}\)
    Explanation.
    +
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(30\) and \(10\) is \(30\text{.}\) +
    The first fraction already has the common denominator.
    Look at the second fraction \(\displaystyle{\frac{1}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{1}{10} \amp = \frac{1 \cdot 3}{10 \cdot 3}\\ \amp = \frac{3}{30} \end{aligned}}\)
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} +\frac{11}{30} - \frac{1}{10} +\amp = \frac{11}{30} - \frac{3}{30} \\ +\amp = \frac{11 - 3}{30} \\ +\amp = \frac{8}{30} \amp \hbox{We can reduce this by dividing 2 into both the numerator and the denominator. }\\ -\amp = \frac{1}{2} +\amp = \frac{4}{15} \end{aligned} }\)

    70.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{\frac{11}{30} - \frac{5}{6}}\) +
    Subtract: \(\displaystyle{\frac{5}{24} - \frac{5}{6}}\)
    -
    Answer.
    \(-{\frac{7}{15}}\)
    Explanation.
    -
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(30\) and \(6\) is \(30\text{.}\) -
    The first fraction already has the common denominator.
    Look at the second fraction \(\displaystyle{\frac{5}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{5}{6} \amp = \frac{5 \cdot 5}{6 \cdot 5}\\ \amp = \frac{25}{30} \end{aligned}}\)
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{11}{30} - \frac{5}{6} -\amp = \frac{11}{30} - \frac{25}{30} \\ -\amp = \frac{11 - 25}{30} \\ -\amp = \frac{-14}{30} \amp \hbox{We can reduce this by dividing 2 into both the numerator and the denominator. +
    Answer.
    \(-{\frac{5}{8}}\)
    Explanation.
    +
    To subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(24\) and \(6\) is \(24\text{.}\) +
    The first fraction already has the common denominator.
    Look at the second fraction \(\displaystyle{\frac{5}{6}}\text{.}\) To change the denominator to \(24\text{,}\) we multiply \(4\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}\frac{5}{6} \amp = \frac{5 \cdot 4}{6 \cdot 4}\\ \amp = \frac{20}{24} \end{aligned}}\)
    Finally, we can do the subtraction. The whole process is:
    \(\displaystyle{\begin{aligned} +\frac{5}{24} - \frac{5}{6} +\amp = \frac{5}{24} - \frac{20}{24} \\ +\amp = \frac{5 - 20}{24} \\ +\amp = \frac{-15}{24} \amp \hbox{We can reduce this by dividing 3 into both the numerator and the denominator. }\\ -\amp = -\frac{7}{15} +\amp = -\frac{5}{8} \end{aligned} }\)

    71.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{-\frac{5}{6}-\frac{7}{10}}\) +
    Subtract: \(\displaystyle{-\frac{3}{10}-\frac{1}{6}}\)
    -
    Answer.
    \(-{\frac{23}{15}}\)
    Explanation.
    -
    To add or subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{5}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{5}{6} \amp = -\frac{5 \cdot 5}{6 \cdot 5}\\ \amp = -\frac{25}{30} \end{aligned}}\)
    Similarly, we change the denominator of the second fraction to \(30\) by:
    \(\displaystyle{ \begin{aligned}-\frac{7}{10} \amp = -\frac{7 \cdot 3}{10 \cdot 3}\\ \amp = -\frac{21}{30} \end{aligned}}\)
    Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} --\frac{5}{6}-\frac{7}{10} \amp = -\frac{25}{30}-\frac{21}{30} \\ -\amp = \frac{-25+(-21)}{30} \\ -\amp = -\frac{46}{30} \amp \hbox{We can reduce this by dividing both the numerator and denominator by 2. +
    Answer.
    \(-{\frac{7}{15}}\)
    Explanation.
    +
    To add or subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{3}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{3}{10} \amp = -\frac{3 \cdot 3}{10 \cdot 3}\\ \amp = -\frac{9}{30} \end{aligned}}\)
    Similarly, we change the denominator of the second fraction to \(30\) by:
    \(\displaystyle{ \begin{aligned}-\frac{1}{6} \amp = -\frac{1 \cdot 5}{6 \cdot 5}\\ \amp = -\frac{5}{30} \end{aligned}}\)
    Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} +-\frac{3}{10}-\frac{1}{6} \amp = -\frac{9}{30}-\frac{5}{30} \\ +\amp = \frac{-9+(-5)}{30} \\ +\amp = -\frac{14}{30} \amp \hbox{We can reduce this by dividing both the numerator and denominator by 2. }\\ -\amp = -\frac{23}{15} +\amp = -\frac{7}{15} \end{aligned} }\)

    72.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{-\frac{1}{10}-\frac{5}{6}}\) +
    Subtract: \(\displaystyle{-\frac{1}{10}-\frac{5}{6}}\)
    -
    Answer.
    \(-{\frac{14}{15}}\)
    Explanation.
    -
    To add or subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{1}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{1}{10} \amp = -\frac{1 \cdot 3}{10 \cdot 3}\\ \amp = -\frac{3}{30} \end{aligned}}\)
    Similarly, we change the denominator of the second fraction to \(30\) by:
    \(\displaystyle{ \begin{aligned}-\frac{5}{6} \amp = -\frac{5 \cdot 5}{6 \cdot 5}\\ \amp = -\frac{25}{30} \end{aligned}}\)
    Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} +
    Answer.
    \(-{\frac{14}{15}}\)
    Explanation.
    +
    To add or subtract two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{1}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{ \begin{aligned}-\frac{1}{10} \amp = -\frac{1 \cdot 3}{10 \cdot 3}\\ \amp = -\frac{3}{30} \end{aligned}}\)
    Similarly, we change the denominator of the second fraction to \(30\) by:
    \(\displaystyle{ \begin{aligned}-\frac{5}{6} \amp = -\frac{5 \cdot 5}{6 \cdot 5}\\ \amp = -\frac{25}{30} \end{aligned}}\)
    Finally, we add or subtract the numerators and keep the common denominator unchanged. The whole process is:
    \(\displaystyle{\begin{aligned} -\frac{1}{10}-\frac{5}{6} \amp = -\frac{3}{30}-\frac{25}{30} \\ \amp = \frac{-3+(-25)}{30} \\ \amp = -\frac{28}{30} \amp \hbox{We can reduce this by dividing both the numerator and denominator by 2. @@ -2368,16 +2381,16 @@

    Adding/Subtracting Fractions.

    73.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{-\frac{3}{10} - \left(-\frac{5}{6}\right)}\) +
    Subtract: \(\displaystyle{-\frac{3}{10} - \left(-\frac{5}{6}\right)}\)
    -
    Answer.
    \({\frac{8}{15}}\)
    Explanation.
    -
    When we subtract a negative number, first we change those two negative signs to addition:
    \(\displaystyle{ -\frac{3}{10} - \left(-\frac{5}{6}\right) = -\frac{3}{10} + \frac{5}{6} }\)
    Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{3}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned}-\frac{3}{10}\amp = -\frac{3 \cdot 3}{10 \cdot 3}\\ \amp = -\frac{9}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) -
    \(\displaystyle{\begin{aligned}\frac{5}{6} \amp = \frac{5 \cdot 5}{6 \cdot 5}\\ \amp = \frac{25}{30}\end{aligned} }\)
    Now we can add those two fractions.
    \(\displaystyle{\begin{aligned} +
    Answer.
    \({\frac{8}{15}}\)
    Explanation.
    +
    When we subtract a negative number, first we change those two negative signs to addition:
    \(\displaystyle{ -\frac{3}{10} - \left(-\frac{5}{6}\right) = -\frac{3}{10} + \frac{5}{6} }\)
    Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(10\) and \(6\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{3}{10}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(3\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned}-\frac{3}{10}\amp = -\frac{3 \cdot 3}{10 \cdot 3}\\ \amp = -\frac{9}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) +
    \(\displaystyle{\begin{aligned}\frac{5}{6} \amp = \frac{5 \cdot 5}{6 \cdot 5}\\ \amp = \frac{25}{30}\end{aligned} }\)
    Now we can add those two fractions.
    \(\displaystyle{\begin{aligned} -\frac{3}{10} + \frac{5}{6} \amp = -\frac{9}{30} + \frac{25}{30} \\ \amp = \frac{-9 + 25}{30} \\ @@ -2390,16 +2403,16 @@

    Adding/Subtracting Fractions.

    74.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{-\frac{5}{6} - \left(-\frac{3}{10}\right)}\) +
    Subtract: \(\displaystyle{-\frac{5}{6} - \left(-\frac{3}{10}\right)}\)
    -
    Answer.
    \(-{\frac{8}{15}}\)
    Explanation.
    -
    When we subtract a negative number, first we change those two negative signs to addition:
    \(\displaystyle{ -\frac{5}{6} - \left(-\frac{3}{10}\right) = -\frac{5}{6} + \frac{3}{10} }\)
    Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) -
    Look at the first fraction \(\displaystyle{-\frac{5}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned}-\frac{5}{6}\amp = -\frac{5 \cdot 5}{6 \cdot 5}\\ \amp = -\frac{25}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) -
    \(\displaystyle{\begin{aligned}\frac{3}{10} \amp = \frac{3 \cdot 3}{10 \cdot 3}\\ \amp = \frac{9}{30}\end{aligned} }\)
    Now we can add those two fractions.
    \(\displaystyle{\begin{aligned} +
    Answer.
    \(-{\frac{8}{15}}\)
    Explanation.
    +
    When we subtract a negative number, first we change those two negative signs to addition:
    \(\displaystyle{ -\frac{5}{6} - \left(-\frac{3}{10}\right) = -\frac{5}{6} + \frac{3}{10} }\)
    Next, we will do fraction addition. To add two fractions with different denominators, we first have to find a common denominator.
    In this case, the least common multiple of \(6\) and \(10\) is \(30\text{.}\) +
    Look at the first fraction \(\displaystyle{-\frac{5}{6}}\text{.}\) To change the denominator to \(30\text{,}\) we multiply \(5\) at both the top and bottom, and we have:
    \(\displaystyle{\begin{aligned}-\frac{5}{6}\amp = -\frac{5 \cdot 5}{6 \cdot 5}\\ \amp = -\frac{25}{30} \end{aligned}}\)
    Similarly, we change the second fraction’s denominator to \(30\text{:}\) +
    \(\displaystyle{\begin{aligned}\frac{3}{10} \amp = \frac{3 \cdot 3}{10 \cdot 3}\\ \amp = \frac{9}{30}\end{aligned} }\)
    Now we can add those two fractions.
    \(\displaystyle{\begin{aligned} -\frac{5}{6} + \frac{3}{10} \amp = -\frac{25}{30} + \frac{9}{30} \\ \amp = \frac{-25 + 9}{30} \\ @@ -2412,39 +2425,39 @@

    Adding/Subtracting Fractions.

    75.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{ -2 - \frac{20}{7}}\) +
    Subtract: \(\displaystyle{ 1 - \frac{16}{5}}\)
    -
    Answer.
    \(-{\frac{34}{7}}\)
    Explanation.
    -
    When doing arithmetic with fractions, it helps to rewrite any integers as fractions:
    \(\displaystyle{ -2 = \frac{-2}{1} }\)
    Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator \(7\text{.}\) We will change the first fraction to have the common denominator:
    \(\displaystyle{ \begin{aligned}\frac{-2}{1} \amp = \frac{-2 \cdot 7}{1 \cdot 7}\\ \amp = \frac{-14}{7} \end{aligned}}\)
    Finally, we subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} --2 - \frac{20}{7} -\amp = \frac{-14}{7} - \frac{20}{7} \\ -\amp = \frac{-14 - 20}{7} \\ -\amp = -\frac{34}{7} +
    Answer.
    \(-{\frac{11}{5}}\)
    Explanation.
    +
    When doing arithmetic with fractions, it helps to rewrite any integers as fractions:
    \(\displaystyle{ 1 = \frac{1}{1} }\)
    Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator \(5\text{.}\) We will change the first fraction to have the common denominator:
    \(\displaystyle{ \begin{aligned}\frac{1}{1} \amp = \frac{1 \cdot 5}{1 \cdot 5}\\ \amp = \frac{5}{5} \end{aligned}}\)
    Finally, we subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} +1 - \frac{16}{5} +\amp = \frac{5}{5} - \frac{16}{5} \\ +\amp = \frac{5 - 16}{5} \\ +\amp = -\frac{11}{5} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{ -\frac{34}{7}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{ -\frac{11}{5}}\text{.}\)

    76.

    -
    -
    +
    +
    -
    Subtract: \(\displaystyle{ -1 - \frac{15}{4}}\) +
    Subtract: \(\displaystyle{ 2 - \frac{21}{2}}\)
    -
    Answer.
    \(-{\frac{19}{4}}\)
    Explanation.
    -
    When doing arithmetic with fractions, it helps to rewrite any integers as fractions:
    \(\displaystyle{ -1 = \frac{-1}{1} }\)
    Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator \(4\text{.}\) We will change the first fraction to have the common denominator:
    \(\displaystyle{ \begin{aligned}\frac{-1}{1} \amp = \frac{-1 \cdot 4}{1 \cdot 4}\\ \amp = \frac{-4}{4} \end{aligned}}\)
    Finally, we subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} --1 - \frac{15}{4} -\amp = \frac{-4}{4} - \frac{15}{4} \\ -\amp = \frac{-4 - 15}{4} \\ -\amp = -\frac{19}{4} +
    Answer.
    \(-{\frac{17}{2}}\)
    Explanation.
    +
    When doing arithmetic with fractions, it helps to rewrite any integers as fractions:
    \(\displaystyle{ 2 = \frac{2}{1} }\)
    Next, to subtract a fraction, we need to find a common denominator. In this case, it is simply the second denominator \(2\text{.}\) We will change the first fraction to have the common denominator:
    \(\displaystyle{ \begin{aligned}\frac{2}{1} \amp = \frac{2 \cdot 2}{1 \cdot 2}\\ \amp = \frac{4}{2} \end{aligned}}\)
    Finally, we subtract the numerators and keep the denominator unchanged.
    \(\displaystyle{\begin{aligned} +2 - \frac{21}{2} +\amp = \frac{4}{2} - \frac{21}{2} \\ +\amp = \frac{4 - 21}{2} \\ +\amp = -\frac{17}{2} \end{aligned} -}\)
    The answer to this question is \(\displaystyle{ -\frac{19}{4}}\text{.}\) +}\)
    The answer to this question is \(\displaystyle{ -\frac{17}{2}}\text{.}\)
    @@ -2456,491 +2469,493 @@

    Adding/Subtracting Fractions.

    Applications.

    77.

    -
    -
    +
    +
    -
    Kurt walked \({{\frac{2}{9}}}\) of a mile in the morning, and then walked \({{\frac{3}{11}}}\) of a mile in the afternoon. How far did Kurt walk altogether?
    Kurt walked a total of of a mile.
    +
    Thanh walked \({{\frac{3}{10}}}\) of a mile in the morning, and then walked \({{\frac{1}{7}}}\) of a mile in the afternoon. How far did Thanh walk altogether?
    Thanh walked a total of of a mile.
    -
    Answer.
    \({\frac{49}{99}}\)
    Explanation.
    -
    This is an addition problem.
    \(\displaystyle{ +
    Answer.
    \({\frac{31}{70}}\)
    Explanation.
    +
    This is an addition problem.
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{2}{9}}} + {{\frac{3}{11}}} -\amp = \frac{2 \cdot 11}{9 \cdot 11}+\frac{3 \cdot 9}{11 \cdot 9} \\ -\amp = \frac{22}{99} + \frac{27}{99} \\ -\amp = \frac{22+27}{99} \\ -\amp = {{\frac{49}{99}}} +{{\frac{3}{10}}} + {{\frac{1}{7}}} +\amp = \frac{3 \cdot 7}{10 \cdot 7}+\frac{1 \cdot 10}{7 \cdot 10} \\ +\amp = \frac{21}{70} + \frac{10}{70} \\ +\amp = \frac{21+10}{70} \\ +\amp = {{\frac{31}{70}}} \end{aligned} -}\)
    Kurt walked a total of \({{\frac{49}{99}}}\) miles altogether.
    +}\)
    Thanh walked a total of \({{\frac{31}{70}}}\) miles altogether.

    78.

    -
    -
    +
    +
    -
    Jerry walked \({{\frac{3}{10}}}\) of a mile in the morning, and then walked \({{\frac{5}{12}}}\) of a mile in the afternoon. How far did Jerry walk altogether?
    Jerry walked a total of of a mile.
    +
    Blake walked \({{\frac{2}{11}}}\) of a mile in the morning, and then walked \({{\frac{1}{10}}}\) of a mile in the afternoon. How far did Blake walk altogether?
    Blake walked a total of of a mile.
    -
    Answer.
    \({\frac{43}{60}}\)
    Explanation.
    -
    This is an addition problem.
    \(\displaystyle{ +
    Answer.
    \({\frac{31}{110}}\)
    Explanation.
    +
    This is an addition problem.
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{3}{10}}} + {{\frac{5}{12}}} -\amp = \frac{3 \cdot 6}{10 \cdot 6}+\frac{5 \cdot 5}{12 \cdot 5} \\ -\amp = \frac{18}{60} + \frac{25}{60} \\ -\amp = \frac{18+25}{60} \\ -\amp = {{\frac{43}{60}}} +{{\frac{2}{11}}} + {{\frac{1}{10}}} +\amp = \frac{2 \cdot 10}{11 \cdot 10}+\frac{1 \cdot 11}{10 \cdot 11} \\ +\amp = \frac{20}{110} + \frac{11}{110} \\ +\amp = \frac{20+11}{110} \\ +\amp = {{\frac{31}{110}}} \end{aligned} -}\)
    Jerry walked a total of \({{\frac{43}{60}}}\) miles altogether.
    +}\)
    Blake walked a total of \({{\frac{31}{110}}}\) miles altogether.

    79.

    -
    -
    +
    +
    -
    Samantha and Randi are sharing a pizza. Samantha ate \({{\frac{1}{9}}}\) of the pizza, and Randi ate \({{\frac{1}{6}}}\) of the pizza. How much of the pizza was eaten in total?
    They ate of the pizza.
    +
    Diane and Peter are sharing a pizza. Diane ate \({{\frac{1}{10}}}\) of the pizza, and Peter ate \({{\frac{1}{6}}}\) of the pizza. How much of the pizza was eaten in total?
    They ate of the pizza.
    -
    Answer.
    \({\frac{5}{18}}\)
    Explanation.
    -
    This is an addition problem.
    \(\displaystyle{ +
    Answer.
    \({\frac{4}{15}}\)
    Explanation.
    +
    This is an addition problem.
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{1}{9}}} + {{\frac{1}{6}}} -\amp = \frac{1 \cdot 2}{9 \cdot 2}+\frac{1 \cdot 3}{6 \cdot 3} \\ -\amp = \frac{2}{18} + \frac{3}{18} \\ -\amp = \frac{2+3}{18} \\ -\amp = {{\frac{5}{18}}} +{{\frac{1}{10}}} + {{\frac{1}{6}}} +\amp = \frac{1 \cdot 3}{10 \cdot 3}+\frac{1 \cdot 5}{6 \cdot 5} \\ +\amp = \frac{3}{30} + \frac{5}{30} \\ +\amp = \frac{3+5}{30} \\ +\amp = \frac{8}{30} \\ +\amp = {{\frac{4}{15}}} \end{aligned} -}\)
    They ate \({{\frac{5}{18}}}\) of the pizza.
    +}\)
    They ate \({{\frac{4}{15}}}\) of the pizza.

    80.

    -
    -
    +
    +
    -
    A trail’s total length is \({{\frac{27}{40}}}\) of a mile. It has two legs. The first leg is \({{\frac{3}{8}}}\) of a mile long. How long is the second leg?
    The second leg is of a mile in length.
    +
    A trail’s total length is \({{\frac{13}{40}}}\) of a mile. It has two legs. The first leg is \({{\frac{1}{8}}}\) of a mile long. How long is the second leg?
    The second leg is of a mile in length.
    -
    Answer.
    \({\frac{3}{10}}\)
    Explanation.
    -
    This is a subtraction problem.
    \(\displaystyle{ +
    Answer.
    \({\frac{1}{5}}\)
    Explanation.
    +
    This is a subtraction problem.
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{27}{40}}} - {{\frac{3}{8}}} -\amp = \frac{27}{40}-\frac{3 \cdot 5}{8 \cdot 5} \\ -\amp = \frac{27}{40} - \frac{15}{40} \\ -\amp = \frac{27-15}{40} \\ -\amp = \frac{12}{40} \\ -\amp = {{\frac{3}{10}}} +{{\frac{13}{40}}} - {{\frac{1}{8}}} +\amp = \frac{13}{40}-\frac{1 \cdot 5}{8 \cdot 5} \\ +\amp = \frac{13}{40} - \frac{5}{40} \\ +\amp = \frac{13-5}{40} \\ +\amp = \frac{8}{40} \\ +\amp = {{\frac{1}{5}}} \end{aligned} -}\)
    The second leg is \({{\frac{3}{10}}}\) of a mile in length.
    +}\)
    The second leg is \({{\frac{1}{5}}}\) of a mile in length.

    81.

    -
    -
    +
    +
    -
    A trail’s total length is \({{\frac{19}{45}}}\) of a mile. It has two legs. The first leg is \({{\frac{2}{9}}}\) of a mile long. How long is the second leg?
    The second leg is of a mile in length.
    +
    A trail’s total length is \({{\frac{11}{30}}}\) of a mile. It has two legs. The first leg is \({{\frac{1}{5}}}\) of a mile long. How long is the second leg?
    The second leg is of a mile in length.
    -
    Answer.
    \({\frac{1}{5}}\)
    Explanation.
    -
    This is a subtraction problem.
    \(\displaystyle{ +
    Answer.
    \({\frac{1}{6}}\)
    Explanation.
    +
    This is a subtraction problem.
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{19}{45}}} - {{\frac{2}{9}}} -\amp = \frac{19}{45}-\frac{2 \cdot 5}{9 \cdot 5} \\ -\amp = \frac{19}{45} - \frac{10}{45} \\ -\amp = \frac{19-10}{45} \\ -\amp = \frac{9}{45} \\ -\amp = {{\frac{1}{5}}} +{{\frac{11}{30}}} - {{\frac{1}{5}}} +\amp = \frac{11}{30}-\frac{1 \cdot 6}{5 \cdot 6} \\ +\amp = \frac{11}{30} - \frac{6}{30} \\ +\amp = \frac{11-6}{30} \\ +\amp = \frac{5}{30} \\ +\amp = {{\frac{1}{6}}} \end{aligned} -}\)
    The second leg is \({{\frac{1}{5}}}\) of a mile in length.
    +}\)
    The second leg is \({{\frac{1}{6}}}\) of a mile in length.

    82.

    -
    -
    +
    +
    -
    Carl is participating in a running event. In the first hour, he completed \({{\frac{1}{8}}}\) of the total distance. After another hour, in total he had completed \({{\frac{13}{40}}}\) of the total distance.
    What fraction of the total distance did Carl complete during the second hour?
    Carl completed of the distance during the second hour.
    +
    Huynh is participating in a running event. In the first hour, he completed \({{\frac{1}{10}}}\) of the total distance. After another hour, in total he had completed \({{\frac{4}{15}}}\) of the total distance.
    What fraction of the total distance did Huynh complete during the second hour?
    Huynh completed of the distance during the second hour.
    -
    Answer.
    \({\frac{1}{5}}\)
    Explanation.
    -
    This is a subtraction problem.
    \(\displaystyle{ +
    Answer.
    \({\frac{1}{6}}\)
    Explanation.
    +
    This is a subtraction problem.
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{13}{40}}} - {{\frac{1}{8}}} -\amp = \frac{13}{40}-\frac{1 \cdot 5}{8 \cdot 5} \\ -\amp = \frac{13}{40} - \frac{5}{40} \\ -\amp = \frac{13-5}{40} \\ -\amp = \frac{8}{40} \\ -\amp = {{\frac{1}{5}}} +{{\frac{4}{15}}} - {{\frac{1}{10}}} +\amp = \frac{4 \cdot 2}{15 \cdot 2}-\frac{1 \cdot 3}{10 \cdot 3} \\ +\amp = \frac{8}{30} - \frac{3}{30} \\ +\amp = \frac{8-3}{30} \\ +\amp = \frac{5}{30} \\ +\amp = {{\frac{1}{6}}} \end{aligned} -}\)
    Carl completed \({{\frac{1}{5}}}\) of the distance during the second hour.
    +}\)
    Huynh completed \({{\frac{1}{6}}}\) of the distance during the second hour.

    83.

    -
    -
    +
    +
    -
    The pie chart represents a school’s student population.
    Together, white and black students make up of the school’s population.
    +
    The pie chart represents a school’s student population.
    Together, white and black students make up of the school’s population.
    -
    Answer.
    \({\frac{1}{2}}\)
    Explanation.
    -
    By the pie chart, the school has \({{\frac{1}{3}}}\) white students, and \({{\frac{1}{6}}}\) black students.
    We will use addition to find the total portion of white and black students at this school.
    \(\displaystyle{\begin{aligned}[t] -\amp \phantom{{}=}{{\frac{1}{3}}}+{{\frac{1}{6}}} \\ -\amp = \frac{1 \cdot 2}{3 \cdot 2} + \frac{1}{6} \\ -\amp = \frac{2}{6} + \frac{1}{6} \\ -\amp = \frac{2+1}{6} \\ -\amp = \frac{3}{6} \\ \amp = \frac{1}{2} \\ +
    Answer.
    \({\frac{13}{24}}\)
    Explanation.
    +
    By the pie chart, the school has \({{\frac{1}{6}}}\) white students, and \({{\frac{3}{8}}}\) black students.
    We will use addition to find the total portion of white and black students at this school.
    \(\displaystyle{\begin{aligned}[t] +\amp \phantom{{}=}{{\frac{1}{6}}}+{{\frac{3}{8}}} \\ +\amp = \frac{1 \cdot 4}{6 \cdot 4} + \frac{3 \cdot 4}{8 \cdot 4} \\ +\amp = \frac{4}{24} + \frac{9}{24} \\ +\amp = \frac{4+9}{24} \\ +\amp = \frac{13}{24} \\ \end{aligned} -}\)
    Together, white and black students make up \({{\frac{1}{2}}}\) of the school’s population.
    +}\)
    Together, white and black students make up \({{\frac{13}{24}}}\) of the school’s population.

    84.

    -
    -
    +
    +
    -
    Each page of a book is \({5 {\textstyle\frac{1}{6}}}\) inches in height, and consists of a header (a top margin), a footer (a bottom margin), and the middle part (the body). The header is \({{\frac{7}{9}}}\) of an inch thick and the middle part is \({4 {\textstyle\frac{1}{9}}}\) inches from top to bottom.
    What is the thickness of the footer?
    The footer is of an inch thick.
    +
    Each page of a book is \({7 {\textstyle\frac{1}{2}}}\) inches in height, and consists of a header (a top margin), a footer (a bottom margin), and the middle part (the body). The header is \({{\frac{2}{7}}}\) of an inch thick and the middle part is \({6 {\textstyle\frac{6}{7}}}\) inches from top to bottom.
    What is the thickness of the footer?
    The footer is of an inch thick.
    -
    Answer.
    \({\frac{5}{18}}\)
    Explanation.
    -
    From a page’s total height, we subtract the thickness of the header and the height of the body to find the thickness of the footer:
    \(\displaystyle{ +
    Answer.
    \({\frac{5}{14}}\)
    Explanation.
    +
    From a page’s total height, we subtract the thickness of the header and the height of the body to find the thickness of the footer:
    \(\displaystyle{ \begin{aligned}[t] -{5 {\textstyle\frac{1}{6}}} - {{\frac{7}{9}}} - {4 {\textstyle\frac{1}{9}}} -\amp = 5 + \frac{1}{6} -{{\frac{7}{9}}} - 4 - {{\frac{1}{9}}} \\ -\amp = 1 + \frac{1}{6} -{{\frac{7}{9}}} - {{\frac{1}{9}}} \\ -\amp = \frac{1}{1} + \frac{1}{6} -{{\frac{7}{9}}} - {{\frac{1}{9}}} \\ -\amp = \frac{1 \cdot 18}{1 \cdot 18} + \frac{1\cdot3}{6\cdot3} - \frac{7\cdot2}{9\cdot2} - \frac{1\cdot2}{9\cdot2} \\ -\amp = \frac{18}{18} + \frac{3}{18} - \frac{14}{18} - \frac{2}{18} \\ -\amp = \frac{18+3-14-2}{18} \\ -\amp = {{\frac{5}{18}}} +{7 {\textstyle\frac{1}{2}}} - {{\frac{2}{7}}} - {6 {\textstyle\frac{6}{7}}} +\amp = 7 + \frac{1}{2} -{{\frac{2}{7}}} - 6 - {{\frac{6}{7}}} \\ +\amp = 1 + \frac{1}{2} -{{\frac{2}{7}}} - {{\frac{6}{7}}} \\ +\amp = \frac{1}{1} + \frac{1}{2} -{{\frac{2}{7}}} - {{\frac{6}{7}}} \\ +\amp = \frac{1 \cdot 14}{1 \cdot 14} + \frac{1\cdot7}{2\cdot7} - \frac{2\cdot2}{7\cdot2} - \frac{6\cdot2}{7\cdot2} \\ +\amp = \frac{14}{14} + \frac{7}{14} - \frac{4}{14} - \frac{12}{14} \\ +\amp = \frac{14+7-4-12}{14} \\ +\amp = {{\frac{5}{14}}} \end{aligned} -}\)
    The footer is \({{\frac{5}{18}}}\) inches thick.
    +}\)
    The footer is \({{\frac{5}{14}}}\) inches thick.

    85.

    -
    -
    +
    +
    -
    Sherial and Wendy are sharing a pizza. Sherial ate \({{\frac{2}{7}}}\) of the pizza, and Wendy ate \({{\frac{1}{8}}}\) of the pizza. How much more pizza did Sherial eat than Wendy?
    Sherial ate more of the pizza than Wendy ate.
    +
    Penelope and Michele are sharing a pizza. Penelope ate \({{\frac{3}{8}}}\) of the pizza, and Michele ate \({{\frac{1}{9}}}\) of the pizza. How much more pizza did Penelope eat than Michele?
    Penelope ate more of the pizza than Michele ate.
    -
    Answer.
    \({\frac{9}{56}}\)
    Explanation.
    -
    To find the difference of two numbers, we use subtraction:
    \(\displaystyle{ +
    Answer.
    \({\frac{19}{72}}\)
    Explanation.
    +
    To find the difference of two numbers, we use subtraction:
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{2}{7}}} - {{\frac{1}{8}}} -\amp = \frac{2 \cdot 8}{7 \cdot 8}-\frac{1 \cdot 7}{8 \cdot 7} \\ -\amp = \frac{16}{56} - \frac{7}{56} \\ -\amp = \frac{16-7}{56} \\ -\amp = {{\frac{9}{56}}} +{{\frac{3}{8}}} - {{\frac{1}{9}}} +\amp = \frac{3 \cdot 9}{8 \cdot 9}-\frac{1 \cdot 8}{9 \cdot 8} \\ +\amp = \frac{27}{72} - \frac{8}{72} \\ +\amp = \frac{27-8}{72} \\ +\amp = {{\frac{19}{72}}} \end{aligned} -}\)
    Sherial ate \({{\frac{9}{56}}}\) more of the pizza than Wendy ate.
    +}\)
    Penelope ate \({{\frac{19}{72}}}\) more of the pizza than Michele ate.

    86.

    -
    -
    +
    +
    -
    Hayden and Maygen are sharing a pizza. Hayden ate \({{\frac{3}{8}}}\) of the pizza, and Maygen ate \({{\frac{1}{6}}}\) of the pizza. How much more pizza did Hayden eat than Maygen?
    Hayden ate more of the pizza than Maygen ate.
    +
    Samantha and Anthony are sharing a pizza. Samantha ate \({{\frac{2}{9}}}\) of the pizza, and Anthony ate \({{\frac{1}{5}}}\) of the pizza. How much more pizza did Samantha eat than Anthony?
    Samantha ate more of the pizza than Anthony ate.
    -
    Answer.
    \({\frac{5}{24}}\)
    Explanation.
    -
    To find the difference of two numbers, we use subtraction:
    \(\displaystyle{ +
    Answer.
    \({\frac{1}{45}}\)
    Explanation.
    +
    To find the difference of two numbers, we use subtraction:
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{3}{8}}} - {{\frac{1}{6}}} -\amp = \frac{3 \cdot 3}{8 \cdot 3}-\frac{1 \cdot 4}{6 \cdot 4} \\ -\amp = \frac{9}{24} - \frac{4}{24} \\ -\amp = \frac{9-4}{24} \\ -\amp = {{\frac{5}{24}}} +{{\frac{2}{9}}} - {{\frac{1}{5}}} +\amp = \frac{2 \cdot 5}{9 \cdot 5}-\frac{1 \cdot 9}{5 \cdot 9} \\ +\amp = \frac{10}{45} - \frac{9}{45} \\ +\amp = \frac{10-9}{45} \\ +\amp = {{\frac{1}{45}}} \end{aligned} -}\)
    Hayden ate \({{\frac{5}{24}}}\) more of the pizza than Maygen ate.
    +}\)
    Samantha ate \({{\frac{1}{45}}}\) more of the pizza than Anthony ate.

    87.

    -
    -
    +
    +
    -
    A school had a fund-raising event. The revenue came from three resources: ticket sales, auction sales, and donations. Ticket sales account for \({{\frac{5}{8}}}\) of the total revenue; auction sales account for \({{\frac{1}{5}}}\) of the total revenue. What fraction of the revenue came from donations?
    +
    A school had a fund-raising event. The revenue came from three resources: ticket sales, auction sales, and donations. Ticket sales account for \({{\frac{7}{10}}}\) of the total revenue; auction sales account for \({{\frac{1}{6}}}\) of the total revenue. What fraction of the revenue came from donations?
    of the revenue came from donations.
    -
    Answer.
    \({\frac{7}{40}}\)
    Explanation.
    -
    The revenue consists of \(3\) parts. The total revenue is like a big \(1\text{;}\) ticket sales account for \({{\frac{5}{8}}}\text{;}\) auction sales account for \({{\frac{1}{5}}}\text{.}\) To find the missing piece (from donations), we use subtraction.
    Note that a common denominator of \(1\text{,}\) \({{\frac{5}{8}}}\) and \({{\frac{1}{5}}}\) is \(40\text{.}\) -
    \(\displaystyle{\begin{aligned}[t] -1 - {{\frac{5}{8}}} - {{\frac{1}{5}}} -\amp = \frac{40}{40} - \frac{5\cdot5}{8\cdot5} - \frac{1\cdot8}{5\cdot8} \\ -\amp = \frac{40-25-8}{40} \\ -\amp = {{\frac{7}{40}}} +
    Answer.
    \({\frac{2}{15}}\)
    Explanation.
    +
    The revenue consists of \(3\) parts. The total revenue is like a big \(1\text{;}\) ticket sales account for \({{\frac{7}{10}}}\text{;}\) auction sales account for \({{\frac{1}{6}}}\text{.}\) To find the missing piece (from donations), we use subtraction.
    Note that a common denominator of \(1\text{,}\) \({{\frac{7}{10}}}\) and \({{\frac{1}{6}}}\) is \(30\text{.}\) +
    \(\displaystyle{\begin{aligned}[t] +1 - {{\frac{7}{10}}} - {{\frac{1}{6}}} +\amp = \frac{30}{30} - \frac{7\cdot3}{10\cdot3} - \frac{1\cdot5}{6\cdot5} \\ +\amp = \frac{30-21-5}{30} \\ +\amp = \frac{4}{30} \\ +\amp = {{\frac{2}{15}}} \end{aligned} -}\)
    So \({{\frac{7}{40}}}\) of the revenue came from donations.
    +}\)
    So \({{\frac{2}{15}}}\) of the revenue came from donations.

    88.

    -
    -
    +
    +
    -
    A few years back, a car was purchased for \({\$18{,}600}\text{.}\) Today it is worth \({{\frac{1}{6}}}\) of its original value. What is the car’s current value?
    The car’s current value is .
    +
    A few years back, a car was purchased for \({\$28{,}800}\text{.}\) Today it is worth \({{\frac{1}{6}}}\) of its original value. What is the car’s current value?
    The car’s current value is .
    -
    Answer.
    \(\$3{,}100\)
    Explanation.
    -
    When we use the word “of” in situations like “\({{\frac{1}{6}}}\) of \({\$18{,}600}\text{,}\)” we can translate “of” into the multiplication symbol.
    In this problem, to find \({{\frac{1}{6}}}\) of \({\$18{,}600}\text{,}\) we do:
    \(\displaystyle{ +
    Answer.
    \(\$4{,}800\)
    Explanation.
    +
    When we use the word “of” in situations like “\({{\frac{1}{6}}}\) of \({\$28{,}800}\text{,}\)” we can translate “of” into the multiplication symbol.
    In this problem, to find \({{\frac{1}{6}}}\) of \({\$28{,}800}\text{,}\) we do:
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{1}{6}}} \cdot 18600 -\amp = {{\frac{1}{6}}} \cdot \frac{18600}{1} \\ -\amp = \frac{1}{1} \cdot \frac{3100}{1} \\ -\amp = 3100 +{{\frac{1}{6}}} \cdot 28800 +\amp = {{\frac{1}{6}}} \cdot \frac{28800}{1} \\ +\amp = \frac{1}{1} \cdot \frac{4800}{1} \\ +\amp = 4800 \end{aligned} -}\)
    The car is worth \({\$3{,}100}\) now.
    +}\)
    The car is worth \({\$4{,}800}\) now.

    89.

    -
    -
    +
    +
    -
    A few years back, a car was purchased for \({\$27{,}000}\text{.}\) Today it is worth \({{\frac{1}{6}}}\) of its original value. What is the car’s current value?
    The car’s current value is .
    +
    A few years back, a car was purchased for \({\$12{,}300}\text{.}\) Today it is worth \({{\frac{1}{3}}}\) of its original value. What is the car’s current value?
    The car’s current value is .
    -
    Answer.
    \(\$4{,}500\)
    Explanation.
    -
    When we use the word “of” in situations like “\({{\frac{1}{6}}}\) of \({\$27{,}000}\text{,}\)” we can translate “of” into the multiplication symbol.
    In this problem, to find \({{\frac{1}{6}}}\) of \({\$27{,}000}\text{,}\) we do:
    \(\displaystyle{ +
    Answer.
    \(\$4{,}100\)
    Explanation.
    +
    When we use the word “of” in situations like “\({{\frac{1}{3}}}\) of \({\$12{,}300}\text{,}\)” we can translate “of” into the multiplication symbol.
    In this problem, to find \({{\frac{1}{3}}}\) of \({\$12{,}300}\text{,}\) we do:
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{1}{6}}} \cdot 27000 -\amp = {{\frac{1}{6}}} \cdot \frac{27000}{1} \\ -\amp = \frac{1}{1} \cdot \frac{4500}{1} \\ -\amp = 4500 +{{\frac{1}{3}}} \cdot 12300 +\amp = {{\frac{1}{3}}} \cdot \frac{12300}{1} \\ +\amp = \frac{1}{1} \cdot \frac{4100}{1} \\ +\amp = 4100 \end{aligned} -}\)
    The car is worth \({\$4{,}500}\) now.
    +}\)
    The car is worth \({\$4{,}100}\) now.

    90.

    -
    -
    +
    +
    -
    The pie chart represents a school’s student population.
    +
    The pie chart represents a school’s student population.
    more of the school is white students than black students.
    -
    Answer.
    \({\frac{1}{6}}\)
    Explanation.
    -
    By the pie chart, the school has \({{\frac{1}{3}}}\) white students, and \({{\frac{1}{6}}}\) black students.
    We will use subtraction to find the difference:
    \(\displaystyle{\begin{aligned}[t] -\amp \phantom{{}=}{{\frac{1}{3}}}-{{\frac{1}{6}}} \\ -\amp = \frac{1 \cdot 2}{3 \cdot 2} - \frac{1}{6} \\ -\amp = \frac{2}{6} - \frac{1}{6} \\ -\amp = \frac{2-1}{6} \\ -\amp = \frac{1}{6} \\ +
    Answer.
    \({\frac{1}{12}}\)
    Explanation.
    +
    By the pie chart, the school has \({{\frac{1}{4}}}\) white students, and \({{\frac{1}{6}}}\) black students.
    We will use subtraction to find the difference:
    \(\displaystyle{\begin{aligned}[t] +\amp \phantom{{}=}{{\frac{1}{4}}}-{{\frac{1}{6}}} \\ +\amp = \frac{1 \cdot 3}{4 \cdot 3} - \frac{1 \cdot 3}{6 \cdot 3} \\ +\amp = \frac{3}{12} - \frac{2}{12} \\ +\amp = \frac{3-2}{12} \\ +\amp = \frac{1}{12} \\ \end{aligned} }\)

    91.

    -
    -
    +
    +
    -
    A town has \(150\) residents in total, of which \({{\frac{2}{3}}}\) are white/Caucasian Americans. How many white/Caucasian Americans reside in this town?
    There are white/Caucasian Americans residing in this town.
    +
    A town has \(200\) residents in total, of which \({{\frac{3}{4}}}\) are Asian Americans. How many Asian Americans reside in this town?
    There are Asian Americans residing in this town.
    -
    Answer.
    \(100\)
    Explanation.
    -
    When we use the word “of” in expressions like “\({{\frac{2}{3}}}\) of \(150\text{,}\)” we can translate “of” into the multiplication symbol.
    In this problem, to find \({{\frac{2}{3}}}\) of \(150\text{,}\) we compute:
    \(\displaystyle{ +
    Answer.
    \(150\)
    Explanation.
    +
    When we use the word “of” in expressions like “\({{\frac{3}{4}}}\) of \(200\text{,}\)” we can translate “of” into the multiplication symbol.
    In this problem, to find \({{\frac{3}{4}}}\) of \(200\text{,}\) we compute:
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{2}{3}}} \cdot 150 -\amp = {{\frac{2}{3}}} \cdot \frac{150}{1} \\ -\amp = \frac{2}{1} \cdot \frac{50}{1} \\ -\amp = \frac{{100}}{1} \\ -\amp = {100} +{{\frac{3}{4}}} \cdot 200 +\amp = {{\frac{3}{4}}} \cdot \frac{200}{1} \\ +\amp = \frac{3}{1} \cdot \frac{50}{1} \\ +\amp = \frac{{150}}{1} \\ +\amp = {150} \end{aligned} -}\)
    There are \({100}\) white/Caucasian Americans residing in this town.
    +}\)
    There are \({150}\) Asian Americans residing in this town.

    92.

    -
    -
    +
    +
    -
    A company received a grant, and decided to spend \({{\frac{5}{12}}}\) of this grant in research and development next year. Out of the money set aside for research and development, \({{\frac{2}{3}}}\) will be used to buy new equipment. What fraction of the grant will be used to buy new equipment?
    +
    A company received a grant, and decided to spend \({{\frac{13}{18}}}\) of this grant in research and development next year. Out of the money set aside for research and development, \({{\frac{8}{13}}}\) will be used to buy new equipment. What fraction of the grant will be used to buy new equipment?
    of the grant will be used to buy new equipment.
    -
    Answer.
    \({\frac{5}{18}}\)
    Explanation.
    -
    For this problem, we are trying to find \({{\frac{2}{3}}}\) of \({{\frac{5}{12}}}\text{.}\) The word “of” implies multiplication:
    \(\displaystyle{ +
    Answer.
    \({\frac{4}{9}}\)
    Explanation.
    +
    For this problem, we are trying to find \({{\frac{8}{13}}}\) of \({{\frac{13}{18}}}\text{.}\) The word “of” implies multiplication:
    \(\displaystyle{ \begin{aligned}[t] -\frac{5}{12} \cdot \frac{2}{3} -\amp = \frac{5}{6} \cdot \frac{1}{3} \\ -\amp = \frac{5\cdot1}{6\cdot3} \\ -\amp = {{\frac{5}{18}}} +\frac{13}{18} \cdot \frac{8}{13} +\amp = \frac{1}{9} \cdot \frac{4}{1} \\ +\amp = \frac{1\cdot4}{9\cdot1} \\ +\amp = {{\frac{4}{9}}} \end{aligned} -}\)
    So \({{\frac{5}{18}}}\) of the grant will be used to buy new equipment.
    +}\)
    So \({{\frac{4}{9}}}\) of the grant will be used to buy new equipment.

    93.

    -
    -
    +
    +
    -
    A food bank just received \(20\) kilograms of emergency food. Each family in need is to receive \({{\frac{2}{5}}}\) kilograms of food. How many families can be served with the \(20\) kilograms of food?
    - families can be served with the \(20\) kilograms of food.
    +
    A food bank just received \(38\) kilograms of emergency food. Each family in need is to receive \({{\frac{2}{5}}}\) kilograms of food. How many families can be served with the \(38\) kilograms of food?
    + families can be served with the \(38\) kilograms of food.
    -
    Answer.
    \(50\)
    Explanation.
    -
    We can address this problem as repeatedly taking away \({{\frac{2}{5}}}\) kilograms from \(20\) kilograms, which implies a division problem:
    \(\displaystyle{ +
    Answer.
    \(95\)
    Explanation.
    +
    We can address this problem as repeatedly taking away \({{\frac{2}{5}}}\) kilograms from \(38\) kilograms, which implies a division problem:
    \(\displaystyle{ \begin{aligned}[t] -20 \div {{\frac{2}{5}}} -\amp = \frac{20}{1} \cdot \frac{5}{2} \\ -\amp = \frac{10}{1} \cdot \frac{5}{1} \\ -\amp = \frac{10\cdot5}{1 \cdot 1} \\ -\amp = {50} +38 \div {{\frac{2}{5}}} +\amp = \frac{38}{1} \cdot \frac{5}{2} \\ +\amp = \frac{19}{1} \cdot \frac{5}{1} \\ +\amp = \frac{19\cdot5}{1 \cdot 1} \\ +\amp = {95} \end{aligned} -}\)
    -\({50}\) families can be served with the \(20\) kilograms of food.
    +}\)
    +\({95}\) families can be served with the \(38\) kilograms of food.

    94.

    -
    -
    +
    +
    -
    A construction team maintains a \({85}\)-mile-long sewage pipe. Each day, the team can cover \({{\frac{5}{6}}}\) of a mile. How many days will it take the team to complete the maintenance of the entire sewage pipe?
    It will take the team days to complete maintaining the entire sewage pipe.
    +
    A construction team maintains a \({90}\)-mile-long sewage pipe. Each day, the team can cover \({{\frac{5}{8}}}\) of a mile. How many days will it take the team to complete the maintenance of the entire sewage pipe?
    It will take the team days to complete maintaining the entire sewage pipe.
    -
    Answer.
    \(102\)
    Explanation.
    -
    In this problem, the team maintains \({{\frac{5}{6}}}\) miles of sewage pipe every day, until they complete all \({85}\) miles. This is like repeatedly taking away \({{\frac{5}{6}}}\) miles from \({85}\) miles, a division problem.
    \(\displaystyle{ +
    Answer.
    \(144\)
    Explanation.
    +
    In this problem, the team maintains \({{\frac{5}{8}}}\) miles of sewage pipe every day, until they complete all \({90}\) miles. This is like repeatedly taking away \({{\frac{5}{8}}}\) miles from \({90}\) miles, a division problem.
    \(\displaystyle{ \begin{aligned}[t] -{85} \div {{\frac{5}{6}}} -\amp = \frac{{85}}{1} \div \frac{5}{6} \\ -\amp = \frac{{85}}{1} \cdot \frac{6}{5} \\ -\amp = \frac{{17}}{1} \cdot \frac{6}{1} \\ -\amp = \frac{{17} \cdot 6}{1 \cdot 1} \\ -\amp = {102} +{90} \div {{\frac{5}{8}}} +\amp = \frac{{90}}{1} \div \frac{5}{8} \\ +\amp = \frac{{90}}{1} \cdot \frac{8}{5} \\ +\amp = \frac{{18}}{1} \cdot \frac{8}{1} \\ +\amp = \frac{{18} \cdot 8}{1 \cdot 1} \\ +\amp = {144} \end{aligned} -}\)
    It will take the team \({102}\) days to complete maintaining the entire sewage pipe.
    +}\)
    It will take the team \({144}\) days to complete maintaining the entire sewage pipe.

    95.

    -
    -
    +
    +
    -
    A child is stacking up tiles. Each tile’s height is \({{\frac{2}{3}}}\) of a centimeter. How many layers of tiles are needed to reach \({12}\) centimeters in total height?
    To reach the total height of \({12}\) centimeters, layers of tiles are needed.
    +
    A child is stacking up tiles. Each tile’s height is \({{\frac{3}{4}}}\) of a centimeter. How many layers of tiles are needed to reach \({15}\) centimeters in total height?
    To reach the total height of \({15}\) centimeters, layers of tiles are needed.
    -
    Answer.
    \(18\)
    Explanation.
    -
    In this problem, each layer of tile is \({{\frac{2}{3}}}\) centimeters tall. The child will keep stacking up layers until the total height reaches \({12}\) centimeters. This is like asking how many “\({{\frac{2}{3}}}\) centimeters” are there in \({12}\) centimeters, a division problem.
    \(\displaystyle{ +
    Answer.
    \(20\)
    Explanation.
    +
    In this problem, each layer of tile is \({{\frac{3}{4}}}\) centimeters tall. The child will keep stacking up layers until the total height reaches \({15}\) centimeters. This is like asking how many “\({{\frac{3}{4}}}\) centimeters” are there in \({15}\) centimeters, a division problem.
    \(\displaystyle{ \begin{aligned}[t] -{12} \div {{\frac{2}{3}}} -\amp = \frac{{12}}{1} \div \frac{2}{3} \\ -\amp = \frac{{12}}{1} \cdot \frac{3}{2} \\ -\amp = \frac{{6}}{1} \cdot \frac{3}{1} \\ -\amp = \frac{{6} \cdot 3}{1 \cdot 1} \\ -\amp = {18} +{15} \div {{\frac{3}{4}}} +\amp = \frac{{15}}{1} \div \frac{3}{4} \\ +\amp = \frac{{15}}{1} \cdot \frac{4}{3} \\ +\amp = \frac{{5}}{1} \cdot \frac{4}{1} \\ +\amp = \frac{{5} \cdot 4}{1 \cdot 1} \\ +\amp = {20} \end{aligned} -}\)
    To reach the total height of \({12}\) centimeters, \({18}\) layers of tiles are needed.
    +}\)
    To reach the total height of \({15}\) centimeters, \({20}\) layers of tiles are needed.

    96.

    -
    -
    +
    +
    -
    A restaurant made \(350\) cups of pudding for a festival.
    Customers at the festival will be served \({{\frac{1}{6}}}\) of a cup of pudding per serving. How many customers can the restaurant serve at the festival with the \(350\) cups of pudding?
    The restaurant can serve customers at the festival with the \(350\) cups of pudding.
    +
    A restaurant made \(450\) cups of pudding for a festival.
    Customers at the festival will be served \({{\frac{1}{5}}}\) of a cup of pudding per serving. How many customers can the restaurant serve at the festival with the \(450\) cups of pudding?
    The restaurant can serve customers at the festival with the \(450\) cups of pudding.
    -
    Answer.
    \(2100\)
    Explanation.
    -
    There are a total of \(350\) cups of pudding, with each serving being \({{\frac{1}{6}}}\) cup. To find how many servings can be served, we need to find how many “\({{\frac{1}{6}}}\) cups” are there in \(350\) cups, a division problem:
    \(\displaystyle{ +
    Answer.
    \(2250\)
    Explanation.
    +
    There are a total of \(450\) cups of pudding, with each serving being \({{\frac{1}{5}}}\) cup. To find how many servings can be served, we need to find how many “\({{\frac{1}{5}}}\) cups” are there in \(450\) cups, a division problem:
    \(\displaystyle{ \begin{aligned}[t] -350 \div {{\frac{1}{6}}} -\amp = \frac{350}{1} \div \frac{1}{6} \\ -\amp = \frac{350}{1} \cdot \frac{6}{1} \\ -\amp = \frac{350\cdot6}{1 \cdot 1} \\ -\amp = \frac{2100}{1} \\ -\amp = {2100} +450 \div {{\frac{1}{5}}} +\amp = \frac{450}{1} \div \frac{1}{5} \\ +\amp = \frac{450}{1} \cdot \frac{5}{1} \\ +\amp = \frac{450\cdot5}{1 \cdot 1} \\ +\amp = \frac{2250}{1} \\ +\amp = {2250} \end{aligned} -}\)
    The restaurant can serve \({2100}\) customers at the festival with the \(350\) cups of pudding.
    Shortcut
    Since each serving has \({{\frac{1}{6}}}\) cup, we know each cup has \(6\) servings. So \(350\) cups has \(350\cdot6={2100}\) servings.
    +}\)
    The restaurant can serve \({2250}\) customers at the festival with the \(450\) cups of pudding.
    Shortcut
    Since each serving has \({{\frac{1}{5}}}\) cup, we know each cup has \(5\) servings. So \(450\) cups has \(450\cdot5={2250}\) servings.

    97.

    -
    -
    +
    +
    -
    A \(2\times4\) piece of lumber in your garage is \({64 {\textstyle\frac{11}{32}}}\) inches long. A second \(2\times4\) is \({39 {\textstyle\frac{31}{32}}}\) inches long. If you lay them end to end, what will the total length be?
    The total length will be inches.
    +
    A \(2\times4\) piece of lumber in your garage is \({57 {\textstyle\frac{27}{32}}}\) inches long. A second \(2\times4\) is \({59 {\textstyle\frac{29}{32}}}\) inches long. If you lay them end to end, what will the total length be?
    The total length will be inches.
    -
    Answer.
    \(104 {\textstyle\frac{5}{16}}\)
    Explanation.
    -
    To find the total length in inches, we add the two lengths together.
    \(\begin{aligned} -{64 {\textstyle\frac{11}{32}}}+{39 {\textstyle\frac{31}{32}}} \amp = 64+\frac{11}{32}+39+\frac{31}{32}\\ -\amp = 103+\frac{11}{32}+\frac{31}{32}\\ -\amp = 103+{1 {\textstyle\frac{5}{16}}}\\ -\amp = {104 {\textstyle\frac{5}{16}}} -\end{aligned}\)
    So the total length is \({104 {\textstyle\frac{5}{16}}}\) inches.
    +
    Answer.
    \(117 {\textstyle\frac{3}{4}}\)
    Explanation.
    +
    To find the total length in inches, we add the two lengths together.
    \(\begin{aligned} +{57 {\textstyle\frac{27}{32}}}+{59 {\textstyle\frac{29}{32}}} \amp = 57+\frac{27}{32}+59+\frac{29}{32}\\ +\amp = 116+\frac{27}{32}+\frac{29}{32}\\ +\amp = 116+{1 {\textstyle\frac{3}{4}}}\\ +\amp = {117 {\textstyle\frac{3}{4}}} +\end{aligned}\)
    So the total length is \({117 {\textstyle\frac{3}{4}}}\) inches.

    98.

    -
    -
    +
    +
    -
    A \(2\times4\) piece of lumber in your garage is \({40 {\textstyle\frac{9}{32}}}\) inches long. A second \(2\times4\) is \({69 {\textstyle\frac{7}{16}}}\) inches long. If you lay them end to end, what will the total length be?
    The total length will be inches.
    +
    A \(2\times4\) piece of lumber in your garage is \({33 {\textstyle\frac{3}{4}}}\) inches long. A second \(2\times4\) is \({47 {\textstyle\frac{3}{8}}}\) inches long. If you lay them end to end, what will the total length be?
    The total length will be inches.
    -
    Answer.
    \(109 {\textstyle\frac{23}{32}}\)
    Explanation.
    -
    To find the total length in inches, we add the two lengths together.
    \(\begin{aligned} -{40 {\textstyle\frac{9}{32}}}+{69 {\textstyle\frac{7}{16}}} \amp = 40+\frac{9}{32}+69+\frac{7}{16}\\ -\amp = 109+\frac{9}{32}+\frac{7}{16}\\ -\amp = 109+{{\frac{23}{32}}}\\ -\amp = {109 {\textstyle\frac{23}{32}}} -\end{aligned}\)
    So the total length is \({109 {\textstyle\frac{23}{32}}}\) inches.
    +
    Answer.
    \(81 {\textstyle\frac{1}{8}}\)
    Explanation.
    +
    To find the total length in inches, we add the two lengths together.
    \(\begin{aligned} +{33 {\textstyle\frac{3}{4}}}+{47 {\textstyle\frac{3}{8}}} \amp = 33+\frac{3}{4}+47+\frac{3}{8}\\ +\amp = 80+\frac{3}{4}+\frac{3}{8}\\ +\amp = 80+{1 {\textstyle\frac{1}{8}}}\\ +\amp = {81 {\textstyle\frac{1}{8}}} +\end{aligned}\)
    So the total length is \({81 {\textstyle\frac{1}{8}}}\) inches.

    99.

    -
    -
    +
    +
    -
    Each page of a book consists of a header, a footer and the middle part. The header is \({{\frac{3}{10}}}\) inches in height; the footer is \({{\frac{13}{20}}}\) inches in height; and the middle part is \({5 {\textstyle\frac{3}{10}}}\) inches in height.
    What is the total height of each page in this book? Use mixed number in your answer if needed.
    Each page in this book is inches in height.
    +
    Each page of a book consists of a header, a footer and the middle part. The header is \({{\frac{5}{9}}}\) inches in height; the footer is \({{\frac{7}{18}}}\) inches in height; and the middle part is \({4 {\textstyle\frac{2}{9}}}\) inches in height.
    What is the total height of each page in this book? Use mixed number in your answer if needed.
    Each page in this book is inches in height.
    -
    Answer.
    \(6 {\textstyle\frac{1}{4}}\)
    Explanation.
    -
    We add up the height of the header, middle part and footer to find the total height of a page:
    \(\displaystyle{ +
    Answer.
    \(5 {\textstyle\frac{1}{6}}\)
    Explanation.
    +
    We add up the height of the header, middle part and footer to find the total height of a page:
    \(\displaystyle{ \begin{aligned}[t] -{{\frac{3}{10}}} + {5 {\textstyle\frac{3}{10}}} + {{\frac{13}{20}}} -\amp = {{\frac{3}{10}}} + 5 + {{\frac{3}{10}}} + {{\frac{13}{20}}} \\ -\amp = 5 + {{\frac{3}{10}}} + {{\frac{3}{10}}} + {{\frac{13}{20}}} \\ -\amp = 5 + \frac{3\cdot2}{10\cdot2} + \frac{3\cdot2}{10\cdot2} + \frac{13}{20} \\ -\amp = 5 + \frac{6}{20} + \frac{6}{20} + \frac{13}{20} \\ -\amp = 5 + \frac{6+6+13}{20} \\ -\amp = 5 + \frac{25}{20} \\ -\amp = 5 + \frac{5}{4} \\ -\amp = 5 + 1 \frac{1}{4} \\ -\amp = {6 {\textstyle\frac{1}{4}}} +{{\frac{5}{9}}} + {4 {\textstyle\frac{2}{9}}} + {{\frac{7}{18}}} +\amp = {{\frac{5}{9}}} + 4 + {{\frac{2}{9}}} + {{\frac{7}{18}}} \\ +\amp = 4 + {{\frac{5}{9}}} + {{\frac{2}{9}}} + {{\frac{7}{18}}} \\ +\amp = 4 + \frac{5\cdot2}{9\cdot2} + \frac{2\cdot2}{9\cdot2} + \frac{7}{18} \\ +\amp = 4 + \frac{10}{18} + \frac{4}{18} + \frac{7}{18} \\ +\amp = 4 + \frac{10+4+7}{18} \\ +\amp = 4 + \frac{21}{18} \\ +\amp = 4 + \frac{7}{6} \\ +\amp = 4 + 1 \frac{1}{6} \\ +\amp = {5 {\textstyle\frac{1}{6}}} \end{aligned} -}\)
    Each page in this book is \({6 {\textstyle\frac{1}{4}}}\) inches in height.
    +}\)
    Each page in this book is \({5 {\textstyle\frac{1}{6}}}\) inches in height.

    100.

    -
    -
    +
    +
    -
    To pave the road on Ellis Street, the crew used \(3{{\frac{1}{3}}}\) tons of cement on the first day, and used \(5{{\frac{5}{6}}}\) tons on the second day. How many tons of cement were used in all?
    +
    To pave the road on Ellis Street, the crew used \(4{{\frac{3}{4}}}\) tons of cement on the first day, and used \(2{{\frac{7}{8}}}\) tons on the second day. How many tons of cement were used in all?
    tons of cement were used in all.
    -
    Answer.
    \(9 {\textstyle\frac{1}{6}}\)
    Explanation.
    -
    This problem is obviously an addition problem.
    To add mixed number, we break each mixed number into an integer and a fraction, and then add up integers and fractions separately.
    \(\displaystyle{\begin{aligned}[t] -3{{\frac{1}{3}}} + 5{{\frac{5}{6}}} -\amp = 3 + {{\frac{1}{3}}} + 5 + {{\frac{5}{6}}} \\ -\amp = 3 + 5 + {{\frac{1}{3}}} + {{\frac{5}{6}}} \\ -\amp = 8 + \frac{1 \cdot 2}{3 \cdot 2} + \frac{5}{6} \\ -\amp = 8 + \frac{2}{6} + \frac{5}{6} \\ -\amp = 8 + \frac{7}{6} \\ -\amp = 8 + {1 {\textstyle\frac{1}{6}}} \\ -\amp = {9 {\textstyle\frac{1}{6}}} +
    Answer.
    \(7 {\textstyle\frac{5}{8}}\)
    Explanation.
    +
    This problem is obviously an addition problem.
    To add mixed number, we break each mixed number into an integer and a fraction, and then add up integers and fractions separately.
    \(\displaystyle{\begin{aligned}[t] +4{{\frac{3}{4}}} + 2{{\frac{7}{8}}} +\amp = 4 + {{\frac{3}{4}}} + 2 + {{\frac{7}{8}}} \\ +\amp = 4 + 2 + {{\frac{3}{4}}} + {{\frac{7}{8}}} \\ +\amp = 6 + \frac{3 \cdot 2}{4 \cdot 2} + \frac{7}{8} \\ +\amp = 6 + \frac{6}{8} + \frac{7}{8} \\ +\amp = 6 + \frac{13}{8} \\ +\amp = 6 + {1 {\textstyle\frac{5}{8}}} \\ +\amp = {7 {\textstyle\frac{5}{8}}} \end{aligned} -}\)
    -\({9 {\textstyle\frac{1}{6}}}\) tons of cement were used in all.
    +}\)
    +\({7 {\textstyle\frac{5}{8}}}\) tons of cement were used in all.

    101.

    -
    -
    +
    +
    -
    When driving on a high way, noticed a sign saying exit to Johnstown is \(2{{\frac{3}{4}}}\) miles away, while exit to Jerrystown is \(4{{\frac{1}{2}}}\) miles away. How far is Johnstown from Jerrystown?
    Johnstown and Jerrystown are miles apart.
    +
    When driving on a high way, noticed a sign saying exit to Johnstown is \(1{{\frac{3}{4}}}\) miles away, while exit to Jerrystown is \(3{{\frac{1}{2}}}\) miles away. How far is Johnstown from Jerrystown?
    Johnstown and Jerrystown are miles apart.
    -
    Answer.
    \(1 {\textstyle\frac{3}{4}}\)
    Explanation.
    -
    To find the distance between Johnstown and Jerrystown, we need to find the difference between their distance. This implies subtraction. We could treat subtraction as “adding a negative”:
    \(\displaystyle{\begin{aligned}[t] -4{{\frac{1}{2}}} -2{{\frac{3}{4}}} -\amp = 4{{\frac{1}{2}}} + (-2{{\frac{3}{4}}}) \\ -\amp = 4 + {{\frac{1}{2}}} + (-2) + (-{{\frac{3}{4}}}) \\ -\amp = 4 + (-2) + {{\frac{1}{2}}} + (-{{\frac{3}{4}}}) \\ +
    Answer.
    \(1 {\textstyle\frac{3}{4}}\)
    Explanation.
    +
    To find the distance between Johnstown and Jerrystown, we need to find the difference between their distance. This implies subtraction. We could treat subtraction as “adding a negative”:
    \(\displaystyle{\begin{aligned}[t] +3{{\frac{1}{2}}} -1{{\frac{3}{4}}} +\amp = 3{{\frac{1}{2}}} + (-1{{\frac{3}{4}}}) \\ +\amp = 3 + {{\frac{1}{2}}} + (-1) + (-{{\frac{3}{4}}}) \\ +\amp = 3 + (-1) + {{\frac{1}{2}}} + (-{{\frac{3}{4}}}) \\ \amp = 2 + \frac{1 \cdot 2}{2 \cdot 2} + (-\frac{3}{4}) \\ \amp = 2 + \frac{2}{4} + (-\frac{3}{4}) \\ \amp = 2 + \frac{2+(-3)}{4} \\ \amp = 2 + ({-{\frac{1}{4}}}) \end{aligned} -}\)
    From here, there are two methods to continue.
    Method 1
    We can “split” \(1\) from the integer, and do \(1 + ({-{\frac{1}{4}}})\text{:}\) -
    \(\displaystyle{\begin{aligned}[t] +}\)
    From here, there are two methods to continue.
    Method 1
    We can “split” \(1\) from the integer, and do \(1 + ({-{\frac{1}{4}}})\text{:}\) +
    \(\displaystyle{\begin{aligned}[t] \amp {\phantom{{}=}} 2 + ({-{\frac{1}{4}}}) \\ \amp = 1 + 1 + ({-{\frac{1}{4}}}) \\ \amp = 1 + \frac{4}{4} + ({-{\frac{1}{4}}}) \\ @@ -2948,9 +2963,9 @@

    Applications.

    \amp = 1 + \frac{3}{4} \\ \amp = {1 {\textstyle\frac{3}{4}}} \end{aligned} -}\)
    Note that we changed \(1\) to \(\frac{4}{4}\text{.}\) -
    Method2
    We could simply change the whole number \(2\) into a fraction without “splitting” \(1\text{:}\) -
    \(\displaystyle{\begin{aligned}[t] +}\)
    Note that we changed \(1\) to \(\frac{4}{4}\text{.}\) +
    Method2
    We could simply change the whole number \(2\) into a fraction without “splitting” \(1\text{:}\) +
    \(\displaystyle{\begin{aligned}[t] \amp {\phantom{{}=}} 2 + ({-{\frac{1}{4}}}) \\ \amp = \frac{2}{1} + ({-{\frac{1}{4}}}) \\ \amp = \frac{2\cdot4}{1\cdot4} + ({-{\frac{1}{4}}}) \\ @@ -2959,25 +2974,25 @@

    Applications.

    \amp = \frac{7}{4} \\ \amp = {1 {\textstyle\frac{3}{4}}} \end{aligned} -}\)
    ran \({1 {\textstyle\frac{3}{4}}}\) more miles than .
    +}\)
    ran \({1 {\textstyle\frac{3}{4}}}\) more miles than .

    102.

    -
    -
    +
    +
    -
    A cake recipe needs \({2 {\textstyle\frac{1}{4}}}\) cups of flour. Using this recipe, to bake \(9\) cakes, how many cups of flour are needed?
    To bake \(9\) cakes, cups of flour are needed.
    +
    A cake recipe needs \({1 {\textstyle\frac{1}{5}}}\) cups of flour. Using this recipe, to bake \(8\) cakes, how many cups of flour are needed?
    To bake \(8\) cakes, cups of flour are needed.
    -
    Answer.
    \(20 {\textstyle\frac{1}{4}}\)
    Explanation.
    -
    Each cake needs \({2 {\textstyle\frac{1}{4}}}\) cups of flour. To find how many cups of flour are needed to bake \(9\) cakes, we use multiplication:
    \(\displaystyle{ +
    Answer.
    \(9 {\textstyle\frac{3}{5}}\)
    Explanation.
    +
    Each cake needs \({1 {\textstyle\frac{1}{5}}}\) cups of flour. To find how many cups of flour are needed to bake \(8\) cakes, we use multiplication:
    \(\displaystyle{ \begin{aligned}[t] -{2 {\textstyle\frac{1}{4}}} \cdot 9 -\amp = \frac{9}{4} \cdot \frac{9}{1} \\ -\amp = \frac{81}{4} \\ -\amp = {20 {\textstyle\frac{1}{4}}} +{1 {\textstyle\frac{1}{5}}} \cdot 8 +\amp = \frac{6}{5} \cdot \frac{8}{1} \\ +\amp = \frac{48}{5} \\ +\amp = {9 {\textstyle\frac{3}{5}}} \end{aligned} -}\)
    To bake \(9\) cakes, \({20 {\textstyle\frac{1}{4}}}\) cups of flour are needed.
    +}\)
    To bake \(8\) cakes, \({9 {\textstyle\frac{3}{5}}}\) cups of flour are needed.
    @@ -3016,17 +3031,17 @@

    Sketching Fractions.

    Challenge.

    107.

    -
    -
    +
    +
    -
    Given that \(a\neq0\text{,}\) simplify \(\displaystyle{}{\frac{4}{a}+\frac{8}{a}}\text{.}\) +
    Given that \(a\neq0\text{,}\) simplify \(\displaystyle{}{\frac{6}{a}+\frac{6}{a}}\text{.}\)
    -
    Answer.
    \(\frac{12}{a}\)
    Explanation.
    -
    Since the fractions have the same denominator, we can just add numerators and keep that denominator.
    +
    Answer.
    \(\frac{12}{a}\)
    Explanation.
    +
    Since the fractions have the same denominator, we can just add numerators and keep that denominator.
    \begin{equation*} \begin{aligned} -{\frac{4}{a}+\frac{8}{a}} \amp= \frac{4+8}{a}\\ +{\frac{6}{a}+\frac{6}{a}} \amp= \frac{6+6}{a}\\ \amp = {\frac{12}{a}} \end{aligned} \end{equation*} @@ -3035,20 +3050,20 @@

    Challenge.

    108.

    -
    -
    +
    +
    -
    Given that \(a\neq0\text{,}\) simplify \(\displaystyle{}{\frac{5}{a}+\frac{5}{2a}}\text{.}\) +
    Given that \(a\neq0\text{,}\) simplify \(\displaystyle{}{\frac{7}{a}+\frac{3}{2a}}\text{.}\)
    -
    Answer.
    \(\frac{15}{2a}\)
    Explanation.
    -
    Since the fractions don’t have the same denominator, we have to make them have like denominators. Multiply the first fraction by \(\frac{2}{2}\text{.}\) -
    +
    Answer.
    \(\frac{17}{2a}\)
    Explanation.
    +
    Since the fractions don’t have the same denominator, we have to make them have like denominators. Multiply the first fraction by \(\frac{2}{2}\text{.}\) +
    \begin{equation*} \begin{aligned} -{\frac{5}{a}+\frac{5}{2a}} \amp = {\frac{5}{a}}\cdot \frac{2}{2} + {\frac{5}{2a}}\\ -\amp = \frac{10}{2a} + {\frac{5}{2a}}\\ -\amp = {\frac{15}{2a}} +{\frac{7}{a}+\frac{3}{2a}} \amp = {\frac{7}{a}}\cdot \frac{2}{2} + {\frac{3}{2a}}\\ +\amp = \frac{14}{2a} + {\frac{3}{2a}}\\ +\amp = {\frac{17}{2a}} \end{aligned} \end{equation*}
    @@ -3056,20 +3071,20 @@

    Challenge.

    109.

    -
    -
    +
    +
    -
    Given that \(a\neq0\text{,}\) simplify \(\displaystyle{}{\frac{6}{a}-\frac{2}{5a}}\text{.}\) +
    Given that \(a\neq0\text{,}\) simplify \(\displaystyle{}{\frac{8}{a}-\frac{9}{5a}}\text{.}\)
    -
    Answer.
    \(\frac{28}{5a}\)
    Explanation.
    -
    Since the fractions don’t have the same denominator, we have to make them have like denominators. Multiply the first fraction by \(\frac{5}{5}\text{.}\) -
    +
    Answer.
    \(\frac{31}{5a}\)
    Explanation.
    +
    Since the fractions don’t have the same denominator, we have to make them have like denominators. Multiply the first fraction by \(\frac{5}{5}\text{.}\) +
    \begin{equation*} \begin{aligned} -{\frac{6}{a}-\frac{2}{5a}} \amp = {\frac{6}{a}}\cdot \frac{5}{5} - {\frac{2}{5a}}\\ -\amp = \frac{30}{5a} - {\frac{2}{5a}}\\ -\amp = {\frac{28}{5a}} +{\frac{8}{a}-\frac{9}{5a}} \amp = {\frac{8}{a}}\cdot \frac{5}{5} - {\frac{9}{5a}}\\ +\amp = \frac{40}{5a} - {\frac{9}{5a}}\\ +\amp = {\frac{31}{5a}} \end{aligned} \end{equation*}
    diff --git a/section-graphing-equations.html b/section-graphing-equations.html index a876594fd..cacb4cdf1 100644 --- a/section-graphing-equations.html +++ b/section-graphing-equations.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-horizontal-vertical-parallel-and-perpendicular-lines.html b/section-horizontal-vertical-parallel-and-perpendicular-lines.html index cb7672416..2861a7c83 100644 --- a/section-horizontal-vertical-parallel-and-perpendicular-lines.html +++ b/section-horizontal-vertical-parallel-and-perpendicular-lines.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -599,7 +613,7 @@

    Search Results:

    • Subsection 3.8.1 Horizontal Lines and Vertical Lines

    -
    We learned in Section 7 that all lines can be written in standard form. When either \(A\) or \(B\) equal \(0\text{,}\) we end up with a horizontal or vertical line, as we will soon see. Let’s take the standard form line equation, \(Ax+Bx=C\text{,}\) and one at a time let \(A=0\) and \(B=0\) and simplify each equation.
    +
    We learned in Section 7 that all lines can be written in standard form. When either \(A\) or \(B\) equal \(0\text{,}\) we end up with a horizontal or vertical line, as we will soon see. Let’s take the standard form line equation, \(Ax+By=C\text{,}\) and one at a time let \(A=0\) and \(B=0\) and simplify each equation.
    \begin{align*} @@ -710,12 +724,12 @@

    Search Results:

    We use horizontal lines to model scenarios where there is no change in \(y\)-values, like when Kato stopped for \(12\) hours (he deserved a rest)!

    Checkpoint 3.8.11. Plotting Points.

    -
    -
    +
    +
    -
    Suppose you need to plot the equation \(y=-4.25\text{.}\) Since the equation is in “\(y=\)” form, you decide to make a table of points. Fill out some points for this table.
    +
    Suppose you need to plot the equation \(y=-4.25\text{.}\) Since the equation is in “\(y=\)” form, you decide to make a table of points. Fill out some points for this table.
    - + @@ -739,10 +753,10 @@

    Search Results:

    \(x\)
    \(x\)
    		 \(y\)
    -
    Explanation.
    -
    We can use whatever values for \(x\) that we like, as long as they are all different. The equation tells us the \(y\)-value has to be \(-4.25\) each time.
    +
    Explanation.
    +
    We can use whatever values for \(x\) that we like, as long as they are all different. The equation tells us the \(y\)-value has to be \(-4.25\) each time.
    - + @@ -765,7 +779,7 @@

    Search Results:

    -
    \(x\)
    \(x\)
    		 \(y\)
    \(2\) \(-4.25\)
    Now that we have a table, we could use its values to assist with plotting the line.
    +
    Now that we have a table, we could use its values to assist with plotting the line.

    @@ -785,12 +799,12 @@

    Search Results:

    Be careful not to mix up “no slope” (which means “its slope is undefined”) with “has slope \(0\text{.}\)” If a line has slope \(0\text{,}\) it does have a slope.
    If you are familiar with NBA basketball, some players wear number \(0\text{.}\) That’s not the same thing as “not having a number”. This is similar to the situation with having slope \(0\) versus not having slope.

    Checkpoint 3.8.16. Plotting Points.

    -
    -
    +
    +
    -
    Suppose you need to plot the equation \(x=3.14\text{.}\) You decide to try making a table of points. Fill out some points for this table.
    +
    Suppose you need to plot the equation \(x=3.14\text{.}\) You decide to try making a table of points. Fill out some points for this table.
    - + @@ -814,10 +828,10 @@

    Search Results:

    \(x\)
    \(x\)
    		 \(y\)
    -
    Explanation.
    -
    Since the equation says \(x\) is always the number \(3.14\text{,}\) we have to use this for the \(x\) value in all the points. This is different from how we would plot a “\(y=\)” equation, where we would use several different \(x\)-values. We can use whatever values for \(y\) that we like, as long as they are all different.
    +
    Explanation.
    +
    Since the equation says \(x\) is always the number \(3.14\text{,}\) we have to use this for the \(x\) value in all the points. This is different from how we would plot a “\(y=\)” equation, where we would use several different \(x\)-values. We can use whatever values for \(y\) that we like, as long as they are all different.
    - + @@ -840,7 +854,7 @@

    Search Results:

    -
    \(x\)
    \(x\)
    		 \(y\)
    \(3.14\) \(2\)
    The reason we made a table was to help with plotting the line.
    +
    The reason we made a table was to help with plotting the line.

    @@ -933,13 +947,13 @@

    Search Results:

    Any two vertical lines are parallel to each other. For two non-vertical lines, they are parallel if and only if they have the same slope.

    Checkpoint 3.8.23.

    -
    -
    +
    +
    -
    A line \(\ell\) is parallel to the line with equation \(y=17.2x-340.9\text{,}\) but \(\ell\) has \(y\)-intercept at \((0,128.2)\text{.}\) What is an equation for \(\ell\text{?}\) +
    A line \(\ell\) is parallel to the line with equation \(y=17.2x-340.9\text{,}\) but \(\ell\) has \(y\)-intercept at \((0,128.2)\text{.}\) What is an equation for \(\ell\text{?}\)
    -
    Explanation.
    -
    Parallel lines have the same slope, and the slope of \(y=17.2x-340.9\) is \(17.2\text{.}\) So \(\ell\) has slope \(17.2\text{.}\) And we have been given that \(\ell\)’s \(y\)-intercept is at \((0,128.2)\text{.}\) So we can use slope-intercept form to write its equation as
    +
    Explanation.
    +
    Parallel lines have the same slope, and the slope of \(y=17.2x-340.9\) is \(17.2\text{.}\) So \(\ell\) has slope \(17.2\text{.}\) And we have been given that \(\ell\)’s \(y\)-intercept is at \((0,128.2)\text{.}\) So we can use slope-intercept form to write its equation as
    \begin{equation*} y=17.2x+128.2 \text{.} @@ -950,13 +964,13 @@

    Search Results:

    Checkpoint 3.8.24.

    -
    -
    +
    +
    -
    A line \(\kappa\) is parallel to the line with equation \(y=-3.5x+17\text{,}\) but \(\kappa\) passes through the point \((-12,23)\text{.}\) What is an equation for \(\kappa\text{?}\) +
    A line \(\kappa\) is parallel to the line with equation \(y=-3.5x+17\text{,}\) but \(\kappa\) passes through the point \((-12,23)\text{.}\) What is an equation for \(\kappa\text{?}\)
    -
    Explanation.
    -
    Parallel lines have the same slope, and the slope of \(y=-3.5x+17\) is \(-3.5\text{.}\) So \(\kappa\) has slope \(-3.5\text{.}\) And we know a point that \(\kappa\) passes through, so we can use point-slope form to write its equation as
    +
    Explanation.
    +
    Parallel lines have the same slope, and the slope of \(y=-3.5x+17\) is \(-3.5\text{.}\) So \(\kappa\) has slope \(-3.5\text{.}\) And we know a point that \(\kappa\) passes through, so we can use point-slope form to write its equation as
    \begin{equation*} y=-3.5(x+12)+23 \text{.} @@ -1013,13 +1027,13 @@

    Search Results:

    Checkpoint 3.8.31.

    -
    -
    +
    +
    -
    Line \(A\) and Line \(B\) are perpendicular. Line \(A\)’s equation is \(2x+3y=12\text{.}\) Line \(B\) passes through the point \((4,-3)\text{.}\) Find an equation for Line \(B\text{.}\) +
    Line \(A\) and Line \(B\) are perpendicular. Line \(A\)’s equation is \(2x+3y=12\text{.}\) Line \(B\) passes through the point \((4,-3)\text{.}\) Find an equation for Line \(B\text{.}\)
    -
    Explanation.
    -
    First, we will find Line \(A\)’s slope by rewriting its equation from standard form to slope-intercept form:
    +
    Explanation.
    +
    First, we will find Line \(A\)’s slope by rewriting its equation from standard form to slope-intercept form:
    \begin{equation*} \begin{aligned} 2x+3y\amp=12\\ @@ -1029,7 +1043,7 @@

    Search Results:

    y\amp=-\frac{2}{3}x+4 \end{aligned} \end{equation*} -
    So Line \(A\)’s slope is \(-\frac{2}{3}\text{.}\) Since Line \(B\) is perpendicular to Line \(A\text{,}\) its slope is \(-\frac{1}{-\frac{2}{3}}=\frac{3}{2}\text{.}\) It’s also given that Line \(B\) passes through \((4,-3)\text{,}\) so we can write Line \(B\)’s point-slope form equation:
    +
    So Line \(A\)’s slope is \(-\frac{2}{3}\text{.}\) Since Line \(B\) is perpendicular to Line \(A\text{,}\) its slope is \(-\frac{1}{-\frac{2}{3}}=\frac{3}{2}\text{.}\) It’s also given that Line \(B\) passes through \((4,-3)\text{,}\) so we can write Line \(B\)’s point-slope form equation:
    \begin{equation*} \begin{aligned} y\amp=m(x-x_0)+y_0\\ @@ -1054,82 +1068,97 @@

    Search Results:

    Exercise Group.
    1.
    -
    -
    +
    +
    -
    +
    Evaluate the following expressions. If the answer is undefined, you may answer with DNE (meaning “does not exist”).
      -
    1. -\(\displaystyle{ \frac{4}{0} = }\) +
    2. +\(\displaystyle{ \frac{0}{3} = }\)
      3. -
    3. -\(\displaystyle{ \frac{0}{4} = }\) +
    4. +\(\displaystyle{ \frac{3}{0} = }\)
    -
    Answer 1.
    \(\text{DNE}\)
    Answer 2.
    \(0\)
    Explanation.
    -
    Let’s observe a pattern first. Recall that multiplication and division are inverse operations, so we have:
    \(\displaystyle{\begin{aligned}[t] +
    Answer 1.
    \(0\)
    Answer 2.
    \(\text{DNE}\)
    Explanation.
    +
    Let’s observe a pattern first. Recall that multiplication and division are inverse operations, so we have:
    \(\displaystyle{\begin{aligned}[t] \frac{12}{3}\amp =4 \amp 3 \cdot 4 \amp = 12 \\ \frac{20}{2}\amp =10 \amp 2 \cdot 10 \amp = 20 \\ \frac{20}{4}\amp =5 \amp 4 \cdot 5 \amp = 20 \\ \amp \vdots\amp \amp \vdots \end{aligned} -}\)
    -\(0\) in numerator
    Now let’s think about \(0\text{.}\) Assume that \(\frac{0}{4} = x\text{.}\) By the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] -\frac{0}{4}\amp =x \amp 4 \cdot x \amp = 0 \\ +}\)
    +\(0\) in numerator
    Now let’s think about \(0\text{.}\) Assume that \(\frac{0}{3} = x\text{.}\) By the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] +\frac{0}{3}\amp =x \amp 3 \cdot x \amp = 0 \\ \end{aligned} -}\)
    Is there any number for \(x\) that will make \(4 \cdot x = 0\) work? Yes! When \(x=0\text{,}\) we have \(4 \cdot 0 = 0\text{.}\) This is why \(\frac{0}{4}=0\text{.}\) We can write:
    \(\displaystyle{\begin{aligned}[t] -\frac{0}{4}\amp =0 \amp 4 \cdot 0 \amp = 0 \\ +}\)
    Is there any number for \(x\) that will make \(3 \cdot x = 0\) work? Yes! When \(x=0\text{,}\) we have \(3 \cdot 0 = 0\text{.}\) This is why \(\frac{0}{3}=0\text{.}\) We can write:
    \(\displaystyle{\begin{aligned}[t] +\frac{0}{3}\amp =0 \amp 3 \cdot 0 \amp = 0 \\ \end{aligned} -}\)
    -\(0\) in denominator
    Assume that \(\frac{4}{0} = x\text{.}\) Then by the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] -\frac{4}{0}\amp =x \amp 0 \cdot x \amp = 4 \\ +}\)
    +\(0\) in denominator
    Assume that \(\frac{3}{0} = x\text{.}\) Then by the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] +\frac{3}{0}\amp =x \amp 0 \cdot x \amp = 3 \\ \end{aligned} -}\)
    Is there any number for \(x\) that will make \(0 \cdot x = 4\) work? No! This is because the product of \(0\) and any number is \(0\text{,}\) and would never be \(4\text{.}\) This is why we say \(\frac{4}{0}\) is undefined.
    Answers:
    \(\displaystyle{ \frac{0}{4}=0 \text{ and } \frac{4}{0} \text{ is undefined.} }\)
    +}\)
    Is there any number for \(x\) that will make \(0 \cdot x = 3\) work? No! This is because the product of \(0\) and any number is \(0\text{,}\) and would never be \(3\text{.}\) This is why we say \(\frac{3}{0}\) is undefined.
    Answers:
    \(\displaystyle{ \frac{0}{3}=0 \text{ and } \frac{3}{0} \text{ is undefined.} }\)
    2.
    -
    -
    +
    +
    -
    +
    Evaluate the following expressions. If the answer is undefined, you may answer with DNE (meaning “does not exist”).
      -
    1. -\(\displaystyle{ \frac{0}{4} = }\) -
      2. -
    2. +
    3. \(\displaystyle{ \frac{4}{0} = }\)
      4. +
    4. +\(\displaystyle{ \frac{0}{4} = }\) +
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(\text{DNE}\)
    Explanation.
    -
    Let’s observe a pattern first. Recall that multiplication and division are inverse operations, so we have:
    \(\displaystyle{\begin{aligned}[t] +
    Answer 1.
    \(\text{DNE}\)
    Answer 2.
    \(0\)
    Explanation.
    +
    Let’s observe a pattern first. Recall that multiplication and division are inverse operations, so we have:
    \(\displaystyle{\begin{aligned}[t] \frac{12}{3}\amp =4 \amp 3 \cdot 4 \amp = 12 \\ \frac{20}{2}\amp =10 \amp 2 \cdot 10 \amp = 20 \\ \frac{20}{4}\amp =5 \amp 4 \cdot 5 \amp = 20 \\ \amp \vdots\amp \amp \vdots \end{aligned} -}\)
    -\(0\) in numerator
    Now let’s think about \(0\text{.}\) Assume that \(\frac{0}{4} = x\text{.}\) By the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] +}\)
    +\(0\) in numerator
    Now let’s think about \(0\text{.}\) Assume that \(\frac{0}{4} = x\text{.}\) By the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] \frac{0}{4}\amp =x \amp 4 \cdot x \amp = 0 \\ \end{aligned} -}\)
    Is there any number for \(x\) that will make \(4 \cdot x = 0\) work? Yes! When \(x=0\text{,}\) we have \(4 \cdot 0 = 0\text{.}\) This is why \(\frac{0}{4}=0\text{.}\) We can write:
    \(\displaystyle{\begin{aligned}[t] +}\)
    Is there any number for \(x\) that will make \(4 \cdot x = 0\) work? Yes! When \(x=0\text{,}\) we have \(4 \cdot 0 = 0\text{.}\) This is why \(\frac{0}{4}=0\text{.}\) We can write:
    \(\displaystyle{\begin{aligned}[t] \frac{0}{4}\amp =0 \amp 4 \cdot 0 \amp = 0 \\ \end{aligned} -}\)
    -\(0\) in denominator
    Assume that \(\frac{4}{0} = x\text{.}\) Then by the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] +}\)
    +\(0\) in denominator
    Assume that \(\frac{4}{0} = x\text{.}\) Then by the pattern we observed, we would have:
    \(\displaystyle{\begin{aligned}[t] \frac{4}{0}\amp =x \amp 0 \cdot x \amp = 4 \\ \end{aligned} -}\)
    Is there any number for \(x\) that will make \(0 \cdot x = 4\) work? No! This is because the product of \(0\) and any number is \(0\text{,}\) and would never be \(4\text{.}\) This is why we say \(\frac{4}{0}\) is undefined.
    Answers:
    \(\displaystyle{ \frac{0}{4}=0 \text{ and } \frac{4}{0} \text{ is undefined.} }\)
    +}\)
    Is there any number for \(x\) that will make \(0 \cdot x = 4\) work? No! This is because the product of \(0\) and any number is \(0\text{,}\) and would never be \(4\text{.}\) This is why we say \(\frac{4}{0}\) is undefined.
    Answers:
    \(\displaystyle{ \frac{0}{4}=0 \text{ and } \frac{4}{0} \text{ is undefined.} }\)
    3.
    +
    +
    +
    +
    A line passes through the points \((2,-3)\) and \((-2,-3)\text{.}\) Find this line’s slope.
    +
    +
    Answer.
    \(0\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (2,-3) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-2,-3) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-3 + 3}{-2-(2)}\\ +\amp =\frac{0}{-4}\\ +\amp ={0} \end{aligned}}\)
    This is a special line which is parallel to the \(x\)-axis.
    So the line’s slope is \({0}\text{.}\) +
    +
    +
    +
    +
    4.
    @@ -1144,22 +1173,20 @@
    Exercise Group.
    -
    4.
    -
    +
    5.
    +
    -
    A line passes through the points \((4,2)\) and \((-3,2)\text{.}\) Find this line’s slope.
    +
    A line passes through the points \((2,-2)\) and \((2,3)\text{.}\) Find this line’s slope.
    -
    Answer.
    \(0\)
    Explanation.
    -
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (4,2) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-3,2) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{2-2}{-3-4}\\ -\amp =\frac{0}{-7}\\ -\amp ={0} \end{aligned}}\)
    This is a special line which is parallel to the \(x\)-axis.
    So the line’s slope is \({0}\text{.}\) -
    +
    Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (2,-2) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (2,3) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{3-(-2)}{2-2}\\ +\amp =\frac{5}{0} \end{aligned}}\)
    Since we cannot divide by \(0\text{,}\) this line’s slope does not exist. This is a special line which is parallel to the \(y\)-axis; it is a vertical line.
    So the line’s slope Does Not Exist.
    -
    5.
    +
    6.
    @@ -1172,20 +1199,28 @@
    Exercise Group.
    -
    6.
    -
    +
    7.
    +
    -
    A line passes through the points \((7,-1)\) and \((7,2)\text{.}\) Find this line’s slope.
    -
    -
    Answer.
    \(\text{DNE}\hbox{ or }\text{NONE}\)
    Explanation.
    -
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (7,-1) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (7,2) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these numbers into the corresponding variables in the slope formula:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{2-(-1)}{7-7}\\ -\amp =\frac{3}{0} \end{aligned}}\)
    Since we cannot divide by \(0\text{,}\) this line’s slope does not exist. This is a special line which is parallel to the \(y\)-axis; it is a vertical line.
    So the line’s slope Does Not Exist.
    -
    +
    +
    Consider the equation:
    +
    +\begin{equation*} +y=1 +\end{equation*}
    +
    Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
      +
    • \(\displaystyle (0,8)\)
      • +
    • \(\displaystyle (-8,1)\)
      • +
    • \(\displaystyle (9,1)\)
      • +
    • \(\displaystyle (1,6)\)
      • +
    +
    Explanation.
    +
    This equation is only asks for the \(y\)-value to equal \(1\text{.}\) So any point whose \(y\)-coordinate is \(1\) is a solution, regardless of its \(x\)-value.
    So \((9,1)\) and \((-8,1)\) are solutions to the given equation, while \((1,6)\) and \((0,8)\) are not solutions.
    +
    -
    7.
    +
    8.
    @@ -1206,28 +1241,34 @@
    Exercise Group.
    This equation is only asks for the \(y\)-value to equal \(1\text{.}\) So any point whose \(y\)-coordinate is \(1\) is a solution, regardless of its \(x\)-value.
    So \((10,1)\) and \((-2,1)\) are solutions to the given equation, while \((1,5)\) and \((0,9)\) are not solutions.
    -
    8.
    -
    +
    9.
    +
    Consider the equation:
    \begin{equation*} -y=1 +x+1=0 \end{equation*}
    Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
      -
    • \(\displaystyle (1,3)\)
      • -
    • \(\displaystyle (0,7)\)
      • -
    • \(\displaystyle (-4,1)\)
      • -
    • \(\displaystyle (2,1)\)
      • +
    • \(\displaystyle (-1,0)\)
      • +
    • \(\displaystyle (-1,1)\)
      • +
    • \(\displaystyle (0,-9)\)
      • +
    • \(\displaystyle (1,-1)\)
    Explanation.
    -
    This equation is only asks for the \(y\)-value to equal \(1\text{.}\) So any point whose \(y\)-coordinate is \(1\) is a solution, regardless of its \(x\)-value.
    So \((2,1)\) and \((-4,1)\) are solutions to the given equation, while \((1,3)\) and \((0,7)\) are not solutions.
    +
    We substitute in each ordered pair’s \(x\)- and \(y\)-values into the equation \(x+1=0\text{,}\) and see whether the resulting equation is true.
    In two cases, \((-1,1)\) and \((-1,0)\text{,}\) we have true equations:
    \(\begin{aligned} +-1+1 \amp \stackrel{?}{=} 0 \amp -1+1 \amp \stackrel{?}{=} 0\\ +0 \amp \stackrel{?}{=} 0 \amp 0 \amp \stackrel{?}{=} 0 +\end{aligned}\)
    So \((-1,1)\) and \((-1,0)\) are solutions to the given equation.
    In the other two cases, \((1,-1)\) and \((0,-9)\text{,}\) we have false equations:
    \(\begin{aligned} +1+1 \amp \stackrel{?}{=} 0 \amp 0+1 \amp \stackrel{?}{=} 0\\ +2 \amp \stackrel{?}{=} 0 \amp 1 \amp \stackrel{?}{=} 0 +\end{aligned}\)
    So \((1,-1)\) and \((0,-9)\) are not solutions.
    -
    9.
    +
    10.
    @@ -1254,33 +1295,6 @@
    Exercise Group.
    \end{aligned}\)
    So \((1,-1)\) and \((0,-5)\) are not solutions.
    -
    10.
    -
    -
    -
    -
    -
    Consider the equation:
    -
    -\begin{equation*} -x+1=0 -\end{equation*} -
    -
    Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
      -
    • \(\displaystyle (-1,2)\)
      • -
    • \(\displaystyle (0,-7)\)
      • -
    • \(\displaystyle (-1,0)\)
      • -
    • \(\displaystyle (1,-1)\)
      • -
    -
    Explanation.
    -
    We substitute in each ordered pair’s \(x\)- and \(y\)-values into the equation \(x+1=0\text{,}\) and see whether the resulting equation is true.
    In two cases, \((-1,2)\) and \((-1,0)\text{,}\) we have true equations:
    \(\begin{aligned} --1+1 \amp \stackrel{?}{=} 0 \amp -1+1 \amp \stackrel{?}{=} 0\\ -0 \amp \stackrel{?}{=} 0 \amp 0 \amp \stackrel{?}{=} 0 -\end{aligned}\)
    So \((-1,2)\) and \((-1,0)\) are solutions to the given equation.
    In the other two cases, \((1,-1)\) and \((0,-7)\text{,}\) we have false equations:
    \(\begin{aligned} -1+1 \amp \stackrel{?}{=} 0 \amp 0+1 \amp \stackrel{?}{=} 0\\ -2 \amp \stackrel{?}{=} 0 \amp 1 \amp \stackrel{?}{=} 0 -\end{aligned}\)
    So \((1,-1)\) and \((0,-7)\) are not solutions.
    -
    -

    Skills Practice

    @@ -1288,10 +1302,10 @@
    Exercise Group.
    Tables for Horizontal and Vertical Lines.
    11.
    -
    -
    +
    +
    -
    Fill out this table for the equation \(y=4\text{.}\) The first row is an example.
    +
    Fill out this table for the equation \(y=3\text{.}\) The first row is an example.
    @@ -1299,8 +1313,8 @@
    Tables for Horizontal and Vertical Lines
    - - + + @@ -1327,10 +1341,10 @@
    Tables for Horizontal and Vertical Lines
    -
    \(x\) \(y\)
    \(-3\)\(4\)\(\left(-3,4\right)\)\(3\)\(\left(-3,3\right)\)
    \(-2\)
    Values of \(x\) and \(y\) satisfying the equation \(y=4\) +
    Values of \(x\) and \(y\) satisfying the equation \(y=3\)
    -
    Answer 1.
    \(4\)
    Answer 2.
    \(\left(-2,4\right)\)
    Answer 3.
    \(4\)
    Answer 4.
    \(\left(-1,4\right)\)
    Answer 5.
    \(4\)
    Answer 6.
    \(\left(0,4\right)\)
    Answer 7.
    \(4\)
    Answer 8.
    \(\left(1,4\right)\)
    Answer 9.
    \(4\)
    Answer 10.
    \(\left(2,4\right)\)
    Explanation.
    +
    Answer 1.
    \(3\)
    Answer 2.
    \(\left(-2,3\right)\)
    Answer 3.
    \(3\)
    Answer 4.
    \(\left(-1,3\right)\)
    Answer 5.
    \(3\)
    Answer 6.
    \(\left(0,3\right)\)
    Answer 7.
    \(3\)
    Answer 8.
    \(\left(1,3\right)\)
    Answer 9.
    \(3\)
    Answer 10.
    \(\left(2,3\right)\)
    Explanation.
    @@ -1339,39 +1353,39 @@
    Tables for Horizontal and Vertical Lines
    - - + + - - + + - - + + - - + + - - + + -
    \(x\)
    \(-2\)\(y=4\)\(\left(-2,4\right)\)\(y=3\)\(\left(-2,3\right)\)
    \(-1\)\(y=4\)\(\left(-1,4\right)\)\(y=3\)\(\left(-1,3\right)\)
    \(0\)\(y=4\)\(\left(0,4\right)\)\(y=3\)\(\left(0,3\right)\)
    \(1\)\(y=4\)\(\left(1,4\right)\)\(y=3\)\(\left(1,3\right)\)
    \(2\)\(y=4\)\(\left(2,4\right)\)\(y=3\)\(\left(2,3\right)\)
    Values of \(x\) and \(y\) satisfying the equation \(y=4\) +
    Values of \(x\) and \(y\) satisfying the equation \(y=3\)
    12.
    -
    -
    +
    +
    -
    Fill out this table for the equation \(y=5\text{.}\) The first row is an example.
    +
    Fill out this table for the equation \(y=4\text{.}\) The first row is an example.
    @@ -1379,8 +1393,8 @@
    Tables for Horizontal and Vertical Lines
    - - + + @@ -1407,10 +1421,10 @@
    Tables for Horizontal and Vertical Lines
    -
    \(x\) \(y\)
    \(-3\)\(5\)\(\left(-3,5\right)\)\(4\)\(\left(-3,4\right)\)
    \(-2\)
    Values of \(x\) and \(y\) satisfying the equation \(y=5\) +
    Values of \(x\) and \(y\) satisfying the equation \(y=4\)
    -
    Answer 1.
    \(5\)
    Answer 2.
    \(\left(-2,5\right)\)
    Answer 3.
    \(5\)
    Answer 4.
    \(\left(-1,5\right)\)
    Answer 5.
    \(5\)
    Answer 6.
    \(\left(0,5\right)\)
    Answer 7.
    \(5\)
    Answer 8.
    \(\left(1,5\right)\)
    Answer 9.
    \(5\)
    Answer 10.
    \(\left(2,5\right)\)
    Explanation.
    +
    Answer 1.
    \(4\)
    Answer 2.
    \(\left(-2,4\right)\)
    Answer 3.
    \(4\)
    Answer 4.
    \(\left(-1,4\right)\)
    Answer 5.
    \(4\)
    Answer 6.
    \(\left(0,4\right)\)
    Answer 7.
    \(4\)
    Answer 8.
    \(\left(1,4\right)\)
    Answer 9.
    \(4\)
    Answer 10.
    \(\left(2,4\right)\)
    Explanation.
    @@ -1419,48 +1433,48 @@
    Tables for Horizontal and Vertical Lines
    - - + + - - + + - - + + - - + + - - + + -
    \(x\)
    \(-2\)\(y=5\)\(\left(-2,5\right)\)\(y=4\)\(\left(-2,4\right)\)
    \(-1\)\(y=5\)\(\left(-1,5\right)\)\(y=4\)\(\left(-1,4\right)\)
    \(0\)\(y=5\)\(\left(0,5\right)\)\(y=4\)\(\left(0,4\right)\)
    \(1\)\(y=5\)\(\left(1,5\right)\)\(y=4\)\(\left(1,4\right)\)
    \(2\)\(y=5\)\(\left(2,5\right)\)\(y=4\)\(\left(2,4\right)\)
    Values of \(x\) and \(y\) satisfying the equation \(y=5\) +
    Values of \(x\) and \(y\) satisfying the equation \(y=4\)
    13.
    -
    -
    +
    +
    -
    Fill out this table for the equation \(x=-5\text{.}\) The first row is an example.
    +
    Fill out this table for the equation \(x=-6\text{.}\) The first row is an example.
    - + - + @@ -1487,10 +1501,10 @@
    Tables for Horizontal and Vertical Lines
    -
    \(x\) \(y\) Points
    \(-5\)\(-6\) \(-3\)\(\left(-5,-3\right)\)\(\left(-6,-3\right)\)
    \(2\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-5\) +
    Values of \(x\) and \(y\) satisfying the equation \(x=-6\)
    -
    Answer 1.
    \(-5\)
    Answer 2.
    \(\left(-5,-2\right)\)
    Answer 3.
    \(-5\)
    Answer 4.
    \(\left(-5,-1\right)\)
    Answer 5.
    \(-5\)
    Answer 6.
    \(\left(-5,0\right)\)
    Answer 7.
    \(-5\)
    Answer 8.
    \(\left(-5,1\right)\)
    Answer 9.
    \(-5\)
    Answer 10.
    \(\left(-5,2\right)\)
    Explanation.
    +
    Answer 1.
    \(-6\)
    Answer 2.
    \(\left(-6,-2\right)\)
    Answer 3.
    \(-6\)
    Answer 4.
    \(\left(-6,-1\right)\)
    Answer 5.
    \(-6\)
    Answer 6.
    \(\left(-6,0\right)\)
    Answer 7.
    \(-6\)
    Answer 8.
    \(\left(-6,1\right)\)
    Answer 9.
    \(-6\)
    Answer 10.
    \(\left(-6,2\right)\)
    Explanation.
    @@ -1498,49 +1512,49 @@
    Tables for Horizontal and Vertical Lines
    - + - + - + - + - + - + - + - + - + - + -
    \(x\)Points
    \(-5\)\(-6\) \(-2\)\(\left(-5,-2\right)\)\(\left(-6,-2\right)\)
    \(-5\)\(-6\) \(-1\)\(\left(-5,-1\right)\)\(\left(-6,-1\right)\)
    \(-5\)\(-6\) \(0\)\(\left(-5,0\right)\)\(\left(-6,0\right)\)
    \(-5\)\(-6\) \(1\)\(\left(-5,1\right)\)\(\left(-6,1\right)\)
    \(-5\)\(-6\) \(2\)\(\left(-5,2\right)\)\(\left(-6,2\right)\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-5\) +
    Values of \(x\) and \(y\) satisfying the equation \(x=-6\)
    14.
    -
    -
    +
    +
    -
    Fill out this table for the equation \(x=-3\text{.}\) The first row is an example.
    +
    Fill out this table for the equation \(x=-5\text{.}\) The first row is an example.
    + - - + @@ -1567,10 +1581,10 @@
    Tables for Horizontal and Vertical Lines
    -
    \(x\) \(y\) Points
    \(-5\) \(-3\)\(-3\)\(\left(-3,-3\right)\)\(\left(-5,-3\right)\)
    \(2\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-3\) +
    Values of \(x\) and \(y\) satisfying the equation \(x=-5\)
    -
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(-3,-2\right)\)
    Answer 3.
    \(-3\)
    Answer 4.
    \(\left(-3,-1\right)\)
    Answer 5.
    \(-3\)
    Answer 6.
    \(\left(-3,0\right)\)
    Answer 7.
    \(-3\)
    Answer 8.
    \(\left(-3,1\right)\)
    Answer 9.
    \(-3\)
    Answer 10.
    \(\left(-3,2\right)\)
    Explanation.
    +
    Answer 1.
    \(-5\)
    Answer 2.
    \(\left(-5,-2\right)\)
    Answer 3.
    \(-5\)
    Answer 4.
    \(\left(-5,-1\right)\)
    Answer 5.
    \(-5\)
    Answer 6.
    \(\left(-5,0\right)\)
    Answer 7.
    \(-5\)
    Answer 8.
    \(\left(-5,1\right)\)
    Answer 9.
    \(-5\)
    Answer 10.
    \(\left(-5,2\right)\)
    Explanation.
    @@ -1578,31 +1592,31 @@
    Tables for Horizontal and Vertical Lines
    - + - + - + - + - + - + - + - + - + - + -
    \(x\)Points
    \(-3\)\(-5\) \(-2\)\(\left(-3,-2\right)\)\(\left(-5,-2\right)\)
    \(-3\)\(-5\) \(-1\)\(\left(-3,-1\right)\)\(\left(-5,-1\right)\)
    \(-3\)\(-5\) \(0\)\(\left(-3,0\right)\)\(\left(-5,0\right)\)
    \(-3\)\(-5\) \(1\)\(\left(-3,1\right)\)\(\left(-5,1\right)\)
    \(-3\)\(-5\) \(2\)\(\left(-3,2\right)\)\(\left(-5,2\right)\)
    Values of \(x\) and \(y\) satisfying the equation \(x=-3\) +
    Values of \(x\) and \(y\) satisfying the equation \(x=-5\)
    @@ -1614,6 +1628,19 @@
    Tables for Horizontal and Vertical Lines
    Line Equations.
    15.
    +
    +
    +
    +
    A line’s graph is shown. Write an equation for the line.
    +
    +
    Answer.
    \(y = 2\)
    Explanation.
    +
    This line is horizontal, passing \((0,2)\) and \((1,2)\text{.}\) This implies its slope is \(0\text{.}\) +
    If a line’s slope is \(0\text{,}\) its equation has the form \(y=b\text{.}\) By the graph, we can see that its \(y\)-intercept is \(2\text{,}\) so this line’s equation is \(y=2\text{.}\) +
    +
    +
    +
    +
    16.
    @@ -1626,20 +1653,19 @@
    Line Equations.
    -
    16.
    -
    +
    17.
    +
    A line’s graph is shown. Write an equation for the line.
    -
    Answer.
    \(y = 4\)
    Explanation.
    -
    This line is horizontal, passing \((0,4)\) and \((1,4)\text{.}\) This implies its slope is \(0\text{.}\) -
    If a line’s slope is \(0\text{,}\) its equation has the form \(y=b\text{.}\) By the graph, we can see that its \(y\)-intercept is \(4\text{,}\) so this line’s equation is \(y=4\text{.}\) +
    Answer.
    \(x = 4\)
    Explanation.
    +
    This line is vertical, passing \((4,0)\) and \((4,1)\text{.}\) This implies its slope is undefined.
    If a line is vertical (which is the same thing as it having an undefined slope), its equation has the form \(x=a\text{.}\) By the graph, we can see it’s \(x\)-intercept is \(4\text{,}\) so this line’s equation is \(x=4\text{.}\)
    -
    17.
    +
    18.
    @@ -1651,18 +1677,6 @@
    Line Equations.
    -
    18.
    -
    -
    -
    -
    A line’s graph is shown. Write an equation for the line.
    -
    -
    Answer.
    \(x = -3\)
    Explanation.
    -
    This line is vertical, passing \((-3,0)\) and \((-3,1)\text{.}\) This implies its slope is undefined.
    If a line is vertical (which is the same thing as it having an undefined slope), its equation has the form \(x=a\text{.}\) By the graph, we can see it’s \(x\)-intercept is \(-3\text{,}\) so this line’s equation is \(x=-3\text{.}\) -
    -
    -
    -
    @@ -1670,6 +1684,22 @@
    Line Equations.
    Exercise Group.
    19.
    +
    +
    +
    +
    A line passes through the points \((3,-7)\) and \((-2,-7)\text{.}\) Find an equation for this line.
    An equation for this line is .
    +
    +
    Answer.
    \(y = -7\)
    Explanation.
    +
    To find an equation for a line, we first try to find its slope. To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (3,-7) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-2,-7) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-7-(-7)}{-2-3}\\ +\amp =\frac{0}{-5}\\ +\amp =0\end{aligned} }\)
    This line’s slope is \(0\text{,}\) implying that it is horizontal and that it has an equation in the form \(y=c\text{.}\) +
    In both of the given points \((3,-7)\) and \((-2,-7)\text{,}\) we can see their \(y\) value is \(-7\text{.}\) Thus this line’s equation must be \(y=-7\text{.}\) +
    +
    +
    +
    +
    20.
    @@ -1685,23 +1715,23 @@
    Exercise Group.
    -
    20.
    -
    +
    21.
    +
    -
    A line passes through the points \((-4,-3)\) and \((-1,-3)\text{.}\) Find an equation for this line.
    An equation for this line is .
    +
    A line passes through the points \((-3,-4)\) and \((-3,-1)\text{.}\) Find an equation for this line.
    An equation for this line is .
    -
    Answer.
    \(y = -3\)
    Explanation.
    -
    To find an equation for a line, we first try to find its slope. To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-4,-3) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-1,-3) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{-3-(-3)}{-1-(-4)}\\ -\amp =\frac{0}{3}\\ -\amp =0\end{aligned} }\)
    This line’s slope is \(0\text{,}\) implying that it is horizontal and that it has an equation in the form \(y=c\text{.}\) -
    In both of the given points \((-4,-3)\) and \((-1,-3)\text{,}\) we can see their \(y\) value is \(-3\text{.}\) Thus this line’s equation must be \(y=-3\text{.}\) +
    Answer.
    \(x = -3\)
    Explanation.
    +
    To find an equation for a line, we first try to find its slope. To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-3,-4) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (-3,-1) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-1-(-4)}{-3-(-3)}\\ +\amp =\frac{3}{0}\\ +\amp \phantom{={}}\text{this is an undefined quantity}\end{aligned} }\)
    This line’s slope is undefined, implying that it is vertical and that it has an equation in the form \(x=c\text{.}\) +
    In both of the given points \((-3,-4)\) and \((-3,-1)\text{,}\) we can see their \(x\) value is \(-3\text{.}\) Thus this line’s equation must be \(x=-3\text{.}\)
    -
    21.
    +
    22.
    @@ -1717,22 +1747,6 @@
    Exercise Group.
    -
    22.
    -
    -
    -
    -
    A line passes through the points \((2,-1)\) and \((2,5)\text{.}\) Find an equation for this line.
    An equation for this line is .
    -
    -
    Answer.
    \(x = 2\)
    Explanation.
    -
    To find an equation for a line, we first try to find its slope. To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    First, we mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (2,-1) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (2,5) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{5-(-1)}{2-2}\\ -\amp =\frac{6}{0}\\ -\amp \phantom{={}}\text{this is an undefined quantity}\end{aligned} }\)
    This line’s slope is undefined, implying that it is vertical and that it has an equation in the form \(x=c\text{.}\) -
    In both of the given points \((2,-1)\) and \((2,5)\text{,}\) we can see their \(x\) value is \(2\text{.}\) Thus this line’s equation must be \(x=2\text{.}\) -
    -
    -
    -
    @@ -1740,12 +1754,12 @@
    Exercise Group.
    Intercepts.
    23.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -x = 4 +x = 2 \end{equation*}
    @@ -1771,11 +1785,11 @@
    Intercepts.
    -\(x\)-intercept and \(y\)-intercept of the line \(x=4\) +\(x\)-intercept and \(y\)-intercept of the line \(x=2\)
    -
    Answer 1.
    \(\text{NONE}\)
    Answer 2.
    \(\text{NONE}\)
    Answer 3.
    \(\text{NONE}\)
    Answer 4.
    \(4\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(4,0\right)\)
    Explanation.
    -
    The line \(x=4\) is special in that it is parallel to the \(y\)-axis. Thus it does not have any \(y\)-intercept.
    The line \(x=4\) crosses the \(x\)-axis at \((4,0)\text{,}\) which is its \(x\)-intercept.
    The entries for the table are:
    +
    Answer 1.
    \(\text{NONE}\)
    Answer 2.
    \(\text{NONE}\)
    Answer 3.
    \(\text{NONE}\)
    Answer 4.
    \(2\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(2,0\right)\)
    Explanation.
    +
    The line \(x=2\) is special in that it is parallel to the \(y\)-axis. Thus it does not have any \(y\)-intercept.
    The line \(x=2\) crosses the \(x\)-axis at \((2,0)\text{,}\) which is its \(x\)-intercept.
    The entries for the table are:
    - + - +
    @@ -1794,23 +1808,23 @@
    Intercepts.
    \(x\)-intercept\(4\)\(2\) \(0\)\((4,0)\)\((2,0)\)
    -\(x\)-intercept and \(y\)-intercept of the line \(x=4\) +\(x\)-intercept and \(y\)-intercept of the line \(x=2\)
    24.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -x = 7 +x = 4 \end{equation*}
    @@ -1836,11 +1850,11 @@
    Intercepts.
    -\(x\)-intercept and \(y\)-intercept of the line \(x=7\) +\(x\)-intercept and \(y\)-intercept of the line \(x=4\)
    -
    Answer 1.
    \(\text{NONE}\)
    Answer 2.
    \(\text{NONE}\)
    Answer 3.
    \(\text{NONE}\)
    Answer 4.
    \(7\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(7,0\right)\)
    Explanation.
    -
    The line \(x=7\) is special in that it is parallel to the \(y\)-axis. Thus it does not have any \(y\)-intercept.
    The line \(x=7\) crosses the \(x\)-axis at \((7,0)\text{,}\) which is its \(x\)-intercept.
    The entries for the table are:
    +
    Answer 1.
    \(\text{NONE}\)
    Answer 2.
    \(\text{NONE}\)
    Answer 3.
    \(\text{NONE}\)
    Answer 4.
    \(4\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(4,0\right)\)
    Explanation.
    +
    The line \(x=4\) is special in that it is parallel to the \(y\)-axis. Thus it does not have any \(y\)-intercept.
    The line \(x=4\) crosses the \(x\)-axis at \((4,0)\text{,}\) which is its \(x\)-intercept.
    The entries for the table are:
    - + - +
    @@ -1859,23 +1873,23 @@
    Intercepts.
    \(x\)-intercept\(7\)\(4\) \(0\)\((7,0)\)\((4,0)\)
    -\(x\)-intercept and \(y\)-intercept of the line \(x=7\) +\(x\)-intercept and \(y\)-intercept of the line \(x=4\)
    25.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -y = 9 +y = 7 \end{equation*}
    @@ -1901,11 +1915,11 @@
    Intercepts.
    -\(x\)-intercept and \(y\)-intercept of the line \(y=9\) +\(x\)-intercept and \(y\)-intercept of the line \(y=7\)
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(9\)
    Answer 3.
    \(\left(0,9\right)\)
    Answer 4.
    \(\text{NONE}\)
    Answer 5.
    \(\text{NONE}\)
    Answer 6.
    \(\text{NONE}\)
    Explanation.
    -
    The line \(y=9\) is special in that it is parallel to the \(x\)-axis. Thus it does not have any \(x\)-intercept.
    The line \(y=9\) crosses the \(y\)-axis at \((0,9)\text{,}\) which is its \(y\)-intercept.
    The entries for the table are:
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(7\)
    Answer 3.
    \(\left(0,7\right)\)
    Answer 4.
    \(\text{NONE}\)
    Answer 5.
    \(\text{NONE}\)
    Answer 6.
    \(\text{NONE}\)
    Explanation.
    +
    The line \(y=7\) is special in that it is parallel to the \(x\)-axis. Thus it does not have any \(x\)-intercept.
    The line \(y=7\) crosses the \(y\)-axis at \((0,7)\text{,}\) which is its \(y\)-intercept.
    The entries for the table are:
    - - + +
    @@ -1918,8 +1932,8 @@
    Intercepts.
    		 \(y\)-intercept \(0\)\(9\)\(\left(0,9\right)\)\(7\)\(\left(0,7\right)\)
    @@ -1929,18 +1943,18 @@
    Intercepts.
    		\(\text{NONE}\)
    -\(x\)-intercept and \(y\)-intercept of the line \(y=9\) +\(x\)-intercept and \(y\)-intercept of the line \(y=7\)
    26.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -y = -10 +y = 9 \end{equation*}
    @@ -1966,11 +1980,11 @@
    Intercepts.
    -\(x\)-intercept and \(y\)-intercept of the line \(y=-10\) +\(x\)-intercept and \(y\)-intercept of the line \(y=9\)
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(-10\)
    Answer 3.
    \(\left(0,-10\right)\)
    Answer 4.
    \(\text{NONE}\)
    Answer 5.
    \(\text{NONE}\)
    Answer 6.
    \(\text{NONE}\)
    Explanation.
    -
    The line \(y=-10\) is special in that it is parallel to the \(x\)-axis. Thus it does not have any \(x\)-intercept.
    The line \(y=-10\) crosses the \(y\)-axis at \((0,-10)\text{,}\) which is its \(y\)-intercept.
    The entries for the table are:
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(9\)
    Answer 3.
    \(\left(0,9\right)\)
    Answer 4.
    \(\text{NONE}\)
    Answer 5.
    \(\text{NONE}\)
    Answer 6.
    \(\text{NONE}\)
    Explanation.
    +
    The line \(y=9\) is special in that it is parallel to the \(x\)-axis. Thus it does not have any \(x\)-intercept.
    The line \(y=9\) crosses the \(y\)-axis at \((0,9)\text{,}\) which is its \(y\)-intercept.
    The entries for the table are:
    - - + +
    @@ -1983,8 +1997,8 @@
    Intercepts.
    		 \(y\)-intercept \(0\)\(-10\)\(\left(0,-10\right)\)\(9\)\(\left(0,9\right)\)
    @@ -1994,7 +2008,7 @@
    Intercepts.
    		\(\text{NONE}\)
    -\(x\)-intercept and \(y\)-intercept of the line \(y=-10\) +\(x\)-intercept and \(y\)-intercept of the line \(y=9\)
    @@ -2024,120 +2038,139 @@
    Graphs of Horizontal and Vertical Lines.
    Parallel or Perpendicular?.
    31.
    -
    +
    +
    +
    +
    Line \(m\) passes points \((2,-8)\) and \((-3,-13)\text{.}\) +
    Line \(n\) passes points \((4,5)\) and \((5,6)\text{.}\) +
    These two lines are .
    +
    +
    Answer.
    \(\text{parallel}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-13-(-8)}{-3-2}\\ +\amp =\frac{-5}{-5}\\ \amp =1 \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{6-5}{5-4}\\ +\amp =\frac{1}{1}\\ +\amp =1 \end{aligned}}\)
    Since these two lines have the same slope, they are parallel to each other.
    +
    +
    +
    +
    32.
    +
    -
    Line \(m\) passes points \((-5,-1)\) and \((-1,3)\text{.}\) -
    Line \(n\) passes points \((-5,-11)\) and \((5,-1)\text{.}\) -
    These two lines are .
    +
    Line \(m\) passes points \((-14,-5)\) and \((-21,-3)\text{.}\) +
    Line \(n\) passes points \((-14,14)\) and \((7,8)\text{.}\) +
    These two linea are .
    Answer.
    \(\text{parallel}\)
    Explanation.
    -
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{3-(-1)}{-1-(-5)}\\ -\amp =\frac{4}{4}\\ \amp =1 \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{-1-(-11)}{5-(-5)}\\ -\amp =\frac{10}{10}\\ -\amp =1 \end{aligned}}\)
    Since these two lines have the same slope, they are parallel to each other.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-3-(-5)}{-21-(-14)}\\ +\amp =\frac{2}{-7}\\ +\amp =-\frac{2}{7} \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{8-14}{7-(-14)}\\ +\amp =\frac{-6}{21}\\ +\amp =-\frac{2}{7} \end{aligned}}\)
    Since these two lines have the same slope, they are parallel to each other.
    -
    32.
    -
    +
    33.
    +
    -
    Line \(m\) passes points \((-10,8)\) and \((-15,12)\text{.}\) -
    Line \(n\) passes points \((20,-11)\) and \((25,-15)\text{.}\) -
    These two linea are .
    +
    Line \(m\) passes points \((-6,-1)\) and \((-9,0)\text{.}\) +
    Line \(n\) passes points \((4,17)\) and \((5,20)\text{.}\) +
    These two lines are .
    -
    Answer.
    \(\text{parallel}\)
    Explanation.
    +
    Answer.
    \(\text{perpendicular}\)
    Explanation.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{12-8}{-15-(-10)}\\ -\amp =\frac{4}{-5}\\ -\amp =-\frac{4}{5} \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{-15-(-11)}{25-20}\\ -\amp =\frac{-4}{5}\\ -\amp =-\frac{4}{5} \end{aligned}}\)
    Since these two lines have the same slope, they are parallel to each other.
    +\amp =\frac{0-(-1)}{-9-(-6)}\\ +\amp =\frac{1}{-3}\\ +\amp =-\frac{1}{3} \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{20-17}{5-4}\\ +\amp =\frac{3}{1}\\ +\amp =3 \end{aligned}}\)
    If the product of two lines’ slopes is \(-1\text{,}\) these two lines are perpendicular.
    Note that \(\displaystyle{ 3 \cdot \left(-\frac{1}{3}\right) = -1 }\text{.}\) This implies these two lines are perpendicular.
    -
    33.
    -
    +
    34.
    +
    -
    Line \(m\) passes points \((9,7)\) and \((-6,12)\text{.}\) -
    Line \(n\) passes points \((-4,-21)\) and \((-1,-12)\text{.}\) +
    Line \(m\) passes points \((27,-3)\) and \((-18,17)\text{.}\) +
    Line \(n\) passes points \((-8,-27)\) and \((-4,-18)\text{.}\)
    These two lines are .
    Answer.
    \(\text{perpendicular}\)
    Explanation.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{12-7}{-6-9}\\ -\amp =\frac{5}{-15}\\ -\amp =-\frac{1}{3} \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{-12-(-21)}{-1-(-4)}\\ -\amp =\frac{9}{3}\\ -\amp =3 \end{aligned}}\)
    If the product of two lines’ slopes is \(-1\text{,}\) these two lines are perpendicular.
    Note that \(\displaystyle{ 3 \cdot \left(-\frac{1}{3}\right) = -1 }\text{.}\) This implies these two lines are perpendicular.
    +\amp =\frac{17-(-3)}{-18-27}\\ +\amp =\frac{20}{-45}\\ +\amp =-\frac{4}{9} \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-18-(-27)}{-4-(-8)}\\ +\amp =\frac{9}{4}\\ +\amp =\frac{9}{4} \end{aligned}}\)
    If the product of two lines’ slopes is \(-1\text{,}\) these two lines are perpendicular.
    Note that \(\displaystyle{ -\frac{4}{9} \cdot \frac{9}{4} = -1 }\text{.}\) This implies these two lines are perpendicular.
    -
    34.
    -
    +
    35.
    +
    -
    Line \(m\) passes points \((7,-8)\) and \((-14,4)\text{.}\) -
    Line \(n\) passes points \((-8,-13)\) and \((8,15)\text{.}\) +
    Line \(m\) passes points \((-2,9)\) and \((1,0)\text{.}\) +
    Line \(n\) passes points \((-3,-2)\) and \((1,2)\text{.}\)
    These two lines are .
    -
    Answer.
    \(\text{perpendicular}\)
    Explanation.
    -
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{4-(-8)}{-14-7}\\ -\amp =\frac{12}{-21}\\ -\amp =-\frac{4}{7} \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{15-(-13)}{8-(-8)}\\ -\amp =\frac{28}{16}\\ -\amp =\frac{7}{4} \end{aligned}}\)
    If the product of two lines’ slopes is \(-1\text{,}\) these two lines are perpendicular.
    Note that \(\displaystyle{ -\frac{4}{7} \cdot \frac{7}{4} = -1 }\text{.}\) This implies these two lines are perpendicular.
    +
    Answer.
    \(\text{neither parallel nor perpendicular}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{0-9}{1-(-2)}\\ +\amp =\frac{-9}{3}=-3 \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{2-(-2)}{1-(-3)}\\ +\amp =\frac{4}{4}\\ +\amp =1 \end{aligned}}\)
    Since \(-3 \neq 1\text{,}\) these two lines are not parallel.
    Since \(-3 \cdot 1 \neq -1\text{,}\) these two lines are not perpendicular.
    The correct answer is these two lines are neither parallel nor perpendicular.
    -
    35.
    -
    +
    36.
    +
    -
    Line \(m\) passes points \((3,-13)\) and \((-2,2)\text{.}\) -
    Line \(n\) passes points \((-2,-12)\) and \((5,-5)\text{.}\) +
    Line \(m\) passes points \((-4,2)\) and \((5,2)\text{.}\) +
    Line \(n\) passes points \((-7,-4)\) and \((-10,-4)\text{.}\)
    These two lines are .
    -
    Answer.
    \(\text{neither parallel nor perpendicular}\)
    Explanation.
    -
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{2-(-13)}{-2-3}\\ -\amp =\frac{15}{-5}=-3 \end{aligned}}\)
    For line \(n\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{-5-(-12)}{5-(-2)}\\ -\amp =\frac{7}{7}\\ -\amp =1 \end{aligned}}\)
    Since \(-3 \neq 1\text{,}\) these two lines are not parallel.
    Since \(-3 \cdot 1 \neq -1\text{,}\) these two lines are not perpendicular.
    The correct answer is these two lines are neither parallel nor perpendicular.
    +
    Answer.
    \(\text{parallel}\)
    Explanation.
    +
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{2-2}{5-(-4)}\\ +\amp =\frac{0}{9}\\ +\amp =0 \end{aligned}}\)
    This is a special line, \(y=2\text{.}\) It is horizontal with slope being \(0\text{.}\) +
    For line \(n\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{-4-(-4)}{-10-(-7)}\\ +\amp =\frac{0}{-3}\\ +\amp =0 \end{aligned}}\)
    This is a special line, \(y=-4\text{.}\) It is horizontal with slope being \(0\text{.}\) +
    Since both lines are horizontal, these two lines are parallel.
    -
    36.
    -
    +
    37.
    +
    -
    Line \(m\) passes points \((9,4)\) and \((-7,4)\text{.}\) -
    Line \(n\) passes points \((-7,-10)\) and \((1,-10)\text{.}\) +
    Line \(m\) passes points \((4,9)\) and \((4,-7)\text{.}\) +
    Line \(n\) passes points \((-10,-7)\) and \((-10,1)\text{.}\)
    These two lines are .
    Answer.
    \(\text{parallel}\)
    Explanation.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{4-4}{-7-9}\\ -\amp =\frac{0}{-16}\\ -\amp =0 \end{aligned}}\)
    This is a special line, \(y=4\text{.}\) It is horizontal with slope being \(0\text{.}\) -
    For line \(n\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{-10-(-10)}{1-(-7)}\\ -\amp =\frac{0}{8}\\ -\amp =0 \end{aligned}}\)
    This is a special line, \(y=-10\text{.}\) It is horizontal with slope being \(0\text{.}\) -
    Since both lines are horizontal, these two lines are parallel.
    +\amp =\frac{-7-9}{4-4}\\ +\amp =\frac{-16}{0}\\ +\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=4\text{.}\) It is vertical with an undefined slope.
    For line \(n\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ +\amp =\frac{1-(-7)}{-10-(-10)}\\ +\amp =\frac{8}{0}\\ +\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=-10\text{.}\) It is vertical with an undefined slope.
    Since both lines are vertical, these two lines are parallel.
    -
    37.
    +
    38.
    @@ -2156,25 +2189,6 @@
    Parallel or Perpendicular?.
    -
    38.
    -
    -
    -
    -
    Line \(m\) passes points \((9,-5)\) and \((9,10)\text{.}\) -
    Line \(n\) passes points \((0,-8)\) and \((0,1)\text{.}\) -
    These two lines are .
    -
    -
    Answer.
    \(\text{parallel}\)
    Explanation.
    -
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    For line \(m\text{,}\) we have:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{10-(-5)}{9-9}\\ -\amp =\frac{15}{0}\\ -\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=9\text{.}\) It is vertical with an undefined slope.
    For line \(n\text{,}\) we have:
    \(\displaystyle{\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ -\amp =\frac{1-(-8)}{0-0}\\ -\amp =\frac{9}{0}\\ -\amp =\text{undefined} \end{aligned}}\)
    This is a special line, \(x=0\text{.}\) It is vertical with an undefined slope.
    Since both lines are vertical, these two lines are parallel.
    -
    -
    -
    @@ -2182,6 +2196,19 @@
    Parallel or Perpendicular?.
    Parallel and Perpendicular Line Equations.
    39.
    +
    +
    +
    +
    A line passes through the point \((10,-5)\text{,}\) and it’s parallel to the line \(y=9\text{.}\) Find an equation for this line.
    +
    +
    Answer.
    \(y = -5\)
    Explanation.
    +
    Since \(y=9\) is horizontal, the unknown line must also be horizontal (because they are parallel). Assume its equation is \(y=b\text{.}\) +
    Since the unknown line passes through \((10,-5)\text{,}\) \(b\)’s value must be \(-5\text{.}\) The unknown line’s equation is \({y = -5}\) +
    +
    +
    +
    +
    40.
    @@ -2194,20 +2221,20 @@
    Parallel and Perpendicular Line Equation
    -
    40.
    -
    +
    41.
    +
    -
    A line passes through the point \((7,1)\text{,}\) and it’s parallel to the line \(y=-8\text{.}\) Find an equation for this line.
    +
    A line passes through the point \((1,7)\text{,}\) and it’s parallel to the line \(x=-8\text{.}\) Find an equation for this line.
    -
    Answer.
    \(y = 1\)
    Explanation.
    -
    Since \(y=-8\) is horizontal, the unknown line must also be horizontal (because they are parallel). Assume its equation is \(y=b\text{.}\) -
    Since the unknown line passes through \((7,1)\text{,}\) \(b\)’s value must be \(1\text{.}\) The unknown line’s equation is \({y = 1}\) +
    Answer.
    \(x = 1\)
    Explanation.
    +
    Since \(x=-8\) is vertical, the unknown line must also be vertical (because they are parallel). Assume its equation is \(x=a\text{.}\) +
    Since the unknown line passes through \((1,7)\text{,}\) \(a\)’s value must be \(1\text{.}\) The unknown line’s equation is \({x = 1}\)
    -
    41.
    +
    42.
    @@ -2220,20 +2247,26 @@
    Parallel and Perpendicular Line Equation
    -
    42.
    -
    +
    43.
    +
    -
    A line passes through the point \((8,3)\text{,}\) and it’s parallel to the line \(x=-3\text{.}\) Find an equation for this line.
    +
    Line \(k\) has the equation \(y={-2x+4}\text{.}\) +
    Line \(\ell\) is parallel to line \(k\text{,}\) but passes through the point \((-1,{5})\text{.}\) +
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -
    Answer.
    \(x = 8\)
    Explanation.
    -
    Since \(x=-3\) is vertical, the unknown line must also be vertical (because they are parallel). Assume its equation is \(x=a\text{.}\) -
    Since the unknown line passes through \((8,3)\text{,}\) \(a\)’s value must be \(8\text{.}\) The unknown line’s equation is \({x = 8}\) -
    +
    Answer 1.
    \(y = -2\mathopen{}\left(x+1\right)+5\)
    Answer 2.
    \(y = -2x+3\)
    Explanation.
    +
    When two lines are parallel, they have the same slope. Since line \(k\)’s slope is \(-2\text{,}\) line \(\ell\)’s slope is also \(-2\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-1,{5})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -2\mathopen{}\left(x+1\right)+5}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {-2\mathopen{}\left(x+1\right)+5}\\ +y \amp = -2x + 2(-1) +{5}\\ +y \amp = {-2x+3} +\end{aligned} +}\)
    -
    43.
    +
    44.
    @@ -2252,26 +2285,26 @@
    Parallel and Perpendicular Line Equation
    -
    44.
    -
    +
    45.
    +
    -
    Line \(k\) has the equation \(y={-4x+4}\text{.}\) -
    Line \(\ell\) is parallel to line \(k\text{,}\) but passes through the point \((-4,{7})\text{.}\) +
    Line \(k\) has the equation \(y={-{\frac{6}{5}}x-9}\text{.}\) +
    Line \(\ell\) is parallel to line \(k\text{,}\) but passes through the point \((-3,{{\frac{38}{5}}})\text{.}\)
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -
    Answer 1.
    \(y = -4\mathopen{}\left(x+4\right)+7\)
    Answer 2.
    \(y = -4x-9\)
    Explanation.
    -
    When two lines are parallel, they have the same slope. Since line \(k\)’s slope is \(-4\text{,}\) line \(\ell\)’s slope is also \(-4\text{.}\) -
    Now we know line \(\ell\)’s slope, and a point on it: \((-4,{7})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -4\mathopen{}\left(x+4\right)+7}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} -y \amp = {-4\mathopen{}\left(x+4\right)+7}\\ -y \amp = -4x + 4(-4) +{7}\\ -y \amp = {-4x-9} +
    Answer 1.
    \(y = -{\frac{6}{5}}\mathopen{}\left(x+3\right)+{\frac{38}{5}}\)
    Answer 2.
    \(y = -{\frac{6}{5}}x+4\)
    Explanation.
    +
    When two lines are parallel, they have the same slope. Since line \(k\)’s slope is \({-{\frac{6}{5}}}\text{,}\) line \(\ell\)’s slope is also \({-{\frac{6}{5}}}\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-3,{{\frac{38}{5}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{6}{5}}\mathopen{}\left(x+3\right)+{\frac{38}{5}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {-{\frac{6}{5}}\mathopen{}\left(x+3\right)+{\frac{38}{5}}}\\ +y \amp = {-{\frac{6}{5}}}x - {-{\frac{6}{5}}}(-3) +{{\frac{38}{5}}}\\ +y \amp = {-{\frac{6}{5}}x+4} \end{aligned} }\)
    -
    45.
    +
    46.
    @@ -2290,98 +2323,79 @@
    Parallel and Perpendicular Line Equation
    -
    46.
    -
    -
    -
    -
    Line \(k\) has the equation \(y={-{\frac{8}{5}}x-5}\text{.}\) -
    Line \(\ell\) is parallel to line \(k\text{,}\) but passes through the point \((-3,{{\frac{14}{5}}})\text{.}\) -
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{14}{5}}\)
    Answer 2.
    \(y = -{\frac{8}{5}}x-2\)
    Explanation.
    -
    When two lines are parallel, they have the same slope. Since line \(k\)’s slope is \({-{\frac{8}{5}}}\text{,}\) line \(\ell\)’s slope is also \({-{\frac{8}{5}}}\text{.}\) -
    Now we know line \(\ell\)’s slope, and a point on it: \((-3,{{\frac{14}{5}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{14}{5}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} -y \amp = {-{\frac{8}{5}}\mathopen{}\left(x+3\right)+{\frac{14}{5}}}\\ -y \amp = {-{\frac{8}{5}}}x - {-{\frac{8}{5}}}(-3) +{{\frac{14}{5}}}\\ -y \amp = {-{\frac{8}{5}}x-2} -\end{aligned} -}\)
    -
    -
    -
    47.
    -
    -
    +
    +
    -
    Line \(k\) has the equation \(y={x-8}\text{.}\) -
    Line \(\ell\) is perpendicular to line \(k\text{,}\) and passes through the point \((5,{-2})\text{.}\) -
    Find an equation for line \(\ell\text{.}\) +
    Line \(k\) has the equation \(y={x-1}\text{.}\) +
    Line \(\ell\) is perpendicular to line \(k\text{,}\) and passes through the point \((-5,{3})\text{.}\) +
    Find an equation for line \(\ell\text{.}\)
    -
    Answer.
    \(y = -x+3\)
    Explanation.
    -
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) Since line \(k\)’s slope is \(1\text{,}\) line \(\ell\)’s slope must be \(-1\text{.}\) -
    Now we know line \(\ell\)’s slope, and a point on it: \((5,{-2})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -\left(x-5\right)-2}}\)
    This simplifies to:
    \(\displaystyle{{y = -x+3}}\)
    +
    Answer.
    \(y = -x-2\)
    Explanation.
    +
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) Since line \(k\)’s slope is \(1\text{,}\) line \(\ell\)’s slope must be \(-1\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-5,{3})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -\left(x+5\right)+3}}\)
    This simplifies to:
    \(\displaystyle{{y = -x-2}}\)
    48.
    -
    -
    +
    +
    -
    Line \(k\) has the equation \(y={2x-4}\text{.}\) -
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-3,{{\frac{7}{2}}})\text{.}\) -
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    +
    Line \(k\) has the equation \(y={-6x+3}\text{.}\) +
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-3,{{\frac{9}{2}}})\text{.}\) +
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -
    Answer 1.
    \(y = -{\frac{1}{2}}\mathopen{}\left(x+3\right)+{\frac{7}{2}}\)
    Answer 2.
    \(y = -{\frac{1}{2}}x+2\)
    Explanation.
    -
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) It’s given that line \(k\)’s slope is \(2\text{.}\) So the slope of \(\ell\) must be \({-{\frac{1}{2}}}\text{.}\) -
    Now we know line \(\ell\)’s slope, and a point on it: \((-3,{{\frac{7}{2}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{1}{2}}\mathopen{}\left(x+3\right)+{\frac{7}{2}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} -y \amp = {-{\frac{1}{2}}\mathopen{}\left(x+3\right)+{\frac{7}{2}}}\\ -y \amp = {-{\frac{1}{2}}}x - {-{\frac{1}{2}}}(-3) +{{\frac{7}{2}}}\\ -y \amp = {-{\frac{1}{2}}x+2} +
    Answer 1.
    \(y = {\frac{1}{6}}\mathopen{}\left(x+3\right)+{\frac{9}{2}}\)
    Answer 2.
    \(y = {\frac{1}{6}}x+5\)
    Explanation.
    +
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) It’s given that line \(k\)’s slope is \(-6\text{.}\) So the slope of \(\ell\) must be \({{\frac{1}{6}}}\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-3,{{\frac{9}{2}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = {\frac{1}{6}}\mathopen{}\left(x+3\right)+{\frac{9}{2}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {{\frac{1}{6}}\mathopen{}\left(x+3\right)+{\frac{9}{2}}}\\ +y \amp = {{\frac{1}{6}}}x - {{\frac{1}{6}}}(-3) +{{\frac{9}{2}}}\\ +y \amp = {{\frac{1}{6}}x+5} \end{aligned} }\)
    49.
    -
    -
    +
    +
    -
    Line \(k\)’s equation is \(y={{\frac{3}{5}}x-1}\text{.}\) -
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-3,{10})\text{.}\) -
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    +
    Line \(k\)’s equation is \(y={{\frac{2}{3}}x-5}\text{.}\) +
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((1,{-{\frac{9}{2}}})\text{.}\) +
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -
    Answer 1.
    \(y = -{\frac{5}{3}}\mathopen{}\left(x+3\right)+10\)
    Answer 2.
    \(y = -{\frac{5}{3}}x+5\)
    Explanation.
    -
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) It’s given that line \(k\)’s slope is \({{\frac{3}{5}}}\text{.}\) So the slope of \(\ell\) must be \({-{\frac{5}{3}}}\text{.}\) -
    Now we know line \(\ell\)’s slope, and a point on it: \((-3,{10})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{5}{3}}\mathopen{}\left(x+3\right)+10}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} -y \amp = {-{\frac{5}{3}}\mathopen{}\left(x+3\right)+10}\\ -y \amp = {-{\frac{5}{3}}}x - {-{\frac{5}{3}}}(-3) +{10}\\ -y \amp = {-{\frac{5}{3}}x+5} +
    Answer 1.
    \(y = -{\frac{3}{2}}\mathopen{}\left(x-1\right)+-{\frac{9}{2}}\)
    Answer 2.
    \(y = -{\frac{3}{2}}x-3\)
    Explanation.
    +
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) It’s given that line \(k\)’s slope is \({{\frac{2}{3}}}\text{.}\) So the slope of \(\ell\) must be \({-{\frac{3}{2}}}\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((1,{-{\frac{9}{2}}})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = -{\frac{3}{2}}\mathopen{}\left(x-1\right) - {\frac{9}{2}}}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {-{\frac{3}{2}}\mathopen{}\left(x-1\right) - {\frac{9}{2}}}\\ +y \amp = {-{\frac{3}{2}}}x - {-{\frac{3}{2}}}(1) +{-{\frac{9}{2}}}\\ +y \amp = {-{\frac{3}{2}}x-3} \end{aligned} }\)
    50.
    -
    -
    +
    +
    -
    Line \(k\) has the equation \({x+2y}=10\text{.}\) -
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-5,{-14})\text{.}\) -
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    +
    Line \(k\) has the equation \({x+5y}=5\text{.}\) +
    Line \(\ell\) is perpendicular to line \(k\) and passes through the point \((-5,{-26})\text{.}\) +
    Find an equation for line \(\ell\) in both point-slope form and slope-intercept form.
    An equation for \(\ell\) in point-slope form is: .
    An equation for \(\ell\) in slope-intercept form is: .
    -
    Answer 1.
    \(y = 2\mathopen{}\left(x+5\right)-14\)
    Answer 2.
    \(y = 2x-4\)
    Explanation.
    -
    First, we find line \(k\)’s slope:
    \(\displaystyle{\begin{aligned} -{x+2y} \amp = 10 \\ +
    Answer 1.
    \(y = 5\mathopen{}\left(x+5\right)-26\)
    Answer 2.
    \(y = 5x-1\)
    Explanation.
    +
    First, we find line \(k\)’s slope:
    \(\displaystyle{\begin{aligned} +{x+5y} \amp = 5 \\ \amp \vdots\\ -y \amp = {-{\frac{1}{2}}x+5} +y \amp = {-{\frac{1}{5}}x+1} \end{aligned} -}\)
    We can see line \(k\)’s slope is \({-{\frac{1}{2}}}\text{.}\) -
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) We just found that line \(k\)’s slope is \({-{\frac{1}{2}}}\text{.}\) So the slope ov \(\ell\) must be \({2}\text{.}\) -
    Now we know line \(\ell\)’s slope, and a point on it: \((-5,{-14})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = 2\mathopen{}\left(x+5\right)-14}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} -y \amp = {2\mathopen{}\left(x+5\right)-14}\\ -y \amp = {2}x - {2}(-5) +{-14}\\ -y \amp = {2x-4} +}\)
    We can see line \(k\)’s slope is \({-{\frac{1}{5}}}\text{.}\) +
    When two lines are perpendicular, the product of their slopes is \(-1\text{.}\) We just found that line \(k\)’s slope is \({-{\frac{1}{5}}}\text{.}\) So the slope ov \(\ell\) must be \({5}\text{.}\) +
    Now we know line \(\ell\)’s slope, and a point on it: \((-5,{-26})\text{.}\) We substitute these numbers into the generic formula for a point-slope equation for a line:
    \(\displaystyle{y=m(x-x_{1})+y_{1}}\)
    and we have:
    \(\displaystyle{{y = 5\mathopen{}\left(x+5\right)-26}}\)
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and expand and simplify the right side.
    \(\displaystyle{\begin{aligned} +y \amp = {5\mathopen{}\left(x+5\right)-26}\\ +y \amp = {5}x - {5}(-5) +{-26}\\ +y \amp = {5x-1} \end{aligned} }\)
    @@ -2391,14 +2405,14 @@
    Parallel and Perpendicular Line Equation

    Challenge

    51.
    -
    -
    +
    +
    -
    Prove that a triangle with vertices at the points \((1, 1)\text{,}\) \((-4, 4)\text{,}\) and \((-3, 0)\) is a right triangle.
    -
    Explanation.
    -
    The slope between \((1,1)\) and \((-3,0)\) is \(\frac{0-1}{-3 - (-1)}\) or \(\frac{1}{4} \text{.}\) -
    The slope between \((-3,0)\) and \((-4,4)\) is \(\frac{4-0}{-4 - (-3)}\) or \(\frac{-4}{1} \text{.}\) -
    Since \(\frac{1}{4}\) and \(\frac{-4}{1}\) are negative reciprocals, those two sides are perpendicular. Thus, the triangle is a right triangle.
    +
    Prove that a triangle with vertices at the points \((1, 1)\text{,}\) \((-4, 4)\text{,}\) and \((-3, 0)\) is a right triangle.
    +
    Explanation.
    +
    The slope between \((1,1)\) and \((-3,0)\) is \(\frac{0-1}{-3 - (-1)}\) or \(\frac{1}{4} \text{.}\) +
    The slope between \((-3,0)\) and \((-4,4)\) is \(\frac{4-0}{-4 - (-3)}\) or \(\frac{-4}{1} \text{.}\) +
    Since \(\frac{1}{4}\) and \(\frac{-4}{1}\) are negative reciprocals, those two sides are perpendicular. Thus, the triangle is a right triangle.
    diff --git a/section-isolating-a-linear-variable.html b/section-isolating-a-linear-variable.html index 38bde802c..6b29de930 100644 --- a/section-isolating-a-linear-variable.html +++ b/section-isolating-a-linear-variable.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-modeling-with-equations-and-inequalities.html b/section-modeling-with-equations-and-inequalities.html index 94504bd5e..ef5c38e31 100644 --- a/section-modeling-with-equations-and-inequalities.html +++ b/section-modeling-with-equations-and-inequalities.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-order-of-operations.html b/section-order-of-operations.html index e8bc6ebac..798d6b32f 100644 --- a/section-order-of-operations.html +++ b/section-order-of-operations.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -605,10 +619,10 @@

    Search Results:

    If we find \(3^2\text{,}\) we have the area of one of the small square garden plots on the left. Then if we double that, we have \(2\cdot\left(3^2\right)\text{,}\) the area of the entire left garden plot.
    But if we find \((2\cdot3)^2\text{,}\) then first we are doubling \(3\text{.}\) So we are getting the area of a large square garden plot whose sides are twice as long. We end up with the area of the entire garden plot on the right.
    The point is that these amounts are different.
    -

    +

    Checkpoint A.5.4.

    -
    +
    Calculate the value of \(30-((2+3)\cdot2)\text{,}\) respecting the order that the grouping symbols are telling you to execute the arithmetic operations.
    @@ -717,10 +731,10 @@

    Search Results:

    Remark A.5.7.

    -
    There are several different ways to write multiplication. We can use the symbols \(\cdot\text{,}\) \(\times\text{,}\) and \(*\) to mean multiplication. We can also write two things right next to each other with no symbol in between them to mean multiplication. That is what is happening in Item f, where the \(4\) is written right next to the \((2)^3\) with no symbol in between.
    Using a symbol for multiplication is called “explicit multiplication” and not writing any symbol at all is called “implicit multiplication”. For this textbook, explicit and implicit multiplication have the same priority in the order of operations. However there are some conventions out in the real world where implicit multiplication has a higher priority in the order of operations than explicit multiplication. You may have seen memes with expressions like \(6\div2(3)\) that play on how the real world has more than one convention for the order of operations.

    +
    There are several different ways to write multiplication. We can use the symbols \(\cdot\text{,}\) \(\times\text{,}\) and \(*\) to mean multiplication. We can also write two things right next to each other with no symbol in between them to mean multiplication. That is what is happening in Item f, where the \(4\) is written right next to the \((2)^3\) with no symbol in between.
    Using a symbol for multiplication is called “explicit multiplication” and not writing any symbol at all is called “implicit multiplication”. For this textbook, explicit and implicit multiplication have the same priority in the order of operations. However there are some conventions out in the real world where implicit multiplication has a higher priority in the order of operations than explicit multiplication. You may have seen memes with expressions like \(6\div2(3)\) that play on how the real world has more than one convention for the order of operations.

    Checkpoint A.5.8. Practice with order of operations.

    -
    +
    Simplify this expression one step at a time, using the order of operations.

    (a)

    @@ -765,10 +779,10 @@

    Search Results:

    \end{equation*}
    -

    +

    Checkpoint A.5.9.

    -
    +
    Simplify \(24\div(15\div 3+1)+2\text{.}\) @@ -787,10 +801,10 @@

    Search Results:

    -

    +

    Checkpoint A.5.10.

    -
    +
    Simplify \((20-4^2)\div(4-6)^3\text{.}\) @@ -875,10 +889,10 @@

    Search Results:

    • -

    +

    Checkpoint A.5.12. Implied Grouping.

    -
    +
    Use the order of operations to evaluate
    @@ -902,10 +916,10 @@

    Search Results:

    -

    +

    Checkpoint A.5.13.

    -
    +
    Simplify \(\dfrac{2\abs{9-15}+1}{\sqrt{(-5)^2+12^2}}\text{.}\) @@ -958,7 +972,7 @@

    Search Results:

    Checkpoint A.5.15. Negating and Raising to Powers.

    -
    +
    Compute the following. In each part, the first expression asks you to exponentiate and then negate the result. The second expression has a negative number raised to a power.

    (a)

    @@ -989,117 +1003,117 @@

    Search Results:

    Exercise Group.

    Use the order of operations to simplify the expression.
    -
    1.
    -
    +
    1.
    +
    -
    \(7 - 2 + 3\)
    -
    Answer.
    \(8\)
    +
    \(9 - 7 + 8\)
    +
    Answer.
    \(10\)
    -
    2.
    -
    +
    2.
    +
    -
    \(8 + 6\cdot5\)
    -
    Answer.
    \(38\)
    +
    \(2 + 6\cdot5\)
    +
    Answer.
    \(32\)
    -
    3.
    -
    +
    3.
    +
    -
    \(9 - 4\cdot8\)
    -
    Answer.
    \(-23\)
    +
    \(3 - 2\cdot8\)
    +
    Answer.
    \(-13\)
    -
    4.
    -
    +
    4.
    +
    -
    \(2 + 9(5)\)
    -
    Answer.
    \(47\)
    +
    \(4 + 8(3)\)
    +
    Answer.
    \(28\)
    -
    5.
    -
    +
    5.
    +
    -
    \(3 - 7(8)\)
    -
    Answer.
    \(-53\)
    +
    \(4 - 5(7)\)
    +
    Answer.
    \(-31\)
    -
    6.
    -
    +
    6.
    +
    -
    \(4 + 3\div3\)
    +
    \(2 + 9\div3\)
    Answer.
    \(5\)
    -
    7.
    -
    +
    7.
    +
    -
    \(9 - 24\div6\)
    -
    Answer.
    \(5\)
    +
    \(8 - 8\div4\)
    +
    Answer.
    \(6\)
    -
    8.
    -
    +
    8.
    +
    -
    \(5 + 4/2\)
    +
    \(4 + 27/9\)
    Answer.
    \(7\)
    -
    9.
    -
    +
    9.
    +
    -
    \(6 - 15/5\)
    -
    Answer.
    \(3\)
    +
    \(8 - 24/4\)
    +
    Answer.
    \(2\)
    -
    10.
    -
    +
    10.
    +
    -
    \(180 \div 15/3\)
    -
    Answer.
    \(4\)
    +
    \(180 \div 12/3\)
    +
    Answer.
    \(5\)
    -
    11.
    -
    +
    11.
    +
    -
    \(4 + 3^{4}\)
    -
    Answer.
    \(85\)
    +
    \(2 + 5^{2}\)
    +
    Answer.
    \(27\)
    -
    12.
    -
    +
    12.
    +
    -
    \(7 - 2^{4}\)
    -
    Answer.
    \(-9\)
    +
    \(6 - 10^{2}\)
    +
    Answer.
    \(-94\)
    -
    13.
    -
    +
    13.
    +
    -
    \(2 \cdot 9^{2}\)
    -
    Answer.
    \(162\)
    +
    \(9 \cdot 7^{2}\)
    +
    Answer.
    \(441\)
    -
    14.
    -
    +
    14.
    +
    -
    \(96 \div 4^{2}\)
    -
    Answer.
    \(6\)
    +
    \(32 \div 2^{3}\)
    +
    Answer.
    \(4\)
    -
    15.
    -
    +
    15.
    +
    -
    \(36 / 2^{2}\)
    -
    Answer.
    \(9\)
    +
    \(216 / 3^{3}\)
    +
    Answer.
    \(8\)
    -
    16.
    -
    +
    16.
    +
    -
    \(4 \cdot 4\mathbin{^\wedge}3\)
    -
    Answer.
    \(256\)
    +
    \(3 \cdot 3\mathbin{^\wedge}3\)
    +
    Answer.
    \(81\)
    @@ -1108,187 +1122,187 @@

    Exercise Group.

    Exercise Group.

    Use the order of operations to simplify the expression that has grouping symbols.
    -
    17.
    -
    +
    17.
    +
    -
    \(5 - ( 8 + 4 )\)
    -
    Answer.
    \(-7\)
    +
    \(4 - [ 7 + 9 ]\)
    +
    Answer.
    \(-12\)
    -
    18.
    -
    +
    18.
    +
    -
    \(2 - [ 4 - 3 ]\)
    -
    Answer.
    \(1\)
    +
    \(9 - [ 2 - 6 ]\)
    +
    Answer.
    \(13\)
    -
    19.
    -
    +
    19.
    +
    -
    \(8 \cdot [ 6 + 9 ]\)
    -
    Answer.
    \(120\)
    +
    \(6 \cdot [ 5 + 3 ]\)
    +
    Answer.
    \(48\)
    -
    20.
    -
    +
    20.
    +
    -
    \([5 + 2] \cdot 7\)
    -
    Answer.
    \(49\)
    +
    \((3 + 9) \cdot 2\)
    +
    Answer.
    \(24\)
    -
    21.
    -
    +
    21.
    +
    -
    \(2 \cdot [ 6 - 4 ]\)
    -
    Answer.
    \(4\)
    +
    \(8 \cdot ( 4 - 7 )\)
    +
    Answer.
    \(-24\)
    -
    22.
    -
    +
    22.
    +
    -
    \((7 - 9) \cdot 2\)
    +
    \((6 - 7) \cdot 4\)
    Answer.
    \(-4\)
    -
    23.
    -
    +
    23.
    +
    -
    \(5 ( 3 + 8 )\)
    -
    Answer.
    \(55\)
    +
    \(3 ( 2 + 5 )\)
    +
    Answer.
    \(21\)
    -
    24.
    -
    +
    24.
    +
    -
    \((2 + 7) 5\)
    -
    Answer.
    \(45\)
    +
    \((8 + 5) 9\)
    +
    Answer.
    \(117\)
    -
    25.
    -
    +
    25.
    +
    -
    \(7 ( 2 - 4 )\)
    -
    Answer.
    \(-14\)
    +
    \(5 [ 9 - 6 ]\)
    +
    Answer.
    \(15\)
    -
    26.
    -
    +
    26.
    +
    -
    \((4 - 6) 9\)
    -
    Answer.
    \(-18\)
    +
    \([3 - 5] 6\)
    +
    Answer.
    \(-12\)
    -
    27.
    -
    +
    27.
    +
    -
    \(72 \div [ 5 + 3 ]\)
    -
    Answer.
    \(9\)
    +
    \(56 \div [ 1 + 6 ]\)
    +
    Answer.
    \(8\)
    -
    28.
    -
    +
    28.
    +
    -
    \([23 + 12] \div 5\)
    -
    Answer.
    \(7\)
    +
    \([5 + 19] \div 4\)
    +
    Answer.
    \(6\)
    -
    29.
    -
    +
    29.
    +
    -
    \(40 \div [ 9 - 1 ]\)
    -
    Answer.
    \(5\)
    +
    \(21 \div ( 10 - 3 )\)
    +
    Answer.
    \(3\)
    -
    30.
    -
    +
    30.
    +
    -
    \([40 - 4] \div 4\)
    -
    Answer.
    \(9\)
    +
    \((26 - 2) \div 3\)
    +
    Answer.
    \(8\)
    -
    31.
    -
    +
    31.
    +
    -
    \(210 \div ( 6 \cdot 7 )\)
    -
    Answer.
    \(5\)
    +
    \(270 \div ( 5 \cdot 6 )\)
    +
    Answer.
    \(9\)
    -
    32.
    -
    +
    32.
    +
    -
    \(80 \div ( 8 \div 2 )\)
    -
    Answer.
    \(20\)
    +
    \(1134 \div ( 18 \div 9 )\)
    +
    Answer.
    \(567\)
    -
    33.
    -
    +
    33.
    +
    \(( 1 + 1 ) ^{4}\)
    Answer.
    \(16\)
    -
    34.
    -
    +
    34.
    +
    -
    \(( 7 - 3 ) ^{3}\)
    +
    \([ 5 - 1 ] ^{3}\)
    Answer.
    \(64\)
    -
    35.
    -
    +
    35.
    +
    -
    \(( 3 \cdot 2 ) ^{2}\)
    -
    Answer.
    \(36\)
    +
    \([ 4 \cdot 2 ] ^{2}\)
    +
    Answer.
    \(64\)
    -
    36.
    -
    +
    36.
    +
    -
    \([ 6 \div 2 ] ^{4}\)
    -
    Answer.
    \(81\)
    +
    \([ 12 \div 4 ] ^{3}\)
    +
    Answer.
    \(27\)
    -
    37.
    -
    +
    37.
    +
    -
    \(-[ 6 + 2 ]\)
    -
    Answer.
    \(-8\)
    +
    \(-[ 4 + 9 ]\)
    +
    Answer.
    \(-13\)
    -
    38.
    -
    +
    38.
    +
    -
    \(-[ 3 - 6 ]\)
    -
    Answer.
    \(3\)
    +
    \(-( 9 - 4 )\)
    +
    Answer.
    \(-5\)
    -
    39.
    -
    +
    39.
    +
    -
    \(-[ 8 \cdot 9 ]\)
    -
    Answer.
    \(-72\)
    +
    \(-( 6 \cdot 8 )\)
    +
    Answer.
    \(-48\)
    -
    40.
    -
    +
    40.
    +
    -
    \(-( 20 \div 4 )\)
    -
    Answer.
    \(-5\)
    +
    \(-( 12 \div 3 )\)
    +
    Answer.
    \(-4\)
    -
    41.
    -
    +
    41.
    +
    -
    \(-( 9 ^{2})\)
    +
    \(-( 3 ^{4})\)
    Answer.
    \(-81\)
    -
    42.
    -
    +
    42.
    +
    -
    \(( -2 )^{4}\)
    -
    Answer.
    \(16\)
    +
    \(( -2 )^{3}\)
    +
    Answer.
    \(-8\)
    @@ -1297,193 +1311,193 @@

    Exercise Group.

    Exercise Group.

    Use the order of operations to simplify the expression that has absolute value or implied grouping.
    -
    43.
    -
    +
    43.
    +
    -
    \(4 - \abs{5 - 7}\)
    -
    Answer.
    \(2\)
    +
    \(6 - \abs{3 - 5}\)
    +
    Answer.
    \(4\)
    -
    44.
    -
    +
    44.
    +
    -
    \(5 \cdot \abs{2 - 3}\)
    -
    Answer.
    \(5\)
    +
    \(7 \cdot \abs{9 - 8}\)
    +
    Answer.
    \(7\)
    -
    45.
    -
    +
    45.
    +
    -
    \(\abs{6 - 8} \cdot 4\)
    -
    Answer.
    \(8\)
    +
    \(\abs{8 - 5} \cdot 3\)
    +
    Answer.
    \(9\)
    -
    46.
    -
    +
    46.
    +
    -
    \(7 \abs{4 - 9}\)
    -
    Answer.
    \(35\)
    +
    \(9 \abs{3 - 7}\)
    +
    Answer.
    \(36\)
    -
    47.
    -
    +
    47.
    +
    -
    \(\abs{8 - 2} 4\)
    -
    Answer.
    \(24\)
    +
    \(\abs{2 - 8} 3\)
    +
    Answer.
    \(18\)
    -
    48.
    -
    +
    48.
    +
    -
    \(63 \div \abs{0 - 7}\)
    -
    Answer.
    \(9\)
    +
    \(18 \div \abs{11 - 5}\)
    +
    Answer.
    \(3\)
    -
    49.
    -
    +
    49.
    +
    -
    \(\abs{-16 - 2} \div 6\)
    -
    Answer.
    \(3\)
    +
    \(\abs{21 - 9} \div 3\)
    +
    Answer.
    \(4\)
    -
    50.
    -
    +
    50.
    +
    -
    \(\abs{15 - 5}^{2}\)
    -
    Answer.
    \(100\)
    +
    \(\abs{0 - 4}^{3}\)
    +
    Answer.
    \(64\)
    -
    51.
    -
    +
    51.
    +
    -
    \(-\abs{4 + 7}\)
    +
    \(-\abs{5 + 6}\)
    Answer.
    \(-11\)
    -
    52.
    -
    +
    52.
    +
    -
    \(-\abs{4 - 5}\)
    -
    Answer.
    \(-1\)
    +
    \(-\abs{6 - 2}\)
    +
    Answer.
    \(-4\)
    -
    53.
    -
    +
    53.
    +
    -
    \(\sqrt{16+9}+11\)
    -
    Answer.
    \(16\)
    +
    \(\sqrt{42+7}+15\)
    +
    Answer.
    \(22\)
    -
    54.
    -
    +
    54.
    +
    -
    \(6+\sqrt{33+3}\)
    -
    Answer.
    \(12\)
    +
    \(5+\sqrt{62+2}\)
    +
    Answer.
    \(13\)
    -
    55.
    -
    +
    55.
    +
    -
    \(\sqrt{67-3}-15\)
    -
    Answer.
    \(-7\)
    +
    \(\sqrt{101-1}-19\)
    +
    Answer.
    \(-9\)
    -
    56.
    -
    +
    56.
    +
    -
    \(9-\sqrt{83-2}\)
    -
    Answer.
    \(0\)
    +
    \(7-\sqrt{10-9}\)
    +
    Answer.
    \(6\)
    -
    57.
    -
    +
    57.
    +
    -
    \(6\sqrt{95+5}\)
    -
    Answer.
    \(60\)
    +
    \(4\sqrt{-1+5}\)
    +
    Answer.
    \(8\)
    -
    58.
    -
    +
    58.
    +
    -
    \(3\sqrt{10-9}\)
    -
    Answer.
    \(3\)
    +
    \(2\sqrt{17-8}\)
    +
    Answer.
    \(6\)
    -
    59.
    -
    +
    59.
    +
    -
    \(4\sqrt{18\cdot2}\)
    -
    Answer.
    \(24\)
    +
    \(3\sqrt{12\cdot3}\)
    +
    Answer.
    \(18\)
    -
    60.
    -
    +
    60.
    +
    -
    \(5\sqrt{189\div7}\)
    -
    Answer.
    \(25.9808\)
    +
    \(9\sqrt{144\div3}\)
    +
    Answer.
    \(62.3538\)
    -
    61.
    -
    +
    61.
    +
    -
    \(\dfrac{96 \div 8}{2 \cdot 3} \cdot 4\)
    -
    Answer.
    \(8\)
    +
    \(\dfrac{149 - 5}{48 \div 4} + 8\)
    +
    Answer.
    \(20\)
    62.
    -
    -
    +
    +
    -
    \(\dfrac{192 \div 2}{18 - 6} \cdot 7\)
    -
    Answer.
    \(56\)
    +
    \(\dfrac{43 - 7}{-2 + 8} + 3\)
    +
    Answer.
    \(9\)
    63.
    -
    -
    +
    +
    -
    \(\dfrac{144 \div 2}{4 \cdot 3} \cdot 3\)
    -
    Answer.
    \(18\)
    +
    \(6 + \dfrac{34 - 2}{32 \div 4}\)
    +
    Answer.
    \(10\)
    64.
    -
    -
    +
    +
    -
    \(6 + \dfrac{2 \cdot 9}{14 - 8}\)
    -
    Answer.
    \(9\)
    +
    \(\dfrac{92 + 4}{15 - 7} - 9\)
    +
    Answer.
    \(3\)
    -
    65.
    -
    +
    65.
    +
    -
    \(9^{18 \div 9} + 7\)
    -
    Answer.
    \(88\)
    +
    \(3 - 2^{2 \cdot 2}\)
    +
    Answer.
    \(-13\)
    66.
    -
    -
    +
    +
    -
    \(10^{-4 + 5} \cdot 7\)
    -
    Answer.
    \(70\)
    +
    \(3 + 3^{-3 + 4}\)
    +
    Answer.
    \(6\)
    67.
    -
    -
    +
    +
    -
    \(2^{10 - 9} - 7\)
    -
    Answer.
    \(-5\)
    +
    \(3 + 4^{11 - 8}\)
    +
    Answer.
    \(67\)
    68.
    -
    -
    +
    +
    -
    \(3^{-2 + 5} + 7\)
    -
    Answer.
    \(34\)
    +
    \(3 \cdot 5^{8 \div 4}\)
    +
    Answer.
    \(75\)

    Challenge

    -

    69.

    -
    +

    69.

    +
    In this challenge, your job is to create expressions, using addition, subtraction, multiplication, and parentheses. You may use the numbers, \(1, 2, 3\text{,}\) and \(4\) in your expression, using each number only once. For example, you could make the expression: \(1+2 \cdot 3 - 4\text{.}\) diff --git a/section-percentages.html b/section-percentages.html index 94b567bf6..7b394d46d 100644 --- a/section-percentages.html +++ b/section-percentages.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -651,42 +665,42 @@

    Search Results:

    Checkpoint A.4.3.
    Solve each equation from Example 2.
    -
    -
    +
    +
    -

    (a)

    -
    How much is \(30\%\) of \(\$24.00\text{?}\) -
    +

    (a)

    +
    How much is \(30\%\) of \(\$24.00\text{?}\) +
    \begin{equation*} x=0.3\cdot24 \end{equation*} -
    +
    \(x\) is .
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] x\amp=0.3\cdot24\\ x\amp=8 -\end{aligned}\)

    (b)

    -
    +\end{aligned}\)

    • (b)

    +
    \(\$7.20\) is what percent of \(\$24.00\text{?}\) -
    +
    \begin{equation*} 7.2=P\cdot24 \end{equation*} -
    +
    \(P\) is .
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] 7.2\amp=P\cdot24\\ \divideunder{7.2}{24}\amp=\divideunder{P\cdot24}{24}\\ 0.3\amp=P -\end{aligned}\)

    (c)

    -
    -\(\$7.20\) is \(30\%\) of how much money?
    +\end{aligned}\)

    (c)

    +
    +\(\$7.20\) is \(30\%\) of how much money?
    \begin{equation*} 7.2=0.3\cdot x \end{equation*} -
    +
    \(x\) is .
    -
    Explanation.
    \(\begin{aligned}[t] +
    Explanation.
    \(\begin{aligned}[t] 7.2\amp=0.3\cdot x\\ \divideunder{7.2}{0.3}\amp=\divideunder{0.3\cdot x}{0.3}\\ 24\amp=x @@ -770,12 +784,12 @@

    Search Results:

    Checkpoint A.4.10.

    -
    -
    +
    +
    -
    Alexis sells cars for a living, and earns \(28\%\) of the dealership’s sales profit as commission. In a certain month, she plans to earn \(\$2200\) in commissions. How much total sales profit does she need to bring in for the dealership?
    Alexis needs to bring in in sales profit.
    -
    Explanation.
    -
    Be careful that you do not calculate \(28\%\) of \(\$2200\text{.}\) That might be what a student would do who doesn’t thoroughly read the question. If you have ever trained yourself to quickly find numbers in word problems and substitute them into formulas, you must unlearn this. The issue is that \(\$2200\) is not the dealership’s sales profit, and if you mistakenly multiply \(0.28\cdot2200=616\text{,}\) then \(\$616\) makes no sense as an answer to this question. How could Alexis bring in only \(\$616\) of sales profit, and earn \(\$2200\) in commission?
    We can translate the problem into “\(\$2200\) is \(28\%\) of what?” Letting \(x\) be the sales profit for the dealership (in dollars), we can write and solve the equation:
    +
    Alexis sells cars for a living, and earns \(28\%\) of the dealership’s sales profit as commission. In a certain month, she plans to earn \(\$2200\) in commissions. How much total sales profit does she need to bring in for the dealership?
    Alexis needs to bring in in sales profit.
    +
    Explanation.
    +
    Be careful that you do not calculate \(28\%\) of \(\$2200\text{.}\) That might be what a student would do who doesn’t thoroughly read the question. If you have ever trained yourself to quickly find numbers in word problems and substitute them into formulas, you must unlearn this. The issue is that \(\$2200\) is not the dealership’s sales profit, and if you mistakenly multiply \(0.28\cdot2200=616\text{,}\) then \(\$616\) makes no sense as an answer to this question. How could Alexis bring in only \(\$616\) of sales profit, and earn \(\$2200\) in commission?
    We can translate the problem into “\(\$2200\) is \(28\%\) of what?” Letting \(x\) be the sales profit for the dealership (in dollars), we can write and solve the equation:
    \begin{equation*} \begin{aligned} \pinover{2200}{\$2200}\amp\pinover{=}{is}\pinover{0.28}{28\%}\pinover{\cdot}{of}\pinover{x}{what}\\ @@ -784,7 +798,7 @@

    Search Results:

    x\amp\approx7857.14 \end{aligned} \end{equation*} -
    To earn \(\$2200\) in commission, Alexis needs to bring in approximately \(\$7857.14\) of sales profit for the dealership.
    +
    To earn \(\$2200\) in commission, Alexis needs to bring in approximately \(\$7857.14\) of sales profit for the dealership.

    @@ -810,19 +824,19 @@

    Search Results:

    Checkpoint A.4.13.

    -
    -
    +
    +
    -
    Last month, a full tank of gas for a car you drive cost you \(\$40.00\text{.}\) You hear on the news that gas prices have risen by \(12\%\text{.}\) By how much, in dollars, has the cost of a full tank gone up?
    A full tank of gas now costs more than it did last month.
    -
    Explanation.
    -
    Let \(x\) represent the amount of increase. We can set up and solve the equation:
    +
    Last month, a full tank of gas for a car you drive cost you \(\$40.00\text{.}\) You hear on the news that gas prices have risen by \(12\%\text{.}\) By how much, in dollars, has the cost of a full tank gone up?
    A full tank of gas now costs more than it did last month.
    +
    Explanation.
    +
    Let \(x\) represent the amount of increase. We can set up and solve the equation:
    \begin{equation*} \begin{aligned} \pinover{0.12}{12\%}\pinover{\cdot}{of}\pinover{40}{old\ cost}\amp\pinover{=}{is}\pinover{x}{how\ much}\\ 4.8\amp=x \end{aligned} \end{equation*} -
    A full tank now costs \(\$4.80\) more than it did last month.
    +
    A full tank now costs \(\$4.80\) more than it did last month.

    @@ -847,122 +861,122 @@

    Search Results:

    Review and Warmup.

    1.

    -
    -
    +
    +
    -
    Write the following percentages as decimals.
      -
    1. -\(\displaystyle{ 19\% = }\) +
      Write the following percentages as decimals.
        +
      1. +\(\displaystyle{ 12\% = }\)
        2. -
      2. -\(\displaystyle{ 55\% = }\) +
      3. +\(\displaystyle{ 53\% = }\)
      -
      Answer 1.
      \(0.19\)
      Answer 2.
      \(0.55\)
      Explanation.
      -
      To change a percentage to a decimal, we move the decimal point to the left twice.
      \(\displaystyle{ 19\% = {0.19} }\)
      \(\displaystyle{ 55\% = {0.55} }\)
      +
      Answer 1.
      \(0.12\)
      Answer 2.
      \(0.53\)
      Explanation.
      +
      To change a percentage to a decimal, we move the decimal point to the left twice.
      \(\displaystyle{ 12\% = {0.12} }\)
      \(\displaystyle{ 53\% = {0.53} }\)

    2.

    -
    -
    +
    +
    -
    Write the following percentages as decimals.
      -
    1. -\(\displaystyle{ 11\% = }\) +
      Write the following percentages as decimals.
        +
      1. +\(\displaystyle{ 13\% = }\)
        2. -
      2. -\(\displaystyle{ 52\% = }\) +
      3. +\(\displaystyle{ 59\% = }\)
      -
      Answer 1.
      \(0.11\)
      Answer 2.
      \(0.52\)
      Explanation.
      -
      To change a percentage to a decimal, we move the decimal point to the left twice.
      \(\displaystyle{ 11\% = {0.11} }\)
      \(\displaystyle{ 52\% = {0.52} }\)
      +
      Answer 1.
      \(0.13\)
      Answer 2.
      \(0.59\)
      Explanation.
      +
      To change a percentage to a decimal, we move the decimal point to the left twice.
      \(\displaystyle{ 13\% = {0.13} }\)
      \(\displaystyle{ 59\% = {0.59} }\)

    3.

    -
    -
    +
    +
    -
    Write the following decimals as percentages.
      -
    1. -\(\displaystyle{ 0.22 = }\) +
      Write the following decimals as percentages.
        +
      1. +\(\displaystyle{ 0.24 = }\)
        2. -
      2. -\(\displaystyle{ 0.68 = }\) +
      3. +\(\displaystyle{ 0.66 = }\)
      -
      Answer 1.
      \(22.0\%\)
      Answer 2.
      \(68.0\%\)
      Explanation.
      -
      To change a decimal to a percentage, we move the decimal point to the right twice.
      \(0.22 = 22\%\)
      \(0.68 = 68\%\)
      +
      Answer 1.
      \(24.0\%\)
      Answer 2.
      \(66.0\%\)
      Explanation.
      +
      To change a decimal to a percentage, we move the decimal point to the right twice.
      \(0.24 = 24\%\)
      \(0.66 = 66\%\)

    4.

    -
    -
    +
    +
    -
    Write the following decimals as percentages.
      -
    1. -\(\displaystyle{ 0.23 = }\) +
      Write the following decimals as percentages.
        +
      1. +\(\displaystyle{ 0.25 = }\)
        2. -
      2. -\(\displaystyle{ 0.65 = }\) +
      3. +\(\displaystyle{ 0.63 = }\)
      -
      Answer 1.
      \(23.0\%\)
      Answer 2.
      \(65.0\%\)
      Explanation.
      -
      To change a decimal to a percentage, we move the decimal point to the right twice.
      \(0.23 = 23\%\)
      \(0.65 = 65\%\)
      +
      Answer 1.
      \(25.0\%\)
      Answer 2.
      \(63.0\%\)
      Explanation.
      +
      To change a decimal to a percentage, we move the decimal point to the right twice.
      \(0.25 = 25\%\)
      \(0.63 = 63\%\)

    5.

    -
    -
    +
    +
    -
    Write the following percentages as decimals.
      -
    1. -\(\displaystyle{ 5\% = }\) +
      Write the following percentages as decimals.
        +
      1. +\(\displaystyle{ 7\% = }\)
        2. -
      2. -\(\displaystyle{ 50\% = }\) +
      3. +\(\displaystyle{ 70\% = }\)
        4. -
      4. +
      5. \(\displaystyle{ 100\% = }\)
        6. -
      6. -\(\displaystyle{ 500\% = }\) +
      7. +\(\displaystyle{ 700\% = }\)
      -
      Answer 1.
      \(0.05\)
      Answer 2.
      \(0.5\)
      Answer 3.
      \(1\)
      Answer 4.
      \(5\)
      Explanation.
      -
      To change a percentage to a decimal, we move the decimal to the left twice.
      \(\displaystyle{ 5\% = {0.05} }\)
      \(\displaystyle{ 50\% = {0.5} }\)
      \(\displaystyle{ 100\% = 1 }\)
      \(\displaystyle{ 500\% = {5} }\)
      +
      Answer 1.
      \(0.07\)
      Answer 2.
      \(0.7\)
      Answer 3.
      \(1\)
      Answer 4.
      \(7\)
      Explanation.
      +
      To change a percentage to a decimal, we move the decimal to the left twice.
      \(\displaystyle{ 7\% = {0.07} }\)
      \(\displaystyle{ 70\% = {0.7} }\)
      \(\displaystyle{ 100\% = 1 }\)
      \(\displaystyle{ 700\% = {7} }\)

    6.

    -
    -
    +
    +
    -
    Write the following percentages as decimals.
      -
    1. -\(\displaystyle{ 6\% = }\) +
      Write the following percentages as decimals.
        +
      1. +\(\displaystyle{ 7\% = }\)
        2. -
      2. -\(\displaystyle{ 60\% = }\) +
      3. +\(\displaystyle{ 70\% = }\)
        4. -
      4. +
      5. \(\displaystyle{ 100\% = }\)
        6. -
      6. -\(\displaystyle{ 600\% = }\) +
      7. +\(\displaystyle{ 700\% = }\)
      -
      Answer 1.
      \(0.06\)
      Answer 2.
      \(0.6\)
      Answer 3.
      \(1\)
      Answer 4.
      \(6\)
      Explanation.
      -
      To change a percentage to a decimal, we move the decimal to the left twice.
      \(\displaystyle{ 6\% = {0.06} }\)
      \(\displaystyle{ 60\% = {0.6} }\)
      \(\displaystyle{ 100\% = 1 }\)
      \(\displaystyle{ 600\% = {6} }\)
      +
      Answer 1.
      \(0.07\)
      Answer 2.
      \(0.7\)
      Answer 3.
      \(1\)
      Answer 4.
      \(7\)
      Explanation.
      +
      To change a percentage to a decimal, we move the decimal to the left twice.
      \(\displaystyle{ 7\% = {0.07} }\)
      \(\displaystyle{ 70\% = {0.7} }\)
      \(\displaystyle{ 100\% = 1 }\)
      \(\displaystyle{ 700\% = {7} }\)
      @@ -973,134 +987,134 @@

      Review and Warmup.

      Exercise Group.

      7.

      -
      -
      +
      +
      -
      Write the following decimals as percentages.
        -
      1. -\(\displaystyle{ 0.07 = }\) +
        Write the following decimals as percentages.
          +
        1. +\(\displaystyle{ 0.08 = }\)
          2. -
        2. -\(\displaystyle{ 0.7 = }\) +
        3. +\(\displaystyle{ 0.8 = }\)
          4. -
        4. -\(\displaystyle{ 7 = }\) +
        5. +\(\displaystyle{ 8 = }\)
          6. -
        6. +
        7. \(\displaystyle{ 1 = }\)
        -
        Answer 1.
        \(7.0\%\)
        Answer 2.
        \(70.0\%\)
        Answer 3.
        \(700.0\%\)
        Answer 4.
        \(100.0\%\)
        Explanation.
        -
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 0.07 = 7\% }\)
        \(\displaystyle{ 0.7 = 70\% }\)
        \(\displaystyle{ 7 = 700\% }\)
        \(\displaystyle{ 1 = 100\% }\)
        +
        Answer 1.
        \(8.0\%\)
        Answer 2.
        \(80.0\%\)
        Answer 3.
        \(800.0\%\)
        Answer 4.
        \(100.0\%\)
        Explanation.
        +
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 0.08 = 8\% }\)
        \(\displaystyle{ 0.8 = 80\% }\)
        \(\displaystyle{ 8 = 800\% }\)
        \(\displaystyle{ 1 = 100\% }\)

      8.

      -
      -
      +
      +
      -
      Write the following decimals as percentages.
        -
      1. -\(\displaystyle{ 0.08 = }\) +
        Write the following decimals as percentages.
          +
        1. +\(\displaystyle{ 0.09 = }\)
          2. -
        2. -\(\displaystyle{ 0.8 = }\) +
        3. +\(\displaystyle{ 0.9 = }\)
          4. -
        4. -\(\displaystyle{ 8 = }\) +
        5. +\(\displaystyle{ 9 = }\)
          6. -
        6. +
        7. \(\displaystyle{ 1 = }\)
        -
        Answer 1.
        \(8.0\%\)
        Answer 2.
        \(80.0\%\)
        Answer 3.
        \(800.0\%\)
        Answer 4.
        \(100.0\%\)
        Explanation.
        -
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 0.08 = 8\% }\)
        \(\displaystyle{ 0.8 = 80\% }\)
        \(\displaystyle{ 8 = 800\% }\)
        \(\displaystyle{ 1 = 100\% }\)
        +
        Answer 1.
        \(9.0\%\)
        Answer 2.
        \(90.0\%\)
        Answer 3.
        \(900.0\%\)
        Answer 4.
        \(100.0\%\)
        Explanation.
        +
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 0.09 = 9\% }\)
        \(\displaystyle{ 0.9 = 90\% }\)
        \(\displaystyle{ 9 = 900\% }\)
        \(\displaystyle{ 1 = 100\% }\)

      9.

      -
      -
      +
      +
      -
      Write the following decimals as percentages.
        -
      1. -\(\displaystyle{ 8.75 = }\) +
        Write the following decimals as percentages.
          +
        1. +\(\displaystyle{ 1.54 = }\)
          2. -
        2. -\(\displaystyle{ 0.875 = }\) +
        3. +\(\displaystyle{ 0.154 = }\)
          4. -
        4. -\(\displaystyle{ 0.0875 = }\) +
        5. +\(\displaystyle{ 0.0154 = }\)
        -
        Answer 1.
        \(875.00\%\)
        Answer 2.
        \(87.50\%\)
        Answer 3.
        \(8.75\%\)
        Explanation.
        -
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 8.75 = 875\% }\)
        \(\displaystyle{ 0.875 = 87.5\% }\)
        \(\displaystyle{ 0.0875 = 8.75\% }\)
        +
        Answer 1.
        \(154.00\%\)
        Answer 2.
        \(15.40\%\)
        Answer 3.
        \(1.54\%\)
        Explanation.
        +
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 1.54 = 154\% }\)
        \(\displaystyle{ 0.154 = 15.4\% }\)
        \(\displaystyle{ 0.0154 = 1.54\% }\)

      10.

      -
      -
      +
      +
      -
      Write the following decimals as percentages.
        -
      1. -\(\displaystyle{ 9.39 = }\) +
        Write the following decimals as percentages.
          +
        1. +\(\displaystyle{ 2.17 = }\)
          2. -
        2. -\(\displaystyle{ 0.939 = }\) +
        3. +\(\displaystyle{ 0.217 = }\)
          4. -
        4. -\(\displaystyle{ 0.0939 = }\) +
        5. +\(\displaystyle{ 0.0217 = }\)
        -
        Answer 1.
        \(939.00\%\)
        Answer 2.
        \(93.90\%\)
        Answer 3.
        \(9.39\%\)
        Explanation.
        -
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 9.39 = 939\% }\)
        \(\displaystyle{ 0.939 = 93.9\% }\)
        \(\displaystyle{ 0.0939 = 9.39\% }\)
        +
        Answer 1.
        \(217.00\%\)
        Answer 2.
        \(21.70\%\)
        Answer 3.
        \(2.17\%\)
        Explanation.
        +
        To change a decimal to a percentage, we move the decimal point to the right twice.
        \(\displaystyle{ 2.17 = 217\% }\)
        \(\displaystyle{ 0.217 = 21.7\% }\)
        \(\displaystyle{ 0.0217 = 2.17\% }\)

      11.

      -
      -
      +
      +
      -
      Write the following percentages as decimals.
        -
      1. -\(\displaystyle{ 104\% = }\) +
        Write the following percentages as decimals.
          +
        1. +\(\displaystyle{ 382\% = }\)
          2. -
        2. -\(\displaystyle{ 10.4\% = }\) +
        3. +\(\displaystyle{ 38.2\% = }\)
          4. -
        4. -\(\displaystyle{ 1.04\% = }\) +
        5. +\(\displaystyle{ 3.82\% = }\)
        -
        Answer 1.
        \(1.04\)
        Answer 2.
        \(0.104\)
        Answer 3.
        \(0.0104\)
        Explanation.
        -
        To change a percentage to a decimal, we move the decimal to the left twice.
        \(\displaystyle{ 104\% = {1.04} }\)
        \(\displaystyle{ 10.4\% = {0.104} }\)
        \(\displaystyle{ 1.04\% = {0.0104} }\)
        +
        Answer 1.
        \(3.82\)
        Answer 2.
        \(0.382\)
        Answer 3.
        \(0.0382\)
        Explanation.
        +
        To change a percentage to a decimal, we move the decimal to the left twice.
        \(\displaystyle{ 382\% = {3.82} }\)
        \(\displaystyle{ 38.2\% = {0.382} }\)
        \(\displaystyle{ 3.82\% = {0.0382} }\)

      12.

      -
      -
      +
      +
      -
      Write the following percentages as decimals.
        -
      1. -\(\displaystyle{ 267\% = }\) +
        Write the following percentages as decimals.
          +
        1. +\(\displaystyle{ 446\% = }\)
          2. -
        2. -\(\displaystyle{ 26.7\% = }\) +
        3. +\(\displaystyle{ 44.6\% = }\)
          4. -
        4. -\(\displaystyle{ 2.67\% = }\) +
        5. +\(\displaystyle{ 4.46\% = }\)
        -
        Answer 1.
        \(2.67\)
        Answer 2.
        \(0.267\)
        Answer 3.
        \(0.0267\)
        Explanation.
        -
        To change a percentage to a decimal, we move the decimal to the left twice.
        \(\displaystyle{ 267\% = {2.67} }\)
        \(\displaystyle{ 26.7\% = {0.267} }\)
        \(\displaystyle{ 2.67\% = {0.0267} }\)
        +
        Answer 1.
        \(4.46\)
        Answer 2.
        \(0.446\)
        Answer 3.
        \(0.0446\)
        Explanation.
        +
        To change a percentage to a decimal, we move the decimal to the left twice.
        \(\displaystyle{ 446\% = {4.46} }\)
        \(\displaystyle{ 44.6\% = {0.446} }\)
        \(\displaystyle{ 4.46\% = {0.0446} }\)
        @@ -1111,463 +1125,463 @@

        Exercise Group.

        Basic Percentage Calculation.

        13.

        -
        -
        -
        -
        -\(4\%\) of \(380\) is .
        -
        -
        Answer.
        \(15.2\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(4\%\) of \(380\) is \(x\text{,}\) so “\(4\) out of \(100\)” corresponds to “\(x\) out of \(380\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{4}{100} \amp = \frac{x}{380} \\ -100x \amp = 4 \cdot 380 \\ -100x \amp = 1520 \\ -\frac{100x}{100} \amp = \frac{1520}{100} \\ -x \amp = 15.2 -\end{aligned} -}\)
        -\(4\%\) of \(380\) is \(15.2\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(4\%\) of \(380\text{?}\) Assume \(x\) is \(4\%\) of \(380\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        +\(2\%\) of \(560\) is .
        +
        +
        Answer.
        \(11.2\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(2\%\) of \(560\) is \(x\text{,}\) so “\(2\) out of \(100\)” corresponds to “\(x\) out of \(560\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{2}{100} \amp = \frac{x}{560} \\ +100x \amp = 2 \cdot 560 \\ +100x \amp = 1120 \\ +\frac{100x}{100} \amp = \frac{1120}{100} \\ +x \amp = 11.2 +\end{aligned} +}\)
        +\(2\%\) of \(560\) is \(11.2\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(2\%\) of \(560\text{?}\) Assume \(x\) is \(2\%\) of \(560\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.04 \cdot 380 \\ -\amp = 15.2 -\end{aligned} -}\)
        -\(4\%\) of \(380\) is \(15.2\text{.}\) -
        Method 3
        In the sentence “\(4\%\) of \(380\) is what,”
          -
        • “what” is the percentage,
          • -
        • -\(4\%\) is the rate,
          • -
        • -\(380\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 4\% \cdot 380 = 0.04 \cdot 380 = 15.2 }\)
        -\(4\%\) of \(380\) is \(15.2\text{.}\) +x \amp = 0.02 \cdot 560 \\ +\amp = 11.2 +\end{aligned} +}\)
        +\(2\%\) of \(560\) is \(11.2\text{.}\) +
        Method 3
        In the sentence “\(2\%\) of \(560\) is what,”
          +
        • “what” is the percentage,
          • +
        • +\(2\%\) is the rate,
          • +
        • +\(560\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 2\% \cdot 560 = 0.02 \cdot 560 = 11.2 }\)
        +\(2\%\) of \(560\) is \(11.2\text{.}\)

        14.

        -
        -
        -
        -
        -\(9\%\) of \(470\) is .
        -
        -
        Answer.
        \(42.3\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(9\%\) of \(470\) is \(x\text{,}\) so “\(9\) out of \(100\)” corresponds to “\(x\) out of \(470\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{9}{100} \amp = \frac{x}{470} \\ -100x \amp = 9 \cdot 470 \\ -100x \amp = 4230 \\ -\frac{100x}{100} \amp = \frac{4230}{100} \\ -x \amp = 42.3 -\end{aligned} -}\)
        -\(9\%\) of \(470\) is \(42.3\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(9\%\) of \(470\text{?}\) Assume \(x\) is \(9\%\) of \(470\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        +\(8\%\) of \(660\) is .
        +
        +
        Answer.
        \(52.8\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(8\%\) of \(660\) is \(x\text{,}\) so “\(8\) out of \(100\)” corresponds to “\(x\) out of \(660\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{8}{100} \amp = \frac{x}{660} \\ +100x \amp = 8 \cdot 660 \\ +100x \amp = 5280 \\ +\frac{100x}{100} \amp = \frac{5280}{100} \\ +x \amp = 52.8 +\end{aligned} +}\)
        +\(8\%\) of \(660\) is \(52.8\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(8\%\) of \(660\text{?}\) Assume \(x\) is \(8\%\) of \(660\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.09 \cdot 470 \\ -\amp = 42.3 -\end{aligned} -}\)
        -\(9\%\) of \(470\) is \(42.3\text{.}\) -
        Method 3
        In the sentence “\(9\%\) of \(470\) is what,”
          -
        • “what” is the percentage,
          • -
        • -\(9\%\) is the rate,
          • -
        • -\(470\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 9\% \cdot 470 = 0.09 \cdot 470 = 42.3 }\)
        -\(9\%\) of \(470\) is \(42.3\text{.}\) +x \amp = 0.08 \cdot 660 \\ +\amp = 52.8 +\end{aligned} +}\)
        +\(8\%\) of \(660\) is \(52.8\text{.}\) +
        Method 3
        In the sentence “\(8\%\) of \(660\) is what,”
          +
        • “what” is the percentage,
          • +
        • +\(8\%\) is the rate,
          • +
        • +\(660\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 8\% \cdot 660 = 0.08 \cdot 660 = 52.8 }\)
        +\(8\%\) of \(660\) is \(52.8\text{.}\)

        15.

        -
        -
        -
        -
        -\(70\%\) of \(570\) is .
        -
        -
        Answer.
        \(399\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(70\%\) of \(570\) is \(x\text{,}\) so “\(70\) out of \(100\)” corresponds to “\(x\) out of \(570\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{70}{100} \amp = \frac{x}{570} \\ -100x \amp = 70 \cdot 570 \\ -100x \amp = 39900 \\ -\frac{100x}{100} \amp = \frac{39900}{100} \\ -x \amp = 399 -\end{aligned} -}\)
        -\(70\%\) of \(570\) is \(399\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(70\%\) of \(570\text{?}\) Assume \(x\) is \(70\%\) of \(570\text{.}\) We have:
        \(\displaystyle{ -\begin{aligned} -x \amp = 0.7 \cdot 570 \\ -\amp = 399 +
        +
        +
        +
        +\(50\%\) of \(760\) is .
        +
        +
        Answer.
        \(380\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(50\%\) of \(760\) is \(x\text{,}\) so “\(50\) out of \(100\)” corresponds to “\(x\) out of \(760\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{50}{100} \amp = \frac{x}{760} \\ +100x \amp = 50 \cdot 760 \\ +100x \amp = 38000 \\ +\frac{100x}{100} \amp = \frac{38000}{100} \\ +x \amp = 380 \end{aligned} -}\)
        -\(70\%\) of \(570\) is \(399\text{.}\) -
        Method 3
        In the sentence “\(70\%\) of \(570\) is what,”
          -
        • “what” is the percentage,
          • -
        • -\(70\%\) is the rate,
          • -
        • -\(570\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 70\% \cdot 570 = 0.7 \cdot 570 = 399 }\)
        -\(70\%\) of \(570\) is \(399\text{.}\) +}\)
        +\(50\%\) of \(760\) is \(380\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(50\%\) of \(760\text{?}\) Assume \(x\) is \(50\%\) of \(760\text{.}\) We have:
        \(\displaystyle{ +\begin{aligned} +x \amp = 0.5 \cdot 760 \\ +\amp = 380 +\end{aligned} +}\)
        +\(50\%\) of \(760\) is \(380\text{.}\) +
        Method 3
        In the sentence “\(50\%\) of \(760\) is what,”
          +
        • “what” is the percentage,
          • +
        • +\(50\%\) is the rate,
          • +
        • +\(760\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 50\% \cdot 760 = 0.5 \cdot 760 = 380 }\)
        +\(50\%\) of \(760\) is \(380\text{.}\)

        16.

        -
        -
        -
        -
        -\(40\%\) of \(670\) is .
        -
        -
        Answer.
        \(268\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(40\%\) of \(670\) is \(x\text{,}\) so “\(40\) out of \(100\)” corresponds to “\(x\) out of \(670\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{40}{100} \amp = \frac{x}{670} \\ -100x \amp = 40 \cdot 670 \\ -100x \amp = 26800 \\ -\frac{100x}{100} \amp = \frac{26800}{100} \\ -x \amp = 268 -\end{aligned} -}\)
        -\(40\%\) of \(670\) is \(268\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(40\%\) of \(670\text{?}\) Assume \(x\) is \(40\%\) of \(670\text{.}\) We have:
        \(\displaystyle{ -\begin{aligned} -x \amp = 0.4 \cdot 670 \\ -\amp = 268 +
        +
        +
        +
        +\(20\%\) of \(860\) is .
        +
        +
        Answer.
        \(172\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(20\%\) of \(860\) is \(x\text{,}\) so “\(20\) out of \(100\)” corresponds to “\(x\) out of \(860\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{20}{100} \amp = \frac{x}{860} \\ +100x \amp = 20 \cdot 860 \\ +100x \amp = 17200 \\ +\frac{100x}{100} \amp = \frac{17200}{100} \\ +x \amp = 172 \end{aligned} -}\)
        -\(40\%\) of \(670\) is \(268\text{.}\) -
        Method 3
        In the sentence “\(40\%\) of \(670\) is what,”
          -
        • “what” is the percentage,
          • -
        • -\(40\%\) is the rate,
          • -
        • -\(670\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 40\% \cdot 670 = 0.4 \cdot 670 = 268 }\)
        -\(40\%\) of \(670\) is \(268\text{.}\) +}\)
        +\(20\%\) of \(860\) is \(172\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(20\%\) of \(860\text{?}\) Assume \(x\) is \(20\%\) of \(860\text{.}\) We have:
        \(\displaystyle{ +\begin{aligned} +x \amp = 0.2 \cdot 860 \\ +\amp = 172 +\end{aligned} +}\)
        +\(20\%\) of \(860\) is \(172\text{.}\) +
        Method 3
        In the sentence “\(20\%\) of \(860\) is what,”
          +
        • “what” is the percentage,
          • +
        • +\(20\%\) is the rate,
          • +
        • +\(860\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 20\% \cdot 860 = 0.2 \cdot 860 = 172 }\)
        +\(20\%\) of \(860\) is \(172\text{.}\)

        17.

        -
        -
        -
        -
        -\(860\%\) of \(770\) is .
        -
        -
        Answer.
        \(6622\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(860\%\) of \(770\) is \(x\text{,}\) so “\(860\) out of \(100\)” corresponds to “\(x\) out of \(770\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{860}{100} \amp = \frac{x}{770} \\ -100x \amp = 860 \cdot 770 \\ -100x \amp = 662200 \\ -\frac{100x}{100} \amp = \frac{662200}{100} \\ -x \amp = 6622 -\end{aligned} -}\)
        -\(860\%\) of \(770\) is \(6622\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(860\%\) of \(770\text{?}\) Assume \(x\) is \(860\%\) of \(770\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        +\(710\%\) of \(960\) is .
        +
        +
        Answer.
        \(6816\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(710\%\) of \(960\) is \(x\text{,}\) so “\(710\) out of \(100\)” corresponds to “\(x\) out of \(960\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{710}{100} \amp = \frac{x}{960} \\ +100x \amp = 710 \cdot 960 \\ +100x \amp = 681600 \\ +\frac{100x}{100} \amp = \frac{681600}{100} \\ +x \amp = 6816 +\end{aligned} +}\)
        +\(710\%\) of \(960\) is \(6816\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(710\%\) of \(960\text{?}\) Assume \(x\) is \(710\%\) of \(960\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 8.6 \cdot 770 \\ -\amp = 6622 +x \amp = 7.1 \cdot 960 \\ +\amp = 6816 \end{aligned} -}\)
        -\(860\%\) of \(770\) is \(6622\text{.}\) -
        Method 3
        In the sentence “\(860\%\) of \(770\) is what,”
          -
        • “what” is the percentage,
          • -
        • -\(860\%\) is the rate,
          • -
        • -\(770\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 860\% \cdot 770 = 8.6 \cdot 770 = 6622 }\)
        -\(860\%\) of \(770\) is \(6622\text{.}\) +}\)
        +\(710\%\) of \(960\) is \(6816\text{.}\) +
        Method 3
        In the sentence “\(710\%\) of \(960\) is what,”
          +
        • “what” is the percentage,
          • +
        • +\(710\%\) is the rate,
          • +
        • +\(960\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 710\% \cdot 960 = 7.1 \cdot 960 = 6816 }\)
        +\(710\%\) of \(960\) is \(6816\text{.}\)

        18.

        -
        -
        -
        -
        -\(610\%\) of \(870\) is .
        -
        -
        Answer.
        \(5307\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(610\%\) of \(870\) is \(x\text{,}\) so “\(610\) out of \(100\)” corresponds to “\(x\) out of \(870\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{610}{100} \amp = \frac{x}{870} \\ -100x \amp = 610 \cdot 870 \\ -100x \amp = 530700 \\ -\frac{100x}{100} \amp = \frac{530700}{100} \\ -x \amp = 5307 -\end{aligned} -}\)
        -\(610\%\) of \(870\) is \(5307\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(610\%\) of \(870\text{?}\) Assume \(x\) is \(610\%\) of \(870\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        +\(460\%\) of \(160\) is .
        +
        +
        Answer.
        \(736\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(460\%\) of \(160\) is \(x\text{,}\) so “\(460\) out of \(100\)” corresponds to “\(x\) out of \(160\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{460}{100} \amp = \frac{x}{160} \\ +100x \amp = 460 \cdot 160 \\ +100x \amp = 73600 \\ +\frac{100x}{100} \amp = \frac{73600}{100} \\ +x \amp = 736 +\end{aligned} +}\)
        +\(460\%\) of \(160\) is \(736\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(460\%\) of \(160\text{?}\) Assume \(x\) is \(460\%\) of \(160\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 6.1 \cdot 870 \\ -\amp = 5307 +x \amp = 4.6 \cdot 160 \\ +\amp = 736 \end{aligned} -}\)
        -\(610\%\) of \(870\) is \(5307\text{.}\) -
        Method 3
        In the sentence “\(610\%\) of \(870\) is what,”
          -
        • “what” is the percentage,
          • -
        • -\(610\%\) is the rate,
          • -
        • -\(870\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 610\% \cdot 870 = 6.1 \cdot 870 = 5307 }\)
        -\(610\%\) of \(870\) is \(5307\text{.}\) +}\)
        +\(460\%\) of \(160\) is \(736\text{.}\) +
        Method 3
        In the sentence “\(460\%\) of \(160\) is what,”
          +
        • “what” is the percentage,
          • +
        • +\(460\%\) is the rate,
          • +
        • +\(160\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 460\% \cdot 160 = 4.6 \cdot 160 = 736 }\)
        +\(460\%\) of \(160\) is \(736\text{.}\)

        19.

        -
        -
        +
        +
        -
        -\(31\%\) of is \(300.7\text{.}\) +
        +\(13\%\) of is \(33.8\text{.}\)
        -
        Answer.
        \(970\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(31\%\) of \(x\) is \(300.7\text{,}\) so “\(31\) out of \(100\)” corresponds to “\(300.7\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{31}{100} \amp = \frac{300.7}{x} \\ -31x \amp = 100 \cdot 300.7 \\ -31x \amp = 30070 \\ -\frac{31x}{31} \amp = \frac{30070}{31} \\ -x \amp = 970 +
        Answer.
        \(260\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(13\%\) of \(x\) is \(33.8\text{,}\) so “\(13\) out of \(100\)” corresponds to “\(33.8\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{13}{100} \amp = \frac{33.8}{x} \\ +13x \amp = 100 \cdot 33.8 \\ +13x \amp = 3380 \\ +\frac{13x}{13} \amp = \frac{3380}{13} \\ +x \amp = 260 \end{aligned} -}\)
        -\(31\%\) of \(970\) is \(300.7\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(300.7\) is \(31\%\) of what? Assume \(300.7\) is \(31\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        +\(13\%\) of \(260\) is \(33.8\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(33.8\) is \(13\%\) of what? Assume \(33.8\) is \(13\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -300.7 \amp = 0.31 \cdot x \\ -\frac{300.7}{0.31} \amp = \frac{0.31x}{0.31} \\ -970 \amp = x -\end{aligned} -}\)
        -\(31\%\) of \(970\) is \(300.7\text{.}\) -
        Method 3
        In the sentence “\(31\%\) of what is \(300.7\text{,}\)”
          -
        • -\(300.7\) is the percentage,
          • -
        • -\(31\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{300.7}{31\%} = \frac{300.7}{0.31} = 970 }\)
        -\(31\%\) of \(970\) is \(300.7\text{.}\) +33.8 \amp = 0.13 \cdot x \\ +\frac{33.8}{0.13} \amp = \frac{0.13x}{0.13} \\ +260 \amp = x +\end{aligned} +}\)
        +\(13\%\) of \(260\) is \(33.8\text{.}\) +
        Method 3
        In the sentence “\(13\%\) of what is \(33.8\text{,}\)”
          +
        • +\(33.8\) is the percentage,
          • +
        • +\(13\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{33.8}{13\%} = \frac{33.8}{0.13} = 260 }\)
        +\(13\%\) of \(260\) is \(33.8\text{.}\)

        20.

        -
        -
        +
        +
        -
        -\(90\%\) of is \(162\text{.}\) +
        +\(71\%\) of is \(255.6\text{.}\)
        -
        Answer.
        \(180\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(90\%\) of \(x\) is \(162\text{,}\) so “\(90\) out of \(100\)” corresponds to “\(162\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{90}{100} \amp = \frac{162}{x} \\ -90x \amp = 100 \cdot 162 \\ -90x \amp = 16200 \\ -\frac{90x}{90} \amp = \frac{16200}{90} \\ -x \amp = 180 +
        Answer.
        \(360\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(71\%\) of \(x\) is \(255.6\text{,}\) so “\(71\) out of \(100\)” corresponds to “\(255.6\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{71}{100} \amp = \frac{255.6}{x} \\ +71x \amp = 100 \cdot 255.6 \\ +71x \amp = 25560 \\ +\frac{71x}{71} \amp = \frac{25560}{71} \\ +x \amp = 360 \end{aligned} -}\)
        -\(90\%\) of \(180\) is \(162\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(162\) is \(90\%\) of what? Assume \(162\) is \(90\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        +\(71\%\) of \(360\) is \(255.6\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(255.6\) is \(71\%\) of what? Assume \(255.6\) is \(71\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -162 \amp = 0.9 \cdot x \\ -\frac{162}{0.9} \amp = \frac{0.9x}{0.9} \\ -180 \amp = x +255.6 \amp = 0.71 \cdot x \\ +\frac{255.6}{0.71} \amp = \frac{0.71x}{0.71} \\ +360 \amp = x \end{aligned} -}\)
        -\(90\%\) of \(180\) is \(162\text{.}\) -
        Method 3
        In the sentence “\(90\%\) of what is \(162\text{,}\)”
          -
        • -\(162\) is the percentage,
          • -
        • -\(90\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{162}{90\%} = \frac{162}{0.9} = 180 }\)
        -\(90\%\) of \(180\) is \(162\text{.}\) +}\)
        +\(71\%\) of \(360\) is \(255.6\text{.}\) +
        Method 3
        In the sentence “\(71\%\) of what is \(255.6\text{,}\)”
          +
        • +\(255.6\) is the percentage,
          • +
        • +\(71\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{255.6}{71\%} = \frac{255.6}{0.71} = 360 }\)
        +\(71\%\) of \(360\) is \(255.6\text{.}\)

        21.

        -
        -
        +
        +
        -
        -\(6\%\) of is \(16.2\text{.}\) +
        +\(4\%\) of is \(18.4\text{.}\)
        -
        Answer.
        \(270\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(6\%\) of \(x\) is \(16.2\text{,}\) so “\(6\) out of \(100\)” corresponds to “\(16.2\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{6}{100} \amp = \frac{16.2}{x} \\ -6x \amp = 100 \cdot 16.2 \\ -6x \amp = 1620 \\ -\frac{6x}{6} \amp = \frac{1620}{6} \\ -x \amp = 270 +
        Answer.
        \(460\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(4\%\) of \(x\) is \(18.4\text{,}\) so “\(4\) out of \(100\)” corresponds to “\(18.4\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{4}{100} \amp = \frac{18.4}{x} \\ +4x \amp = 100 \cdot 18.4 \\ +4x \amp = 1840 \\ +\frac{4x}{4} \amp = \frac{1840}{4} \\ +x \amp = 460 \end{aligned} -}\)
        -\(6\%\) of \(270\) is \(16.2\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(16.2\) is \(6\%\) of what? Assume \(16.2\) is \(6\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        +\(4\%\) of \(460\) is \(18.4\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(18.4\) is \(4\%\) of what? Assume \(18.4\) is \(4\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -16.2 \amp = 0.06 \cdot x \\ -\frac{16.2}{0.06} \amp = \frac{0.06x}{0.06} \\ -270 \amp = x -\end{aligned} -}\)
        -\(6\%\) of \(270\) is \(16.2\text{.}\) -
        Method 3
        In the sentence “\(2 \text{ is } 50\% \text{ of } 4\text{,}\)”
          -
        • +18.4 \amp = 0.04 \cdot x \\ +\frac{18.4}{0.04} \amp = \frac{0.04x}{0.04} \\ +460 \amp = x +\end{aligned} +}\)
          +\(4\%\) of \(460\) is \(18.4\text{.}\) +
          Method 3
          In the sentence “\(2 \text{ is } 50\% \text{ of } 4\text{,}\)”
            +
          • \(2\) is the percentage,
            • -
          • +
          • \(50\%\) is the rate,
            • -
          • +
          • \(4\) is the base (following the word “of”).
            • -
          By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
          \(\displaystyle{ \text{base} = \frac{16.2}{6\%} = \frac{16.2}{0.06} = 270 }\)
          -\(6\%\) of \(270\) is \(16.2\text{.}\) -
          Method 3
          In the sentence “\(6\%\) of what is \(16.2\text{,}\)”
            -
          • -\(16.2\) is the percentage,
            • -
          • -\(6\%\) is the rate,
            • -
          • “what” is the base (following the word “of”).
            • -
          By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
          \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{16.2}{6\%} = \frac{16.2}{0.06} = 270 }\)
          -\(6\%\) of \(270\) is \(16.2\text{.}\) +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{18.4}{4\%} = \frac{18.4}{0.04} = 460 }\)
        +\(4\%\) of \(460\) is \(18.4\text{.}\) +
        Method 3
        In the sentence “\(4\%\) of what is \(18.4\text{,}\)”
          +
        • +\(18.4\) is the percentage,
          • +
        • +\(4\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{18.4}{4\%} = \frac{18.4}{0.04} = 460 }\)
        +\(4\%\) of \(460\) is \(18.4\text{.}\)

        22.

        -
        -
        +
        +
        -
        -\(3\%\) of is \(11.1\text{.}\) +
        +\(9\%\) of is \(50.4\text{.}\)
        -
        Answer.
        \(370\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(3\%\) of \(x\) is \(11.1\text{,}\) so “\(3\) out of \(100\)” corresponds to “\(11.1\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{3}{100} \amp = \frac{11.1}{x} \\ -3x \amp = 100 \cdot 11.1 \\ -3x \amp = 1110 \\ -\frac{3x}{3} \amp = \frac{1110}{3} \\ -x \amp = 370 +
        Answer.
        \(560\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(9\%\) of \(x\) is \(50.4\text{,}\) so “\(9\) out of \(100\)” corresponds to “\(50.4\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{9}{100} \amp = \frac{50.4}{x} \\ +9x \amp = 100 \cdot 50.4 \\ +9x \amp = 5040 \\ +\frac{9x}{9} \amp = \frac{5040}{9} \\ +x \amp = 560 \end{aligned} -}\)
        -\(3\%\) of \(370\) is \(11.1\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(11.1\) is \(3\%\) of what? Assume \(11.1\) is \(3\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        +\(9\%\) of \(560\) is \(50.4\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(50.4\) is \(9\%\) of what? Assume \(50.4\) is \(9\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -11.1 \amp = 0.03 \cdot x \\ -\frac{11.1}{0.03} \amp = \frac{0.03x}{0.03} \\ -370 \amp = x -\end{aligned} -}\)
        -\(3\%\) of \(370\) is \(11.1\text{.}\) -
        Method 3
        In the sentence “\(2 \text{ is } 50\% \text{ of } 4\text{,}\)”
          -
        • +50.4 \amp = 0.09 \cdot x \\ +\frac{50.4}{0.09} \amp = \frac{0.09x}{0.09} \\ +560 \amp = x +\end{aligned} +}\)
          +\(9\%\) of \(560\) is \(50.4\text{.}\) +
          Method 3
          In the sentence “\(2 \text{ is } 50\% \text{ of } 4\text{,}\)”
            +
          • \(2\) is the percentage,
            • -
          • +
          • \(50\%\) is the rate,
            • -
          • +
          • \(4\) is the base (following the word “of”).
            • -
          By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
          \(\displaystyle{ \text{base} = \frac{11.1}{3\%} = \frac{11.1}{0.03} = 370 }\)
          -\(3\%\) of \(370\) is \(11.1\text{.}\) -
          Method 3
          In the sentence “\(3\%\) of what is \(11.1\text{,}\)”
            -
          • -\(11.1\) is the percentage,
            • -
          • -\(3\%\) is the rate,
            • -
          • “what” is the base (following the word “of”).
            • -
          By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
          \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{11.1}{3\%} = \frac{11.1}{0.03} = 370 }\)
          -\(3\%\) of \(370\) is \(11.1\text{.}\) +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{50.4}{9\%} = \frac{50.4}{0.09} = 560 }\)
        +\(9\%\) of \(560\) is \(50.4\text{.}\) +
        Method 3
        In the sentence “\(9\%\) of what is \(50.4\text{,}\)”
          +
        • +\(50.4\) is the percentage,
          • +
        • +\(9\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{50.4}{9\%} = \frac{50.4}{0.09} = 560 }\)
        +\(9\%\) of \(560\) is \(50.4\text{.}\)

        23.

        -
        -
        +
        +
        -
        -\(520\%\) of is \(2444\text{.}\) +
        +\(420\%\) of is \(2772\text{.}\)
        -
        Answer.
        \(470\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(520\%\) of \(x\) is \(2444\text{,}\) so “\(520\) out of \(100\)” corresponds to “\(2444\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{520}{100} \amp = \frac{2444}{x} \\ -520x \amp = 100 \cdot 2444 \\ -520x \amp = 244400 \\ -\frac{520x}{520} \amp = \frac{244400}{520} \\ -x \amp = 470 +
        Answer.
        \(660\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(420\%\) of \(x\) is \(2772\text{,}\) so “\(420\) out of \(100\)” corresponds to “\(2772\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{420}{100} \amp = \frac{2772}{x} \\ +420x \amp = 100 \cdot 2772 \\ +420x \amp = 277200 \\ +\frac{420x}{420} \amp = \frac{277200}{420} \\ +x \amp = 660 \end{aligned} -}\)
        -\(520\%\) of \(470\) is \(2444\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(2444\) is \(520\%\) of what? Assume \(2444\) is \(520\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        +\(420\%\) of \(660\) is \(2772\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(2772\) is \(420\%\) of what? Assume \(2772\) is \(420\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -2444 \amp = 5.2 \cdot x \\ -\frac{2444}{5.2} \amp = \frac{5.2x}{5.2} \\ -470 \amp = x +2772 \amp = 4.2 \cdot x \\ +\frac{2772}{4.2} \amp = \frac{4.2x}{4.2} \\ +660 \amp = x \end{aligned} -}\)
        -\(520\%\) of \(470\) is \(2444\text{.}\) -
        Method 3
        In the sentence “\(520\%\) of what is \(2444\text{,}\)”
          -
        • -\(2444\) is the percentage,
          • -
        • -\(520\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{2444}{520\%} = \frac{2444}{5.2} = 470 }\)
        -\(520\%\) of \(470\) is \(2444\text{.}\) +}\)
        +\(420\%\) of \(660\) is \(2772\text{.}\) +
        Method 3
        In the sentence “\(420\%\) of what is \(2772\text{,}\)”
          +
        • +\(2772\) is the percentage,
          • +
        • +\(420\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{2772}{420\%} = \frac{2772}{4.2} = 660 }\)
        +\(420\%\) of \(660\) is \(2772\text{.}\)

        24.

        -
        -
        +
        +
        -
        -\(350\%\) of is \(1995\text{.}\) +
        +\(250\%\) of is \(1875\text{.}\)
        -
        Answer.
        \(570\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem. Assume \(350\%\) of \(x\) is \(1995\text{,}\) so “\(350\) out of \(100\)” corresponds to “\(1995\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{350}{100} \amp = \frac{1995}{x} \\ -350x \amp = 100 \cdot 1995 \\ -350x \amp = 199500 \\ -\frac{350x}{350} \amp = \frac{199500}{350} \\ -x \amp = 570 +
        Answer.
        \(750\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem. Assume \(250\%\) of \(x\) is \(1875\text{,}\) so “\(250\) out of \(100\)” corresponds to “\(1875\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{250}{100} \amp = \frac{1875}{x} \\ +250x \amp = 100 \cdot 1875 \\ +250x \amp = 187500 \\ +\frac{250x}{250} \amp = \frac{187500}{250} \\ +x \amp = 750 \end{aligned} -}\)
        -\(350\%\) of \(570\) is \(1995\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(1995\) is \(350\%\) of what? Assume \(1995\) is \(350\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        +\(250\%\) of \(750\) is \(1875\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(1875\) is \(250\%\) of what? Assume \(1875\) is \(250\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -1995 \amp = 3.5 \cdot x \\ -\frac{1995}{3.5} \amp = \frac{3.5x}{3.5} \\ -570 \amp = x +1875 \amp = 2.5 \cdot x \\ +\frac{1875}{2.5} \amp = \frac{2.5x}{2.5} \\ +750 \amp = x \end{aligned} -}\)
        -\(350\%\) of \(570\) is \(1995\text{.}\) -
        Method 3
        In the sentence “\(350\%\) of what is \(1995\text{,}\)”
          -
        • -\(1995\) is the percentage,
          • -
        • -\(350\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{1995}{350\%} = \frac{1995}{3.5} = 570 }\)
        -\(350\%\) of \(570\) is \(1995\text{.}\) +}\)
        +\(250\%\) of \(750\) is \(1875\text{.}\) +
        Method 3
        In the sentence “\(250\%\) of what is \(1875\text{,}\)”
          +
        • +\(1875\) is the percentage,
          • +
        • +\(250\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{1875}{250\%} = \frac{1875}{2.5} = 750 }\)
        +\(250\%\) of \(750\) is \(1875\text{.}\)
        @@ -1579,231 +1593,231 @@

        Basic Percentage Calculation.

        Exercise Group.

        25.

        -
        -
        -
        -
        Answer with a percent.
        -\(168\) is of \(240\text{.}\) -
        -
        -
        Answer.
        \(70\%\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem.
        Assume \(168\) is \(x\%\) of \(240\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(168\) out of \(240\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{168}{240} \\ -240x \amp = 168 \cdot 100 \\ -240x \amp = 16800 \\ -\frac{240x}{240} \amp = \frac{16800}{240} \\ -x \amp = 70 -\end{aligned} -}\)
        -\(168\) is \(70\%\) of \(240\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(168\) is \(x\) (as a percent) of \(240\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -168 \amp = x \cdot 240 \\ -\frac{168}{240} \amp = \frac{x \cdot 240}{240} \\ -0.7 \amp = x\\ -x \amp = 70\% -\end{aligned} -}\)
        -\(168\) is \(70\%\) of \(240\text{.}\) -
        Method 3
        In the sentence “\(168\) is what percent of \(240\text{,}\)”
          -
        • -\(168\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(240\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{168}{240} = 0.7 = 70\% }\)
        -\(168\) is \(70\%\) of \(240\text{.}\) +
        +
        +
        +
        Answer with a percent.
        +\(752\) is of \(940\text{.}\) +
        +
        +
        Answer.
        \(80\%\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem.
        Assume \(752\) is \(x\%\) of \(940\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(752\) out of \(940\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{752}{940} \\ +940x \amp = 752 \cdot 100 \\ +940x \amp = 75200 \\ +\frac{940x}{940} \amp = \frac{75200}{940} \\ +x \amp = 80 +\end{aligned} +}\)
        +\(752\) is \(80\%\) of \(940\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(752\) is \(x\) (as a percent) of \(940\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +752 \amp = x \cdot 940 \\ +\frac{752}{940} \amp = \frac{x \cdot 940}{940} \\ +0.8 \amp = x\\ +x \amp = 80\% +\end{aligned} +}\)
        +\(752\) is \(80\%\) of \(940\text{.}\) +
        Method 3
        In the sentence “\(752\) is what percent of \(940\text{,}\)”
          +
        • +\(752\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(940\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{752}{940} = 0.8 = 80\% }\)
        +\(752\) is \(80\%\) of \(940\text{.}\)

        26.

        -
        -
        +
        +
        -
        Answer with a percent.
        -\(574\) is of \(820\text{.}\) +
        Answer with a percent.
        +\(567\) is of \(630\text{.}\)
        -
        Answer.
        \(70\%\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem.
        Assume \(574\) is \(x\%\) of \(820\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(574\) out of \(820\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{574}{820} \\ -820x \amp = 574 \cdot 100 \\ -820x \amp = 57400 \\ -\frac{820x}{820} \amp = \frac{57400}{820} \\ -x \amp = 70 +
        Answer.
        \(90\%\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem.
        Assume \(567\) is \(x\%\) of \(630\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(567\) out of \(630\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{567}{630} \\ +630x \amp = 567 \cdot 100 \\ +630x \amp = 56700 \\ +\frac{630x}{630} \amp = \frac{56700}{630} \\ +x \amp = 90 \end{aligned} -}\)
        -\(574\) is \(70\%\) of \(820\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(574\) is \(x\) (as a percent) of \(820\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -574 \amp = x \cdot 820 \\ -\frac{574}{820} \amp = \frac{x \cdot 820}{820} \\ -0.7 \amp = x\\ -x \amp = 70\% +}\)
        +\(567\) is \(90\%\) of \(630\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(567\) is \(x\) (as a percent) of \(630\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +567 \amp = x \cdot 630 \\ +\frac{567}{630} \amp = \frac{x \cdot 630}{630} \\ +0.9 \amp = x\\ +x \amp = 90\% \end{aligned} -}\)
        -\(574\) is \(70\%\) of \(820\text{.}\) -
        Method 3
        In the sentence “\(574\) is what percent of \(820\text{,}\)”
          -
        • -\(574\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(820\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{574}{820} = 0.7 = 70\% }\)
        -\(574\) is \(70\%\) of \(820\text{.}\) +}\)
        +\(567\) is \(90\%\) of \(630\text{.}\) +
        Method 3
        In the sentence “\(567\) is what percent of \(630\text{,}\)”
          +
        • +\(567\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(630\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{567}{630} = 0.9 = 90\% }\)
        +\(567\) is \(90\%\) of \(630\text{.}\)

        27.

        -
        -
        +
        +
        -
        Answer with a percent.
        -\(137.7\) is of \(51\text{.}\) +
        Answer with a percent.
        +\(38.4\) is of \(32\text{.}\)
        -
        Answer.
        \(270\%\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem.
        Assume \(137.7\) is \(x\%\) of \(51\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(137.7\) out of \(51\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{137.7}{51} \\ -51x \amp = 137.7 \cdot 100 \\ -51x \amp = 13770 \\ -\frac{51x}{51} \amp = \frac{13770}{51} \\ -x \amp = 270 +
        Answer.
        \(120\%\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem.
        Assume \(38.4\) is \(x\%\) of \(32\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(38.4\) out of \(32\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{38.4}{32} \\ +32x \amp = 38.4 \cdot 100 \\ +32x \amp = 3840 \\ +\frac{32x}{32} \amp = \frac{3840}{32} \\ +x \amp = 120 \end{aligned} -}\)
        -\(137.7\) is \(270\%\) of \(51\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(137.7\) is \(x\) (as a percent) of \(51\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -137.7 \amp = x \cdot 51 \\ -\frac{137.7}{51} \amp = \frac{x \cdot 51}{51} \\ -2.7 \amp = x\\ -x \amp = 270\% +}\)
        +\(38.4\) is \(120\%\) of \(32\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(38.4\) is \(x\) (as a percent) of \(32\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +38.4 \amp = x \cdot 32 \\ +\frac{38.4}{32} \amp = \frac{x \cdot 32}{32} \\ +1.2 \amp = x\\ +x \amp = 120\% \end{aligned} -}\)
        -\(137.7\) is \(270\%\) of \(51\text{.}\) -
        Method 3
        In the sentence “\(137.7\) is what percent of \(51\text{,}\)”
          -
        • -\(137.7\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(51\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{137.7}{51} = 2.7 = 270\% }\)
        -\(137.7\) is \(270\%\) of \(51\text{.}\) +}\)
        +\(38.4\) is \(120\%\) of \(32\text{.}\) +
        Method 3
        In the sentence “\(38.4\) is what percent of \(32\text{,}\)”
          +
        • +\(38.4\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(32\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{38.4}{32} = 1.2 = 120\% }\)
        +\(38.4\) is \(120\%\) of \(32\text{.}\)

        28.

        -
        -
        +
        +
        -
        Answer with a percent.
        -\(58\) is of \(20\text{.}\) +
        Answer with a percent.
        +\(126\) is of \(90\text{.}\)
        -
        Answer.
        \(290\%\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem.
        Assume \(58\) is \(x\%\) of \(20\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(58\) out of \(20\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{58}{20} \\ -20x \amp = 58 \cdot 100 \\ -20x \amp = 5800 \\ -\frac{20x}{20} \amp = \frac{5800}{20} \\ -x \amp = 290 +
        Answer.
        \(140\%\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem.
        Assume \(126\) is \(x\%\) of \(90\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(126\) out of \(90\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{126}{90} \\ +90x \amp = 126 \cdot 100 \\ +90x \amp = 12600 \\ +\frac{90x}{90} \amp = \frac{12600}{90} \\ +x \amp = 140 \end{aligned} -}\)
        -\(58\) is \(290\%\) of \(20\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(58\) is \(x\) (as a percent) of \(20\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -58 \amp = x \cdot 20 \\ -\frac{58}{20} \amp = \frac{x \cdot 20}{20} \\ -2.9 \amp = x\\ -x \amp = 290\% +}\)
        +\(126\) is \(140\%\) of \(90\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(126\) is \(x\) (as a percent) of \(90\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +126 \amp = x \cdot 90 \\ +\frac{126}{90} \amp = \frac{x \cdot 90}{90} \\ +1.4 \amp = x\\ +x \amp = 140\% \end{aligned} -}\)
        -\(58\) is \(290\%\) of \(20\text{.}\) -
        Method 3
        In the sentence “\(58\) is what percent of \(20\text{,}\)”
          -
        • -\(58\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(20\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{58}{20} = 2.9 = 290\% }\)
        -\(58\) is \(290\%\) of \(20\text{.}\) +}\)
        +\(126\) is \(140\%\) of \(90\text{.}\) +
        Method 3
        In the sentence “\(126\) is what percent of \(90\text{,}\)”
          +
        • +\(126\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(90\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{126}{90} = 1.4 = 140\% }\)
        +\(126\) is \(140\%\) of \(90\text{.}\)

        29.

        -
        -
        +
        +
        -
        Answer with a percent.
        -\(3\) is about of \(76\text{.}\) +
        Answer with a percent.
        +\(7\) is about of \(58\text{.}\)
        -
        Answer.
        \(3.95\%\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem.
        Assume \(3\) is \(x\%\) of \(76\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(3\) out of \(76\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{3}{76} \\ -76x \amp = 3 \cdot 100 \\ -76x \amp = 300 \\ -\frac{76x}{76} \amp = \frac{300}{76} \\ -x \amp \approx 3.94736842105263 \\ -x \amp \approx 3.95 +
        Answer.
        \(12.07\%\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem.
        Assume \(7\) is \(x\%\) of \(58\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(7\) out of \(58\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{7}{58} \\ +58x \amp = 7 \cdot 100 \\ +58x \amp = 700 \\ +\frac{58x}{58} \amp = \frac{700}{58} \\ +x \amp \approx 12.0689655172414 \\ +x \amp \approx 12.07 \end{aligned} -}\)
        -\(3\) is approximately \(3.95\%\) of \(76\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to math should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(3\) is \(x\) (as a percent) of \(76\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -3 \amp = x \cdot 76 \\ -\frac{3}{76} \amp = \frac{x \cdot 76}{76} \\ -0.0394736842105263 \amp \approx x\\ -x \amp \approx 3.95\% +}\)
        +\(7\) is approximately \(12.07\%\) of \(58\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to math should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(7\) is \(x\) (as a percent) of \(58\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +7 \amp = x \cdot 58 \\ +\frac{7}{58} \amp = \frac{x \cdot 58}{58} \\ +0.120689655172414 \amp \approx x\\ +x \amp \approx 12.07\% \end{aligned} -}\)
        -\(3\) is approximately \(3.95\%\) of \(76\text{.}\) -
        Method 3
        In the sentence “\(3\) is what percent of \(76\text{,}\)”
          -
        • -\(3\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(76\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{3}{76} \approx 0.0394736842105263 \approx 3.95\% }\)
        -\(3\) is approximately \(3.95\%\) of \(76\text{.}\) +}\)
        +\(7\) is approximately \(12.07\%\) of \(58\text{.}\) +
        Method 3
        In the sentence “\(7\) is what percent of \(58\text{,}\)”
          +
        • +\(7\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(58\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{7}{58} \approx 0.120689655172414 \approx 12.07\% }\)
        +\(7\) is approximately \(12.07\%\) of \(58\text{.}\)

        30.

        -
        -
        +
        +
        -
        Answer with a percent.
        -\(5\) is about of \(44\text{.}\) +
        Answer with a percent.
        +\(9\) is about of \(27\text{.}\)
        -
        Answer.
        \(11.36\%\)
        Explanation.
        -
        Method 1
        We will use proportion to solve this problem.
        Assume \(5\) is \(x\%\) of \(44\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(5\) out of \(44\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{5}{44} \\ -44x \amp = 5 \cdot 100 \\ -44x \amp = 500 \\ -\frac{44x}{44} \amp = \frac{500}{44} \\ -x \amp \approx 11.3636363636364 \\ -x \amp \approx 11.36 +
        Answer.
        \(33.33\%\)
        Explanation.
        +
        Method 1
        We will use proportion to solve this problem.
        Assume \(9\) is \(x\%\) of \(27\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(9\) out of \(27\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{9}{27} \\ +27x \amp = 9 \cdot 100 \\ +27x \amp = 900 \\ +\frac{27x}{27} \amp = \frac{900}{27} \\ +x \amp \approx 33.3333333333333 \\ +x \amp \approx 33.33 \end{aligned} -}\)
        -\(5\) is approximately \(11.36\%\) of \(44\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to math should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(5\) is \(x\) (as a percent) of \(44\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -5 \amp = x \cdot 44 \\ -\frac{5}{44} \amp = \frac{x \cdot 44}{44} \\ -0.113636363636364 \amp \approx x\\ -x \amp \approx 11.36\% +}\)
        +\(9\) is approximately \(33.33\%\) of \(27\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to math should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(9\) is \(x\) (as a percent) of \(27\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +9 \amp = x \cdot 27 \\ +\frac{9}{27} \amp = \frac{x \cdot 27}{27} \\ +0.333333333333333 \amp \approx x\\ +x \amp \approx 33.33\% \end{aligned} -}\)
        -\(5\) is approximately \(11.36\%\) of \(44\text{.}\) -
        Method 3
        In the sentence “\(5\) is what percent of \(44\text{,}\)”
          -
        • -\(5\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(44\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{5}{44} \approx 0.113636363636364 \approx 11.36\% }\)
        -\(5\) is approximately \(11.36\%\) of \(44\text{.}\) +}\)
        +\(9\) is approximately \(33.33\%\) of \(27\text{.}\) +
        Method 3
        In the sentence “\(9\) is what percent of \(27\text{,}\)”
          +
        • +\(9\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(27\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{9}{27} \approx 0.333333333333333 \approx 33.33\% }\)
        +\(9\) is approximately \(33.33\%\) of \(27\text{.}\)
        @@ -1815,960 +1829,960 @@

        Exercise Group.

        Applications.

        31.

        -
        -
        -
        -
        A town has \(2200\) registered residents. Among them, \(31\%\) were Democrats, \(34\%\) were Republicans. The rest were Independents. How many registered Independents live in this town?
        There are registered Independent residents in this town.
        -
        -
        Answer.
        \(770\)
        Explanation.
        -
        It’s given there are \(31\%\) Democrats, \(34\%\) Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents:
        \(\displaystyle{\begin{aligned}[t] -\amp \phantom{{}=}1-31\%-34\% \\ -\amp =1-0.31-0.34 \\ -\amp =0.35 \\ -\amp = 35\% -\end{aligned} -}\)
        So there are \(35\%\) Independents. Now this problem can be boiled down to this question: What is \(35\%\) of \(2200\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(35\%\) of \(2200\) is \(x\text{,}\) so “\(35\) out of \(100\)” corresponds to “\(x\) out of \(2200\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{35}{100} \amp = \frac{x}{2200} \\ -100x \amp = 35 \cdot 2200 \\ -100x \amp = 77000 \\ -\frac{100x}{100} \amp = \frac{77000}{100} \\ -x \amp = 770 -\end{aligned} -}\)
        There are \(770\) registered Independent residents in this town.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(35\%\) of \(2200\text{?}\) Assume \(x\) is \(35\%\) of \(2200\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        A town has \(3000\) registered residents. Among them, \(38\%\) were Democrats, \(31\%\) were Republicans. The rest were Independents. How many registered Independents live in this town?
        There are registered Independent residents in this town.
        +
        +
        Answer.
        \(930\)
        Explanation.
        +
        It’s given there are \(38\%\) Democrats, \(31\%\) Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents:
        \(\displaystyle{\begin{aligned}[t] +\amp \phantom{{}=}1-38\%-31\% \\ +\amp =1-0.38-0.31 \\ +\amp =0.31 \\ +\amp = 31\% +\end{aligned} +}\)
        So there are \(31\%\) Independents. Now this problem can be boiled down to this question: What is \(31\%\) of \(3000\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(31\%\) of \(3000\) is \(x\text{,}\) so “\(31\) out of \(100\)” corresponds to “\(x\) out of \(3000\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{31}{100} \amp = \frac{x}{3000} \\ +100x \amp = 31 \cdot 3000 \\ +100x \amp = 93000 \\ +\frac{100x}{100} \amp = \frac{93000}{100} \\ +x \amp = 930 +\end{aligned} +}\)
        There are \(930\) registered Independent residents in this town.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(31\%\) of \(3000\text{?}\) Assume \(x\) is \(31\%\) of \(3000\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.35 \cdot 2200 \\ -\amp = 770 +x \amp = 0.31 \cdot 3000 \\ +\amp = 930 \end{aligned} -}\)
        There are \(770\) registered Independent residents in this town.
        Method 3
        In the sentence “What is \(35\%\) of \(2200\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(35\%\) is the rate,
          • -
        • -\(2200\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 35\% \cdot 2200 = 0.35 \cdot 2200 = 770 }\)
        There are \(770\) registered Independent residents in this town.
        +}\)
        There are \(930\) registered Independent residents in this town.
        Method 3
        In the sentence “What is \(31\%\) of \(3000\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(31\%\) is the rate,
          • +
        • +\(3000\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 31\% \cdot 3000 = 0.31 \cdot 3000 = 930 }\)
        There are \(930\) registered Independent residents in this town.

        32.

        -
        -
        -
        -
        A town has \(2600\) registered residents. Among them, \(37\%\) were Democrats, \(23\%\) were Republicans. The rest were Independents. How many registered Independents live in this town?
        There are registered Independent residents in this town.
        -
        -
        Answer.
        \(1040\)
        Explanation.
        -
        It’s given there are \(37\%\) Democrats, \(23\%\) Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents:
        \(\displaystyle{\begin{aligned}[t] -\amp \phantom{{}=}1-37\%-23\% \\ -\amp =1-0.37-0.23 \\ -\amp =0.4 \\ -\amp = 40\% -\end{aligned} -}\)
        So there are \(40\%\) Independents. Now this problem can be boiled down to this question: What is \(40\%\) of \(2600\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(40\%\) of \(2600\) is \(x\text{,}\) so “\(40\) out of \(100\)” corresponds to “\(x\) out of \(2600\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{40}{100} \amp = \frac{x}{2600} \\ -100x \amp = 40 \cdot 2600 \\ -100x \amp = 104000 \\ -\frac{100x}{100} \amp = \frac{104000}{100} \\ -x \amp = 1040 -\end{aligned} -}\)
        There are \(1040\) registered Independent residents in this town.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(40\%\) of \(2600\text{?}\) Assume \(x\) is \(40\%\) of \(2600\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        A town has \(3500\) registered residents. Among them, \(35\%\) were Democrats, \(39\%\) were Republicans. The rest were Independents. How many registered Independents live in this town?
        There are registered Independent residents in this town.
        +
        +
        Answer.
        \(910\)
        Explanation.
        +
        It’s given there are \(35\%\) Democrats, \(39\%\) Republicans, and the rest are Independents. Now we can use subtraction to find the percentage of registered Independent residents:
        \(\displaystyle{\begin{aligned}[t] +\amp \phantom{{}=}1-35\%-39\% \\ +\amp =1-0.35-0.39 \\ +\amp =0.26 \\ +\amp = 26\% +\end{aligned} +}\)
        So there are \(26\%\) Independents. Now this problem can be boiled down to this question: What is \(26\%\) of \(3500\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(26\%\) of \(3500\) is \(x\text{,}\) so “\(26\) out of \(100\)” corresponds to “\(x\) out of \(3500\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{26}{100} \amp = \frac{x}{3500} \\ +100x \amp = 26 \cdot 3500 \\ +100x \amp = 91000 \\ +\frac{100x}{100} \amp = \frac{91000}{100} \\ +x \amp = 910 +\end{aligned} +}\)
        There are \(910\) registered Independent residents in this town.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(26\%\) of \(3500\text{?}\) Assume \(x\) is \(26\%\) of \(3500\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.4 \cdot 2600 \\ -\amp = 1040 +x \amp = 0.26 \cdot 3500 \\ +\amp = 910 \end{aligned} -}\)
        There are \(1040\) registered Independent residents in this town.
        Method 3
        In the sentence “What is \(40\%\) of \(2600\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(40\%\) is the rate,
          • -
        • -\(2600\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 40\% \cdot 2600 = 0.4 \cdot 2600 = 1040 }\)
        There are \(1040\) registered Independent residents in this town.
        +}\)
        There are \(910\) registered Independent residents in this town.
        Method 3
        In the sentence “What is \(26\%\) of \(3500\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(26\%\) is the rate,
          • +
        • +\(3500\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 26\% \cdot 3500 = 0.26 \cdot 3500 = 910 }\)
        There are \(910\) registered Independent residents in this town.

        33.

        -
        -
        +
        +
        -
        Sharell is paying a dinner bill of \({\$36.00}\text{.}\) Sharell plans to pay \(14\%\) in tips. How much tip will Sharell pay?
        Sharell will pay in tip.
        +
        Eric is paying a dinner bill of \({\$42.00}\text{.}\) Eric plans to pay \(11\%\) in tips. How much tip will Eric pay?
        Eric will pay in tip.
        -
        Answer.
        \(\$5.04\)
        Explanation.
        -
        This problem can be boiled down to this question: What is \(14\%\) of \(36\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(14\%\) of \(36\) is \(x\text{,}\) so “\(14\) out of \(100\)” corresponds to “\(x\) out of \(36\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{14}{100} \amp = \frac{x}{36} \\ -100x \amp = 14 \cdot 36 \\ -100x \amp = 504 \\ -\frac{100x}{100} \amp = \frac{504}{100} \\ -x \amp = 5.04 +
        Answer.
        \(\$4.62\)
        Explanation.
        +
        This problem can be boiled down to this question: What is \(11\%\) of \(42\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(11\%\) of \(42\) is \(x\text{,}\) so “\(11\) out of \(100\)” corresponds to “\(x\) out of \(42\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{11}{100} \amp = \frac{x}{42} \\ +100x \amp = 11 \cdot 42 \\ +100x \amp = 462 \\ +\frac{100x}{100} \amp = \frac{462}{100} \\ +x \amp = 4.62 \end{aligned} -}\)
        Sharell will pay \({\$5.04}\) in tip.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(14\%\) of \(36\text{?}\) Assume \(x\) is \(14\%\) of \(36\text{.}\) We have:
        \(\displaystyle{ +}\)
        Eric will pay \({\$4.62}\) in tip.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(11\%\) of \(42\text{?}\) Assume \(x\) is \(11\%\) of \(42\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.14 \cdot 36 \\ -\amp = 5.04 -\end{aligned} -}\)
        Sharell will pay \({\$5.04}\) in tip.
        Method 3
        In the sentence “What is \(14\%\) of \(36\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(14\%\) is the rate,
          • -
        • -\(36\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 14\% \cdot 36 = 0.14 \cdot 36 = 5.04 }\)
        Sharell will pay \({\$5.04}\) in tip.
        +x \amp = 0.11 \cdot 42 \\ +\amp = 4.62 +\end{aligned} +}\)
        Eric will pay \({\$4.62}\) in tip.
        Method 3
        In the sentence “What is \(11\%\) of \(42\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(11\%\) is the rate,
          • +
        • +\(42\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 11\% \cdot 42 = 0.11 \cdot 42 = 4.62 }\)
        Eric will pay \({\$4.62}\) in tip.

        34.

        -
        -
        +
        +
        -
        Lindsay is paying a dinner bill of \({\$39.00}\text{.}\) Lindsay plans to pay \(10\%\) in tips. How much tip will Lindsay pay?
        Lindsay will pay in tip.
        +
        Amber is paying a dinner bill of \({\$45.00}\text{.}\) Amber plans to pay \(18\%\) in tips. How much tip will Amber pay?
        Amber will pay in tip.
        -
        Answer.
        \(\$3.90\)
        Explanation.
        -
        This problem can be boiled down to this question: What is \(10\%\) of \(39\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(10\%\) of \(39\) is \(x\text{,}\) so “\(10\) out of \(100\)” corresponds to “\(x\) out of \(39\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{10}{100} \amp = \frac{x}{39} \\ -100x \amp = 10 \cdot 39 \\ -100x \amp = 390 \\ -\frac{100x}{100} \amp = \frac{390}{100} \\ -x \amp = 3.9 +
        Answer.
        \(\$8.10\)
        Explanation.
        +
        This problem can be boiled down to this question: What is \(18\%\) of \(45\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(18\%\) of \(45\) is \(x\text{,}\) so “\(18\) out of \(100\)” corresponds to “\(x\) out of \(45\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{18}{100} \amp = \frac{x}{45} \\ +100x \amp = 18 \cdot 45 \\ +100x \amp = 810 \\ +\frac{100x}{100} \amp = \frac{810}{100} \\ +x \amp = 8.1 \end{aligned} -}\)
        Lindsay will pay \({\$3.90}\) in tip.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(10\%\) of \(39\text{?}\) Assume \(x\) is \(10\%\) of \(39\text{.}\) We have:
        \(\displaystyle{ +}\)
        Amber will pay \({\$8.10}\) in tip.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(18\%\) of \(45\text{?}\) Assume \(x\) is \(18\%\) of \(45\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.1 \cdot 39 \\ -\amp = 3.9 +x \amp = 0.18 \cdot 45 \\ +\amp = 8.1 \end{aligned} -}\)
        Lindsay will pay \({\$3.90}\) in tip.
        Method 3
        In the sentence “What is \(10\%\) of \(39\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(10\%\) is the rate,
          • -
        • -\(39\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 10\% \cdot 39 = 0.1 \cdot 39 = 3.9 }\)
        Lindsay will pay \({\$3.90}\) in tip.
        +}\)
        Amber will pay \({\$8.10}\) in tip.
        Method 3
        In the sentence “What is \(18\%\) of \(45\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(18\%\) is the rate,
          • +
        • +\(45\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 18\% \cdot 45 = 0.18 \cdot 45 = 8.1 }\)
        Amber will pay \({\$8.10}\) in tip.

        35.

        -
        -
        +
        +
        -
        Emiliano is paying a dinner bill of \({\$42.00}\text{.}\) Emiliano plans to pay \(17\%\) in tips. How much in total (including bill and tip) will Emiliano pay?
        Emiliano will pay in total (including bill and tip).
        +
        Phil is paying a dinner bill of \({\$49.00}\text{.}\) Phil plans to pay \(15\%\) in tips. How much in total (including bill and tip) will Phil pay?
        Phil will pay in total (including bill and tip).
        -
        Answer.
        \(\$49.14\)
        Explanation.
        -
        First, we need to find how much tip Emiliano paid. This problem can be boiled down to this question: What is \(17\%\) of \(42\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(17\%\) of \(42\) is \(x\text{,}\) so “\(17\) out of \(100\)” corresponds to “\(x\) out of \(42\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{17}{100} \amp = \frac{x}{42} \\ -100x \amp = 17 \cdot 42 \\ -100x \amp = 714 \\ -\frac{100x}{100} \amp = \frac{714}{100} \\ -x \amp = 7.14 +
        Answer.
        \(\$56.35\)
        Explanation.
        +
        First, we need to find how much tip Phil paid. This problem can be boiled down to this question: What is \(15\%\) of \(49\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(15\%\) of \(49\) is \(x\text{,}\) so “\(15\) out of \(100\)” corresponds to “\(x\) out of \(49\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{15}{100} \amp = \frac{x}{49} \\ +100x \amp = 15 \cdot 49 \\ +100x \amp = 735 \\ +\frac{100x}{100} \amp = \frac{735}{100} \\ +x \amp = 7.35 \end{aligned} -}\)
        Emiliano will pay \({\$7.14}\) in tip. In total, he will pay \({\$42.00}+{\$7.14}={\$49.14}\text{.}\) -
        Emiliano will pay \({\$49.14}\) in total (including bill and tip).
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(17\%\) of \(42\text{?}\) Assume \(x\) is \(17\%\) of \(42\text{.}\) We have:
        \(\displaystyle{ +}\)
        Phil will pay \({\$7.35}\) in tip. In total, he will pay \({\$49.00}+{\$7.35}={\$56.35}\text{.}\) +
        Phil will pay \({\$56.35}\) in total (including bill and tip).
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(15\%\) of \(49\text{?}\) Assume \(x\) is \(15\%\) of \(49\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.17 \cdot 42 \\ -\amp = 7.14 -\end{aligned} -}\)
        Emiliano will pay \({\$7.14}\) in tip. In total, he will pay \({\$42.00}+{\$7.14}={\$49.14}\text{.}\) -
        Emiliano will pay \({\$49.14}\) in total (including bill and tip).
        Method 3
        In the sentence “What is \(17\%\) of \(42\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(17\%\) is the rate,
          • -
        • -\(42\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 17\% \cdot 42 = 0.17 \cdot 42 = 7.14 }\)
        Emiliano will pay \({\$7.14}\) in tip. In total, he will pay \({\$42.00}+{\$7.14}={\$49.14}\text{.}\) -
        Emiliano will pay \({\$49.14}\) in total (including bill and tip).
        +x \amp = 0.15 \cdot 49 \\ +\amp = 7.35 +\end{aligned} +}\)
        Phil will pay \({\$7.35}\) in tip. In total, he will pay \({\$49.00}+{\$7.35}={\$56.35}\text{.}\) +
        Phil will pay \({\$56.35}\) in total (including bill and tip).
        Method 3
        In the sentence “What is \(15\%\) of \(49\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(15\%\) is the rate,
          • +
        • +\(49\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 15\% \cdot 49 = 0.15 \cdot 49 = 7.35 }\)
        Phil will pay \({\$7.35}\) in tip. In total, he will pay \({\$49.00}+{\$7.35}={\$56.35}\text{.}\) +
        Phil will pay \({\$56.35}\) in total (including bill and tip).

        36.

        -
        -
        +
        +
        -
        Alyson is paying a dinner bill of \({\$46.00}\text{.}\) Alyson plans to pay \(13\%\) in tips. How much in total (including bill and tip) will Alyson pay?
        Alyson will pay in total (including bill and tip).
        +
        Kim is paying a dinner bill of \({\$21.00}\text{.}\) Kim plans to pay \(11\%\) in tips. How much in total (including bill and tip) will Kim pay?
        Kim will pay in total (including bill and tip).
        -
        Answer.
        \(\$51.98\)
        Explanation.
        -
        First, we need to find how much tip Alyson paid. This problem can be boiled down to this question: What is \(13\%\) of \(46\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(13\%\) of \(46\) is \(x\text{,}\) so “\(13\) out of \(100\)” corresponds to “\(x\) out of \(46\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{13}{100} \amp = \frac{x}{46} \\ -100x \amp = 13 \cdot 46 \\ -100x \amp = 598 \\ -\frac{100x}{100} \amp = \frac{598}{100} \\ -x \amp = 5.98 +
        Answer.
        \(\$23.31\)
        Explanation.
        +
        First, we need to find how much tip Kim paid. This problem can be boiled down to this question: What is \(11\%\) of \(21\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(11\%\) of \(21\) is \(x\text{,}\) so “\(11\) out of \(100\)” corresponds to “\(x\) out of \(21\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{11}{100} \amp = \frac{x}{21} \\ +100x \amp = 11 \cdot 21 \\ +100x \amp = 231 \\ +\frac{100x}{100} \amp = \frac{231}{100} \\ +x \amp = 2.31 \end{aligned} -}\)
        Alyson will pay \({\$5.98}\) in tip. In total, she will pay \({\$46.00}+{\$5.98}={\$51.98}\text{.}\) -
        Alyson will pay \({\$51.98}\) in total (including bill and tip).
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(13\%\) of \(46\text{?}\) Assume \(x\) is \(13\%\) of \(46\text{.}\) We have:
        \(\displaystyle{ +}\)
        Kim will pay \({\$2.31}\) in tip. In total, she will pay \({\$21.00}+{\$2.31}={\$23.31}\text{.}\) +
        Kim will pay \({\$23.31}\) in total (including bill and tip).
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(11\%\) of \(21\text{?}\) Assume \(x\) is \(11\%\) of \(21\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.13 \cdot 46 \\ -\amp = 5.98 +x \amp = 0.11 \cdot 21 \\ +\amp = 2.31 \end{aligned} -}\)
        Alyson will pay \({\$5.98}\) in tip. In total, she will pay \({\$46.00}+{\$5.98}={\$51.98}\text{.}\) -
        Alyson will pay \({\$51.98}\) in total (including bill and tip).
        Method 3
        In the sentence “What is \(13\%\) of \(46\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(13\%\) is the rate,
          • -
        • -\(46\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 13\% \cdot 46 = 0.13 \cdot 46 = 5.98 }\)
        Alyson will pay \({\$5.98}\) in tip. In total, she will pay \({\$46.00}+{\$5.98}={\$51.98}\text{.}\) -
        Alyson will pay \({\$51.98}\) in total (including bill and tip).
        +}\)
        Kim will pay \({\$2.31}\) in tip. In total, she will pay \({\$21.00}+{\$2.31}={\$23.31}\text{.}\) +
        Kim will pay \({\$23.31}\) in total (including bill and tip).
        Method 3
        In the sentence “What is \(11\%\) of \(21\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(11\%\) is the rate,
          • +
        • +\(21\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 11\% \cdot 21 = 0.11 \cdot 21 = 2.31 }\)
        Kim will pay \({\$2.31}\) in tip. In total, she will pay \({\$21.00}+{\$2.31}={\$23.31}\text{.}\) +
        Kim will pay \({\$23.31}\) in total (including bill and tip).

        37.

        -
        -
        -
        -
        A watch’s wholesale price was \({\$500.00}\text{.}\) The retailer marked up the price by \(40\%\text{.}\) What’s the watch’s new price (markup price)?
        The watch’s markup price is .
        -
        -
        Answer.
        \(\$700.00\)
        Explanation.
        -
        First, we need to find the amount of increase in price. It’s given that the watch’s price was marked up by \(40\%\) of its original price, \({\$500.00}\text{.}\) -
        The problem can be boiled down to this question: What is \(40\%\) of \(500\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(40\%\) of \(500\) is \(x\text{,}\) so “\(40\) out of \(100\)” corresponds to “\(x\) out of \(500\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{40}{100} \amp = \frac{x}{500} \\ -100x \amp = 40 \cdot 500 \\ -100x \amp = 20000 \\ -\frac{100x}{100} \amp = \frac{20000}{100} \\ -x \amp = 200 -\end{aligned} -}\)
        The amount of price increase was \({\$200.00}\text{,}\) so the new price is \({\$500.00}+{\$200.00}={\$700.00}\text{.}\) -
        So the watch’s markup price is \({\$700.00}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(40\%\) of \(500\text{?}\) Assume \(x\) is \(40\%\) of \(500\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        A watch’s wholesale price was \({\$240.00}\text{.}\) The retailer marked up the price by \(35\%\text{.}\) What’s the watch’s new price (markup price)?
        The watch’s markup price is .
        +
        +
        Answer.
        \(\$324.00\)
        Explanation.
        +
        First, we need to find the amount of increase in price. It’s given that the watch’s price was marked up by \(35\%\) of its original price, \({\$240.00}\text{.}\) +
        The problem can be boiled down to this question: What is \(35\%\) of \(240\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(35\%\) of \(240\) is \(x\text{,}\) so “\(35\) out of \(100\)” corresponds to “\(x\) out of \(240\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{35}{100} \amp = \frac{x}{240} \\ +100x \amp = 35 \cdot 240 \\ +100x \amp = 8400 \\ +\frac{100x}{100} \amp = \frac{8400}{100} \\ +x \amp = 84 +\end{aligned} +}\)
        The amount of price increase was \({\$84.00}\text{,}\) so the new price is \({\$240.00}+{\$84.00}={\$324.00}\text{.}\) +
        So the watch’s markup price is \({\$324.00}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(35\%\) of \(240\text{?}\) Assume \(x\) is \(35\%\) of \(240\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.4 \cdot 500 \\ -\amp = 200 -\end{aligned} -}\)
        The amount of price increase was \({\$200.00}\text{,}\) so the new price is \({\$500.00}+{\$200.00}={\$700.00}\text{.}\) -
        So the watch’s markup price is \({\$700.00}\text{.}\) -
        Method 3
        In the sentence “What is \(40\%\) of \(500\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(40\%\) is the rate,
          • -
        • -\(500\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 40\% \cdot 500 = 0.4 \cdot 500 = 200 }\)
        The amount of price increase was \({\$200.00}\text{,}\) so the new price is \({\$500.00}+{\$200.00}={\$700.00}\text{.}\) -
        So the watch’s markup price is \({\$700.00}\text{.}\) +x \amp = 0.35 \cdot 240 \\ +\amp = 84 +\end{aligned} +}\)
        The amount of price increase was \({\$84.00}\text{,}\) so the new price is \({\$240.00}+{\$84.00}={\$324.00}\text{.}\) +
        So the watch’s markup price is \({\$324.00}\text{.}\) +
        Method 3
        In the sentence “What is \(35\%\) of \(240\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(35\%\) is the rate,
          • +
        • +\(240\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 35\% \cdot 240 = 0.35 \cdot 240 = 84 }\)
        The amount of price increase was \({\$84.00}\text{,}\) so the new price is \({\$240.00}+{\$84.00}={\$324.00}\text{.}\) +
        So the watch’s markup price is \({\$324.00}\text{.}\)

        38.

        -
        -
        -
        -
        A watch’s wholesale price was \({\$220.00}\text{.}\) The retailer marked up the price by \(35\%\text{.}\) What’s the watch’s new price (markup price)?
        The watch’s markup price is .
        -
        -
        Answer.
        \(\$297.00\)
        Explanation.
        -
        First, we need to find the amount of increase in price. It’s given that the watch’s price was marked up by \(35\%\) of its original price, \({\$220.00}\text{.}\) -
        The problem can be boiled down to this question: What is \(35\%\) of \(220\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(35\%\) of \(220\) is \(x\text{,}\) so “\(35\) out of \(100\)” corresponds to “\(x\) out of \(220\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{35}{100} \amp = \frac{x}{220} \\ -100x \amp = 35 \cdot 220 \\ -100x \amp = 7700 \\ -\frac{100x}{100} \amp = \frac{7700}{100} \\ -x \amp = 77 -\end{aligned} -}\)
        The amount of price increase was \({\$77.00}\text{,}\) so the new price is \({\$220.00}+{\$77.00}={\$297.00}\text{.}\) -
        So the watch’s markup price is \({\$297.00}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(35\%\) of \(220\text{?}\) Assume \(x\) is \(35\%\) of \(220\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        A watch’s wholesale price was \({\$280.00}\text{.}\) The retailer marked up the price by \(30\%\text{.}\) What’s the watch’s new price (markup price)?
        The watch’s markup price is .
        +
        +
        Answer.
        \(\$364.00\)
        Explanation.
        +
        First, we need to find the amount of increase in price. It’s given that the watch’s price was marked up by \(30\%\) of its original price, \({\$280.00}\text{.}\) +
        The problem can be boiled down to this question: What is \(30\%\) of \(280\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(30\%\) of \(280\) is \(x\text{,}\) so “\(30\) out of \(100\)” corresponds to “\(x\) out of \(280\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{30}{100} \amp = \frac{x}{280} \\ +100x \amp = 30 \cdot 280 \\ +100x \amp = 8400 \\ +\frac{100x}{100} \amp = \frac{8400}{100} \\ +x \amp = 84 +\end{aligned} +}\)
        The amount of price increase was \({\$84.00}\text{,}\) so the new price is \({\$280.00}+{\$84.00}={\$364.00}\text{.}\) +
        So the watch’s markup price is \({\$364.00}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(30\%\) of \(280\text{?}\) Assume \(x\) is \(30\%\) of \(280\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.35 \cdot 220 \\ -\amp = 77 -\end{aligned} -}\)
        The amount of price increase was \({\$77.00}\text{,}\) so the new price is \({\$220.00}+{\$77.00}={\$297.00}\text{.}\) -
        So the watch’s markup price is \({\$297.00}\text{.}\) -
        Method 3
        In the sentence “What is \(35\%\) of \(220\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(35\%\) is the rate,
          • -
        • -\(220\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 35\% \cdot 220 = 0.35 \cdot 220 = 77 }\)
        The amount of price increase was \({\$77.00}\text{,}\) so the new price is \({\$220.00}+{\$77.00}={\$297.00}\text{.}\) -
        So the watch’s markup price is \({\$297.00}\text{.}\) +x \amp = 0.3 \cdot 280 \\ +\amp = 84 +\end{aligned} +}\)
        The amount of price increase was \({\$84.00}\text{,}\) so the new price is \({\$280.00}+{\$84.00}={\$364.00}\text{.}\) +
        So the watch’s markup price is \({\$364.00}\text{.}\) +
        Method 3
        In the sentence “What is \(30\%\) of \(280\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(30\%\) is the rate,
          • +
        • +\(280\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 30\% \cdot 280 = 0.3 \cdot 280 = 84 }\)
        The amount of price increase was \({\$84.00}\text{,}\) so the new price is \({\$280.00}+{\$84.00}={\$364.00}\text{.}\) +
        So the watch’s markup price is \({\$364.00}\text{.}\)

        39.

        -
        -
        +
        +
        -
        In the past few seasons’ basketball games, Connor attempted \(220\) free throws, and made \(66\) of them. What percent of free throws did Connor make?
        Connor made of free throws in the past few seasons.
        +
        In the past few seasons’ basketball games, Michele attempted \(130\) free throws, and made \(65\) of them. What percent of free throws did Michele make?
        Michele made of free throws in the past few seasons.
        -
        Answer.
        \(30\%\)
        Explanation.
        -
        This problem can be boiled down to this question: \(66\) is what percent of \(220\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(66\) is \(x\%\) of \(220\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(66\) out of \(220\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{66}{220} \\ -220x \amp = 66 \cdot 100 \\ -220x \amp = 6600 \\ -\frac{220x}{220} \amp = \frac{6600}{220} \\ -x \amp = 30 +
        Answer.
        \(50\%\)
        Explanation.
        +
        This problem can be boiled down to this question: \(65\) is what percent of \(130\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(65\) is \(x\%\) of \(130\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(65\) out of \(130\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{65}{130} \\ +130x \amp = 65 \cdot 100 \\ +130x \amp = 6500 \\ +\frac{130x}{130} \amp = \frac{6500}{130} \\ +x \amp = 50 \end{aligned} -}\)
        Connor made \(30\%\) of free throws in the past few seasons.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(66\) is \(x\) (as a percent) of \(220\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -66 \amp = x \cdot 220 \\ -\frac{66}{220} \amp = \frac{x \cdot 220}{220} \\ -0.3 \amp = x\\ -x \amp = 30\% +}\)
        Michele made \(50\%\) of free throws in the past few seasons.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(65\) is \(x\) (as a percent) of \(130\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +65 \amp = x \cdot 130 \\ +\frac{65}{130} \amp = \frac{x \cdot 130}{130} \\ +0.5 \amp = x\\ +x \amp = 50\% \end{aligned} -}\)
        Connor made \(30\%\) of free throws in the past few seasons.
        Method 3
        In the sentence “\(66\) is what percent of \(220\text{,}\)”
          -
        • -\(66\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(220\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{66}{220} = 0.3 = 30\% }\)
        Connor made \(30\%\) of free throws in the past few seasons.
        +}\)
        Michele made \(50\%\) of free throws in the past few seasons.
        Method 3
        In the sentence “\(65\) is what percent of \(130\text{,}\)”
          +
        • +\(65\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(130\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{65}{130} = 0.5 = 50\% }\)
        Michele made \(50\%\) of free throws in the past few seasons.

        40.

        -
        -
        +
        +
        -
        In the past few seasons’ basketball games, Stephen attempted \(470\) free throws, and made \(188\) of them. What percent of free throws did Stephen make?
        Stephen made of free throws in the past few seasons.
        +
        In the past few seasons’ basketball games, Grant attempted \(390\) free throws, and made \(234\) of them. What percent of free throws did Grant make?
        Grant made of free throws in the past few seasons.
        -
        Answer.
        \(40\%\)
        Explanation.
        -
        This problem can be boiled down to this question: \(188\) is what percent of \(470\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(188\) is \(x\%\) of \(470\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(188\) out of \(470\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{188}{470} \\ -470x \amp = 188 \cdot 100 \\ -470x \amp = 18800 \\ -\frac{470x}{470} \amp = \frac{18800}{470} \\ -x \amp = 40 +
        Answer.
        \(60\%\)
        Explanation.
        +
        This problem can be boiled down to this question: \(234\) is what percent of \(390\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(234\) is \(x\%\) of \(390\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(234\) out of \(390\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{234}{390} \\ +390x \amp = 234 \cdot 100 \\ +390x \amp = 23400 \\ +\frac{390x}{390} \amp = \frac{23400}{390} \\ +x \amp = 60 \end{aligned} -}\)
        Stephen made \(40\%\) of free throws in the past few seasons.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(188\) is \(x\) (as a percent) of \(470\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -188 \amp = x \cdot 470 \\ -\frac{188}{470} \amp = \frac{x \cdot 470}{470} \\ -0.4 \amp = x\\ -x \amp = 40\% +}\)
        Grant made \(60\%\) of free throws in the past few seasons.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(234\) is \(x\) (as a percent) of \(390\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +234 \amp = x \cdot 390 \\ +\frac{234}{390} \amp = \frac{x \cdot 390}{390} \\ +0.6 \amp = x\\ +x \amp = 60\% \end{aligned} -}\)
        Stephen made \(40\%\) of free throws in the past few seasons.
        Method 3
        In the sentence “\(188\) is what percent of \(470\text{,}\)”
          -
        • -\(188\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(470\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{188}{470} = 0.4 = 40\% }\)
        Stephen made \(40\%\) of free throws in the past few seasons.
        +}\)
        Grant made \(60\%\) of free throws in the past few seasons.
        Method 3
        In the sentence “\(234\) is what percent of \(390\text{,}\)”
          +
        • +\(234\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(390\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{234}{390} = 0.6 = 60\% }\)
        Grant made \(60\%\) of free throws in the past few seasons.

        41.

        -
        -
        -
        -
        A painting is on sale at \({\$440.00}\text{.}\) Its original price was \({\$550.00}\text{.}\) What percentage is this off its original price?
        The painting was off its original price.
        -
        -
        Answer.
        \(20\%\)
        Explanation.
        -
        The price changed from \({\$550.00}\) to \({\$440.00}\text{,}\) implying the markdown was \({\$550.00} - {\$440.00} = {\$110.00}\text{.}\) -
        Now this problem can be boiled down to this question: \(110\) is what percent of \(550\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(110\) is \(x\%\) of \(550\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(110\) out of \(550\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{110}{550} \\ -550x \amp = 110 \cdot 100 \\ -550x \amp = 11000 \\ -\frac{550x}{550} \amp = \frac{11000}{550} \\ -x \amp = 20 +
        +
        +
        +
        A painting is on sale at \({\$552.50}\text{.}\) Its original price was \({\$650.00}\text{.}\) What percentage is this off its original price?
        The painting was off its original price.
        +
        +
        Answer.
        \(15\%\)
        Explanation.
        +
        The price changed from \({\$650.00}\) to \({\$552.50}\text{,}\) implying the markdown was \({\$650.00} - {\$552.50} = {\$97.50}\text{.}\) +
        Now this problem can be boiled down to this question: \(97.5\) is what percent of \(650\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(97.5\) is \(x\%\) of \(650\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(97.5\) out of \(650\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{97.5}{650} \\ +650x \amp = 97.5 \cdot 100 \\ +650x \amp = 9750 \\ +\frac{650x}{650} \amp = \frac{9750}{650} \\ +x \amp = 15 \end{aligned} -}\)
        The painting was \(20\%\) off its original price.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(110\) is \(x\) (as a percent) of \(550\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -110 \amp = x \cdot 550 \\ -\frac{110}{550} \amp = \frac{x \cdot 550}{550} \\ -0.2 \amp = x\\ -x \amp = 20\% +}\)
        The painting was \(15\%\) off its original price.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(97.5\) is \(x\) (as a percent) of \(650\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +97.5 \amp = x \cdot 650 \\ +\frac{97.5}{650} \amp = \frac{x \cdot 650}{650} \\ +0.15 \amp = x\\ +x \amp = 15\% \end{aligned} -}\)
        The painting was \(20\%\) off its original price.
        Method 3
        In the sentence “\(110\) is what percent of \(550\text{,}\)”
          -
        • -\(110\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(550\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{110}{550} = 0.2 = 20\% }\)
        The painting was \(20\%\) off its original price.
        +}\)
        The painting was \(15\%\) off its original price.
        Method 3
        In the sentence “\(97.5\) is what percent of \(650\text{,}\)”
          +
        • +\(97.5\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(650\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{97.5}{650} = 0.15 = 15\% }\)
        The painting was \(15\%\) off its original price.

        42.

        -
        -
        +
        +
        -
        A painting is on sale at \({\$510.00}\text{.}\) Its original price was \({\$600.00}\text{.}\) What percentage is this off its original price?
        The painting was off its original price.
        +
        A painting is on sale at \({\$630.00}\text{.}\) Its original price was \({\$700.00}\text{.}\) What percentage is this off its original price?
        The painting was off its original price.
        -
        Answer.
        \(15\%\)
        Explanation.
        -
        The price changed from \({\$600.00}\) to \({\$510.00}\text{,}\) implying the markdown was \({\$600.00} - {\$510.00} = {\$90.00}\text{.}\) -
        Now this problem can be boiled down to this question: \(90\) is what percent of \(600\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(90\) is \(x\%\) of \(600\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(90\) out of \(600\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{90}{600} \\ -600x \amp = 90 \cdot 100 \\ -600x \amp = 9000 \\ -\frac{600x}{600} \amp = \frac{9000}{600} \\ -x \amp = 15 +
        Answer.
        \(10\%\)
        Explanation.
        +
        The price changed from \({\$700.00}\) to \({\$630.00}\text{,}\) implying the markdown was \({\$700.00} - {\$630.00} = {\$70.00}\text{.}\) +
        Now this problem can be boiled down to this question: \(70\) is what percent of \(700\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(70\) is \(x\%\) of \(700\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(70\) out of \(700\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{70}{700} \\ +700x \amp = 70 \cdot 100 \\ +700x \amp = 7000 \\ +\frac{700x}{700} \amp = \frac{7000}{700} \\ +x \amp = 10 \end{aligned} -}\)
        The painting was \(15\%\) off its original price.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(90\) is \(x\) (as a percent) of \(600\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -90 \amp = x \cdot 600 \\ -\frac{90}{600} \amp = \frac{x \cdot 600}{600} \\ -0.15 \amp = x\\ -x \amp = 15\% +}\)
        The painting was \(10\%\) off its original price.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(70\) is \(x\) (as a percent) of \(700\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +70 \amp = x \cdot 700 \\ +\frac{70}{700} \amp = \frac{x \cdot 700}{700} \\ +0.1 \amp = x\\ +x \amp = 10\% \end{aligned} -}\)
        The painting was \(15\%\) off its original price.
        Method 3
        In the sentence “\(90\) is what percent of \(600\text{,}\)”
          -
        • -\(90\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(600\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{90}{600} = 0.15 = 15\% }\)
        The painting was \(15\%\) off its original price.
        +}\)
        The painting was \(10\%\) off its original price.
        Method 3
        In the sentence “\(70\) is what percent of \(700\text{,}\)”
          +
        • +\(70\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(700\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{70}{700} = 0.1 = 10\% }\)
        The painting was \(10\%\) off its original price.

        43.

        -
        -
        -
        -
        The pie chart represents a collector’s collection of signatures from various artists.
        If the collector has a total of \(1350\) signatures, there are signatures by Sting.
        -
        -
        Answer.
        \(270\)
        Explanation.
        -
        We can use subtraction to find the percent of Sting signatures:
        \(\displaystyle{\begin{aligned}[t] -\amp \phantom{{}=}1-36\%-24\%-10\%-10\% \\ -\amp =1-0.36-0.24-0.1-0.1 \\ -\amp =0.2 \\ -\amp = 20\% -\end{aligned} -}\)
        So there are \(20\%\) Sting signatures. Now this problem can be boiled down to this question: What is \(20\%\) of \(1350\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(20\%\) of \(1350\) is \(x\text{,}\) so “\(20\) out of \(100\)” corresponds to “\(x\) out of \(1350\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{20}{100} \amp = \frac{x}{1350} \\ -100x \amp = 20 \cdot 1350 \\ -100x \amp = 27000 \\ -\frac{100x}{100} \amp = \frac{27000}{100} \\ -x \amp = 270 -\end{aligned} -}\)
        There are \(270\) signatures by Sting in this collection.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(20\%\) of \(1350\text{?}\) Assume \(x\) is \(20\%\) of \(1350\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        The pie chart represents a collector’s collection of signatures from various artists.
        If the collector has a total of \(1750\) signatures, there are signatures by Sting.
        +
        +
        Answer.
        \(245\)
        Explanation.
        +
        We can use subtraction to find the percent of Sting signatures:
        \(\displaystyle{\begin{aligned}[t] +\amp \phantom{{}=}1-32\%-22\%-16\%-16\% \\ +\amp =1-0.32-0.22-0.16-0.16 \\ +\amp =0.14 \\ +\amp = 14\% +\end{aligned} +}\)
        So there are \(14\%\) Sting signatures. Now this problem can be boiled down to this question: What is \(14\%\) of \(1750\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(14\%\) of \(1750\) is \(x\text{,}\) so “\(14\) out of \(100\)” corresponds to “\(x\) out of \(1750\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{14}{100} \amp = \frac{x}{1750} \\ +100x \amp = 14 \cdot 1750 \\ +100x \amp = 24500 \\ +\frac{100x}{100} \amp = \frac{24500}{100} \\ +x \amp = 245 +\end{aligned} +}\)
        There are \(245\) signatures by Sting in this collection.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(14\%\) of \(1750\text{?}\) Assume \(x\) is \(14\%\) of \(1750\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.2 \cdot 1350 \\ -\amp = 270 +x \amp = 0.14 \cdot 1750 \\ +\amp = 245 \end{aligned} -}\)
        There are \(270\) signatures by Sting in this collection.
        Method 3
        In the sentence “What is \(20\%\) of \(1350\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(20\%\) is the rate,
          • -
        • -\(1350\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 20\% \cdot 1350 = 0.2 \cdot 1350 = 270 }\)
        There are \(270\) signatures by Sting in this collection.
        +}\)
        There are \(245\) signatures by Sting in this collection.
        Method 3
        In the sentence “What is \(14\%\) of \(1750\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(14\%\) is the rate,
          • +
        • +\(1750\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 14\% \cdot 1750 = 0.14 \cdot 1750 = 245 }\)
        There are \(245\) signatures by Sting in this collection.

        44.

        -
        -
        -
        -
        The pie chart represents a collector’s collection of signatures from various artists.
        If the collector has a total of \(1550\) signatures, there are signatures by Sting.
        -
        -
        Answer.
        \(434\)
        Explanation.
        -
        We can use subtraction to find the percent of Sting signatures:
        \(\displaystyle{\begin{aligned}[t] -\amp \phantom{{}=}1-28\%-16\%-18\%-10\% \\ -\amp =1-0.28-0.16-0.18-0.1 \\ -\amp =0.28 \\ -\amp = 28\% -\end{aligned} -}\)
        So there are \(28\%\) Sting signatures. Now this problem can be boiled down to this question: What is \(28\%\) of \(1550\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(28\%\) of \(1550\) is \(x\text{,}\) so “\(28\) out of \(100\)” corresponds to “\(x\) out of \(1550\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{28}{100} \amp = \frac{x}{1550} \\ -100x \amp = 28 \cdot 1550 \\ -100x \amp = 43400 \\ -\frac{100x}{100} \amp = \frac{43400}{100} \\ -x \amp = 434 -\end{aligned} -}\)
        There are \(434\) signatures by Sting in this collection.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(28\%\) of \(1550\text{?}\) Assume \(x\) is \(28\%\) of \(1550\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        The pie chart represents a collector’s collection of signatures from various artists.
        If the collector has a total of \(1950\) signatures, there are signatures by Sting.
        +
        +
        Answer.
        \(429\)
        Explanation.
        +
        We can use subtraction to find the percent of Sting signatures:
        \(\displaystyle{\begin{aligned}[t] +\amp \phantom{{}=}1-24\%-26\%-12\%-16\% \\ +\amp =1-0.24-0.26-0.12-0.16 \\ +\amp =0.22 \\ +\amp = 22\% +\end{aligned} +}\)
        So there are \(22\%\) Sting signatures. Now this problem can be boiled down to this question: What is \(22\%\) of \(1950\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(22\%\) of \(1950\) is \(x\text{,}\) so “\(22\) out of \(100\)” corresponds to “\(x\) out of \(1950\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{22}{100} \amp = \frac{x}{1950} \\ +100x \amp = 22 \cdot 1950 \\ +100x \amp = 42900 \\ +\frac{100x}{100} \amp = \frac{42900}{100} \\ +x \amp = 429 +\end{aligned} +}\)
        There are \(429\) signatures by Sting in this collection.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: What is \(22\%\) of \(1950\text{?}\) Assume \(x\) is \(22\%\) of \(1950\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -x \amp = 0.28 \cdot 1550 \\ -\amp = 434 +x \amp = 0.22 \cdot 1950 \\ +\amp = 429 \end{aligned} -}\)
        There are \(434\) signatures by Sting in this collection.
        Method 3
        In the sentence “What is \(28\%\) of \(1550\text{,}\)”
          -
        • “what” is the percentage,
          • -
        • -\(28\%\) is the rate,
          • -
        • -\(1550\) is the base (following the word “of”).
          • -
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 28\% \cdot 1550 = 0.28 \cdot 1550 = 434 }\)
        There are \(434\) signatures by Sting in this collection.
        +}\)
        There are \(429\) signatures by Sting in this collection.
        Method 3
        In the sentence “What is \(22\%\) of \(1950\text{,}\)”
          +
        • “what” is the percentage,
          • +
        • +\(22\%\) is the rate,
          • +
        • +\(1950\) is the base (following the word “of”).
          • +
        By the formula \(\text{percentage} = \text{rate} \cdot \text{base}\text{,}\) we do a multiplication to solve the problem:
        \(\displaystyle{ \text{percentage } = \text{rate} \cdot \text{base} = 22\% \cdot 1950 = 0.22 \cdot 1950 = 429 }\)
        There are \(429\) signatures by Sting in this collection.

        45.

        -
        -
        +
        +
        -
        In the last election, \(38\%\) of a county’s residents, or \(16682\) people, turned out to vote. How many residents live in this county?
        This county has residents.
        +
        In the last election, \(69\%\) of a county’s residents, or \(8418\) people, turned out to vote. How many residents live in this county?
        This county has residents.
        -
        Answer.
        \(43900\)
        Explanation.
        -
        This problem can be boiled down to this question: \(16682\) is \(38\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(16682\) is \(38\%\) of \(x\text{,}\) so “\(38\) out of \(100\)” corresponds to “\(16682\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{38}{100} \amp = \frac{16682}{x} \\ -38x \amp = 100 \cdot 16682 \\ -38x \amp = 1668200 \\ -\frac{38x}{38} \amp = \frac{1668200}{38} \\ -x \amp = 43900 +
        Answer.
        \(12200\)
        Explanation.
        +
        This problem can be boiled down to this question: \(8418\) is \(69\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(8418\) is \(69\%\) of \(x\text{,}\) so “\(69\) out of \(100\)” corresponds to “\(8418\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{69}{100} \amp = \frac{8418}{x} \\ +69x \amp = 100 \cdot 8418 \\ +69x \amp = 841800 \\ +\frac{69x}{69} \amp = \frac{841800}{69} \\ +x \amp = 12200 \end{aligned} -}\)
        This county has \(43900\) residents.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(16682\) is \(38\%\) of what? Assume \(16682\) is \(38\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        This county has \(12200\) residents.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(8418\) is \(69\%\) of what? Assume \(8418\) is \(69\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -16682 \amp = 0.38 \cdot x \\ -\frac{16682}{0.38} \amp = \frac{0.38x}{0.38} \\ -43900 \amp = x +8418 \amp = 0.69 \cdot x \\ +\frac{8418}{0.69} \amp = \frac{0.69x}{0.69} \\ +12200 \amp = x \end{aligned} -}\)
        This county has \(43900\) residents.
        Method 3
        In the sentence “\(16682\) is \(38\%\) of what,”
          -
        • -\(16682\) is the percentage,
          • -
        • -\(38\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{16682}{38\%} = \frac{16682}{0.38} = 43900 }\)
        This county has \(43900\) residents.
        +}\)
        This county has \(12200\) residents.
        Method 3
        In the sentence “\(8418\) is \(69\%\) of what,”
          +
        • +\(8418\) is the percentage,
          • +
        • +\(69\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{8418}{69\%} = \frac{8418}{0.69} = 12200 }\)
        This county has \(12200\) residents.

        46.

        -
        -
        +
        +
        -
        In the last election, \(64\%\) of a county’s residents, or \(30976\) people, turned out to vote. How many residents live in this county?
        This county has residents.
        +
        In the last election, \(55\%\) of a county’s residents, or \(9130\) people, turned out to vote. How many residents live in this county?
        This county has residents.
        -
        Answer.
        \(48400\)
        Explanation.
        -
        This problem can be boiled down to this question: \(30976\) is \(64\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(30976\) is \(64\%\) of \(x\text{,}\) so “\(64\) out of \(100\)” corresponds to “\(30976\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{64}{100} \amp = \frac{30976}{x} \\ -64x \amp = 100 \cdot 30976 \\ -64x \amp = 3097600 \\ -\frac{64x}{64} \amp = \frac{3097600}{64} \\ -x \amp = 48400 +
        Answer.
        \(16600\)
        Explanation.
        +
        This problem can be boiled down to this question: \(9130\) is \(55\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(9130\) is \(55\%\) of \(x\text{,}\) so “\(55\) out of \(100\)” corresponds to “\(9130\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{55}{100} \amp = \frac{9130}{x} \\ +55x \amp = 100 \cdot 9130 \\ +55x \amp = 913000 \\ +\frac{55x}{55} \amp = \frac{913000}{55} \\ +x \amp = 16600 \end{aligned} -}\)
        This county has \(48400\) residents.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(30976\) is \(64\%\) of what? Assume \(30976\) is \(64\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        This county has \(16600\) residents.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(9130\) is \(55\%\) of what? Assume \(9130\) is \(55\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -30976 \amp = 0.64 \cdot x \\ -\frac{30976}{0.64} \amp = \frac{0.64x}{0.64} \\ -48400 \amp = x +9130 \amp = 0.55 \cdot x \\ +\frac{9130}{0.55} \amp = \frac{0.55x}{0.55} \\ +16600 \amp = x \end{aligned} -}\)
        This county has \(48400\) residents.
        Method 3
        In the sentence “\(30976\) is \(64\%\) of what,”
          -
        • -\(30976\) is the percentage,
          • -
        • -\(64\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{30976}{64\%} = \frac{30976}{0.64} = 48400 }\)
        This county has \(48400\) residents.
        +}\)
        This county has \(16600\) residents.
        Method 3
        In the sentence “\(9130\) is \(55\%\) of what,”
          +
        • +\(9130\) is the percentage,
          • +
        • +\(55\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{9130}{55\%} = \frac{9130}{0.55} = 16600 }\)
        This county has \(16600\) residents.

        47.

        -
        -
        -
        -
        -\(42.84\) grams of pure alcohol was used to produce a bottle of \(20.4\%\) alcohol solution. What is the weight of the solution in grams?
        The alcohol solution weighs .
        -
        -
        Answer.
        \(210\ {\rm g}\)
        Explanation.
        -
        This problem can be boiled down to this question: \(42.84\) is \(20.4\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(42.84\) is \(20.4\%\) of \(x\text{,}\) so “\(20.4\) out of \(100\)” corresponds to “\(42.84\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{20.4}{100} \amp = \frac{42.84}{x} \\ -20.4x \amp = 100 \cdot 42.84 \\ -20.4x \amp = 4284 \\ -\frac{20.4x}{20.4} \amp = \frac{4284}{20.4} \\ -x \amp = 210 -\end{aligned} -}\)
        The alcohol solution weighs \({210\ {\rm g}}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(42.84\) is \(20.4\%\) of what? Assume \(42.84\) is \(20.4\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        +\(40.5\) grams of pure alcohol was used to produce a bottle of \(16.2\%\) alcohol solution. What is the weight of the solution in grams?
        The alcohol solution weighs .
        +
        +
        Answer.
        \(250\ {\rm g}\)
        Explanation.
        +
        This problem can be boiled down to this question: \(40.5\) is \(16.2\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(40.5\) is \(16.2\%\) of \(x\text{,}\) so “\(16.2\) out of \(100\)” corresponds to “\(40.5\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{16.2}{100} \amp = \frac{40.5}{x} \\ +16.2x \amp = 100 \cdot 40.5 \\ +16.2x \amp = 4050 \\ +\frac{16.2x}{16.2} \amp = \frac{4050}{16.2} \\ +x \amp = 250 +\end{aligned} +}\)
        The alcohol solution weighs \({250\ {\rm g}}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(40.5\) is \(16.2\%\) of what? Assume \(40.5\) is \(16.2\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -42.84 \amp = 0.204 \cdot x \\ -\frac{42.84}{0.204} \amp = \frac{0.204x}{0.204} \\ -210 \amp = x +40.5 \amp = 0.162 \cdot x \\ +\frac{40.5}{0.162} \amp = \frac{0.162x}{0.162} \\ +250 \amp = x \end{aligned} -}\)
        The alcohol solution weighs \({210\ {\rm g}}\text{.}\) -
        Method 3
        In the sentence “\(42.84\) is \(20.4\%\) of what,”
          -
        • -\(42.84\) is the percentage,
          • -
        • -\(20.4\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{42.84}{20.4\%} = \frac{42.84}{0.204} = 210 }\)
        The alcohol solution weighs \({210\ {\rm g}}\text{.}\) +}\)
        The alcohol solution weighs \({250\ {\rm g}}\text{.}\) +
        Method 3
        In the sentence “\(40.5\) is \(16.2\%\) of what,”
          +
        • +\(40.5\) is the percentage,
          • +
        • +\(16.2\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{40.5}{16.2\%} = \frac{40.5}{0.162} = 250 }\)
        The alcohol solution weighs \({250\ {\rm g}}\text{.}\)

        48.

        -
        -
        -
        -
        -\(31.51\) grams of pure alcohol was used to produce a bottle of \(13.7\%\) alcohol solution. What is the weight of the solution in grams?
        The alcohol solution weighs .
        -
        -
        Answer.
        \(230\ {\rm g}\)
        Explanation.
        -
        This problem can be boiled down to this question: \(31.51\) is \(13.7\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(31.51\) is \(13.7\%\) of \(x\text{,}\) so “\(13.7\) out of \(100\)” corresponds to “\(31.51\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{13.7}{100} \amp = \frac{31.51}{x} \\ -13.7x \amp = 100 \cdot 31.51 \\ -13.7x \amp = 3151 \\ -\frac{13.7x}{13.7} \amp = \frac{3151}{13.7} \\ -x \amp = 230 -\end{aligned} -}\)
        The alcohol solution weighs \({230\ {\rm g}}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(31.51\) is \(13.7\%\) of what? Assume \(31.51\) is \(13.7\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +
        +
        +
        +
        +\(82.88\) grams of pure alcohol was used to produce a bottle of \(29.6\%\) alcohol solution. What is the weight of the solution in grams?
        The alcohol solution weighs .
        +
        +
        Answer.
        \(280\ {\rm g}\)
        Explanation.
        +
        This problem can be boiled down to this question: \(82.88\) is \(29.6\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(82.88\) is \(29.6\%\) of \(x\text{,}\) so “\(29.6\) out of \(100\)” corresponds to “\(82.88\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{29.6}{100} \amp = \frac{82.88}{x} \\ +29.6x \amp = 100 \cdot 82.88 \\ +29.6x \amp = 8288 \\ +\frac{29.6x}{29.6} \amp = \frac{8288}{29.6} \\ +x \amp = 280 +\end{aligned} +}\)
        The alcohol solution weighs \({280\ {\rm g}}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(82.88\) is \(29.6\%\) of what? Assume \(82.88\) is \(29.6\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -31.51 \amp = 0.137 \cdot x \\ -\frac{31.51}{0.137} \amp = \frac{0.137x}{0.137} \\ -230 \amp = x +82.88 \amp = 0.296 \cdot x \\ +\frac{82.88}{0.296} \amp = \frac{0.296x}{0.296} \\ +280 \amp = x \end{aligned} -}\)
        The alcohol solution weighs \({230\ {\rm g}}\text{.}\) -
        Method 3
        In the sentence “\(31.51\) is \(13.7\%\) of what,”
          -
        • -\(31.51\) is the percentage,
          • -
        • -\(13.7\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{31.51}{13.7\%} = \frac{31.51}{0.137} = 230 }\)
        The alcohol solution weighs \({230\ {\rm g}}\text{.}\) +}\)
        The alcohol solution weighs \({280\ {\rm g}}\text{.}\) +
        Method 3
        In the sentence “\(82.88\) is \(29.6\%\) of what,”
          +
        • +\(82.88\) is the percentage,
          • +
        • +\(29.6\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{82.88}{29.6\%} = \frac{82.88}{0.296} = 280 }\)
        The alcohol solution weighs \({280\ {\rm g}}\text{.}\)

        49.

        -
        -
        +
        +
        -
        Hayden paid a dinner and left \(18\%\text{,}\) or \({\$5.22}\text{,}\) in tips. How much was the original bill (without counting the tip)?
        The original bill (not including the tip) was .
        +
        Shane paid a dinner and left \(16\%\text{,}\) or \({\$5.60}\text{,}\) in tips. How much was the original bill (without counting the tip)?
        The original bill (not including the tip) was .
        -
        Answer.
        \(\$29.00\)
        Explanation.
        -
        This problem can be boiled down to this question: \(5.22\) is \(18\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(5.22\) is \(18\%\) of \(x\text{,}\) so “\(18\) out of \(100\)” corresponds to “\(5.22\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{18}{100} \amp = \frac{5.22}{x} \\ -18x \amp = 100 \cdot 5.22 \\ -18x \amp = 522 \\ -\frac{18x}{18} \amp = \frac{522}{18} \\ -x \amp = 29 +
        Answer.
        \(\$35.00\)
        Explanation.
        +
        This problem can be boiled down to this question: \(5.6\) is \(16\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(5.6\) is \(16\%\) of \(x\text{,}\) so “\(16\) out of \(100\)” corresponds to “\(5.6\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{16}{100} \amp = \frac{5.6}{x} \\ +16x \amp = 100 \cdot 5.6 \\ +16x \amp = 560 \\ +\frac{16x}{16} \amp = \frac{560}{16} \\ +x \amp = 35 \end{aligned} -}\)
        The original bill (not including the tip) was \({\$29.00}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(5.22\) is \(18\%\) of what? Assume \(5.22\) is \(18\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        The original bill (not including the tip) was \({\$35.00}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(5.6\) is \(16\%\) of what? Assume \(5.6\) is \(16\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -5.22 \amp = 0.18 \cdot x \\ -\frac{5.22}{0.18} \amp = \frac{0.18x}{0.18} \\ -29 \amp = x -\end{aligned} -}\)
        The original bill (not including the tip) was \({\$29.00}\text{.}\) -
        Method 3
        In the sentence “\(5.22\) is \(18\%\) of what,”
          -
        • -\(5.22\) is the percentage,
          • -
        • -\(18\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{5.22}{18\%} = \frac{5.22}{0.18} = 29 }\)
        The original bill (not including the tip) was \({\$29.00}\text{.}\) +5.6 \amp = 0.16 \cdot x \\ +\frac{5.6}{0.16} \amp = \frac{0.16x}{0.16} \\ +35 \amp = x +\end{aligned} +}\)
        The original bill (not including the tip) was \({\$35.00}\text{.}\) +
        Method 3
        In the sentence “\(5.6\) is \(16\%\) of what,”
          +
        • +\(5.6\) is the percentage,
          • +
        • +\(16\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{5.6}{16\%} = \frac{5.6}{0.16} = 35 }\)
        The original bill (not including the tip) was \({\$35.00}\text{.}\)

        50.

        -
        -
        +
        +
        -
        Carmen paid a dinner and left \(15\%\text{,}\) or \({\$4.80}\text{,}\) in tips. How much was the original bill (without counting the tip)?
        The original bill (not including the tip) was .
        +
        Kenji paid a dinner and left \(12\%\text{,}\) or \({\$4.56}\text{,}\) in tips. How much was the original bill (without counting the tip)?
        The original bill (not including the tip) was .
        -
        Answer.
        \(\$32.00\)
        Explanation.
        -
        This problem can be boiled down to this question: \(4.8\) is \(15\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(4.8\) is \(15\%\) of \(x\text{,}\) so “\(15\) out of \(100\)” corresponds to “\(4.8\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{15}{100} \amp = \frac{4.8}{x} \\ -15x \amp = 100 \cdot 4.8 \\ -15x \amp = 480 \\ -\frac{15x}{15} \amp = \frac{480}{15} \\ -x \amp = 32 +
        Answer.
        \(\$38.00\)
        Explanation.
        +
        This problem can be boiled down to this question: \(4.56\) is \(12\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(4.56\) is \(12\%\) of \(x\text{,}\) so “\(12\) out of \(100\)” corresponds to “\(4.56\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{12}{100} \amp = \frac{4.56}{x} \\ +12x \amp = 100 \cdot 4.56 \\ +12x \amp = 456 \\ +\frac{12x}{12} \amp = \frac{456}{12} \\ +x \amp = 38 \end{aligned} -}\)
        The original bill (not including the tip) was \({\$32.00}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(4.8\) is \(15\%\) of what? Assume \(4.8\) is \(15\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        The original bill (not including the tip) was \({\$38.00}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(4.56\) is \(12\%\) of what? Assume \(4.56\) is \(12\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -4.8 \amp = 0.15 \cdot x \\ -\frac{4.8}{0.15} \amp = \frac{0.15x}{0.15} \\ -32 \amp = x -\end{aligned} -}\)
        The original bill (not including the tip) was \({\$32.00}\text{.}\) -
        Method 3
        In the sentence “\(4.8\) is \(15\%\) of what,”
          -
        • -\(4.8\) is the percentage,
          • -
        • -\(15\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{4.8}{15\%} = \frac{4.8}{0.15} = 32 }\)
        The original bill (not including the tip) was \({\$32.00}\text{.}\) +4.56 \amp = 0.12 \cdot x \\ +\frac{4.56}{0.12} \amp = \frac{0.12x}{0.12} \\ +38 \amp = x +\end{aligned} +}\)
        The original bill (not including the tip) was \({\$38.00}\text{.}\) +
        Method 3
        In the sentence “\(4.56\) is \(12\%\) of what,”
          +
        • +\(4.56\) is the percentage,
          • +
        • +\(12\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{4.56}{12\%} = \frac{4.56}{0.12} = 38 }\)
        The original bill (not including the tip) was \({\$38.00}\text{.}\)

        51.

        -
        -
        +
        +
        -
        Selena sells cars for a living. Each month, she earns \({\$2{,}000.00}\) of base pay, plus a certain percentage of commission from her sales.
        One month, Selena made \({\$60{,}500.00}\) in sales, and earned a total of \({\$3{,}258.40}\) in that month (including base pay and commission). What percent commission did Selena earn?
        Selena earned in commission.
        +
        Diane sells cars for a living. Each month, she earns \({\$1{,}400.00}\) of base pay, plus a certain percentage of commission from her sales.
        One month, Diane made \({\$68{,}800.00}\) in sales, and earned a total of \({\$5{,}417.92}\) in that month (including base pay and commission). What percent commission did Diane earn?
        Diane earned in commission.
        -
        Answer.
        \(2.08\%\)
        Explanation.
        -
        Selena’s pay is made up of base pay and commission. In that month, Selena earned a total of \({\$3{,}258.40}\text{,}\) with \({\$2{,}000.00}\) of base pay. This implies Selena earned \({\$3{,}258.40}-{\$2{,}000.00}={\$1{,}258.40}\) in commission, out of \({\$60{,}500.00}\) in sales.
        Now the problem can be boiled down to this question: \(1258.4\) is what percent of \(60500\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(1258.4\) is \(x\%\) of \(60500\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(1258.4\) out of \(60500\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{1258.4}{60500} \\ -60500x \amp = 1258.4 \cdot 100 \\ -60500x \amp = 125840 \\ -\frac{60500x}{60500} \amp = \frac{125840}{60500} \\ -x \amp = 2.08 +
        Answer.
        \(5.84\%\)
        Explanation.
        +
        Diane’s pay is made up of base pay and commission. In that month, Diane earned a total of \({\$5{,}417.92}\text{,}\) with \({\$1{,}400.00}\) of base pay. This implies Diane earned \({\$5{,}417.92}-{\$1{,}400.00}={\$4{,}017.92}\) in commission, out of \({\$68{,}800.00}\) in sales.
        Now the problem can be boiled down to this question: \(4017.92\) is what percent of \(68800\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(4017.92\) is \(x\%\) of \(68800\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(4017.92\) out of \(68800\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{4017.92}{68800} \\ +68800x \amp = 4017.92 \cdot 100 \\ +68800x \amp = 401792 \\ +\frac{68800x}{68800} \amp = \frac{401792}{68800} \\ +x \amp = 5.84 \end{aligned} -}\)
        Selena earned \(2.08\%\) in commission.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(1258.4\) is \(x\) (as a percent) of \(60500\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -1258.4 \amp = x \cdot 60500 \\ -\frac{1258.4}{60500} \amp = \frac{x \cdot 60500}{60500} \\ -0.0208 \amp = x\\ -x \amp = 2.08\% +}\)
        Diane earned \(5.84\%\) in commission.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(4017.92\) is \(x\) (as a percent) of \(68800\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +4017.92 \amp = x \cdot 68800 \\ +\frac{4017.92}{68800} \amp = \frac{x \cdot 68800}{68800} \\ +0.0584 \amp = x\\ +x \amp = 5.84\% \end{aligned} -}\)
        Selena earned \(2.08\%\) in commission.
        Method 3
        In the sentence “\(1258.4\) is what percent of \(60500\text{,}\)”
          -
        • -\(1258.4\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(60500\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{1258.4}{60500} = 0.0208 = 2.08\% }\)
        Selena earned \(2.08\%\) in commission.
        +}\)
        Diane earned \(5.84\%\) in commission.
        Method 3
        In the sentence “\(4017.92\) is what percent of \(68800\text{,}\)”
          +
        • +\(4017.92\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(68800\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{4017.92}{68800} = 0.0584 = 5.84\% }\)
        Diane earned \(5.84\%\) in commission.

        52.

        -
        -
        +
        +
        -
        Thanh sells cars for a living. Each month, he earns \({\$2{,}000.00}\) of base pay, plus a certain percentage of commission from his sales.
        One month, Thanh made \({\$64{,}900.00}\) in sales, and earned a total of \({\$5{,}543.54}\) in that month (including base pay and commission). What percent commission did Thanh earn?
        Thanh earned in commission.
        +
        Emiliano sells cars for a living. Each month, he earns \({\$1{,}400.00}\) of base pay, plus a certain percentage of commission from his sales.
        One month, Emiliano made \({\$73{,}300.00}\) in sales, and earned a total of \({\$4{,}485.93}\) in that month (including base pay and commission). What percent commission did Emiliano earn?
        Emiliano earned in commission.
        -
        Answer.
        \(5.46\%\)
        Explanation.
        -
        Thanh’s pay is made up of base pay and commission. In that month, Thanh earned a total of \({\$5{,}543.54}\text{,}\) with \({\$2{,}000.00}\) of base pay. This implies Thanh earned \({\$5{,}543.54}-{\$2{,}000.00}={\$3{,}543.54}\) in commission, out of \({\$64{,}900.00}\) in sales.
        Now the problem can be boiled down to this question: \(3543.54\) is what percent of \(64900\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(3543.54\) is \(x\%\) of \(64900\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(3543.54\) out of \(64900\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{3543.54}{64900} \\ -64900x \amp = 3543.54 \cdot 100 \\ -64900x \amp = 354354 \\ -\frac{64900x}{64900} \amp = \frac{354354}{64900} \\ -x \amp = 5.46 +
        Answer.
        \(4.21\%\)
        Explanation.
        +
        Emiliano’s pay is made up of base pay and commission. In that month, Emiliano earned a total of \({\$4{,}485.93}\text{,}\) with \({\$1{,}400.00}\) of base pay. This implies Emiliano earned \({\$4{,}485.93}-{\$1{,}400.00}={\$3{,}085.93}\) in commission, out of \({\$73{,}300.00}\) in sales.
        Now the problem can be boiled down to this question: \(3085.93\) is what percent of \(73300\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(3085.93\) is \(x\%\) of \(73300\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(3085.93\) out of \(73300\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{3085.93}{73300} \\ +73300x \amp = 3085.93 \cdot 100 \\ +73300x \amp = 308593 \\ +\frac{73300x}{73300} \amp = \frac{308593}{73300} \\ +x \amp = 4.21 \end{aligned} -}\)
        Thanh earned \(5.46\%\) in commission.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(3543.54\) is \(x\) (as a percent) of \(64900\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -3543.54 \amp = x \cdot 64900 \\ -\frac{3543.54}{64900} \amp = \frac{x \cdot 64900}{64900} \\ -0.0546 \amp = x\\ -x \amp = 5.46\% +}\)
        Emiliano earned \(4.21\%\) in commission.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(3085.93\) is \(x\) (as a percent) of \(73300\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +3085.93 \amp = x \cdot 73300 \\ +\frac{3085.93}{73300} \amp = \frac{x \cdot 73300}{73300} \\ +0.0421 \amp = x\\ +x \amp = 4.21\% \end{aligned} -}\)
        Thanh earned \(5.46\%\) in commission.
        Method 3
        In the sentence “\(3543.54\) is what percent of \(64900\text{,}\)”
          -
        • -\(3543.54\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(64900\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{3543.54}{64900} = 0.0546 = 5.46\% }\)
        Thanh earned \(5.46\%\) in commission.
        +}\)
        Emiliano earned \(4.21\%\) in commission.
        Method 3
        In the sentence “\(3085.93\) is what percent of \(73300\text{,}\)”
          +
        • +\(3085.93\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(73300\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{3085.93}{73300} = 0.0421 = 4.21\% }\)
        Emiliano earned \(4.21\%\) in commission.

        53.

        -
        -
        +
        +
        -
        The following is a nutrition fact label from a certain macaroni and cheese box.
        The highlighted row means each serving of macaroni and cheese in this box contains \({10.5\ {\rm g}}\) of fat, which is \(14\%\) of an average person’s daily intake of fat. What’s the recommended daily intake of fat for an average person?
        The recommended daily intake of fat for an average person is .
        +
        The following is a nutrition fact label from a certain macaroni and cheese box.
        The highlighted row means each serving of macaroni and cheese in this box contains \({9.6\ {\rm g}}\) of fat, which is \(12\%\) of an average person’s daily intake of fat. What’s the recommended daily intake of fat for an average person?
        The recommended daily intake of fat for an average person is .
        -
        Answer.
        \(75\ {\rm g}\)
        Explanation.
        -
        This problem can be boiled down to this question: \(10.5\) is \(14\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(10.5\) is \(14\%\) of \(x\text{,}\) so “\(14\) out of \(100\)” corresponds to “\(10.5\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{14}{100} \amp = \frac{10.5}{x} \\ -14x \amp = 100 \cdot 10.5 \\ -14x \amp = 1050 \\ -\frac{14x}{14} \amp = \frac{1050}{14} \\ -x \amp = 75 +
        Answer.
        \(80\ {\rm g}\)
        Explanation.
        +
        This problem can be boiled down to this question: \(9.6\) is \(12\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(9.6\) is \(12\%\) of \(x\text{,}\) so “\(12\) out of \(100\)” corresponds to “\(9.6\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{12}{100} \amp = \frac{9.6}{x} \\ +12x \amp = 100 \cdot 9.6 \\ +12x \amp = 960 \\ +\frac{12x}{12} \amp = \frac{960}{12} \\ +x \amp = 80 \end{aligned} -}\)
        The recommended daily intake of fat for an average person is \({75\ {\rm g}}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(10.5\) is \(14\%\) of what? Assume \(10.5\) is \(14\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        The recommended daily intake of fat for an average person is \({80\ {\rm g}}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(9.6\) is \(12\%\) of what? Assume \(9.6\) is \(12\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -10.5 \amp = 0.14 \cdot x \\ -\frac{10.5}{0.14} \amp = \frac{0.14x}{0.14} \\ -75 \amp = x -\end{aligned} -}\)
        The recommended daily intake of fat for an average person is \({75\ {\rm g}}\text{.}\) -
        Method 3
        In the sentence “\(10.5\) is \(14\%\) of what,”
          -
        • -\(10.5\) is the percentage,
          • -
        • -\(14\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{10.5}{14\%} = \frac{10.5}{0.14} = 75 }\)
        The recommended daily intake of fat for an average person is \({75\ {\rm g}}\text{.}\) +9.6 \amp = 0.12 \cdot x \\ +\frac{9.6}{0.12} \amp = \frac{0.12x}{0.12} \\ +80 \amp = x +\end{aligned} +}\)
        The recommended daily intake of fat for an average person is \({80\ {\rm g}}\text{.}\) +
        Method 3
        In the sentence “\(9.6\) is \(12\%\) of what,”
          +
        • +\(9.6\) is the percentage,
          • +
        • +\(12\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{9.6}{12\%} = \frac{9.6}{0.12} = 80 }\)
        The recommended daily intake of fat for an average person is \({80\ {\rm g}}\text{.}\)

        54.

        -
        -
        +
        +
        -
        The following is a nutrition fact label from a certain macaroni and cheese box.
        The highlighted row means each serving of macaroni and cheese in this box contains \({7.5\ {\rm g}}\) of fat, which is \(10\%\) of an average person’s daily intake of fat. What’s the recommended daily intake of fat for an average person?
        The recommended daily intake of fat for an average person is .
        +
        The following is a nutrition fact label from a certain macaroni and cheese box.
        The highlighted row means each serving of macaroni and cheese in this box contains \({10\ {\rm g}}\) of fat, which is \(20\%\) of an average person’s daily intake of fat. What’s the recommended daily intake of fat for an average person?
        The recommended daily intake of fat for an average person is .
        -
        Answer.
        \(75\ {\rm g}\)
        Explanation.
        -
        This problem can be boiled down to this question: \(7.5\) is \(10\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(7.5\) is \(10\%\) of \(x\text{,}\) so “\(10\) out of \(100\)” corresponds to “\(7.5\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{10}{100} \amp = \frac{7.5}{x} \\ -10x \amp = 100 \cdot 7.5 \\ -10x \amp = 750 \\ -\frac{10x}{10} \amp = \frac{750}{10} \\ -x \amp = 75 +
        Answer.
        \(50\ {\rm g}\)
        Explanation.
        +
        This problem can be boiled down to this question: \(10\) is \(20\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(10\) is \(20\%\) of \(x\text{,}\) so “\(20\) out of \(100\)” corresponds to “\(10\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{20}{100} \amp = \frac{10}{x} \\ +20x \amp = 100 \cdot 10 \\ +20x \amp = 1000 \\ +\frac{20x}{20} \amp = \frac{1000}{20} \\ +x \amp = 50 \end{aligned} -}\)
        The recommended daily intake of fat for an average person is \({75\ {\rm g}}\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(7.5\) is \(10\%\) of what? Assume \(7.5\) is \(10\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        The recommended daily intake of fat for an average person is \({50\ {\rm g}}\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(10\) is \(20\%\) of what? Assume \(10\) is \(20\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -7.5 \amp = 0.1 \cdot x \\ -\frac{7.5}{0.1} \amp = \frac{0.1x}{0.1} \\ -75 \amp = x -\end{aligned} -}\)
        The recommended daily intake of fat for an average person is \({75\ {\rm g}}\text{.}\) -
        Method 3
        In the sentence “\(7.5\) is \(10\%\) of what,”
          -
        • -\(7.5\) is the percentage,
          • -
        • -\(10\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{7.5}{10\%} = \frac{7.5}{0.1} = 75 }\)
        The recommended daily intake of fat for an average person is \({75\ {\rm g}}\text{.}\) +10 \amp = 0.2 \cdot x \\ +\frac{10}{0.2} \amp = \frac{0.2x}{0.2} \\ +50 \amp = x +\end{aligned} +}\)
        The recommended daily intake of fat for an average person is \({50\ {\rm g}}\text{.}\) +
        Method 3
        In the sentence “\(10\) is \(20\%\) of what,”
          +
        • +\(10\) is the percentage,
          • +
        • +\(20\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{10}{20\%} = \frac{10}{0.2} = 50 }\)
        The recommended daily intake of fat for an average person is \({50\ {\rm g}}\text{.}\)

        55.

        -
        -
        +
        +
        -
        A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey.
        What percent of students ride Bus #1?
        Approximately of students ride Bus #1.
        +
        A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey.
        What percent of students ride Bus #1?
        Approximately of students ride Bus #1.
        -
        Answer.
        \(23\%\)
        Explanation.
        -
        First, we find the total number of students who participated in this survey:
        \(\displaystyle{ 175+144+140+149+166=774 }\)
        Now, this problem can be boiled down to this question: \(175\) is what percent of \(774\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(175\) is \(x\%\) of \(774\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(175\) out of \(774\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{175}{774} \\ -774x \amp = 175 \cdot 100 \\ -774x \amp = 17500 \\ -\frac{774x}{774} \amp = \frac{17500}{774} \\ -x \amp \approx 22.609819121447 \\ -x \amp \approx 23\% +
        Answer.
        \(15\%\)
        Explanation.
        +
        First, we find the total number of students who participated in this survey:
        \(\displaystyle{ 71+116+119+80+101=487 }\)
        Now, this problem can be boiled down to this question: \(71\) is what percent of \(487\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(71\) is \(x\%\) of \(487\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(71\) out of \(487\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{71}{487} \\ +487x \amp = 71 \cdot 100 \\ +487x \amp = 7100 \\ +\frac{487x}{487} \amp = \frac{7100}{487} \\ +x \amp \approx 14.5790554414784 \\ +x \amp \approx 15\% \end{aligned} -}\)
        Approximately \(23\) of students ride Bus #1.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(175\) is \(x\) (as a percent) of \(774\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -175 \amp = x \cdot 774 \\ -\frac{175}{774} \amp = \frac{x \cdot 774}{774} \\ -0.22609819121447 \amp \approx x \\ -x \amp \approx 23\% +}\)
        Approximately \(15\) of students ride Bus #1.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(71\) is \(x\) (as a percent) of \(487\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +71 \amp = x \cdot 487 \\ +\frac{71}{487} \amp = \frac{x \cdot 487}{487} \\ +0.145790554414784 \amp \approx x \\ +x \amp \approx 15\% \end{aligned} -}\)
        Approximately \(23\) of students ride Bus #1.
        Method 3
        In the sentence “\(175\) is what percent of \(774\text{,}\)”
          -
        • -\(175\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(774\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{175}{774} \approx 0.22609819121447 \approx 23\% }\)
        Approximately \(23\) of students ride Bus #1.
        +}\)
        Approximately \(15\) of students ride Bus #1.
        Method 3
        In the sentence “\(71\) is what percent of \(487\text{,}\)”
          +
        • +\(71\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(487\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{71}{487} \approx 0.145790554414784 \approx 15\% }\)
        Approximately \(15\) of students ride Bus #1.

        56.

        -
        -
        +
        +
        -
        A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey.
        What percent of students ride Bus #1?
        Approximately of students ride Bus #1.
        +
        A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey.
        What percent of students ride Bus #1?
        Approximately of students ride Bus #1.
        -
        Answer.
        \(12\%\)
        Explanation.
        -
        First, we find the total number of students who participated in this survey:
        \(\displaystyle{ 58+98+63+113+164=496 }\)
        Now, this problem can be boiled down to this question: \(58\) is what percent of \(496\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(58\) is \(x\%\) of \(496\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(58\) out of \(496\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{58}{496} \\ -496x \amp = 58 \cdot 100 \\ -496x \amp = 5800 \\ -\frac{496x}{496} \amp = \frac{5800}{496} \\ -x \amp \approx 11.6935483870968 \\ -x \amp \approx 12\% +
        Answer.
        \(14\%\)
        Explanation.
        +
        First, we find the total number of students who participated in this survey:
        \(\displaystyle{ 85+71+174+175+99=604 }\)
        Now, this problem can be boiled down to this question: \(85\) is what percent of \(604\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(85\) is \(x\%\) of \(604\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(85\) out of \(604\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{85}{604} \\ +604x \amp = 85 \cdot 100 \\ +604x \amp = 8500 \\ +\frac{604x}{604} \amp = \frac{8500}{604} \\ +x \amp \approx 14.0728476821192 \\ +x \amp \approx 14\% \end{aligned} -}\)
        Approximately \(12\) of students ride Bus #1.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(58\) is \(x\) (as a percent) of \(496\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -58 \amp = x \cdot 496 \\ -\frac{58}{496} \amp = \frac{x \cdot 496}{496} \\ -0.116935483870968 \amp \approx x \\ -x \amp \approx 12\% +}\)
        Approximately \(14\) of students ride Bus #1.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(85\) is \(x\) (as a percent) of \(604\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +85 \amp = x \cdot 604 \\ +\frac{85}{604} \amp = \frac{x \cdot 604}{604} \\ +0.140728476821192 \amp \approx x \\ +x \amp \approx 14\% \end{aligned} -}\)
        Approximately \(12\) of students ride Bus #1.
        Method 3
        In the sentence “\(58\) is what percent of \(496\text{,}\)”
          -
        • -\(58\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(496\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{58}{496} \approx 0.116935483870968 \approx 12\% }\)
        Approximately \(12\) of students ride Bus #1.
        +}\)
        Approximately \(14\) of students ride Bus #1.
        Method 3
        In the sentence “\(85\) is what percent of \(604\text{,}\)”
          +
        • +\(85\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(604\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{85}{604} \approx 0.140728476821192 \approx 14\% }\)
        Approximately \(14\) of students ride Bus #1.

        57.

        -
        -
        +
        +
        -
        Scot earned \({\$8.50}\) of interest from a mutual fund, which was \(0.05\%\) of his total investment. How much money did Scot invest into this mutual fund?
        Scot invested in this mutual fund.
        +
        Irene earned \({\$213.36}\) of interest from a mutual fund, which was \(0.84\%\) of his total investment. How much money did Irene invest into this mutual fund?
        Irene invested in this mutual fund.
        -
        Answer.
        \(\$17{,}000.00\)
        Explanation.
        -
        This problem can be boiled down to this question: \(8.5\) is \(0.05\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(8.5\) is \(0.05\%\) of \(x\text{,}\) so “\(0.05\) out of \(100\)” corresponds to “\(8.5\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{0.05}{100} \amp = \frac{8.5}{x} \\ -0.05x \amp = 100 \cdot 8.5 \\ -0.05x \amp = 850 \\ -\frac{0.05x}{0.05} \amp = \frac{850}{0.05} \\ -x \amp = 17000 +
        Answer.
        \(\$25{,}400.00\)
        Explanation.
        +
        This problem can be boiled down to this question: \(213.36\) is \(0.84\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(213.36\) is \(0.84\%\) of \(x\text{,}\) so “\(0.84\) out of \(100\)” corresponds to “\(213.36\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{0.84}{100} \amp = \frac{213.36}{x} \\ +0.84x \amp = 100 \cdot 213.36 \\ +0.84x \amp = 21336 \\ +\frac{0.84x}{0.84} \amp = \frac{21336}{0.84} \\ +x \amp = 25400 \end{aligned} -}\)
        Scot invested \({\$17{,}000.00}\) in this mutual fund.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(8.5\) is \(0.05\%\) of what? Assume \(8.5\) is \(0.05\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        Irene invested \({\$25{,}400.00}\) in this mutual fund.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(213.36\) is \(0.84\%\) of what? Assume \(213.36\) is \(0.84\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -8.5 \amp = 0.0005 \cdot x \\ -\frac{8.5}{0.0005} \amp = \frac{0.0005x}{0.0005} \\ -17000 \amp = x +213.36 \amp = 0.0084 \cdot x \\ +\frac{213.36}{0.0084} \amp = \frac{0.0084x}{0.0084} \\ +25400 \amp = x \end{aligned} -}\)
        Scot invested \({\$17{,}000.00}\) in this mutual fund.
        Method 3
        In the sentence “\(8.5\) is \(0.05\%\) of what,”
          -
        • -\(8.5\) is the percentage,
          • -
        • -\(0.05\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{8.5}{0.05\%} = \frac{8.5}{0.0005} = 17000 }\)
        Scot invested \({\$17{,}000.00}\) in this mutual fund.
        +}\)
        Irene invested \({\$25{,}400.00}\) in this mutual fund.
        Method 3
        In the sentence “\(213.36\) is \(0.84\%\) of what,”
          +
        • +\(213.36\) is the percentage,
          • +
        • +\(0.84\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{213.36}{0.84\%} = \frac{213.36}{0.0084} = 25400 }\)
        Irene invested \({\$25{,}400.00}\) in this mutual fund.

        58.

        -
        -
        +
        +
        -
        Sharell earned \({\$148.35}\) of interest from a mutual fund, which was \(0.69\%\) of his total investment. How much money did Sharell invest into this mutual fund?
        Sharell invested in this mutual fund.
        +
        Farshad earned \({\$143.04}\) of interest from a mutual fund, which was \(0.48\%\) of his total investment. How much money did Farshad invest into this mutual fund?
        Farshad invested in this mutual fund.
        -
        Answer.
        \(\$21{,}500.00\)
        Explanation.
        -
        This problem can be boiled down to this question: \(148.35\) is \(0.69\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(148.35\) is \(0.69\%\) of \(x\text{,}\) so “\(0.69\) out of \(100\)” corresponds to “\(148.35\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{0.69}{100} \amp = \frac{148.35}{x} \\ -0.69x \amp = 100 \cdot 148.35 \\ -0.69x \amp = 14835 \\ -\frac{0.69x}{0.69} \amp = \frac{14835}{0.69} \\ -x \amp = 21500 +
        Answer.
        \(\$29{,}800.00\)
        Explanation.
        +
        This problem can be boiled down to this question: \(143.04\) is \(0.48\%\) of what? We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem. Assume \(143.04\) is \(0.48\%\) of \(x\text{,}\) so “\(0.48\) out of \(100\)” corresponds to “\(143.04\) out of \(x\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{0.48}{100} \amp = \frac{143.04}{x} \\ +0.48x \amp = 100 \cdot 143.04 \\ +0.48x \amp = 14304 \\ +\frac{0.48x}{0.48} \amp = \frac{14304}{0.48} \\ +x \amp = 29800 \end{aligned} -}\)
        Sharell invested \({\$21{,}500.00}\) in this mutual fund.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(148.35\) is \(0.69\%\) of what? Assume \(148.35\) is \(0.69\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ +}\)
        Farshad invested \({\$29{,}800.00}\) in this mutual fund.
        Method 2
        We will use the percentage formula to solve this problem. This translation from English to math may help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \iff 2 = 0.5 \cdot 4\)
        The question is: \(143.04\) is \(0.48\%\) of what? Assume \(143.04\) is \(0.48\%\) of \(x\text{.}\) We have:
        \(\displaystyle{ \begin{aligned} -148.35 \amp = 0.0069 \cdot x \\ -\frac{148.35}{0.0069} \amp = \frac{0.0069x}{0.0069} \\ -21500 \amp = x +143.04 \amp = 0.0048 \cdot x \\ +\frac{143.04}{0.0048} \amp = \frac{0.0048x}{0.0048} \\ +29800 \amp = x \end{aligned} -}\)
        Sharell invested \({\$21{,}500.00}\) in this mutual fund.
        Method 3
        In the sentence “\(148.35\) is \(0.69\%\) of what,”
          -
        • -\(148.35\) is the percentage,
          • -
        • -\(0.69\%\) is the rate,
          • -
        • “what” is the base (following the word “of”).
          • -
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{148.35}{0.69\%} = \frac{148.35}{0.0069} = 21500 }\)
        Sharell invested \({\$21{,}500.00}\) in this mutual fund.
        +}\)
        Farshad invested \({\$29{,}800.00}\) in this mutual fund.
        Method 3
        In the sentence “\(143.04\) is \(0.48\%\) of what,”
          +
        • +\(143.04\) is the percentage,
          • +
        • +\(0.48\%\) is the rate,
          • +
        • “what” is the base (following the word “of”).
          • +
        By the formula \(\text{base} = \frac{\text{percentage}}{\text{rate}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{base} = \frac{\text{percentage}}{\text{rate}} = \frac{143.04}{0.48\%} = \frac{143.04}{0.0048} = 29800 }\)
        Farshad invested \({\$29{,}800.00}\) in this mutual fund.

        59.

        -
        -
        +
        +
        -
        A town has \(2600\) registered residents. Among them, there are \(858\) Democrats and \(728\) Republicans. The rest are Independents. What percentage of registered voters in this town are Independents?
        In this town, of all registered voters are Independents.
        +
        A town has \(3400\) registered residents. Among them, there are \(1088\) Democrats and \(816\) Republicans. The rest are Independents. What percentage of registered voters in this town are Independents?
        In this town, of all registered voters are Independents.
        -
        Answer.
        \(39\%\)
        Explanation.
        -
        Out of \(2600\) registered voters, there are \(858\) Democrats and \(728\) Republicans. This implies there are \(2600-858-728=1014\) Independents.
        Now this problem can be boiled down to this question: \(1014\) is what percent of \(2600\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(1014\) is \(x\%\) of \(2600\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(1014\) out of \(2600\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{1014}{2600} \\ -2600x \amp = 1014 \cdot 100 \\ -2600x \amp = 101400 \\ -\frac{2600x}{2600} \amp = \frac{101400}{2600} \\ -x \amp = 39 +
        Answer.
        \(44\%\)
        Explanation.
        +
        Out of \(3400\) registered voters, there are \(1088\) Democrats and \(816\) Republicans. This implies there are \(3400-1088-816=1496\) Independents.
        Now this problem can be boiled down to this question: \(1496\) is what percent of \(3400\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(1496\) is \(x\%\) of \(3400\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(1496\) out of \(3400\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{1496}{3400} \\ +3400x \amp = 1496 \cdot 100 \\ +3400x \amp = 149600 \\ +\frac{3400x}{3400} \amp = \frac{149600}{3400} \\ +x \amp = 44 \end{aligned} -}\)
        In this town, \(39\%\) of all registered voters are Independents.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(1014\) is \(x\) (as a percent) of \(2600\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -1014 \amp = x \cdot 2600 \\ -\frac{1014}{2600} \amp = \frac{x \cdot 2600}{2600} \\ -0.39 \amp = x\\ -x \amp = 39\% +}\)
        In this town, \(44\%\) of all registered voters are Independents.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(1496\) is \(x\) (as a percent) of \(3400\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +1496 \amp = x \cdot 3400 \\ +\frac{1496}{3400} \amp = \frac{x \cdot 3400}{3400} \\ +0.44 \amp = x\\ +x \amp = 44\% \end{aligned} -}\)
        In this town, \(39\%\) of all registered voters are Independents.
        Method 3
        In the sentence “\(1014\) is what percent of \(2600\text{,}\)”
          -
        • -\(1014\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(2600\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{1014}{2600} = 0.39 = 39\% }\)
        In this town, \(39\%\) of all registered voters are Independents.
        +}\)
        In this town, \(44\%\) of all registered voters are Independents.
        Method 3
        In the sentence “\(1496\) is what percent of \(3400\text{,}\)”
          +
        • +\(1496\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(3400\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{1496}{3400} = 0.44 = 44\% }\)
        In this town, \(44\%\) of all registered voters are Independents.

        60.

        -
        -
        +
        +
        -
        A town has \(3000\) registered residents. Among them, there are \(1170\) Democrats and \(1080\) Republicans. The rest are Independents. What percentage of registered voters in this town are Independents?
        In this town, of all registered voters are Independents.
        -
        -
        Answer.
        \(25\%\)
        Explanation.
        -
        Out of \(3000\) registered voters, there are \(1170\) Democrats and \(1080\) Republicans. This implies there are \(3000-1170-1080=750\) Independents.
        Now this problem can be boiled down to this question: \(750\) is what percent of \(3000\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(750\) is \(x\%\) of \(3000\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(750\) out of \(3000\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{750}{3000} \\ -3000x \amp = 750 \cdot 100 \\ -3000x \amp = 75000 \\ -\frac{3000x}{3000} \amp = \frac{75000}{3000} \\ -x \amp = 25 -\end{aligned} -}\)
        In this town, \(25\%\) of all registered voters are Independents.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(750\) is \(x\) (as a percent) of \(3000\text{.}\) We have:
        \(\displaystyle{\begin{aligned} -750 \amp = x \cdot 3000 \\ -\frac{750}{3000} \amp = \frac{x \cdot 3000}{3000} \\ -0.25 \amp = x\\ -x \amp = 25\% -\end{aligned} -}\)
        In this town, \(25\%\) of all registered voters are Independents.
        Method 3
        In the sentence “\(750\) is what percent of \(3000\text{,}\)”
          -
        • -\(750\) is the percentage,
          • -
        • “what percent” is the rate,
          • -
        • -\(3000\) is the base (following the word “of”).
          • -
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{750}{3000} = 0.25 = 25\% }\)
        In this town, \(25\%\) of all registered voters are Independents.
        +
        A town has \(3900\) registered residents. Among them, there are \(1443\) Democrats and \(1287\) Republicans. The rest are Independents. What percentage of registered voters in this town are Independents?
        In this town, of all registered voters are Independents.
        +
        +
        Answer.
        \(30\%\)
        Explanation.
        +
        Out of \(3900\) registered voters, there are \(1443\) Democrats and \(1287\) Republicans. This implies there are \(3900-1443-1287=1170\) Independents.
        Now this problem can be boiled down to this question: \(1170\) is what percent of \(3900\text{?}\) We will show multiple methods to solve this problem.
        Method 1
        We will use proportion to solve this problem.
        Assume \(1170\) is \(x\%\) of \(3900\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(1170\) out of \(3900\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{1170}{3900} \\ +3900x \amp = 1170 \cdot 100 \\ +3900x \amp = 117000 \\ +\frac{3900x}{3900} \amp = \frac{117000}{3900} \\ +x \amp = 30 +\end{aligned} +}\)
        In this town, \(30\%\) of all registered voters are Independents.
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Assume \(1170\) is \(x\) (as a percent) of \(3900\text{.}\) We have:
        \(\displaystyle{\begin{aligned} +1170 \amp = x \cdot 3900 \\ +\frac{1170}{3900} \amp = \frac{x \cdot 3900}{3900} \\ +0.3 \amp = x\\ +x \amp = 30\% +\end{aligned} +}\)
        In this town, \(30\%\) of all registered voters are Independents.
        Method 3
        In the sentence “\(1170\) is what percent of \(3900\text{,}\)”
          +
        • +\(1170\) is the percentage,
          • +
        • “what percent” is the rate,
          • +
        • +\(3900\) is the base (following the word “of”).
          • +
        By the formula \(\text{rate} = \frac{\text{percentage}}{\text{base}}\text{,}\) we do a division to solve the problem:
        \(\displaystyle{ \text{rate} = \frac{\text{percentage}}{\text{base}} = \frac{1170}{3900} = 0.3 = 30\% }\)
        In this town, \(30\%\) of all registered voters are Independents.
        @@ -2779,353 +2793,353 @@

        Applications.

        Percent Increase/Decrease.

        61.

        -
        -
        -
        -
        The population of cats in a shelter decreased from \(140\) to \(98\text{.}\) What is the percentage decrease of the shelter’s cat population?
        The percentage decrease is .
        -
        -
        Answer.
        \(30\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(98-140=42\text{.}\) -
        Next, since we started with \(140\) cats, we need to ask: \(42\) is what percent of \(140\text{?}\) -
        Assume \(42\) is \(x\%\) of \(140\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(42\) out of \(140\)”.
        Method 1
        We will use proportion to solve this problem.
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{42}{140} \\ -140x \amp = 42 \cdot 100 \\ -140x \amp = 4200 \\ -\frac{140x}{140} \amp = \frac{4200}{140} \\ -x \amp = 30 +
        +
        +
        +
        The population of cats in a shelter decreased from \(180\) to \(144\text{.}\) What is the percentage decrease of the shelter’s cat population?
        The percentage decrease is .
        +
        +
        Answer.
        \(20\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(144-180=36\text{.}\) +
        Next, since we started with \(180\) cats, we need to ask: \(36\) is what percent of \(180\text{?}\) +
        Assume \(36\) is \(x\%\) of \(180\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(36\) out of \(180\)”.
        Method 1
        We will use proportion to solve this problem.
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{36}{180} \\ +180x \amp = 36 \cdot 100 \\ +180x \amp = 3600 \\ +\frac{180x}{180} \amp = \frac{3600}{180} \\ +x \amp = 20 \end{aligned} -}\)
        The percentage decrease of the shelter’s cat population is \(30\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let \(x\) be the unknown percentage, and let the decrease \(42\) be \(x\) (percent) of \(140\text{.}\) That means:
        \(\begin{aligned} -42 \amp = x \cdot 140 \\ -\frac{42}{140} \amp = \frac{x \cdot 140}{140} \\ -0.3 \amp = x \\ -30\% \amp = x \\ -\end{aligned}\)
        The percentage decrease of the shelter’s cat population is \(30\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{98}{140} = 0.7 = 70\% }\)
        So the new number is \(70\%\) of the original number, implying the percentage decrease is \(100\% - 70\% = 30\%\text{.}\) -
        The percentage decrease of the shelter’s cat population is \(30\%\text{.}\) +}\)
        The percentage decrease of the shelter’s cat population is \(20\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let \(x\) be the unknown percentage, and let the decrease \(36\) be \(x\) (percent) of \(180\text{.}\) That means:
        \(\begin{aligned} +36 \amp = x \cdot 180 \\ +\frac{36}{180} \amp = \frac{x \cdot 180}{180} \\ +0.2 \amp = x \\ +20\% \amp = x \\ +\end{aligned}\)
        The percentage decrease of the shelter’s cat population is \(20\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{144}{180} = 0.8 = 80\% }\)
        So the new number is \(80\%\) of the original number, implying the percentage decrease is \(100\% - 80\% = 20\%\text{.}\) +
        The percentage decrease of the shelter’s cat population is \(20\%\text{.}\)

        62.

        -
        -
        +
        +
        -
        The population of cats in a shelter decreased from \(160\) to \(136\text{.}\) What is the percentage decrease of the shelter’s cat population?
        The percentage decrease is .
        +
        The population of cats in a shelter decreased from \(200\) to \(190\text{.}\) What is the percentage decrease of the shelter’s cat population?
        The percentage decrease is .
        -
        Answer.
        \(15\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(136-160=24\text{.}\) -
        Next, since we started with \(160\) cats, we need to ask: \(24\) is what percent of \(160\text{?}\) -
        Assume \(24\) is \(x\%\) of \(160\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(24\) out of \(160\)”.
        Method 1
        We will use proportion to solve this problem.
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{24}{160} \\ -160x \amp = 24 \cdot 100 \\ -160x \amp = 2400 \\ -\frac{160x}{160} \amp = \frac{2400}{160} \\ -x \amp = 15 +
        Answer.
        \(5\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(190-200=10\text{.}\) +
        Next, since we started with \(200\) cats, we need to ask: \(10\) is what percent of \(200\text{?}\) +
        Assume \(10\) is \(x\%\) of \(200\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(10\) out of \(200\)”.
        Method 1
        We will use proportion to solve this problem.
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{10}{200} \\ +200x \amp = 10 \cdot 100 \\ +200x \amp = 1000 \\ +\frac{200x}{200} \amp = \frac{1000}{200} \\ +x \amp = 5 \end{aligned} -}\)
        The percentage decrease of the shelter’s cat population is \(15\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let \(x\) be the unknown percentage, and let the decrease \(24\) be \(x\) (percent) of \(160\text{.}\) That means:
        \(\begin{aligned} -24 \amp = x \cdot 160 \\ -\frac{24}{160} \amp = \frac{x \cdot 160}{160} \\ -0.15 \amp = x \\ -15\% \amp = x \\ -\end{aligned}\)
        The percentage decrease of the shelter’s cat population is \(15\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{136}{160} = 0.85 = 85\% }\)
        So the new number is \(85\%\) of the original number, implying the percentage decrease is \(100\% - 85\% = 15\%\text{.}\) -
        The percentage decrease of the shelter’s cat population is \(15\%\text{.}\) +}\)
        The percentage decrease of the shelter’s cat population is \(5\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let \(x\) be the unknown percentage, and let the decrease \(10\) be \(x\) (percent) of \(200\text{.}\) That means:
        \(\begin{aligned} +10 \amp = x \cdot 200 \\ +\frac{10}{200} \amp = \frac{x \cdot 200}{200} \\ +0.05 \amp = x \\ +5\% \amp = x \\ +\end{aligned}\)
        The percentage decrease of the shelter’s cat population is \(5\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{190}{200} = 0.95 = 95\% }\)
        So the new number is \(95\%\) of the original number, implying the percentage decrease is \(100\% - 95\% = 5\%\text{.}\) +
        The percentage decrease of the shelter’s cat population is \(5\%\text{.}\)

        63.

        -
        -
        +
        +
        -
        The population of cats in a shelter increased from \(69\) to \(89\text{.}\) What is the percentage increase of the shelter’s cat population?
        The percentage increase is approximately .
        +
        The population of cats in a shelter increased from \(13\) to \(29\text{.}\) What is the percentage increase of the shelter’s cat population?
        The percentage increase is approximately .
        -
        Answer.
        \(28.99\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of increase is \(89-69\text{,}\) which is \(20\text{.}\) -
        Next, since we started with \(69\) cats, we need to ask: \(20\) is what percent of \(69\text{?}\) -
        Method 1
        We will use proportion to solve this problem.
        Assume \(20\) is \(x\%\) of \(69\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(20\) out of \(69\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{20}{69} \\ -69x \amp = 20 \cdot 100 \\ -69x \amp = 2000 \\ -\frac{69x}{69} \amp = \frac{2000}{69} \\ -x \amp \approx 28.9855072463768 \\ -x \amp \approx 28.99 +
        Answer.
        \(123.08\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of increase is \(29-13\text{,}\) which is \(16\text{.}\) +
        Next, since we started with \(13\) cats, we need to ask: \(16\) is what percent of \(13\text{?}\) +
        Method 1
        We will use proportion to solve this problem.
        Assume \(16\) is \(x\%\) of \(13\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(16\) out of \(13\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{16}{13} \\ +13x \amp = 16 \cdot 100 \\ +13x \amp = 1600 \\ +\frac{13x}{13} \amp = \frac{1600}{13} \\ +x \amp \approx 123.076923076923 \\ +x \amp \approx 123.08 \end{aligned} -}\)
        The percentage increase of the shelter’s cat population is approximately \(28.99\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the increase \(20\) be \(x\) (as a percent) of \(69\text{.}\) That means:
        \(\begin{aligned} -20 \amp = x \cdot 69 \\ -\frac{20}{69} \amp = \frac{x \cdot 69}{69} \\ -0.289855072463768 \amp \approx x\\ -x \amp \approx 28.99\%\\ -\end{aligned}\)
        The percentage increase of the shelter’s cat population is approximately \(28.99\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{89}{69} \approx 1.28985507246377 \approx 128.99\% }\)
        So the new number is approximately \(128.99\%\) of the original number, implying the percentage increase is approximately \(128.99\% - 100\% = 28.99\%\text{.}\) -
        The percentage increase of the shelter’s cat population is approximately \(28.99\%\text{.}\) +}\)
        The percentage increase of the shelter’s cat population is approximately \(123.08\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the increase \(16\) be \(x\) (as a percent) of \(13\text{.}\) That means:
        \(\begin{aligned} +16 \amp = x \cdot 13 \\ +\frac{16}{13} \amp = \frac{x \cdot 13}{13} \\ +1.23076923076923 \amp \approx x\\ +x \amp \approx 123.08\%\\ +\end{aligned}\)
        The percentage increase of the shelter’s cat population is approximately \(123.08\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{29}{13} \approx 2.23076923076923 \approx 223.08\% }\)
        So the new number is approximately \(223.08\%\) of the original number, implying the percentage increase is approximately \(223.08\% - 100\% = 123.08\%\text{.}\) +
        The percentage increase of the shelter’s cat population is approximately \(123.08\%\text{.}\)

        64.

        -
        -
        +
        +
        -
        The population of cats in a shelter increased from \(77\) to \(91\text{.}\) What is the percentage increase of the shelter’s cat population?
        The percentage increase is approximately .
        +
        The population of cats in a shelter increased from \(21\) to \(32\text{.}\) What is the percentage increase of the shelter’s cat population?
        The percentage increase is approximately .
        -
        Answer.
        \(18.18\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of increase is \(91-77\text{,}\) which is \(14\text{.}\) -
        Next, since we started with \(77\) cats, we need to ask: \(14\) is what percent of \(77\text{?}\) -
        Method 1
        We will use proportion to solve this problem.
        Assume \(14\) is \(x\%\) of \(77\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(14\) out of \(77\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{14}{77} \\ -77x \amp = 14 \cdot 100 \\ -77x \amp = 1400 \\ -\frac{77x}{77} \amp = \frac{1400}{77} \\ -x \amp \approx 18.1818181818182 \\ -x \amp \approx 18.18 +
        Answer.
        \(52.38\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of increase is \(32-21\text{,}\) which is \(11\text{.}\) +
        Next, since we started with \(21\) cats, we need to ask: \(11\) is what percent of \(21\text{?}\) +
        Method 1
        We will use proportion to solve this problem.
        Assume \(11\) is \(x\%\) of \(21\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(11\) out of \(21\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{11}{21} \\ +21x \amp = 11 \cdot 100 \\ +21x \amp = 1100 \\ +\frac{21x}{21} \amp = \frac{1100}{21} \\ +x \amp \approx 52.3809523809524 \\ +x \amp \approx 52.38 \end{aligned} -}\)
        The percentage increase of the shelter’s cat population is approximately \(18.18\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the increase \(14\) be \(x\) (as a percent) of \(77\text{.}\) That means:
        \(\begin{aligned} -14 \amp = x \cdot 77 \\ -\frac{14}{77} \amp = \frac{x \cdot 77}{77} \\ -0.181818181818182 \amp \approx x\\ -x \amp \approx 18.18\%\\ -\end{aligned}\)
        The percentage increase of the shelter’s cat population is approximately \(18.18\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{91}{77} \approx 1.18181818181818 \approx 118.18\% }\)
        So the new number is approximately \(118.18\%\) of the original number, implying the percentage increase is approximately \(118.18\% - 100\% = 18.18\%\text{.}\) -
        The percentage increase of the shelter’s cat population is approximately \(18.18\%\text{.}\) +}\)
        The percentage increase of the shelter’s cat population is approximately \(52.38\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the increase \(11\) be \(x\) (as a percent) of \(21\text{.}\) That means:
        \(\begin{aligned} +11 \amp = x \cdot 21 \\ +\frac{11}{21} \amp = \frac{x \cdot 21}{21} \\ +0.523809523809524 \amp \approx x\\ +x \amp \approx 52.38\%\\ +\end{aligned}\)
        The percentage increase of the shelter’s cat population is approximately \(52.38\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{32}{21} \approx 1.52380952380952 \approx 152.38\% }\)
        So the new number is approximately \(152.38\%\) of the original number, implying the percentage increase is approximately \(152.38\% - 100\% = 52.38\%\text{.}\) +
        The percentage increase of the shelter’s cat population is approximately \(52.38\%\text{.}\)

        65.

        -
        -
        +
        +
        -
        Last year, a small town’s population was \(510\text{.}\) This year, the population decreased to \(506\text{.}\) What is the percentage decrease?
        The percentage decrease of the town’s population was approximately .
        +
        Last year, a small town’s population was \(580\text{.}\) This year, the population decreased to \(574\text{.}\) What is the percentage decrease?
        The percentage decrease of the town’s population was approximately .
        -
        Answer.
        \(0.78\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(510-506=4\text{.}\) -
        Since the city’s population was initially \(510\text{,}\) we need to ask: \(4\) is what percent of \(510\text{?}\) -
        Method 1
        We will use proportion to solve this problem.
        Assume \(4\) is \(x\%\) of \(510\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(4\) out of \(510\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{4}{510} \\ -510x \amp = 4 \cdot 100 \\ -510x \amp = 400 \\ -\frac{510x}{510} \amp = \frac{400}{510} \\ -x \amp \approx 0.784313725490196 \\ -x \amp \approx 0.78 +
        Answer.
        \(1.03\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(580-574=6\text{.}\) +
        Since the city’s population was initially \(580\text{,}\) we need to ask: \(6\) is what percent of \(580\text{?}\) +
        Method 1
        We will use proportion to solve this problem.
        Assume \(6\) is \(x\%\) of \(580\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(6\) out of \(580\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{6}{580} \\ +580x \amp = 6 \cdot 100 \\ +580x \amp = 600 \\ +\frac{580x}{580} \amp = \frac{600}{580} \\ +x \amp \approx 1.03448275862069 \\ +x \amp \approx 1.03 \end{aligned} -}\)
        The percentage decrease of the town’s population was approximately \(0.78\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the decrease \(4\) be \(x\) (as a percent) of \(510\text{.}\) That means:
        \(\begin{aligned} -4 \amp = x \cdot 510 \\ -\frac{4}{510} \amp = \frac{x \cdot 510}{510} \\ -0.00784313725490196 \amp \approx x\\ -x \amp \approx 0.78\% \\ -\end{aligned}\)
        The percentage decrease of the town’s population was approximately \(0.78\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{506}{510} \approx 0.992156862745098 \approx 99.22\% }\)
        So the new number is \(99.22\%\) of the original number, implying the percentage decrease is \(100\% - 99.22\% = 0.78\%\text{.}\) -
        The percentage decrease of the town’s population was approximately \(0.78\%\text{.}\) +}\)
        The percentage decrease of the town’s population was approximately \(1.03\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the decrease \(6\) be \(x\) (as a percent) of \(580\text{.}\) That means:
        \(\begin{aligned} +6 \amp = x \cdot 580 \\ +\frac{6}{580} \amp = \frac{x \cdot 580}{580} \\ +0.0103448275862069 \amp \approx x\\ +x \amp \approx 1.03\% \\ +\end{aligned}\)
        The percentage decrease of the town’s population was approximately \(1.03\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{574}{580} \approx 0.989655172413793 \approx 98.97\% }\)
        So the new number is \(98.97\%\) of the original number, implying the percentage decrease is \(100\% - 98.97\% = 1.03\%\text{.}\) +
        The percentage decrease of the town’s population was approximately \(1.03\%\text{.}\)

        66.

        -
        -
        +
        +
        -
        Last year, a small town’s population was \(550\text{.}\) This year, the population decreased to \(549\text{.}\) What is the percentage decrease?
        The percentage decrease of the town’s population was approximately .
        +
        Last year, a small town’s population was \(610\text{.}\) This year, the population decreased to \(608\text{.}\) What is the percentage decrease?
        The percentage decrease of the town’s population was approximately .
        -
        Answer.
        \(0.18\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(550-549=1\text{.}\) -
        Since the city’s population was initially \(550\text{,}\) we need to ask: \(1\) is what percent of \(550\text{?}\) -
        Method 1
        We will use proportion to solve this problem.
        Assume \(1\) is \(x\%\) of \(550\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(1\) out of \(550\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{1}{550} \\ -550x \amp = 1 \cdot 100 \\ -550x \amp = 100 \\ -\frac{550x}{550} \amp = \frac{100}{550} \\ -x \amp \approx 0.181818181818182 \\ -x \amp \approx 0.18 +
        Answer.
        \(0.33\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of increase/decrease by doing a simple subtraction calculation, and then we find the percentage increase/decrease.
        In this problem, the amount of decrease is \(610-608=2\text{.}\) +
        Since the city’s population was initially \(610\text{,}\) we need to ask: \(2\) is what percent of \(610\text{?}\) +
        Method 1
        We will use proportion to solve this problem.
        Assume \(2\) is \(x\%\) of \(610\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(2\) out of \(610\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{2}{610} \\ +610x \amp = 2 \cdot 100 \\ +610x \amp = 200 \\ +\frac{610x}{610} \amp = \frac{200}{610} \\ +x \amp \approx 0.327868852459016 \\ +x \amp \approx 0.33 \end{aligned} -}\)
        The percentage decrease of the town’s population was approximately \(0.18\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the decrease \(1\) be \(x\) (as a percent) of \(550\text{.}\) That means:
        \(\begin{aligned} -1 \amp = x \cdot 550 \\ -\frac{1}{550} \amp = \frac{x \cdot 550}{550} \\ -0.00181818181818182 \amp \approx x\\ -x \amp \approx 0.18\% \\ -\end{aligned}\)
        The percentage decrease of the town’s population was approximately \(0.18\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{549}{550} \approx 0.998181818181818 \approx 99.82\% }\)
        So the new number is \(99.82\%\) of the original number, implying the percentage decrease is \(100\% - 99.82\% = 0.18\%\text{.}\) -
        The percentage decrease of the town’s population was approximately \(0.18\%\text{.}\) +}\)
        The percentage decrease of the town’s population was approximately \(0.33\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the decrease \(2\) be \(x\) (as a percent) of \(610\text{.}\) That means:
        \(\begin{aligned} +2 \amp = x \cdot 610 \\ +\frac{2}{610} \amp = \frac{x \cdot 610}{610} \\ +0.00327868852459016 \amp \approx x\\ +x \amp \approx 0.33\% \\ +\end{aligned}\)
        The percentage decrease of the town’s population was approximately \(0.33\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{608}{610} \approx 0.99672131147541 \approx 99.67\% }\)
        So the new number is \(99.67\%\) of the original number, implying the percentage decrease is \(100\% - 99.67\% = 0.33\%\text{.}\) +
        The percentage decrease of the town’s population was approximately \(0.33\%\text{.}\)

        67.

        -
        -
        +
        +
        -
        Your salary used to be \({\$41{,}000}\) per year.
        You had to take a \(3\%\) pay cut. After the cut, your salary was per year.
        Then, you earned a \(3\%\) raise. After the raise, your salary was per year.
        +
        Your salary used to be \({\$37{,}000}\) per year.
        You had to take a \(3\%\) pay cut. After the cut, your salary was per year.
        Then, you earned a \(3\%\) raise. After the raise, your salary was per year.
        -
        Answer 1.
        \(\$39{,}770\)
        Answer 2.
        \(\$40{,}963.10\)
        Explanation.
        -
        Question 1
        Your original salary was \({\$41{,}000}\) per year. The amount of cut was \(3\%\) of \({\$41{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{41000} \amp = \frac{3}{100} \\ -100x \amp = 41000\cdot3 \\ -100x \amp = 123000 \\ -\frac{100x}{100} \amp = \frac{123000}{100} \\ -x \amp = 1230 +
        Answer 1.
        \(\$35{,}890\)
        Answer 2.
        \(\$36{,}966.70\)
        Explanation.
        +
        Question 1
        Your original salary was \({\$37{,}000}\) per year. The amount of cut was \(3\%\) of \({\$37{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{37000} \amp = \frac{3}{100} \\ +100x \amp = 37000\cdot3 \\ +100x \amp = 111000 \\ +\frac{100x}{100} \amp = \frac{111000}{100} \\ +x \amp = 1110 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of increase:
        \(\displaystyle{ 0.03 \cdot {\$41{,}000} = {\$1{,}230} }\)
        After the cut, your salary became \({\$41{,}000}-{\$1{,}230} = {\$39{,}770}\) dollars per year.
        Question 2
        Next, the amount of pay raise was \(3\%\) of \({\$39{,}770}\text{.}\) Notice that it’s incorrect to find \(3\%\) of \({\$41{,}000}\text{,}\) because the annual salary has changed from \({\$41{,}000}\) to \({\$39{,}770}\text{.}\) -
        We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was \(x\) dollars. We have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{39770} \amp = \frac{3}{100} \\ -100x \amp = 39770\cdot3 \\ -100x \amp = 119310 \\ -\frac{100x}{100} \amp = \frac{119310}{100} \\ -x \amp = 1193.1 +}\)
        Or we can use the percentage formula to find the amount of increase:
        \(\displaystyle{ 0.03 \cdot {\$37{,}000} = {\$1{,}110} }\)
        After the cut, your salary became \({\$37{,}000}-{\$1{,}110} = {\$35{,}890}\) dollars per year.
        Question 2
        Next, the amount of pay raise was \(3\%\) of \({\$35{,}890}\text{.}\) Notice that it’s incorrect to find \(3\%\) of \({\$37{,}000}\text{,}\) because the annual salary has changed from \({\$37{,}000}\) to \({\$35{,}890}\text{.}\) +
        We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was \(x\) dollars. We have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{35890} \amp = \frac{3}{100} \\ +100x \amp = 35890\cdot3 \\ +100x \amp = 107670 \\ +\frac{100x}{100} \amp = \frac{107670}{100} \\ +x \amp = 1076.7 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of pay cut:
        \(\displaystyle{ 0.03 \cdot {\$39{,}770} = {\$1{,}193.10} }\)
        After the raise, your salary became \({\$39{,}770}+{\$1{,}193.10} = {\$40{,}963.10}\) per year.
        After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut).
        +}\)
        Or we can use the percentage formula to find the amount of pay cut:
        \(\displaystyle{ 0.03 \cdot {\$35{,}890} = {\$1{,}076.70} }\)
        After the raise, your salary became \({\$35{,}890}+{\$1{,}076.70} = {\$36{,}966.70}\) per year.
        After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut).

        68.

        -
        -
        +
        +
        -
        Your salary used to be \({\$34{,}000}\) per year.
        You had to take a \(3\%\) pay cut. After the cut, your salary was per year.
        Then, you earned a \(3\%\) raise. After the raise, your salary was per year.
        +
        Your salary used to be \({\$50{,}000}\) per year.
        You had to take a \(4\%\) pay cut. After the cut, your salary was per year.
        Then, you earned a \(4\%\) raise. After the raise, your salary was per year.
        -
        Answer 1.
        \(\$32{,}980\)
        Answer 2.
        \(\$33{,}969.40\)
        Explanation.
        -
        Question 1
        Your original salary was \({\$34{,}000}\) per year. The amount of cut was \(3\%\) of \({\$34{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{34000} \amp = \frac{3}{100} \\ -100x \amp = 34000\cdot3 \\ -100x \amp = 102000 \\ -\frac{100x}{100} \amp = \frac{102000}{100} \\ -x \amp = 1020 +
        Answer 1.
        \(\$48{,}000\)
        Answer 2.
        \(\$49{,}920\)
        Explanation.
        +
        Question 1
        Your original salary was \({\$50{,}000}\) per year. The amount of cut was \(4\%\) of \({\$50{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{50000} \amp = \frac{4}{100} \\ +100x \amp = 50000\cdot4 \\ +100x \amp = 200000 \\ +\frac{100x}{100} \amp = \frac{200000}{100} \\ +x \amp = 2000 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of increase:
        \(\displaystyle{ 0.03 \cdot {\$34{,}000} = {\$1{,}020} }\)
        After the cut, your salary became \({\$34{,}000}-{\$1{,}020} = {\$32{,}980}\) dollars per year.
        Question 2
        Next, the amount of pay raise was \(3\%\) of \({\$32{,}980}\text{.}\) Notice that it’s incorrect to find \(3\%\) of \({\$34{,}000}\text{,}\) because the annual salary has changed from \({\$34{,}000}\) to \({\$32{,}980}\text{.}\) -
        We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was \(x\) dollars. We have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{32980} \amp = \frac{3}{100} \\ -100x \amp = 32980\cdot3 \\ -100x \amp = 98940 \\ -\frac{100x}{100} \amp = \frac{98940}{100} \\ -x \amp = 989.4 +}\)
        Or we can use the percentage formula to find the amount of increase:
        \(\displaystyle{ 0.04 \cdot {\$50{,}000} = {\$2{,}000} }\)
        After the cut, your salary became \({\$50{,}000}-{\$2{,}000} = {\$48{,}000}\) dollars per year.
        Question 2
        Next, the amount of pay raise was \(4\%\) of \({\$48{,}000}\text{.}\) Notice that it’s incorrect to find \(4\%\) of \({\$50{,}000}\text{,}\) because the annual salary has changed from \({\$50{,}000}\) to \({\$48{,}000}\text{.}\) +
        We will first use proportion to find the amount of pay cut. Assume the amount of pay cut was \(x\) dollars. We have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{48000} \amp = \frac{4}{100} \\ +100x \amp = 48000\cdot4 \\ +100x \amp = 192000 \\ +\frac{100x}{100} \amp = \frac{192000}{100} \\ +x \amp = 1920 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of pay cut:
        \(\displaystyle{ 0.03 \cdot {\$32{,}980} = {\$989.40} }\)
        After the raise, your salary became \({\$32{,}980}+{\$989.40} = {\$33{,}969.40}\) per year.
        After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut).
        +}\)
        Or we can use the percentage formula to find the amount of pay cut:
        \(\displaystyle{ 0.04 \cdot {\$48{,}000} = {\$1{,}920} }\)
        After the raise, your salary became \({\$48{,}000}+{\$1{,}920} = {\$49{,}920}\) per year.
        After a cut and a raise of the same rate, your salary decreased! This is because the rate of cut was based on a bigger salary, while the rate of raise was based on a smaller salary (after the cut).

        69.

        -
        -
        +
        +
        -
        A house was bought two years ago at the price of \({\$440{,}000}\text{.}\) Each year, the house’s value decreased by \(5\%\text{.}\) What’s the house’s value this year?
        The house’s value this year is .
        +
        A house was bought two years ago at the price of \({\$360{,}000}\text{.}\) Each year, the house’s value decreased by \(6\%\text{.}\) What’s the house’s value this year?
        The house’s value this year is .
        -
        Answer.
        \(\$397{,}100\)
        Explanation.
        -
        The house’s value two years ago was \({\$440{,}000}\text{.}\) After one year, the house’s value decreased by \(5\%\text{.}\) -
        The amount of decrease was \(5\%\) of \({\$440{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{440000} \amp = \frac{5}{100} \\ -100x \amp = 440000\cdot5 \\ -100x \amp = 2200000 \\ -\frac{100x}{100} \amp = \frac{2200000}{100} \\ -x \amp = 22000 +
        Answer.
        \(\$318{,}096\)
        Explanation.
        +
        The house’s value two years ago was \({\$360{,}000}\text{.}\) After one year, the house’s value decreased by \(6\%\text{.}\) +
        The amount of decrease was \(6\%\) of \({\$360{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{360000} \amp = \frac{6}{100} \\ +100x \amp = 360000\cdot6 \\ +100x \amp = 2160000 \\ +\frac{100x}{100} \amp = \frac{2160000}{100} \\ +x \amp = 21600 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.05 \cdot 440000 = 22000 }\)
        After one year, the house’s value became \({\$440{,}000}-{\$22{,}000}={\$418{,}000}\text{.}\) -
        After another year, the house’s value decreased by another \(5\%\text{.}\) -
        The amount of decrease was \(5\%\) of \({\$418{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{418000} \amp = \frac{5}{100} \\ -100x \amp = 418000\cdot5 \\ -100x \amp = 2090000 \\ -\frac{100x}{100} \amp = \frac{2090000}{100} \\ -x \amp = 20900 +}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.06 \cdot 360000 = 21600 }\)
        After one year, the house’s value became \({\$360{,}000}-{\$21{,}600}={\$338{,}400}\text{.}\) +
        After another year, the house’s value decreased by another \(6\%\text{.}\) +
        The amount of decrease was \(6\%\) of \({\$338{,}400}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{338400} \amp = \frac{6}{100} \\ +100x \amp = 338400\cdot6 \\ +100x \amp = 2030400 \\ +\frac{100x}{100} \amp = \frac{2030400}{100} \\ +x \amp = 20304 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.05 \cdot 418000 = 20900 }\)
        After two years, the house’s value became \({\$418{,}000}-{\$20{,}900}={\$397{,}100}\text{.}\) -
        The house’s value this year is \({\$397{,}100}\text{.}\) +}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.06 \cdot 338400 = 20304 }\)
        After two years, the house’s value became \({\$338{,}400}-{\$20{,}304}={\$318{,}096}\text{.}\) +
        The house’s value this year is \({\$318{,}096}\text{.}\)

        70.

        -
        -
        +
        +
        -
        A house was bought two years ago at the price of \({\$300{,}000}\text{.}\) Each year, the house’s value decreased by \(6\%\text{.}\) What’s the house’s value this year?
        The house’s value this year is .
        +
        A house was bought two years ago at the price of \({\$220{,}000}\text{.}\) Each year, the house’s value decreased by \(7\%\text{.}\) What’s the house’s value this year?
        The house’s value this year is .
        -
        Answer.
        \(\$265{,}080\)
        Explanation.
        -
        The house’s value two years ago was \({\$300{,}000}\text{.}\) After one year, the house’s value decreased by \(6\%\text{.}\) -
        The amount of decrease was \(6\%\) of \({\$300{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{300000} \amp = \frac{6}{100} \\ -100x \amp = 300000\cdot6 \\ -100x \amp = 1800000 \\ -\frac{100x}{100} \amp = \frac{1800000}{100} \\ -x \amp = 18000 +
        Answer.
        \(\$190{,}278\)
        Explanation.
        +
        The house’s value two years ago was \({\$220{,}000}\text{.}\) After one year, the house’s value decreased by \(7\%\text{.}\) +
        The amount of decrease was \(7\%\) of \({\$220{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{220000} \amp = \frac{7}{100} \\ +100x \amp = 220000\cdot7 \\ +100x \amp = 1540000 \\ +\frac{100x}{100} \amp = \frac{1540000}{100} \\ +x \amp = 15400 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.06 \cdot 300000 = 18000 }\)
        After one year, the house’s value became \({\$300{,}000}-{\$18{,}000}={\$282{,}000}\text{.}\) -
        After another year, the house’s value decreased by another \(6\%\text{.}\) -
        The amount of decrease was \(6\%\) of \({\$282{,}000}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{282000} \amp = \frac{6}{100} \\ -100x \amp = 282000\cdot6 \\ -100x \amp = 1692000 \\ -\frac{100x}{100} \amp = \frac{1692000}{100} \\ -x \amp = 16920 +}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.07 \cdot 220000 = 15400 }\)
        After one year, the house’s value became \({\$220{,}000}-{\$15{,}400}={\$204{,}600}\text{.}\) +
        After another year, the house’s value decreased by another \(7\%\text{.}\) +
        The amount of decrease was \(7\%\) of \({\$204{,}600}\text{.}\) There are two methods to find this amount.
        We can use proportion. Let \(x\) dollars be the amount of decrease, then we have:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{204600} \amp = \frac{7}{100} \\ +100x \amp = 204600\cdot7 \\ +100x \amp = 1432200 \\ +\frac{100x}{100} \amp = \frac{1432200}{100} \\ +x \amp = 14322 \end{aligned} -}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.06 \cdot 282000 = 16920 }\)
        After two years, the house’s value became \({\$282{,}000}-{\$16{,}920}={\$265{,}080}\text{.}\) -
        The house’s value this year is \({\$265{,}080}\text{.}\) +}\)
        Or we can use the percentage formula to find the amount of decrease:
        \(\displaystyle{ 0.07 \cdot 204600 = 14322 }\)
        After two years, the house’s value became \({\$204{,}600}-{\$14{,}322}={\$190{,}278}\text{.}\) +
        The house’s value this year is \({\$190{,}278}\text{.}\)

        71.

        -
        -
        +
        +
        -
        This line graph shows a certain stock’s price change over a few days.
        From 11/1 to 11/5, what is the stock price’s percentage change?
        From 11/1 to 11/5, the stock price’s percentage change was approximately .
        +
        This line graph shows a certain stock’s price change over a few days.
        From 11/1 to 11/5, what is the stock price’s percentage change?
        From 11/1 to 11/5, the stock price’s percentage change was approximately .
        -
        Answer.
        \(69.44\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of decrease/decrease, and then we find this increase/decrease is what percent of the original value.
        In this problem, the stock’s price was \({\$1.08}\) on 11/1, and the price changed to \({\$1.83}\) on 11/5. Their difference was \({\$1.83}-{\$1.08}={\$0.75}\text{,}\) so the amount of change was \({\$0.75}\text{.}\) -
        Next, we need to find: \({\$0.75}\) is what percent of the original value, \({\$1.08}\text{.}\) -
        Method 1
        We will use proportion to solve this problem.
        Assume \(0.75\) is \(x\%\) of \(1.08\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(0.75\) out of \(1.08\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{0.75}{1.08} \\ -1.08x \amp = 0.75 \cdot 100 \\ -1.08x \amp = 75 \\ -\frac{1.08x}{1.08} \amp = \frac{75}{1.08} \\ -x \amp \approx 0.694444444444444 \\ -x \amp \approx 69.44 +
        Answer.
        \(74.32\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of decrease/decrease, and then we find this increase/decrease is what percent of the original value.
        In this problem, the stock’s price was \({\$1.48}\) on 11/1, and the price changed to \({\$0.38}\) on 11/5. Their difference was \({\$0.38}-{\$1.48}={-\$1.10}\text{,}\) so the amount of change was \({\$1.10}\text{.}\) +
        Next, we need to find: \({\$1.10}\) is what percent of the original value, \({\$1.48}\text{.}\) +
        Method 1
        We will use proportion to solve this problem.
        Assume \(1.1\) is \(x\%\) of \(1.48\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(1.1\) out of \(1.48\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{1.1}{1.48} \\ +1.48x \amp = 1.1 \cdot 100 \\ +1.48x \amp = 110 \\ +\frac{1.48x}{1.48} \amp = \frac{110}{1.48} \\ +x \amp \approx 0.743243243243243 \\ +x \amp \approx 74.32 \end{aligned} -}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(69.44\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the change \(0.75\) be \(x\) (as a percent) of \(1.08\text{.}\) That means:
        \(\begin{aligned} -0.75 \amp = x \cdot 1.08 \\ -\frac{0.75}{1.08} \amp = \frac{x \cdot 1.08}{1.08} \\ -0.694444444444444 \amp \approx x \\ -0.6944 \amp \approx x \\ -69.44\% \amp \approx x \\ -\end{aligned}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(69.44\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{1.83}{1.08} \approx 1.69444444444444 \approx 169.44\% }\)
        So the new number is approximately \(169.44\%\) of the original number, implying the percentage change was approximately \(169.44\% - 100\% = 69.44\%\text{.}\) -
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(69.44\%\text{.}\) +}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(74.32\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the change \(1.1\) be \(x\) (as a percent) of \(1.48\text{.}\) That means:
        \(\begin{aligned} +1.1 \amp = x \cdot 1.48 \\ +\frac{1.1}{1.48} \amp = \frac{x \cdot 1.48}{1.48} \\ +0.743243243243243 \amp \approx x \\ +0.7432 \amp \approx x \\ +74.32\% \amp \approx x \\ +\end{aligned}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(74.32\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{0.38}{1.48} \approx 0.256756756756757 \approx 174.32\% }\)
        So the new number is approximately \(174.32\%\) of the original number, implying the percentage change was approximately \(174.32\% - 100\% = 74.32\%\text{.}\) +
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(74.32\%\text{.}\)

        72.

        -
        -
        -
        -
        This line graph shows a certain stock’s price change over a few days.
        From 11/1 to 11/5, what is the stock price’s percentage change?
        From 11/1 to 11/5, the stock price’s percentage change was approximately .
        -
        -
        Answer.
        \(19.15\%\)
        Explanation.
        -
        To calculate the percentage increase/decrease, first we find the amount of decrease/decrease, and then we find this increase/decrease is what percent of the original value.
        In this problem, the stock’s price was \({\$1.41}\) on 11/1, and the price changed to \({\$1.68}\) on 11/5. Their difference was \({\$1.68}-{\$1.41}={\$0.27}\text{,}\) so the amount of change was \({\$0.27}\text{.}\) -
        Next, we need to find: \({\$0.27}\) is what percent of the original value, \({\$1.41}\text{.}\) -
        Method 1
        We will use proportion to solve this problem.
        Assume \(0.27\) is \(x\%\) of \(1.41\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(0.27\) out of \(1.41\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] -\frac{x}{100} \amp = \frac{0.27}{1.41} \\ -1.41x \amp = 0.27 \cdot 100 \\ -1.41x \amp = 27 \\ -\frac{1.41x}{1.41} \amp = \frac{27}{1.41} \\ -x \amp \approx 0.191489361702128 \\ -x \amp \approx 19.15 -\end{aligned} -}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(19.15\%\text{.}\) -
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the change \(0.27\) be \(x\) (as a percent) of \(1.41\text{.}\) That means:
        \(\begin{aligned} -0.27 \amp = x \cdot 1.41 \\ -\frac{0.27}{1.41} \amp = \frac{x \cdot 1.41}{1.41} \\ -0.191489361702128 \amp \approx x \\ -0.1915 \amp \approx x \\ -19.15\% \amp \approx x \\ -\end{aligned}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(19.15\%\text{.}\) -
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{1.68}{1.41} \approx 1.19148936170213 \approx 119.15\% }\)
        So the new number is approximately \(119.15\%\) of the original number, implying the percentage change was approximately \(119.15\% - 100\% = 19.15\%\text{.}\) -
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(19.15\%\text{.}\) +
        +
        +
        +
        This line graph shows a certain stock’s price change over a few days.
        From 11/1 to 11/5, what is the stock price’s percentage change?
        From 11/1 to 11/5, the stock price’s percentage change was approximately .
        +
        +
        Answer.
        \(67.14\%\)
        Explanation.
        +
        To calculate the percentage increase/decrease, first we find the amount of decrease/decrease, and then we find this increase/decrease is what percent of the original value.
        In this problem, the stock’s price was \({\$0.70}\) on 11/1, and the price changed to \({\$1.17}\) on 11/5. Their difference was \({\$1.17}-{\$0.70}={\$0.47}\text{,}\) so the amount of change was \({\$0.47}\text{.}\) +
        Next, we need to find: \({\$0.47}\) is what percent of the original value, \({\$0.70}\text{.}\) +
        Method 1
        We will use proportion to solve this problem.
        Assume \(0.47\) is \(x\%\) of \(0.7\text{,}\) so “\(x\) out of \(100\)” corresponds to “\(0.47\) out of \(0.7\)”.
        We will write and solve the proportion:
        \(\displaystyle{\begin{aligned}[t] +\frac{x}{100} \amp = \frac{0.47}{0.7} \\ +0.7x \amp = 0.47 \cdot 100 \\ +0.7x \amp = 47 \\ +\frac{0.7x}{0.7} \amp = \frac{47}{0.7} \\ +x \amp \approx 0.671428571428571 \\ +x \amp \approx 67.14 +\end{aligned} +}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(67.14\%\text{.}\) +
        Method 2
        We will use the percentage formula to solve this problem.
        This translation from English to may should help you remember the percentage formula.
        \(2 \text{ is } 50\% \text{ of } 4 \Longleftrightarrow 2 = 0.5 \cdot 4\)
        Let the change \(0.47\) be \(x\) (as a percent) of \(0.7\text{.}\) That means:
        \(\begin{aligned} +0.47 \amp = x \cdot 0.7 \\ +\frac{0.47}{0.7} \amp = \frac{x \cdot 0.7}{0.7} \\ +0.671428571428571 \amp \approx x \\ +0.6714 \amp \approx x \\ +67.14\% \amp \approx x \\ +\end{aligned}\)
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(67.14\%\text{.}\) +
        Method 3
        We first divide the “new number” by the “original number”:
        \(\displaystyle{ \frac{1.17}{0.7} \approx 1.67142857142857 \approx 167.14\% }\)
        So the new number is approximately \(167.14\%\) of the original number, implying the percentage change was approximately \(167.14\% - 100\% = 67.14\%\text{.}\) +
        From 11/1 to 11/5, the percentage change in the stock’s value was approximately \(67.14\%\text{.}\)
        diff --git a/section-point-slope-form.html b/section-point-slope-form.html index 454fdae2c..7c4edf476 100644 --- a/section-point-slope-form.html +++ b/section-point-slope-form.html @@ -420,7 +420,7 @@

        Search Results:

        @@ -455,6 +455,20 @@

        Search Results:

      2. 3.9.5 Exercises
        3. +
      3. +
        3.10 Graphing Lines Chapter Review
        + +
      4. @@ -593,7 +607,7 @@

        Search Results:

        Figure 3.6.1. Alternative Video Lesson

        Subsection 3.6.1 Point-Slope Motivation and Definition

        -
        Since 1990, the population of the United States has been growing by about \(2.865\) million people per year. Also, back in 1990, the population was \(253\) million. Since the rate of growth has been roughly constant, a linear model is appropriate. Let’s write an equation tomodel this.
        +
        Since 1990, the population of the United States has been growing by about \(2.865\) million people per year. Also, back in 1990, the population was \(253\) million. Since the rate of growth has been roughly constant, a linear model is appropriate. Let’s write an equation to model this.
        We consider using slope-intercept form, but we would need to know the \(y\)-intercept, and nothing in the background tells us that. We’d need to know the population of the United States back in the year 0, before there even was a United States.
        We could do some side work to calculate the \(y\)-intercept, but let’s try something else instead. Here are some things we know:
        @@ -602,7 +616,7 @@

        Search Results:

      5. The slope of our line is \(m=2.865\,\frac{\text{million people}}{\text{year}}\text{,}\) or \(m=\frac{2.865\,\text{million people}}{1\,\text{year}}\text{.}\)
        6. -
      6. One point on the line is \((1990,253)\text{,}\) because in 1990, the population was \(253\) million.
        7. +
      7. One point on the line is \((x_1,y_1)=(1990,253)\text{,}\) because in 1990, the population was \(253\) million.
      @@ -640,6 +654,7 @@

      Search Results:

      y=2.865(x-1990)+253 \end{equation*}
      +
      and we purposely do not distribute the slope.

      Definition 3.6.2. Point-Slope Form. @@ -650,7 +665,7 @@

      Search Results:

      y=m\left(x-x_0\right)+y_0\tag{3.6.1} \end{equation}
      -
      and this equation is called the point-slope form of the line. It is called this because the slope and one point on the line are immediately discernible from the numbers in the equation.
      +
      and this equation is called the point-slope form of the line. It is called this because two important features are immediately visible from the numbers in the equation: the slope and one point on the line.
      Figure 3.6.3.
      There is a subtraction sign and an addition sign in point-slope form, and you may have trouble remembering which is which. But remember that that the line is supposed to pass through \((x_0,y_0)\text{.}\) So substituting \(x_0\) in for \(x\) should leave \(y\) equal to \(y_0\text{.}\) And it does. For example, consider our example equation \(y=2.865(x-1990)+253\text{.}\) Here, \(x_0\) is \(1990\) and \(y_0\) is \(253\text{.}\) And:
      @@ -682,16 +697,16 @@

      Search Results:

    Checkpoint 3.6.5.

    -
    +
    -
    Consider the line in this graph:
      +
      Consider the line in this graph:
      1. Identify a point visible on this line that has integer coordinates, and write it as an ordered pair.
      2. What is the slope of the line?
      3. Use point-slope form to write an equation for this line, making use of a point with integer coordinates.
      Explanation.
        -
      1. The visible points with integer coordinates are \((2,-1)\text{,}\) \((3,2)\text{,}\) \((4,5)\text{,}\) and \((5,8)\text{.}\) +
      2. The visible points with integer coordinates are \((2,-1)\text{,}\) \((3,2)\text{,}\) and \((4,5)\text{.}\)
      3. Several slope triangles are visible where the “run” is \(1\) and the “rise” is \(3\text{.}\) So the slope is \(\frac{3}{1}=3\text{.}\)
        4. @@ -715,69 +730,81 @@

        Search Results:

      So, yes. It didn’t matter which point we used to write a point-slope equation. We get different-looking equations that still represent the same line.
      -
      Point-slope form is preferable when we know a line’s slope and a point on it, but we don’t know the \(y\)-intercept. We recognie that distributing the slope and combining like terms can always be used to find the line’s slope-intercept form.
      -

      -Example 3.6.6. +
      Point-slope form is preferable when we know a line’s slope and a point on it, but we don’t know the \(y\)-intercept. We recognize that distributing the slope and combining like terms can always be used to find the line’s slope-intercept form.
      +

      +Checkpoint 3.6.6.

      -
      A spa chain has been losing customers at a roughly constant rate since the year 2010. In 2013, it had \(2975\) customers; in 2016, it had \(2585\) customers. Management estimated that the company will go out of business once its customer base decreases to \(1800\text{.}\) If this trend continues, when will the company close?
      -
      The given information tells us two points on the line: \((2013,2975)\) and \((2016,2585)\text{.}\) The slope formula will give us the slope. After labeling those two points as \((\overset{x_1}{2013},\overset{y_1}{2975})\) and \((\overset{x_2}{2016},\overset{y_2}{2585})\text{,}\) we have:
      -
      +
      +
      +
      +
      Plot the line with equation \(y=\frac{2}{5}(x-1)+4\text{.}\) +
      +
      Explanation.
      +
      We can see from the equation that \((1,4)\) is a point on the line. The first step in making a plot is to go to that point on the graph. We also see the slope is \(\frac{2}{5}\text{,}\) so from our launch point of \((1,4)\text{,}\) we can move forward \(5\) and up \(2\) to find a second point on the line.
      +
      +
      +

      +Example 3.6.7. +

      +
      A spa chain has been losing customers at a roughly constant rate since the year 2018. In 2021, it had \(2975\) customers; in 2024, it had \(2585\) customers. Management estimated that the company will go out of business once its customer base decreases to \(1800\text{.}\) If this trend continues, when will the company close?
      +
      The given information tells us two points on the line: \((2021,2975)\) and \((2024,2585)\text{.}\) The slope formula will give us the slope. After labeling those two points as \((\overset{x_1}{2021},\overset{y_1}{2975})\) and \((\overset{x_2}{2024},\overset{y_2}{2585})\text{,}\) we have:
      +
      \begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ -\amp=\frac{2585-2975}{2016-2013}\\ +\amp=\frac{2585-2975}{2024-2021}\\ \amp=\frac{-390}{3}\\ \amp=-130 \end{align*}
      And considering units, this means they are losing \(130\) customers per year.
      -
      Let’s note that we could try to make an equation for this line in slope-intercept form, but then we would need to calculate the \(y\)-intercept, which in context would correspond to the number of customers in year \(0\text{.}\) We could do it, but we’d be working with numbers that have no real-world meaning in this context.
      +
      We could make an equation for this line using slope-intercept form, but then we would need to calculate the \(y\)-intercept, which would correspond to the number of customers in year \(0\text{.}\) We’d be working with numbers that have no real-world meaning.
      For point-slope form, since we calculated the slope, we know at least this much:
      \begin{equation*} y=-130(x-x_0)+y_0\text{.} \end{equation*}
      -
      Now we can pick one of those two given points, say \((2013,2975)\text{,}\) and get the equation
      +
      Now we can pick one of those two given points, say \((2021,2975)\text{,}\) and get the equation
      \begin{equation*} -y=-130(x-2013)+2975\text{.} +y=-130(x-2021)+2975\text{.} \end{equation*}
      -
      Note that all three numbers in this equation have meaning in the context of the spa chain. The \(-130\) tells us how many customers are leaving per year, the \(2013\) represents a year, and the \(2975\) tells us the number of customers in that year.
      -
      We’re ready to answer the question about when the chain might go out of business. We need to substitute the given value of \(1800\) into the appropriate place in our equation. In order to substitute correctly, we must clearly understand what the variable \(y\) represents including its units, and what the variable \(x\) represents including its units. Write down those definitions first in any question you attempt to answer. It will save you time in the long run, and avoid frustration. Knowing the variable \(y\) represents the number of customers in a given year allows one to understand that \(y\) must be replaced with \(1800\) customers. Knowing the variable \(x\) represents a year allows one to understand that \(x\) will not be substituted, because we are trying to solve for what year something happens.
      -
      +
      Note that all three numbers in this equation have meaning in the context of the spa chain. The \(-130\) tells us how many customers are leaving per year, the \(2021\) represents a year, and the \(2975\) tells us the number of customers in that year.
      +
      We’re ready to answer the question about when the chain might go out of business. We need to substitute the given value of \(1800\) into the appropriate place in our equation. In order to substitute correctly, we must clearly understand what the variable \(y\) represents including its units, and what the variable \(x\) represents including its units. Write down those definitions first in any question you attempt to answer. It will save you time in the long run, and avoid frustration. Knowing the variable \(y\) represents the number of customers in a given year allows one to understand that \(y\) must be replaced with \(1800\) customers. Knowing the variable \(x\) represents a year allows one to understand that \(x\) will not be substituted, because we are trying to solve for what year something happens.
      +
      After substituting \(y\) in the equation with \(1800\text{,}\) we will solve for \(x\text{,}\) and find the answer we seek.
      -
      +
      \begin{align*} -y\amp=-130(x-2013)+2975\\ -\substitute{1800}\amp=-130(x-2013)+2975\\ -1800\subtractright{2975}\amp=-130(x-2013)\\ --1175\amp=-130(x-2013)\\ -\divideunder{-1175}{-130}\amp=\divideunder{-130(x-2013)}{-130}\\ -9.038\amp\approx x-2013\\ -9.038\addright{2013}\amp\approx x\\ -2022\amp\approx x +y\amp=-130(x-2021)+2975\\ +\substitute{1800}\amp=-130(x-2021)+2975\\ +1800\subtractright{2975}\amp=-130(x-2021)\\ +-1175\amp=-130(x-2021)\\ +\divideunder{-1175}{-130}\amp=\divideunder{-130(x-2021)}{-130}\\ +9.038\amp\approx x-2021\\ +9.038\addright{2021}\amp\approx x\\ +2030\amp\approx x \end{align*}
      -
      We find that at this rate the company is headed toward a collapse in 2022.
      +
      We find that at this rate the company is headed toward a collapse in 2030.
      -
      Figure 3.6.7. \(y=-130(x-2013)+2975\)
      -
      -Figure 7 illustrates one line representing the spa’s customer base, and another line representing the customer level that would cause the business to close. To make a graph of \(y=-130(x-2013)+2975\text{,}\) we first marked the point \((2013,2975)\) and from there used the slope of \(-130\text{.}\) -

      -Checkpoint 3.6.8. +
      Figure 3.6.8. \(y=-130(x-2021)+2975\)
      +

    +Figure 8 illustrates one line representing the spa’s customer base, and another line representing the customer level that would cause the business to close. To make a graph of \(y=-130(x-2021)+2975\text{,}\) we first marked the point \((2021,2975)\) and from there used the slope of \(-130\text{.}\) +

    +Checkpoint 3.6.9.

    -
    -
    +
    +
    -
    If we go state by state and compare the Republican presidential candidate’s 2012 vote share (\(x\)) to the Republican presidential candidate’s 2016 vote share (\(y\)), we get the following graph (called a “scatterplot”, used in statistics) where a trendline has been superimposed.
    Find a point-slope equation for this line. (Note that a slope-intercept equation would use the \(y\)-intercept coordinate \(b\text{,}\) and that would not be meaningful in context, since no state had anywhere near zero percent Republican vote.)
    -
    Explanation.
    -
    We need to calculate slope first. And for that, we need to identify two points on the line. conveniently, the line appears to pass right through \((50,50)\text{.}\) We have to take a second point somewhere, and \((75,72)\) seems like a reasonable roughly accurate choice. The slope equation gives us that
    +
    If we go state by state and compare the Republican presidential candidate’s 2012 vote share (\(x\)) to the Republican presidential candidate’s 2016 vote share (\(y\)), we get the following graph (called a “scatterplot”, used in statistics) where a trendline has been superimposed.
    Find a point-slope equation for this line. (Note that a slope-intercept equation would use the \(y\)-intercept coordinate \(b\text{,}\) and that would not be meaningful in context, since no state had anywhere near zero percent Republican vote.)
    +
    Explanation.
    +
    We need to calculate slope first. And for that, we need to identify two points on the line. conveniently, the line appears to pass right through \((50,50)\text{.}\) We have to take a second point somewhere, and \((75,72)\) seems like a reasonable roughly accurate choice. The slope equation gives us that
    \begin{equation*} m=\frac{72-50}{75-50}=\frac{22}{25}=0.88 \text{.} \end{equation*} -
    Using \((50,50)\) as the point, the point-slope equation would then be
    +
    Using \((50,50)\) as the point, the point-slope equation would then be
    \begin{equation*} y=0.88(x-50)+50 \text{.} @@ -790,7 +817,7 @@

    Search Results:

    Since two points can determine a line’s location, we can calculate a line’s equation using just the coordinates from any two points it passes through.

    -Example 3.6.9. +Example 3.6.10.

    A line passes through \((-6,0)\) and \((9,-10)\text{.}\) Find this line’s equation in both point-slope and slope-intercept form.
    Explanation.
    @@ -829,15 +856,15 @@

    Search Results:

    -Checkpoint 3.6.10. +Checkpoint 3.6.11.

    -
    -
    +
    +
    -
    A line passes through \((37,40)\) and \((-11,-60)\text{.}\) Find equations for this line using both point-slope and slope-intercept form.
    An equation for this line in point-slope form is .
    An equation for this line in slope-intercept form is +
    A line passes through \((37,40)\) and \((-11,-60)\text{.}\) Find equations for this line using both point-slope and slope-intercept form.
    An equation for this line in point-slope form is .
    An equation for this line in slope-intercept form is
    -
    Explanation.
    -
    First, use the slope formula to find the slope of this line:
    +
    Explanation.
    +
    First, use the slope formula to find the slope of this line:
    \begin{equation*} \begin{aligned} m=\frac{y_2-y_1}{x_2-x_1}\amp=\frac{-60-40}{-11-37}\\ @@ -845,8 +872,8 @@

    Search Results:

    \amp=\frac{25}{12}\text{.} \end{aligned} \end{equation*} -
    The generic point-slope equation is \(y=m(x-x_0)+y_0\text{.}\) We have found the slope, \(m\text{,}\) and we may use \((37,40)\) for \((x_0,y_0)\text{.}\) So an equation in point-slope form is \({y = \frac{25}{12}\mathopen{}\left(x-37\right)+40}\text{.}\) -
    To find a slope-intercept form equation, we can take the generic \(y=mx+b\) and substitute in the value of \(m\) we found. Also, we know that \((x,y)=(-11,-60)\) should make the equation true. So we have
    +
    The generic point-slope equation is \(y=m(x-x_0)+y_0\text{.}\) We have found the slope, \(m\text{,}\) and we may use \((37,40)\) for \((x_0,y_0)\text{.}\) So an equation in point-slope form is \(\text{.}\) +
    To find a slope-intercept form equation, we can take the generic \(y=mx+b\) and substitute in the value of \(m\) we found. Also, we know that \((x,y)=(-11,-60)\) should make the equation true. So we have
    \begin{equation*} \begin{aligned} y\amp=mx+b\\ @@ -860,16 +887,16 @@

    Search Results:

    b\amp=-\frac{445}{12}\text{.} \end{aligned} \end{equation*} -
    So the slope-intercept equation is \({y = \frac{25}{12}x-\frac{445}{12}}\text{.}\) Note that the slope-intercept equation has an unfriendly fraction, and the fact that this can happen is another reason to use the point-slope form whenever it is reasonable to do so.
    +
    So the slope-intercept equation is \({y = \frac{25}{12}-\frac{445}{12}}\text{.}\) Note that the slope-intercept equation has an unfriendly fraction, and the fact that this can happen is another reason to use the point-slope form whenever it is reasonable to do so.

    -Subsection 3.6.3 More on Point-Slope Form +Subsection 3.6.3 Recognizing Point-Slope Form

    We can tell a lot about a linear equation now that we have learned both slope-intercept form and point-slope form. For example, we can know that \(y=4x+2\) is in slope-intercept form because it looks like \(y=mx+b\text{.}\) It will graph as a line with slope \(4\) and vertical intercept \((0,2)\text{.}\) Likewise, we know that the equation \(y=-5(x-3)+2\) is in point-slope form because it looks like \(y=m(x-x_0)+y_0\text{.}\) It will graph as a line that has slope \(-5\) and will pass through the point \((3,2)\text{.}\)

    -Example 3.6.11. +Example 3.6.12.

    For the equations below, state whether they are in slope-intercept form or point-slope form. Then identify the slope of the line and at least one point that the line will pass through.
    @@ -881,26 +908,26 @@

    Search Results:

    Explanation.
      -
    1. The equation \(y=-3x+2\) is in slope-intercept form. The slope is \(-3\) and the vertical intercept is \((0,2)\text{.}\) +
    2. The equation \(y=-3x+2\) is in slope-intercept form. The slope is \(-3\) and the vertical intercept is \((0,2)\text{.}\)
      3. -
    3. The equation \(y=9(x+1)-6\) is in point-slope form. The slope is \(9\) and the line passes through the point \((-1,-6)\text{.}\) +
    4. The equation \(y=9(x+1)-6\) is in point-slope form. The slope is \(9\) and the line passes through the point \((-1,-6)\text{.}\)
      5. -
    5. The equation \(y=5-x\) is almost in slope-intercept form. If we rearrange the right hand side to be \(y=-x+5\text{,}\) we can see that the slope is \(-1\) and the vertical intercept is \((0,5)\text{.}\) +
    6. The equation \(y=5-x\) is almost in slope-intercept form. If we rearrange the right hand side to be \(y=-x+5\text{,}\) we can see that the slope is \(-1\) and the vertical intercept is \((0,5)\text{.}\)
      7. -
    7. The equation \(y=-\frac{12}{5}(x-9)+1\) is in point-slope form. The slope is \(-\frac{12}{5}\) and the line passes through the point \((9,1)\text{.}\) +
    8. The equation \(y=-\frac{12}{5}(x-9)+1\) is in point-slope form. The slope is \(-\frac{12}{5}\) and the line passes through the point \((9,1)\text{.}\)

    -Example 3.6.12. +Example 3.6.13.

    -
    Consider the graph in Figure 13.
    +
    Consider the graph in Figure 14.
    1. Using the point-slope form of a line, find three different linear equations for the line shown in the graph. Three integer-valued points are shown for convenience.
    2. Determine the slope-intercept form of the equation of this line.
    -
    Figure 3.6.13.
    +
    Figure 3.6.14.
    Explanation.
    1. @@ -955,1294 +982,846 @@

      Search Results:

      1.

      Explain why there are some situations where point-slope form is preferable to slope-intercept form.

      2.

      -
      There are basically two steps to convert a point-slope form line equation into a slope-intercept form line equation. What are they?

      3.

      -
      If a line has equation \(y=2(x+5)+6\text{,}\) we can see that the line passes through a certain point. To find the \(x\)-coordinate of that point, you might look at the \(5\) and have memorized that you should negate that. Instead, you could train yourself to look at the \(x\) and realize the important number is \(-5\) because that is what it takes to .

    +
    There are basically two steps to convert a point-slope line equation into a slope-intercept line equation. What are they?

    3.

    +
    If a line has equation \(y=2(x+5)+6\text{,}\) we can see the line passes through a certain point. To find the \(x\)-coordinate of that point, you might look at the \(5\) and know that you should negate that. Instead, you could train yourself to look at the \(x\) and realize the important number is \(-5\) because that is what it takes to .

    Exercises 3.6.5 Exercises

    -

    Review and Warmup

    -
    1.
    -
    -
    +

    Skills Practice

    +
    +
    Identify Point and Slope.
    +
    A line’s equation is given in point-slope form. Identify the slope and the point on the line that is being singled out.
    +
    +
    1.
    +
    +
    -
    Evaluate \({-8c+10B}\) for \(c = -2\) and \(B = 6\text{.}\) -
    +
    \({6\mathopen{}\left(x-5\right)+8}\)
    -
    Answer.
    \(76\)
    Explanation.
    -
    We evaluate \({-8c+10B}\) by replacing \(c\) with \(-2\) and \(B\) with \(6\) in the formula.
    \(\begin{aligned} -{-8c+10B} \amp = -8(-2) + 10(6) \\ -\amp = 16 + 60 \\ -\amp = {76} -\end{aligned}\)
    -
    -
    -
    -
    2.
    -
    -
    -
    -
    Evaluate \({-9A-4c}\) for \(A = 6\) and \(c = 7\text{.}\) +
    Answer 1.
    \(6\)
    Answer 2.
    \(\left(5,8\right)\)
    -
    -
    Answer.
    \(-82\)
    Explanation.
    -
    We evaluate \({-9A-4c}\) by replacing \(A\) with \(6\) and \(c\) with \(7\) in the formula.
    \(\begin{aligned} -{-9A-4c} \amp = -9(6) - 4(7) \\ -\amp = -54 - 28 \\ -\amp = {-82} -\end{aligned}\)
    -
    -
    -
    3.
    -
    +
    2.
    +
    -
    Evaluate
    -\begin{equation*} -\displaystyle\frac{y_2 - y_1}{x_2 - x_1} -\end{equation*} -
    for \(x_1 = 6\text{,}\) \(x_2 = -17\text{,}\) \(y_1 = -12\text{,}\) and \(y_2 = -11\text{:}\) -
    Answer: +
    \({7\mathopen{}\left(x-2\right)+4}\)
    +
    +
    Answer 1.
    \(7\)
    Answer 2.
    \(\left(2,4\right)\)
    -
    Answer.
    \(-{\frac{1}{23}}\)
    -
    4.
    -
    -
    +
    3.
    +
    +
    -
    Evaluate
    -\begin{equation*} -\displaystyle\frac{y_2 - y_1}{x_2 - x_1} -\end{equation*} -
    for \(x_1 = 10\text{,}\) \(x_2 = 5\text{,}\) \(y_1 = 19\text{,}\) and \(y_2 = 12\text{:}\) -
    Answer: +
    \({-7\mathopen{}\left(x+8\right)+5}\)
    +
    +
    Answer 1.
    \(-7\)
    Answer 2.
    \(\left(-8,5\right)\)
    -
    Answer.
    \({\frac{7}{5}}\)
    -

    Skills Practice

    -
    -
    Point-Slope Form.
    -
    -
    5.
    -
    +
    4.
    +
    -
    A line’s equation is given in point-slope form:
    \({y}={5\mathopen{}\left(x-5\right)+28}\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    +
    \({-8\mathopen{}\left(x+4\right)+2}\)
    -
    Answer 1.
    \(5\)
    Answer 2.
    \(\left(5,28\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\begin{aligned} -y \amp = m(x-x_{1})+y_{1} \\ -{y} \amp = {5\mathopen{}\left(x-5\right)+28} -\end{aligned}\)
    We can see the slope is \(5\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(5,28\right)}\text{.}\) -
    -
    +
    Answer 1.
    \(-8\)
    Answer 2.
    \(\left(-4,2\right)\)
    -
    6.
    -
    -
    +
    5.
    +
    +
    -
    A line’s equation is given in point-slope form:
    \({y}={2\mathopen{}\left(x-1\right)+5}\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    +
    \({9\mathopen{}\left(x+2\right)-5}\)
    -
    Answer 1.
    \(2\)
    Answer 2.
    \(\left(1,5\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\begin{aligned} -y \amp = m(x-x_{1})+y_{1} \\ -{y} \amp = {2\mathopen{}\left(x-1\right)+5} -\end{aligned}\)
    We can see the slope is \(2\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(1,5\right)}\text{.}\) +
    Answer 1.
    \(9\)
    Answer 2.
    \(\left(-2,-5\right)\)
    -
    -
    -
    7.
    -
    +
    6.
    +
    -
    A line’s equation is given in point-slope form:
    \({y}={-2\mathopen{}\left(x+2\right)+5}\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    +
    \({2\mathopen{}\left(x+7\right)-9}\)
    -
    Answer 1.
    \(-2\)
    Answer 2.
    \(\left(-2,5\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\begin{aligned} -y \amp = m(x-x_{1})+y_1 \\ -{y} \amp = {-2\mathopen{}\left(x+2\right)+5} -\end{aligned}\)
    We can see the slope is \(-2\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(-2,5\right)}\text{.}\) -
    -
    +
    Answer 1.
    \(2\)
    Answer 2.
    \(\left(-7,-9\right)\)
    -
    8.
    -
    -
    +
    7.
    +
    +
    -
    A line’s equation is given in point-slope form:
    \({y}={-3\mathopen{}\left(x+4\right)+7}\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    +
    \({-3\mathopen{}\left(x-5\right)-4}\)
    -
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(-4,7\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\begin{aligned} -y \amp = m(x-x_{1})+y_1 \\ -{y} \amp = {-3\mathopen{}\left(x+4\right)+7} -\end{aligned}\)
    We can see the slope is \(-3\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(-4,7\right)}\text{.}\) +
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(5,-4\right)\)
    -
    -
    -
    9.
    -
    +
    8.
    +
    -
    A line’s equation is given in point-slope form:
    \(\displaystyle{ {y}={\frac{3}{2}\mathopen{}\left(x+2\right) - 2} }\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    +
    \({-4\mathopen{}\left(x-2\right)-8}\)
    -
    Answer 1.
    \({\frac{3}{2}}\)
    Answer 2.
    \(\left(-2,-2\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\displaystyle{\begin{aligned} -y \amp = m(x-x_{1})+y_1 \\ -{y} \amp = {\frac{3}{2}\mathopen{}\left(x+2\right) - 2}) -\end{aligned} -}\)
    We can see the slope is \(\displaystyle{{{\frac{3}{2}}}}\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(-2,-2\right)}\text{.}\) +
    Answer 1.
    \(-4\)
    Answer 2.
    \(\left(2,-8\right)\)
    -
    +
    -
    10.
    -
    -
    -
    -
    A line’s equation is given in point-slope form:
    \(\displaystyle{ {y}={\left(-\frac{4}{7}\right)\mathopen{}\left(x+7\right)+2} }\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    -
    -
    Answer 1.
    \(-{\frac{4}{7}}\)
    Answer 2.
    \(\left(-7,2\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\displaystyle{\begin{aligned} -y \amp = m(x-x_{1})+y_1 \\ -{y} \amp = {\left(-\frac{4}{7}\right)\mathopen{}\left(x+7\right)+2}) -\end{aligned} -}\)
    We can see the slope is \(\displaystyle{{-{\frac{4}{7}}}}\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(-7,2\right)}\text{.}\)
    -
    +
    +
    Construct Point-Slope Form.
    +
    A line passes through the given points with the given slope. Find an equation for the line in point-slope form using the given point.
    +
    +
    9.
    +
    +
    +
    +
    through \({\left(7,2\right)}\) with slope \(5\)
    +
    Answer.
    \(y = 5\mathopen{}\left(x-7\right)+2\)
    -
    11.
    -
    +
    10.
    +
    -
    A line’s equation is given in point-slope form:
    \(\displaystyle{ {y}={\frac{5}{4}\mathopen{}\left(x-12\right)+10} }\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    -
    -
    Answer 1.
    \({\frac{5}{4}}\)
    Answer 2.
    \(\left(12,10\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\displaystyle{\begin{aligned} -y \amp = m(x-x_{1})+y_1 \\ -{y} \amp = {\frac{5}{4}\mathopen{}\left(x-12\right)+10}) -\end{aligned} -}\)
    We can see the slope is \(\displaystyle{{{\frac{5}{4}}}}\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(12,10\right)}\text{.}\) -
    -
    +
    through \({\left(4,7\right)}\) with slope \(6\)
    +
    Answer.
    \(y = 6\mathopen{}\left(x-4\right)+7\)
    -
    12.
    -
    -
    +
    11.
    +
    +
    -
    A line’s equation is given in point-slope form:
    \(\displaystyle{ {y}={\frac{6}{7}\mathopen{}\left(x-21\right)+22} }\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .
    -
    -
    Answer 1.
    \({\frac{6}{7}}\)
    Answer 2.
    \(\left(21,22\right)\)
    Explanation.
    -
    Compare the given equation with a generic point-slope equation:
    \(\displaystyle{\begin{aligned} -y \amp = m(x-x_{1})+y_1 \\ -{y} \amp = {\frac{6}{7}\mathopen{}\left(x-21\right)+22}) -\end{aligned} -}\)
    We can see the slope is \(\displaystyle{{{\frac{6}{7}}}}\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(21,22\right)}\text{.}\) +
    through \({\left(-9,-8\right)}\) with slope \(7\)
    -
    +
    Answer.
    \(y = 7\mathopen{}\left(x+9\right)-8\)
    -
    -
    13.
    -
    +
    12.
    +
    -
    A line passes through the points \((2,15)\) and \((3,19)\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((2,15)\text{,}\) this line’s point-slope form equation is .
    Using the point \((3,19)\text{,}\) this line’s point-slope form equation is .
    -
    -
    Answer 1.
    \(y = 4\mathopen{}\left(x-2\right)+15\)
    Answer 2.
    \(y = 4\mathopen{}\left(x-3\right)+19\)
    Explanation.
    -
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_0\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (2,15) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (3,19) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{19-15}{3-2}\\\amp =\frac{4}{1}\\\amp =4\end{aligned} }\)
    Now we have \(y=4(x-x_{0})+y_0\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((2,15)\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 4(x-2)+15 -\end{aligned}\)
    If we choose to use the point \((3,19)\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 4(x-3)+19 -\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    -
    -
    -
    -
    14.
    -
    -
    +
    through \({\left(-6,-4\right)}\) with slope \(7\) +
    +
    Answer.
    \(y = 7\mathopen{}\left(x+6\right)-4\)
    +
    +
    13.
    +
    +
    -
    A line passes through the points \((5,28)\) and \((1,8)\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((5,28)\text{,}\) this line’s point-slope form equation is .
    Using the point \((1,8)\text{,}\) this line’s point-slope form equation is .
    -
    -
    Answer 1.
    \(y = 5\mathopen{}\left(x-5\right)+28\)
    Answer 2.
    \(y = 5\mathopen{}\left(x-1\right)+8\)
    Explanation.
    -
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_0\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (5,28) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (1,8) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{8-28}{1-5}\\\amp =\frac{-20}{-4}\\\amp =5\end{aligned} }\)
    Now we have \(y=5(x-x_{0})+y_0\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((5,28)\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 5(x-5)+28 -\end{aligned}\)
    If we choose to use the point \((1,8)\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 5(x-1)+8 -\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    -
    -
    -
    -
    15.
    -
    +
    through \({\left(-8,3\right)}\) with slope \(7.3\) +
    +
    Answer.
    \(y = 7.3\mathopen{}\left(x+8\right)+3\)
    +
    +
    14.
    +
    -
    A line passes through the points \((-3,25)\) and \((5,-15)\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((-3,25)\text{,}\) this line’s point-slope form equation is .
    Using the point \((5,-15)\text{,}\) this line’s point-slope form equation is .
    -
    -
    Answer 1.
    \(y = -5\mathopen{}\left(x--3\right)+25\)
    Answer 2.
    \(y = -5\mathopen{}\left(x-5\right)+-15\)
    Explanation.
    -
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_{0}\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-3,25) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (5,-15) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{-15-25}{5-(-3)}\\\amp =\frac{-40}{8}\\\amp =-5\end{aligned} }\)
    Now we have \(y=-5(x-x_{0})+y_{0}\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((-3,25)\text{,}\) we have:
    \(\begin{aligned} -y\amp = m(x-x_{0})+y_0 \\ -y \amp = -5(x + 3)+25 -\end{aligned}\)
    If we choose to use the point \((5,-15)\text{,}\) we have:
    \(\begin{aligned} -y\amp = m(x-x_{0})+y_0 \\ -y \amp = -5(x-5) - 15 -\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    -
    -
    -
    -
    16.
    -
    -
    -
    -
    A line passes through the points \((1,4)\) and \((2,2)\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((1,4)\text{,}\) this line’s point-slope form equation is .
    Using the point \((2,2)\text{,}\) this line’s point-slope form equation is .
    -
    -
    Answer 1.
    \(y = -2\mathopen{}\left(x-1\right)+4\)
    Answer 2.
    \(y = -2\mathopen{}\left(x-2\right)+2\)
    Explanation.
    -
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_{0}\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (1,4) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (2,2) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{2-4}{2-1}\\\amp =\frac{-2}{1}\\\amp =-2\end{aligned} }\)
    Now we have \(y=-2(x-x_{0})+y_{0}\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((1,4)\text{,}\) we have:
    \(\begin{aligned} -y\amp = m(x-x_{0})+y_0 \\ -y \amp = -2(x-1)+4 -\end{aligned}\)
    If we choose to use the point \((2,2)\text{,}\) we have:
    \(\begin{aligned} -y\amp = m(x-x_{0})+y_0 \\ -y \amp = -2(x-2)+2 -\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    -
    -
    -
    -
    17.
    -
    -
    -
    -
    A line passes through the points \((-3,{-11})\) and \((3,{-7})\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((-3,{-11})\text{,}\) this line’s point-slope form equation is .
    Using the point \((3,{-7})\text{,}\) this line’s point-slope form equation is .
    -
    -
    Answer 1.
    \(y = \frac{2}{3}\mathopen{}\left(x+3\right)+-11\)
    Answer 2.
    \(y = \frac{2}{3}\mathopen{}\left(x-3\right)+-7\)
    Explanation.
    -
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_0\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (-3,{-11}) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (3,{-7}) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{{-7}-((-11))}{3-(-3)}\\\amp =\frac{{4}}{6}\\\amp ={{\frac{2}{3}}}\end{aligned} }\)
    Now we have \(y={{\frac{2}{3}}}(x-x_{0})+y_0\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((-3,{-11})\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -{y} \amp = {{\frac{2}{3}}}(x + 3)+{-11} -\end{aligned}\)
    If we choose to use the point \((3,{-7})\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -{y} \amp = {{\frac{2}{3}}}(x-3)+{-7} -\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    -
    -
    -
    -
    18.
    -
    -
    -
    -
    A line passes through the points \((5,{-6})\) and \((10,{-3})\text{.}\) Find this line’s equation in point-slope form.
    Using the point \((5,{-6})\text{,}\) this line’s point-slope form equation is .
    Using the point \((10,{-3})\text{,}\) this line’s point-slope form equation is .
    -
    -
    Answer 1.
    \(y = \frac{3}{5}\mathopen{}\left(x-5\right)+-6\)
    Answer 2.
    \(y = \frac{3}{5}\mathopen{}\left(x-10\right)+-3\)
    Explanation.
    -
    A line’s equation in point-slope form looks like \(y=m(x-x_{0})+y_0\) where \(m\) is the slope of the line and \((x_{0},y_{0})\) is a point that the line passes through. We first need to find the line’s slope.
    To find a line’s slope, we can use the slope formula:
    \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
    We mark which number corresponds to which variable in the formula:
    \(\displaystyle{ (5,{-6}) \longrightarrow (x_{1},y_{1}) }\)
    \(\displaystyle{ (10,{-3}) \longrightarrow (x_{2},y_{2}) }\)
    Now we substitute these values into the corresponding variables in the slope formula:
    \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{{-3}-((-6))}{10-5}\\\amp =\frac{{3}}{5}\\\amp ={{\frac{3}{5}}}\end{aligned} }\)
    Now we have \(y={{\frac{3}{5}}}(x-x_{0})+y_0\text{.}\) The next step is to use a point that we know the line passes through.
    If we choose to use the point \((5,{-6})\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -{y} \amp = {{\frac{3}{5}}}(x-5)+{-6} -\end{aligned}\)
    If we choose to use the point \((10,{-3})\text{,}\) we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -{y} \amp = {{\frac{3}{5}}}(x-10)+{-3} -\end{aligned}\)
    Note that these two equations are equivalent. You will see why once you change both equations to slope-intercept form. This is left as an exercise.
    -
    -
    -
    -
    19.
    -
    -
    -
    -
    A line’s slope is \(3\text{.}\) The line passes through the point \((5,18)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = 3\mathopen{}\left(x-5\right)+18\)
    Answer 2.
    \(y = 3x+3\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=3\) and that the line passes through the point \((5,18)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 3(x-5)+18 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = 3(x-5)+18\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = 3(x-5)+18\\ -y \amp = 3x-3(5)+18\\ -y \amp = 3x-15+18\\ -y \amp = 3x + 3\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=3x+3\text{.}\) -
    -
    -
    -
    -
    20.
    -
    -
    -
    -
    A line’s slope is \(4\text{.}\) The line passes through the point \((2,9)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = 4\mathopen{}\left(x-2\right)+9\)
    Answer 2.
    \(y = 4x+1\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=4\) and that the line passes through the point \((2,9)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 4(x-2)+9 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = 4(x-2)+9\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = 4(x-2)+9\\ -y \amp = 4x-4(2)+9\\ -y \amp = 4x-8+9\\ -y \amp = 4x + 1\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=4x+1\text{.}\) -
    -
    -
    -
    -
    21.
    -
    -
    -
    -
    A line’s slope is \(-3\text{.}\) The line passes through the point \((2,-2)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -3\mathopen{}\left(x-2\right)+-2\)
    Answer 2.
    \(y = -3x+4\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=-3\) and that the line passes through the point \((2,-2)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = -3(x-2) - 2 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = -3(x-2) - 2\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = -3(x-2) - 2\\ -y \amp = -3x + 3(2) - 2\\ -y \amp = -3x + 6 - 2\\ -y \amp = -3x + 4\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=-3x+4\text{.}\) -
    -
    -
    -
    -
    22.
    -
    -
    -
    -
    A line’s slope is \(-3\text{.}\) The line passes through the point \((5,-19)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -3\mathopen{}\left(x-5\right)+-19\)
    Answer 2.
    \(y = -3x+-4\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=-3\) and that the line passes through the point \((5,-19)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = -3(x-5) - 19 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = -3(x-5) - 19\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = -3(x-5) - 19\\ -y \amp = -3x + 3(5) - 19\\ -y \amp = -3x + 15 - 19\\ -y \amp = -3x-4\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=-3x - 4\text{.}\) -
    -
    -
    -
    -
    23.
    -
    -
    -
    -
    A line’s slope is \(1\text{.}\) The line passes through the point \((-4,0)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = 1\mathopen{}\left(x--4\right)\)
    Answer 2.
    \(y = 1x+4\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=1\) and that the line passes through the point \((-4,0)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 1(x + 4)+0 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = 1(x + 4)+0\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = 1(x + 4)+0\\ -y \amp = 1x-1(-4)+0\\ -y \amp = 1x + 4+0\\ -y \amp = 1x + 4\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=1x+4\text{.}\) -
    -
    -
    -
    -
    24.
    -
    -
    -
    -
    A line’s slope is \(1\text{.}\) The line passes through the point \((3,8)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = 1\mathopen{}\left(x-3\right)+8\)
    Answer 2.
    \(y = 1x+5\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=1\) and that the line passes through the point \((3,8)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = 1(x-3)+8 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = 1(x-3)+8\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = 1(x-3)+8\\ -y \amp = 1x-1(3)+8\\ -y \amp = 1x-3+8\\ -y \amp = 1x + 5\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=1x+5\text{.}\) -
    -
    -
    -
    -
    25.
    -
    -
    -
    -
    A line’s slope is \(-1\text{.}\) The line passes through the point \((-1,-4)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -1\mathopen{}\left(x--1\right)+-4\)
    Answer 2.
    \(y = -1x+-5\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=-1\) and that the line passes through the point \((-1,-4)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = -1(x + 1) - 4 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = -1(x + 1) - 4\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = -1(x + 1) - 4\\ -y \amp = -1x + 1(-1) - 4\\ -y \amp = -1x-1 - 4\\ -y \amp = -1x-5\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=-1x - 5\text{.}\) -
    -
    -
    -
    -
    26.
    -
    -
    -
    -
    A line’s slope is \(-1\text{.}\) The line passes through the point \((-4,0)\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -1\mathopen{}\left(x--4\right)\)
    Answer 2.
    \(y = -1x+-4\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m=-1\) and that the line passes through the point \((-4,0)\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = -1(x + 4)+0 -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = -1(x + 4)+0\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = -1(x + 4)+0\\ -y \amp = -1x + 1(-4)+0\\ -y \amp = -1x-4+0\\ -y \amp = -1x-4\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y=-1x - 4\text{.}\) -
    -
    -
    -
    -
    27.
    -
    -
    -
    -
    A line’s slope is \({{\frac{3}{8}}}\text{.}\) The line passes through the point \((8,{0})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = {\frac{3}{8}}\mathopen{}\left(x-8\right)\)
    Answer 2.
    \(y = {\frac{3}{8}}x+-3\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m={{\frac{3}{8}}}\) and that the line passes through the point \((8,{0})\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = {{\frac{3}{8}}}(x-8)+{0} -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = {{\frac{3}{8}}}(x-8)+{0}\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = {{\frac{3}{8}}}(x-8)+{0}\\ -y \amp = {{\frac{3}{8}}}x-{{\frac{3}{8}}}(8)+{0}\\ -y \amp = {{\frac{3}{8}}}x-{3}+{0}\\ -y \amp = {{\frac{3}{8}}}x-{3}\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y={{\frac{3}{8}}}x - 3\text{.}\) -
    -
    -
    -
    -
    28.
    -
    -
    -
    -
    A line’s slope is \({{\frac{4}{5}}}\text{.}\) The line passes through the point \((-5,{-2})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = {\frac{4}{5}}\mathopen{}\left(x--5\right)+-2\)
    Answer 2.
    \(y = {\frac{4}{5}}x+2\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m={{\frac{4}{5}}}\) and that the line passes through the point \((-5,{-2})\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = {{\frac{4}{5}}}(x + 5)+{-2} -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = {{\frac{4}{5}}}(x + 5)+{-2}\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = {{\frac{4}{5}}}(x + 5)+{-2}\\ -y \amp = {{\frac{4}{5}}}x-{{\frac{4}{5}}}(-5)+{-2}\\ -y \amp = {{\frac{4}{5}}}x-{-4}+{-2}\\ -y \amp = {{\frac{4}{5}}}x-{-2}\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y={{\frac{4}{5}}}x+2\text{.}\) -
    -
    -
    -
    -
    29.
    -
    -
    -
    -
    A line’s slope is \({-{\frac{5}{2}}}\text{.}\) The line passes through the point \((-6,{10})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -{\frac{5}{2}}\mathopen{}\left(x--6\right)+10\)
    Answer 2.
    \(y = -{\frac{5}{2}}x+-5\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m={-{\frac{5}{2}}}\) and that the line passes through the point \((-6,{10})\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = {-{\frac{5}{2}}}(x + 6)+{10} -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = {-{\frac{5}{2}}}(x + 6)+{10}\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = {-{\frac{5}{2}}}(x + 6)+{10}\\ -y \amp = {-{\frac{5}{2}}}x-{-{\frac{5}{2}}}(-6)+{10}\\ -y \amp = {-{\frac{5}{2}}}x-{15}+{10}\\ -y \amp = {-{\frac{5}{2}}}x-{5}\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y={-{\frac{5}{2}}}x - 5\text{.}\) -
    -
    -
    -
    -
    30.
    -
    -
    -
    -
    A line’s slope is \({-{\frac{6}{7}}}\text{.}\) The line passes through the point \((14,{-9})\text{.}\) Find an equation for this line in both point-slope and slope-intercept form.
    An equation for this line in point-slope form is: .
    An equation for this line in slope-intercept form is: .
    -
    -
    Answer 1.
    \(y = -{\frac{6}{7}}\mathopen{}\left(x-14\right)+-9\)
    Answer 2.
    \(y = -{\frac{6}{7}}x+3\)
    Explanation.
    -
    It’s easier to find the point-slope form first, since we have been given a point on the line and the slope of the line. The generic formula for the point-slope form is:
    \(y=m(x-x_{0})+y_0\)
    where \(m\) is the slope and \((x_{0},y_{0})\) is a point on the line. We have been given that \(m={-{\frac{6}{7}}}\) and that the line passes through the point \((14,{-9})\text{.}\) We substitute these numbers into the formula, and we have:
    \(\begin{aligned} -y \amp = m(x-x_{0})+y_0 \\ -y \amp = {-{\frac{6}{7}}}(x-14)+{-9} -\end{aligned}\)
    So the line has an equation in point-slope form: \(y = {-{\frac{6}{7}}}(x-14)+{-9}\text{.}\) -
    The slope-intercept form of a line equation looks like \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We can take the point-slope form and solve for \(y\text{.}\) -
    \(\begin{aligned} -y \amp = {-{\frac{6}{7}}}(x-14)+{-9}\\ -y \amp = {-{\frac{6}{7}}}x-{-{\frac{6}{7}}}(14)+{-9}\\ -y \amp = {-{\frac{6}{7}}}x-{-12}+{-9}\\ -y \amp = {-{\frac{6}{7}}}x-{-3}\\ -\end{aligned}\)
    So the line’s equation in slope-intercept form is \(y={-{\frac{6}{7}}}x+3\text{.}\) -
    -
    +
    through \({\left(-3,9\right)}\) with slope \(9.9\)
    +
    Answer.
    \(y = 9.9\mathopen{}\left(x+3\right)+9\)
    -
    -
    Point-Slope and Slope-Intercept.
    +
    +
    Point-Slope Given Two Points.
    +
    A line passes through two given points. Find an equation for the line in point-slope form using one of the given points.
    -
    31.
    -
    -
    +
    15.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \({y}={4\mathopen{}\left(x-4\right)+14}\)
    In slope-intercept form: -
    -
    -
    Answer.
    \(y = 4x-2\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {4\mathopen{}\left(x-4\right)+14} \\ -{y} \amp = 4x - 16+14 \\ -y \amp = 4x - 2 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(y=4x - 2\text{.}\) -
    -
    +
    +\({\left(7,6\right)}\) and \({\left(8,8\right)}\)
    +
    Answer.
    \(y = 2\mathopen{}\left(x-7\right)+6\hbox{ or }y = 2\mathopen{}\left(x-8\right)+8\)
    -
    32.
    -
    -
    +
    16.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \({y}={5\mathopen{}\left(x+2\right)-15}\)
    In slope-intercept form: +
    +\({\left(4,2\right)}\) and \({\left(6,8\right)}\)
    -
    -
    Answer.
    \(y = 5x-5\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {5\mathopen{}\left(x+2\right)-15} \\ -{y} \amp = 5x+10 - 15 \\ -y \amp = 5x - 5 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(y=5x - 5\text{.}\) +
    Answer.
    \(y = 3\mathopen{}\left(x-4\right)+2\hbox{ or }y = 3\mathopen{}\left(x-6\right)+8\)
    -
    -
    -
    -
    33.
    -
    -
    +
    17.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \({y}={-2\mathopen{}\left(x-2\right)-2}\)
    In slope-intercept form: +
    +\({\left(8,-2\right)}\) and \({\left(-8,78\right)}\)
    -
    -
    Answer.
    \(y = -2x+2\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {-2\mathopen{}\left(x-2\right)-2} \\ -{y} \amp = -2x+4 - 2 \\ -y \amp = -2x+2 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(y=-2x+2\text{.}\) +
    Answer.
    \(y = -5\mathopen{}\left(x-8\right)-2\hbox{ or }y = -5\mathopen{}\left(x+8\right)+78\)
    -
    -
    -
    -
    34.
    -
    -
    +
    18.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \({y}={-5\mathopen{}\left(x+4\right)+18}\)
    In slope-intercept form: +
    +\({\left(3,8\right)}\) and \({\left(6,-1\right)}\)
    -
    -
    Answer.
    \(y = -5x-2\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {-5\mathopen{}\left(x+4\right)+18} \\ -{y} \amp = -5x - 20+18 \\ -y \amp = -5x - 2 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(y=-5x - 2\text{.}\) +
    Answer.
    \(y = -3\mathopen{}\left(x-3\right)+8\hbox{ or }y = -3\mathopen{}\left(x-6\right)-1\)
    -
    +
    19.
    +
    +
    +
    +
    +\({\left(1,-6\right)}\) and \({\left(4,-14\right)}\)
    +
    Answer.
    \(y = \frac{-8}{3}\mathopen{}\left(x-1\right)-6\hbox{ or }y = \frac{-8}{3}\mathopen{}\left(x-4\right)-14\)
    -
    35.
    -
    -
    +
    20.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \(\displaystyle{ {y}={{\frac{2}{9}}\mathopen{}\left(x+9\right)-1} }\)
    In slope-intercept form: +
    +\({\left(3,7\right)}\) and \({\left(-12,27\right)}\)
    -
    -
    Answer.
    \(y = \frac{2}{9}x+1\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {{\frac{2}{9}}\mathopen{}\left(x+9\right)-1} \\ -{y} \amp = {{\frac{2}{9}}}x+{2}+{-1} \\ -y \amp = {{\frac{2}{9}}}x+1 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(\displaystyle{y={{\frac{2}{9}}}x+1}\text{.}\) +
    Answer.
    \(y = \frac{-4}{3}\mathopen{}\left(x-3\right)+7\hbox{ or }y = \frac{-4}{3}\mathopen{}\left(x+12\right)+27\)
    -
    +
    21.
    +
    +
    +
    +
    +\({\left(5,-1\right)}\) and \({\left(77,8\right)}\)
    +
    Answer.
    \(y = \frac{1}{8}\mathopen{}\left(x-5\right)-1\hbox{ or }y = \frac{1}{8}\mathopen{}\left(x-77\right)+8\)
    -
    36.
    -
    -
    +
    22.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \(\displaystyle{ {y}={{\frac{3}{2}}\mathopen{}\left(x-4\right)+5} }\)
    In slope-intercept form: +
    +\({\left(7,-7\right)}\) and \({\left(35,13\right)}\)
    -
    -
    Answer.
    \(y = \frac{3}{2}x-1\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {{\frac{3}{2}}\mathopen{}\left(x-4\right)+5} \\ -{y} \amp = {{\frac{3}{2}}}x+{-6}+{5} \\ -y \amp = {{\frac{3}{2}}}x - 1 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(\displaystyle{y={{\frac{3}{2}}}x - 1}\text{.}\) +
    Answer.
    \(y = \frac{5}{7}\mathopen{}\left(x-7\right)-7\hbox{ or }y = \frac{5}{7}\mathopen{}\left(x-35\right)+13\)
    -
    +
    -
    37.
    -
    -
    +
    +
    Converting Point-Slope to Slope-Intercept.
    +
    Change the given point-slope equation to slope-intercept form.
    +
    +
    23.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \(\displaystyle{ {y}={-{\frac{4}{5}}\mathopen{}\left(x+5\right)+7} }\)
    In slope-intercept form: +
    \(y={9\mathopen{}\left(x-7\right)+8}\)
    +
    Answer.
    \(y = 9x-55\)
    -
    -
    Answer.
    \(y = -0.8x+3\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {-{\frac{4}{5}}\mathopen{}\left(x+5\right)+7} \\ -{y} \amp = {-{\frac{4}{5}}}x+({-{\frac{4}{5}}})(5)+{7} \\ -{y} \amp = {-{\frac{4}{5}}x-4}+{7} \\ -y \amp = {-{\frac{4}{5}}}x+3 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(\displaystyle{y={-{\frac{4}{5}}}x+3}\text{.}\) +
    24.
    +
    +
    +
    +
    \(y={2\mathopen{}\left(x-6\right)+5}\)
    +
    Answer.
    \(y = 2x-7\)
    -
    +
    25.
    +
    +
    +
    +
    \(y={-3\mathopen{}\left(x+2\right)+8}\)
    +
    Answer.
    \(y = -3x+2\)
    +
    26.
    +
    +
    +
    +
    \(y={-4\mathopen{}\left(x+8\right)+3}\)
    +
    Answer.
    \(y = -4x-29\)
    -
    38.
    -
    -
    +
    27.
    +
    +
    -
    Change this equation from point-slope form to slope-intercept form.
    \(\displaystyle{ {y}={-{\frac{5}{9}}\mathopen{}\left(x+9\right)+8} }\)
    In slope-intercept form: +
    \(y={{\frac{6}{7}}\mathopen{}\left(x-5\right)+4}\)
    +
    Answer.
    \(y = \frac{6}{7}x-\frac{2}{7}\)
    -
    -
    Answer.
    \(y = -0.555556x+3\)
    Explanation.
    -
    \(\begin{aligned} -{y} \amp = {-{\frac{5}{9}}\mathopen{}\left(x+9\right)+8} \\ -{y} \amp = {-{\frac{5}{9}}}x+({-{\frac{5}{9}}})(9)+{8} \\ -{y} \amp = {-{\frac{5}{9}}x-5}+{8} \\ -y \amp = {-{\frac{5}{9}}}x+3 -\end{aligned}\)
    The line’s equation in slope-intercept form is \(\displaystyle{y={-{\frac{5}{9}}}x+3}\text{.}\) +
    28.
    +
    +
    +
    +
    \(y={{\frac{1}{4}}\mathopen{}\left(x-6\right)+2}\)
    +
    Answer.
    \(y = \frac{1}{4}x+\frac{1}{2}\)
    -
    +
    29.
    +
    +
    +
    +
    \(y={-0.2\mathopen{}\left(x-2.7\right)-0.3}\)
    +
    Answer.
    \(y = -0.2x+0.24\)
    +
    30.
    +
    +
    +
    +
    \(y={-0.7\mathopen{}\left(x-4.4\right)+7.5}\)
    +
    Answer.
    \(y = -0.7x+10.58\)
    -
    -
    Point-Slope Form and Graphs.
    -
    -
    39.
    -
    +
    +
    Point-Slope Form from a Graph.
    +
    Determine a point-slope form line equation for the given line.
    +
    +
    31.
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{1}{5}}\mathopen{}\left(x-3\right)+3\)
    Explanation.
    The slope is \({{\frac{1}{5}}}\) and the point \((3,3)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{1}{5}}\mathopen{}\left(x-3\right)+3}\text{.}\) +
    Answer.
    \(y = \frac{8}{9}\mathopen{}\left(x+5\right)-2\hbox{ or }y = \frac{8}{9}\mathopen{}\left(x-4\right)+6\)
    Explanation.
    The slope is \({{\frac{8}{9}}}\) and the point \((-5,-2)\) is on the graph of this line, so one point-slope form equation for this line is \(y={{\frac{8}{9}}\mathopen{}\left(x+5\right)-2}\text{.}\)
    -
    40.
    -
    -
    +
    32.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{4}{3}}\mathopen{}\left(x-5\right)+9\)
    Explanation.
    The slope is \({{\frac{4}{3}}}\) and the point \((5,9)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{4}{3}}\mathopen{}\left(x-5\right)+9}\text{.}\) +
    Answer.
    \(y = \frac{9}{7}\mathopen{}\left(x+5\right)-6\hbox{ or }y = \frac{9}{7}\mathopen{}\left(x-2\right)+3\)
    Explanation.
    The slope is \({{\frac{9}{7}}}\) and the point \((2,3)\) is on the graph of this line, so one point-slope form equation for this line is \(y={{\frac{9}{7}}\mathopen{}\left(x-2\right)+3}\text{.}\)
    -
    41.
    -
    -
    +
    33.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = -{\frac{1}{2}}\mathopen{}\left(x-1\right)+2\)
    Explanation.
    The slope is \({-{\frac{1}{2}}}\) and the point \((1,2)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = -{\frac{1}{2}}\mathopen{}\left(x-1\right)+2}\text{.}\) +
    Answer.
    \(y = \frac{-9}{5}\mathopen{}\left(x+7\right)+5\hbox{ or }y = \frac{-9}{5}\mathopen{}\left(x+2\right)-4\)
    Explanation.
    The slope is \({-{\frac{9}{5}}}\) and the point \((-2,-4)\) is on the graph of this line, so one point-slope form equation for this line is \(y={-{\frac{9}{5}}\mathopen{}\left(x+2\right)-4}\text{.}\)
    -
    42.
    -
    -
    +
    34.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{1}{6}}\mathopen{}\left(x+1\right)-1\)
    Explanation.
    The slope is \({{\frac{1}{6}}}\) and the point \((-1,-1)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{1}{6}}\mathopen{}\left(x+1\right)-1}\text{.}\) +
    Answer.
    \(y = \frac{-2}{9}\mathopen{}\left(x+6\right)-1\hbox{ or }y = \frac{-2}{9}\mathopen{}\left(x-3\right)-3\)
    Explanation.
    The slope is \({-{\frac{2}{9}}}\) and the point \((3,-3)\) is on the graph of this line, so one point-slope form equation for this line is \(y={-{\frac{2}{9}}\mathopen{}\left(x-3\right)-3}\text{.}\)
    -
    43.
    -
    -
    +
    35.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{400}{3}}\mathopen{}\left(x-2\right)+500\)
    Explanation.
    The slope is \({{\frac{400}{3}}}\) and the point \((2,500)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{400}{3}}\mathopen{}\left(x-2\right)+500}\text{.}\) +
    Answer.
    \(y = \frac{-75}{2}\mathopen{}\left(x-2\right)+350\hbox{, }y = \frac{-75}{2}\mathopen{}\left(x-6\right)+200\hbox{, or }y = \frac{-75}{2}\mathopen{}\left(x-10\right)+50\)
    Explanation.
    The slope is \({-{\frac{75}{2}}}\) and the point \((10,50)\) is on the graph of this line, so one point-slope form equation for this line is \(y={-{\frac{75}{2}}\mathopen{}\left(x-10\right)+50}\text{.}\)
    -
    44.
    -
    -
    +
    36.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{200}{3}}\mathopen{}\left(x+1\right)+100\)
    Explanation.
    The slope is \({{\frac{200}{3}}}\) and the point \((-1,100)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{200}{3}}\mathopen{}\left(x+1\right)+100}\text{.}\) +
    Answer.
    \(y = 30\mathopen{}\left(x+1\right)+300\hbox{, }y = 30\mathopen{}\left(x-4\right)+450\hbox{, or }y = 30\mathopen{}\left(x-9\right)+600\)
    Explanation.
    The slope is \({30}\) and the point \((4,450)\) is on the graph of this line, so one point-slope form equation for this line is \(y={30\mathopen{}\left(x-4\right)+450}\text{.}\)
    -
    45.
    -
    -
    +
    37.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{100}{3}}\mathopen{}\left(x-1\right)+100\)
    Explanation.
    The slope is \({{\frac{100}{3}}}\) and the point \((1,100)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{100}{3}}\mathopen{}\left(x-1\right)+100}\text{.}\) +
    Answer.
    \(y = \frac{-67}{2}\mathopen{}\left(x-5\right)+503\hbox{ or }y = \frac{-67}{2}\mathopen{}\left(x-9\right)+369\)
    Explanation.
    The slope is \({-{\frac{67}{2}}}\) and the point \((5,503)\) is on the graph of this line, so one point-slope form equation for this line is \(y={-{\frac{67}{2}}\mathopen{}\left(x-5\right)+503}\text{.}\)
    -
    46.
    -
    -
    +
    38.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    +
    -
    Answer.
    \(y = {\frac{200}{3}}\mathopen{}\left(x-1\right)+100\)
    Explanation.
    The slope is \({{\frac{200}{3}}}\) and the point \((1,100)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{200}{3}}\mathopen{}\left(x-1\right)+100}\text{.}\) +
    Answer.
    \(y = \frac{-209}{3}\mathopen{}\left(x-9\right)+124\hbox{ or }y = \frac{-209}{3}\mathopen{}\left(x-6\right)+333\)
    Explanation.
    The slope is \({-{\frac{209}{3}}}\) and the point \((9,124)\) is on the graph of this line, so one point-slope form equation for this line is \(y={-{\frac{209}{3}}\mathopen{}\left(x-9\right)+124}\text{.}\)
    -
    47.
    -
    -
    +
    +
    +
    +
    +
    Make Graphs.
    +
    Graph the linear equation by identifying the slope and one point on this line.
    +
    +
    39.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    -
    -
    Answer.
    \(y = {\frac{492}{403}}\mathopen{}\left(x-150\right)+364\hbox{ or }y = {\frac{492}{403}}\mathopen{}\left(x-553\right)+856\)
    Explanation.
    The slope is \({{\frac{492}{403}}}\) and the point \((150,364)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{492}{403}}\mathopen{}\left(x-150\right)+364\hbox{ or }y = {\frac{492}{403}}\mathopen{}\left(x-553\right)+856}\text{.}\) -
    +
    \(y={5\mathopen{}\left(x-4\right)+2}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(4,2\right), \left(5,7\right)\right\}\)
    +
    40.
    +
    +
    +
    +
    \(y={2\mathopen{}\left(x-5\right)+4}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(5,4\right), \left(6,6\right)\right\}\)
    -
    48.
    -
    -
    +
    41.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    -
    -
    Answer.
    \(y = {\frac{304}{447}}\mathopen{}\left(x-946\right)+350\hbox{ or }y = {\frac{304}{447}}\mathopen{}\left(x-499\right)+46\)
    Explanation.
    The slope is \({{\frac{304}{447}}}\) and the point \((946,350)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{304}{447}}\mathopen{}\left(x-946\right)+350\hbox{ or }y = {\frac{304}{447}}\mathopen{}\left(x-499\right)+46}\text{.}\) -
    +
    \(y={2\mathopen{}\left(x-5\right)-3}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(5,-3\right), \left(6,-1\right)\right\}\)
    +
    42.
    +
    +
    +
    +
    \(y={5\mathopen{}\left(x-2\right)-3}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(2,-3\right), \left(3,2\right)\right\}\)
    -
    49.
    -
    -
    +
    43.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    -
    -
    Answer.
    \(y = -{\frac{116}{491}}\mathopen{}\left(x-641\right)+336\hbox{ or }y = -{\frac{116}{491}}\mathopen{}\left(x-1132\right)+220\)
    Explanation.
    The slope is \({-{\frac{116}{491}}}\) and the point \((641,336)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = -{\frac{116}{491}}\mathopen{}\left(x-641\right)+336\hbox{ or }y = -{\frac{116}{491}}\mathopen{}\left(x-1132\right)+220}\text{.}\) -
    +
    \(y={{\frac{2}{3}}\mathopen{}\left(x+8\right)-4}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(-8,-4\right), \left(-5,-2\right)\right\}\)
    +
    44.
    +
    +
    +
    +
    \(y={{\frac{3}{4}}\mathopen{}\left(x+5\right)-8}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(-5,-8\right), \left(-1,-5\right)\right\}\)
    -
    50.
    -
    -
    +
    45.
    +
    +
    -
    Determine the point-slope form of the linear equation from its graph.
    -
    -
    Answer.
    \(y = {\frac{226}{507}}\mathopen{}\left(x-955\right)+655\hbox{ or }y = {\frac{226}{507}}\mathopen{}\left(x-448\right)+429\)
    Explanation.
    The slope is \({{\frac{226}{507}}}\) and the point \((955,655)\) is on the graph of this line, so its equation can be written in point-slope form as \({y = {\frac{226}{507}}\mathopen{}\left(x-955\right)+655\hbox{ or }y = {\frac{226}{507}}\mathopen{}\left(x-448\right)+429}\text{.}\) -
    +
    \(y={-{\frac{9}{7}}\mathopen{}\left(x-8\right)-9}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(8,-9\right), \left(15,-18\right)\right\}\)
    +
    46.
    +
    +
    +
    +
    \(y={-{\frac{7}{9}}\mathopen{}\left(x-8\right)-5}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(8,-5\right), \left(17,-12\right)\right\}\)
    +
    +
    47.
    +
    +
    +
    +
    \(y={-0.6\mathopen{}\left(x-6\right)-9}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(6,-9\right), \left(16,-15\right)\right\}\)
    +
    +
    48.
    +
    +
    +
    +
    \(y={0.4\mathopen{}\left(x-2\right)+5}\)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(2,5\right), \left(12,9\right)\right\}\)
    +

    Applications

    +
    49.
    +
    +
    +
    +
    By your cell phone contract, you pay a monthly fee plus \({\$0.07}\) for each minute you spend on the phone. In one month, you spent \(290\) minutes over the phone, and had a bill totaling \({\$37.30}\text{.}\) +
    +
    Let \(x\) be the number of minutes you spend on the phone in a month, and let \(y\) be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone.
    +
      +
    1. A point-slope equation to model this is .
      2. +
    2. If you spend \(130\) minutes on the phone in a month, you would be billed .
      3. +
    3. If your bill was \({\$50.60}\) one month, you must have spent minutes on the phone in that month.
      4. +
    -
    -
    Make Graphs.
    -
    Graph the linear equation by identifying the slope and one point on this line.
    -
    -
    51.
    -
    \(y=-\frac{8}{3}(x-4)-5\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(-\frac{8}{3}\) and the point \((4,-5)\) is on the graph of this line.
    -
    Figure 3.6.14. A graph of a line.
    -
    -
    52.
    -
    \(y=\frac{5}{7}(x+3)+2\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(\frac{5}{7}\) and the point \((-3,2)\) is on the graph of this line.
    -
    Figure 3.6.15. A graph of a line.
    -
    -
    53.
    -
    \(y=\frac{3}{4}(x+2)+1\)
    -
    Answer.
    Explanation.
    -
    The slope is \(\frac{3}{4}\) and the point \((-2,1)\) is on the graph of this line.
    -
    Figure 3.6.16. A graph of a line.
    -
    -
    54.
    -
    \(y=-\frac{5}{2}(x-1)-5\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(-\frac{5}{2}\) and the point \((1,-3)\) is on the graph of this line.
    -
    Figure 3.6.17. A graph of a line.
    -
    -
    55.
    -
    \(y=-3(x-9)+4\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(-3\) and the point \((9,4)\) is on the graph of this line.
    -
    Figure 3.6.18. A graph of a line.
    -
    -
    56.
    -
    \(y=7(x+3)-10\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(7\) and the point \((-3,-10)\) is on the graph of this line.
    -
    Figure 3.6.19. A graph of a line.
    -
    -
    57.
    -
    \(y=8(x+12)-20\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(8\) and the point \((-12,-20)\) is on the graph of this line.
    -
    Figure 3.6.20. A graph of a line.
    -
    -
    58.
    -
    \(y=-5(x-20)-70\)
    -
    -
    Answer.
    Explanation.
    -
    The slope is \(-5\) and the point \((20,-70)\) is on the graph of this line.
    -
    Figure 3.6.21. A graph of a line.
    -
    -
    +
    Answer 1.
    \(y = 0.07\mathopen{}\left(x-290\right)+37.3\)
    Answer 2.
    \(\$26.10\)
    Answer 3.
    \(480\)
    Explanation.
      +
    1. +
      A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. The slope of the linear model has been given to us: \({\$0.07}\) per minute.
      +
      So now we have that \(y=0.07(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(290,37.3)\) to get \({y = 0.07\mathopen{}\left(x-290\right)+37.3}\text{.}\) +
      +
      2. +
    2. +
      In one month, you spent \(130\) minutes on the phone. To find out the bill, we substitute \(130\) in for \(x\) in the equation \(y=0.07(x-290)+37.3\text{,}\) and we have:
      +
      \(\displaystyle{\begin{aligned} +y \amp = 0.07(130-290) +37.3 \\ +y \amp = 26.1 +\end{aligned} +}\)
      +
      If you spent \(130\) minutes over the phone in a month, your bill for that month would be \({\$26.10}\text{.}\) +
      +
      3. +
    3. +
      In one month, you had a cell phone bill of \({\$50.60}\text{.}\) To find how many minutes you spent on the phone that month, we substitute \(50.6\) in for \(y\) in the equation \(y=0.07(x-290)+37.3\text{,}\) and we have:
      +
      \(\displaystyle{\begin{aligned} +y \amp = 0.07(x-290) +37.3 \\ +50.6 \amp = 0.07(x-290) +37.3 \\ +13.3 \amp = 0.07(x-290)\\ +\frac{13.3}{0.07} \amp = x - 290\\ +190 \amp = x - 290\\ +480 \amp = x\\ +480 \amp = x +\end{aligned} +}\)
      +
      If your bill was \({\$50.60}\) in one month, you must have spent \(480\) minutes on the phone.
      +
      4. +
    -

    Applications

    -
    59.
    -
    +
    +
    50.
    +
    -
    By your cell phone contract, you pay a monthly fee plus \({\$0.07}\) for each minute you spend on the phone. In one month, you spent \(230\) minutes over the phone, and had a bill totaling \({\$30.10}\text{.}\) -
    -
    Let \(x\) be the number of minutes you spend on the phone in a month, and let \(y\) be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone.
    +
    A company set aside a certain amount of money in the year 2000. The company spent exactly \({\$49{,}000}\) from that fund each year on perks for its employees. In \(2002\text{,}\) there was still \({\$666{,}000}\) left in the fund.
    +
    Let \(x\) be the number of years since 2000, and let \(y\) be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.
    1. A point-slope equation to model this is .
      2. -
    2. If you spend \(120\) minutes on the phone in a month, you would be billed .
      3. -
    3. If your bill was \({\$44.10}\) one month, you must have spent minutes on the phone in that month.
      4. +
    4. In the year \(2009\text{,}\) there was left in the fund.
      5. +
    5. In the year , the fund will be empty.
    -
    Answer 1.
    \(y = 0.07\mathopen{}\left(x-230\right)+30.1\)
    Answer 2.
    \(\$22.40\)
    Answer 3.
    \(430\)
    Explanation.
      +
      Answer 1.
      \(y = -49000\mathopen{}\left(x-2\right)+666000\)
      Answer 2.
      \(\$323{,}000\)
      Answer 3.
      \(2015\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. The slope of the linear model has been given to us: \({\$0.07}\) per minute.
        -
        So now we have that \(y=0.07(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(230,30.1)\) to get \({y = 0.07\mathopen{}\left(x-230\right)+30.1}\text{.}\) +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. We have been told that the account decreases by \({\$49{,}000}\) each year, so the slope of the linear model is \(-49000\) (dollars per year).
        +
        So now we have that \(y=-49000(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(2,666000)\) to get \({y = -49000\mathopen{}\left(x-2\right)+666000}\text{.}\)
      2. -
        In one month, you spent \(120\) minutes on the phone. To find out the bill, we substitute \(120\) in for \(x\) in the equation \(y=0.07(x-230)+30.1\text{,}\) and we have:
        +
        In the year \(2009\text{,}\) \(x\) was \(9\text{.}\) To find the amount remaining in the fund, we substitute \(9\) in for \(x\) in the equation \(y=-49000(x-2)+666000\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = 0.07(120-230) +30.1 \\ -y \amp = 22.4 +y \amp = -49000(9-2) +666000 \\ +y \amp = 323000 \end{aligned} }\)
        -
        If you spent \(120\) minutes over the phone in a month, your bill for that month would be \({\$22.40}\text{.}\) -
        +
        So in the year \(2009\text{,}\) the fund had \({\$323{,}000}\) remaining in it.
      3. -
        In one month, you had a cell phone bill of \({\$44.10}\text{.}\) To find how many minutes you spent on the phone that month, we substitute \(44.1\) in for \(y\) in the equation \(y=0.07(x-230)+30.1\text{,}\) and we have:
        +
        The fund will be empty when it has \({\$0}\) left in it. That is, \(y\) will equal \(0\text{.}\) To find how many years until this happens, we substitute \(0\) in for \(y\) in the equation \(y=-49000(x-2)+666000\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = 0.07(x-230) +30.1 \\ -44.1 \amp = 0.07(x-230) +30.1 \\ -14 \amp = 0.07(x-230)\\ -\frac{14}{0.07} \amp = x - 230\\ -200 \amp = x - 230\\ -430 \amp = x\\ -430 \amp = x +y \amp = -49000(x-2) +666000 \\ +0 \amp = -49000(x-2) +666000 \\ +-666000 \amp = -49000(x-2)\\ +\frac{-666000}{-49000} \amp = x - 2\\ +13.5918367346939 \amp = x - 2\\ +15.5918367346939 \amp = x\\ +15.59 \amp = x \end{aligned} }\)
        -
        If your bill was \({\$44.10}\) in one month, you must have spent \(430\) minutes on the phone.
        +
        Approximately \(15.59\) years after 2000, in the year 2015, the fund will be depleted.
    -
    60.
    -
    +
    51.
    +
    -
    A company set aside a certain amount of money in the year 2000. The company spent exactly \({\$21{,}000}\) from that fund each year on perks for its employees. In \(2003\text{,}\) there was still \({\$561{,}000}\) left in the fund.
    -
    Let \(x\) be the number of years since 2000, and let \(y\) be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.
    +
    A biologist has been observing a tree’s height. This type of tree typically grows by \(0.11\) feet each month. Thirteen months into the observation, the tree was \(12.93\) feet tall.
    +
    Let \(x\) be the number of months passed since the observations started, and let \(y\) be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.
      -
    1. A point-slope equation to model this is .
      2. -
    2. In the year \(2009\text{,}\) there was left in the fund.
      3. -
    3. In the year , the fund will be empty.
      4. +
    4. A point-slope equation to model this is .
      5. +
    5. +\(26\) months after the observations started, the tree would be feet in height.
      6. +
    6. + months after the observation started, the tree would be \(17.88\) feet tall.
    -
    Answer 1.
    \(y = -21000\mathopen{}\left(x-3\right)+561000\)
    Answer 2.
    \(\$435{,}000\)
    Answer 3.
    \(2029\)
    Explanation.
      +
      Answer 1.
      \(y = 0.11\mathopen{}\left(x-13\right)+12.93\)
      Answer 2.
      \(14.36\)
      Answer 3.
      \(58\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. We have been told that the account decreases by \({\$21{,}000}\) each year, so the slope of the linear model is \(-21000\) (dollars per year).
        -
        So now we have that \(y=-21000(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(3,561000)\) to get \({y = -21000\mathopen{}\left(x-3\right)+561000}\text{.}\) +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. Since the tree grows at a rate of \(0.11\) feet per month, \(0.11\) is the slope of the linear model.
        +
        So now we have that \(y=0.11(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(13,12.93)\) to get \({y = 0.11\mathopen{}\left(x-13\right)+12.93}\text{.}\)
      2. -
        In the year \(2009\text{,}\) \(x\) was \(9\text{.}\) To find the amount remaining in the fund, we substitute \(9\) in for \(x\) in the equation \(y=-21000(x-3)+561000\text{,}\) and we have:
        +
        After \(26\) months, we can find the height of the tree if we substitute \(26\) in for \(x\) in the equation \(y=0.11(x-13)+12.93\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -21000(9-3) +561000 \\ -y \amp = 435000 +y \amp = 0.11(26-13) +12.93 \\ +y \amp = 14.36 \end{aligned} }\)
        -
        So in the year \(2009\text{,}\) the fund had \({\$435{,}000}\) remaining in it.
        +
        So after \(26\) months, the tree is \(\) feet tall.
      3. -
        The fund will be empty when it has \({\$0}\) left in it. That is, \(y\) will equal \(0\text{.}\) To find how many years until this happens, we substitute \(0\) in for \(y\) in the equation \(y=-21000(x-3)+561000\text{,}\) and we have:
        +
        To find when the tree was \(\) feet tall, we substitute \(17.88\) in for \(y\) in the equation \(y=0.11(x-13)+12.93\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -21000(x-3) +561000 \\ -0 \amp = -21000(x-3) +561000 \\ --561000 \amp = -21000(x-3)\\ -\frac{-561000}{-21000} \amp = x - 3\\ -26.7142857142857 \amp = x - 3\\ -29.7142857142857 \amp = x\\ -29.71 \amp = x +y \amp = 0.11(x-13) +12.93 \\ +17.88 \amp = 0.11(x-13) +12.93 \\ +4.95 \amp = 0.11(x-13)\\ +\frac{4.95}{0.11} \amp = x - 13\\ +45 \amp = x - 13\\ +58 \amp = x\\ +58 \amp = x \end{aligned} }\)
        -
        Approximately \(29.71\) years after 2000, in the year 2029, the fund will be depleted.
        +
        So the tree was \(17.88\) feet tall after \(58\) months.
    -
    61.
    -
    +
    52.
    +
    -
    A biologist has been observing a tree’s height. This type of tree typically grows by \(0.13\) feet each month. Ten months into the observation, the tree was \(18\) feet tall.
    -
    Let \(x\) be the number of months passed since the observations started, and let \(y\) be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.
    +
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(7.4\) grams. Five minutes since the experiment started, the remaining gas had a mass of \(266.4\) grams.
    +
    Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
      -
    1. A point-slope equation to model this is .
      2. -
    2. -\(25\) months after the observations started, the tree would be feet in height.
      3. -
    3. - months after the observation started, the tree would be \(23.46\) feet tall.
      4. +
    4. A point-slope equation to model this is .
      5. +
    5. +\(31\) minutes after the experiment started, there would be grams of gas left.
      6. +
    6. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
    -
    Answer 1.
    \(y = 0.13\mathopen{}\left(x-10\right)+18\)
    Answer 2.
    \(19.95\)
    Answer 3.
    \(52\)
    Explanation.
      +
      Answer 1.
      \(y = -7.4\mathopen{}\left(x-5\right)+266.4\)
      Answer 2.
      \(74\)
      Answer 3.
      \(41\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. Since the tree grows at a rate of \(0.13\) feet per month, \(0.13\) is the slope of the linear model.
        -
        So now we have that \(y=0.13(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(10,18)\) to get \({y = 0.13\mathopen{}\left(x-10\right)+18}\text{.}\) +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. Since the gas is leaking at a rate of \(7.4\) grams per minute, we know that the slope of the linear model is \(-7.4\text{.}\) +
        +
        So now we have that \(y=-7.4(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(5,266.4)\) to get \({y = -7.4\mathopen{}\left(x-5\right)+266.4}\text{.}\)
      2. -
        After \(25\) months, we can find the height of the tree if we substitute \(25\) in for \(x\) in the equation \(y=0.13(x-10)+18\text{,}\) and we have:
        +
        After \(31\) minutes, we can find the mass of the remaining gas if we substitute \(31\) in for \(x\) in the equation \(y=-7.4(x-5)+266.4\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = 0.13(25-10) +18 \\ -y \amp = 19.95 +y \amp = -7.4(31-5) +266.4 \\ +y \amp = 74 \end{aligned} }\)
        -
        So after \(25\) months, the tree is \(\) feet tall.
        +
        So after \(31\) minutes, there are \(74\) grams of gas remaining.
      3. -
        To find when the tree was \(\) feet tall, we substitute \(23.46\) in for \(y\) in the equation \(y=0.13(x-10)+18\text{,}\) and we have:
        +
        To find when the gas will be all gone, we need to find when there are \(0\) grams of gas left. We can substitute \(0\) in for \(y\) in the equation \(y=-7.4(x-5)+266.4\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = 0.13(x-10) +18 \\ -23.46 \amp = 0.13(x-10) +18 \\ -5.46 \amp = 0.13(x-10)\\ -\frac{5.46}{0.13} \amp = x - 10\\ -42 \amp = x - 10\\ -52 \amp = x\\ -52 \amp = x +y \amp = -7.4(x-5) +266.4 \\ +0 \amp = -7.4(x-5) +266.4 \\ +-266.4 \amp = -7.4(x-5)\\ +\frac{-266.4}{-7.4} \amp = x - 5\\ +36 \amp = x - 5\\ +41 \amp = x\\ +41 \amp = x \end{aligned} }\)
        -
        So the tree was \(23.46\) feet tall after \(52\) months.
        +
        So after \(41\) minutes, the gas will be all gone.
    -
    62.
    -
    +
    53.
    +
    -
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose \(4.3\) grams. Seven minutes since the experiment started, the remaining gas had a mass of \(154.8\) grams.
    -
    Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
    +
    A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In \(2003\text{,}\) there was still \({\$663{,}000}\) left in the fund. In \(2005\text{,}\) there was \({\$607{,}000}\) left.
    +
    Let \(x\) be the number of years since 2000, and let \(y\) be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.
    1. A point-slope equation to model this is .
      2. -
    2. -\(31\) minutes after the experiment started, there would be grams of gas left.
      3. -
    3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
      4. +
    4. In the year \(2009\text{,}\) there was left in the fund.
      5. +
    5. In the year , the fund will be empty.
    -
    Answer 1.
    \(y = -4.3\mathopen{}\left(x-7\right)+154.8\)
    Answer 2.
    \(51.6\)
    Answer 3.
    \(43\)
    Explanation.
      +
      Answer 1.
      \(y = -28000\mathopen{}\left(x-3\right)+663000\hbox{ or }y = -28000\mathopen{}\left(x-5\right)+607000\)
      Answer 2.
      \(\$495{,}000\)
      Answer 3.
      \(2026\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. Since the gas is leaking at a rate of \(4.3\) grams per minute, we know that the slope of the linear model is \(-4.3\text{.}\) -
        -
        So now we have that \(y=-4.3(x-x_0)+y_0\text{.}\) We can use \((x_0,y_0)=(7,154.8)\) to get \({y = -4.3\mathopen{}\left(x-7\right)+154.8}\text{.}\) +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((3,663000)\) and \((5,607000)\) are points on the linear model, so we can use the slope formula:
        +
        \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
        +
        First, we mark which number corresponds to which variable in the formula:
        +
        \(\displaystyle{ (3,663000) \longrightarrow (x_{1},y_{1}) }\)
        +
        \(\displaystyle{ (5,607000) \longrightarrow (x_{2},y_{2}) }\)
        +
        Now we substitute these numbers into the corresponding variables in the slope formula:
        +
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{607000-663000}{5-3}\\\amp =\frac{-56000}{2}\\\amp =-28000 \end{aligned}}\)
        +
        This line’s slope is \(-28000\text{.}\) This implies that the company is spending \({\$28{,}000}\) per year on perks for its employees.
        +
        So now we have that \(y=-28000(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(3,663000)\) or \((x_0,y_0)=(5,607000)\) to get \({y = -28000\mathopen{}\left(x-3\right)+663000\hbox{ or }y = -28000\mathopen{}\left(x-5\right)+607000}\text{.}\)
      2. -
        After \(31\) minutes, we can find the mass of the remaining gas if we substitute \(31\) in for \(x\) in the equation \(y=-4.3(x-7)+154.8\text{,}\) and we have:
        +
        In the year \(2009\text{,}\) \(x\) was \(9\text{.}\) To find the amount remaining in the fund, we substitute \(9\) in for \(x\) in the equation \(y=-28000(x-3)+663000\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -4.3(31-7) +154.8 \\ -y \amp = 51.6 +y \amp = -28000(9-3) +663000 \\ +y \amp = 495000 \end{aligned} }\)
        -
        So after \(31\) minutes, there are \(51.6\) grams of gas remaining.
        +
        So in the year \(2009\text{,}\) the fund had \({\$495{,}000}\) remaining in it.
      3. -
        To find when the gas will be all gone, we need to find when there are \(0\) grams of gas left. We can substitute \(0\) in for \(y\) in the equation \(y=-4.3(x-7)+154.8\text{,}\) and we have:
        +
        The fund will be empty when it has \({\$0}\) left in it. That is, \(y\) will equal \(0\text{.}\) To find how many years until this happens, we substitute \(0\) in for \(y\) in the equation \(y=-28000(x-3)+663000\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -4.3(x-7) +154.8 \\ -0 \amp = -4.3(x-7) +154.8 \\ --154.8 \amp = -4.3(x-7)\\ -\frac{-154.8}{-4.3} \amp = x - 7\\ -36 \amp = x - 7\\ -43 \amp = x\\ -43 \amp = x +y \amp = -28000(x-3) +663000 \\ +0 \amp = -28000(x-3) +663000 \\ +-663000 \amp = -28000(x-3)\\ +\frac{-663000}{-28000} \amp = x - 3\\ +23.6785714285714 \amp = x - 3\\ +26.6785714285714 \amp = x\\ +26.68 \amp = x \end{aligned} }\)
        -
        So after \(43\) minutes, the gas will be all gone.
        +
        Approximately \(26.68\) years after 2000, in the year 2026, the fund will be depleted.
    -
    63.
    -
    +
    54.
    +
    -
    A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In \(2004\text{,}\) there was still \({\$479{,}000}\) left in the fund. In \(2007\text{,}\) there was \({\$383{,}000}\) left.
    -
    Let \(x\) be the number of years since 2000, and let \(y\) be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.
    +
    By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent \(290\) minutes on the phone, and paid \({\$21.15}\text{.}\) In another month, you spent \(390\) minutes on the phone, and paid \({\$24.65}\text{.}\) +
    +
    Let \(x\) be the number of minutes you talk over the phone in a month, and let \(y\) be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone.
    1. A point-slope equation to model this is .
      2. -
    2. In the year \(2009\text{,}\) there was left in the fund.
      3. -
    3. In the year , the fund will be empty.
      4. +
    4. If you spent \(120\) minutes over the phone in a month, you would pay .
      5. +
    5. If in a month, you paid \({\$26.05}\) of cell phone bill, you must have spent minutes on the phone in that month.
    -
    Answer 1.
    \(y = -32000\mathopen{}\left(x-4\right)+479000\hbox{ or }y = -32000\mathopen{}\left(x-7\right)+383000\)
    Answer 2.
    \(\$319{,}000\)
    Answer 3.
    \(2018\)
    Explanation.
      +
      Answer 1.
      \(y = 0.035\mathopen{}\left(x-290\right)+21.15\hbox{ or }y = 0.035\mathopen{}\left(x-390\right)+24.65\)
      Answer 2.
      \(\$15.20\)
      Answer 3.
      \(430\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((4,479000)\) and \((7,383000)\) are points on the linear model, so we can use the slope formula:
        +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((290,21.15)\) and \((390,24.65)\) are points on the linear model, so we can use the slope formula:
        \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
        First, we mark which number corresponds to which variable in the formula:
        -
        \(\displaystyle{ (4,479000) \longrightarrow (x_{1},y_{1}) }\)
        -
        \(\displaystyle{ (7,383000) \longrightarrow (x_{2},y_{2}) }\)
        +
        \(\displaystyle{ (290,21.15) \longrightarrow (x_{1},y_{1}) }\)
        +
        \(\displaystyle{ (390,24.65) \longrightarrow (x_{2},y_{2}) }\)
        Now we substitute these numbers into the corresponding variables in the slope formula:
        -
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{383000-479000}{7-4}\\\amp =\frac{-96000}{3}\\\amp =-32000 \end{aligned}}\)
        -
        This line’s slope is \(-32000\text{.}\) This implies that the company is spending \({\$32{,}000}\) per year on perks for its employees.
        -
        So now we have that \(y=-32000(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(4,479000)\) or \((x_0,y_0)=(7,383000)\) to get \({y = -32000\mathopen{}\left(x-4\right)+479000\hbox{ or }y = -32000\mathopen{}\left(x-7\right)+383000}\text{.}\) +
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{24.65-21.15}{390-290}\\\amp =\frac{3.5}{100}\\\amp =0.035 \end{aligned}}\)
        +
        This line’s slope is \(0.035\text{.}\) This implies you have to pay \({\$0.04}\) per minute that you use the phone.
        +
        So now we have that \(y=0.035(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(290,21.15)\) or \((x_0,y_0)=(390,24.65)\) to get \({y = 0.035\mathopen{}\left(x-290\right)+21.15\hbox{ or }y = 0.035\mathopen{}\left(x-390\right)+24.65}\text{.}\)
      2. -
        In the year \(2009\text{,}\) \(x\) was \(9\text{.}\) To find the amount remaining in the fund, we substitute \(9\) in for \(x\) in the equation \(y=-32000(x-4)+479000\text{,}\) and we have:
        +
        In one month, you spent \(120\) minutes on the phone. To find out the bill, we substitute \(120\) in for \(x\) in the equation \(y=0.035(x-290)+21.15\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -32000(9-4) +479000 \\ -y \amp = 319000 +y \amp = 0.035(120-290) +21.15 \\ +y \amp = 15.2 \end{aligned} }\)
        -
        So in the year \(2009\text{,}\) the fund had \({\$319{,}000}\) remaining in it.
        +
        If you spent \(120\) minutes over the phone in a month, your bill for that month would be \({\$15.20}\text{.}\) +
      3. -
        The fund will be empty when it has \({\$0}\) left in it. That is, \(y\) will equal \(0\text{.}\) To find how many years until this happens, we substitute \(0\) in for \(y\) in the equation \(y=-32000(x-4)+479000\text{,}\) and we have:
        +
        In one month, you had a cell phone bill of \({\$26.05}\text{.}\) To find how many minutes you spent on the phone that month, we substitute \(26.05\) in for \(y\) in the equation \(y=0.035(x-290)+21.15\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -32000(x-4) +479000 \\ -0 \amp = -32000(x-4) +479000 \\ --479000 \amp = -32000(x-4)\\ -\frac{-479000}{-32000} \amp = x - 4\\ -14.96875 \amp = x - 4\\ -18.96875 \amp = x\\ -18.97 \amp = x +y \amp = 0.035(x-290) +21.15 \\ +26.05 \amp = 0.035(x-290) +21.15 \\ +4.9 \amp = 0.035(x-290)\\ +\frac{4.9}{0.035} \amp = x - 290\\ +140 \amp = x - 290\\ +430 \amp = x\\ +430 \amp = x \end{aligned} }\)
        -
        Approximately \(18.97\) years after 2000, in the year 2018, the fund will be depleted.
        +
        If your bill was \({\$26.05}\) in one month, you must have spent \(430\) minutes on the phone.
    -
    64.
    -
    +
    55.
    +
    -
    By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent \(240\) minutes on the phone, and paid \({\$27.80}\text{.}\) In another month, you spent \(360\) minutes on the phone, and paid \({\$33.20}\text{.}\) -
    -
    Let \(x\) be the number of minutes you talk over the phone in a month, and let \(y\) be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone.
    +
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way.
    Six minutes since the experiment started, the gas had a mass of \(273\) grams.
    Fifteen minutes since the experiment started, the gas had a mass of \(210\) grams.
    +
    Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
      -
    1. A point-slope equation to model this is .
      2. -
    2. If you spent \(120\) minutes over the phone in a month, you would pay .
      3. -
    3. If in a month, you paid \({\$38.60}\) of cell phone bill, you must have spent minutes on the phone in that month.
      4. +
    4. A point-slope equation to model this is .
      5. +
    5. +\(31\) minutes after the experiment started, there would be grams of gas left.
      6. +
    6. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
    -
    Answer 1.
    \(y = 0.045\mathopen{}\left(x-240\right)+27.8\hbox{ or }y = 0.045\mathopen{}\left(x-360\right)+33.2\)
    Answer 2.
    \(\$22.40\)
    Answer 3.
    \(480\)
    Explanation.
      +
      Answer 1.
      \(y = -7\mathopen{}\left(x-6\right)+273\hbox{ or }y = -7\mathopen{}\left(x-15\right)+210\)
      Answer 2.
      \(98\)
      Answer 3.
      \(45\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((240,27.8)\) and \((360,33.2)\) are points on the linear model, so we can use the slope formula:
        +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((6,273)\) and \((15,210)\) are points on the linear model, so we can use the slope formula:
        \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
        First, we mark which number corresponds to which variable in the formula:
        -
        \(\displaystyle{ (240,27.8) \longrightarrow (x_{1},y_{1}) }\)
        -
        \(\displaystyle{ (360,33.2) \longrightarrow (x_{2},y_{2}) }\)
        +
        \(\displaystyle{ (6,273) \longrightarrow (x_{1},y_{1}) }\)
        +
        \(\displaystyle{ (15,210) \longrightarrow (x_{2},y_{2}) }\)
        Now we substitute these numbers into the corresponding variables in the slope formula:
        -
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{33.2-27.8}{360-240}\\\amp =\frac{5.40000000000001}{120}\\\amp =0.045 \end{aligned}}\)
        -
        This line’s slope is \(0.045\text{.}\) This implies you have to pay \({\$0.04}\) per minute that you use the phone.
        -
        So now we have that \(y=0.045(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(240,27.8)\) or \((x_0,y_0)=(360,33.2)\) to get \({y = 0.045\mathopen{}\left(x-240\right)+27.8\hbox{ or }y = 0.045\mathopen{}\left(x-360\right)+33.2}\text{.}\) +
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{210-273}{15-6}\\\amp =\frac{-63}{9}\\\amp =-7 \end{aligned}}\)
        +
        This line’s slope is \(-7\text{.}\) This implies that the gas is leaking with a rate of \(7\) grams per minute.
        +
        So now we have that \(y=-7(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(6,273)\) or \((x_0,y_0)=(15,210)\) to get \({y = -7\mathopen{}\left(x-6\right)+273\hbox{ or }y = -7\mathopen{}\left(x-15\right)+210}\text{.}\)
      2. -
        In one month, you spent \(120\) minutes on the phone. To find out the bill, we substitute \(120\) in for \(x\) in the equation \(y=0.045(x-240)+27.8\text{,}\) and we have:
        +
        After \(31\) minutes, we can find the mass of the remaining gas if we substitute \(31\) in for \(x\) in the equation \(y=-7(x-6)+273\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = 0.045(120-240) +27.8 \\ -y \amp = 22.4 +y \amp = -7(31-6) +273 \\ +y \amp = 98 \end{aligned} }\)
        -
        If you spent \(120\) minutes over the phone in a month, your bill for that month would be \({\$22.40}\text{.}\) -
        +
        So after \(31\) minutes, there are \(98\) grams of gas remaining.
      3. -
        In one month, you had a cell phone bill of \({\$38.60}\text{.}\) To find how many minutes you spent on the phone that month, we substitute \(38.6\) in for \(y\) in the equation \(y=0.045(x-240)+27.8\text{,}\) and we have:
        +
        To find when the gas will be all gone, we need to find when there are \(0\) grams of gas left. We can substitute \(0\) in for \(y\) in the equation \(y=-7(x-6)+273\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = 0.045(x-240) +27.8 \\ -38.6 \amp = 0.045(x-240) +27.8 \\ -10.8 \amp = 0.045(x-240)\\ -\frac{10.8}{0.045} \amp = x - 240\\ -240 \amp = x - 240\\ -480 \amp = x\\ -480 \amp = x +y \amp = -7(x-6) +273 \\ +0 \amp = -7(x-6) +273 \\ +-273 \amp = -7(x-6)\\ +\frac{-273}{-7} \amp = x - 6\\ +39 \amp = x - 6\\ +45 \amp = x\\ +45 \amp = x \end{aligned} }\)
        -
        If your bill was \({\$38.60}\) in one month, you must have spent \(480\) minutes on the phone.
        +
        So after \(45\) minutes, the gas will be all gone.
    -
    65.
    -
    +
    56.
    +
    -
    Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way.
    Nine minutes since the experiment started, the gas had a mass of \(144.3\) grams.
    Thirteen minutes since the experiment started, the gas had a mass of \(128.7\) grams.
    -
    Let \(x\) be the number of minutes that have passed since the experiment started, and let \(y\) be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
    +
    A biologist has been observing a tree’s height. \(14\) months into the observation, the tree was \(16.68\) feet tall. \(17\) months into the observation, the tree was \(17.34\) feet tall.
    +
    Let \(x\) be the number of months passed since the observations started, and let \(y\) be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.
      -
    1. A point-slope equation to model this is .
      2. -
    2. -\(31\) minutes after the experiment started, there would be grams of gas left.
      3. -
    3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.
      4. +
    4. A point-slope equation to model this is .
      5. +
    5. +\(25\) months after the observations started, the tree would be feet in height.
      6. +
    6. + months after the observation started, the tree would be \(25.04\) feet tall.
    -
    Answer 1.
    \(y = -3.9\mathopen{}\left(x-9\right)+144.3\hbox{ or }y = -3.9\mathopen{}\left(x-13\right)+128.7\)
    Answer 2.
    \(58.5\)
    Answer 3.
    \(46\)
    Explanation.
      +
      Answer 1.
      \(y = 0.22\mathopen{}\left(x-14\right)+16.68\hbox{ or }y = 0.22\mathopen{}\left(x-17\right)+17.34\)
      Answer 2.
      \(19.1\)
      Answer 3.
      \(52\)
      Explanation.
      1. -
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((9,144.3)\) and \((13,128.7)\) are points on the linear model, so we can use the slope formula:
        +
        A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((14,16.68)\) and \((17,17.34)\) are points on the linear model, so we can use the slope formula:
        \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
        First, we mark which number corresponds to which variable in the formula:
        -
        \(\displaystyle{ (9,144.3) \longrightarrow (x_{1},y_{1}) }\)
        -
        \(\displaystyle{ (13,128.7) \longrightarrow (x_{2},y_{2}) }\)
        +
        \(\displaystyle{ (14,16.68) \longrightarrow (x_{1},y_{1}) }\)
        +
        \(\displaystyle{ (17,17.34) \longrightarrow (x_{2},y_{2}) }\)
        Now we substitute these numbers into the corresponding variables in the slope formula:
        -
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{128.7-144.3}{13-9}\\\amp =\frac{-15.6}{4}\\\amp =-3.9 \end{aligned}}\)
        -
        This line’s slope is \(-3.9\text{.}\) This implies that the gas is leaking with a rate of \(3.9\) grams per minute.
        -
        So now we have that \(y=-3.9(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(9,144.3)\) or \((x_0,y_0)=(13,128.7)\) to get \({y = -3.9\mathopen{}\left(x-9\right)+144.3\hbox{ or }y = -3.9\mathopen{}\left(x-13\right)+128.7}\text{.}\) +
        \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{17.34-16.68}{17-14}\\\amp =\frac{0.66}{3}\\\amp =0.22 \end{aligned}}\)
        +
        This line’s slope is \(0.22\text{.}\) This implies that the tree is growing by \(0.22\) feet each month.
        +
        So now we have that \(y=0.22(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(14,16.68)\) or \((x_0,y_0)=(17,17.34)\) to get \({y = 0.22\mathopen{}\left(x-14\right)+16.68\hbox{ or }y = 0.22\mathopen{}\left(x-17\right)+17.34}\text{.}\)
      2. -
        After \(31\) minutes, we can find the mass of the remaining gas if we substitute \(31\) in for \(x\) in the equation \(y=-3.9(x-9)+144.3\text{,}\) and we have:
        +
        After \(25\) months, we can find the height of the tree if we substitute \(25\) in for \(x\) in the equation \(y=0.22(x-14)+16.68\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -3.9(31-9) +144.3 \\ -y \amp = 58.5 +y \amp = 0.22(25-14) +16.68 \\ +y \amp = 19.1 \end{aligned} }\)
        -
        So after \(31\) minutes, there are \(58.5\) grams of gas remaining.
        +
        So after \(25\) months, the tree is \(19.1\) feet tall.
      3. -
        To find when the gas will be all gone, we need to find when there are \(0\) grams of gas left. We can substitute \(0\) in for \(y\) in the equation \(y=-3.9(x-9)+144.3\text{,}\) and we have:
        +
        To find when the tree was \(25.04\) feet tall, we substitute \(25.04\) in for \(y\) in the equation \(y=0.22(x-14)+16.68\text{,}\) and we have:
        \(\displaystyle{\begin{aligned} -y \amp = -3.9(x-9) +144.3 \\ -0 \amp = -3.9(x-9) +144.3 \\ --144.3 \amp = -3.9(x-9)\\ -\frac{-144.3}{-3.9} \amp = x - 9\\ -37 \amp = x - 9\\ -46 \amp = x\\ -46 \amp = x -\end{aligned} -}\)
        -
        So after \(46\) minutes, the gas will be all gone.
        -
        4. -
      -
    -
    -
    66.
    -
    -
    -
    -
    A biologist has been observing a tree’s height. \(10\) months into the observation, the tree was \(21.4\) feet tall. \(20\) months into the observation, the tree was \(23.9\) feet tall.
    -
    Let \(x\) be the number of months passed since the observations started, and let \(y\) be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.
    -
      -
    1. A point-slope equation to model this is .
      2. -
    2. -\(25\) months after the observations started, the tree would be feet in height.
      3. -
    3. - months after the observation started, the tree would be \(33.4\) feet tall.
      4. -
    -
    -
    -
    Answer 1.
    \(y = 0.25\mathopen{}\left(x-10\right)+21.4\hbox{ or }y = 0.25\mathopen{}\left(x-20\right)+23.9\)
    Answer 2.
    \(25.15\)
    Answer 3.
    \(58\)
    Explanation.
      -
    1. -
      A line’s equation in point-slope form looks like \(y=m(x-x_0)+y_0\text{,}\) where \(m\) is the slope, and \((x_0,y_0)\) is some point we know to be on the line. We first need to find the line’s slope. To find the slope, we need two points on the line. We have been given that \((10,21.4)\) and \((20,23.9)\) are points on the linear model, so we can use the slope formula:
      -
      \(\displaystyle{ \text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} }\)
      -
      First, we mark which number corresponds to which variable in the formula:
      -
      \(\displaystyle{ (10,21.4) \longrightarrow (x_{1},y_{1}) }\)
      -
      \(\displaystyle{ (20,23.9) \longrightarrow (x_{2},y_{2}) }\)
      -
      Now we substitute these numbers into the corresponding variables in the slope formula:
      -
      \(\displaystyle{ \begin{aligned}\text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\amp =\frac{23.9-21.4}{20-10}\\\amp =\frac{2.5}{10}\\\amp =0.25 \end{aligned}}\)
      -
      This line’s slope is \(0.25\text{.}\) This implies that the tree is growing by \(0.25\) feet each month.
      -
      So now we have that \(y=0.25(x-x_0)+y_0\text{.}\) We can either use \((x_0,y_0)=(10,21.4)\) or \((x_0,y_0)=(20,23.9)\) to get \({y = 0.25\mathopen{}\left(x-10\right)+21.4\hbox{ or }y = 0.25\mathopen{}\left(x-20\right)+23.9}\text{.}\) -
      -
      2. -
    2. -
      After \(25\) months, we can find the height of the tree if we substitute \(25\) in for \(x\) in the equation \(y=0.25(x-10)+21.4\text{,}\) and we have:
      -
      \(\displaystyle{\begin{aligned} -y \amp = 0.25(25-10) +21.4 \\ -y \amp = 25.15 -\end{aligned} -}\)
      -
      So after \(25\) months, the tree is \(25.15\) feet tall.
      -
      3. -
    3. -
      To find when the tree was \(33.4\) feet tall, we substitute \(33.4\) in for \(y\) in the equation \(y=0.25(x-10)+21.4\text{,}\) and we have:
      -
      \(\displaystyle{\begin{aligned} -y \amp = 0.25(x-10) +21.4 \\ -33.4 \amp = 0.25(x-10) +21.4 \\ -12 \amp = 0.25(x-10)\\ -\frac{12}{0.25} \amp = x - 10\\ -48 \amp = x - 10\\ -58 \amp = x\\ -58 \amp = x +y \amp = 0.22(x-14) +16.68 \\ +25.04 \amp = 0.22(x-14) +16.68 \\ +8.36 \amp = 0.22(x-14)\\ +\frac{8.36}{0.22} \amp = x - 14\\ +38 \amp = x - 14\\ +52 \amp = x\\ +52 \amp = x \end{aligned} }\)
      -
      So the tree was \(33.4\) feet tall after \(58\) months.
      +
      So the tree was \(25.04\) feet tall after \(52\) months.
    diff --git a/section-set-notation-and-types-of-numbers.html b/section-set-notation-and-types-of-numbers.html index 11f9d5307..2c9c4b637 100644 --- a/section-set-notation-and-types-of-numbers.html +++ b/section-set-notation-and-types-of-numbers.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -608,14 +622,14 @@

    Search Results:

    Checkpoint A.6.2. Set Notation.
    Practice using (and not using) set notation.
    -
    -
    +
    +
    -
    According to Google, the three most common error codes from visiting a web site are 403, 404, and 500.

    (a)

    -
    Without knowing which error code is most common, express this set mathematically.
    -
    Explanation.
    Since we only have to describe a collection of three numbers and their order doesn’t matter, we can write {403,404,500}.

    (b)

    -
    Error code 500 is the most common. Error code 403 is the least common of these three. And that leaves 404 in the middle. Express the error codes in a mathematical way that appreciates how frequently they happen, from most often to least often.
    -
    Explanation.
    Now we must describe the same three numbers and we want readers to know that the order we are writing the numbers matters. We can write (500,404,403).
    +
    According to Google, the three most common error codes from visiting a web site are 403, 404, and 500.

    (a)

    +
    Without knowing which error code is most common, express this set mathematically.
    +
    Explanation.
    Since we only have to describe a collection of three numbers and their order doesn’t matter, we can write {403,404,500}.

    (b)

    +
    Error code 500 is the most common. Error code 403 is the least common of these three. And that leaves 404 in the middle. Express the error codes in a mathematical way that appreciates how frequently they happen, from most often to least often.
    +
    Explanation.
    Now we must describe the same three numbers and we want readers to know that the order we are writing the numbers matters. We can write (500,404,403).

    • Subsection A.6.2 Different Number Sets @@ -678,44 +692,44 @@

    Search Results:

    Checkpoint A.6.6.

    -
    -
    +
    +
    -
      -
    1. Give an example of a whole number that is not an integer.
      2. -
    2. Give an example of an integer that is not a whole number.
      3. -
    3. Give an example of a rational number that is not an integer.
      4. -
    4. Give an example of a irrational number.
      5. -
    5. Give an example of a irrational number that is also an integer.
      6. +
        +
      1. Give an example of a whole number that is not an integer.
        2. +
      2. Give an example of an integer that is not a whole number.
        3. +
      3. Give an example of a rational number that is not an integer.
        4. +
      4. Give an example of a irrational number.
        5. +
      5. Give an example of a irrational number that is also an integer.
      -
      Explanation.
        -
      1. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question.
        2. -
      2. Any negative integer, like \(-1\text{,}\) is not a whole number, but is an integer.
        3. -
      3. Any terminating decimal, like \(1.2\text{,}\) is a rational number, but is not an integer.
        4. -
      4. +
        Explanation.
          +
        1. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question.
          2. +
        2. Any negative integer, like \(-1\text{,}\) is not a whole number, but is an integer.
          3. +
        3. Any terminating decimal, like \(1.2\text{,}\) is a rational number, but is not an integer.
          4. +
        4. \(\pi\) is the easiest number to remember as an irrational number. Another constant worth knowing is \(e\approx2.718\text{.}\) Finally, the square root of most integers are irrational, like \(\sqrt{2}\) and \(\sqrt{3}\text{.}\)
          5. -
        5. All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers.
          6. +
        6. All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers.

    Checkpoint A.6.7.

    In the introduction, we mentioned that the different types of numbers are appropriate in different situation. Which number set do you think is most appropriate in each of the following situations?
    -
    -
    +
    +
    -

    (a)

    -
    The number of people in a math class that play the ukulele.
    This number is best considered as a .
    -
    Explanation.
    The number of people who play the ukulele could be \(0,1,2,\dots\text{,}\) so the whole numbers are the appropriate set.

    (b)

    -
    The hypotenuse’s length in a given right triangle.
    This number is best considered as a .
    -
    Explanation.
    A hypotenuse’s length could be \(1\text{,}\) \(1.2\text{,}\) \(\sqrt{2}\) (which is irrational), or any other positive number. So the real numbers are the appropriate set.

    (c)

    -
    The proportion of people in a math class that have a cat.
    This number is best considered as a .
    -
    Explanation.
    This proportion will be a ratio of integers, as both the total number of people in the class and the number of people who have a cat are integers. So the rational numbers are the appropriate set.

    (d)

    -
    The number of people in the room with you who have the same birthday as you.
    This number is best considered as a .
    -
    Explanation.
    We know that the number of people must be a counting number, and since you are in the room with yourself, there is at least one person in that room with your birthday. So the natural numbers are the appropriate set.

    (e)

    -
    The total revenue (in dollars) generated for ticket sales at a Timbers soccer game.
    This number is best considered as a .
    -
    Explanation.
    The total revenue will be some number of dollars and cents, such as \(\$631{,}897.15\text{,}\) which is a terminating decimal and thus a rational number. So the rational numbers are the appropriate set.
    +

    (a)

    +
    The number of people in a math class that play the ukulele.
    This number is best considered as a .
    +
    Explanation.
    The number of people who play the ukulele could be \(0,1,2,\dots\text{,}\) so the whole numbers are the appropriate set.

    (b)

    +
    The hypotenuse’s length in a given right triangle.
    This number is best considered as a .
    +
    Explanation.
    A hypotenuse’s length could be \(1\text{,}\) \(1.2\text{,}\) \(\sqrt{2}\) (which is irrational), or any other positive number. So the real numbers are the appropriate set.

    (c)

    +
    The proportion of people in a math class that have a cat.
    This number is best considered as a .
    +
    Explanation.
    This proportion will be a ratio of integers, as both the total number of people in the class and the number of people who have a cat are integers. So the rational numbers are the appropriate set.

    (d)

    +
    The number of people in the room with you who have the same birthday as you.
    This number is best considered as a .
    +
    Explanation.
    We know that the number of people must be a counting number, and since you are in the room with yourself, there is at least one person in that room with your birthday. So the natural numbers are the appropriate set.

    (e)

    +
    The total revenue (in dollars) generated for ticket sales at a Timbers soccer game.
    This number is best considered as a .
    +
    Explanation.
    The total revenue will be some number of dollars and cents, such as \(\$631{,}897.15\text{,}\) which is a terminating decimal and thus a rational number. So the rational numbers are the appropriate set.

    Subsection A.6.3 Converting Repeating Decimals to Fractions @@ -747,15 +761,15 @@

    Search Results:

    Checkpoint A.6.8.

    -
    -
    +
    +
    -

    (a)

    -
    Write the rational number \(0.772772772\ldots\) as a fraction.
    -
    Explanation.
    The three-digit number \(772\) repeats after the decimal. So we will make use of the three-digit denominator \(999\text{.}\) And we have \(\frac{772}{999}\text{.}\) -

    (b)

    -
    Write the rational number \(0.69696969\ldots\) as a fraction.
    -
    Explanation.
    The two-digit number \(69\) repeats after the decimal. So we will make use of the two-digit denominator \(99\text{.}\) And we have \(\frac{69}{99}\text{.}\) But this fraction can be reduced to \(\frac{23}{33}\text{.}\) +

    (a)

    +
    Write the rational number \(0.772772772\ldots\) as a fraction.
    +
    Explanation.
    The three-digit number \(772\) repeats after the decimal. So we will make use of the three-digit denominator \(999\text{.}\) And we have \(\frac{772}{999}\text{.}\) +

    (b)

    +
    Write the rational number \(0.69696969\ldots\) as a fraction.
    +
    Explanation.
    The two-digit number \(69\) repeats after the decimal. So we will make use of the two-digit denominator \(99\text{.}\) And we have \(\frac{69}{99}\text{.}\) But this fraction can be reduced to \(\frac{23}{33}\text{.}\)
    Converting a repeating decimal to a fraction is not always quite this straightforward. There are complications if the number takes a few digits before it begins repeating. For your interest, here is one example on how to do that.
    @@ -793,79 +807,79 @@

    Search Results:

    Review and Warmup.

    1.

    -
    -
    +
    +
    -
    Write the decimal number as a fraction.
    -\(0.85\) = +
    Write the decimal number as a fraction.
    +\(0.15\) =
    -
    Answer.
    \({\frac{17}{20}}\)
    Explanation.
    -\(0.85\) is read as eighty-five hundredth, so it can be written as \(\frac{85}{100}\text{,}\) which can be reduced to \({{\frac{17}{20}}}\) by dividing \(5\) in both the numerator and denominator.
    +
    Answer.
    \({\frac{3}{20}}\)
    Explanation.
    +\(0.15\) is read as fifteen hundredth, so it can be written as \(\frac{15}{100}\text{,}\) which can be reduced to \({{\frac{3}{20}}}\) by dividing \(5\) in both the numerator and denominator.

    2.

    -
    -
    +
    +
    -
    Write the decimal number as a fraction.
    -\(0.95\) = +
    Write the decimal number as a fraction.
    +\(0.25\) =
    -
    Answer.
    \({\frac{19}{20}}\)
    Explanation.
    -\(0.95\) is read as ninety-five hundredth, so it can be written as \(\frac{95}{100}\text{,}\) which can be reduced to \({{\frac{19}{20}}}\) by dividing \(5\) in both the numerator and denominator.
    +
    Answer.
    \({\frac{1}{4}}\)
    Explanation.
    +\(0.25\) is read as twenty-five hundredth, so it can be written as \(\frac{25}{100}\text{,}\) which can be reduced to \({{\frac{1}{4}}}\) by dividing \(25\) in both the numerator and denominator.

    3.

    -
    -
    +
    +
    -
    Write the decimal number as a fraction.
    -\(1.95\) = +
    Write the decimal number as a fraction.
    +\(3.75\) =
    -
    Answer.
    \(1 {\textstyle\frac{19}{20}}\)
    Explanation.
    -
    The integer part of the decimal \(1.95\) will not change. We will only change the decimal part--0.95.
    -\(0.95\) is read as ninety-five hundredth, so it can be written as \(\frac{95}{100}\text{,}\) which can be reduced to \({{\frac{19}{20}}}\) by dividing \(5\) in both the numerator and denominator.
    \(\displaystyle{ 1.95 = {1 {\textstyle\frac{19}{20}}} }\)
    +
    Answer.
    \(3 {\textstyle\frac{3}{4}}\)
    Explanation.
    +
    The integer part of the decimal \(3.75\) will not change. We will only change the decimal part--0.75.
    +\(0.75\) is read as seventy-five hundredth, so it can be written as \(\frac{75}{100}\text{,}\) which can be reduced to \({{\frac{3}{4}}}\) by dividing \(25\) in both the numerator and denominator.
    \(\displaystyle{ 3.75 = {3 {\textstyle\frac{3}{4}}} }\)

    4.

    -
    -
    +
    +
    -
    Write the decimal number as a fraction.
    -\(2.65\) = +
    Write the decimal number as a fraction.
    +\(4.45\) =
    -
    Answer.
    \(2 {\textstyle\frac{13}{20}}\)
    Explanation.
    -
    The integer part of the decimal \(2.65\) will not change. We will only change the decimal part--0.65.
    -\(0.65\) is read as sixty-five hundredth, so it can be written as \(\frac{65}{100}\text{,}\) which can be reduced to \({{\frac{13}{20}}}\) by dividing \(5\) in both the numerator and denominator.
    \(\displaystyle{ 2.65 = {2 {\textstyle\frac{13}{20}}} }\)
    +
    Answer.
    \(4 {\textstyle\frac{9}{20}}\)
    Explanation.
    +
    The integer part of the decimal \(4.45\) will not change. We will only change the decimal part--0.45.
    +\(0.45\) is read as forty-five hundredth, so it can be written as \(\frac{45}{100}\text{,}\) which can be reduced to \({{\frac{9}{20}}}\) by dividing \(5\) in both the numerator and denominator.
    \(\displaystyle{ 4.45 = {4 {\textstyle\frac{9}{20}}} }\)

    5.

    -
    -
    +
    +
    -
    Write the decimal number as a fraction.
    -\(0.322\) = +
    Write the decimal number as a fraction.
    +\(0.502\) =
    -
    Answer.
    \({\frac{161}{500}}\)
    Explanation.
    -\(0.322\) is read as 322 thousandths, so it can be written as \(\frac{322}{1000}\text{,}\) which can be reduced to \({{\frac{161}{500}}}\) by dividing \(2\) in both the numerator and denominator.
    +
    Answer.
    \({\frac{251}{500}}\)
    Explanation.
    +\(0.502\) is read as 502 thousandths, so it can be written as \(\frac{502}{1000}\text{,}\) which can be reduced to \({{\frac{251}{500}}}\) by dividing \(2\) in both the numerator and denominator.

    6.

    -
    -
    +
    +
    -
    Write the decimal number as a fraction.
    -\(0.496\) = +
    Write the decimal number as a fraction.
    +\(0.664\) =
    -
    Answer.
    \({\frac{62}{125}}\)
    Explanation.
    -\(0.496\) is read as 496 thousandths, so it can be written as \(\frac{496}{1000}\text{,}\) which can be reduced to \({{\frac{62}{125}}}\) by dividing \(8\) in both the numerator and denominator.
    +
    Answer.
    \({\frac{83}{125}}\)
    Explanation.
    +\(0.664\) is read as 664 thousandths, so it can be written as \(\frac{664}{1000}\text{,}\) which can be reduced to \({{\frac{83}{125}}}\) by dividing \(8\) in both the numerator and denominator.
    @@ -875,43 +889,43 @@

    Review and Warmup.

    Exercise Group.

    7.

    -
    -
    +
    +
    -
    Write the fraction as a decimal number.
      -
    1. -\({{\frac{1}{8}}}\) = +
      Write the fraction as a decimal number.
        +
      1. +\({{\frac{17}{20}}}\) =
        2. -
      2. -\({{\frac{15}{16}}}\) = +
      3. +\({{\frac{3}{8}}}\) =
      -
      Answer 1.
      \(0.125\)
      Answer 2.
      \(0.9375\)
      Explanation.
      -
      To change a fraction to decimal, we divide the numerator by the denominator.
        -
      1. \(\displaystyle {{\frac{1}{8}}} = 1 \div 8 = 0.125\)
        2. -
      2. \(\displaystyle {{\frac{15}{16}}} = 15 \div 16 = 0.9375\)
        3. +
        Answer 1.
        \(0.85\)
        Answer 2.
        \(0.375\)
        Explanation.
        +
        To change a fraction to decimal, we divide the numerator by the denominator.
          +
        1. \(\displaystyle {{\frac{17}{20}}} = 17 \div 20 = 0.85\)
          2. +
        2. \(\displaystyle {{\frac{3}{8}}} = 3 \div 8 = 0.375\)

    8.

    -
    -
    +
    +
    -
    Write the fraction as a decimal number.
      -
    1. -\({{\frac{7}{16}}}\) = +
      Write the fraction as a decimal number.
        +
      1. +\({{\frac{3}{20}}}\) =
        2. -
      2. -\({{\frac{3}{5}}}\) = +
      3. +\({{\frac{1}{4}}}\) =
      -
      Answer 1.
      \(0.4375\)
      Answer 2.
      \(0.6\)
      Explanation.
      -
      To change a fraction to decimal, we divide the numerator by the denominator.
        -
      1. \(\displaystyle {{\frac{7}{16}}} = 7 \div 16 = 0.4375\)
        2. -
      2. \(\displaystyle {{\frac{3}{5}}} = 3 \div 5 = 0.6\)
        3. +
        Answer 1.
        \(0.15\)
        Answer 2.
        \(0.25\)
        Explanation.
        +
        To change a fraction to decimal, we divide the numerator by the denominator.
          +
        1. \(\displaystyle {{\frac{3}{20}}} = 3 \div 20 = 0.15\)
          2. +
        2. \(\displaystyle {{\frac{1}{4}}} = 1 \div 4 = 0.25\)
      @@ -923,43 +937,43 @@

      Exercise Group.

      Exercise Group.

      9.

      -
      -
      +
      +
      -
      Write the mixed number as a decimal number.
        -
      1. -\({4 {\textstyle\frac{17}{20}}}\) = +
        Write the mixed number as a decimal number.
          +
        1. +\({1 {\textstyle\frac{18}{25}}}\) =
          2. -
        2. -\({6 {\textstyle\frac{13}{25}}}\) = +
        3. +\({3 {\textstyle\frac{13}{16}}}\) =
        -
        Answer 1.
        \(4.85\)
        Answer 2.
        \(6.52\)
        Explanation.
        -
        To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number.
          -
        1. \(\displaystyle {4 {\textstyle\frac{17}{20}}} = 4 + 17 \div 20 = 4.85\)
          2. -
        2. \(\displaystyle {6 {\textstyle\frac{13}{25}}} = 6 + 13 \div 25 = 6.52\)
          3. +
          Answer 1.
          \(1.72\)
          Answer 2.
          \(3.8125\)
          Explanation.
          +
          To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number.
            +
          1. \(\displaystyle {1 {\textstyle\frac{18}{25}}} = 1 + 18 \div 25 = 1.72\)
            2. +
          2. \(\displaystyle {3 {\textstyle\frac{13}{16}}} = 3 + 13 \div 16 = 3.8125\)

      10.

      -
      -
      +
      +
      -
      Write the mixed number as a decimal number.
        -
      1. -\({3 {\textstyle\frac{3}{20}}}\) = +
        Write the mixed number as a decimal number.
          +
        1. +\({9 {\textstyle\frac{1}{4}}}\) =
          2. -
        2. -\({6 {\textstyle\frac{11}{16}}}\) = +
        3. +\({7 {\textstyle\frac{3}{5}}}\) =
        -
        Answer 1.
        \(3.15\)
        Answer 2.
        \(6.6875\)
        Explanation.
        -
        To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number.
          -
        1. \(\displaystyle {3 {\textstyle\frac{3}{20}}} = 3 + 3 \div 20 = 3.15\)
          2. -
        2. \(\displaystyle {6 {\textstyle\frac{11}{16}}} = 6 + 11 \div 16 = 6.6875\)
          3. +
          Answer 1.
          \(9.25\)
          Answer 2.
          \(7.6\)
          Explanation.
          +
          To change a fraction to decimal, we divide the numerator by the denominator. We can simply copy over the integer part of a mixed number.
            +
          1. \(\displaystyle {9 {\textstyle\frac{1}{4}}} = 9 + 1 \div 4 = 9.25\)
            2. +
          2. \(\displaystyle {7 {\textstyle\frac{3}{5}}} = 7 + 3 \div 5 = 7.6\)
        @@ -971,32 +985,32 @@

        Exercise Group.

        Set Notation.

        11.

        -
        -
        +
        +
        -
        There are two numbers that you can square to get \(81\text{.}\) Express this collection of two numbers using set notation.
        -
        Answer.
        \(\left\{-9,9\right\}\)
        +
        There are two numbers that you can square to get \(4\text{.}\) Express this collection of two numbers using set notation.
        +
        Answer.
        \(\left\{-2,2\right\}\)

        12.

        -
        -
        +
        +
        -
        There are four positive, even, one-digit numbers. Express this collection of four numbers using set notation.
        -
        Answer.
        \(\left\{2,4,6,8\right\}\)
        +
        There are four positive, even, one-digit numbers. Express this collection of four numbers using set notation.
        +
        Answer.
        \(\left\{2,4,6,8\right\}\)

        13.

        -
        -
        +
        +
        -
        There are six two-digit perfect square numbers. Express this collection of six numbers using set notation.
        -
        Answer.
        \(\left\{16,25,36,49,64,81\right\}\)
        +
        There are six two-digit perfect square numbers. Express this collection of six numbers using set notation.
        +
        Answer.
        \(\left\{16,25,36,49,64,81\right\}\)

        14.

        -
        -
        +
        +
        -
        There is a set of three small positive integers where you can square all three numbers, then add the results, and get \(41\text{.}\) Express this collection of three numbers using set notation.
        -
        Answer.
        \(\left\{1,2,6\right\}\)
        +
        There is a set of three small positive integers where you can square all three numbers, then add the results, and get \(70\text{.}\) Express this collection of three numbers using set notation.
        +
        Answer.
        \(\left\{3,5,6\right\}\)
        @@ -1005,36 +1019,36 @@

        Set Notation.

        Types of Numbers.

        15.

        -
        -
        +
        +
        -
        Which of the following are whole numbers? There may be more than one correct answer.
          -
        • \(\displaystyle -84094\)
          • -
        • \(\displaystyle -6\)
          • -
        • \(\displaystyle 78\)
          • -
        • \(\displaystyle \sqrt{5}\)
          • -
        • \(\displaystyle 8.9\overline{20}\)
          • -
        • \(\displaystyle 38225\)
          • -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle -4.995\)
          • +
          Which of the following are whole numbers? There may be more than one correct answer.
            +
          • \(\displaystyle 0\)
            • +
          • \(\displaystyle 6.5\overline{15}\)
            • +
          • \(\displaystyle -7\)
            • +
          • \(\displaystyle -46155\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle -8.269\)
            • +
          • \(\displaystyle \sqrt{25}\)
            • +
          • \(\displaystyle \sqrt{11}\)
          -
          Explanation.
          To be a whole number, you have to be one of the numbers \(0,1,2,\ldots\text{.}\) Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is a whole number, since \(\sqrt{4}=2\text{.}\) So the correct answers are CF.
          +
          Explanation.
          To be a whole number, you have to be one of the numbers \(0,1,2,\ldots\text{.}\) Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is a whole number, since \(\sqrt{4}=2\text{.}\) So the correct answers are AG.

        16.

        -
        -
        +
        +
        -
        Which of the following are whole numbers? There may be more than one correct answer.
          -
        • \(\displaystyle -30468\)
          • -
        • \(\displaystyle -7.101001000100001\ldots\)
          • -
        • \(\displaystyle -3.445\)
          • -
        • \(\displaystyle -7\)
          • -
        • \(\displaystyle 43\)
          • -
        • \(\displaystyle 7.4\overline{28}\)
          • -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle \sqrt{25}\)
          • +
          Which of the following are whole numbers? There may be more than one correct answer.
            +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle 8.531\)
            • +
          • \(\displaystyle 64101\)
            • +
          • \(\displaystyle -6885\)
            • +
          • \(\displaystyle -8\)
            • +
          • \(\displaystyle \sqrt{36}\)
            • +
          • \(\displaystyle \sqrt{2}\)
            • +
          • \(\displaystyle {\frac{9}{34}}\)
          -
          Explanation.
          To be a whole number, you have to be one of the numbers \(0,1,2,\ldots\text{.}\) Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is a whole number, since \(\sqrt{4}=2\text{.}\) So the correct answers are EH.
          +
          Explanation.
          To be a whole number, you have to be one of the numbers \(0,1,2,\ldots\text{.}\) Sometimes whole numbers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is a whole number, since \(\sqrt{4}=2\text{.}\) So the correct answers are CF.
        @@ -1043,36 +1057,36 @@

        Types of Numbers.

        Exercise Group.

        17.

        -
        -
        +
        +
        -
        Which of the following are integers? There may be more than one correct answer.
          -
        • \(\displaystyle 3.101001000100001\ldots\)
          • -
        • \(\displaystyle -6\)
          • -
        • \(\displaystyle 9\)
          • -
        • \(\displaystyle -5.4\overline{9}\)
          • -
        • \(\displaystyle -66308\)
          • -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle 5.337\)
          • -
        • \(\displaystyle 0\)
          • +
          Which of the following are integers? There may be more than one correct answer.
            +
          • \(\displaystyle \sqrt{49}\)
            • +
          • \(\displaystyle {\frac{4}{89}}\)
            • +
          • \(\displaystyle -28368\)
            • +
          • \(\displaystyle 5848\)
            • +
          • \(\displaystyle -4.849\)
            • +
          • \(\displaystyle -6.101001000100001\ldots\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle -7\)
          -
          Explanation.
          To be an integer, you have to be one of the numbers \(\ldots,-2,-1,0,1,2,\ldots\text{.}\) Sometimes integers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is an integer, since \(\sqrt{4}=2\text{.}\) So the correct answers are BCEH.
          +
          Explanation.
          To be an integer, you have to be one of the numbers \(\ldots,-2,-1,0,1,2,\ldots\text{.}\) Sometimes integers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is an integer, since \(\sqrt{4}=2\text{.}\) So the correct answers are ACDH.

        18.

        -
        -
        +
        +
        -
        Which of the following are integers? There may be more than one correct answer.
          -
        • \(\displaystyle -5.8\overline{96}\)
          • -
        • \(\displaystyle -9.101001000100001\ldots\)
          • -
        • \(\displaystyle -57415\)
          • -
        • \(\displaystyle 74\)
          • -
        • \(\displaystyle {\frac{5}{37}}\)
          • -
        • \(\displaystyle -2\)
          • -
        • \(\displaystyle \sqrt{6}\)
          • -
        • \(\displaystyle \sqrt{36}\)
          • +
          Which of the following are integers? There may be more than one correct answer.
            +
          • \(\displaystyle -3\)
            • +
          • \(\displaystyle -3.101001000100001\ldots\)
            • +
          • \(\displaystyle 7.469\)
            • +
          • \(\displaystyle -19475\)
            • +
          • \(\displaystyle 53\)
            • +
          • \(\displaystyle \sqrt{100}\)
            • +
          • \(\displaystyle -5.6\overline{79}\)
            • +
          • \(\displaystyle \pi\)
          -
          Explanation.
          To be an integer, you have to be one of the numbers \(\ldots,-2,-1,0,1,2,\ldots\text{.}\) Sometimes integers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is an integer, since \(\sqrt{4}=2\text{.}\) So the correct answers are CDFH.
          +
          Explanation.
          To be an integer, you have to be one of the numbers \(\ldots,-2,-1,0,1,2,\ldots\text{.}\) Sometimes integers are explicitly written this way, and sometimes they are hidden. For example \(\sqrt{4}\) is an integer, since \(\sqrt{4}=2\text{.}\) So the correct answers are ADEF.
        @@ -1081,36 +1095,36 @@

        Exercise Group.

        Exercise Group.

        19.

        -
        -
        +
        +
        -
        Which of the following are rational numbers? There may be more than one correct answer.
          -
        • \(\displaystyle 39\)
          • -
        • \(\displaystyle \sqrt{11}\)
          • -
        • \(\displaystyle -6\)
          • -
        • \(\displaystyle 6.6\overline{85}\)
          • -
        • \(\displaystyle 7.225\)
          • -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle 5017\)
          • -
        • \(\displaystyle -48522\)
          • +
          Which of the following are rational numbers? There may be more than one correct answer.
            +
          • \(\displaystyle \sqrt{4}\)
            • +
          • \(\displaystyle 0\)
            • +
          • \(\displaystyle -10582\)
            • +
          • \(\displaystyle -7\)
            • +
          • \(\displaystyle -{\frac{2}{95}}\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle 4.343\)
            • +
          • \(\displaystyle -5.101001000100001\ldots\)
          -
          Explanation.
          To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) So the correct answers are ACDEGH.
          +
          Explanation.
          To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) So the correct answers are ABCDEG.

        20.

        -
        -
        +
        +
        -
        Which of the following are rational numbers? There may be more than one correct answer.
          -
        • \(\displaystyle 0.680999999999999\)
          • -
        • \(\displaystyle \sqrt{100}\)
          • -
        • \(\displaystyle -4\)
          • -
        • \(\displaystyle -67968\)
          • -
        • \(\displaystyle 5\)
          • -
        • \(\displaystyle \sqrt{13}\)
          • -
        • \(\displaystyle -0.699999999999999\overline{63}\)
          • -
        • \(\displaystyle 4.101001000100001\ldots\)
          • +
          Which of the following are rational numbers? There may be more than one correct answer.
            +
          • \(\displaystyle \sqrt{2}\)
            • +
          • \(\displaystyle 30893\)
            • +
          • \(\displaystyle -8.927\)
            • +
          • \(\displaystyle \sqrt{4}\)
            • +
          • \(\displaystyle -1\)
            • +
          • \(\displaystyle -24217\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle 3.4\overline{32}\)
          -
          Explanation.
          To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) So the correct answers are ABCDEG.
          +
          Explanation.
          To be an rational number, you either have to be a fraction with whole numbers for the numerator and denominator, or you have to be a number with a terminating decimal expression, or you have to be a number with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) So the correct answers are BCDEFH.
        @@ -1119,36 +1133,36 @@

        Exercise Group.

        Exercise Group.

        21.

        -
        -
        +
        +
        -
        Which of the following are irrational numbers? There may be more than one correct answer.
          -
        • \(\displaystyle -{\frac{5}{12}}\)
          • -
        • \(\displaystyle -8.715\)
          • -
        • \(\displaystyle -30735\)
          • -
        • \(\displaystyle 0\)
          • -
        • \(\displaystyle 69\)
          • -
        • \(\displaystyle -6\)
          • -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle \sqrt{2}\)
          • +
          Which of the following are irrational numbers? There may be more than one correct answer.
            +
          • \(\displaystyle 3.885\)
            • +
          • \(\displaystyle -7\)
            • +
          • \(\displaystyle 72541\)
            • +
          • \(\displaystyle 0.101001000100001\ldots\)
            • +
          • \(\displaystyle {\frac{9}{50}}\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle \sqrt{9}\)
            • +
          • \(\displaystyle -92796\)
          -
          Explanation.
          To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) The irrational numbers here are are GH.
          +
          Explanation.
          To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) The irrational numbers here are are DF.

        22.

        -
        -
        +
        +
        -
        Which of the following are irrational numbers? There may be more than one correct answer.
          -
        • \(\displaystyle \sqrt{2}\)
          • -
        • \(\displaystyle -8.025\)
          • -
        • \(\displaystyle \sqrt{4}\)
          • -
        • \(\displaystyle 35\)
          • -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle -21842\)
          • -
        • \(\displaystyle {\frac{8}{67}}\)
          • -
        • \(\displaystyle -2\)
          • +
          Which of the following are irrational numbers? There may be more than one correct answer.
            +
          • \(\displaystyle -6.101001000100001\ldots\)
            • +
          • \(\displaystyle -8.961\)
            • +
          • \(\displaystyle \sqrt{16}\)
            • +
          • \(\displaystyle -{\frac{4}{79}}\)
            • +
          • \(\displaystyle \sqrt{13}\)
            • +
          • \(\displaystyle -3\)
            • +
          • \(\displaystyle 14\)
            • +
          • \(\displaystyle -83903\)
          -
          Explanation.
          To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) The irrational numbers here are are AE.
          +
          Explanation.
          To be an irrational number, you simply have to not be a rational number. A rational number is a number that can be written as a fraction with whole numbers for the numerator and denominator, or with a terminating decimal expression, or with a repeating decimal expression. (Actually, these descriptions are a little redundant.) Sometimes rational numbers are explicitly written in one of these ways, and sometimes they are hidden. For example \(\sqrt{0.16}\) is a rational number, since \(\sqrt{0.16}=0.4\text{.}\) The irrational numbers here are are AE.
        @@ -1157,36 +1171,36 @@

        Exercise Group.

        Exercise Group.

        23.

        -
        -
        +
        +
        -
        Which of the following are real numbers? There may be more than one correct answer.
          -
        • \(\displaystyle \pi\)
          • -
        • \(\displaystyle 71710\)
          • -
        • \(\displaystyle -{\frac{9}{43}}\)
          • -
        • \(\displaystyle 2.101001000100001\ldots\)
          • -
        • \(\displaystyle 1\)
          • -
        • \(\displaystyle -8.2\overline{27}\)
          • -
        • \(\displaystyle -6\)
          • -
        • \(\displaystyle -12949\)
          • +
          Which of the following are real numbers? There may be more than one correct answer.
            +
          • \(\displaystyle 1.535\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle -2342\)
            • +
          • \(\displaystyle -{\frac{8}{37}}\)
            • +
          • \(\displaystyle -7\)
            • +
          • \(\displaystyle \sqrt{5}\)
            • +
          • \(\displaystyle 0\)
            • +
          • \(\displaystyle \sqrt{25}\)
          -
          Explanation.
          All of these numbers are real numbers, since they can all be located on a number line.
          +
          Explanation.
          All of these numbers are real numbers, since they can all be located on a number line.

        24.

        -
        -
        +
        +
        -
        Which of the following are real numbers? There may be more than one correct answer.
          -
        • \(\displaystyle 0\)
          • -
        • \(\displaystyle \sqrt{3}\)
          • -
        • \(\displaystyle -4056\)
          • -
        • \(\displaystyle 65\)
          • -
        • \(\displaystyle -{\frac{8}{11}}\)
          • -
        • \(\displaystyle 5.717\)
          • -
        • \(\displaystyle 3.101001000100001\ldots\)
          • -
        • \(\displaystyle -2\)
          • +
          Which of the following are real numbers? There may be more than one correct answer.
            +
          • \(\displaystyle -3\)
            • +
          • \(\displaystyle 6.8\overline{72}\)
            • +
          • \(\displaystyle -66116\)
            • +
          • \(\displaystyle 97586\)
            • +
          • \(\displaystyle -{\frac{9}{20}}\)
            • +
          • \(\displaystyle \pi\)
            • +
          • \(\displaystyle -9.101001000100001\ldots\)
            • +
          • \(\displaystyle \sqrt{25}\)
          -
          Explanation.
          All of these numbers are real numbers, since they can all be located on a number line.
          +
          Explanation.
          All of these numbers are real numbers, since they can all be located on a number line.
        @@ -1195,87 +1209,87 @@

        Exercise Group.

        Exercise Group.

        25.

        -
        -
        +
        +
        -
        Determine the validity of each statement by selecting True or False.
          -
        1. The number \(0\) is an integer
          2. -
        2. The number \(\sqrt{\frac{4}{16}}\) is an integer, but not a whole number
          3. -
        3. The number \(\frac{-\sqrt{11}}{37\sqrt{11}}\) is irrational
          4. -
        4. The number \(\pi\) is irrational
          5. -
        5. The number \(36\) is an integer, but not a whole number
          6. +
          Determine the validity of each statement by selecting True or False.
            +
          1. The number \(\frac{-\sqrt{37}}{37\sqrt{37}}\) is irrational
            2. +
          2. The number \(0\) is an integer
            3. +
          3. The number \(14\) is an integer, but not a whole number
            4. +
          4. The number \(\pi\) is irrational
            5. +
          5. The number \(\sqrt{(-14)^2}\) is irrational

        26.

        -
        -
        +
        +
        -
        Determine the validity of each statement by selecting True or False.
          -
        1. The number \(\frac{13}{20}\) is rational, but not a natural number
          2. -
        2. The number \(\sqrt{5^2}\) is a real number, but not a rational number
          3. -
        3. The number \(-13\) is an integer that is also a natural number
          4. -
        4. The number \(\sqrt{\frac{25}{81}}\) is rational, but not an integer
          5. -
        5. The number \(\frac{0}{5}\) is rational, but not a whole number
          6. +
          Determine the validity of each statement by selecting True or False.
            +
          1. The number \(\frac{0}{4}\) is rational, but not a whole number
            2. +
          2. The number \(\sqrt{20}\) is a real number, but not a rational number
            3. +
          3. The number \(\sqrt{23}\) is a real number, but not an irrational number
            4. +
          4. The number \(\sqrt{41^2}\) is a real number, but not a rational number
            5. +
          5. The number \(0.20002000200020002000...\) is rational

        27.

        -
        -
        +
        +
        -
        In each situation, which number set do you think is most appropriate?
          -
        1. -
          The number of dogs a student has owned throughout their lifetime.
          -
          This number is best considered as a .
          +
          In each situation, which number set do you think is most appropriate?
            +
          1. +
            The number of dogs a student has owned throughout their lifetime.
            +
            This number is best considered as a .
            2. -
          2. -
            The difference between the projected annual expenditures and the actual annual expenditures for a given company.
            -
            This number is best considered as a .
            +
          3. +
            The difference between the projected annual expenditures and the actual annual expenditures for a given company.
            +
            This number is best considered as a .
            4. -
          4. -
            The length around swimming pool in the shape of a half circle with radius \(10\,\text{ft}\text{.}\) +
          5. +
            The length around swimming pool in the shape of a half circle with radius \(10\,\text{ft}\text{.}\)
            -
            This number is best considered as a .
            +
            This number is best considered as a .
            6. -
          6. -
            The proportion of students at a college who own a car.
            -
            This number is best considered as a .
            +
          7. +
            The proportion of students at a college who own a car.
            +
            This number is best considered as a .
            8. -
          8. -
            The width of a sheet of paper, in inches.
            -
            This number is best considered as a .
            +
          9. +
            The width of a sheet of paper, in inches.
            +
            This number is best considered as a .
            10. -
          10. -
            The number of people eating in a non-empty restaurant.
            -
            This number is best considered as a .
            +
          11. +
            The number of people eating in a non-empty restaurant.
            +
            This number is best considered as a .
          -
          Answer 1.
          \(\text{whole number}\)
          Answer 2.
          \(\text{rational number}\)
          Answer 3.
          \(\text{irrational number}\)
          Answer 4.
          \(\text{rational number}\)
          Answer 5.
          \(\text{rational number}\)
          Answer 6.
          \(\text{natural number}\)
          +
          Answer 1.
          \(\text{whole number}\)
          Answer 2.
          \(\text{rational number}\)
          Answer 3.
          \(\text{irrational number}\)
          Answer 4.
          \(\text{rational number}\)
          Answer 5.
          \(\text{rational number}\)
          Answer 6.
          \(\text{natural number}\)

        28.

        -
        -
        +
        +
        -
          -
        1. Give an example of a whole number that is not an integer.
          2. -
        2. Give an example of an integer that is not a whole number.
          3. -
        3. Give an example of a rational number that is not an integer.
          4. -
        4. Give an example of a irrational number.
          5. -
        5. Give an example of a irrational number that is also an integer.
          6. +
            +
          1. Give an example of a whole number that is not an integer.
            2. +
          2. Give an example of an integer that is not a whole number.
            3. +
          3. Give an example of a rational number that is not an integer.
            4. +
          4. Give an example of a irrational number.
            5. +
          5. Give an example of a irrational number that is also an integer.
          -
          Answer 1.
          \(\text{DNE}\hbox{ or }\text{NONE}\)
          Answer 2.
          \(\text{any of }-1, -2, -3,\ldots\)
          Answer 3.
          \(\text{many possibilities: }\frac{1}{2}, 2.3, \ldots\)
          Answer 4.
          \(\text{many possibilities: }\sqrt{2}, \pi, \ldots\)
          Answer 5.
          \(\text{DNE}\hbox{ or }\text{NONE}\)
          Explanation.
            -
          1. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question.
            2. -
          2. Any negative integer, like \(-1\text{,}\) is not a whole number, but is an integer.
            3. -
          3. Any terminating decimal, like \(1.2\text{,}\) is a rational number, but is not an integer.
            4. -
          4. +
            Answer 1.
            \(\text{DNE}\hbox{ or }\text{NONE}\)
            Answer 2.
            \(\text{any of }-1, -2, -3,\ldots\)
            Answer 3.
            \(\text{many possibilities: }\frac{1}{2}, 2.3, \ldots\)
            Answer 4.
            \(\text{many possibilities: }\sqrt{2}, \pi, \ldots\)
            Answer 5.
            \(\text{DNE}\hbox{ or }\text{NONE}\)
            Explanation.
              +
            1. Since all whole numbers belong to integers, we cannot write any whole number which is not an integer. Type DNE (does not exist) for this question.
              2. +
            2. Any negative integer, like \(-1\text{,}\) is not a whole number, but is an integer.
              3. +
            3. Any terminating decimal, like \(1.2\text{,}\) is a rational number, but is not an integer.
              4. +
            4. \(\pi\) is the easiest number to remember as an irrational number. Another constant worth knowing is \(e\approx2.718\text{.}\) Finally, the square root of most integers are irrational, like \(\sqrt{2}\) and \(\sqrt{3}\text{.}\)
              5. -
            5. All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers.
              6. +
            6. All irrational numbers are non-repeating and non-terminating decimals. No irrational numbers are integers.
          @@ -1283,46 +1297,46 @@

          Exercise Group.

          Writing Decimals as Fractions.

          29.

          -
          -
          +
          +
          -
          Write the rational number \(9.09\) as a fraction.
          -
          Answer.
          \({\frac{909}{100}}\)
          +
          Write the rational number \(1.27\) as a fraction.
          +
          Answer.
          \({\frac{127}{100}}\)

          30.

          -
          -
          +
          +
          -
          Write the rational number \(11.646\) as a fraction.
          -
          Answer.
          \({\frac{5823}{500}}\)
          +
          Write the rational number \(30.284\) as a fraction.
          +
          Answer.
          \({\frac{7571}{250}}\)

          31.

          -
          -
          +
          +
          -
          Write the rational number \(0.\overline{22}=0.2222\ldots\) as a fraction.
          -
          Answer.
          \({\frac{2}{9}}\)
          +
          Write the rational number \(0.\overline{31}=0.3131\ldots\) as a fraction.
          +
          Answer.
          \({\frac{31}{99}}\)

          32.

          -
          -
          +
          +
          -
          Write the rational number \(0.\overline{248}=0.248248\ldots\) as a fraction.
          -
          Answer.
          \({\frac{248}{999}}\)
          +
          Write the rational number \(0.\overline{446}=0.446446\ldots\) as a fraction.
          +
          Answer.
          \({\frac{446}{999}}\)

          33.

          -
          -
          +
          +
          -
          Write the rational number \(8.7\overline{35}=8.73535\ldots\) as a fraction.
          -
          Answer.
          \({\frac{8963}{990}}\)
          +
          Write the rational number \(7.2\overline{53}=7.25353\ldots\) as a fraction.
          +
          Answer.
          \({\frac{3829}{495}}\)

          34.

          -
          -
          +
          +
          -
          Write the rational number \(3.5\overline{422}=3.5422422\ldots\) as a fraction.
          -
          Answer.
          \({\frac{7837}{1998}}\)
          +
          Write the rational number \(2.9\overline{579}=2.9579579\ldots\) as a fraction.
          +
          Answer.
          \({\frac{11587}{3330}}\)
          @@ -1330,13 +1344,13 @@

          Writing Decimals as Fractions.

          Challenge.

          35.

          -
          -
          +
          +
          -
          Imagine making up a number with the following pattern. After the decimal point, write the natural numbers 1, 2, 3, 4, 5, etc. The decimal digits will extend forever with this pattern: \(0.12345\ldots\text{.}\) -
          Is the number a rational number or an irrational number?
          +
          Imagine making up a number with the following pattern. After the decimal point, write the natural numbers 1, 2, 3, 4, 5, etc. The decimal digits will extend forever with this pattern: \(0.12345\ldots\text{.}\) +
          Is the number a rational number or an irrational number?
          -
          Answer.
          \(\text{irrational}\)
          Explanation.
          The number is irrational. The decimal will never end and it will never repeat.
          +
          Answer.
          \(\text{irrational}\)
          Explanation.
          The number is irrational. The decimal will never end and it will never repeat.
          diff --git a/section-slope-intercept-form.html b/section-slope-intercept-form.html index 36f4f37e0..3813a7968 100644 --- a/section-slope-intercept-form.html +++ b/section-slope-intercept-form.html @@ -420,7 +420,7 @@

          Search Results:

          @@ -455,6 +455,20 @@

          Search Results:

        6. 3.9.5 Exercises
          7. +
        7. +
          3.10 Graphing Lines Chapter Review
          + +
        8. diff --git a/section-slope.html b/section-slope.html index 0113c0741..254c1a83c 100644 --- a/section-slope.html +++ b/section-slope.html @@ -420,7 +420,7 @@

          Search Results:

          @@ -455,6 +455,20 @@

          Search Results:

        9. 3.9.5 Exercises
          10. +
        10. +
          3.10 Graphing Lines Chapter Review
          + +
        11. diff --git a/section-solving-a-system-by-graphing.html b/section-solving-a-system-by-graphing.html index 1f3bb64db..307db5da6 100644 --- a/section-solving-a-system-by-graphing.html +++ b/section-solving-a-system-by-graphing.html @@ -420,7 +420,7 @@

          Search Results:

          @@ -455,6 +455,20 @@

          Search Results:

        12. 3.9.5 Exercises
          13. +
        13. +
          3.10 Graphing Lines Chapter Review
          + +
        14. @@ -680,11 +694,11 @@

          Search Results:

      Checkpoint 4.1.8.

      -
      -
      +
      +
      -
      Determine the solution to the system of equations graphed below.
      -
      Explanation.
      The two lines intersect where \(x=3\) and \(y=2\text{,}\) so the solution is the point \((3,2)\text{.}\) +
      Determine the solution to the system of equations graphed below.
      +
      Explanation.
      The two lines intersect where \(x=3\) and \(y=2\text{,}\) so the solution is the point \((3,2)\text{.}\)
      Now let’s solve a system where we need to make the graph ourselves.
      @@ -720,10 +734,10 @@

      Search Results:

    Checkpoint 4.1.11.

    -
    -
    +
    +
    -
    Solve the following system of equations by graphing.
    +
    Solve the following system of equations by graphing.
    \begin{equation*} \left\{ \begin{alignedat}{4} @@ -733,8 +747,8 @@

    Search Results:

    \right. \end{equation*}
    -
    Explanation.
    -
    Both of these equations are written in slope-intercept form. The first equation, \(y=\frac{2}{3}x-5\text{,}\) has slope \(\frac{2}{3}\) and \(y\)-intercept \((0,-5)\text{.}\) The second equation, \(y=-\frac{1}{2}x+2\text{,}\) has slope \(-\frac{1}{2}\) and \(y\)-intercept \((0,2)\text{.}\) We can use this information to graph both lines.
    a Cartesian grid with two intersecting lines; one line has a y-intercept of -5 and a slope of 2/3; the other line has a y-intercept of 2 and a slope of -1/2
    It appears that the two lines intersect where \(x=6\) and \(y=-1\text{,}\) so the solution to the system of equations would be the point \((6,-1)\text{.}\) However we should check that \((6,-1)\) actually works as a solution to each of the original equations.
    +
    Explanation.
    +
    Both of these equations are written in slope-intercept form. The first equation, \(y=\frac{2}{3}x-5\text{,}\) has slope \(\frac{2}{3}\) and \(y\)-intercept \((0,-5)\text{.}\) The second equation, \(y=-\frac{1}{2}x+2\text{,}\) has slope \(-\frac{1}{2}\) and \(y\)-intercept \((0,2)\text{.}\) We can use this information to graph both lines.
    a Cartesian grid with two intersecting lines; one line has a y-intercept of -5 and a slope of 2/3; the other line has a y-intercept of 2 and a slope of -1/2
    It appears that the two lines intersect where \(x=6\) and \(y=-1\text{,}\) so the solution to the system of equations would be the point \((6,-1)\text{.}\) However we should check that \((6,-1)\) actually works as a solution to each of the original equations.
    \begin{equation*} \begin{aligned} y\amp=\frac{2}{3}x-5\amp y\amp=-\frac{1}{2}x+2\\ @@ -742,16 +756,16 @@

    Search Results:

    -1\amp\confirm{=}4-5\amp -1\amp\confirm{=}-3+2 \end{aligned} \end{equation*} -
    This verifies that \((6,-1)\) is the solution.
    +
    This verifies that \((6,-1)\) is the solution.

    Checkpoint 4.1.12.

    -
    -
    +
    +
    -
    Solve the following system of equations by graphing. Note that the equations are in standard form. To review plotting a line equation that is in standard form, see Section 3.7.
    +
    Solve the following system of equations by graphing. Note that the equations are in standard form. To review plotting a line equation that is in standard form, see Section 3.7.
    \begin{equation*} \left\{ \begin{alignedat}{4} @@ -761,8 +775,8 @@

    Search Results:

    \right. \end{equation*}
    -
    Explanation.
    -
    Both of these equations are written in standard form. The first equation, \(x-3y=-12\text{,}\) has \(x\)-intercept at \((-12,0)\) and \(y\)-intercept at \((0,4)\text{.}\) The second equation, \(x+y=-4\text{,}\) has \(x\)-intercept at \((-4,0)\) and \(y\)-intercept at \((0,-4)\text{.}\) We can use this information to graph both lines.
    a Cartesian grid with two intersecting lines; one line has a x-intercept of -12 and y-intercept of 4; the other line has x-intercept of -4 and y-intercept of -4
    It appears that the two lines intersect where \(x=-6\) and \(y=2\text{,}\) so the solution to the system of equations would be the point \((-6,2)\text{.}\) However we should check that \((-6,2)\) actually works as a solution to each of the original equations.
    +
    Explanation.
    +
    Both of these equations are written in standard form. The first equation, \(x-3y=-12\text{,}\) has \(x\)-intercept at \((-12,0)\) and \(y\)-intercept at \((0,4)\text{.}\) The second equation, \(x+y=-4\text{,}\) has \(x\)-intercept at \((-4,0)\) and \(y\)-intercept at \((0,-4)\text{.}\) We can use this information to graph both lines.
    a Cartesian grid with two intersecting lines; one line has a x-intercept of -12 and y-intercept of 4; the other line has x-intercept of -4 and y-intercept of -4
    It appears that the two lines intersect where \(x=-6\) and \(y=2\text{,}\) so the solution to the system of equations would be the point \((-6,2)\text{.}\) However we should check that \((-6,2)\) actually works as a solution to each of the original equations.
    \begin{equation*} \begin{aligned} x-3y\amp=-12\amp x+y\amp=-4\\ @@ -770,16 +784,16 @@

    Search Results:

    -6-6\amp\confirm{=}-12\amp -4\amp\confirm{=}-4 \end{aligned} \end{equation*} -
    This verifies that \((-6,2)\) is the solution.
    +
    This verifies that \((-6,2)\) is the solution.

    Checkpoint 4.1.13.

    -
    -
    +
    +
    -
    Solve the following system of equations by graphing. Note that the equations are in point-slope form. To review plotting a line equation that is in point-slope form, see Section 3.6.
    +
    Solve the following system of equations by graphing. Note that the equations are in point-slope form. To review plotting a line equation that is in point-slope form, see Section 3.6.
    \begin{equation*} \left\{ \begin{alignedat}{3} @@ -789,8 +803,8 @@

    Search Results:

    \right. \end{equation*}
    -
    Explanation.
    -
    Both of these equations are written in point-slope form. The first equation, \(y=3(x-2)+1\text{,}\) passes through \((2,1)\) and has slope \(3\text{.}\) The second equation, \(y=-\frac{1}{2}(x+1)-1\text{,}\) passes through \((-1,-1)\) and has slope \(-\frac{1}{2}\text{.}\) We can use this information to graph both lines.
    a Cartesian grid with two intersecting lines; one line passes through (2,1) with slope 3; the other line passes through (-1,-1) with slope -1/2
    It appears that the two lines intersect where \(x=1\) and \(y=-2\text{,}\) so the solution to the system of equations would be the point \((1,-2)\text{.}\) However we should check that \((1,-2)\) actually works as a solution to each of the original equations.
    +
    Explanation.
    +
    Both of these equations are written in point-slope form. The first equation, \(y=3(x-2)+1\text{,}\) passes through \((2,1)\) and has slope \(3\text{.}\) The second equation, \(y=-\frac{1}{2}(x+1)-1\text{,}\) passes through \((-1,-1)\) and has slope \(-\frac{1}{2}\text{.}\) We can use this information to graph both lines.
    a Cartesian grid with two intersecting lines; one line passes through (2,1) with slope 3; the other line passes through (-1,-1) with slope -1/2
    It appears that the two lines intersect where \(x=1\) and \(y=-2\text{,}\) so the solution to the system of equations would be the point \((1,-2)\text{.}\) However we should check that \((1,-2)\) actually works as a solution to each of the original equations.
    \begin{equation*} \begin{aligned} y\amp=3(x-2)+1\amp y\amp=-\frac{1}{2}(x+1)-1\\ @@ -798,25 +812,25 @@

    Search Results:

    -2\amp\confirm{=}3(-1)+1\amp -2\amp\confirm{=}-\frac{1}{2}(2)-1 \end{aligned} \end{equation*} -
    This verifies that \((1,-2)\) is the solution.
    +
    This verifies that \((1,-2)\) is the solution.

    Checkpoint 4.1.14.

    -
    -
    +
    +
    -
    A college has a north campus and a south campus. The north campus has \(16\) thousand students, but enrollment has been declining by \(0.8\) thousand students per year. The newer south campus has only \(2\) thousand students, but enrollment has been increasing by \(0.6\) thousand students per year. If these trends continue, in how many years would the two campuses have the same number of students? Write a system of equations for this scenario, then solve it graphically to find the answer.
    (a)
    -
    Write two equations that form a system for this scenario.
    -
    Explanation.
    -
    The north campus has initial population value \(16\) and decreases with rate \(0.8\) thousand students per year. So one equation is \(y=16-0.8t\text{.}\) -
    The south campus has initial population value \(2\) and increases with rate \(0.6\) thousand students per year. So the other equation is \(y=2+0.6t\text{.}\) -
    -
    (b)
    -
    Plot the lines for the system of two equations.
    -
    Explanation.
    -
    For plotting, it may be helpful to write the system with the decimals converted to fractions:
    +
    A college has a north campus and a south campus. The north campus has \(16\) thousand students, but enrollment has been declining by \(0.8\) thousand students per year. The newer south campus has only \(2\) thousand students, but enrollment has been increasing by \(0.6\) thousand students per year. If these trends continue, in how many years would the two campuses have the same number of students? Write a system of equations for this scenario, then solve it graphically to find the answer.
    (a)
    +
    Write two equations that form a system for this scenario.
    +
    Explanation.
    +
    The north campus has initial population value \(16\) and decreases with rate \(0.8\) thousand students per year. So one equation is \(y=16-0.8t\text{.}\) +
    The south campus has initial population value \(2\) and increases with rate \(0.6\) thousand students per year. So the other equation is \(y=2+0.6t\text{.}\) +
    +
    (b)
    +
    Plot the lines for the system of two equations.
    +
    Explanation.
    +
    For plotting, it may be helpful to write the system with the decimals converted to fractions:
    \begin{equation*} \left\{ \begin{alignedat}{3} @@ -825,10 +839,10 @@

    Search Results:

    \end{alignedat} \right. \end{equation*} -
    So we see that we can use \((0,16)\) and \((0,2)\) as \(y\)-intercepts. And then for the first line, use slope triangles where we move \(5\) units to the right and then \(4\) units down. While for the second line, use slope triangles where we move \(5\) units to the right and then \(3\) units up.
    a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8)
    -
    (c)
    -
    Based on the graph, how long will it be until the two campuses have the same number of students?
    How many students will each campus have at that time?
    -
    Explanation.
    The lines cross at \((10,8)\text{.}\) So after \(10\) years, each campus will have \(8\) thousand students.
    +
    So we see that we can use \((0,16)\) and \((0,2)\) as \(y\)-intercepts. And then for the first line, use slope triangles where we move \(5\) units to the right and then \(4\) units down. While for the second line, use slope triangles where we move \(5\) units to the right and then \(3\) units up.
    a Cartesian grid with two intersecting lines; one line passes through (0,16) with slope -0.8; the other line passes through (0,2) with slope 0.6; the lines cross at (10,8)
    +
    (c)
    +
    Based on the graph, how long will it be until the two campuses have the same number of students?
    How many students will each campus have at that time?
    +
    Explanation.
    The lines cross at \((10,8)\text{.}\) So after \(10\) years, each campus will have \(8\) thousand students.

    Subsection 4.1.3 Special Systems of Equations @@ -895,10 +909,10 @@

    Search Results:

    Notice that for a system of equations with infinite solutions like Example 18, we didn’t say that everything is a solution. It’s only the points that are on that coinciding line that are solutions. It would be incorrect to say that the solution set is “all real numbers” or as “all ordered pairs”.

    Checkpoint 4.1.22.

    -
    -
    +
    +
    -
    Solve the following system of equations by graphing.
    +
    Solve the following system of equations by graphing.
    \begin{equation*} \left\{ \begin{alignedat}{3} @@ -908,17 +922,17 @@

    Search Results:

    \right. \end{equation*}
    -
    Explanation.
    -
    If we graph these lines, we find they are the same line.
    a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0)
    So there are infinitely many solutions: all points \((x_1,y_1)\) on that common line.
    +
    Explanation.
    +
    If we graph these lines, we find they are the same line.
    a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0)
    So there are infinitely many solutions: all points \((x_1,y_1)\) on that common line.

    Checkpoint 4.1.23.

    -
    -
    +
    +
    -
    Solve the following system of equations by graphing.
    +
    Solve the following system of equations by graphing.
    \begin{equation*} \left\{ \begin{alignedat}{3} @@ -928,8 +942,8 @@

    Search Results:

    \right. \end{equation*}
    -
    Explanation.
    -
    If we graph these lines, we find they are parallel and never cross.
    a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0)
    So there are no solutions to this system.
    +
    Explanation.
    +
    If we graph these lines, we find they are parallel and never cross.
    a coordinate plot of two lines that are coinciding; they each have a y-intercept of (0,-5) and an x-intercept of (8,0)
    So there are no solutions to this system.
    We can summarize the possibilities for what a soltuion set to a system of two linear equations in two variables.
    @@ -956,34 +970,34 @@
    Exercise Group.
    Select the equations/inequalities that are linear with one variable.
    1.
    -
    -
    +
    +
    -
      -
    • \(\displaystyle 2\pi r\geq10\pi \)
      • -
    • \(\displaystyle 9-y\neq4\)
      • -
    • \(\displaystyle V+2q^{2}=25\)
      • -
    • \(\displaystyle 2-8q^{2}\lt27\)
      • -
    • \(\displaystyle \sqrt{8.5V+2}\geq9\)
      • -
    • \(\displaystyle 2.7V\lt-1.4\)
      • -
    • None of the above
      • +
        +
      • \(\displaystyle 0.9z=7.3\)
        • +
      • \(\displaystyle 2\pi r\lt20\pi \)
        • +
      • \(\displaystyle 6p+4r^{2}=8\)
        • +
      • \(\displaystyle 1-5y^{2}\geq20\)
        • +
      • \(\displaystyle 7r-13=-7\)
        • +
      • \(\displaystyle \sqrt{4.5V-1}=8\)
        • +
      • None of the above
      -
      Answer.
      \(\text{Choice 1, Choice 2, Choice 6}\)
      +
      Answer.
      \(\text{Choice 1, Choice 2, Choice 5}\)
    2.
    -
    -
    +
    +
    -
      -
    • \(\displaystyle 2zq=-96\)
      • -
    • \(\displaystyle \pi r^{2}=30\pi \)
      • -
    • \(\displaystyle 7p+2=30\)
      • -
    • \(\displaystyle \left|-7.9r-6\right|\geq-60\)
      • -
    • \(\displaystyle y^{2}+r^{2}=-56\)
      • -
    • \(\displaystyle r\sqrt{28}=-20\)
      • -
    • None of the above
      • +
        +
      • \(\displaystyle \pi r^{2}\leq17\pi \)
        • +
      • \(\displaystyle \left|9-5.3V\right|\gt36\)
        • +
      • \(\displaystyle z\sqrt{27}\leq58\)
        • +
      • \(\displaystyle 17z+6=-86\)
        • +
      • \(\displaystyle 6rz\leq59\)
        • +
      • \(\displaystyle x^{2}+V^{2}=-85\)
        • +
      • None of the above
      -
      Answer.
      \(\text{Choice 3, Choice 6}\)
      +
      Answer.
      \(\text{Choice 3, Choice 4}\)
    @@ -993,76 +1007,76 @@
    Check a Possible Solution to a System.
    Check if the given point is a solution to the given system of two linear equations.
    3.
    -
    +
    -
    Is \({\left(-1,-4\right)}\) a solution?
    \(\left\{ +
    Is \({\left(3,-8\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{10}{3}}\mathopen{}\left(x-2\right)+6}\\ -y \amp= {-{\frac{13}{6}}\mathopen{}\left(x+7\right)+9} +y \amp= {-{\frac{6}{5}}\mathopen{}\left(x+2\right)-2}\\ +y \amp= {-15\mathopen{}\left(x-2\right)+7} \end{alignedat} \right.\)
    Answer.
    \(\text{Yes, it is a solution.}\)
    4.
    -
    -
    +
    +
    -
    Is \({\left(1,7\right)}\) a solution?
    \(\left\{ +
    Is \({\left(5,3\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{1}{5}}\mathopen{}\left(x+9\right)+5}\\ -y \amp= {-{\frac{3}{2}}\mathopen{}\left(x-7\right)-2} +y \amp= {-{\frac{7}{2}}\mathopen{}\left(x-7\right)-4}\\ +y \amp= {{\frac{5}{9}}\mathopen{}\left(x+4\right)-2} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{Yes, it is a solution.}\)
    +
    Answer.
    \(\text{Yes, it is a solution.}\)
    5.
    -
    +
    -
    Is \({\left(-4,4\right)}\) a solution?
    \(\left\{ +
    Is \({\left(-5,-5\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -\amp {x+2y = 2}\\ -\amp {3x-2y = -18} +\amp {-x+2y = -4}\\ +\amp {5x-3y = -15} \end{alignedat} \right.\)
    Answer.
    \(\text{No, it is not a solution.}\)
    6.
    -
    -
    +
    +
    -
    Is \({\left(8,8\right)}\) a solution?
    \(\left\{ +
    Is \({\left(-1,3\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -\amp {-4x+3y = -12}\\ -\amp {2x-3y = -6} +\amp {x-y = -5}\\ +\amp {2x-y = -6} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{No, it is not a solution.}\)
    +
    Answer.
    \(\text{No, it is not a solution.}\)
    7.
    -
    +
    -
    Is \({\left(0,-5\right)}\) a solution?
    \(\left\{ +
    Is \({\left(0,-8\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -y\amp= {{\frac{10}{7}}x-5}\\ -y\amp= {2x-9} +y\amp= {-{\frac{11}{7}}x-8}\\ +y\amp= {-{\frac{3}{7}}x} \end{alignedat} \right.\)
    Answer.
    \(\text{No, it is not a solution.}\)
    8.
    -
    -
    +
    +
    -
    Is \({\left(8,7\right)}\) a solution?
    \(\left\{ +
    Is \({\left(0,1\right)}\) a solution?
    \(\left\{ \begin{alignedat}{3} -y\amp= {{\frac{3}{8}}x+4}\\ -y\amp= {{\frac{1}{2}}x+5} +y\amp= {{\frac{4}{5}}x+1}\\ +y\amp= {-{\frac{3}{5}}x-6} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{No, it is not a solution.}\)
    +
    Answer.
    \(\text{No, it is not a solution.}\)
    @@ -1072,46 +1086,46 @@
    Identify Solution from Graph.
    The graph of two lines represents a system of two linear equations. Use the graph to identify the solution to the system.
    9.
    -
    +
    -
    a coordinate plot of two lines that cross at (-4, -3)
    -
    Answer.
    \(\left(-4,-3\right)\)
    +
    a coordinate plot of two lines that cross at (-2, 5)
    +
    Answer.
    \(\left(-2,5\right)\)
    10.
    -
    -
    +
    +
    -
    a coordinate plot of two lines that cross at (-3, 4)
    -
    Answer.
    \(\left(-3,4\right)\)
    +
    a coordinate plot of two lines that cross at (-1, 2)
    +
    Answer.
    \(\left(-1,2\right)\)
    11.
    -
    +
    -
    a coordinate plot of two lines that cross at (-2, 3)
    -
    Answer.
    \(\left(-2,3\right)\)
    +
    a coordinate plot of two lines that cross at (1, -3)
    +
    Answer.
    \(\left(1,-3\right)\)
    12.
    -
    -
    +
    +
    -
    a coordinate plot of two lines that cross at (-1, -4)
    -
    Answer.
    \(\left(-1,-4\right)\)
    +
    a coordinate plot of two lines that cross at (2, 5)
    +
    Answer.
    \(\left(2,5\right)\)
    13.
    -
    +
    a coordinate plot of two parallel lines
    Answer.
    \(\text{no solutions}\)
    14.
    -
    -
    +
    +
    -
    a coordinate plot of two parallel lines
    -
    Answer.
    \(\text{no solutions}\)
    +
    a coordinate plot of two parallel lines
    +
    Answer.
    \(\text{no solutions}\)
    @@ -1121,13 +1135,13 @@
    See How Many Solutions.
    Simply by looking closely at the linear equations, determine how many solutions the system will have. And state whether the system is dependeent, inconsistent, or neither.
    15.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{7}{9}}\mathopen{}\left(x-2\right)+6}\\ -y \amp= {{\frac{1}{5}}x+3} +y \amp= {{\frac{1}{2}}\mathopen{}\left(x+1\right)-5}\\ +y \amp= {{\frac{8}{9}}x-6} \end{alignedat} \right.\)
    @@ -1135,27 +1149,27 @@
    See How Many Solutions.
    16.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{9}{2}}\mathopen{}\left(x+9\right)+1}\\ -y \amp= {{\frac{7}{5}}x+4} +y \amp= {{\frac{1}{7}}\mathopen{}\left(x-7\right)+9}\\ +y \amp= {{\frac{5}{6}}x-7} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\text{One}\)
    Answer 2.
    \(\text{None of the above}\)
    +
    Answer 1.
    \(\text{One}\)
    Answer 2.
    \(\text{None of the above}\)
    17.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-4x-9}\\ -y \amp= {-4\mathopen{}\left(x+2\right)-1} +y \amp= {-{\frac{2}{7}}x-5}\\ +y \amp= {-{\frac{2}{7}}\mathopen{}\left(x+7\right)-3} \end{alignedat} \right.\)
    @@ -1163,27 +1177,27 @@
    See How Many Solutions.
    18.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-{\frac{14}{9}}x-7}\\ -y \amp= {-{\frac{14}{9}}\mathopen{}\left(x+9\right)+7} +y \amp= {x-4}\\ +y \amp= {x-7+3} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\text{Infinitely many}\)
    Answer 2.
    \(\text{Dependent}\)
    +
    Answer 1.
    \(\text{Infinitely many}\)
    Answer 2.
    \(\text{Dependent}\)
    19.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{1}{4}}x-5}\\ -y \amp= {{\frac{1}{4}}x+5} +y \amp= {-6x-2}\\ +y \amp= {-6x-6} \end{alignedat} \right.\)
    @@ -1191,17 +1205,17 @@
    See How Many Solutions.
    20.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-{\frac{7}{4}}x-3}\\ -y \amp= {-{\frac{7}{4}}x-1} +y \amp= {-{\frac{1}{7}}x+1}\\ +y \amp= {-{\frac{1}{7}}x+8} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\text{Zero}\)
    Answer 2.
    \(\text{Inconsistent}\)
    +
    Answer 1.
    \(\text{Zero}\)
    Answer 2.
    \(\text{Inconsistent}\)
    @@ -1212,337 +1226,337 @@
    Solve a System.
    Solve the given system of linear equations graphically.
    21.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-16x-8}\\ -y \amp= {-14x-6} +y \amp= {-x+8}\\ +y \amp= {{\frac{4}{3}}x+1} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-8\right), \left(1,-24\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,-6\right), \left(1,-20\right)\right\}\)
    Answer 2.
    \(\left(-1,8\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-1, 8)
    So the only solution is \((-1,8)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,8\right), \left(1,7\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,1\right), \left(3,5\right)\right\}\)
    Answer 2.
    \(\left(3,5\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (3, 5)
    So the only solution is \((3,5)\text{.}\)
    22.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{3}{2}}x}\\ -y \amp= {-2x+7} +y \amp= {{\frac{1}{5}}x-3}\\ +y \amp= {{\frac{2}{5}}x-4} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(2,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,7\right), \left(1,5\right)\right\}\)
    Answer 2.
    \(\left(2,3\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (2, 3)
    So the only solution is \((2,3)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-3\right), \left(5,-2\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,-4\right), \left(5,-2\right)\right\}\)
    Answer 2.
    \(\left(5,-2\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (5, -2)
    So the only solution is \((5,-2)\text{.}\)
    23.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{6}{5}}\mathopen{}\left(x-8\right)+2}\\ -y \amp= {-4x+8} +y \amp= {{\frac{1}{3}}\mathopen{}\left(x-4\right)-8}\\ +y \amp= {-{\frac{13}{16}}\mathopen{}\left(x+9\right)+6} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(8,2\right), \left(13,8\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,8\right), \left(1,4\right)\right\}\)
    Answer 2.
    \(\left(3,-4\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (3, -4)
    So the only solution is \((3,-4)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(4,-8\right), \left(7,-7\right)\right\}, \left\{\text{line}, \text{solid}, \left(-9,6\right), \left(7,-7\right)\right\}\)
    Answer 2.
    \(\left(7,-7\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (7, -7)
    So the only solution is \((7,-7)\text{.}\)
    24.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{7}{8}}\mathopen{}\left(x+3\right)+1}\\ -y \amp= {{\frac{11}{9}}\mathopen{}\left(x+4\right)-3} +y \amp= {{\frac{11}{14}}\mathopen{}\left(x+6\right)-8}\\ +y \amp= {2\mathopen{}\left(x-5\right)-3} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-3,1\right), \left(5,8\right)\right\}, \left\{\text{line}, \text{solid}, \left(-4,-3\right), \left(5,8\right)\right\}\)
    Answer 2.
    \(\left(5,8\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (5, 8)
    So the only solution is \((5,8)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-6,-8\right), \left(8,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(5,-3\right), \left(6,-1\right)\right\}\)
    Answer 2.
    \(\left(8,3\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (8, 3)
    So the only solution is \((8,3)\text{.}\)
    25.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{1}{3}}\mathopen{}\left(x-4\right)}\\ -y \amp= {{\frac{1}{3}}\mathopen{}\left(x+9\right)+8} +y \amp= {-{\frac{8}{9}}\mathopen{}\left(x-2\right)-9}\\ +y \amp= {-{\frac{8}{9}}x+6} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(4,0\right), \left(7,1\right)\right\}, \left\{\text{line}, \text{solid}, \left(-9,8\right), \left(-6,9\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    -
    If we graph these lines, we find they are parallel and do not cross.
    a coordinate plot of two lines that cross at (7, 1)
    So there are no solutions.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(2,-9\right), \left(11,-17\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,6\right), \left(9,-2\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    If we graph these lines, we find they are parallel and do not cross.
    a coordinate plot of two lines that cross at (-7, -1)
    So there are no solutions.
    26.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{4}{3}}\mathopen{}\left(x+5\right)+1}\\ -y \amp= {{\frac{4}{3}}\mathopen{}\left(x-5\right)-2} +y \amp= {{\frac{1}{4}}\mathopen{}\left(x+9\right)-9}\\ +y \amp= {{\frac{1}{4}}\mathopen{}\left(x+4\right)-2} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-5,1\right), \left(-2,5\right)\right\}, \left\{\text{line}, \text{solid}, \left(5,-2\right), \left(8,2\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    -
    If we graph these lines, we find they are parallel and do not cross.
    a coordinate plot of two lines that cross at (-8, -3)
    So there are no solutions.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-9,-9\right), \left(-5,-8\right)\right\}, \left\{\text{line}, \text{solid}, \left(-4,-2\right), \left(0,-1\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    If we graph these lines, we find they are parallel and do not cross.
    a coordinate plot of two lines that cross at (-5, -8)
    So there are no solutions.
    27.
    -
    +
    \(\left\{ \begin{alignedat}{3} -\amp {5x-4y = -20}\\ -\amp {x-y = -3} +\amp {3x-2y = -6}\\ +\amp {-x+4y = -8} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-4,0\right), \left(0,5\right)\right\}, \left\{\text{line}, \text{solid}, \left(-3,0\right), \left(0,3\right)\right\}\)
    Answer 2.
    \(\left(-8,-5\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-8, -5)
    So the only solution is \((-8,-5)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-2,0\right), \left(0,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(8,0\right), \left(0,-2\right)\right\}\)
    Answer 2.
    \(\left(-4,-3\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-4, -3)
    So the only solution is \((-4,-3)\text{.}\)
    28.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp {2x-3y = -6}\\ -\amp {-2x+9y = -18} +\amp {-2x-y = 2}\\ +\amp {6x+y = 6} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-3,0\right), \left(0,2\right)\right\}, \left\{\text{line}, \text{solid}, \left(9,0\right), \left(0,-2\right)\right\}\)
    Answer 2.
    \(\left(-9,-4\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-9, -4)
    So the only solution is \((-9,-4)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-1,0\right), \left(0,-2\right)\right\}, \left\{\text{line}, \text{solid}, \left(1,0\right), \left(0,6\right)\right\}\)
    Answer 2.
    \(\left(2,-6\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (2, -6)
    So the only solution is \((2,-6)\text{.}\)
    29.
    -
    +
    \(\left\{ \begin{alignedat}{3} -\amp y={x+8}\\ -\amp {-2x-y = 4} +\amp y={{\frac{7}{2}}x-2}\\ +\amp {-5x+y = -5} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-2,0\right), \left(0,-4\right)\right\}, \left\{\text{line}, \text{solid}, -44, \left(0,8\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(1,0\right), \left(0,-5\right)\right\}, \left\{\text{line}, \text{solid}, 25, \left(0,-2\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    If we graph these lines, we find they are parallel and never cross.
    a coordinate plot of two parallel lines
    So the there are no solutions.
    30.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp y={6x+2}\\ -\amp {4x-y = -4} +\amp y={{\frac{7}{4}}x-8}\\ +\amp {x+2y = 2} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-1,0\right), \left(0,4\right)\right\}, \left\{\text{line}, \text{solid}, 18, \left(0,2\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    -
    If we graph these lines, we find they are parallel and never cross.
    a coordinate plot of two parallel lines
    So the there are no solutions.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(2,0\right), \left(0,1\right)\right\}, \left\{\text{line}, \text{solid}, 3, \left(0,-8\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    If we graph these lines, we find they are parallel and never cross.
    a coordinate plot of two parallel lines
    So the there are no solutions.
    31.
    -
    +
    \(\left\{ \begin{alignedat}{3} -\amp {-x+y = -1}\\ -\amp y={x-3} +\amp {-x+y = -3}\\ +\amp y={x-5} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(1,0\right), \left(0,-1\right)\right\}, \left\{\text{line}, \text{solid}, \left(1,-2\right), \left(0,-3\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(3,0\right), \left(0,-3\right)\right\}, \left\{\text{line}, \text{solid}, \left(3,-2\right), \left(0,-5\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (, )
    So the only solution is \(no solutions\text{.}\)
    32.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp {-2x+y = -4}\\ -\amp y={2x+3} +\amp {5x+4y = 20}\\ +\amp y={-{\frac{5}{4}}x-1} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(2,0\right), \left(0,-4\right)\right\}, \left\{\text{line}, \text{solid}, \left(2,7\right), \left(0,3\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (, )
    So the only solution is \(no solutions\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(4,0\right), \left(0,5\right)\right\}, \left\{\text{line}, \text{solid}, \left(4,-6\right), \left(0,-1\right)\right\}\)
    Answer 2.
    \(\text{no solutions}\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (, )
    So the only solution is \(no solutions\text{.}\)
    33.
    -
    +
    \(\left\{ \begin{alignedat}{3} -\amp y={-{\frac{10}{9}}\mathopen{}\left(x-6\right)-2}\\ -\amp {4x+3y = 12} +\amp y={7\mathopen{}\left(x+4\right)+9}\\ +\amp {x+5y = 5} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(3,0\right), \left(0,4\right)\right\}, \left\{\text{line}, \text{solid}, \left(6,-2\right), \left(-3,8\right)\right\}\)
    Answer 2.
    \(\left(-3,8\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-3, 8)
    So the only solution is \((-3,8)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(5,0\right), \left(0,1\right)\right\}, \left\{\text{line}, \text{solid}, \left(-4,9\right), \left(-5,2\right)\right\}\)
    Answer 2.
    \(\left(-5,2\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-5, 2)
    So the only solution is \((-5,2)\text{.}\)
    34.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp y={{\frac{1}{5}}\mathopen{}\left(x-1\right)-1}\\ -\amp {-x+4y = -4} +\amp y={{\frac{9}{11}}\mathopen{}\left(x-9\right)+8}\\ +\amp {-x-2y = 4} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(4,0\right), \left(0,-1\right)\right\}, \left\{\text{line}, \text{solid}, \left(1,-1\right), \left(-4,-2\right)\right\}\)
    Answer 2.
    \(\left(-4,-2\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-4, (-2))
    So the only solution is \((-4,-2)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-4,0\right), \left(0,-2\right)\right\}, \left\{\text{line}, \text{solid}, \left(9,8\right), \left(-2,-1\right)\right\}\)
    Answer 2.
    \(\left(-2,-1\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-2, (-1))
    So the only solution is \((-2,-1)\text{.}\)
    35.
    -
    +
    \(\left\{ \begin{alignedat}{3} -\amp {-3x-5y = 15}\\ -\amp y={-{\frac{3}{5}}\mathopen{}\left(x-5\right)-6} +\amp {5x-3y = -15}\\ +\amp y={{\frac{5}{3}}\mathopen{}\left(x+6\right)-5} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-5,0\right), \left(0,-3\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-3,0\right), \left(0,5\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (, )
    So the only solution is \(infinitely many solutions\text{.}\)
    36.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -\amp {5x-2y = -10}\\ -\amp y={{\frac{5}{2}}\mathopen{}\left(x+4\right)-5} +\amp {x-4y = -4}\\ +\amp y={{\frac{1}{4}}\mathopen{}\left(x-8\right)+3} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-2,0\right), \left(0,5\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (, )
    So the only solution is \(infinitely many solutions\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(-4,0\right), \left(0,1\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (, )
    So the only solution is \(infinitely many solutions\text{.}\)
    37.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{6}{5}}x+5}\\ -y \amp= {-{\frac{1}{9}}\mathopen{}\left(x-4\right)-2} +y \amp= {{\frac{7}{2}}x+2}\\ +y \amp= {-{\frac{13}{4}}\mathopen{}\left(x+6\right)+8} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,5\right), \left(5,11\right)\right\}, \left\{\text{line}, \text{solid}, \left(4,-2\right), \left(13,-3\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-5, -1)
    So the only solution is \(infinitely many solutions\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,2\right), \left(2,9\right)\right\}, \left\{\text{line}, \text{solid}, \left(-6,8\right), \left(-2,-5\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-2, -5)
    So the only solution is \(infinitely many solutions\text{.}\)
    38.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {{\frac{2}{3}}x-5}\\ -y \amp= {{\frac{5}{2}}\mathopen{}\left(x+1\right)-2} +y \amp= {16x-9}\\ +y \amp= {-{\frac{1}{7}}\mathopen{}\left(x-8\right)+6} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-5\right), \left(3,-3\right)\right\}, \left\{\text{line}, \text{solid}, \left(-1,-2\right), \left(1,3\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-3, -7)
    So the only solution is \(infinitely many solutions\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-9\right), \left(1,7\right)\right\}, \left\{\text{line}, \text{solid}, \left(8,6\right), \left(15,5\right)\right\}\)
    Answer 2.
    \(\text{infinitely many solutions}\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (1, 7)
    So the only solution is \(infinitely many solutions\text{.}\)
    39.
    -
    +
    \(\left\{ \begin{alignedat}{3} -y \amp= {-4x+1}\\ -y \amp= {-4\mathopen{}\left(x+1\right)+5} +y \amp= {{\frac{1}{3}}x-2}\\ +y \amp= {{\frac{1}{3}}\mathopen{}\left(x-3\right)-1} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,1\right), \left(1,-3\right)\right\}\)
    Answer 2.
    \(\left(-1,5\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (-1, 5)
    So the only solution is \((-1,5)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-2\right), \left(3,-1\right)\right\}\)
    Answer 2.
    \(\left(3,-1\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (3, -1)
    So the only solution is \((3,-1)\text{.}\)
    40.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{3} -y \amp= {6x-9}\\ -y \amp= {6\mathopen{}\left(x-1\right)-3} +y \amp= {-{\frac{13}{5}}x+7}\\ +y \amp= {-{\frac{13}{5}}\mathopen{}\left(x-5\right)-6} \end{alignedat} \right.\)
    -
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,-9\right), \left(1,-3\right)\right\}\)
    Answer 2.
    \(\left(1,-3\right)\)
    Explanation.
    -
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (1, -3)
    So the only solution is \((1,-3)\text{.}\) +
    Answer 1.
    \(\left\{\text{line}, \text{solid}, \left(0,7\right), \left(5,-6\right)\right\}\)
    Answer 2.
    \(\left(5,-6\right)\)
    Explanation.
    +
    If we graph these lines, we find they cross.
    a coordinate plot of two lines that cross at (5, -6)
    So the only solution is \((5,-6)\text{.}\)
    @@ -1551,94 +1565,94 @@
    Solve a System.

    Applications

    41.
    -
    +
    -
    Ezekiel is organizing an office lunch party. His budget for beverages is \({\$141.75}\) and he will try to spend all of it. He assume that \(36.5\) people will attend, and there should be two beverages avaiable per person. Ezekiel has decided to order cans of a fancy beverage that cost \({\$2.25}\) each, and cans of a cheaper beverage that cost \({\$1.75}\) each. How many of each should he order?
    (a)
    +
    Rhys is organizing an office lunch party. His budget for beverages is \({\$42}\) and he will try to spend all of it. He assume that \(8.5\) people will attend, and there should be two beverages avaiable per person. Rhys has decided to order cans of a fancy beverage that cost \({\$3}\) each, and cans of a cheaper beverage that cost \({\$2}\) each. How many of each should he order?
    (a)
    Write two equations that form a system for this scenario.
    -
    Answer 1.
    \(2.25x+1.75y = 141.75\)
    Answer 2.
    \(x+y = 73\)
    Explanation.
    -
    Since there are \(36.5\) people and there should be two drinks per person, one equation is \({2.25x+1.75y = 141.75}\text{.}\) -
    Since the first type costs \({\$2.25}\) each, the second type costs \({\$1.75}\) each, and the total should be \({\$141.75}\text{,}\) another equation is \({x+y = 73}\text{.}\) +
    Answer 1.
    \(3x+2y = 42\)
    Answer 2.
    \(x+y = 17\)
    Explanation.
    +
    Since there are \(8.5\) people and there should be two drinks per person, one equation is \({3x+2y = 42}\text{.}\) +
    Since the first type costs \({\$3}\) each, the second type costs \({\$2}\) each, and the total should be \({\$42}\text{,}\) another equation is \({x+y = 17}\text{.}\)
    (b)
    Plot the lines for the system of two equations.
    -
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,81\right), \left(63,0\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,73\right), \left(73,0\right)\right\}\)
    Explanation.
    -
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,141.75/1.75) and (141.75/2.25,0); the other line passes through (0,2*36.5) and (2*36.5,0); the lines cross at (28,45)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,21\right), \left(14,0\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,17\right), \left(17,0\right)\right\}\)
    Explanation.
    +
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,42/2) and (42/3,0); the other line passes through (0,2*8.5) and (2*8.5,0); the lines cross at (8,9)
    (c)
    -
    Based on the graph, how many of the more expensive beverage should Ezekiel buy?
    How many of the less expensive beverage should Ezekiel buy?
    +
    Based on the graph, how many of the more expensive beverage should Rhys buy?
    How many of the less expensive beverage should Rhys buy?
    -
    Answer 1.
    \(28\)
    Answer 2.
    \(45\)
    Explanation.
    The lines cross at \((28,45)\text{.}\) So Ezekiel should buy 28 of the more expensive drink, and 45 of the less expensive drink.
    +
    Answer 1.
    \(8\)
    Answer 2.
    \(9\)
    Explanation.
    The lines cross at \((8,9)\text{.}\) So Rhys should buy 8 of the more expensive drink, and 9 of the less expensive drink.
    42.
    -
    +
    -
    At the local hardware store, Rhys bought a hammer and four boxes of nails. The total cost was \({\$36}\text{.}\) The hammer costs \({\$16}\) more than a box of nails. How much does a hammer cost and how much does one box of nails cost?
    (a)
    +
    At the local hardware store, Marshall bought a hammer and four boxes of nails. The total cost was \({\$52}\text{.}\) The hammer costs \({\$37}\) more than a box of nails. How much does a hammer cost and how much does one box of nails cost?
    (a)
    Write two equations that form a system for this scenario.
    -
    Answer 1.
    \(x+4y = 36\)
    Answer 2.
    \(x = y+16\)
    Explanation.
    -
    Since the total cost of the hammer and four boxes of nails is \({\$36}\text{,}\) one equation is \({x+4y = 36}\text{.}\) -
    Since the hammer costs \({\$16}\) more than a box of nails, another equation is \({x = y+16}\text{.}\) +
    Answer 1.
    \(x+4y = 52\)
    Answer 2.
    \(x = y+37\)
    Explanation.
    +
    Since the total cost of the hammer and four boxes of nails is \({\$52}\text{,}\) one equation is \({x+4y = 52}\text{.}\) +
    Since the hammer costs \({\$37}\) more than a box of nails, another equation is \({x = y+37}\text{.}\)
    (b)
    Plot the lines for the system of two equations.
    -
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,9\right), \left(36,0\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,-16\right), \left(16,0\right)\right\}\)
    Explanation.
    -
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,36/4) and (36,0); the other line passes through (0,-16) and (16,0); the lines cross at (20,4)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,13\right), \left(52,0\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,-37\right), \left(37,0\right)\right\}\)
    Explanation.
    +
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,52/4) and (52,0); the other line passes through (0,-37) and (37,0); the lines cross at (40,3)
    (c)
    Based on the graph, how much does the hammer cost?
    How much does a box of nails cost?
    -
    Answer 1.
    \(\$20\)
    Answer 2.
    \(\$4\)
    Explanation.
    The lines cross at \((20,4)\text{.}\) So a hammer costs \({\$20}\text{,}\) and a box of nails costs $4.
    +
    Answer 1.
    \(\$40\)
    Answer 2.
    \(\$3\)
    Explanation.
    The lines cross at \((40,3)\text{.}\) So a hammer costs \({\$40}\text{,}\) and a box of nails costs $3.
    43.
    -
    +
    -
    Donavan enters a grassy field from some dense trees. His dog is standing out in the field, \({75\ {\rm ft}}\) away. As soon as they see each other, they start running toward each other. Donavan runs with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) and his dog runs with speed \({12\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) How long will it be until they meet?
    (a)
    +
    Aryana enters a grassy field from some dense trees. Her dog is standing out in the field, \({55\ {\rm ft}}\) away. As soon as they see each other, they start running toward each other. Aryana runs with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) and her dog runs with speed \({8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) How long will it be until they meet?
    (a)
    Write two equations that form a system for this scenario.
    -
    Answer 1.
    \(d = 3t\)
    Answer 2.
    \(d = 75+\left(-12\right)t\)
    Explanation.
    -
    Donavan starts out at position \(d=0\text{,}\) running with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) So one equation is \({d = 3t}\text{.}\) -
    The dog starts running with speed \({12\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) but getting closer to Donavan. So we’ll use a negative rate of change. And the dog starts out \({75\ {\rm ft}}\) away. So the other equation is \({d = 75+\left(-12\right)t}\text{.}\) +
    Answer 1.
    \(d = 3t\)
    Answer 2.
    \(d = 55+\left(-8\right)t\)
    Explanation.
    +
    Aryana starts out at position \(d=0\text{,}\) running with speed \({3\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) So one equation is \({d = 3t}\text{.}\) +
    The dog starts running with speed \({8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\) but getting closer to Aryana. So we’ll use a negative rate of change. And the dog starts out \({55\ {\rm ft}}\) away. So the other equation is \({d = 55+\left(-8\right)t}\text{.}\)
    (b)
    Plot the lines for the system of two equations.
    -
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,75\right), \left(1,63\right)\right\}\)
    Explanation.
    -
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,75) with slope 3; the other line passes through (0,0) with slope -12; the lines cross at (5,15)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,3\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,55\right), \left(1,47\right)\right\}\)
    Explanation.
    +
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,55) with slope 3; the other line passes through (0,0) with slope -8; the lines cross at (5,15)
    (c)
    -
    Based on the graph, how long will it be until they meet?
    How far will that be from the place where Donavan started?
    +
    Based on the graph, how long will it be until they meet?
    How far will that be from the place where Aryana started?
    -
    Answer 1.
    \(5\ {\rm s}\)
    Answer 2.
    \(15\ {\rm m}\)
    Explanation.
    The lines cross at \((5,15)\text{.}\) So after \(5\) seconds, they will meet 15 feet away from where Donavan started.
    +
    Answer 1.
    \(5\ {\rm s}\)
    Answer 2.
    \(15\ {\rm m}\)
    Explanation.
    The lines cross at \((5,15)\text{.}\) So after \(5\) seconds, they will meet 15 feet away from where Aryana started.
    44.
    -
    +
    -
    A cyclist riding at \({25\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\) rides past a dog. A moment later, when the bicycle is \({50\ {\rm ft}}\) away, the dog begins to chase the bicycle at a speed of \({30\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\text{.}\) Assuming the cyclist does not change course or speed, how long will it be until the dog catches up to the bicycle?
    (a)
    +
    A cyclist riding at \({26\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\) rides past a dog. A moment later, when the bicycle is \({40\ {\rm ft}}\) away, the dog begins to chase the bicycle at a speed of \({30\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\text{.}\) Assuming the cyclist does not change course or speed, how long will it be until the dog catches up to the bicycle?
    (a)
    Write two equations that form a system for this scenario.
    -
    Answer 1.
    \(d = 25t+50\)
    Answer 2.
    \(d = 30t\)
    Explanation.
    -
    The bicycle starts out \({50\ {\rm ft}}\) away and is traveling at \({25\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\text{.}\) So one equation is \({d = 25t+50}\text{.}\) +
    Answer 1.
    \(d = 26t+40\)
    Answer 2.
    \(d = 30t\)
    Explanation.
    +
    The bicycle starts out \({40\ {\rm ft}}\) away and is traveling at \({26\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\text{.}\) So one equation is \({d = 26t+40}\text{.}\)
    The dog starts running at time \(t=0\) with speed \({30\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}}\text{.}\) So the other equation is \({d = 30t}\text{.}\)
    (b)
    Plot the lines for the system of two equations.
    -
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,50\right), \left(1,75\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,30\right)\right\}\)
    Explanation.
    -
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,50) with slope 25; the other line passes through (0,0) with slope 30; the lines cross at (10,300)
    +
    Answer.
    \(\left\{\text{line}, \text{solid}, \left(0,40\right), \left(1,66\right)\right\}, \left\{\text{line}, \text{solid}, \left(0,0\right), \left(1,30\right)\right\}\)
    Explanation.
    +
    Plotting the two slope-intercept equations:
    a Cartesian grid with two intersecting lines; one line passes through (0,40) with slope 26; the other line passes through (0,0) with slope 30; the lines cross at (10,300)
    (c)
    Based on the graph, how long will it be until the dog catches up with the bicycle?
    How far will that be from the place where the dog started running?
    diff --git a/section-solving-multistep-linear-equations.html b/section-solving-multistep-linear-equations.html index 0dda3e215..8d5b4c341 100644 --- a/section-solving-multistep-linear-equations.html +++ b/section-solving-multistep-linear-equations.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-solving-multistep-linear-inequalities.html b/section-solving-multistep-linear-inequalities.html index 9448568f8..24c1d0dbf 100644 --- a/section-solving-multistep-linear-inequalities.html +++ b/section-solving-multistep-linear-inequalities.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-solving-one-step-equations.html b/section-solving-one-step-equations.html index 4a66b32a6..f5c48c102 100644 --- a/section-solving-one-step-equations.html +++ b/section-solving-one-step-equations.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-solving-one-step-inequalities.html b/section-solving-one-step-inequalities.html index 23d832799..d6e9fb744 100644 --- a/section-solving-one-step-inequalities.html +++ b/section-solving-one-step-inequalities.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-special-solution-sets.html b/section-special-solution-sets.html index 9e1eb4516..636ffb198 100644 --- a/section-special-solution-sets.html +++ b/section-special-solution-sets.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • diff --git a/section-standard-form.html b/section-standard-form.html index aebab88bc..e02f74fde 100644 --- a/section-standard-form.html +++ b/section-standard-form.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -588,13 +602,14 @@

    Search Results:

    • -
    We’ve seen that a linear relationship can be expressed with an equation in slope-intercept form or with an equation in point-slope form. There is a third form that you can use to write line equations. It’s known as standard form.
    +
    We’ve seen that a linear relationship can be expressed with an equation in slope-intercept form or with an equation in point-slope form. There is a third form that you can use to write line equations. It’s known as “standard form”.
    Figure 3.7.1. Alternative Video Lesson

    Subsection 3.7.1 Standard Form Definition

    -
    Imagine trying to gather donations to pay for a \(\$10{,}000\) medical procedure you cannot afford. Oversimplifying the mathematics a bit, suppose that there were only two types of donors in the world: those who will donate \(\$20\) and those who will donate \(\$100\text{.}\) How many of each, or what combination, do you need to reach the funding goal? As in, if \(x\) people donate \(\$20\) and \(y\) people donate \(\$100\text{,}\) what numbers could \(x\) and \(y\) be? The donors of the first type have collectively donated \(20x\) dollars, and the donors of the second type have collectively donated \(100y\text{.}\) Reflect on the meaning of \(20x\) and \(100y\text{.}\) Make sure you understand their meaning before reading on.
    +
    Imagine trying to gather donations to pay for a \(\$10{,}000\) medical procedure you cannot afford. Oversimplifying the mathematics a bit, suppose that there were only two types of donors in the world: those who will donate \(\$20\) and those who will donate \(\$100\text{.}\) How many of each, or what combination, do you need to reach the funding goal? That is, if \(x\) people donate \(\$20\) and \(y\) people donate \(\$100\text{,}\) what numbers could \(x\) and \(y\) be to meet your goal? The donors of the first type have collectively donated \(20x\) dollars, and the donors of the second type have collectively donated \(100y\text{.}\) +
    -
    So altogether you’d need
    +
    To reach \(\$10{,}000\text{,}\) altogether you’d need to have
    \begin{equation*} 20x+100y=10000 @@ -604,61 +619,20 @@

    Search Results:

    Definition 3.7.2. Standard Form. -

    +
    It is always possible to write an equation for a line in the form
    \begin{equation} Ax+By=C\tag{3.7.1} \end{equation}
    -
    where \(A\text{,}\) \(B\text{,}\) and \(C\) are three numbers (each of which might be \(0\text{,}\) although at least one of \(A\) and \(B\) must be nonzero). This form of a line equation is called standard form. In the context of an application, the meaning of \(A\text{,}\) \(B\text{,}\) and \(C\) depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations).
    +
    where \(A\text{,}\) \(B\text{,}\) and \(C\) are three numbers (each of which might be \(0\text{,}\) although at least one of \(A\) and \(B\) must be nonzero). This form of a line equation is called standard form. In the context of an application, the meaning of \(A\text{,}\) \(B\text{,}\) and \(C\) depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines (which cannot be written using slope-intercept or point-slope form equations).

    Checkpoint 3.7.3.

    -
    -
    -
    -
    For each of the following equations, identify what form they are in.
    - - - - - - - - - - - - - - - - - - - - - - - - -
    \(2.7x+3.4y=-82\)
    \(y=\frac{2}{7}(x-3)+\frac{1}{10}\)
    \(12x-3=y+2\)
    \(y=x^2+5\)
    \(x-y=10\)
    \(y=4x+1\)
    -
    Explanation.
    -
    -\(2.7x+3.4y=-82\) is in standard form, with \(A=2.7\text{,}\) \(B=3.4\text{,}\) and \(C=-82\text{.}\) -
    -\(y=\frac{2}{7}(x-3)+\frac{1}{10}\) is in point-slope form, with slope \(\frac{2}{7}\text{,}\) and passing through \(\left(3,\frac{1}{10}\right)\text{.}\) -
    -\(12x-3=y+2\) is linear, but not in any of the forms we have studied. Using algebra, you can rearrange it to read \(y=12x-5\text{.}\) -
    -\(y=x^2+5\) is not linear. The exponent on \(x\) is a dead giveaway.
    -\(x-y=10\) is in standard form, with \(A=1\text{,}\) \(B=-1\text{,}\) and \(C=10\text{.}\) -
    -\(y=4x+1\) is in slope-intercept form, with slope \(4\) and \(y\)-intercept at \((0,1)\text{.}\) -
    -
    -
    +
    +
    +
    PTX:ERROR: WeBWorK problem webwork-1289 with seed 1289 is either empty or failed to compile Use -a to halt with full PG and returned content
    Returning to the example with donations for the medical procedure, let’s examine the equation
    @@ -838,25 +812,25 @@

    Search Results:

    +
    If a line’s \(x\)-intercept is at \((r,0)\) and its \(y\)-intercept is at \((0,b)\text{,}\) then the slope of the line is \(-\frac{b}{r}\text{.}\) (Unless the line passes through the origin, in which case both \(r\) and \(b\) equal \(0\text{,}\) and then this fraction is undefined. And the slope of the line could be anything.)

    Checkpoint 3.7.11.

    -
    -
    +
    +
    -
    +
    Consider the line with equation \(2x+4.3y=\frac{1000}{99}\text{.}\)
      -
    1. What is its \(x\)-intercept?
      2. -
    2. What is its \(y\)-intercept?
      3. -
    3. What is its slope?
      4. +
    4. What is its \(x\)-intercept?
      5. +
    5. What is its \(y\)-intercept?
      6. +
    6. What is its slope?
    -
    Explanation.
      -
    1. -
      To find the \(x\)-intercept:
      -
      +
      Explanation.
        +
      1. +
        To find the \(x\)-intercept:
        +
        \begin{equation*} \begin{aligned} 2x+4.3y\amp=\frac{1000}{99}\\ @@ -866,12 +840,12 @@

        Search Results:

        \end{aligned} \end{equation*}
        -
        So the \(x\)-intercept is at \(\left(\frac{500}{99},0\right)\text{.}\) +
        So the \(x\)-intercept is at \(\left(\frac{500}{99},0\right)\text{.}\)
        2. -
      2. -
        To find the \(y\)-intercept:
        -
        +
      3. +
        To find the \(y\)-intercept:
        +
        \begin{equation*} \begin{aligned} 2x+4.3y\amp=\frac{1000}{99}\\ @@ -882,12 +856,12 @@

        Search Results:

        \end{aligned} \end{equation*}
        -
        So the \(y\)-intercept is at about \((0,2.349)\text{.}\) +
        So the \(y\)-intercept is at about \((0,2.349)\text{.}\)
        4. -
      4. -
        Since we have the \(x\)- and \(y\)-intercepts, we can calculate the slope:
        -
        +
      5. +
        Since we have the \(x\)- and \(y\)-intercepts, we can calculate the slope:
        +
        \begin{equation*} m\approx-\frac{2.349}{\frac{500}{99}}=-\frac{2.349\cdot99}{500}\approx-0.4561 \text{.} @@ -998,6 +972,25 @@

        Search Results:

        Review and Warmup

        1.
        +
        +
        +
        +
        \({-45x-5y}={10}\)
        +
        +
        Answer.
        \(y = -9x-2\)
        Explanation.
        +\begin{equation*} +\begin{aligned} +{-45x-5y} \amp = {10}\\ +{-45x-5y+45x} \amp = {10+45x}\\ +{-5y} \amp = {10+45x}\\ +{\frac{-5y}{-5}} \amp = {\frac{10+45x}{-5}}\\ +{y} \amp = {-9x-2} +\end{aligned} +\end{equation*} +
        +
        +
        +
        2.
        @@ -1016,26 +1009,26 @@

        Search Results:

      -
    2.
    -
    +
    3.
    +
    -
    \({12x+2y}={30}\)
    +
    \({8x+3y}={-54}\)
    -
    Answer.
    \(y = -6x+15\)
    Explanation.
    +
    Answer.
    \(y = -{\frac{8}{3}}x-18\)
    Explanation.
    \begin{equation*} \begin{aligned} -{12x+2y} \amp = {30}\\ -{12x+2y-12x} \amp = {30-12x}\\ -{2y} \amp = {30-12x}\\ -{\frac{2y}{2}} \amp = {\frac{30-12x}{2}}\\ -{y} \amp = {-6x+15} +{8x+3y} \amp = {-54}\\ +{8x+3y-8x} \amp = {-54-8x}\\ +{3y} \amp = {-54-8x}\\ +{\frac{3y}{3}} \amp = {\frac{-54-8x}{3}}\\ +{y} \amp = {-{\frac{8}{3}}x-18} \end{aligned} \end{equation*}
    -
    3.
    +
    4.
    @@ -1054,26 +1047,26 @@

    Search Results:

    -
    4.
    -
    +
    5.
    +
    -
    \({-7x+5y}={-95}\)
    +
    \({-65x+47y}={-481}\)
    -
    Answer.
    \(y = {\frac{7}{5}}x-19\)
    Explanation.
    +
    Answer.
    \(y = {\frac{65}{47}}x+-{\frac{481}{47}}\)
    Explanation.
    \begin{equation*} \begin{aligned} -{-7x+5y} \amp = {-95}\\ -{-7x+5y+7x} \amp = {-95+7x}\\ -{5y} \amp = {-95+7x}\\ -{\frac{5y}{5}} \amp = {\frac{-95+7x}{5}}\\ -{y} \amp = {{\frac{7}{5}}x-19} +{-65x+47y} \amp = {-481}\\ +{-65x+47y+65x} \amp = {-481+65x}\\ +{47y} \amp = {-481+65x}\\ +{\frac{47y}{47}} \amp = {\frac{-481+65x}{47}}\\ +{y} \amp = {{\frac{65}{47}}x - {\frac{481}{47}}} \end{aligned} \end{equation*}
    -
    5.
    +
    6.
    @@ -1092,189 +1085,191 @@

    Search Results:

    -
    6.
    -
    +

    Skills Practice

    +
    7.
    +
    -
    \({54x-63y}={493}\)
    +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -{7}x+y= -6 }\text{.}\) +
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer.
    \(y = {\frac{6}{7}}x+-{\frac{493}{63}}\)
    Explanation.
    -\begin{equation*} -\begin{aligned} -{54x-63y} \amp = {493}\\ -{54x-63y-54x} \amp = {493-54x}\\ -{-63y} \amp = {493-54x}\\ -{\frac{-63y}{-63}} \amp = {\frac{493-54x}{-63}}\\ -{y} \amp = {{\frac{6}{7}}x - {\frac{493}{63}}} +
    Answer 1.
    \(7\)
    Answer 2.
    \(\left(0,-6\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.
    In this problem, the line’s equation is given as \(\displaystyle{ -{7}x+y= -6 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\) +
    \(\displaystyle{\begin{aligned} +-{7}x+y \amp = -6 \\ +-{7}x+y\mathbf{{}+{7}x} \amp = -6\mathbf{{}+{7}x} \\ +y \amp = {7}x - 6 \end{aligned} -\end{equation*} -
    +}\)
    Now we can see the line’s slope is \({7}\text{,}\) and its \(y\)-intercept has coordinates \((0,-6)\text{.}\)
    +
    -

    Skills Practice

    -
    7.
    -
    +
    +
    8.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -{8}x+y= 7 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -x-y= 5 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(8\)
    Answer 2.
    \(\left(0,7\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.
    In this problem, the line’s equation is given as \(\displaystyle{ -{8}x+y= 7 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\) +
    Answer 1.
    \(-1\)
    Answer 2.
    \(\left(0,-5\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.
    In this problem, the line’s equation is given as \(\displaystyle{ -x-y= 5 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\)
    \(\displaystyle{\begin{aligned} --{8}x+y \amp = 7 \\ --{8}x+y\mathbf{{}+{8}x} \amp = 7\mathbf{{}+{8}x} \\ -y \amp = {8}x+7 +-x-y \amp = 5 \\ +-x-y\mathbf{{}+x} \amp = 5\mathbf{{}+x} \\ +-y \amp = x+5 \\ +(-1) \cdot (-y) \amp = (-1) \cdot (x+ 5) \\ +y \amp = (-1) \cdot x+(-1) \cdot 5 \\ +y \amp = -x-5 \\ \end{aligned} -}\)
    Now we can see the line’s slope is \({8}\text{,}\) and its \(y\)-intercept has coordinates \((0,7)\text{.}\) +}\)
    Now we can see the line’s slope is \({-1}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,-5\right)}\text{.}\)
    -
    8.
    -
    +
    9.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ -x-y= 7 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 15x+5y= -10 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(-1\)
    Answer 2.
    \(\left(0,-7\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.
    In this problem, the line’s equation is given as \(\displaystyle{ -x-y= 7 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\) -
    \(\displaystyle{\begin{aligned} --x-y \amp = 7 \\ --x-y\mathbf{{}+x} \amp = 7\mathbf{{}+x} \\ --y \amp = x+7 \\ -(-1) \cdot (-y) \amp = (-1) \cdot (x+ 7) \\ -y \amp = (-1) \cdot x+(-1) \cdot 7 \\ -y \amp = -x-7 \\ +
    Answer 1.
    \(-3\)
    Answer 2.
    \(\left(0,-2\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 15x+5y=-10 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +15x+5y \amp = -10 \\ +15x+5y\mathbf{{} -15x} \amp = -10\mathbf{{} -15x} \\ +5y \amp = -15x - 10 \\ +\frac{5y}{5} \amp = \frac{-15x}{5} + \frac{-10}{5} \\ +y \amp = {-3}x - 2 \end{aligned} -}\)
    Now we can see the line’s slope is \({-1}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,-7\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \({-3}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,-2\right)}\text{.}\)
    -
    9.
    -
    +
    10.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 5x+5y= 15 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 5x-5y=15 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(-1\)
    Answer 2.
    \(\left(0,3\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 5x+5y=15 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -5x+5y \amp = 15 \\ -5x+5y\mathbf{{} -5x} \amp = 15\mathbf{{} -5x} \\ -5y \amp = -5x+15 \\ -\frac{5y}{5} \amp = \frac{-5x}{5} + \frac{15}{5} \\ -y \amp = {-1}x + 3 +
    Answer 1.
    \(1\)
    Answer 2.
    \(\left(0,-3\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 5x-5y=15 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +5x-5y \amp = 15 \\ +5x-5y\mathbf{{}-5x} \amp = 15\mathbf{{}-5x} \\ +-5y \amp = -5x+15 \\ +\frac{-5y}{-5} \amp = \frac{-5x}{-5} + \frac{15}{-5} \\ +y \amp = {1}x - 3 \end{aligned} -}\)
    Now we can see the line’s slope is \({-1}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,3\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \({1}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,-3\right)}\text{.}\)
    -
    10.
    -
    +
    11.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 10x-2y=-8 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ x+2y=6 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(5\)
    Answer 2.
    \(\left(0,4\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 10x-2y=-8 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -10x-2y \amp = -8 \\ -10x-2y\mathbf{{}-10x} \amp = -8\mathbf{{}-10x} \\ --2y \amp = -10x - 8 \\ -\frac{-2y}{-2} \amp = \frac{-10x}{-2} + \frac{-8}{-2} \\ -y \amp = {5}x + 4 +
    Answer 1.
    \(-{\frac{1}{2}}\)
    Answer 2.
    \(\left(0,3\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ x+2y=6 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +x+2y \amp = 6 \\ +x+2y\mathbf{{}-x} \amp = 6\mathbf{{}-x} \\ +2y \amp = -x+6 \\ +\frac{2y}{2} \amp = \frac{-x}{2} + \frac{6}{2} \\ +y \amp = -\frac{1}{2}x + 3 \end{aligned} -}\)
    Now we can see the line’s slope is \({5}\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,4\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \(\displaystyle{ -\frac{1}{2} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,3\right)}\text{.}\)
    -
    11.
    -
    +
    12.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ x+2y=2 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 5x+3y=3 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(-{\frac{1}{2}}\)
    Answer 2.
    \(\left(0,1\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ x+2y=2 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -x+2y \amp = 2 \\ -x+2y\mathbf{{}-x} \amp = 2\mathbf{{}-x} \\ -2y \amp = -x+2 \\ -\frac{2y}{2} \amp = \frac{-x}{2} + \frac{2}{2} \\ -y \amp = -\frac{1}{2}x + 1 +
    Answer 1.
    \(-{\frac{5}{3}}\)
    Answer 2.
    \(\left(0,1\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 5x+3y=3 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +5x+3y \amp = 3 \\ +5x+3y\mathbf{{}-5 x} \amp = 3\mathbf{{}-5 x} \\ +3y \amp = -5x+3 \\ +\frac{3y}{3} \amp = \frac{-5x}{3} + \frac{3}{3} \\ +y \amp = -\frac{5}{3}x + 1 \end{aligned} -}\)
    Now we can see the line’s slope is \(\displaystyle{ -\frac{1}{2} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,1\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \(\displaystyle{ -\frac{5}{3} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,1\right)}\text{.}\)
    -
    12.
    -
    +
    13.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 2x+3y=15 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 2x-3y=15 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(-{\frac{2}{3}}\)
    Answer 2.
    \(\left(0,5\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 2x+3y=15 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -2x+3y \amp = 15 \\ -2x+3y\mathbf{{}-2 x} \amp = 15\mathbf{{}-2 x} \\ -3y \amp = -2x+15 \\ -\frac{3y}{3} \amp = \frac{-2x}{3} + \frac{15}{3} \\ -y \amp = -\frac{2}{3}x + 5 +
    Answer 1.
    \({\frac{2}{3}}\)
    Answer 2.
    \(\left(0,-5\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 2x-3y=15 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +2x-3y \amp = 15 \\ +2x-3y\mathbf{{}-2x} \amp = 15\mathbf{{}-2x} \\ +-3y \amp = -2x+15 \\ +\frac{-3y}{-3} \amp = \frac{-2x}{-3} + \frac{15}{-3} \\ +y \amp = \frac{2}{3}x - 5 \end{aligned} -}\)
    Now we can see the line’s slope is \(\displaystyle{ -\frac{2}{3} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,5\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \(\displaystyle{ \frac{2}{3} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,-5\right)}\text{.}\)
    -
    13.
    -
    +
    14.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 7x-4y=-4 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 9x+15y=30 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \({\frac{7}{4}}\)
    Answer 2.
    \(\left(0,1\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 7x-4y=-4 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -7x-4y \amp = -4 \\ -7x-4y\mathbf{{}-7x} \amp = -4\mathbf{{}-7x} \\ --4y \amp = -7x - 4 \\ -\frac{-4y}{-4} \amp = \frac{-7x}{-4} + \frac{-4}{-4} \\ -y \amp = \frac{7}{4}x + 1 +
    Answer 1.
    \(-{\frac{3}{5}}\)
    Answer 2.
    \(\left(0,2\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 9x+15y=30 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +9x+15y \amp = 30 \\ +9x+15y\mathbf{{}-9x} \amp = 30\mathbf{{}-9x} \\ +15y \amp = -9x+30 \\ +\frac{15y}{15} \amp = \frac{-9x}{15} + \frac{30}{15} \\ +y \amp = -\frac{9}{15}x + 2 \\ +y \amp = -\frac{3}{5}x + 2 \end{aligned} -}\)
    Now we can see the line’s slope is \(\displaystyle{ \frac{7}{4} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,1\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \(\displaystyle{ -\frac{3}{5} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,2\right)}\text{.}\)
    -
    14.
    -
    +
    15.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 12x+16y=48 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 12x-16y=0 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \(-{\frac{3}{4}}\)
    Answer 2.
    \(\left(0,3\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 12x+16y=48 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -12x+16y \amp = 48 \\ -12x+16y\mathbf{{}-12x} \amp = 48\mathbf{{}-12x} \\ -16y \amp = -12x+48 \\ -\frac{16y}{16} \amp = \frac{-12x}{16} + \frac{48}{16} \\ -y \amp = -\frac{12}{16}x + 3 \\ -y \amp = -\frac{3}{4}x + 3 +
    Answer 1.
    \({\frac{3}{4}}\)
    Answer 2.
    \(\left(0,0\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 12x-16y=0 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +12x-16y \amp = 0 \\ +12x-16y\mathbf{{}-12x} \amp = 0\mathbf{{}-12x} \\ +-16y \amp = -12x \\ +\frac{-16y}{-16} \amp = \frac{-12x}{-16} \\ +y \amp = \frac{12}{16}x \\ +y \amp = \frac{3}{4}x \end{aligned} -}\)
    Now we can see the line’s slope is \(\displaystyle{ -\frac{3}{4} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,3\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \(\displaystyle{ \frac{3}{4} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,0\right)}\text{.}\)
    -
    15.
    +
    16.
    @@ -1295,28 +1290,28 @@

    Search Results:

    -
    16.
    -
    +
    17.
    +
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 20x-8y=0 }\text{.}\) +
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 20x+8y=5 }\text{.}\)
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    Answer 1.
    \({\frac{5}{2}}\)
    Answer 2.
    \(\left(0,0\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 20x-8y=0 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -20x-8y \amp = 0 \\ -20x-8y\mathbf{{}-20x} \amp = 0\mathbf{{}-20x} \\ --8y \amp = -20x \\ -\frac{-8y}{-8} \amp = \frac{-20x}{-8} \\ -y \amp = \frac{20}{8}x \\ -y \amp = \frac{5}{2}x +
    Answer 1.
    \(-{\frac{5}{2}}\)
    Answer 2.
    \(\left(0,\frac{5}{8}\right)\)
    Explanation.
    +
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 20x+8y=5 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} +20x+8y \amp = 5 \\ +20x+8y\mathbf{{}-20x} \amp = 5\mathbf{{}-20x} \\ +8y \amp = -20x +5 \\ +\frac{8y}{8} \amp = \frac{-20x}{8} + \frac{5}{8}\\ +y \amp = -\frac{20}{8}x + \frac{5}{8}\\ +y \amp = -\frac{5}{2}x + \frac{5}{8} \end{aligned} -}\)
    Now we can see the line’s slope is \(\displaystyle{ \frac{5}{2} }\text{,}\) and its \(y\)-intercept has coordinates \({\left(0,0\right)}\text{.}\) +}\)
    Now we can see the line’s slope is \(-\frac{5}{2}\text{,}\) and its \(y\)-intercept has coordinates \(\left(0,\frac{5}{8}\right)\text{.}\)
    -
    17.
    +
    18.
    @@ -1337,112 +1332,91 @@

    Search Results:

    -
    18.
    -
    -
    -
    -
    Find the line’s slope and \(y\)-intercept.
    A line has equation \(\displaystyle{ 24x+20y=3 }\text{.}\) -
    This line’s slope is .
    This line’s \(y\)-intercept is .
    -
    -
    Answer 1.
    \(-{\frac{6}{5}}\)
    Answer 2.
    \(\left(0,\frac{3}{20}\right)\)
    Explanation.
    -
    When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form.
    In this problem, the line’s equation is given in the standard form \(\displaystyle{ 24x+20y=3 }\text{.}\) If we algebraically rearrange the equation from standard form to slope-intercept form \(y=mx+b\text{,}\) it will be easy to see the slope and \(y\)-intercept.
    \(\displaystyle{\begin{aligned} -24x+20y \amp = 3 \\ -24x+20y\mathbf{{}-24x} \amp = 3\mathbf{{}-24x} \\ -20y \amp = -24x +3 \\ -\frac{20y}{20} \amp = \frac{-24x}{20} + \frac{3}{20}\\ -y \amp = -\frac{24}{20}x + \frac{3}{20}\\ -y \amp = -\frac{6}{5}x + \frac{3}{20} -\end{aligned} -}\)
    Now we can see the line’s slope is \(-\frac{6}{5}\text{,}\) and its \(y\)-intercept has coordinates \(\left(0,\frac{3}{20}\right)\text{.}\) -
    -
    -
    -
    Converting to Standard Form.
    -
    19.
    -
    +
    19.
    +
    -
    Rewrite \({y}={2x+3}\) in standard form.
    +
    Rewrite \({y}={9x-9}\) in standard form.
    -
    Answer.
    \(2x-y=-3\)
    Explanation.
    +
    Answer.
    \(9x-y=9\)
    Explanation.
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    \begin{equation*} \begin{aligned} -{y}\amp={2x+3}\\ -{y}\subtractright{2x}\amp={2x+3}\subtractright{2x}\\ -{-2x+y}\amp=3\\ -(-1)({-2x+y})\amp=(-1)(3)\\ -{2x-y}\amp=-3 +{y}\amp={9x-9}\\ +{y}\subtractright{9x}\amp={9x-9}\subtractright{9x}\\ +{-9x+y}\amp=-9\\ +(-1)({-9x+y})\amp=(-1)(-9)\\ +{9x-y}\amp=9 \end{aligned} \end{equation*} -
    Note that \({-2x+y}=3\) is already in standard form, but it’s better to get rid of the leading negative sign in an equation.
    +
    Note that \({-9x+y}=-9\) is already in standard form, but it’s better to get rid of the leading negative sign in an equation.
    20.
    -
    -
    +
    +
    -
    Rewrite \({y}={3x-3}\) in standard form.
    +
    Rewrite \({y}={2x+3}\) in standard form.
    -
    Answer.
    \(3x-y=3\)
    Explanation.
    -
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    +
    Answer.
    \(2x-y=-3\)
    Explanation.
    +
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    \begin{equation*} \begin{aligned} -{y}\amp={3x-3}\\ -{y}\subtractright{3x}\amp={3x-3}\subtractright{3x}\\ -{-3x+y}\amp=-3\\ -(-1)({-3x+y})\amp=(-1)(-3)\\ -{3x-y}\amp=3 +{y}\amp={2x+3}\\ +{y}\subtractright{2x}\amp={2x+3}\subtractright{2x}\\ +{-2x+y}\amp=3\\ +(-1)({-2x+y})\amp=(-1)(3)\\ +{2x-y}\amp=-3 \end{aligned} \end{equation*} -
    Note that \({-3x+y}=-3\) is already in standard form, but it’s better to get rid of the leading negative sign in an equation.
    +
    Note that \({-2x+y}=3\) is already in standard form, but it’s better to get rid of the leading negative sign in an equation.
    21.
    -
    -
    +
    +
    -
    Rewrite \({y}={\frac{4}{9}x+4}\) in standard form.
    +
    Rewrite \({y}={\frac{3}{4}x - 3}\) in standard form.
    -
    Answer.
    \(4x-9y=-36\)
    Explanation.
    -
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    +
    Answer.
    \(3x-4y=12\)
    Explanation.
    +
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    \begin{equation*} \begin{aligned} -{y}\amp={\frac{4}{9}x+4}\\ -\multiplyleft{9}{y}\amp=\multiplyleft{9}\left({\frac{4}{9}x+4}\right)\\ -9y\amp=9\left( \frac{4}{9}x \right)+9(4)\\ -9y\amp=4x+36\\ -9y\subtractright{4x}\amp=4x+36\subtractright{4x}\\ -{-4x+9y}\amp=36\\ -\multiplyleft{-1}({-4x+9y})\amp=\multiplyleft{-1}(36)\\ -{4x-9y}\amp=-36 +{y}\amp={\frac{3}{4}x - 3}\\ +\multiplyleft{4}{y}\amp=\multiplyleft{4}\left({\frac{3}{4}x - 3}\right)\\ +4y\amp=4\left( \frac{3}{4}x \right)+4(-3)\\ +4y\amp=3x - 12\\ +4y\subtractright{3x}\amp=3x - 12\subtractright{3x}\\ +{-3x+4y}\amp=-12\\ +\multiplyleft{-1}({-3x+4y})\amp=\multiplyleft{-1}(-12)\\ +{3x-4y}\amp=12 \end{aligned} \end{equation*} -
    Note that \({-4x+9y}=36\) is already in standard form, but it’s better to get rid of the leading negative sign in an equation.
    +
    Note that \({-3x+4y}=-12\) is already in standard form, but it’s better to get rid of the leading negative sign in an equation.
    22.
    -
    -
    +
    +
    -
    Rewrite \(y=-\frac{5}{7}x - 7\) in standard form.
    +
    Rewrite \(y=-\frac{4}{9}x+4\) in standard form.
    -
    Answer.
    \(5x+7y=-49\)
    Explanation.
    -
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    +
    Answer.
    \(4x+9y=36\)
    Explanation.
    +
    A line’s standard form is \(Ax+By=C\text{.}\) We need to move \(x\) and \(y\) terms to the same side of the equals sign.
    \begin{equation*} \begin{aligned} -y\amp=-\frac{5}{7}x - 7\\ -\multiplyleft{7}y\amp=\multiplyleft{7}\left(-\frac{5}{7}x - 7\right)\\ -7y\amp=7\left( -\frac{5}{7}x \right)+7(-7)\\ -7y\amp=-5x - 49\\ -7y\addright{5x}\amp=-5x - 49\addright{5x}\\ -{5x+7y}\amp=-49 +y\amp=-\frac{4}{9}x+4\\ +\multiplyleft{9}y\amp=\multiplyleft{9}\left(-\frac{4}{9}x+4\right)\\ +9y\amp=9\left( -\frac{4}{9}x \right)+9(4)\\ +9y\amp=-4x+36\\ +9y\addright{4x}\amp=-4x+36\addright{4x}\\ +{4x+9y}\amp=36 \end{aligned} \end{equation*}
    @@ -1456,12 +1430,12 @@
    Converting to Standard Form.
    Graphs and Standard Form.
    23.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -5 x + 4 y = 40 +5 x + 6 y = 30 \end{equation*}
    @@ -1487,25 +1461,25 @@
    Graphs and Standard Form.
    -\(x\)-intercept and \(y\)-intercept of the line \(5 x + 4 y = 40\) +\(x\)-intercept and \(y\)-intercept of the line \(5 x + 6 y = 30\)
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(10\)
    Answer 3.
    \(\left(0,10\right)\)
    Answer 4.
    \(8\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(8,0\right)\)
    Explanation.
    -
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \(\begin{aligned} -5 x+4 y \amp = 40\\ -5 (0)+4 y \amp = 40\\ -4 y \amp = 40 \\ -\frac{4 y}{4} \amp = \frac{40}{4}\\ -y \amp = 10 -\end{aligned}\)
    This line’s \(y\)-intercept is \((0,10)\text{.}\) -
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \(\begin{aligned} -5 x+4 y \amp = 40\\ -5 x+4 (0) \amp = 40\\ -5 x \amp = 40\\ -\frac{5 x}{5} \amp = \frac{40}{5}\\ -x \amp = 8 -\end{aligned}\)
    The line’s \(x\)-intercept is \((8,0)\text{.}\) -
    The entries for the table are:
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(5\)
    Answer 3.
    \(\left(0,5\right)\)
    Answer 4.
    \(6\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(6,0\right)\)
    Explanation.
    +
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \(\begin{aligned} +5 x+6 y \amp = 30\\ +5 (0)+6 y \amp = 30\\ +6 y \amp = 30 \\ +\frac{6 y}{6} \amp = \frac{30}{6}\\ +y \amp = 5 +\end{aligned}\)
    This line’s \(y\)-intercept is \((0,5)\text{.}\) +
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \(\begin{aligned} +5 x+6 y \amp = 30\\ +5 x+6 (0) \amp = 30\\ +5 x \amp = 30\\ +\frac{5 x}{5} \amp = \frac{30}{5}\\ +x \amp = 6 +\end{aligned}\)
    The line’s \(x\)-intercept is \((6,0)\text{.}\) +
    The entries for the table are:
    - - + + - + - +
    @@ -1518,29 +1492,29 @@
    Graphs and Standard Form.
    		 \(y\)-intercept \(0\)\(10\)\((0,10)\)\(5\)\((0,5)\)
    \(x\)-intercept\(8\)\(6\) \(0\)\((8,0)\)\((6,0)\)
    -\(x\)-intercept and \(y\)-intercept of the line \(5 x + 4 y = 40\) +\(x\)-intercept and \(y\)-intercept of the line \(5 x + 6 y = 30\)
    24.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -6 x + 7 y = -126 +4 x + 3 y = -24 \end{equation*}
    @@ -1566,33 +1540,33 @@
    Graphs and Standard Form.
    -\(x\)-intercept and \(y\)-intercept of the line \(6 x+7 y=-126\) +\(x\)-intercept and \(y\)-intercept of the line \(4 x+3 y=-24\)
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(-18\)
    Answer 3.
    \(\left(0,-18\right)\)
    Answer 4.
    \(-21\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-21,0\right)\)
    Explanation.
    -
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(-8\)
    Answer 3.
    \(\left(0,-8\right)\)
    Answer 4.
    \(-6\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-6,0\right)\)
    Explanation.
    +
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \begin{equation*} \begin{aligned} -6 x+7 y \amp = -126\\ -6 (0)+7 y \amp = -126\\ -7 y \amp = -126 \\ -\frac{7 y}{7} \amp = \frac{-126}{7}\\ -y \amp = -18 +4 x+3 y \amp = -24\\ +4 (0)+3 y \amp = -24\\ +3 y \amp = -24 \\ +\frac{3 y}{3} \amp = \frac{-24}{3}\\ +y \amp = -8 \end{aligned} \end{equation*} -
    This line’s \(y\)-intercept is \((0,-18)\text{.}\) -
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    +
    This line’s \(y\)-intercept is \((0,-8)\text{.}\) +
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \begin{equation*} \begin{aligned} -6 x+7 y \amp = -126\\ -6 x+7 (0) \amp = -126\\ -6 x \amp = -126\\ -\frac{6 x}{6} \amp = \frac{-126}{6}\\ -x \amp = -21 +4 x+3 y \amp = -24\\ +4 x+3 (0) \amp = -24\\ +4 x \amp = -24\\ +\frac{4 x}{4} \amp = \frac{-24}{4}\\ +x \amp = -6 \end{aligned} \end{equation*} -
    The line’s \(x\)-intercept is \((-21,0)\text{.}\) -
    The entries for the table are:
    +
    The line’s \(x\)-intercept is \((-6,0)\text{.}\) +
    The entries for the table are:
    - - + + - + - +
    @@ -1605,29 +1579,29 @@
    Graphs and Standard Form.
    		 \(y\)-intercept \(0\)\(-18\)\((0,-18)\)\(-8\)\((0,-8)\)
    \(x\)-intercept\(-21\)\(-6\) \(0\)\((-21,0)\)\((-6,0)\)
    -\(x\)-intercept and \(y\)-intercept of the line \(6 x+7 y=-126\) +\(x\)-intercept and \(y\)-intercept of the line \(4 x+3 y=-24\)
    25.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -6 x - 5 y = -60 +6 x - 7 y = -126 \end{equation*}
    @@ -1653,25 +1627,25 @@
    Graphs and Standard Form.
    -\(x\)-intercept and \(y\)-intercept of the line \(6 x - 5 y=-60\) +\(x\)-intercept and \(y\)-intercept of the line \(6 x - 7 y=-126\)
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(12\)
    Answer 3.
    \(\left(0,12\right)\)
    Answer 4.
    \(-10\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-10,0\right)\)
    Explanation.
    -
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \(\begin{aligned} -6 x - 5 y \amp = -60\\ -6 (0) - 5 y \amp = -60\\ --5 y \amp = -60 \\ -\frac{-5 y}{-5} \amp = \frac{-60}{-5}\\ -y \amp = 12 -\end{aligned}\)
    This line’s \(y\)-intercept is \((0,12)\text{.}\) -
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \(\begin{aligned} -6 x - 5 y \amp = -60\\ -6 x - 5 (0) \amp = -60\\ -6 x \amp = -60\\ -\frac{6 x}{6} \amp = \frac{-60}{6}\\ -x \amp = -10 -\end{aligned}\)
    The line’s \(x\)-intercept is \((-10,0)\text{.}\) -
    The entries for the table are:
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(18\)
    Answer 3.
    \(\left(0,18\right)\)
    Answer 4.
    \(-21\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-21,0\right)\)
    Explanation.
    +
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \(\begin{aligned} +6 x - 7 y \amp = -126\\ +6 (0) - 7 y \amp = -126\\ +-7 y \amp = -126 \\ +\frac{-7 y}{-7} \amp = \frac{-126}{-7}\\ +y \amp = 18 +\end{aligned}\)
    This line’s \(y\)-intercept is \((0,18)\text{.}\) +
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \(\begin{aligned} +6 x - 7 y \amp = -126\\ +6 x - 7 (0) \amp = -126\\ +6 x \amp = -126\\ +\frac{6 x}{6} \amp = \frac{-126}{6}\\ +x \amp = -21 +\end{aligned}\)
    The line’s \(x\)-intercept is \((-21,0)\text{.}\) +
    The entries for the table are:
    - - + + - + - +
    @@ -1684,29 +1658,29 @@
    Graphs and Standard Form.
    		 \(y\)-intercept \(0\)\(12\)\((0,12)\)\(18\)\((0,18)\)
    \(x\)-intercept\(-10\)\(-21\) \(0\)\((-10,0)\)\((-21,0)\)
    -\(x\)-intercept and \(y\)-intercept of the line \(6 x - 5 y=-60\) +\(x\)-intercept and \(y\)-intercept of the line \(6 x - 7 y=-126\)
    26.
    -
    -
    +
    +
    -
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    +
    Find the \(y\)-intercept and \(x\)-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
    \begin{equation*} -x - 7 y = -7 +x - 7 y = -14 \end{equation*}
    @@ -1732,23 +1706,23 @@
    Graphs and Standard Form.
    -\(x\)-intercept and \(y\)-intercept of the line \(x - 7 y = -7\) +\(x\)-intercept and \(y\)-intercept of the line \(x - 7 y = -14\)
    -
    Answer 1.
    \(0\)
    Answer 2.
    \(1\)
    Answer 3.
    \(\left(0,1\right)\)
    Answer 4.
    \(-7\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-7,0\right)\)
    Explanation.
    -
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \(\begin{aligned} -x - 7 y \amp = -7\\ -0 - 7 y \amp = -7\\ --7 y \amp = -7 \\ -\frac{-7 y}{-7} \amp = \frac{-7}{-7}\\ -y \amp = 1 -\end{aligned}\)
    This line’s \(y\)-intercept is \((0,1)\text{.}\) -
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \(\begin{aligned} -x - 7 y \amp = -7\\ -x - 7 (0) \amp = -7\\ -x \amp = -7\\ -\end{aligned}\)
    The line’s \(x\)-intercept is \((-7,0)\text{.}\) -
    The entries for the table are:
    +
    Answer 1.
    \(0\)
    Answer 2.
    \(2\)
    Answer 3.
    \(\left(0,2\right)\)
    Answer 4.
    \(-14\)
    Answer 5.
    \(0\)
    Answer 6.
    \(\left(-14,0\right)\)
    Explanation.
    +
    A line’s \(y\)-intercept is on the \(y\)-axis, implying that its \(x\)-value must be \(0\text{.}\) To find a line’s \(y\)-intercept, we substitute in \(x=0\text{.}\) In this problem we have:
    \(\begin{aligned} +x - 7 y \amp = -14\\ +0 - 7 y \amp = -14\\ +-7 y \amp = -14 \\ +\frac{-7 y}{-7} \amp = \frac{-14}{-7}\\ +y \amp = 2 +\end{aligned}\)
    This line’s \(y\)-intercept is \((0,2)\text{.}\) +
    Next, a line’s \(x\)-intercept is on the \(x\)-axis, implying that its \(y\)-value must be \(0\text{.}\) To find a line’s \(x\)-intercept, we substitute in \(y=0\text{.}\) In this problem we have:
    \(\begin{aligned} +x - 7 y \amp = -14\\ +x - 7 (0) \amp = -14\\ +x \amp = -14\\ +\end{aligned}\)
    The line’s \(x\)-intercept is \((-14,0)\text{.}\) +
    The entries for the table are:
    - - + + - + - +
    @@ -1761,18 +1735,18 @@
    Graphs and Standard Form.
    		 \(y\)-intercept \(0\)\(1\)\((0,1)\)\(2\)\((0,2)\)
    \(x\)-intercept\(-7\)\(-14\) \(0\)\((-7,0)\)\((-14,0)\)
    -\(x\)-intercept and \(y\)-intercept of the line \(x - 7 y = -7\) +\(x\)-intercept and \(y\)-intercept of the line \(x - 7 y = -14\)
    @@ -1868,158 +1842,158 @@
    Exercise Group.
    Interpreting Intercepts in Context.
    43.
    -
    +
    +
    +
    +
    Daniel is buying some tea bags and some sugar bags. Each tea bag costs \(4\) cents, and each sugar bag costs \(5\) cents. He can spend a total of \({\$1.00}\text{.}\) +
    Assume Daniel will purchase \(x\) tea bags and \(y\) sugar bags. Use a linear equation to model the number of tea bags and sugar bags he can purchase.
    Find this line’s \(x\)-intercept, and interpret its meaning in this context.
      +
    • +A. The x-intercept is (0, 20). It implies Daniel can purchase 20 sugar bags with no tea bags.
      • +
    • +B. The x-intercept is (0,25). It implies Daniel can purchase 25 sugar bags with no tea bags.
      • +
    • +C. The x-intercept is (25,0). It implies Daniel can purchase 25 tea bags with no sugar bags.
      • +
    • +D. The x-intercept is (20,0). It implies Daniel can purchase 20 tea bags with no sugar bags.
      • +
    +
    +
    Answer.
    \(\text{C}\)
    Explanation.
    +
    Assume Daniel will purchase \(x\) tea bags and \(y\) sugar bags.
    Since each tea bag costs \(4\) cents, \(x\) tea bags would cost \(4 x\) cents.
    Similarly, \(y\) sugar bags would cost \(5 y\) cents.
    Since Daniel can spend a total of \({\$1.00}\text{,}\) or \(100\) cents, we can write the equation:
    \(\displaystyle{ 4x+5y=100 }\)
    A line’s \(x\)-intercept looks like \((x,0)\text{.}\) To find its \(x\) intercept, we substitute \(y\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} +4x+5y \amp = 100 \\ +4x+5(0) \amp = 100 \\ +4x \amp = 100 \\ +\frac{4x}{4} \amp = \frac{100}{4} \\ +x \amp = 25 +\end{aligned} +}\)
    The line’s \(x\)-intercept is \((25,0)\text{.}\) +
    The correct solution is C.
    +
    +
    +
    +
    44.
    +
    Casandra is buying some tea bags and some sugar bags. Each tea bag costs \(10\) cents, and each sugar bag costs \(4\) cents. She can spend a total of \({\$6.00}\text{.}\) -
    Assume Casandra will purchase \(x\) tea bags and \(y\) sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase.
    Find this line’s \(x\)-intercept, and interpret its meaning in this context.
      +
    Assume Casandra will purchase \(x\) tea bags and \(y\) sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase.
    Find this line’s \(y\)-intercept, and interpret its meaning in this context.
    • -A. The x-intercept is (150,0). It implies Casandra can purchase 150 tea bags with no sugar bags.
      • +A. The y-intercept is (0,60). It implies Casandra can purchase 60 sugar bags with no tea bags.
  • -B. The x-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags.
    • +B. The y-intercept is (150,0). It implies Casandra can purchase 150 tea bags with no sugar bags.
  • -C. The x-intercept is (60,0). It implies Casandra can purchase 60 tea bags with no sugar bags.
    • +C. The y-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags.
  • -D. The x-intercept is (0,60). It implies Casandra can purchase 60 sugar bags with no tea bags.
    • +D. The y-intercept is (60,0). It implies Casandra can purchase 60 tea bags with no sugar bags.
    Answer.
    \(\text{C}\)
    Explanation.
    -
    Assume Casandra will purchase \(x\) tea bags and \(y\) sugar bags.
    Since each tea bag costs \(10\) cents, \(x\) tea bags would cost \(10 x\) cents.
    Similarly, \(y\) sugar bags would cost \(4 y\) cents.
    Since Casandra can spend a total of \({\$6.00}\text{,}\) or \(600\) cents, we can write the equation:
    \(\displaystyle{ 10x+4y=600 }\)
    A line’s \(x\)-intercept looks like \((x,0)\text{.}\) To find its \(x\) intercept, we substitute \(y\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} +
    Assume Casandra will purchase \(x\) tea bags and \(y\) sugar bags.
    Since each tea bag costs \(10\) cents, \(x\) tea bags would cost \(10 x\) cents.
    Similarly, \(y\) sugar bags would cost \(4 y\) cents.
    Since Casandra can spend a total of \({\$6.00}\text{,}\) or \(600\) cents, we can write the equation:
    \(\displaystyle{ 10x+4y=600 }\)
    A line’s \(y\)-intercept looks like \((0,y)\text{.}\) To find its \(y\) intercept, we substitute \(x\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} 10x+4y \amp = 600 \\ -10x+4(0) \amp = 600 \\ -10x \amp = 600 \\ -\frac{10x}{10} \amp = \frac{600}{10} \\ -x \amp = 60 +10(0)+4y \amp = 600 \\ +4y \amp = 600 \\ +\frac{4x}{4} \amp = \frac{600}{4} \\ +x \amp = 150 \end{aligned} -}\)
    The line’s \(x\)-intercept is \((60,0)\text{.}\) -
    The correct solution is C.
    +}\)
    The line’s \(y\)-intercept is \((0,150)\text{.}\) +
    The correct solution is: The y-intercept is (0, 150). It implies Casandra can purchase 150 sugar bags with no tea bags.
    -
    44.
    -
    +
    45.
    +
    -
    Page is buying some tea bags and some sugar bags. Each tea bag costs \(2\) cents, and each sugar bag costs \(6\) cents. She can spend a total of \({\$0.90}\text{.}\) -
    Assume Page will purchase \(x\) tea bags and \(y\) sugar bags. Use a linear equation to model the number of tea bags and sugar bags she can purchase.
    Find this line’s \(y\)-intercept, and interpret its meaning in this context.
      +
      An engine’s tank can hold \(175\) gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming \(5\) gallons of gas per hour.
      Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes.
      Find this line’s \(x\)-intercept, and interpret its meaning in this context.
      • -A. The y-intercept is (0,45). It implies Page can purchase 45 sugar bags with no tea bags.
        • +A. The x-intercept is (175,0). It implies the engine will run out of gas 175 hours after its tank was refilled.
    • -B. The y-intercept is (15,0). It implies Page can purchase 15 tea bags with no sugar bags.
      • +B. The x-intercept is (0,175). It implies the engine started with 175 gallons of gas in its tank.
  • -C. The y-intercept is (0, 15). It implies Page can purchase 15 sugar bags with no tea bags.
    • +C. The x-intercept is (0,35). It implies the engine started with 35 gallons of gas in its tank.
  • -D. The y-intercept is (45,0). It implies Page can purchase 45 tea bags with no sugar bags.
    • +D. The x-intercept is (35,0). It implies the engine will run out of gas 35 hours after its tank was refilled.
    -
    Answer.
    \(\text{C}\)
    Explanation.
    -
    Assume Page will purchase \(x\) tea bags and \(y\) sugar bags.
    Since each tea bag costs \(2\) cents, \(x\) tea bags would cost \(2 x\) cents.
    Similarly, \(y\) sugar bags would cost \(6 y\) cents.
    Since Page can spend a total of \({\$0.90}\text{,}\) or \(90\) cents, we can write the equation:
    \(\displaystyle{ 2x+6y=90 }\)
    A line’s \(y\)-intercept looks like \((0,y)\text{.}\) To find its \(y\) intercept, we substitute \(x\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} -2x+6y \amp = 90 \\ -2(0)+6y \amp = 90 \\ -6y \amp = 90 \\ -\frac{6x}{6} \amp = \frac{90}{6} \\ -x \amp = 15 +
    Answer.
    \(\text{D}\)
    Explanation.
    +
    Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is:
    \(\displaystyle{ y = -5x+175 }\)
    A line’s \(x\)-intercept looks like \((x,0)\text{.}\) To find its \(x\) intercept, we substitute \(y\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} +y \amp = -5x+175 \\ +0 \amp = -5x+175 \\ +-175 \amp = -5x \\ +\frac{-175}{-5} \amp = \frac{-5x}{-5} \\ +35 \amp = x \end{aligned} -}\)
    The line’s \(y\)-intercept is \((0,15)\text{.}\) -
    The correct solution is: The y-intercept is (0, 15). It implies Page can purchase 15 sugar bags with no tea bags.
    +}\)
    The line’s \(x\)-intercept is \((35,0)\text{.}\) +
    The correct solution is: The x-intercept is (35,0). It implies the engine will run out of gas 35 hours after its tank was refilled.
    -
    45.
    -
    +
    46.
    +
    -
    An engine’s tank can hold \(150\) gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming \(5\) gallons of gas per hour.
    Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes.
    Find this line’s \(x\)-intercept, and interpret its meaning in this context.
      +
      An engine’s tank can hold \(150\) gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming \(5\) gallons of gas per hour.
      Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes.
      Find this line’s \(y\)-intercept, and interpret its meaning in this context.
      • -A. The x-intercept is (150,0). It implies the engine will run out of gas 150 hours after its tank was refilled.
        • +A. The y-intercept is (0,30). It implies the engine started with 30 gallons of gas in its tank.
    • -B. The x-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled.
      • +B. The y-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank.
  • -C. The x-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank.
    • +C. The y-intercept is (150,0). It implies the engine will run out of gas 150 hours after its tank was refilled.
  • -D. The x-intercept is (0,30). It implies the engine started with 30 gallons of gas in its tank.
    • +D. The y-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled.
    Answer.
    \(\text{B}\)
    Explanation.
    -
    Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is:
    \(\displaystyle{ y = -5x+150 }\)
    A line’s \(x\)-intercept looks like \((x,0)\text{.}\) To find its \(x\) intercept, we substitute \(y\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} -y \amp = -5x+150 \\ -0 \amp = -5x+150 \\ --150 \amp = -5x \\ -\frac{-150}{-5} \amp = \frac{-5x}{-5} \\ -30 \amp = x -\end{aligned} -}\)
    The line’s \(x\)-intercept is \((30,0)\text{.}\) -
    The correct solution is: The x-intercept is (30,0). It implies the engine will run out of gas 30 hours after its tank was refilled.
    +
    Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is:
    \(\displaystyle{ y = -5x+150 }\)
    Since the equation is in slope-intercept mode, we can see its \(y\)-intercept is \((0,150)\text{.}\) +
    The correct solution is: The y-intercept is (0,150). It implies the engine started with 150 gallons of gas in its tank.
    -
    46.
    -
    +
    47.
    +
    -
    An engine’s tank can hold \(180\) gallons of gasoline. It was refilled with a full tank, and has been running without breaks, consuming \(4\) gallons of gas per hour.
    Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. Use a linear equation to model the amount of gas in the tank as time passes.
    Find this line’s \(y\)-intercept, and interpret its meaning in this context.
      +
      A new car of a certain model costs \({\$52{,}800.00}\text{.}\) According to Blue Book, its value decreases by \({\$2{,}200.00}\) every year.
      Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. Use a linear equation to model the car’s value.
      Find this line’s \(x\)-intercept, and interpret its meaning in this context.
      • -A. The y-intercept is (45,0). It implies the engine will run out of gas 45 hours after its tank was refilled.
        • +A. The x-intercept is (0,24). It implies the car would have no more value 24 years since its purchase.
    • -B. The y-intercept is (0,180). It implies the engine started with 180 gallons of gas in its tank.
      • +B. The x-intercept is (24,0). It implies the car would have no more value 24 years since its purchase.
  • -C. The y-intercept is (180,0). It implies the engine will run out of gas 180 hours after its tank was refilled.
    • +C. The x-intercept is (0,52800). It implies the car’s initial value was 52800.
  • -D. The y-intercept is (0,45). It implies the engine started with 45 gallons of gas in its tank.
    • +D. The x-intercept is (52800,0). It implies the car’s initial value was 52800.
    Answer.
    \(\text{B}\)
    Explanation.
    -
    Assume the engine has been running for \(x\) hours since its tank was refilled, and assume there are \(y\) gallons of gas left in the tank. A linear equation to model the amount of gas in the tank is:
    \(\displaystyle{ y = -4x+180 }\)
    Since the equation is in slope-intercept mode, we can see its \(y\)-intercept is \((0,180)\text{.}\) -
    The correct solution is: The y-intercept is (0,180). It implies the engine started with 180 gallons of gas in its tank.
    +
    Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. A linear equation to model the car’s value is:
    \(\displaystyle{ y = -2200x+52800 }\)
    A line’s \(x\)-intercept looks like \((x,0)\text{.}\) To find its \(x\) intercept, we substitute \(y\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} +y \amp = -2200x+52800 \\ +0 \amp = -2200x+52800 \\ +-52800 \amp = -2200x \\ +\frac{-52800}{-2200} \amp = \frac{-2200x}{-2200} \\ +24 \amp = x +\end{aligned} +}\)
    The line’s \(x\)-intercept is \((24,0)\text{.}\) +
    The correct solution is: The x-intercept is (24,0). It implies the car would have no more value 24 years since its purchase.
    -
    47.
    -
    +
    48.
    +
    -
    A new car of a certain model costs \({\$44{,}000.00}\text{.}\) According to Blue Book, its value decreases by \({\$2{,}200.00}\) every year.
    Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. Use a linear equation to model the car’s value.
    Find this line’s \(x\)-intercept, and interpret its meaning in this context.
      +
      A new car of a certain model costs \({\$44{,}000.00}\text{.}\) According to Blue Book, its value decreases by \({\$2{,}200.00}\) every year.
      Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. Use a linear equation to model the car’s value.
      Find this line’s \(y\)-intercept, and interpret its meaning in this context.
      • -A. The x-intercept is (0,44000). It implies the car’s initial value was 44000.
        • +A. The y-intercept is (44000,0). It implies the car’s initial value was 44000.
    • -B. The x-intercept is (44000,0). It implies the car’s initial value was 44000.
      • +B. The y-intercept is (0,20). It implies the car would have no more value 20 years since its purchase.
  • -C. The x-intercept is (0,20). It implies the car would have no more value 20 years since its purchase.
    • +C. The y-intercept is (20,0). It implies the car would have no more value 20 years since its purchase.
  • -D. The x-intercept is (20,0). It implies the car would have no more value 20 years since its purchase.
    • +D. The y-intercept is (0,44000). It implies the car’s initial value was 44000.
    Answer.
    \(\text{D}\)
    Explanation.
    -
    Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. A linear equation to model the car’s value is:
    \(\displaystyle{ y = -2200x+44000 }\)
    A line’s \(x\)-intercept looks like \((x,0)\text{.}\) To find its \(x\) intercept, we substitute \(y\) in the equation with \(0\text{,}\) and we have:
    \(\displaystyle{\begin{aligned} -y \amp = -2200x+44000 \\ -0 \amp = -2200x+44000 \\ --44000 \amp = -2200x \\ -\frac{-44000}{-2200} \amp = \frac{-2200x}{-2200} \\ -20 \amp = x -\end{aligned} -}\)
    The line’s \(x\)-intercept is \((20,0)\text{.}\) -
    The correct solution is: The x-intercept is (20,0). It implies the car would have no more value 20 years since its purchase.
    -
    -
    - -
    48.
    -
    -
    -
    -
    A new car of a certain model costs \({\$35{,}200.00}\text{.}\) According to Blue Book, its value decreases by \({\$2{,}200.00}\) every year.
    Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. Use a linear equation to model the car’s value.
    Find this line’s \(y\)-intercept, and interpret its meaning in this context.
      -
    • -A. The y-intercept is (16,0). It implies the car would have no more value 16 years since its purchase.
      • -
    • -B. The y-intercept is (0,16). It implies the car would have no more value 16 years since its purchase.
      • -
    • -C. The y-intercept is (35200,0). It implies the car’s initial value was 35200.
      • -
    • -D. The y-intercept is (0,35200). It implies the car’s initial value was 35200.
      • -
    -
    -
    Answer.
    \(\text{D}\)
    Explanation.
    -
    Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. A linear equation to model the car’s value is:
    \(\displaystyle{ y = -2200x+35200 }\)
    Since this line is in its slope-intercept mode, we can see its \(y\)-intercept is \((0,35200)\text{.}\) -
    The correct solution is: The y-intercept is (0,35200). It implies the car’s initial value was 35200.
    +
    Assume \(x\) years since its purchase, the car’s value is \(y\) dollars. A linear equation to model the car’s value is:
    \(\displaystyle{ y = -2200x+44000 }\)
    Since this line is in its slope-intercept mode, we can see its \(y\)-intercept is \((0,44000)\text{.}\) +
    The correct solution is: The y-intercept is (0,44000). It implies the car’s initial value was 44000.
    @@ -2027,21 +2001,21 @@
    Interpreting Intercepts in Context.

    Challenge

    49.
    -
    -
    +
    +
    -
    +
    Fill in the variables \(A\text{,}\) \(B\text{,}\) and \(C\) in \(Ax + By = C\) with the numbers \(6, -7\) and \(14\text{.}\) You may only use each number once.
    +7\) and \(9\text{.}\) You may only use each number once.
      -
    1. For the steepest possible slope, \(A\) must be , \(B\) must be , and \(C\) must be .
      2. -
    2. For the shallowest possible slope, \(A\) must be , \(B\) must be , and \(C\) must be .
      3. +
    3. For the steepest possible slope, \(A\) must be , \(B\) must be , and \(C\) must be .
      4. +
    4. For the shallowest possible slope, \(A\) must be , \(B\) must be , and \(C\) must be .
    -
    Answer 1.
    \(14\)
    Answer 2.
    \(6\)
    Answer 3.
    \(7\)
    Answer 4.
    \(6\)
    Answer 5.
    \(14\)
    Answer 6.
    \(7\)
    Explanation.
      -
    1. Let’s put \(Ax + By = C\) into slope-intercept form. Then, the equation becomes \(y = \frac{-A}{B}x + \frac{C}{B} \text{.}\) To make the steepest slope possible, we must maximize the fraction \(\frac{A}{B}\text{.}\) The possibilities are \(\frac{6}{7}, \frac{6}{14}, \frac{7}{6}, \frac{7}{14}, \frac{14}{6}\text{,}\) and \(\frac{14}{7} \text{.}\) The biggest of these is \(\frac{14}{6} \text{.}\) Thus, \(A\) must equal 14, \(B\) must equal 6, and \(C\) must equal 7.
      2. -
    2. To make the shallowest slope possible, we must minimize the fraction \(\frac{A}{B}\text{.}\) The smallest of the possible fractions is \(\frac{6}{14} \text{.}\) Thus, \(A\) must equal 6, \(B\) must equal 14, and \(C\) must equal 7.
      3. +
      Answer 1.
      \(9\)
      Answer 2.
      \(6\)
      Answer 3.
      \(7\)
      Answer 4.
      \(6\)
      Answer 5.
      \(9\)
      Answer 6.
      \(7\)
      Explanation.
        +
      1. Let’s put \(Ax + By = C\) into slope-intercept form. Then, the equation becomes \(y = \frac{-A}{B}x + \frac{C}{B} \text{.}\) To make the steepest slope possible, we must maximize the fraction \(\frac{A}{B}\text{.}\) The possibilities are \(\frac{6}{7}, \frac{6}{9}, \frac{7}{6}, \frac{7}{9}, \frac{9}{6}\text{,}\) and \(\frac{9}{7} \text{.}\) The biggest of these is \(\frac{9}{6} \text{.}\) Thus, \(A\) must equal 9, \(B\) must equal 6, and \(C\) must equal 7.
        2. +
      2. To make the shallowest slope possible, we must minimize the fraction \(\frac{A}{B}\text{.}\) The smallest of the possible fractions is \(\frac{6}{9} \text{.}\) Thus, \(A\) must equal 6, \(B\) must equal 9, and \(C\) must equal 7.
    diff --git a/section-substitution.html b/section-substitution.html index e69f4d8b1..a7457c1c9 100644 --- a/section-substitution.html +++ b/section-substitution.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +
  • @@ -595,10 +609,10 @@

    Search Results:

    Example 4.2.2. The Interview. -

    Once upon a time, the New York Times
     1 
    nyti.ms/2pupebT
    published an article about the movie, The Interview. It included the following quote:
    -
    -The Interview generated roughly \(\$15\) million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.
    Sony did not say how much of that total represented \(\$6\) digital rentals versus \(\$15\) sales. The studio said there were about two million transactions overall.
    -
    Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog
     2 
    www.joeydevilla.com/2014/12/31/
    that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
    First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.
    +
    Once upon a time, the New York Times
     1 
    nyti.ms/2pupebT
    published an article about the movie, The Interview. It included the following quote:
    +
    +The Interview generated roughly \(\$15\) million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.
    Sony did not say how much of that total represented \(\$6\) digital rentals versus \(\$15\) sales. The studio said there were about two million transactions overall.
    +
    Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in his blog
     2 
    www.joeydevilla.com/2014/12/31/
    that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
    First, we will define variables. There are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.
    If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both \(r\) and \(s\) are measured in “transactions”. The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two million. Our equation is:
    \begin{equation*} @@ -611,7 +625,7 @@

    Search Results:

    r+s=2{,}000{,}000 \end{equation*}
    -
    +
    The price of each rental was \(\$6\text{.}\) That means the problem has given us a rate of \(6\,\frac{\text{dollars}}{\text{transaction}}\) to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by \(r\text{,}\) and that would give us the total number of dollars generated from rentals. This is \(6r\text{.}\) Similarly, the price of each sale was \(\$15\text{,}\) so the revenue from sales was \(15s\text{.}\) The total revenue was \(\$15\) million, which we can represent with this equation:
    \begin{equation*} @@ -624,7 +638,7 @@

    Search Results:

    6r+15s=15{,}000{,}000 \end{equation*}
    -
    +
    Here is our system of equations, with the commas removed:
    \begin{equation*} @@ -637,17 +651,17 @@

    Search Results:

    \end{equation*}
    To solve the system, we will use a method called substitution. The idea is to use one equation to isolate \(r\text{.}\) Then substitute this for the “\(r\)” that’s in the other equation. This leaves you with one equation where the only variable is \(s\text{.}\) And we can handle that directly.
    -
    +
    The first equation from the system is an easy one to isolate \(r\text{:}\)
    -
    +
    \begin{align*} r+s \amp=2\,000\,000\amp\text{(the system's first equation)}\\ r \amp=2\,000\,000-s \end{align*}
    This tells us that the expression \(2\,000\,000-s\) is equal to \(r\text{,}\) so we can substitute that in place of \(r\) in the second equation:
    -
    +
    \begin{align*} 6r+15s \amp=15\,000\,000\amp\text{(the system's second equation)}\\ 6(\substitute{2\,000\,000-s})+15s \amp=15\,000\,000\\ @@ -662,10 +676,10 @@

    Search Results:

    s \amp= 333\,333.\overline{3} \end{align*}
    -
    +
    At this point, we know that \(s=333\,333.\overline{3}\text{.}\) This tells us that out of the \(2\) million transactions, roughly \(333{,}333\) were from online sales. Recall that we isolated \(r\) previously and found \(r=2\,000\,000-s\text{.}\)
    -
    +
    \begin{align*} r \amp=2\,000\,000-s\\ r \amp=2\,000\,000-\substitute{333\,333.\overline{3}}\\ @@ -673,15 +687,15 @@

    Search Results:

    \end{align*}
    In summary, there were roughly \(333{,}333\) ssales and roughly \(1{,}666{,}667\) rentals.
    -

    +

    • Remark 4.2.3.

    -
    In Example 2, we rounded the solution values because only whole numbers make sense in the context of the problem. It was especially acceptable to round because the original numbers we worked with were already rounded. In fact, it would be OK to round even more to something like \(s=330{,}000\) and \(r=1{,}670{,}000\) as long as we communicate clearly that we rounded.
    In other exercises where there is no context, it is not OK to round. Solutions should be communicated with their exact values.

    +
    In Example 2, we rounded the solution values because only whole numbers make sense in the context of the problem. It was especially acceptable to round because the original numbers we worked with were already rounded. In fact, it would be OK to round even more to something like \(s=330{,}000\) and \(r=1{,}670{,}000\) as long as we communicate clearly that we rounded.
    In other exercises where there is no context, it is not OK to round. Solutions should be communicated with their exact values.

    Example 4.2.4.

    -
    +
    Solve the system of equations using substitution:
    -
    +
    \begin{align*} \left\{ \begin{alignedat}{4} @@ -692,18 +706,18 @@

    Search Results:

    \end{align*}
    -
    Explanation.
    -
    +
    Explanation.
    +
    To use substitution, we need to isolate one of the variables in one of our equations. Looking over both equations, it will be easiest to isolate \(x\) in the first equation:
    -
    +
    \begin{align*} x+2y\amp=5\\ x\amp= 5-2y \end{align*}
    -
    +
    Next, we substitute \(5-2y\) in for \(x\) in the second equation, giving us a linear equation in only one variable \(y\text{.}\) And this is an equation that we may solve using skills from [cross-reference to target(s) "section-solving-multistep-linear-equations.ptx" missing or not unique].
    -
    +
    \begin{align*} 3x-2y\amp=8\\ 3(\substitute{5-2y})-2y\amp=8\\ @@ -713,9 +727,9 @@

    Search Results:

    y\amp=\frac{7}{8} \end{align*}
    -
    +
    Now that we have the value for \(y\text{,}\) we need to find the value for \(x\text{.}\) We already isolated \(x\text{,}\) and it’s easiest to just use that equation.
    -
    +
    \begin{align*} x\amp= 5-2y\\ x\amp= 5-2\left(\substitute{\frac{7}{8}}\right)\\ @@ -723,9 +737,9 @@

    Search Results:

    x\amp=\frac{20}{4}-\frac{7}{4}=\frac{13}{4} \end{align*}
    -
    +
    At this point we think the solution is the point \(\left(\frac{13}{4},\frac{7}{8}\right)\) or in other words: \(x=\frac{13}{4}\text{,}\) \(y=\frac{7}{8}\text{.}\) It’s only human to make mistakes though, so we should check that the solution actually works. To check it, try using \(x=\frac{13}{4}\text{,}\) \(y=\frac{7}{8}\) in both of the original equations.
    -
    +
    \begin{align*} x+2y\amp=5\amp3x-2y\amp=8\\ \substitute{\frac{13}{4}}+2\left(\substitute{\frac{7}{8}}\right)\amp\wonder{=}5\amp3\left(\substitute{\frac{13}{4}}\right)-2\left(\substitute{\frac{7}{8}}\right)\amp\wonder{=}8\\ @@ -733,12 +747,12 @@

    Search Results:

    \frac{20}{4}\amp\confirm{=}5\amp\frac{32}{4}\amp\confirm{=}8 \end{align*}
    -
    We conclude then that this system of equations is true when \(x=\frac{13}{4}\) and \(y=\frac{7}{8}\text{.}\) The solution is the point \(\left(\frac{13}{4},\frac{7}{8}\right)\) and we write the solution set as \(\left\{\left(\frac{13}{4},\frac{7}{8}\right)\right\}\text{.}\) +
    We conclude then that this system of equations is true when \(x=\frac{13}{4}\) and \(y=\frac{7}{8}\text{.}\) The solution is the point \(\left(\frac{13}{4},\frac{7}{8}\right)\) and we write the solution set as \(\left\{\left(\frac{13}{4},\frac{7}{8}\right)\right\}\text{.}\)

    Checkpoint 4.2.5.

    -
    +
    @@ -785,13 +799,13 @@

    Search Results:

    \end{equation*}
    -
    Sometimes it makes more sense to start the process using the second equation instead of the first. And sometimes it makes more sense to isolate the second variable instead of the first.
    -

    +

    Sometimes it makes more sense to start the process using the second equation instead of the first. And sometimes it makes more sense to isolate the second variable instead of the first.
    +

    Example 4.2.6.

    -
    +
    Solve this system of equations using substitution:
    -
    +
    \begin{align*} \left\{ \begin{alignedat}{4} @@ -802,10 +816,10 @@

    Search Results:

    \end{align*}
    -
    Explanation.
    -
    +
    Explanation.
    +
    We need to isolate one of the variables in one of the equations. Looking over both equations, it will be easiest to isolate \(y\) in the second equation because that is where the coefficient is smallest. We are going to have to divide by some coefficient; it might as well be the smallest one to keep the fraction arithmetic as simple as we can.
    -
    +
    \begin{align*} -5x+2y\amp=11\\ 2y\amp=11+5x\\ @@ -814,9 +828,9 @@

    Search Results:

    \end{align*}
    Note that there are fractions once we’ve isolated \(y\text{.}\) We should take care with the steps that follow to make sure that the fraction arithmetic is correct.
    -
    +
    Substitute \(\frac{11}{2}+\frac{5}{2}x\) in for \(y\) in the first equation, and that leads to having a single linear equation with only one variable. We can solve this as we did in Section 2.3. Note the step in the middle where we clear denominators.
    -
    +
    \begin{align*} 3x-7y\amp=5\\ 3x-7\left(\frac{11}{2}+\frac{5}{2}x\right)\amp=5\\ @@ -828,9 +842,9 @@

    Search Results:

    x\amp=\divideunder{87}{-29}=-3 \end{align*}
    -
    +
    Now that we have the value for \(x\text{,}\) we need to find the value for \(y\text{.}\) We already isolated \(y\text{,}\) and it’s easiest to just use that equation.
    -
    +
    \begin{align*} y\amp=\frac{11}{2}+\frac{5}{2}x\\ y\amp=\frac{11}{2}+\frac{5}{2}(\substitute{-3})\\ @@ -838,24 +852,24 @@

    Search Results:

    y\amp=-\frac{4}{2}=-2 \end{align*}
    -
    +
    To check the solution we think we’ve found, try using \(x=-3\text{,}\) \(y=-2\) in both of the original equations.
    -
    +
    \begin{align*} 3x-7y \amp=5 \amp -5x+2y \amp= 11\\ 3(\substitute{-3})-7(\substitute{-2}) \amp\wonder{=}5 \amp -5(\substitute{-3})+2(\substitute{-2}) \amp\wonder{=} 11\\ -9+14\amp\confirm{=}5\amp 15-4\amp\confirm{=}11 \end{align*}
    -
    We conclude then that this system of equations is true when \(x=-3\) and \(y=-2\text{.}\) The solution is the point \((-3,-2)\) and we write the solution set as \(\{(-3,-2)\}\text{.}\) +
    We conclude then that this system of equations is true when \(x=-3\) and \(y=-2\text{.}\) The solution is the point \((-3,-2)\) and we write the solution set as \(\{(-3,-2)\}\text{.}\)
    -
    A system may start out with fractions among the coefficients. Just as we learned in Section 2.3, the algebra can go more smoothly if we clear the denominators before doing more work.
    -

    +

    A system may start out with fractions among the coefficients. Just as we learned in Section 2.3, the algebra can go more smoothly if we clear the denominators before doing more work.
    +

    Example 4.2.7.

    -
    +
    Solve the system of equations using the substitution method.
    -
    +
    \begin{align*} \left\{ \begin{aligned} @@ -866,11 +880,11 @@

    Search Results:

    \end{align*}
    -
    Explanation.
    -
    When a system of equations has fraction coefficients, it may be helpful to “clear the denomintors”. With each equation, multiply each side by the least common multiple of that equation’s denominators.
    -
    +
    Explanation.
    +
    When a system of equations has fraction coefficients, it may be helpful to “clear the denomintors”. With each equation, multiply each side by the least common multiple of that equation’s denominators.
    +
    In the first equation, the least common multiple of the denominators is \(6\text{,}\) so:
    -
    +
    \begin{align*} \frac{x}{3}-\frac{1}{2}y\amp=\frac{5}{6}\\ \multiplyleft{6}\left(\frac{x}{3}-\frac{1}{2}y\right)\amp=\multiplyleft{6}\frac{5}{6}\\ @@ -878,9 +892,9 @@

    Search Results:

    \end{align*}
    -
    +
    In the second equation, the least common multiple of the denominators is \(4\text{,}\) so:
    -
    +
    \begin{align*} \frac{1}{4}x\amp=\frac{y}{2}+1\\ \multiplyleft{4}\frac{1}{4}x\amp=\multiplyleft{4}\left(\frac{y}{2}+1\right)\\ @@ -888,9 +902,9 @@

    Search Results:

    \end{align*}
    -
    +
    Now we have a system that is equivalent to the original system of equations, but there are no fraction coefficients:
    -
    +
    \begin{align*} \left\{ \begin{aligned} @@ -901,7 +915,7 @@

    Search Results:

    \end{align*}
    The second equation has already isolateed \(x\text{,}\) so we will substitute \(2y+4\) in for \(x\) in the first equation.
    -
    +
    \begin{align*} 2x-3y\amp=5\\ 2(\substitute{2y+4})-3y\amp=5\\ @@ -911,7 +925,7 @@

    Search Results:

    \end{align*}
    And we have solved for \(y\text{.}\) To find \(x\text{,}\) we know \(x=2y+4\text{,}\) so we have:
    -
    +
    \begin{align*} x\amp=2y+4\\ x\amp=2(\substitute{-3})+4\\ @@ -920,13 +934,13 @@

    Search Results:

    The solution is \((-2,-3)\text{.}\) (This should be checked in the original two equations though.)
    -
    Real world applications can lead to a system where both equations are either in point-slope form or slope-intercept form. This is helpful, since it means one variable has already been isolated.
    -

    +

    Real world applications can lead to a system where both equations are either in point-slope form or slope-intercept form. This is helpful, since it means one variable has already been isolated.
    +

    Example 4.2.8.

    -
    +
    Solve the system of equations using the substitution method.
    -
    +
    \begin{align*} \left\{ \begin{aligned} @@ -937,11 +951,11 @@

    Search Results:

    \end{align*}
    -
    Explanation.
    -
    In the first equation, \(y\) is already isolated. So we can substitute \(\frac{5}{6}x+3\) in for \(y\) in the second equation: \(\frac{5}{6}x+3 = \frac23x + 1\text{.}\) -
    +
    Explanation.
    +
    In the first equation, \(y\) is already isolated. So we can substitute \(\frac{5}{6}x+3\) in for \(y\) in the second equation: \(\frac{5}{6}x+3 = \frac23x + 1\text{.}\) +
    Some people prefer to think of this as setting the two right sides equal to each other. That works too: if \(\frac{5}{6}x+3\) is equal to \(y\text{,}\) and \(\frac23x + 1\) is also equal to \(y\text{,}\) then these expressions are equal to each other. Either way:
    -
    +
    \begin{align*} \frac{5}{6}x + 3\amp=\frac23x + 1\\ \multiplyleft{6}\left(\frac{5}{6}x + 3\right)\amp=\multiplyleft{6}\left(\frac23x + 1\right)\\ @@ -951,7 +965,7 @@

    Search Results:

    \end{align*}
    And then:
    -
    +
    \begin{align*} y \amp= \frac{5}{6}(\substitute{-12})+3\\ y \amp= -10+3=-7 @@ -960,7 +974,7 @@

    Search Results:

    And the solution is \((-12,-7)\text{.}\)
    -
    For summary reference, here is the general procedure.
    +
    For summary reference, here is the general procedure.

    Exercises 4.2.5 Exercises

    -

    Review

    Skills Practice

    -
    +

    Skills Practice

    +
    Exercise Group.
    -
    Solve the system of equations using substitution.
    +
    Solve the system of equations using substitution.
    1.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {X = -3r+25} \\ -\amp {X = 4r-17} +\amp {L = 3B-14} \\ +\amp {L = 2B-10} \end{alignedat} \right.\)
    -
    Answer.
    \(r = 6, X = 7\)
    +
    Answer.
    \(B = 4, L = -2\)
    2.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {F = -3w-6} \\ -\amp {F = -w} +\amp {u = 3H+9} \\ +\amp {u = -3H-21} \end{alignedat} \right.\)
    -
    Answer.
    \(w = -3, F = 3\)
    +
    Answer.
    \(H = -5, u = -6\)
    3.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {m = -3\mathopen{}\left(C-3\right)} \\ -\amp {m = 4\mathopen{}\left(C+3\right)-31} +\amp {c = 3\mathopen{}\left(M+4\right)-13} \\ +\amp {c = 2\mathopen{}\left(M-2\right)+5} \end{alignedat} \right.\)
    -
    Answer.
    \(C = 4, m = -3\)
    +
    Answer.
    \(M = 2, c = 5\)
    4.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {V = -3\mathopen{}\left(H-4\right)-33} \\ -\amp {V = -\left(H-1\right)-12} +\amp {I = 3\mathopen{}\left(S+3\right)+14} \\ +\amp {I = -3\mathopen{}\left(S+5\right)-4} \end{alignedat} \right.\)
    -
    Answer.
    \(H = -5, V = -6\)
    +
    Answer.
    \(S = -7, I = 2\)
    5.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {1N+5C = 27} \\ -\amp {4N+\left(-5\right)C = -17} +\amp {3Z+\left(-3\right)r = 9} \\ +\amp {2Z+\left(-1\right)r = 2} \end{alignedat} \right.\)
    -
    Answer.
    \(N = 2, C = 5\)
    +
    Answer.
    \(Z = -1, r = -4\)
    6.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-4T+\left(-1\right)k = 26} \\ -\amp {-2T+1k = 16} +\amp {-1e+\left(-2\right)a = 9} \\ +\amp {-3e+4a = -43} \end{alignedat} \right.\)
    -
    Answer.
    \(T = -7, k = 2\)
    +
    Answer.
    \(e = 5, a = -7\)
    7.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {S = 4a+4} \\ -\amp {4a+4S = -24} +\amp {H = -2k+\left(-4\right)} \\ +\amp {3k+\left(-3\right)H = -24} \end{alignedat} \right.\)
    -
    Answer.
    \(a = -2, S = -4\)
    +
    Answer.
    \(k = -4, H = 4\)
    8.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {B = 4f+\left(-13\right)} \\ -\amp {-2f+\left(-2\right)B = -24} +\amp {p = -2q+4} \\ +\amp {-4q+3p = -18} \end{alignedat} \right.\)
    -
    Answer.
    \(f = 5, B = 7\)
    +
    Answer.
    \(q = 3, p = -2\)
    9.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {i = -4k-12} \\ -\amp {i = 4\mathopen{}\left(k-2\right)+28} +\amp {Y = 2\mathopen{}\left(w-2\right)+12} \\ +\amp {Y = 3w+14} \end{alignedat} \right.\)
    -
    Answer.
    \(k = -4, i = 4\)
    +
    Answer.
    \(w = -6, Y = -4\)
    10.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {Q = -4\mathopen{}\left(r+2\right)+18} \\ -\amp {Q = -2r+4} +\amp {F = 2B+4} \\ +\amp {F = -4\mathopen{}\left(B+1\right)+14} \end{alignedat} \right.\)
    -
    Answer.
    \(r = 3, Q = -2\)
    +
    Answer.
    \(B = 1, F = 6\)
    11.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-8w+\left(-1\right)z = -5} \\ -\amp {16w+2z = -9} +\amp {G+1m = 5} \\ +\amp {0G+4m = 3} \end{alignedat} \right.\)
    Answer.
    \(\text{no solutions}\)
    12.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {1C+9g = -8} \\ -\amp {4C+36g = -4} +\amp {1M+\left(-4\right)W = 5} \\ +\amp {-3M+12W = 8} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{no solutions}\)
    +
    Answer.
    \(\text{no solutions}\)
    13.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {N = -4H+29} \\ -\amp {4H+1N = 29} +\amp {C = -3S+5} \\ +\amp {3S+1C = 5} \end{alignedat} \right.\)
    Answer.
    \(\text{infinitely many solutions}\)
    14.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {v = 2N+4} \\ -\amp {-2N+1v = 4} +\amp {k = 4Z+25} \\ +\amp {-4Z+1k = 25} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    15.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-5T+4e = -48} \\ -\amp {e = -\left(T-\left(-3\right)\right)+\left(-7\right)} +\amp {3e+\left(-4\right)T = 10} \\ +\amp {T = -5\mathopen{}\left(e-3\right)+\left(-6\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(T = 4, e = -7\)
    +
    Answer.
    \(e = 2, T = -1\)
    16.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-5Z+4K = 45} \\ -\amp {K = 2\mathopen{}\left(Z-1\right)+17} +\amp {3k+\left(-3\right)B = -12} \\ +\amp {B = 4\mathopen{}\left(k-\left(-4\right)\right)+9} \end{alignedat} \right.\)
    -
    Answer.
    \(Z = -5, K = 5\)
    +
    Answer.
    \(k = -7, B = -3\)
    17.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {2f+1u = 0} \\ -\amp {u = -2\mathopen{}\left(f-3\right)+\left(-15\right)} +\amp {-2q+1i = 9} \\ +\amp {i = 2\mathopen{}\left(q-2\right)+4} \end{alignedat} \right.\)
    Answer.
    \(\text{no solutions}\)
    18.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-5k+1c = -5} \\ -\amp {c = 5\mathopen{}\left(k-\left(-2\right)\right)+\left(-19\right)} +\amp {4v+1R = 4} \\ +\amp {R = -4\mathopen{}\left(v-\left(-3\right)\right)+12} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{no solutions}\)
    +
    Answer.
    \(\text{no solutions}\)
    19.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-5r+\left(-4\right)J = -23} \\ -\amp {4r+4J = 24} +\amp {2B+2y = -10} \\ +\amp {3B+\left(-4\right)y = -8} \end{alignedat} \right.\)
    -
    Answer.
    \(r = -1, J = 7\)
    +
    Answer.
    \(B = -4, y = -1\)
    20.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-5w+\left(-4\right)r = -33} \\ -\amp {-2w+\left(-5\right)r = -20} +\amp {2G+3g = -12} \\ +\amp {-4G+\left(-2\right)g = 0} \end{alignedat} \right.\)
    -
    Answer.
    \(w = 5, r = 2\)
    +
    Answer.
    \(G = 3, g = -6\)
    21.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {a = 9B-4} \\ -\amp {a = 5B-2} +\amp {N = M-7} \\ +\amp {N = -8M+7} \end{alignedat} \right.\)
    -
    Answer.
    \(B = {\frac{1}{2}}, a = {\frac{1}{2}}\)
    +
    Answer.
    \(M = {\frac{14}{9}}, N = -{\frac{49}{9}}\)
    22.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {G = 9H+3} \\ -\amp {G = 4H-8} +\amp {w = -S} \\ +\amp {w = 7S+2} \end{alignedat} \right.\)
    -
    Answer.
    \(H = -{\frac{11}{5}}, G = -{\frac{84}{5}}\)
    +
    Answer.
    \(S = -{\frac{1}{4}}, w = {\frac{1}{4}}\)
    23.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {20M+4p = -100} \\ -\amp {p = -5\mathopen{}\left(M-\left(-2\right)\right)+\left(-15\right)} +\amp {-6Z+3e = -54} \\ +\amp {e = 2\mathopen{}\left(Z-4\right)+\left(-10\right)} \end{alignedat} \right.\)
    Answer.
    \(\text{infinitely many solutions}\)
    24.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-15T+\left(-3\right)Y = -21} \\ -\amp {Y = -5\mathopen{}\left(T-1\right)+2} +\amp {8e+\left(-4\right)L = 12} \\ +\amp {L = 2\mathopen{}\left(e-5\right)+7} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    25.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {E = 4\mathopen{}\left(Z-8\right)+8} \\ -\amp {E = -3\mathopen{}\left(Z+5\right)+4} +\amp {v = -7\mathopen{}\left(j-5\right)-1} \\ +\amp {v = -3\mathopen{}\left(j-4\right)+4} \end{alignedat} \right.\)
    -
    Answer.
    \(Z = {\frac{13}{7}}, E = -{\frac{116}{7}}\)
    +
    Answer.
    \(j = {\frac{9}{2}}, v = {\frac{5}{2}}\)
    26.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {n = 5\mathopen{}\left(f+3\right)+8} \\ -\amp {n = 2\mathopen{}\left(f-9\right)-5} +\amp {c = -5\mathopen{}\left(q+6\right)-1} \\ +\amp {c = 2q-7} \end{alignedat} \right.\)
    -
    Answer.
    \(f = -{\frac{46}{3}}, n = -{\frac{161}{3}}\)
    +
    Answer.
    \(q = -{\frac{24}{7}}, c = -{\frac{97}{7}}\)
    27.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {9k+1V = 8} \\ -\amp {-1k+\left(-1\right)V = 6} +\amp {-1v+1J = 6} \\ +\amp {v+3J = -5} \end{alignedat} \right.\)
    -
    Answer.
    \(k = {\frac{7}{4}}, V = -{\frac{31}{4}}\)
    +
    Answer.
    \(v = -{\frac{23}{3}}, J = -{\frac{5}{3}}\)
    28.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {-6q+9D = -6} \\ -\amp {1q+\left(-1\right)D = -4} +\amp {-2B+1s = 1} \\ +\amp {-6B+2s = 6} \end{alignedat} \right.\)
    -
    Answer.
    \(q = -14, D = -10\)
    +
    Answer.
    \(B = -2, s = -3\)
    29.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {5w+\left(-2\right)k = -1} \\ -\amp {k = -1w+\left(-7\right)} +\amp {4G+4a = -6} \\ +\amp {a = 1G+\left(-2\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(w = -{\frac{15}{7}}, k = -{\frac{34}{7}}\)
    +
    Answer.
    \(G = {\frac{1}{4}}, a = -{\frac{7}{4}}\)
    30.
    -
    -
    -
    PTX:ERROR: WeBWorK problem webwork-1459 with seed 1459 is either empty or failed to compile Use -a to halt with full PG and returned content
    +
    +
    +
    +
    \(\left\{ +\begin{alignedat}{4} +\amp {-2L+2G = -5} \\ +\amp {G = 3L+6} +\end{alignedat} +\right.\)
    +
    Answer.
    \(L = -{\frac{17}{4}}, G = -{\frac{27}{4}}\)
    +
    31.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {A = 7\mathopen{}\left(H-\left(-2\right)\right)+\left(-6\right)} \\ -\amp {5H+\left(-2\right)A = 1} +\amp {q = 4\mathopen{}\left(S-2\right)+10} \\ +\amp {-8S+4q = -9} \end{alignedat} \right.\)
    -
    Answer.
    \(H = -{\frac{17}{9}}, A = -{\frac{47}{9}}\)
    +
    Answer.
    \(S = -{\frac{17}{8}}, q = -{\frac{13}{2}}\)
    32.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {i = -\left(M-\left(-1\right)\right)+\left(-4\right)} \\ -\amp {-3M+\left(-5\right)i = -9} +\amp {X = -\left(Y-4\right)+2} \\ +\amp {-5Y+\left(-1\right)X = -3} \end{alignedat} \right.\)
    -
    Answer.
    \(M = {\frac{29}{3}}, i = -4\)
    +
    Answer.
    \(Y = {\frac{1}{5}}, X = 2\)
    33.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {-4T+6P = 1} \\ -\amp {-3T+5P = -2} +\amp {-2e+\left(-7\right)F = -2} \\ +\amp {e+\left(-4\right)F = 5} \end{alignedat} \right.\)
    -
    Answer.
    \(T = -{\frac{17}{2}}, P = -{\frac{11}{2}}\)
    +
    Answer.
    \(e = {\frac{43}{8}}, F = -{\frac{5}{4}}\)
    34.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {2Z+5y = -9} \\ -\amp {3Z+4y = -5} +\amp {j+\left(-3\right)p = 9} \\ +\amp {2j+3p = 5} \end{alignedat} \right.\)
    -
    Answer.
    \(Z = {\frac{11}{7}}, y = -{\frac{17}{7}}\)
    +
    Answer.
    \(j = 7, p = -3\)
    35.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {h = -6.4e+5.1} \\ -\amp {h = 0.1e+4} +\amp {W = 7.1q-2.9} \\ +\amp {W = 7.7q+2.3} \end{alignedat} \right.\)
    -
    Answer.
    \(e = -0.169231, h = 6.18308\)
    +
    Answer.
    \(q = 8.66667, W = 58.6333\)
    36.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {N = 0.3k+4.9} \\ -\amp {N = -4.4k-4.1} +\amp {D = -2.2v-3.1} \\ +\amp {D = 3.2v-5.8} \end{alignedat} \right.\)
    -
    Answer.
    \(k = 1.91489, N = 5.47447\)
    +
    Answer.
    \(v = -0.5, D = -2\)
    37.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {7.5q+6.3x = 7.1} \\ -\amp {7.7q+\left(-1\right)x = -3} +\amp {-1A+2.3k = -0.4} \\ +\amp {3.4A+5.1k = 0.4} \end{alignedat} \right.\)
    -
    Answer.
    \(q = -0.210677, x = 1.37779\)
    +
    Answer.
    \(A = 0.229102, k = -0.0743034\)
    38.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {2.8w+6.2e = -4.1} \\ -\amp {5.4w+\left(-1\right)e = 4.3} +\amp {1G+2.2T = 4.5} \\ +\amp {1.1G+1T = 7.7} \end{alignedat} \right.\)
    -
    Answer.
    \(w = 0.62183, e = -0.942117\)
    +
    Answer.
    \(G = 8.76056, T = -1.93662\)
    39.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {L = {\frac{5}{2}}B+{\frac{3}{4}}} \\ -\amp {L = {\frac{2}{5}}B+{\frac{5}{2}}} +\amp {A = {\frac{1}{3}}L - {\frac{1}{3}}} \\ +\amp {A = {\frac{2}{5}}L+{\frac{2}{3}}} \end{alignedat} \right.\)
    -
    Answer.
    \(B = {\frac{5}{6}}, L = {\frac{17}{6}}\)
    +
    Answer.
    \(L = -15, A = -{\frac{16}{3}}\)
    40.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {u = {\frac{1}{4}}H - {\frac{5}{3}}} \\ -\amp {u = {\frac{1}{3}}H+{\frac{1}{4}}} +\amp {i = {\frac{1}{2}}S+{\frac{2}{5}}} \\ +\amp {i = {\frac{3}{5}}S - {\frac{1}{6}}} \end{alignedat} \right.\)
    -
    Answer.
    \(H = -23, u = -{\frac{89}{12}}\)
    +
    Answer.
    \(S = {\frac{17}{3}}, i = {\frac{97}{30}}\)
    41.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {c = {\frac{2}{3}}\mathopen{}\left(M+3\right) - {\frac{3}{4}}} \\ -\amp {c = {\frac{1}{4}}\mathopen{}\left(M+5\right) - {\frac{5}{12}}} +\amp {Q = {\frac{3}{2}}\mathopen{}\left(Y-2\right)+{\frac{8}{3}}} \\ +\amp {Q = {\frac{2}{3}}\mathopen{}\left(Y+2\right)-2} \end{alignedat} \right.\)
    -
    Answer.
    \(M = -1, c = {\frac{7}{12}}\)
    +
    Answer.
    \(Y = -{\frac{2}{5}}, Q = -{\frac{14}{15}}\)
    42.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {I = {\frac{1}{2}}\mathopen{}\left(S-1\right)+{\frac{3}{10}}} \\ -\amp {I = {\frac{3}{5}}\mathopen{}\left(S-5\right)+{\frac{13}{4}}} +\amp {z = {\frac{5}{6}}\mathopen{}\left(e-3\right)+4} \\ +\amp {z = {\frac{5}{2}}\mathopen{}\left(e+3\right) - {\frac{73}{10}}} \end{alignedat} \right.\)
    -
    Answer.
    \(S = -{\frac{9}{2}}, I = -{\frac{49}{20}}\)
    +
    Answer.
    \(e = {\frac{39}{50}}, z = {\frac{43}{20}}\)
    43.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {0.4Z+0.4r = -0.8} \\ -\amp {1Z+1.5r = 0.75} +\amp {0.6j+1g = -1.25} \\ +\amp {0.5j+1.33333g = 0.25} \end{alignedat} \right.\)
    -
    Answer.
    \(Z = -{\frac{15}{2}}, r = {\frac{77}{8}}\)
    +
    Answer.
    \(j = -{\frac{115}{18}}, g = {\frac{115}{36}}\)
    44.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {1.66667e+0.4a = 2.5} \\ -\amp {1e+0.25a = -0.25} +\amp {0.8p+0.333333P = -0.666667} \\ +\amp {1.5p+\left(-1\right)P = 0.4} \end{alignedat} \right.\)
    -
    Answer.
    \(e = {\frac{87}{2}}, a = -{\frac{615}{4}}\)
    +
    Answer.
    \(p = -{\frac{16}{39}}, P = -{\frac{6}{13}}\)
    45.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {0.166667k+1.5H = 1.5} \\ -\amp {1.33333k+12H = -0.8} +\amp {0.333333v+0.333333x = -0.666667} \\ +\amp {0.2v+0.2x = 0.8} \end{alignedat} \right.\)
    Answer.
    \(\text{no solutions}\)
    46.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.8q+\left(-1.33333\right)p = 0.333333} \\ -\amp {1.33333q+\left(-2.22222\right)p = -0.75} +\amp {0.75A+1.66667e = -0.2} \\ +\amp {0.25A+0.555556e = 1.25} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{no solutions}\)
    +
    Answer.
    \(\text{no solutions}\)
    47.
    -
    +
    -
    PTX:ERROR: WeBWorK problem webwork-substitution-fraction-coefficients-standard-slope-intercept with seed 1476 is either empty or failed to compile Use -a to halt with full PG and returned content
    +
    +
    \(\left\{ +\begin{alignedat}{4} +\amp {M = -2G+\left(-1.5\right)} \\ +\amp {1.33333G+\left(-0.666667\right)M = -1.5} +\end{alignedat} +\right.\)
    +
    Answer.
    \(G = -{\frac{15}{16}}, M = -{\frac{3}{2}}\)
    +
    48.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.75B+1F = 0.2} \\ -\amp {F = -3.75B+0.5} +\amp {1L+\left(-0.166667\right)u = 1.25} \\ +\amp {u = 3.33333L+\left(-6.66667\right)} \end{alignedat} \right.\)
    -
    Answer.
    \(B = {\frac{1}{10}}, F = {\frac{7}{36}}\)
    +
    Answer.
    \(L = {\frac{5}{16}}, u = {\frac{99}{8}}\)
    49.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {m = 2.08333\mathopen{}\left(G-5\right)+\left(-8.91667\right)} \\ -\amp {1.33333G+1m = 1.5} +\amp {c = 5\mathopen{}\left(S-2\right)+\left(-7.5\right)} \\ +\amp {0.666667S+0.4c = 1.5} \end{alignedat} \right.\)
    -
    Answer 1.
    \(G = 0, m = {\frac{69}{41}}\)
    Answer 2.
    \(16\)
    +
    Answer 1.
    \(S = {\frac{3}{16}}, c = {\frac{7}{4}}\)
    Answer 2.
    \(16\)
    50.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {W = 1\mathopen{}\left(M-5\right)+\left(-8.75\right)} \\ -\amp {0.666667M+\left(-1\right)W = 0.333333} +\amp {J = 1\mathopen{}\left(Y-5\right)+\left(-8.75\right)} \\ +\amp {1Y+\left(-0.2\right)J = -0.833333} \end{alignedat} \right.\)
    -
    Answer 1.
    \(M = {\frac{41}{4}}, W = 9\)
    Answer 2.
    \(16\)
    +
    Answer 1.
    \(Y = -{\frac{95}{48}}, J = -{\frac{175}{48}}\)
    Answer 2.
    \(16\)
    51.
    -
    +
    \(\left\{ \begin{alignedat}{4} -\amp {2.5S+\left(-0.25\right)C = 0.4} \\ -\amp {0.6S+\left(-0.06\right)C = 0.096} +\amp {0.6d+\left(-0.75\right)t = -0.666667} \\ +\amp {0.333333d+\left(-0.416667\right)t = -0.37037} \end{alignedat} \right.\)
    Answer.
    \(\text{infinitely many solutions}\)
    52.
    -
    -
    +
    +
    -
    \(\left\{ +
    \(\left\{ \begin{alignedat}{4} -\amp {0.2Z+0.6k = 0.333333} \\ -\amp {0.6Z+1.8k = 1} +\amp {0.5j+0.2a = -0.25} \\ +\amp {0.333333j+0.133333a = -0.166667} \end{alignedat} \right.\)
    -
    Answer.
    \(\text{infinitely many solutions}\)
    +
    Answer.
    \(\text{infinitely many solutions}\)
    -

    Applications

    +

    Applications

    53.
    -
    +
    -
    A rectangle’s length is \({10\ {\rm ft}}\) shorter than two times its width. The rectangle’s perimeter is \({161\ {\rm ft}}\text{.}\) Find the rectangle’s length and width.
    +
    A rectangle’s length is \({8\ {\rm ft}}\) shorter than four times its width. The rectangle’s perimeter is \({145\ {\rm ft}}\text{.}\) Find the rectangle’s length and width.
    -
    Answer 1.
    \(L = 2W-10\)
    Answer 2.
    \(2L+2W = 161\)
    Answer 3.
    \(50.3333\ {\rm ft}\)
    Answer 4.
    \(30.1667\ {\rm ft}\)
    +
    Answer 1.
    \(L = 4W-8\)
    Answer 2.
    \(2L+2W = 145\)
    Answer 3.
    \(56.4\ {\rm ft}\)
    Answer 4.
    \(16.1\ {\rm ft}\)
    54.
    -
    +
    -
    A school fund raising event sold a total of \(149\) tickets and generated a total revenue of \({\$621.50}\text{.}\) Each adult ticket cost \({\$5.50}\text{,}\) and each child ticket cost \({\$1}\text{.}\) How many adult tickets and how many child tickets were sold?
    +
    A school fund raising event sold a total of \(150\) tickets and generated a total revenue of \({\$834}\text{.}\) Each adult ticket cost \({\$7}\text{,}\) and each child ticket cost \({\$2.50}\text{.}\) How many adult tickets and how many child tickets were sold?
    -
    Answer 1.
    \(A+C = 149\)
    Answer 2.
    \(5.5A+1C = 621.5\)
    Answer 3.
    \(105\)
    Answer 4.
    \(44\)
    +
    Answer 1.
    \(A+C = 150\)
    Answer 2.
    \(7A+2.5C = 834\)
    Answer 3.
    \(102\)
    Answer 4.
    \(48\)
    55.
    -
    +
    -
    One telecom company charges a monthly fee of \({\$32.95}\) and \({\$5}\) for each Gigabyte (GB) of data transmitted. A rival telecom company charges a monthly fee of \({\$38.95}\) and \({\$3.45}\) for each GB of data transmitted. How many GB of data would you have to use for the monthly bill to be the same for the two providers? And what would that monthly bill be?
    +
    One telecom company charges a monthly fee of \({\$36.95}\) and \({\$4.45}\) for each Gigabyte (GB) of data transmitted. A rival telecom company charges a monthly fee of \({\$46.95}\) and \({\$3.25}\) for each GB of data transmitted. How many GB of data would you have to use for the monthly bill to be the same for the two providers? And what would that monthly bill be?
    -
    Answer 1.
    \(B = 32.95+5D\)
    Answer 2.
    \(B = 38.95+3.45D\)
    Answer 3.
    \(3.87097\ {\rm GB}\)
    Answer 4.
    \(\$52.30\)
    +
    Answer 1.
    \(B = 36.95+4.45D\)
    Answer 2.
    \(B = 46.95+3.25D\)
    Answer 3.
    \(8.33333\ {\rm GB}\)
    Answer 4.
    \(\$74.03\)
    56.
    -
    +
    -
    A local restaurant has two locations. At one location, the revenue this month is \({\$88{,}000}\) but it has been decreasing by \({\$4{,}500}\) per month. At the other location, the annual revenue this month is \({\$57{,}000}\) and it has been increasing by \({\$3{,}000}\) per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be?
    +
    A local restaurant has two locations. At one location, the revenue this month is \({\$93{,}000}\) but it has been decreasing by \({\$3{,}500}\) per month. At the other location, the annual revenue this month is \({\$53{,}000}\) and it has been increasing by \({\$3{,}000}\) per month. How long will it be until the two restaurant locations are taking in the same monthly revenue? And what will that monthly revenue be?
    -
    Answer 1.
    \(R = 88000-4500t\)
    Answer 2.
    \(R = 57000+3000t\)
    Answer 3.
    \(4.13333\ {\rm months}\)
    Answer 4.
    \(\$0\)
    +
    Answer 1.
    \(R = 93000-3500t\)
    Answer 2.
    \(R = 53000+3000t\)
    Answer 3.
    \(6.15385\ {\rm months}\)
    Answer 4.
    \(\$0\)
    57.
    -
    +
    -
    An algebra exam has \(30\) questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth \(2\) points, and short-answer questions, each worth \(7\) points. How many questions are there of each type?
    +
    An algebra exam has \(26\) questions, worth a total of 100 points. There are two types of question on the exam. There are multiple-choice questions each worth \(2\) points, and short-answer questions, each worth \(8\) points. How many questions are there of each type?
    -
    Answer 1.
    \(M+S = 30\)
    Answer 2.
    \(2M+7S = 100\)
    Answer 3.
    \(22\)
    Answer 4.
    \(8\)
    +
    Answer 1.
    \(M+S = 26\)
    Answer 2.
    \(2M+8S = 100\)
    Answer 3.
    \(18\)
    Answer 4.
    \(8\)
    58.
    -
    +
    -
    Lizette invested a total of \({\$4{,}200}\) in two investments. Her savings account pays \({1\%}\) interest annually. A riskier stock investment earned \({6\%}\) at the end of the year. At the end of the year, Lizette earned a total of \({\$197}\) in interest. How much money did she invest in each account?
    +
    Rocky invested a total of \({\$3{,}700}\) in two investments. His savings account pays \({2.5\%}\) interest annually. A riskier stock investment earned \({5.5\%}\) at the end of the year. At the end of the year, Rocky earned a total of \({\$155.50}\) in interest. How much money did he invest in each account?
    -
    Answer 1.
    \(x+y = 4200\)
    Answer 2.
    \(0.01x+0.06y = 197\)
    Answer 3.
    \(\$1{,}100\)
    Answer 4.
    \(\$3{,}100\)
    +
    Answer 1.
    \(x+y = 3700\)
    Answer 2.
    \(0.025x+0.055y = 155.5\)
    Answer 3.
    \(\$1{,}600\)
    Answer 4.
    \(\$2{,}100\)
    59.
    -
    +
    -
    Molly invested a total of \({\$5{,}000}\) in two investments. Her savings account pays \({2.5\%}\) interest annually. A riskier stock investment lost \({3.5\%}\) at the end of the year. At the end of the year, Molly’s total fell from \({\$5{,}000}\) to \({\$4{,}844}\text{.}\) How much money did she invest in each account?
    +
    Tyler invested a total of \({\$3{,}400}\) in two investments. Their savings account pays \({2\%}\) interest annually. A riskier stock investment lost \({7\%}\) at the end of the year. At the end of the year, Tyler’s total fell from \({\$3{,}400}\) to \({\$3{,}227}\text{.}\) How much money did they invest in each account?
    -
    Answer 1.
    \(x+y = 5000\)
    Answer 2.
    \(0.025x-0.035y = -156\)
    Answer 3.
    \(\$1{,}900\)
    Answer 4.
    \(\$3{,}100\)
    +
    Answer 1.
    \(x+y = 3400\)
    Answer 2.
    \(0.02x-0.07y = -173\)
    Answer 3.
    \(\$1{,}300\)
    Answer 4.
    \(\$2{,}100\)
    60.
    -
    +
    -
    Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with \({44\%}\) married. Balsamburg had a population with \({52\%}\) married. After the merge, the new city has a total of \(4500\) residents, with \({49.5\%}\) married. How many residents did each town have before the merge?
    +
    Asterton and Balsamburg were two towns located close to each other, and recently they merged into one city. Asterton had a population with \({40\%}\) Democrats. Balsamburg had a population with \({55\%}\) Democrats. After the merge, the new city has a total of \(4000\) residents, with \({50.5\%}\) Democrats. How many residents did each town have before the merge?
    -
    Answer 1.
    \(A+B = 4500\)
    Answer 2.
    \(0.44A+0.52B = 2227.5\)
    Answer 3.
    \(1406\)
    Answer 4.
    \(3094\)
    +
    Answer 1.
    \(A+B = 4000\)
    Answer 2.
    \(0.4A+0.55B = 2020\)
    Answer 3.
    \(1200\)
    Answer 4.
    \(2800\)
    61.
    -
    +
    -
    Xzavier poured some \({8\%}\) alcohol solution and some \({20\%}\) alcohol solution together into a beaker, and then the beaker had \({500\ {\rm ml}}\) of \({12.8\%}\) alcohol solution. How much of each solution did Xzavier pour into the beaker?
    +
    Colin poured some \({11\%}\) alcohol solution and some \({25\%}\) alcohol solution together into a beaker, and then the beaker had \({580\ {\rm ml}}\) of \({15.3\%}\) alcohol solution. How much of each solution did Colin pour into the beaker?
    -
    Answer 1.
    \(x+y = 500\)
    Answer 2.
    \(0.08x+0.2y = 64\)
    Answer 3.
    \(300\ {\rm ml}\)
    Answer 4.
    \(200\ {\rm ml}\)
    +
    Answer 1.
    \(x+y = 580\)
    Answer 2.
    \(0.11x+0.25y = 88.74\)
    Answer 3.
    \(400\ {\rm ml}\)
    Answer 4.
    \(180\ {\rm ml}\)
    62.
    -
    +
    -
    A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost \({\$1.61}\) per pound, and cashews cost \({\$5.90}\) per pound. The company is targeting a product that will cost them \({\$2.70}\) per pound worth of ingredients. How much of each type of nut should go into a bag?
    +
    A snack company will produce and sell 1-lb bags with a mixture of peanuts and cashews. Peanuts cost \({\$1.65}\) per pound, and cashews cost \({\$5.82}\) per pound. The company is targeting a product that will cost them \({\$2.60}\) per pound worth of ingredients. How much of each type of nut should go into a bag?
    -
    Answer 1.
    \(P+C = 1\)
    Answer 2.
    \(1.61P+5.9C = 2.7\)
    Answer 3.
    \(0.745921\ {\rm lb}\)
    Answer 4.
    \(0.254079\ {\rm lb}\)
    +
    Answer 1.
    \(P+C = 1\)
    Answer 2.
    \(1.65P+5.82C = 2.6\)
    Answer 3.
    \(0.772182\ {\rm lb}\)
    Answer 4.
    \(0.227818\ {\rm lb}\)
    diff --git a/section-summary-of-graphing-lines.html b/section-summary-of-graphing-lines.html index 53d82cc9f..727dbe051 100644 --- a/section-summary-of-graphing-lines.html +++ b/section-summary-of-graphing-lines.html @@ -171,7 +171,7 @@

    Search Results:

    "autoHeight": true, "disableAutoScale": false}, true); -PrevUpNext
    -PrevTopNext +PrevTopNext
    diff --git a/section-variables-and-evaluating-expressions.html b/section-variables-and-evaluating-expressions.html index 974e20d51..4d9995647 100644 --- a/section-variables-and-evaluating-expressions.html +++ b/section-variables-and-evaluating-expressions.html @@ -420,7 +420,7 @@

    Search Results:

    @@ -455,6 +455,20 @@

    Search Results:

  • 3.9.5 Exercises
    • +
  • +
    3.10 Graphing Lines Chapter Review
    + +