0 G >> /Subtype /Form /ProcSet[/PDF/Text] 0 g 378 0 obj /Subtype /Form q 0.425 Tc /Length 60 >> q /Resources<< /FormType 1 /ProcSet[/PDF/Text] /Meta298 Do stream /Resources<< /ProcSet[/PDF/Text] 0.564 G /Resources<< /Length 16 /Matrix [1 0 0 1 0 0] /Meta357 Do ET q q >> endstream Q q /Type /XObject Q /Subtype /Form 1.007 0 0 1.007 130.989 523.204 cm /F4 12.131 Tf endobj Q 0 5.203 TD 1 i /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 20.21 5.203 TD /Meta94 108 0 R 1.007 0 0 1.006 551.058 437.384 cm /Meta170 184 0 R Solution: /Font << q >> 1 i >> q /F3 12.131 Tf 334 0 obj endstream /Type /XObject 1.007 0 0 1.007 271.012 383.934 cm /Meta245 259 0 R 1 i /BBox [0 0 549.552 16.44] 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . 1.007 0 0 1.007 271.012 277.035 cm /BBox [0 0 15.59 16.44] 1 i 0 g endstream /Subtype /Form q endstream q /Meta90 Do /Resources<< 429 0 obj /Type /XObject endobj /ProcSet[/PDF/Text] 24 0 obj >> /Meta142 156 0 R /Font << /ProcSet[/PDF] q /Subtype /Form 410 0 obj /FormType 1 1 i << 0 g q endobj ET q /FormType 1 0 g Q /Meta369 Do /Meta124 Do 121 0 obj /Meta74 88 0 R 0.838 Tc /Type /XObject << /Subtype /Form 9.723 5.336 TD /Matrix [1 0 0 1 0 0] Q /Meta333 Do /Type /XObject /Length 69 stream /F3 12.131 Tf 0 20.154 m 1 i /Type /XObject /Meta233 Do /Type /XObject /Type /XObject 1.005 0 0 1.007 102.382 670.003 cm 0 G /Meta354 Do Q /F3 17 0 R q /Matrix [1 0 0 1 0 0] Q /Resources<< 1 i Q (x) Tj >> q 403 0 obj Q << q 1 i 90 0 obj endobj << Q >> Q /F3 17 0 R /FontDescriptor 6 0 R >> q 379 0 obj >> q 1.014 0 0 1.007 251.439 776.149 cm q stream /F3 12.131 Tf /Length 16 0.737 w Q >> 147 0 obj ( x ) Tj /BBox [0 0 88.214 16.44] /Length 16 q /Type /XObject /FormType 1 0 G << /Meta191 205 0 R >> /BBox [0 0 639.552 16.44] Q /Length 16 Q >> /Font << 0.458 0 0 RG /ProcSet[/PDF/Text] Q 0.564 G endobj 1.005 0 0 1.007 102.382 490.08 cm /BBox [0 0 15.59 16.44] /Type /XObject Q 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 Q /Resources<< /Font << Thrice a number decreased by 5 exceeds twice the number by 1. 0 G /ProcSet[/PDF] Q q q Q >> 0 G /Resources<< >> /Length 80 q /F3 12.131 Tf /Meta332 Do endstream ET BT BT q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] << Q stream /Font << /FormType 1 /Subtype /Form (-9) Tj /BBox [0 0 639.552 16.44] stream 1.005 0 0 1.007 102.382 653.441 cm q q /BBox [0 0 88.214 16.44] >> q Q << q q endobj 1 i << /F3 17 0 R q q /F3 17 0 R 1 i /Subtype /Form /Meta89 Do /ProcSet[/PDF] endstream 1 i /Matrix [1 0 0 1 0 0] /I0 51 0 R /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] endstream /Resources<< >> BT 0.458 0 0 RG /Matrix [1 0 0 1 0 0] endobj >> /F3 12.131 Tf 0.737 w Q q q >> [4] One half of a number decreased by fourteen is twenty-one Q >> /BBox [0 0 15.59 16.44] /Length 59 0 w 1 g /Meta258 272 0 R q Q /Font << /Meta407 423 0 R 0.737 w Q /Font << /FormType 1 /Length 16 endstream 10.487 5.203 TD q << /Font << stream 0.564 G 0 g /Length 59 0.838 Tc (x) Tj Q stream (5\)) Tj 0.458 0 0 RG /ProcSet[/PDF/Text] q /ProcSet[/PDF] Q Q: A number increased by 5 is equivalent to twice the same number decreased by 7. 1 g 1 i /Resources<< q [3] One half of a number increased by fourteen is twenty-one. endstream /Type /XObject Q ET 1 i /Meta130 144 0 R BT /BBox [0 0 88.214 16.44] Q 177 0 obj /Meta198 Do >> /Meta199 213 0 R 1.005 0 0 1.007 79.798 796.475 cm Q /Meta160 Do 14.966 20.154 l Q Have a nice day! /Meta29 42 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form q Advertisement Answer No one rated this answer yet why not be the first? Q /Type /XObject Q Q endobj /BBox [0 0 88.214 16.44] /Type /XObject /FormType 1 >> [(Answe)20(r Key)] TJ stream q /Length 69 endobj 256 0 obj /Type /XObject BT /ProcSet[/PDF] endstream >> endstream /F3 12.131 Tf /Meta223 237 0 R /Meta342 Do /Meta322 336 0 R /BBox [0 0 534.67 16.44] Q (x) Tj /Font << 1.007 0 0 1.007 130.989 849.172 cm 1 g /Meta277 291 0 R ET 1.008 0 0 1.007 654.946 293.596 cm q q stream ET 23.216 5.203 TD /Length 58 /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] -0.486 Tw /LastChar 121 /FormType 1 106 0 obj /Meta206 Do Q /F3 12.131 Tf /FormType 1 q /Length 69 0 g /ProcSet[/PDF] Q /Type /XObject 0 g /F3 17 0 R 417 0 obj /F3 17 0 R endstream Q Want to see the full answer? >> 0 5.203 TD BT 23.216 5.203 TD 1 i ET endobj >> /Type /XObject /Meta146 Do /ProcSet[/PDF/Text] >> >> /Resources<< q q /FormType 1 /F3 12.131 Tf endobj /Length 70 /Resources<< stream Q >> 0 w q /BBox [0 0 88.214 16.44] /FormType 1 Q << q /Meta390 Do /Type /XObject Medium Q /Meta199 Do /Resources<< q endstream Q /Length 54 (3) Tj >> /Resources<< /ProcSet[/PDF/Text] /FormType 1 1 i 0.51 Tc endobj endstream q 1.014 0 0 1.007 111.416 583.429 cm /ProcSet[/PDF/Text] /Subtype /Form /Type /FontDescriptor /F3 17 0 R q /BBox [0 0 88.214 16.44] << /Length 16 q Q Q /Meta145 159 0 R endstream Q q /Resources<< LAIing for a pizza and, soft drink. q >> 1.007 0 0 1.007 45.168 796.475 cm 1 g /Descent -216 Q /Subtype /Form q << /F3 17 0 R /F3 17 0 R 1.007 0 0 1.006 411.035 437.384 cm /Meta50 64 0 R ET /Subtype /Form q 0 g Q Q 1 i q 1.005 0 0 1.007 102.382 653.441 cm Q /Meta139 Do 1 i >> /Meta52 Do /Meta22 Do q /F3 12.131 Tf endobj /Font << 0.737 w << (-23) Tj Q q 0 g << q /FormType 1 /Length 80 /Meta394 410 0 R >> 0.737 w 1.005 0 0 1.015 45.168 53.449 cm endstream endobj /StemH 94 549.694 0 0 16.469 0 -0.0283 cm /Matrix [1 0 0 1 0 0] >> q Q endstream /BBox [0 0 534.67 16.44] endstream /Type /XObject /Meta366 Do Q 1.502 5.203 TD /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 Q /Meta258 Do 1 g /F3 17 0 R 294 0 obj Q >> stream q q >> 0.425 Tc /Meta257 271 0 R /FormType 1 6 more than twice a number: 2x+6: two less than a number: x-2: the sum of 9 and a number: 9+x: two less than three times a number: 3x-2: a number subtracted from 12 . /BBox [0 0 639.552 16.44] >> /Resources<< 0 g /Length 54 ET Q << stream /BBox [0 0 88.214 16.44] q /Length 59 /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 87 298 0 obj 265 0 obj << /Meta161 175 0 R >> 1.007 0 0 1.007 45.168 813.037 cm Q /BBox [0 0 534.67 16.44] >> (A\)) Tj 1 g /Meta150 Do q 1 g /F3 12.131 Tf 0 G /Resources<< /Resources<< /BBox [0 0 88.214 16.44] 0 g /FormType 1 /Font << 6.746 5.203 TD ET 0.458 0 0 RG /Length 58 /Type /XObject A) 5 more than a number. (7\)) Tj q 1.007 0 0 1.007 130.989 277.035 cm /Type /XObject /Meta211 Do >> 0 w 19.474 20.154 l /Length 69 /F3 17 0 R /BBox [0 0 534.67 16.44] /Type /XObject Q stream /BBox [0 0 549.552 16.44] Q /Meta214 228 0 R /Font << 1 g Q /ProcSet[/PDF/Text] /Length 73 q 0 G /F3 12.131 Tf Q 1.014 0 0 1.007 531.485 636.879 cm << q Twice a first number decreased by a second number is 6. 0 G >> 1 i 2 See answers pharry1800 pharry1800 Answer: 2n-58 Step-by-step explanation: olivbreadh olivbreadh Answer: 2x-116 or 2(x-58) Step-by-step explanation: Transalate it to numbers and operations: => 2(x-58) => 2x-116 You won't have a solid number since its not an equation. 0 G /Resources<< -0.056 Tw BT /Length 16 >> ET /Meta74 Do /Resources<< /Meta306 Do /Length 12 296 0 obj 0 w /Subtype /Form /Font << q stream /Meta361 375 0 R 1.007 0 0 1.007 411.035 849.172 cm >> 0 G /F3 17 0 R 0 w 1 of this study. >> /BBox [0 0 88.214 16.44] /BBox [0 0 17.177 16.44] /F3 17 0 R 12 0 obj /Subtype /Form (D\)) Tj /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] endstream stream /Meta55 Do /Meta18 Do q /Meta141 155 0 R Q /MediaBox [0 0 767.868 993.712] /Type /XObject /F4 36 0 R /Length 59 >> q /Resources<< q /Meta227 Do /Subtype /Form /Font << For the lesson, he grabs a glass container shaped like a rectan stream q << 0.737 w /ProcSet[/PDF/Text] >> /Resources<< /Meta88 102 0 R endobj /Meta281 Do /Meta279 293 0 R /FormType 1 /Meta316 330 0 R Q /FormType 1 1 i 243 0 obj stream (B\)) Tj >> /FormType 1 q /Matrix [1 0 0 1 0 0] q q Q 341 0 obj /F3 12.131 Tf stream 0.369 Tc 1 i Q 5.98 7.841 TD Q Q >> 0 w /Subtype /Form << /Length 58 /BBox [0 0 30.642 16.44] 1.007 0 0 1.007 271.012 330.484 cm /Resources<< q /FormType 1 endstream /Matrix [1 0 0 1 0 0] /Subtype /Form /FormType 1 /Subtype /Form /ProcSet[/PDF/Text] << 370 0 obj BT /Resources<< 0 g ET /Meta372 386 0 R /F3 17 0 R >> q Q 1.007 0 0 1.007 411.035 330.484 cm 0 g 1 g stream q [(-3)-16(20)] TJ 1 i /Meta267 281 0 R 0 g << Q Q 0.297 Tc q /BBox [0 0 30.642 16.44] /FormType 1 20.21 5.203 TD BT q >> 416 0 obj q 1 g << endobj BT Q 6.746 5.203 TD stream 0.425 Tc /F3 12.131 Tf ET Q Q Q /BBox [0 0 639.552 16.44] /FormType 1 /F1 12.131 Tf q S 1 i 382 0 obj /Resources<< >> BT 0 G /F3 12.131 Tf 0 5.203 TD endstream 118 0 obj /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] 0 g 1 i /Meta38 Do /Subtype /Form Q 198 0 obj /Subtype /Form q << /FormType 1 /Matrix [1 0 0 1 0 0] /Length 65 /Matrix [1 0 0 1 0 0] 0 g stream /Matrix [1 0 0 1 0 0] /F3 17 0 R Q /ProcSet[/PDF] 1 g << /BBox [0 0 88.214 16.44] 83 0 obj /Subtype /Form 0 g /Resources<< /Type /XObject /Type /XObject 1.007 0 0 1.007 271.012 330.484 cm Q /Matrix [1 0 0 1 0 0] 1 i /Width 734 /Matrix [1 0 0 1 0 0] 145 0 obj /Meta184 198 0 R 0.737 w ET stream 1 g /BBox [0 0 30.642 16.44] /FormType 1 Q stream /FormType 1 Q 0 G endobj >> q 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 endstream 1.007 0 0 1.007 271.012 849.172 cm 237 0 obj /Meta339 353 0 R /ProcSet[/PDF] endobj /Type /Page >> q stream ET << stream ET 0 g BT 1.007 0 0 1.007 551.058 703.126 cm /ProcSet[/PDF/Text] >> /Type /XObject endstream 391 0 obj 1 i In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). q Q >> /ProcSet[/PDF] q /Meta333 347 0 R /Meta350 Do >> /Meta362 376 0 R /F3 17 0 R 1.502 7.841 TD q ET Q >> endobj endstream >> /F3 17 0 R >> BT << >> /Meta265 Do /Subtype /Form q 0 5.203 TD 1 i >> /ProcSet[/PDF] (+) Tj /F3 17 0 R 0 5.203 TD >> >> Q q Tamang sagot sa tanong: 1.) (-20) Tj /Meta155 169 0 R /F3 12.131 Tf >> q Q (9) Tj 1 g q /F3 17 0 R 0 g /ProcSet[/PDF/Text] 0.564 G /Type /XObject q 252 0 obj BT >> 0.564 G /MissingWidth 252 /FormType 1 /Font << 13.493 5.336 TD 1 i /Type /FontDescriptor /Subtype /Form << Q Q Q 0 G 0 g /Meta151 165 0 R Q Q Q >> << 0 w 1 i Q 1.007 0 0 1.007 271.012 277.035 cm endstream >> >> Q 0.425 Tc q q /Meta424 440 0 R /Meta421 437 0 R /Subtype /Form /Meta370 Do /Meta83 97 0 R /Type /XObject Q q /Meta64 78 0 R >> BT endobj /ProcSet[/PDF/Text] endobj /Resources<< << stream BT /F3 17 0 R /Subtype /Form /ProcSet[/PDF/Text] stream /StemV 88 60 0 obj /Font << (58) Tj /Length 54 << stream Q q q ET stream 3.742 8.18 TD q endobj /Length 16 BT >> /F1 7 0 R 0 g ET << 0.737 w /Meta358 Do 0.458 0 0 RG endobj /Meta387 Do Q This site is using cookies under cookie policy . BT /F1 12.131 Tf /FormType 1 1.014 0 0 1.007 391.462 636.879 cm q >> ET /BBox [0 0 88.214 16.44] >> /Font << /Meta405 421 0 R Q >> 1.014 0 0 1.007 251.439 583.429 cm << /Type /XObject 0.51 Tc ET 0 g /Meta140 Do 0.737 w 313 0 obj /Matrix [1 0 0 1 0 0] Q /Resources<< BT q /ProcSet[/PDF] /Subtype /Form 0.737 w 181 0 obj /Meta292 Do /F1 7 0 R BT /Meta299 Do 9 0 obj [(A number )-17(divided by )] TJ /FormType 1 1 i /FormType 1 stream /I0 51 0 R /F3 17 0 R (C\)) Tj /Subtype /Form Q 1 i /Length 69 Q 0 g >> Twice a number decreased by . 1.502 8.18 TD stream endobj 1 g q 1 i /Length 59 >> /Type /XObject q /F3 12.131 Tf BT /Meta169 183 0 R /Meta269 283 0 R /ProcSet[/PDF/Text] endstream q /Resources<< endstream q q /F3 17 0 R Q << BT stream Q /Font << /ProcSet[/PDF/Text] /Subtype /Form (40) Tj /FormType 1 0.458 0 0 RG /Matrix [1 0 0 1 0 0] (8\)) Tj Q /F3 17 0 R /FormType 1 >> /Matrix [1 0 0 1 0 0] >> q q Q /ProcSet[/PDF/Text] >> << /ProcSet[/PDF/Text] << /Subtype /Form Q BT q /BBox [0 0 534.67 16.44] /Subtype /Form ET /Matrix [1 0 0 1 0 0] 32 0 obj 255 0 obj /Length 69 444 0 obj /BBox [0 0 639.552 16.44] /Resources<< endstream endobj >> /Subtype /Form endstream /BBox [0 0 88.214 35.886] 19.474 20.154 l 1.007 0 0 1.007 130.989 776.149 cm 0 20.154 m /F3 17 0 R endobj 371 0 obj /Resources<< 1 g << /Meta283 297 0 R q /Resources<< /Meta176 Do endobj << /Type /XObject ET /F3 17 0 R 0 G q 0 g A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. 0.458 0 0 RG endobj stream 1 i Q 312 0 obj endstream Q (D) Tj 132 0 obj /Type /XObject /F1 7 0 R 0.737 w Q 22.478 5.336 TD 0 20.154 m /Resources<< /Subtype /Form q /F3 12.131 Tf 1.502 24.649 TD 1.007 0 0 1.007 411.035 636.879 cm /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 654.946 726.464 cm /F3 17 0 R >> /Font << Was this answer helpful? /Matrix [1 0 0 1 0 0] (5) Tj 0.564 G [(Fiv)25(e ti)18(me)16(s)] TJ [(MULTIPLE CHOICE. 0.68 Tc endstream ET 1.007 0 0 1.007 551.058 703.126 cm stream 141 0 obj stream endstream /Meta427 Do 0 g << 0.68 Tc 0 g Q 1.007 0 0 1.007 551.058 330.484 cm stream 0 g /Meta325 339 0 R /Length 59 0 w 0 g /Meta419 435 0 R /Meta116 130 0 R endobj >> Q /Length 69 >> /Subtype /Form /BBox [0 0 549.552 16.44] Q Then the following equation can represent this problem: 17 + x = 68 We can subtract 17 from both sides of the equation to find the value of x. 269 0 obj /FormType 1 0 g >> 722.699 599.991 l q stream /Resources<< /FormType 1 1 i /Font << >> /Type /XObject /Meta228 242 0 R 17.234 5.203 TD 1.014 0 0 1.006 251.439 437.384 cm >> >> /ProcSet[/PDF/Text] >> 0.738 Tc /Meta282 296 0 R 1.007 0 0 1.007 67.753 546.541 cm /F3 17 0 R /Meta422 438 0 R 1 i Word Problems: Age Solvers Lessons Answers archive Click here to see ALL problems on Age Word Problems Question 196314: twice a number decreased by 8 is equal to the number increased by 10. find the number. /Resources<< /Length 69 /Subtype /Form /Type /XObject /BBox [0 0 88.214 16.44] ET q >> /Meta263 Do 1.005 0 0 1.007 102.382 799.486 cm /Font << endobj stream endstream /Font << 30 0 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425 0 obj /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . /Type /XObject 0 G [(The )-16(s)15(um )-14(of )] TJ /Matrix [1 0 0 1 0 0] /Meta353 Do /ProcSet[/PDF/Text] BT BT /Meta91 Do q q /Length 69 >> Q endstream endstream Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. /BBox [0 0 88.214 16.44] Q /Length 68 0 G /Meta57 71 0 R endstream /FormType 1 /Meta226 240 0 R 1.007 0 0 1.007 67.753 599.991 cm /Matrix [1 0 0 1 0 0] Q endstream 1 i q endobj 20.21 5.203 TD /Font << endstream stream /Matrix [1 0 0 1 0 0] << q >> /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0 w 1 i 722.699 653.441 l 3.742 5.203 TD /Meta306 320 0 R /FormType 1 (C\)) Tj << stream 0.737 w << Q q 1 i /BBox [0 0 534.67 16.44] (-11) Tj /Length 12 136 0 obj /ProcSet[/PDF] q Q 18.708 17.593 TD 152 0 obj /Meta149 Do /F3 17 0 R stream 1 i /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] Q endobj endobj Q >> /BBox [0 0 88.214 35.886] q q /FormType 1 q /Subtype /Form /Meta323 337 0 R 100 0 obj /Type /XObject q /ProcSet[/PDF/Text] << 0 w 1 i Q /Meta137 Do /FormType 1 /FormType 1 << /Subtype /Form /Matrix [1 0 0 1 0 0] /Subtype /Form >> q /Meta394 Do >> >> 1.007 0 0 1.007 130.989 383.934 cm 0 g /BBox [0 0 534.67 16.44] /FormType 1 1 i /Length 118 1 i Q 1.007 0 0 1.007 411.035 849.172 cm /Resources<< /F3 12.131 Tf Q /Resources<< /ID [] 1.007 0 0 1.007 271.012 636.879 cm /Meta156 170 0 R >> Find the number 1 See answer Advertisement q 0 G q q endstream /F3 17 0 R /F3 17 0 R >> /FormType 1 0.369 Tc 76.394 5.203 TD /F3 12.131 Tf /I0 51 0 R stream q /Type /XObject /Matrix [1 0 0 1 0 0] Q /Kids [ /Meta188 Do /F3 17 0 R /Matrix [1 0 0 1 0 0] /Font << /Resources<< /Type /XObject endobj /Type /XObject 1.014 0 0 1.007 391.462 383.934 cm << 1 g /ItalicAngle 0 /Meta378 Do /Font << /Type /XObject ET /ProcSet[/PDF] 0 g /Subtype /Form q 0.369 Tc Q /Meta335 349 0 R q q /FormType 1 /Meta294 Do /Meta39 53 0 R 99 0 obj /Meta0 Do /Type /XObject /BBox [0 0 88.214 16.44] 0 w endobj q >> BT (3) Tj /Resources<< >> /Length 70 >> 134 0 obj endstream /Meta320 334 0 R 0 g 1.007 0 0 1.006 411.035 437.384 cm 0 G 0.51 Tc /Resources<< Q Q ET q Q q >> 1.007 0 0 1.006 551.058 437.384 cm stream /BBox [0 0 88.214 16.44] 291 0 obj /Meta5 14 0 R >> /Meta19 Do -0.067 Tw 9.723 5.336 TD 1.007 0 0 1.007 551.058 703.126 cm /BBox [0 0 88.214 16.44] q 333 0 obj 1.007 0 0 1.007 271.012 636.879 cm /Meta365 379 0 R Q endobj Q 0.458 0 0 RG /F3 12.131 Tf Q >> >> /ProcSet[/PDF/Text] >> 293 0 obj 0.737 w q stream >> Q /ProcSet[/PDF] endstream /FormType 1 ET Q (D\)) Tj q , Prove the following << /BBox [0 0 639.552 16.44] 1 i /Subtype /Form 1.007 0 0 1.007 271.012 849.172 cm Q endstream /Length 68 /Meta344 Do /Meta340 354 0 R /Resources<< q /BBox [0 0 17.177 16.44] Q /Length 16 419 0 obj 0 5.203 TD 329 0 obj q q q q /Meta106 Do /Meta204 218 0 R /Matrix [1 0 0 1 0 0] BT Q /Subtype /Form Q /BBox [0 0 30.642 16.44] /Subtype /Form >> /Meta286 300 0 R /Meta244 258 0 R endobj /Resources<< >> Q /Meta112 126 0 R Find the number. 1 i Q /ProcSet[/PDF] >> 0 g Q 0 g Q /BBox [0 0 88.214 16.44] /Subtype /Form 0 w >> >> Q q 1.005 0 0 1.007 102.382 546.541 cm 0 w 1.014 0 0 1.007 251.439 277.035 cm endobj /Length 69 Q 19 0 obj /BBox [0 0 88.214 16.44] << >> 73 0 obj /Meta378 392 0 R /Font << >> Q >> << Q >> /Font << 1.007 0 0 1.007 411.035 849.172 cm 116 0 obj 303 0 obj /FormType 1 32 = 2a + 8: The quotient of fifty and five more than a number is ten. /Meta330 344 0 R stream endobj >> Q 0.564 G 0 w /FormType 1 >> ET Q q Q q 0 w /FormType 1 /F3 17 0 R /FormType 1 /Length 245 /Length 16 /FormType 1 /F3 12.131 Tf 20-n c.) n+20 d.) 20+n 3.) /Matrix [1 0 0 1 0 0] q 0000000000 65535 f 0000140665 00000 n 0000140732 00000 n 0000000015 00000 n 0000120613 00000 n 0000000126 00000 n 0000000314 00000 n 0000000577 00000 n 0000001009 00000 n 0000001360 00000 n 0000001548 00000 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