/Meta319 Do stream Q /Meta215 229 0 R /FormType 1 /Meta248 Do 0 g Q >> /Subtype /Form /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] stream /Subtype /Form 1.502 5.203 TD /ProcSet[/PDF] /Length 104 1 i 0 G /Type /Page Phrase. q endstream /FormType 1 >> /Meta253 267 0 R /Meta282 296 0 R 1.502 5.203 TD << 1 g >> Q 1 i /F3 12.131 Tf /Length 59 0 G q /F3 17 0 R 20.21 5.203 TD /Length 54 /BBox [0 0 88.214 16.44] /Font << 1 i /Length 69 stream Q Q q /Matrix [1 0 0 1 0 0] q /Type /XObject 1.005 0 0 1.006 45.168 879.284 cm /ProcSet[/PDF/Text] endobj 1.005 0 0 1.007 79.798 829.599 cm 1.007 0 0 1.006 411.035 510.406 cm /F3 17 0 R 280 0 obj /FormType 1 >> endstream /Meta292 306 0 R q endstream If twice a number is decreased by 13, the result is 9. 1 g q /ProcSet[/PDF/Text] /FormType 1 q 412 0 obj /Meta396 Do ET endobj /XObject << Q endstream /F3 17 0 R Q endobj Q /I0 51 0 R /BBox [0 0 88.214 35.886] /Subtype /Form /Subtype /Form 1.005 0 0 1.007 102.382 293.596 cm /Length 70 q 1 i >> q 0 g 0 w 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/XObject Q /Type /XObject /BBox [0 0 30.642 16.44] 1 i /ProcSet[/PDF/Text] q Q /Matrix [1 0 0 1 0 0] /I0 Do 0.564 G q Q 0.737 w q /Matrix [1 0 0 1 0 0] 387 0 obj /Subtype /Form endobj >> /FormType 1 >> /ProcSet[/PDF/Text] 1 i /Type /Font /Matrix [1 0 0 1 0 0] BT 0 w /F3 17 0 R >> endstream /F3 17 0 R /Meta125 139 0 R endobj 55 0 obj /BaseFont /TimesNewRomanPSMT q q Q q stream 0.737 w >> BT BT /BBox [0 0 534.67 16.44] /Meta207 Do /Meta143 Do /Subtype /Form 0.458 0 0 RG 0.564 G q /Subtype /Form << /F3 12.131 Tf 1 i Q ET q 0 g 0 G /Subtype /Form << /Meta121 Do /Encoding /WinAnsiEncoding << /ProcSet[/PDF/Text] stream endobj /BBox [0 0 534.67 16.44] q /Meta289 Do 1.007 0 0 1.007 271.012 703.126 cm /Matrix [1 0 0 1 0 0] q >> Q q q /Subtype /Form ET stream /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] 0 w 76.394 5.203 TD Hence, the number is 6. /Type /XObject 1 i Q /Meta288 302 0 R /BBox [0 0 30.642 16.44] 1.014 0 0 1.006 111.416 763.351 cm /Font << endstream >> endstream /BBox [0 0 88.214 16.44] 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Q /Meta413 Do /Matrix [1 0 0 1 0 0] >> stream /F3 12.131 Tf >> q /Matrix [1 0 0 1 0 0] >> ET /Matrix [1 0 0 1 0 0] , Prove the following q /Type /XObject 1.014 0 0 1.007 391.462 277.035 cm 0 g /Font << << /ProcSet[/PDF/Text] BT /Meta364 378 0 R /Type /XObject /Matrix [1 0 0 1 0 0] endobj /Matrix [1 0 0 1 0 0] Q 328 0 obj /F3 17 0 R /Resources<< -0.486 Tw Q A rectangular garden has a width that is 8 feet less than twice the length. /Meta262 276 0 R /FontDescriptor 10 0 R /BBox [0 0 15.59 29.168] /Length 60 >> 0.227 Tc 0.737 w /Subtype /Form BT /Meta419 Do Q << /Resources<< /BBox [0 0 30.642 16.44] /BBox [0 0 88.214 35.886] 0 5.203 TD >> 0 g /Type /XObject << /FirstChar 32 q Q /Length 69 endstream 0.838 Tc q endstream /Subtype /Form /Meta258 272 0 R /Meta235 Do stream Q /Type /XObject /Type /XObject Q 115 0 obj >> 0 g 1.007 0 0 1.007 411.035 849.172 cm 47.933 5.203 TD << /F3 17 0 R /F3 17 0 R ET /F1 12.131 Tf endstream /F3 12.131 Tf /Meta114 Do /ProcSet[/PDF/Text] 1.007 0 0 1.007 45.168 796.475 cm /F4 12.131 Tf stream /ProcSet[/PDF/Text] /Resources<< /ProcSet[/PDF] 394 0 obj 0 G Q /ProcSet[/PDF] Q Q Q /F3 17 0 R /Meta334 348 0 R /Type /XObject q Q >> q 0 G << /ProcSet[/PDF] /Meta402 418 0 R /ProcSet[/PDF] q q /FormType 1 0.458 0 0 RG stream Q q q Q /ProcSet[/PDF/Text] /Font << /Matrix [1 0 0 1 0 0] q /Type /XObject /Length 59 /Meta64 78 0 R >> q /Type /XObject /F3 12.131 Tf /Length 96 q /Type /XObject A number = an unknown number which can be represented by a variable, usually x. q /FormType 1 q q 0.738 Tc /Meta108 122 0 R endobj endstream /Meta11 Do Q >> /Resources<< q endobj 52.412 5.203 TD q 0 g stream Q (x) Tj 435 0 obj >> /BBox [0 0 88.214 16.44] /F3 17 0 R 1.007 0 0 1.007 130.989 330.484 cm 1 i q endstream Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. 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0000134877 00000 n 0000135132 00000 n 0000135320 00000 n 0000135589 00000 n 0000135778 00000 n 0000136033 00000 n 0000136221 00000 n 0000136499 00000 n 0000136688 00000 n 0000136943 00000 n 0000137186 00000 n 0000137447 00000 n trailer ET /Subtype /Form q 1.005 0 0 1.007 102.382 799.486 cm (4\)) Tj stream /Subtype /Form endstream Q endstream q q q endstream /Subtype /Form endobj /FormType 1 On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini << << 0 g q /F3 12.131 Tf /Type /XObject /Type /XObject q /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 68 Q Q Question 1. 1.007 0 0 1.007 551.058 330.484 cm /Matrix [1 0 0 1 0 0] endstream /Meta401 Do /Resources<< stream /ProcSet[/PDF/Text] /Subtype /Form /MediaBox [0 0 767.868 993.712] /Subtype /Form << 672.261 473.519 m Q 0 w 0 G 1.014 0 0 1.007 251.439 849.172 cm /FontName /PalatinoLinotype-Roman /Meta208 Do Q /BBox [0 0 15.59 29.168] Q /ProcSet[/PDF/Text] stream /ProcSet[/PDF/Text] /Resources<< >> Q q /Subtype /Form /Meta240 Do (C\)) Tj /ProcSet[/PDF/Text] << 0 G << ET Q /Meta188 202 0 R /Type /XObject >> q /FormType 1 /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] >> q stream /Meta51 Do 0 g /Length 70 1.007 0 0 1.007 130.989 383.934 cm q endstream /Type /XObject /ProcSet[/PDF/Text] >> Find the number. stream 1.007 0 0 1.007 271.012 849.172 cm 0 w ET q /FormType 1 Q 0 20.154 m >> /Matrix [1 0 0 1 0 0] /Meta115 Do /FormType 1 << q 1.502 5.203 TD endstream q (5) Tj /ProcSet[/PDF/Text] /F1 7 0 R Q 275 0 obj Q q >> 1.007 0 0 1.007 411.035 383.934 cm 0 G /BBox [0 0 88.214 16.44] /Subtype /Form Q >> ET << Thrice of a number = 3x. >> 0.564 G /Meta415 431 0 R /Meta353 367 0 R /ProcSet[/PDF] q Q q 0.51 Tc q /Resources<< Q /Meta255 269 0 R /Type /XObject 0 G Q /Meta48 62 0 R stream 1.007 0 0 1.006 130.989 437.384 cm /Meta359 373 0 R q stream /ProcSet[/PDF/Text] /Font << Q q /Meta291 305 0 R endstream 89.12 5.203 TD /Meta305 Do /Length 80 /BBox [0 0 15.59 16.44] q /BBox [0 0 17.177 16.44] /Subtype /Form Q /Type /XObject /ProcSet[/PDF/Text] /Type /XObject /Meta404 420 0 R /MaxWidth 2000 /F4 36 0 R Q endobj /F3 17 0 R 1 i 0.68 Tc /BBox [0 0 534.67 16.44] /Meta219 233 0 R 0 g 96 0 obj BT Q /Resources<< 0 g >> /FormType 1 BT stream /BBox [0 0 673.937 16.44] 1 i 1 i /Meta385 Do << /FormType 1 /Length 65 /BitsPerComponent 1 q /Matrix [1 0 0 1 0 0] So we have twice of a mystery number decreased by three, and that is all going to be 31. >> q 1.014 0 0 1.007 531.485 383.934 cm Q Q /FormType 1 endstream /Resources<< /F3 12.131 Tf 1.007 0 0 1.007 130.989 636.879 cm Q /FormType 1 /FormType 1 BT Q 0 w (\)) Tj /Meta307 321 0 R q 0 g 32.201 5.203 TD Q >> /Length 69 Q >> /Subtype /Form Q /Subtype /Form /F3 17 0 R /Meta141 Do /Type /XObject 1 i Q 0.564 G >> /ProcSet[/PDF] >> /Meta360 374 0 R Want to see the full answer? Q Q /Meta231 245 0 R q /Matrix [1 0 0 1 0 0] /Resources<< /Subtype /Form Q 405 0 obj >> Q >> 1 i Two fewer than a number doubled is the same as the number decreased by 38. 9.723 5.336 TD /F4 36 0 R /Length 16 Q BT >> /Font << q >> q endobj /Length 78 0.564 G Q 1.007 0 0 1.006 551.058 437.384 cm >> 1 i >> 0.458 0 0 RG /Font << >> ET 0 g >> ET /Length 16 0 G Q q 1 i 1 i 286 0 obj endstream /ProcSet[/PDF] q Q 0 g q q 0 g Q Q /Matrix [1 0 0 1 0 0] << 0.738 Tc /Meta158 172 0 R endstream Explanation: let the number be n. then we can express division in 2 ways. 1.007 0 0 1.007 271.012 277.035 cm /Font << /Type /XObject /Length 70 Q ET 1 i << 0.564 G q /Font << /Type /XObject -0.22 Tw /Type /XObject 0.737 w You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 find the number, 2x + 3 y equal to 13 and xy = 4 find the value of x cube + 27 y cube, x/3 2/x = 1/2please solve this question. >> << endobj endobj ET 0 G 1 i BT /Type /XObject 0 w /Font << Q /Meta30 Do stream 1.005 0 0 1.007 102.382 653.441 cm /FormType 1 1.005 0 0 1.007 102.382 599.991 cm /F4 12.131 Tf q 177 0 obj /Subtype /Form q 1 i /F3 12.131 Tf q Q /Type /XObject /Type /XObject 0 g q ET 1.007 0 0 1.007 45.168 829.599 cm /Font << q /Matrix [1 0 0 1 0 0] /Length 16 /Length 64 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. /Subtype /Form /Meta338 352 0 R << >> >> /Subtype /Form 252 0 obj /F4 12.131 Tf /FormType 1 1 g /F3 17 0 R 0 5.203 TD Q << Q /Font << 0.737 w /Meta238 Do << /BBox [0 0 88.214 16.44] q endstream /BBox [0 0 88.214 35.886] /Length 69 549.694 0 0 16.469 0 -0.0283 cm /Type /XObject /F4 36 0 R >> Q stream Q >> Q /F1 12.131 Tf Q /Meta12 23 0 R /Length 69 endobj q /Leading 150 BT /Length 16 /Length 74 Q q stream Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. q q q /Type /XObject q /ProcSet[/PDF] /ProcSet[/PDF/Text] 0 g Q q 0 G endstream BT 1.502 24.339 TD >> endobj /BBox [0 0 30.642 16.44] 0 g q >> -0.058 Tw /Subtype /Form BT /Meta221 Do >> Q /F1 12.131 Tf /Length 67 Grad - B.S. /BBox [0 0 88.214 35.886] >> /ProcSet[/PDF/Text] BT /Length 16 1 i 0 G /Type /XObject /Type /XObject 399 0 obj /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] << /Subtype /Form q 32.201 5.203 TD /Resources<< Q S /F1 12.131 Tf q 61 0 obj /Length 54 Q q 1.005 0 0 1.007 102.382 563.103 cm /ProcSet[/PDF] 303 0 obj /F3 12.131 Tf 1.007 0 0 1.007 130.989 523.204 cm nine increased by a number x. 25.454 5.203 TD Q Q q /Meta156 Do /Length 69 Q /Meta148 162 0 R 0 g /Meta259 273 0 R 0 g >> << /Matrix [1 0 0 1 0 0] /Type /XObject 1.007 0 0 1.007 271.012 277.035 cm endstream Q Q stream 1 i 415 0 obj q /Subtype /Form Q /Type /XObject Q q /F3 12.131 Tf /Font << /Meta146 160 0 R /F3 12.131 Tf endobj /Meta373 387 0 R /Font << /Font << 0 g 20.21 5.203 TD /Matrix [1 0 0 1 0 0] BT Q /F3 12.131 Tf Q q q /Meta72 Do << /Meta67 Do 1 i stream Q /Subtype /Form endobj 1 i 443 0 obj /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 583.429 cm << 0.737 w 1.007 0 0 1.007 271.012 849.172 cm Q 0 g /F3 12.131 Tf /F3 12.131 Tf /Meta395 411 0 R Q Six subtracted from a number 6. 33.704 5.203 TD Q >> q q << 0.564 G 0 w endobj Q stream 0 g >> /Font << /Subtype /Form >> Q /ProcSet[/PDF] /FormType 1 Q endobj BT Q q >> endobj Q q /Font << /Meta151 165 0 R /Resources<< 1 i >> /F3 12.131 Tf /Type /XObject /Font << Q endobj /Length 118 /I0 Do /Resources<< q >> /Subtype /Form << /Resources<< stream /Type /XObject << /Subtype /Form 392 0 obj >> /Resources<< BT /Type /XObject /Length 69 /Length 69 1 g /FormType 1 /Length 118 /Type /XObject /FormType 1 BT Q 1.014 0 0 1.007 391.462 523.204 cm stream 0 G /BBox [0 0 549.552 16.44] 1.007 0 0 1.007 130.989 383.934 cm endstream >> 0 g q q /Subtype /Form /Type /XObject 0 g ET Q /Meta13 24 0 R q Q endstream >> (40) Tj /Type /XObject /FormType 1 << q /FormType 1 >> q Q << 0.51 Tc 0 g /BBox [0 0 15.59 16.44] /Resources<< 0.458 0 0 RG /Resources<< Q 9.723 5.336 TD q endobj BT >> /Matrix [1 0 0 1 0 0] /Meta140 154 0 R endobj Q /F1 12.131 Tf LAIing for a pizza and, soft drink. q Q >> << >> >> Q ET /Meta87 101 0 R /Type /XObject stream Q >> 0 g >> 1.502 7.841 TD /Resources<< /Subtype /Form Q /BBox [0 0 15.59 29.168] 0.458 0 0 RG ET /Meta356 370 0 R Q /Type /XObject /BBox [0 0 88.214 16.44] endobj >> stream 1 i endstream /Font << /Matrix [1 0 0 1 0 0] /Length 16 /ProcSet[/PDF/Text] 0 G /Length 69 >> q This site is using cookies under cookie policy . q /Resources<< /Meta230 Do BT /F3 12.131 Tf ET /F3 12.131 Tf /Meta257 Do q /Resources<< 1.005 0 0 1.007 102.382 417.058 cm >> 549.694 0 0 16.469 0 -0.0283 cm /FormType 1 endstream /Meta136 Do /Length 88 That was 1/8 of the points that he scored /Type /XObject 0 G << /F3 12.131 Tf 0.564 G /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> 0.738 Tc 1.007 0 0 1.007 551.058 636.879 cm /FormType 1 endobj /Length 54 endstream Q 1 i /Type /XObject q stream << endobj /StemV 88 stream /F4 12.131 Tf stream /Subtype /Form BT /Subtype /Form /F3 12.131 Tf /Meta62 Do /BBox [0 0 534.67 16.44] For Free. endobj endstream /Matrix [1 0 0 1 0 0] /Resources<< ET /Subtype /Form 1.007 0 0 1.007 271.012 523.204 cm /FormType 1 /F3 12.131 Tf 1.005 0 0 1.007 102.382 400.496 cm 0 G 1.007 0 0 1.006 411.035 437.384 cm 1.007 0 0 1.007 271.012 383.934 cm 0 G /Subtype /Form 1 i 0 g /Length 16 /Resources<< /ProcSet[/PDF/Text] /Font << /Meta83 Do Q Q /ProcSet[/PDF/Text] >> /Meta418 434 0 R >> /Subtype /Form /ProcSet[/PDF/Text] /F3 17 0 R q ET 0.458 0 0 RG 0 g /Subtype /Form >> endobj Q 60 0 obj 0 w << /Resources<< endstream /FormType 1 /Font << /Matrix [1 0 0 1 0 0] /Meta82 96 0 R q 1.007 0 0 1.007 271.012 636.879 cm Q /BBox [0 0 15.59 16.44] 317 0 obj 0.297 Tc /BBox [0 0 30.642 16.44] BT q endstream q /Subtype /Form 406 0 obj (\)) Tj << /Resources<< q endobj 1 i >> /BBox [0 0 15.59 16.44] /Type /XObject >> 1 g stream q /Subtype /Form Q 1.007 0 0 1.006 551.058 763.351 cm /Meta212 226 0 R endstream 23.216 5.203 TD 293 0 obj endstream q q /Type /XObject /Subtype /Form 0.737 w 1.007 0 0 1.007 130.989 583.429 cm ET >> Q Q /FormType 1 /F1 12.131 Tf 0.37 Tc /ProcSet[/PDF/Text] << >> /AvgWidth 459 Q 259 0 obj << /Resources<< /Meta138 152 0 R Q /Subtype /Form /Subtype /Form endobj /Length 63 Q << /ProcSet[/PDF] /F2 12.131 Tf >> 0 g /F3 17 0 R endstream /Meta0 Do

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