/Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] 1 i 0 G /Subtype /Form /Meta54 68 0 R q 0.68 Tc stream /FormType 1 Q 0.564 G /Length 79 /ProcSet[/PDF] >> >> << >> 1.007 0 0 1.007 271.012 523.204 cm 0.425 Tc 0 w /Type /XObject /Meta292 Do Q endstream >> Q 1.005 0 0 1.007 102.382 653.441 cm << /Subtype /Form Q 239 0 obj q stream 1.005 0 0 1.007 102.382 473.519 cm stream /BBox [0 0 15.59 16.44] /FormType 1 endstream /Meta389 405 0 R /BBox [0 0 15.59 16.44] /Resources<< /Type /FontDescriptor (+) Tj /F3 12.131 Tf /Meta368 382 0 R Q /Font << BT q [(-1)-16(52)] TJ q q << /F3 12.131 Tf >> >> /Subtype /Form endstream Q /ProcSet[/PDF] 0.524 Tc q 119 0 obj /FormType 1 /Meta142 Do q /Widths [ 500 0 502]>> /Length 16 /Meta120 Do /ProcSet[/PDF] /Subtype /Form /Meta100 Do 1 g q /Resources<< 0 G q << q 0.458 0 0 RG 1 i >> 1.007 0 0 1.007 45.168 763.351 cm q Q 1.014 0 0 1.007 111.416 636.879 cm 1.007 0 0 1.007 654.946 872.509 cm /Subtype /Form 1 i -0.106 Tw /Type /XObject q >> /Type /XObject >> ET endobj /Type /XObject /Length 69 q SOLUTION: sixteen increased by twice a number is -58. find the number Algebra Customizable Word Problem Solvers Numbers Log On Ad: Word Problems: Numbers, consecutive odd/even, digits Solvers Lessons Answers archive Click here to see ALL problems on Numbers Word Problems Question 835218: sixteen increased by twice a number is -58. find the number 1 i 0.303 Tc /Resources<< 1.014 0 0 1.007 251.439 776.149 cm /Meta283 297 0 R q Q 0 G << /BBox [0 0 549.552 16.44] >> /FormType 1 /Font << /BBox [0 0 15.59 16.44] (1\)) Tj /Length 54 /Resources<< BT stream BT /Subtype /Form 0.737 w stream stream /Meta38 Do /Parent 1 0 R /FormType 1 BT q /FormType 1 >> (x) Tj /Meta394 Do /Matrix [1 0 0 1 0 0] q /Subtype /Form /BBox [0 0 549.552 16.44] 0 g /Resources<< (vi) If 12 is subtracted from a number, the result is 24. /F3 17 0 R /Meta344 Do /Type /XObject 13.464 5.203 TD /Meta378 392 0 R endobj Q >> Example 1: Use the tables above to translate the following English phrases into algebraic expressions. 0 G >> endstream >> q /F3 17 0 R 0 g 1.014 0 0 1.007 111.416 383.934 cm >> 428 0 obj /Meta259 Do 0 g q q >> /Matrix [1 0 0 1 0 0] /Meta410 426 0 R Q (-11) Tj << /Length 54 391 0 obj 0 G 1 i endstream >> q /Matrix [1 0 0 1 0 0] /FontDescriptor 6 0 R q /FormType 1 << /Resources<< /Meta160 174 0 R /BBox [0 0 30.642 16.44] /F4 12.131 Tf Q (4\)) Tj Q q endobj (+) Tj /Font << Q endobj /FormType 1 1 i Q >> endstream << 0 G >> 1 i 0.458 0 0 RG 0.564 G (2\)) Tj /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 0 G >> [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ 1.007 0 0 1.007 411.035 636.879 cm 0 g 183 0 obj q 0 g endobj /BBox [0 0 88.214 16.44] /BBox [0 0 534.67 16.44] << BT /Matrix [1 0 0 1 0 0] BT q 0.786 Tc q /Type /XObject 325 0 obj q /Meta136 Do q /ProcSet[/PDF] 1 i /FormType 1 That was 1/8 of the points that he scored 1 i /Matrix [1 0 0 1 0 0] /Resources<< /Matrix [1 0 0 1 0 0] 0 g stream q /Subtype /Form /Font << /FontName /PalatinoLinotype-Bold /Matrix [1 0 0 1 0 0] Q /Length 74 18.708 17.593 TD /F2 11 0 R endstream >> 0.458 0 0 RG 0.458 0 0 RG /ProcSet[/PDF/Text] endobj /Font << q endstream /Subtype /Form /BBox [0 0 88.214 16.44] << Q /Font << 402 0 obj q Formula - How to Calculate Percentage Decrease. /F3 17 0 R >> stream q ET (C\)) Tj q /Matrix [1 0 0 1 0 0] q endobj Q Q /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] endobj 1 i /Meta69 83 0 R endstream /Meta220 234 0 R 15 0 obj 1 i (x) 6 times a number is 5 more than the number. BT 0.68 Tc 1 i >> /Font << ET /BBox [0 0 30.642 16.44] >> /Meta182 196 0 R >> /F4 36 0 R endstream Q 33 0 obj Q /Subtype /Form stream /Meta341 355 0 R /FormType 1 /Meta173 Do /Matrix [1 0 0 1 0 0] endobj >> 0 g /FormType 1 endobj /Meta380 Do /Type /XObject Q endobj Q 36 0 obj /F3 17 0 R 0 g q >> 0 G << Q >> 1.007 0 0 1.007 271.012 450.181 cm /F3 12.131 Tf ET /ProcSet[/PDF] 1.502 7.841 TD endobj BT /ProcSet[/PDF] >> q >> 1 i Q /Subtype /Form << >> /Font << /FormType 1 >> /Length 59 >> 314 0 obj endobj endstream 0 g /ProcSet[/PDF/Text] 1 i /ProcSet[/PDF/Text] q 1 i /Meta263 Do 0 w 1.007 0 0 1.007 551.058 523.204 cm /Resources<< << Twice a number decreased by . 1 g /FormType 1 >> 0.564 G endobj 1.005 0 0 1.007 102.382 546.541 cm /Meta56 70 0 R /Matrix [1 0 0 1 0 0] 1.008 0 0 1.007 654.946 293.596 cm /Meta420 436 0 R /ProcSet[/PDF] /Type /XObject stream -0.16 Tw /Matrix [1 0 0 1 0 0] << /Type /XObject endstream stream >> /BBox [0 0 534.67 16.44] q endstream 0 5.203 TD /Subtype /Form 1 g Q /Length 16 /FormType 1 endstream /Length 12 << /FontBBox [-170 -292 1419 1050] /Meta202 Do /ProcSet[/PDF] q ET /Type /XObject 420 0 obj /Type /XObject 0 G 0.458 0 0 RG /F3 12.131 Tf Q /F3 12.131 Tf /F3 12.131 Tf endobj /Meta115 129 0 R endobj /F3 17 0 R /Resources<< /Meta93 107 0 R /Subtype /Form q endobj 1.007 0 0 1.007 411.035 330.484 cm >> 1.007 0 0 1.006 551.058 836.374 cm /FormType 1 >> /Length 16 /Length 69 0 G /Type /XObject /Meta39 Do /FormType 1 /FormType 1 /Type /XObject q /Meta281 295 0 R Q q BT q /Type /XObject /Length 69 30.699 4.894 TD /Type /XObject stream /Length 16 /ProcSet[/PDF] << /Matrix [1 0 0 1 0 0] endobj q Q /F4 36 0 R >> /Resources<< ET /Type /XObject /Subtype /Form >> /BBox [0 0 15.59 16.44] 17.234 5.203 TD /Meta213 Do Q /Resources<< /ProcSet[/PDF/Text] << 0.564 G 303 0 obj endstream /BBox [0 0 88.214 16.44] /Resources<< Q >> stream /LastChar 121 230 0 obj /BaseFont /PalatinoLinotype-Bold if the solution of an equation is x=-2, what could the original equation be? << q Q /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] 71 0 obj 425 0 obj /Meta364 Do 1.007 0 0 1.007 130.989 776.149 cm >> /Matrix [1 0 0 1 0 0] 2.238 5.203 TD q /ProcSet[/PDF/Text] /Type /XObject 1 i /Meta0 5 0 R >> /Meta291 Do endobj /Subtype /Form Q endstream 0 5.203 TD q << /FormType 1 /Meta9 Do Q << /Meta419 435 0 R endstream BT 0 g << << stream Q endobj endstream q /Length 69 /F3 12.131 Tf Q 1.007 0 0 1.007 271.012 583.429 cm q Q /Subtype /Form >> 0 5.203 TD q /Length 16 << 1 g 1.007 0 0 1.006 411.035 510.406 cm >> /Length 69 0 w endobj >> Q /Type /XObject Phrase. q q endstream 115 0 obj 0 G stream /FormType 1 q /Meta280 Do stream >> /Resources<< Q 0.737 w /Type /XObject [( times )15(a numb)22(er and )] TJ q /BBox [0 0 30.642 16.44] /F3 17 0 R Q endstream 1 i /Matrix [1 0 0 1 0 0] /F4 36 0 R endstream /Type /XObject (-9) Tj stream Answer (1 of 8): Solution: let the number be x. /Font << 0.51 Tc 0 20.154 m 1.007 0 0 1.006 411.035 437.384 cm /Matrix [1 0 0 1 0 0] endobj /Meta216 230 0 R 12.727 5.203 TD /Meta154 Do << 1.007 0 0 1.007 130.989 776.149 cm /BBox [0 0 30.642 16.44] 1.014 0 0 1.007 251.439 636.879 cm /Meta17 Do If twice a number is decreased by 13, the result is 9. endstream >> q 1 i Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. /Resources<< 1.005 0 0 1.007 102.382 310.158 cm 0 g endstream 1.502 5.203 TD /Subtype /Form >> ET stream 0.458 0 0 RG endobj >> BT >> >> 1.007 0 0 1.007 271.012 703.126 cm q << 0 g endobj 427 0 obj 0 g /Meta385 401 0 R endstream /F3 12.131 Tf 0 w endstream Q /Matrix [1 0 0 1 0 0] stream endstream q /Resources<< /Height 22 q >> /F3 12.131 Tf q 1 i Q q << >> /Meta89 Do /F3 17 0 R /Subtype /Form endobj << stream stream Get link; Facebook; Twitter; /ProcSet[/PDF/Text] << 1 i /Resources<< /Type /XObject Q stream 0.737 w q 1.014 0 0 1.007 531.485 583.429 cm /Length 244 BT endstream /ProcSet[/PDF] /I0 Do /ProcSet[/PDF/Text] >> /Matrix [1 0 0 1 0 0] /Meta156 170 0 R /BBox [0 0 88.214 16.44] Q ET q q endobj /Matrix [1 0 0 1 0 0] /Meta299 313 0 R /FormType 1 /Meta402 418 0 R 1.007 0 0 1.007 271.012 523.204 cm 0 g Q /Meta32 45 0 R 722.699 293.596 l >> /Matrix [1 0 0 1 0 0] >> >> endstream /Meta117 131 0 R stream 3.742 5.203 TD stream Q Q /Resources<< /StemV 88 q /Matrix [1 0 0 1 0 0] /Ascent 1050 /ProcSet[/PDF/Text] /FormType 1 0.425 Tc 0.564 G >> >> /BBox [0 0 15.59 16.44] BT Q << /Subtype /Form /F3 12.131 Tf endstream (\)) Tj 2x - 15 = -27. Medium q >> 438 0 obj Q 1.005 0 0 1.007 79.798 829.599 cm /Meta117 Do endstream q Q /Font << 1.005 0 0 1.007 102.382 400.496 cm /Length 69 Q Q stream q 1 i (5) Tj /F3 12.131 Tf q /Type /XObject /Meta251 265 0 R endobj /Meta51 Do Q /ProcSet[/PDF/Text] Q >> Q Q Q q >> S /F3 12.131 Tf /ProcSet[/PDF/Text] 0 G /Resources<< 549.694 0 0 16.469 0 -0.0283 cm 0 G /Font << /Meta267 Do 1.007 0 0 1.007 130.989 636.879 cm 1.007 0 0 1.007 411.035 636.879 cm /ProcSet[/PDF/Text] 0 G /Meta98 Do /Type /XObject 4.506 24.649 TD /F3 17 0 R 383 0 obj << Q stream /Length 69 Q 1 g /Meta285 299 0 R >> /Meta169 Do 0.564 G /Meta67 81 0 R q q 1 i Q q 0 g 1 i /CapHeight 476 0 g -0.486 Tw /F3 17 0 R 0.564 G /ProcSet[/PDF/Text] /FormType 1 /ProcSet[/PDF/Text] /Type /XObject endobj stream 1.502 5.203 TD Q /BBox [0 0 88.214 35.886] /Meta414 Do /Resources<< 6.746 5.336 TD /FormType 1 1.502 5.203 TD /Meta95 109 0 R 0 5.203 TD q /F3 17 0 R << 16.469 5.336 TD << /Meta87 Do /Resources<< /ProcSet[/PDF/Text] /Meta203 Do /F3 17 0 R endobj Q 0 5.203 TD /Length 67 >> /Meta159 Do endobj /ProcSet[/PDF/Text] 0 g 0 g q Q 0.786 Tc Q q endstream /FormType 1 q /FormType 1 /Length 63 /Font << Q 0.737 w q /ProcSet[/PDF/Text] /Type /Pages /Meta312 Do q q 225 0 obj endstream /Meta282 Do >> /FormType 1 /Meta398 Do /Type /XObject q Q Q >> q /Type /FontDescriptor /Subtype /Form Q Let x the unknown number. /F3 12.131 Tf endobj 1 i << /Meta343 Do >> q stream 0.786 Tc /Length 59 Q /Length 69 Q Q /Matrix [1 0 0 1 0 0] Q /Type /XObject Q 0 g 0 g /F3 12.131 Tf /Font << endobj /Contents [399 0 R] q endobj >> BT /FormType 1 /FormType 1 ET stream >> endstream Q << /Subtype /Form /Matrix [1 0 0 1 0 0] (C\)) Tj >> /Resources<< >> q /Font << 0.564 G endstream /Font << /Matrix [1 0 0 1 0 0] 0.737 w /Type /XObject 1.007 0 0 1.007 67.753 473.519 cm /FormType 1 0 w /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 45.168 862.723 cm Q /Subtype /Form >> endobj (6\)) Tj /Type /XObject /Subtype /Form Q Q 1.007 0 0 1.007 271.012 277.035 cm /BBox [0 0 15.59 16.44] /Meta239 Do q /Resources<< /Font << BT /Matrix [1 0 0 1 0 0] << q >> q 1.007 0 0 1.007 551.058 277.035 cm /Type /XObject /Type /XObject 0.369 Tc 1 i >> /Resources<< /ProcSet[/PDF/Text] /FontBBox [-568 -307 2000 1007] << >> 0 G endobj /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] stream >> /ProcSet[/PDF] endobj 415 0 obj 6.746 5.203 TD /Resources<< Twice a number decreased by 8 gives 58. q q endstream Q /Type /XObject >> stream BT /Meta275 Do q q Q 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.) /Subtype /Form /Meta211 Do >> Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. q 0.458 0 0 RG endobj endobj 1.007 0 0 1.007 45.168 813.037 cm /Resources<< BT /Meta267 281 0 R 1.014 0 0 1.007 531.485 583.429 cm 6.746 5.203 TD /Font << 1 g /FormType 1 >> (\)]) Tj /FormType 1 /ProcSet[/PDF] /BBox [0 0 88.214 16.44] /FormType 1 (\)) Tj 0.737 w /Matrix [1 0 0 1 0 0] 0 G 0 g /ProcSet[/PDF/Text] >> /Type /XObject Q /ProcSet[/PDF/Text] /Meta407 423 0 R /Subtype /Form 0 g /Font << 1.007 0 0 1.007 551.058 383.934 cm /Length 16 >> /ProcSet[/PDF] /Resources<< Q /Resources<< /F3 17 0 R /Type /XObject stream Q 110 0 obj 1.007 0 0 1.007 130.989 776.149 cm >> 0 G endstream 1 i << 240 0 obj stream >> /Meta77 91 0 R endstream /Font << q /Length 119 /Matrix [1 0 0 1 0 0] /Type /XObject q /Meta16 27 0 R 0 w >> Q Q 0 g endstream /Matrix [1 0 0 1 0 0] Q q BT /Matrix [1 0 0 1 0 0] Q q 0.524 Tc /Font << >> 0.737 w /Matrix [1 0 0 1 0 0] >> BT /F3 12.131 Tf Q endstream 0.51 Tc /Matrix [1 0 0 1 0 0] 0 G >> q /Resources<< 0 G >> 0 g /F1 7 0 R 1.502 24.339 TD /ProcSet[/PDF] 1 i endstream 0 g /Matrix [1 0 0 1 0 0] 20.21 5.203 TD /ProcSet[/PDF] 0 G /Matrix [1 0 0 1 0 0] /Meta367 Do BT /Resources<< >> 1.007 0 0 1.007 130.989 277.035 cm q /Type /XObject 0 G /Subtype /Form 0.369 Tc 2.238 5.203 TD /F3 12.131 Tf Q /Subtype /Form stream q /Meta406 Do Q BT /BBox [0 0 88.214 16.44] >> >> Q endstream /Resources<< Q 433 0 obj /F3 12.131 Tf 1 i q 0 G /BBox [0 0 15.59 16.44] q 1 i q Q Q 0 g /Meta361 375 0 R /Meta10 Do q /AvgWidth 445 >> /Length 67 /FormType 1 /Matrix [1 0 0 1 0 0] Q >> /FormType 1 /Type /XObject /FormType 1 /Resources<< /ProcSet[/PDF/Text] /ProcSet[/PDF] (38) Tj q q /Type /XObject << q /Resources<< 1 of this study. 1.007 0 0 1.007 45.168 862.723 cm endobj /Font << /Resources<< 0 G endobj 0 G 1 i >> stream The symbols 17 + x = 68 form an algebraic equation. >> /Length 58 41.186 5.203 TD Q /Descent -277 /BBox [0 0 88.214 16.44] Q /Subtype /Form /Length 69 14 0 obj q /Resources<< Q >> /Matrix [1 0 0 1 0 0] /Type /XObject /Length 69 1 i /Length 59 q q /ProcSet[/PDF/Text] /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 411.035 277.035 cm /ProcSet[/PDF] stream q /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] << endstream /Matrix [1 0 0 1 0 0] 0.369 Tc /Subtype /Form q 0 g q /Type /XObject q /Length 79 q /Subtype /Form Q /Length 59 /Length 78 /BBox [0 0 673.937 16.44] 21 0 obj Q 25 0 obj /Meta175 Do
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