twice a number decreased by 58

/Matrix [1 0 0 1 0 0] 0 G 0.737 w Q /Font << 0 G 1 i 0 5.203 TD >> /Font << 1.005 0 0 1.006 45.168 879.284 cm /Length 80 /Resources<< (11) Tj /Meta2 9 0 R /Meta118 132 0 R 17.234 5.203 TD 0 g 549.694 0 0 16.469 0 -0.0283 cm q q << Q 0 g ET /BBox [0 0 15.59 16.44] TJ 0 g /F3 17 0 R Q endstream /Meta214 228 0 R 1.007 0 0 1.007 271.012 330.484 cm 1.007 0 0 1.007 551.058 383.934 cm /FormType 1 /Meta250 Do /F3 17 0 R /Subtype /Form BT (x) Tj stream endstream 0 G stream /Length 69 Q /Matrix [1 0 0 1 0 0] /F3 17 0 R /ProcSet[/PDF/Text] -0.099 Tw q q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endstream (x) Tj 1st step. /Resources<< /Matrix [1 0 0 1 0 0] 383 0 obj /Resources<< >> 1 g stream 1 i endstream /Length 69 >> Q Q >> 0.564 G q endstream q /Type /XObject 1 i 1.007 0 0 1.006 130.989 437.384 cm q stream 20.21 5.336 TD q /Subtype /Form stream /FormType 1 Q stream /Subtype /Form 0 G 437 0 obj 1 i BT /Subtype /Form endstream /Font << /F3 12.131 Tf q /FormType 1 /Resources<< Q endstream /Type /XObject /Type /XObject Q 39 0 obj /ProcSet[/PDF/Text] 0 g (D\)) Tj >> Q 1.005 0 0 1.007 102.382 473.519 cm Q /Resources<< endobj 0 g In the problem above, x is a variable. /FormType 1 >> /F3 17 0 R Q /F3 17 0 R /Matrix [1 0 0 1 0 0] << 20.21 5.203 TD /F3 12.131 Tf Q /Resources<< Q Q 1 i 1.007 0 0 1.007 45.168 796.475 cm /FormType 1 ET 1.007 0 0 1.007 45.168 846.161 cm >> 0.458 0 0 RG 1.005 0 0 1.007 102.382 726.464 cm Q 1 i /Resources<< /XObject << /BBox [0 0 15.59 16.44] stream 0 G /Font << /FormType 1 Q /Meta418 Do Q 1 i 94 0 obj 0.737 w q 0 5.203 TD >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] << 1.005 0 0 1.007 102.382 473.519 cm /Subtype /Form /FormType 1 /Resources<< << BT /ProcSet[/PDF] 79 0 obj << (11) Tj 54 0 obj /F3 12.131 Tf Q >> /Length 12 /Font << >> 1.014 0 0 1.007 251.439 383.934 cm q q q endstream stream >> q Q BT << ET (C\)) Tj endstream 1.007 0 0 1.006 551.058 763.351 cm /Meta164 Do (+) Tj /Matrix [1 0 0 1 0 0] Q 57 0 obj endstream /Meta124 138 0 R q /Font << q q 0 g /F1 12.131 Tf /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 411.035 849.172 cm /Type /XObject /Length 54 /ProcSet[/PDF] /Font << /Subtype /Form endstream q Q 0.738 Tc (B) Tj 0 G Q /Resources<< Q q /Length 16 q 1 i /Type /XObject endobj /Font << endstream 9.723 5.336 TD q Q >> /Resources<< /Font << Q /Type /XObject 0.369 Tc Q q q 1.502 5.203 TD /Meta78 92 0 R Grad - B.S. 0 g >> Q endobj BT /FormType 1 /Matrix [1 0 0 1 0 0] 21 0 obj Q /F3 17 0 R /F3 17 0 R endstream 1.005 0 0 1.007 102.382 799.486 cm BT stream Q /BBox [0 0 88.214 35.886] /FormType 1 >> q /ProcSet[/PDF/Text] /Meta158 172 0 R Q >> >> BT 0 w << << /Meta26 Do 0.458 0 0 RG stream /Meta236 Do /BBox [0 0 88.214 16.44] (x) Tj /Meta223 237 0 R /Font << endobj 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 Check out a sample Q&A here. /ProcSet[/PDF/Text] q /Subtype /Form >> >> 53 0 obj stream Q q endobj 20.21 5.203 TD 1 i 6.746 5.203 TD endobj >> 1.007 0 0 1.007 551.058 383.934 cm /Meta11 Do >> /Resources<< 0 g q , Prove the following /Length 69 q /StemH 94 q q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 277.035 cm 1 i /BBox [0 0 534.67 16.44] 1 i ET 0 w ET endobj /Meta422 Do /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] << 353 0 obj /Length 69 /Subtype /Form 0 G q /FormType 1 /ProcSet[/PDF/Text] Q stream /I0 51 0 R % >> ET q /Font << >> >> stream BT stream /FormType 1 0 g >> /Subtype /Form Q ET /Type /XObject Q /Type /XObject 22.478 5.203 TD 1 g 230 0 obj Q Q BT (+) Tj 0.458 0 0 RG 0.68 Tc /Subtype /Form /Resources<< stream the quotient of five and a number 7.) /ProcSet[/PDF/Text] >> /Meta190 204 0 R /Matrix [1 0 0 1 0 0] Q /Font << endobj BT /Length 69 /ProcSet[/PDF/Text] >> /Font << /Meta362 376 0 R Q /ProcSet[/PDF] endobj Q /ProcSet[/PDF] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /FormType 1 (-9) Tj 1.007 0 0 1.007 67.753 653.441 cm 0 g endobj stream S 1 g /BBox [0 0 15.59 29.168] 0 g Q /Meta271 Do /Matrix [1 0 0 1 0 0] (C\)) Tj /Type /XObject /Matrix [1 0 0 1 0 0] /Font << /ProcSet[/PDF/Text] /FormType 1 q << Q 0 w >> /Length 69 q 0.369 Tc /Matrix [1 0 0 1 0 0] endobj endobj >> /F3 12.131 Tf /Length 108 /Matrix [1 0 0 1 0 0] ( x) Tj /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] ET /Resources<< /Meta129 143 0 R << /ProcSet[/PDF/Text] /FormType 1 /Meta39 Do /Font << >> -0.056 Tw >> << Q endstream q 0 g Q endstream /BBox [0 0 17.177 16.44] /Meta245 Do /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] >> S q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 849.172 cm >> endstream /Resources<< Q /FormType 1 >> 0 G Q /Meta410 Do 6.746 5.203 TD 1.007 0 0 1.007 654.946 546.541 cm 0 g 0 G /Resources<< endstream 0.737 w /Type /XObject ET /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] Q /Font << /Resources<< << /Length 59 q /Length 60 Q /Length 69 0.564 G 159 0 obj /BBox [0 0 88.214 16.44] /FormType 1 /Meta398 414 0 R /Meta393 409 0 R 1.007 0 0 1.007 551.058 636.879 cm << q >> 0.786 Tc 1.005 0 0 1.007 102.382 400.496 cm /Type /XObject /Meta45 59 0 R >> /I0 51 0 R << Find an answer to your question Twice a number decreased by 8gives 58. ET 1 i 0 g 0 w /BBox [0 0 88.214 35.886] /Meta301 Do 0.425 Tc /F3 12.131 Tf /Length 12 >> (-) Tj /BBox [0 0 15.59 29.168] (- 8) Tj /Resources<< /Type /XObject >> (13) Tj >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 116 0 obj 0 4.894 TD 1.007 0 0 1.007 411.035 636.879 cm ET /F3 17 0 R q 0.737 w /Type /XObject >> /Resources<< 0.463 Tc 388 0 obj Q /Matrix [1 0 0 1 0 0] /Font << 1 i 398 0 obj /XObject << /StemV 77 /BBox [0 0 88.214 16.44] 0 g >> 1 i /BBox [0 0 30.642 16.44] BT /Font << (B\)) Tj /Font << 0 g Q (-) Tj /ProcSet[/PDF] q Q endobj 0 g Q /FormType 1 q /Matrix [1 0 0 1 0 0] Q /Meta286 300 0 R /Type /XObject /F3 12.131 Tf 1.007 0 0 1.007 551.058 330.484 cm Q >> q a and b or something else.***. stream 0 g /Meta195 209 0 R endobj 0.564 G /FormType 1 /Meta217 231 0 R /Matrix [1 0 0 1 0 0] /F1 7 0 R /Matrix [1 0 0 1 0 0] 1 g Q Q /Subtype /Form q Q (x) 6 times a number is 5 more than the number. /Meta12 Do /Resources<< endstream q endobj /BBox [0 0 15.59 16.44] /Meta159 Do q 0 w endstream << /FormType 1 S /Meta102 116 0 R 0.564 G stream q 213 0 obj 0.271 Tc Two times the sum of a number x and five c.) a number x times the sum of five and two d.) the sum of five times a number x and two 2.) << >> /F3 12.131 Tf 0.564 G /Type /XObject /Subtype /Form stream /Matrix [1 0 0 1 0 0] 160 0 obj q Q 1.007 0 0 1.006 411.035 763.351 cm /Meta231 Do /Type /XObject endobj q endstream ET /Length 118 Q >> /Font << /BBox [0 0 88.214 16.44] 0 5.203 TD /BBox [0 0 88.214 35.886] stream q 1.007 0 0 1.007 130.989 330.484 cm 0 G 0 w q Q Q << 0 g Q /Type /XObject /Length 16 Q /Meta48 62 0 R 0.737 w 0.737 w endstream /Meta232 246 0 R /Resources<< >> BT /Length 189 >> /Subtype /Form /Meta213 Do Q 1 i q -0.021 Tw q 0 g q 12.727 5.203 TD /Meta40 54 0 R Q 0.458 0 0 RG /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /Resources<< 1.005 0 0 1.007 79.798 763.351 cm /ProcSet[/PDF] Q q /Type /XObject The sum of 18 and tour times a number is -6 Find the number. endobj (13) Tj Q /Font << /ProcSet[/PDF/Text] 549.694 0 0 16.469 0 -0.0283 cm /Resources<< >> Q endobj 312 0 obj 170 0 obj Q /ProcSet[/PDF/Text] endobj q /ProcSet[/PDF/Text] /Resources<< /ProcSet[/PDF/Text] /BBox [0 0 673.937 68.796] 1 i /Meta371 385 0 R 20.21 5.203 TD /Length 64 /Resources<< stream q 0.564 G /Meta7 18 0 R ET /F3 12.131 Tf BT endstream 255 0 obj endobj ET /F3 17 0 R /Font << q /Encoding /WinAnsiEncoding q q (A\)) Tj Q 0 G 32.201 5.203 TD /Subtype /Form /Type /XObject ET q stream >> /Subtype /Form /Subtype /Form >> Q << Q /Length 118 >> q /Meta285 Do endstream q q /F3 12.131 Tf /Length 60 q 0.737 w /MaxWidth 2000 /XObject << stream Q /Matrix [1 0 0 1 0 0] For Free. >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 205.199 4.894 TD >> /FormType 1 /Subtype /Form /Meta412 428 0 R Q /Length 54 Q q /ProcSet[/PDF/Text] q /Type /XObject >> 424 0 obj q /Type /XObject /BBox [0 0 30.642 16.44] /F3 17 0 R /BBox [0 0 549.552 16.44] q /FormType 1 >> /Type /XObject endobj /Type /XObject So let's go ahead and identify a v Q /Meta317 331 0 R 1 i endobj 0.564 G /Type /XObject 0 G 1 i /Length 67 /F3 17 0 R << 1.007 0 0 1.007 130.989 383.934 cm /Subtype /Form 67 0 obj ET 0 g /F3 17 0 R /FormType 1 >> /Meta346 360 0 R q q q /Meta259 273 0 R /Subtype /Form Q 1 g >> 0 G >> /Meta188 Do /FormType 1 /F1 7 0 R Q q << q >> BT Q /Font << /Meta209 223 0 R /Font << 140 0 obj 1 i 0 w 411 0 obj /BBox [0 0 88.214 16.44] /Type /XObject /Length 16 /Matrix [1 0 0 1 0 0] stream << /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] 0.425 Tc q 0 g /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 58 decreased by twice Gails age. Q Q Q /Matrix [1 0 0 1 0 0] /Resources<< >> q 1 i 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 /F3 12.131 Tf q q /FormType 1 /BBox [0 0 15.59 29.168] >> /Type /XObject BT << /ProcSet[/PDF/Text] >> Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /FormType 1 q q /Meta53 Do /BBox [0 0 15.59 16.44] /Meta226 240 0 R /Meta180 Do 0 5.203 TD /Resources<< /FormType 1 /Subtype /Form /Subtype /Form /Subtype /Form 1 i /Type /XObject 0.738 Tc See Solution. endobj 0 g 264 0 obj Q q 0 g /BBox [0 0 15.59 16.44] >> >> q q >> /Meta85 99 0 R ET Q q /F3 12.131 Tf 0 g q /F4 36 0 R 1.014 0 0 1.007 111.416 849.172 cm Q Q >> Q /Subtype /Form /Meta278 Do >> q q /Meta248 Do /Meta145 159 0 R 1 i ET endobj 25.454 5.203 TD /Type /XObject stream 13.493 5.336 TD q 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) >> Q /BBox [0 0 88.214 16.44] Q /Length 68 stream Q /FormType 1 /Font << 1 i Q Q /Resources<< Q 26.957 5.203 TD stream /BBox [0 0 88.214 16.44] 1 i 1 i endstream /Meta328 342 0 R 0 w Q 0 w /Type /XObject q /F4 36 0 R Q 0 g /Meta119 Do [(A numb)-16(er subtract)-15(ed from )] TJ ET 1.005 0 0 1.007 102.382 653.441 cm /Subtype /Form endstream 0.737 w /Length 63 q Q endstream stream /Meta94 108 0 R /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g endstream 0 5.203 TD 0 5.203 TD stream /Subtype /Form 0 g 40 0 obj >> /Meta225 239 0 R /FormType 1 /Font << /Subtype /TrueType q 1.014 0 0 1.006 391.462 690.329 cm >> /BBox [0 0 88.214 35.886] 0.382 Tc /Type /XObject Q q Q /Type /XObject 0 g 0 g endobj 0 g stream 0 g BT /Type /XObject << q >> 0 G /F3 12.131 Tf 0 G BT /BBox [0 0 30.642 16.44] 0 G Q 0.369 Tc /ProcSet[/PDF] stream 0.564 G /ProcSet[/PDF/Text] /Meta377 Do Q >> q << /Resources<< /Meta317 Do /Matrix [1 0 0 1 0 0] Q 1 i /Subtype /Form endstream 0 G Q 1.007 0 0 1.007 271.012 450.181 cm /FontDescriptor 10 0 R 1.007 0 0 1.007 130.989 776.149 cm 1 i /Meta122 136 0 R >> (A\)) Tj /ProcSet[/PDF/Text] /Type /XObject q endstream Q /Length 69 Q /Meta170 184 0 R 355 0 obj >> /Matrix [1 0 0 1 0 0] q /Font << 0.564 G /Meta85 Do 1 g q Q ET >> << stream 1 i 0.737 w /Length 12 q Q << 0 G Q Q /Resources<< Q Example 1: Use the tables above to translate the following English phrases into algebraic expressions. endobj /Meta5 Do /Subtype /Form /FormType 1 q endstream 0 g /Meta210 224 0 R 150 0 obj 0 G >> Q q >> /Meta19 30 0 R /Subtype /Form /Type /XObject Q 1.007 0 0 1.007 654.946 872.509 cm >> >> BT /ProcSet[/PDF/Text] 1 i q 0 g >> Q Q /F3 17 0 R /Length 245 >> /Matrix [1 0 0 1 0 0] BT /ProcSet[/PDF/Text] /Type /XObject /Resources<< /FormType 1 /Matrix [1 0 0 1 0 0] Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM q /Font << /XHeight 477 q /Type /XObject endstream 0 G 0.564 G << stream /Subtype /Form endstream 0 g /ProcSet[/PDF] >> endstream Q Q /Resources<< 0.369 Tc 1 i >> Q /Type /XObject q /Length 118 Q 4.506 24.649 TD /Length 60 0 5.203 TD Select the correct mathematical statement for the following equation. >> q Q /FormType 1 Q q Q endstream >> BT /FormType 1 q /Meta392 Do >> q /Matrix [1 0 0 1 0 0] >> 0 w /Subtype /Form Q /Matrix [1 0 0 1 0 0] 1 i /Font << stream /Type /XObject q /Resources<< /ProcSet[/PDF] /Length 59 /ProcSet[/PDF/Text] /F3 17 0 R Q stream (-23) Tj stream /Length 59 /Length 16 (B\)) Tj /Length 12 0 w stream 0 g 1.005 0 0 1.007 45.168 916.925 cm q /BBox [0 0 534.67 16.44] endobj /Length 59 BT q /Meta47 61 0 R q /Length 69 /ProcSet[/PDF] 0.458 0 0 RG >> /BBox [0 0 639.552 16.44] >> /ProcSet[/PDF/Text] stream endstream /FormType 1 q q /I0 Do q 1 g endstream >> /ProcSet[/PDF] >> Q /BBox [0 0 88.214 16.44] q 0 g /Resources<< /Meta338 Do 1 i Q /Meta216 230 0 R 1.005 0 0 1.007 102.382 310.158 cm endobj endobj Question. >> >> /Meta322 Do BT q /F3 12.131 Tf q /Type /XObject 0 g /F4 36 0 R q 1.007 0 0 1.007 411.035 277.035 cm /F3 12.131 Tf Q /Meta346 Do /Meta16 Do Q stream 0.564 G >> /Matrix [1 0 0 1 0 0] q << Q 1 i >> q Q /Meta406 422 0 R stream /Resources<< /Length 69 BT 0.738 Tc Q Q 2x - 15 = -27. ET ET (-20) Tj Q /Length 54 /ProcSet[/PDF] /F1 7 0 R /F4 12.131 Tf << endstream BT /Resources<< 1.007 0 0 1.006 130.989 690.329 cm Q ET Q q q Q /Length 151 >> /FormType 1 /LastChar 121 1 g Double or twice a number means 2x, and triple or thrice a number means 3x. Q q >> endstream endstream >> 0.458 0 0 RG /Subtype /Form q /Matrix [1 0 0 1 0 0] 1 i /Subtype /Form /FormType 1 stream 0.134 Tc >> stream /ProcSet[/PDF/Text] /Meta51 Do /FormType 1 >> /Font << << /Matrix [1 0 0 1 0 0] /F4 12.131 Tf 0.737 w /Meta334 Do ( x) Tj 1 i /Type /XObject >> Q >> 1.014 0 0 1.006 111.416 510.406 cm 1 i /F1 12.131 Tf /Type /XObject q /Length 118 endstream /Meta191 Do 1.007 0 0 1.007 411.035 277.035 cm ET >> 1.005 0 0 1.007 102.382 473.519 cm stream endstream >> Q 0 g ET /Subtype /Form /Matrix [1 0 0 1 0 0] 0 5.203 TD q >> >> /BBox [0 0 88.214 16.44] Q /Subtype /Form 0.37 Tc Q /F3 12.131 Tf stream 0 w /Type /XObject /Font << >> for the season. /FormType 1 0.486 Tc q /ProcSet[/PDF/Text] q /Meta59 Do /Length 245 /Resources<< /Matrix [1 0 0 1 0 0] ET 0 G 0.458 0 0 RG /Meta135 149 0 R Q q /Meta407 Do /FormType 1 endstream /F3 12.131 Tf << /Type /XObject 0.458 0 0 RG q /Matrix [1 0 0 1 0 0] /Length 16 /F3 17 0 R >> 0 w 1 i >> Q ET 0.737 w >> (+) Tj 1 i stream (C\)) Tj 0.564 G 14.23 24.649 TD >> /Meta409 425 0 R The result is 8 less than 10 times the number. /BBox [0 0 639.552 16.44] /FormType 1 /Length 60 q /Matrix [1 0 0 1 0 0] Q Q 0 w q >> 0 g stream /Resources<< 1 i /Meta51 65 0 R /BBox [0 0 639.552 16.44] >> /Length 59 BT 1 i BT >> 0 G Q Q Q BT Q /Resources<< /Font << >> endobj /Matrix [1 0 0 1 0 0] q ET /Subtype /Form ET << >> /Meta112 Do /Meta193 Do /Meta367 Do ET ET /Matrix [1 0 0 1 0 0] /Meta221 235 0 R << 101 0 obj /Meta319 333 0 R /FormType 1 q 0.564 G ET 0 w q /Type /XObject >> >> >> 0 g Q endobj /FormType 1 endstream 0.458 0 0 RG /Meta80 Do /Type /XObject 101.849 5.203 TD /Type /XObject 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 endstream 0 5.203 TD A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. q 369 0 obj /Meta303 Do /Meta354 368 0 R /ProcSet[/PDF/Text] /Resources<< endobj Q /F3 17 0 R stream 265 0 obj /Type /Font Table 1. /F3 17 0 R Q q /Meta20 Do >> Q endstream Q 0.564 G /Subtype /Form q BT /Meta86 100 0 R endstream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] >> Q /ProcSet[/PDF] >> >> q /Type /XObject Q /Meta84 Do 125.064 4.894 TD /Type /XObject << 1.005 0 0 1.007 102.382 347.046 cm /Length 69 /Meta363 Do /F3 17 0 R /Subtype /Form 1.007 0 0 1.006 551.058 763.351 cm /Type /XObject 0 G Q /Font << ET /BBox [0 0 88.214 16.44] 0 G /Length 59 0 w /Length 12 /Subtype /Form 1.014 0 0 1.007 251.439 523.204 cm /Meta177 Do 63 0 obj /Font << 0 G Q /F3 12.131 Tf endstream /FormType 1 /Type /XObject /Meta411 Do 0 g endstream /Type /XObject /Resources<< /F3 17 0 R 205 0 obj << /Meta301 315 0 R Q /F3 17 0 R >> /Meta283 297 0 R 1 g /F3 12.131 Tf Abstract: The aim of the study was to investigate the expression of miR-155 in plasma and peripheral blood mononuclear cells (PBMCs), the effects of miR-155 on the apoptosis rate >> (-9) Tj /FormType 1 /Meta79 Do Q >> /Font << /Meta330 344 0 R /Length 16 q q /Resources<< >> /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0 G There were x cookies at the beginning of a party. 1 i stream /Type /Font ( x) Tj ET 397 0 obj /Subtype /Form /Meta239 253 0 R endstream stream /Resources<< (3\)) Tj >> >> /Subtype /Form /F3 12.131 Tf stream endobj 324 0 obj /Subtype /Form 0 w /ProcSet[/PDF/Text] Q ET BT /BBox [0 0 15.59 16.44] endobj stream /Resources<< /FormType 1 /Resources<< /Subtype /Form stream q Q 0 g -0.16 Tw endstream /Type /XObject /Font << /Meta197 211 0 R endstream 0.68 Tc Q 0 G /Type /XObject /FormType 1 /Length 16 /Resources<< /Resources<< /Length 106 q q -0.092 Tw 18 0 obj >> /Subtype /Form >> /Subtype /Form q stream endstream >> 0.369 Tc /Meta74 Do /Meta25 Do /Resources<< << 0 5.203 TD /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 639.552 16.44] 0 w /Type /XObject /Subtype /Form 32 = 2a + 8: The quotient of fifty and five more than a number is ten. /Type /XObject 0 5.203 TD 1 i stream /BBox [0 0 15.59 16.44] stream 1.007 0 0 1.007 411.035 330.484 cm Q BT /Resources<< q /Meta175 189 0 R /Subtype /Form Q >> /BBox [0 0 15.59 16.44] 1 i 0.458 0 0 RG Q (40) Tj /FormType 1 Explanation: let the number be n. then we can express division in 2 ways. << 136 0 obj 0.458 0 0 RG /ProcSet[/PDF/Text] 0 g endobj q /ProcSet[/PDF/Text] q q Q 1.502 5.203 TD /Resources<< /BBox [0 0 88.214 16.44] q 0 G /FormType 1 0 G /Matrix [1 0 0 1 0 0] 46 0 obj 0 5.203 TD 0 g q 273 0 obj 0 g Q /FormType 1 BT 1.005 0 0 1.007 45.168 889.071 cm ET /ProcSet[/PDF] /Type /XObject 1.014 0 0 1.007 111.416 636.879 cm BT 0 g >> Q q << q (B) Tj 0 w 11.99 8.18 TD /Length 65 0.458 0 0 RG /Meta174 Do << /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] Q 0 G /FormType 1 << << 425 0 obj /FormType 1 /Type /XObject 1 i >> >> S endstream /Meta207 221 0 R 222 0 obj /FormType 1 0.458 0 0 RG stream q endstream Q endobj endobj 285 0 obj 1 i q 1 i 0 g /Subtype /Form /Meta370 384 0 R >> Twice a number decreased by another number: Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. stream /BBox [0 0 88.214 16.44] endstream Q /Matrix [1 0 0 1 0 0] q q /Type /XObject BT /ProcSet[/PDF] /F1 12.131 Tf Q Q /ItalicAngle 0 0 G Q /FormType 1 stream << ET 0 g 0 g q /Length 69 0 g endstream endobj 0.786 Tc 0.564 G >> /Meta320 Do Q Q q /Meta45 Do 1 i 0.524 Tc /Length 127 0.737 w 0.241 Tc Q q endobj /Length 58 stream 197 0 obj 1 i 0 G 0.737 w /Type /XObject /FormType 1 /Font << endstream Q 1.014 0 0 1.007 391.462 330.484 cm >> endobj 0 0 Similar questions Find the number which when decreased by 8% becomes 506. 1.007 0 0 1.007 271.012 523.204 cm Q /Length 69 /Length 59 /F3 12.131 Tf 1.007 0 0 1.007 130.989 523.204 cm /FormType 1 /Resources<< BT stream >> 672.261 653.441 m 0 w << endobj q /Length 16 stream /Subtype /Form Q /Meta331 Do /Type /XObject /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 636.879 cm q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 726.464 cm /F4 12.131 Tf

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