1 i BT /F3 17 0 R 1 g 0 w 1.014 0 0 1.007 251.439 849.172 cm 1. 0.564 G /Type /XObject /F3 17 0 R endstream << Five times the sum of a number and four 7. /Type /XObject Q Q q q /FormType 1 /F3 17 0 R 1.007 0 0 1.007 411.035 583.429 cm stream Q stream /ProcSet[/PDF] /Type /XObject /Font << /F3 17 0 R 0.458 0 0 RG /Type /XObject >> /Matrix [1 0 0 1 0 0] /F3 17 0 R q q Q Q q q /BBox [0 0 88.214 35.886] 1 i endstream ET Q 0 g /Meta306 Do stream BT Q stream Q endstream endstream 1 i /Matrix [1 0 0 1 0 0] /Length 69 >> q q 0 g Q /Meta218 232 0 R Q /Subtype /Form /Type /XObject q >> >> stream 1 i >> /Length 69 /Type /XObject /F3 17 0 R /BBox [0 0 15.59 16.44] << endobj Q q Q q /Type /XObject /ProcSet[/PDF] -0.486 Tw algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. >> /BBox [0 0 17.177 16.44] /Meta46 60 0 R 0 g /FormType 1 Q /BBox [0 0 30.642 16.44] 0 w endstream q 1 i 0.737 w /BBox [0 0 88.214 35.886] 21 0 obj A. >> /ProcSet[/PDF] Q /Resources<< Q 1.014 0 0 1.007 391.462 636.879 cm /BBox [0 0 15.59 16.44] So let's go ahead and identify a v q Q Q ( decreased by ) Tj q /ProcSet[/PDF/Text] >> Q /Resources<< >> q Q Q Q q BT [(-3)-16(20)] TJ Q << 1 i BT /FormType 1 0.737 w /Font << >> (13) Tj >> stream 18.708 17.593 TD /Meta204 218 0 R 0 G The sum of 18 and tour times a number is -6 Find the number. endobj q >> q Twice a number decreased by 58! BT Answer by Mathtut (3670) ( Show Source ): /F3 12.131 Tf 1.014 0 0 1.006 391.462 510.406 cm Q (1\)) Tj >> endstream 286 0 obj >> /Resources<< << 0 G /Subtype /Form /Matrix [1 0 0 1 0 0] q /Font << q 0.737 w 0 g 0 w (5) Tj q /ProcSet[/PDF/Text] >> 1 i 0 G BT /Type /XObject /Length 245 0.737 w 315 0 obj (x ) Tj Q q /Meta129 143 0 R 0 G /Subtype /Form /Subtype /Form /Length 54 /FormType 1 >> stream q Q >> /Widths [ 500 0 502]>> endstream /Length 80 q 0 G /FormType 1 stream /ProcSet[/PDF/Text] /Subtype /Form /FormType 1 0 g endobj /Font << /Length 12 endobj B. /Resources<< >> endobj /FormType 1 << /Meta301 Do /BBox [0 0 88.214 35.886] /Type /XObject 1.007 0 0 1.007 271.012 583.429 cm >> 0 5.203 TD /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] endstream /Resources<< /Type /XObject 390 0 obj q q BT endobj endstream Q /Resources<< >> /BBox [0 0 88.214 16.44] Q >> 0 g Q /Font << /ProcSet[/PDF/Text] /F3 17 0 R /F3 12.131 Tf /BBox [0 0 88.214 16.44] 1 i Find the number. /Length 16 /FormType 1 0.297 Tc /FormType 1 /Subtype /Form endstream /Resources<< endstream /Subtype /Form q >> /FormType 1 q >> /F3 17 0 R /Meta227 241 0 R BT 1.007 0 0 1.007 551.058 703.126 cm >> endobj q q Q /FormType 1 ET stream /F3 17 0 R 433 0 obj /Length 67 0 g >> Q /Length 59 /Meta139 153 0 R ET /FormType 1 Q Q /FormType 1 >> /BBox [0 0 549.552 16.44] /Font << /Length 59 [(A number )-17(divided by )] TJ /Font << Q endstream /Font << /BBox [0 0 88.214 16.44] /FormType 1 q << Q Q /FormType 1 242 0 obj Q 61 0 obj >> q /I0 Do ET 1 i Q /Meta144 Do /FormType 1 >> /Type /XObject 0 w /Meta358 Do BT 1.007 0 0 1.007 271.012 776.149 cm /FormType 1 /Length 78 BT >> /Length 69 /BBox [0 0 17.177 16.44] 0.297 Tc endobj 0 g 445 0 obj Q q /BBox [0 0 88.214 35.886] Q endobj >> /Subtype /Form /F1 7 0 R >> >> q Q Q Q 1 g 1 i /Subtype /Image 1.014 0 0 1.007 391.462 583.429 cm /FormType 1 /Resources<< /Meta221 Do ET endobj /ProcSet[/PDF/Text] /BBox [0 0 15.59 16.44] q Q 0.369 Tc /F4 12.131 Tf ET /Meta207 221 0 R >> /BBox [0 0 88.214 16.44] << ET 1 g q 549.694 0 0 16.469 0 -0.0283 cm q ([x ) Tj For the lesson, he grabs a glass container shaped like a rectan /Resources<< 171 0 obj BT 0 g /Font << /Subtype /Form /ProcSet[/PDF/Text] >> >> q /F3 17 0 R 0 G /Matrix [1 0 0 1 0 0] 0 g stream /BBox [0 0 88.214 16.44] /Flags 32 Q << q ET /F3 12.131 Tf 350 0 obj 89.12 5.203 TD /Type /XObject q << Q Q /Subtype /Form /Subtype /Form q Q >> /ProcSet[/PDF] /Meta116 Do stream /Matrix [1 0 0 1 0 0] /Meta252 Do 1 i /F3 12.131 Tf /F3 12.131 Tf << /Subtype /Form >> 1.502 5.203 TD 0.297 Tc 0 g Q >> 0 G 1.007 0 0 1.007 271.012 523.204 cm /Matrix [1 0 0 1 0 0] Q Q stream /Subtype /Form endobj << Q Q << 0 G Q 0 g >> 1.502 5.203 TD ET stream 0.737 w Q /Meta338 352 0 R /Resources<< 0.458 0 0 RG /Subtype /Form /Resources<< endobj Q ET 1 i /BBox [0 0 15.59 16.44] 9 0 obj stream 0 G >> 0 w /F3 17 0 R 0.285 Tc 1.005 0 0 1.007 102.382 400.496 cm 270 0 obj << /ProcSet[/PDF] /ProcSet[/PDF] stream endstream /FormType 1 >> /FormType 1 (A\)) Tj ET q /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 102.382 473.519 cm q /XObject << 20.21 5.203 TD 0 G /F3 17 0 R Q >> /ProcSet[/PDF] Q Q q 20.975 5.336 TD /F3 12.131 Tf >> << endstream 1.007 0 0 1.007 271.012 583.429 cm q BT q 0 G q /Resources<< 1 i BT q (-) Tj /Resources<< 3 0 obj 0 w stream 0.737 w (\)) Tj stream /ProcSet[/PDF] 0.564 G /BBox [0 0 534.67 16.44] << /Meta105 Do Q (5) Tj /BBox [0 0 88.214 16.44] /Flags 32 0.564 G 0 5.203 TD >> /Type /XObject q /ProcSet[/PDF/Text] /BBox [0 0 88.214 35.886] /Length 69 q >> Q /Length 244 stream 0 g /F1 12.131 Tf Q /Type /XObject q 672.261 347.046 m 0 g /BBox [0 0 88.214 16.44] 0.564 G q << >> >> q q /F4 12.131 Tf /F3 17 0 R Q >> stream << Q 0 G 60 0 obj 0 g 0 g /Length 19882 /Length 54 Q >> /Resources<< Q 0 w 17.234 5.203 TD 1.007 0 0 1.007 271.012 636.879 cm endstream >> q Q >> q >> /Meta258 Do q endstream /ProcSet[/PDF/Text] /Font << [(F)-22(ive)] TJ 1 i 0 G /FormType 1 1 i /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 45.168 889.071 cm /Meta315 Do Q Q q q 1.014 0 0 1.006 251.439 510.406 cm /Meta330 Do >> /Meta280 Do 152 0 obj /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 546.541 cm 1 i BT /Meta60 Do 0.458 0 0 RG q endstream 416 0 obj /FormType 1 BT Q q /Font << /Meta308 Do /Subtype /Form 1.014 0 0 1.007 391.462 849.172 cm q /F1 12.131 Tf Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] /FormType 1 1 i /Length 54 << 0 5.203 TD /Type /XObject 1.005 0 0 1.007 102.382 726.464 cm 1 i Q q Q /FormType 1 q /Meta149 Do /Length 68 (x ) Tj /Type /XObject endobj 0 w >> endstream >> 0 g /Subtype /Form /BBox [0 0 639.552 16.44] q >> BT /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] ET /Subtype /Form Q >> q /Type /XObject 110 0 obj endstream /Subtype /Form 1.502 5.203 TD /Type /XObject 0 g /Subtype /Form [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. 0 w /BBox [0 0 88.214 16.44] q q In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. Q Q /Matrix [1 0 0 1 0 0] Q /Subtype /Form << /Length 59 q q Q /ProcSet[/PDF/Text] /Resources<< (A\)) Tj BT 1.007 0 0 1.007 271.012 523.204 cm Q 1 i /Resources<< Q /Subtype /Form /Font << >> 1 i (3) Tj 98.843 5.203 TD q 0.737 w 0 g >> /Font << endobj /F3 17 0 R /Font << /F1 12.131 Tf /Matrix [1 0 0 1 0 0] /Font << << >> >> Q /Font << >> 1.007 0 0 1.007 654.946 726.464 cm /Length 58 >> endstream >> 1.007 0 0 1.007 45.168 713.666 cm Q /FormType 1 BT 1.007 0 0 1.007 271.012 703.126 cm /Meta42 Do /Matrix [1 0 0 1 0 0] /Flags 32 /Type /XObject endstream 0 G Q >> Q q /Font << /Meta293 307 0 R q stream endstream /BBox [0 0 88.214 16.44] Q endobj (40) Tj >> /Meta311 Do q /F3 17 0 R /Meta54 68 0 R endstream /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] 0 g 180 0 obj endstream >> >> ET 0.458 0 0 RG /Meta15 26 0 R /F3 17 0 R stream /Resources<< q /Meta48 62 0 R 0 G Q /Length 69 << /Type /XObject Q /F3 17 0 R /Type /XObject /ProcSet[/PDF/Text] /Resources<< 0.458 0 0 RG 3.742 5.203 TD /F3 12.131 Tf 0.564 G endstream q q /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 277.035 cm q endstream /Font << /FontName /TimesNewRomanPSMT /FormType 1 << << /Meta215 229 0 R 0 5.203 TD Q /F1 12.131 Tf /Resources<< 1 i /FormType 1 Q q q ET endobj This s problem could be, interpreted either way. 0 G /Font << 1.005 0 0 1.007 102.382 256.709 cm ET /ProcSet[/PDF/Text] /Type /Pages Q >> Q 1.014 0 0 1.007 111.416 330.484 cm q 32.201 5.203 TD BT /BBox [0 0 639.552 16.44] /Matrix [1 0 0 1 0 0] endstream Q (7\)) Tj 0 g q /Font << 0 w /Resources<< 0.458 0 0 RG q endobj 0 G 1.502 5.203 TD /Meta50 64 0 R /Subtype /Form Q stream q endstream /F3 17 0 R /ProcSet[/PDF/Text] ET Kobe scored 85 points in a basketball game. << q /Resources<< ET >> >> q q /BBox [0 0 88.214 16.44] 0 G 1 i endobj 1 g 48 0 obj /Font << /Meta236 Do /Length 99 1.005 0 0 1.015 45.168 53.449 cm /Resources<< /Type /XObject /F4 12.131 Tf Q Q /BBox [0 0 88.214 16.44] Q q Q endobj Q /FormType 1 /Subtype /Form 24 0 obj 0 G >> endobj /Resources<< S /Type /XObject Q /Resources<< q stream /Font << >> 25.454 5.203 TD q q /Font << /Resources<< 5 0 obj << 20.21 5.203 TD stream 0 G /F3 12.131 Tf endobj (D\)) Tj /Length 12 /Subtype /Form /Meta289 303 0 R 326 0 obj << endstream >> /F3 12.131 Tf 1.007 0 0 1.007 67.753 726.464 cm 0 g 1 g << Q /Meta411 427 0 R Q 0 g 1 i /Subtype /Form 1 g 1.007 0 0 1.007 654.946 599.991 cm 1 i Q >> 0 G /Resources<< 1.014 0 0 1.007 531.485 277.035 cm Q q 0.564 G Q Q /Font << /Meta146 Do Q q /F3 17 0 R q Q >> 1 i /Meta81 Do >> /F3 17 0 R -0.029 Tw stream /Meta266 280 0 R /F3 17 0 R 1.502 7.841 TD Q /FormType 1 Q 0 G /Length 16 << >> 1.005 0 0 1.007 102.382 799.486 cm 1 i endstream Q BT /FormType 1 /Subtype /Form Q /Meta77 91 0 R stream (\)) Tj Q /Matrix [1 0 0 1 0 0] 0.737 w /Length 59 Q 0.297 Tc 0 G 1 i Q q /Meta413 Do 1.005 0 0 1.015 45.168 53.449 cm /Matrix [1 0 0 1 0 0] Q q 1.502 5.203 TD /Subtype /Form 0 g Q /ProcSet[/PDF] >> /FormType 1 >> /FormType 1 /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Subtype /Form endobj /Matrix [1 0 0 1 0 0] 0 w Q 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 /Subtype /Form 1 g ET /Meta414 Do 1.007 0 0 1.007 411.035 636.879 cm BT >> 0 w stream /Resources<< Q 0 g 1.014 0 0 1.007 531.485 383.934 cm >> 135 0 obj /Meta158 Do ( x) Tj q 1.014 0 0 1.007 111.416 636.879 cm /Matrix [1 0 0 1 0 0] /Resources<< Q q /Length 80 Q /Meta195 209 0 R 23.216 5.203 TD /XObject << q << /Resources<< 1 i Q stream (-) Tj Q 1.007 0 0 1.007 551.058 330.484 cm /Type /XObject 1.007 0 0 1.007 45.168 763.351 cm /Font << /Type /XObject q /FormType 1 >> << endobj Q 1 g >> /Meta213 Do 0 g q >> /Subtype /Form 0 G /Subtype /Form Q q stream /ProcSet[/PDF] q 0.458 0 0 RG endstream /Type /XObject 1 i /ProcSet[/PDF] endstream /F3 17 0 R ET (3) Tj /FormType 1 1 i endobj /F4 36 0 R q 3.742 5.203 TD /ProcSet[/PDF/Text] /FormType 1 first we change the sentence in formula as the following, i think this is clear &easy to understand .i hope it helps you, This site is using cookies under cookie policy . ET >> 0 G /Resources<< /Subtype /Form endobj /Matrix [1 0 0 1 0 0] /Subtype /Form /F4 36 0 R /FormType 1 /FormType 1 Q 0 g /Subtype /Form /ProcSet[/PDF/Text] q q -0.486 Tw 1 i /Subtype /Form >> >> q Q Q /Meta357 371 0 R 1 i Q /F3 17 0 R /Meta235 Do (2\)) Tj /Subtype /Form >> << /Type /XObject endobj << /Matrix [1 0 0 1 0 0] /Subtype /Form /BBox [0 0 15.59 16.44] endobj >> stream >> /FormType 1 0.738 Tc q << /ProcSet[/PDF/Text] << 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 1 i q Q << Q 35 0 obj /F3 12.131 Tf 0 g Q /Type /XObject /FormType 1 0 w /BBox [0 0 15.59 29.168] BT 0.458 0 0 RG 1 i >> /F3 17 0 R /Matrix [1 0 0 1 0 0] /Meta226 Do Q /Meta282 Do /BBox [0 0 88.214 16.44] 146 0 obj 1 g /Type /XObject /Resources<< q q /FormType 1 << /Meta121 135 0 R ET /F3 12.131 Tf << /Length 69 1.007 0 0 1.007 411.035 636.879 cm q >> for the season. 0.458 0 0 RG >> 0 G 0.458 0 0 RG Q q << /Matrix [1 0 0 1 0 0] q stream Let x the unknown number. /Subtype /Form >> /F1 7 0 R ET Q q 0 G /F3 17 0 R >> /Meta145 Do /Subtype /Form stream q q 0 g /Matrix [1 0 0 1 0 0] q >> /Font << ET endstream 108 0 obj 375 0 obj endobj Q /Subtype /Form 0 G q 27.693 5.203 TD stream q /F3 17 0 R /Matrix [1 0 0 1 0 0] 0 G 0 g q BT /BBox [0 0 88.214 16.44] >> Q /Meta59 Do /Font << 1.007 0 0 1.007 411.035 330.484 cm /FormType 1 q /Length 12 /Length 69 stream 1 i Mat 1 g q Making educational experiences better for everyone. BT /Font << /F4 12.131 Tf /Meta24 37 0 R /FormType 1 endstream >> /Meta21 Do >> 1 i >> 0 w /F1 12.131 Tf /Type /XObject 1.014 0 0 1.007 531.485 450.181 cm (\(x ) Tj endobj /Type /XObject /Length 79 /Font << /Subtype /Form /Meta209 Do /Length 80 q >> 1 i >> << /Type /XObject /Meta168 182 0 R stream /ProcSet[/PDF/Text] q << endobj /Subtype /Form BT /Resources<< /Font << ET /FormType 1 >> 0 G /Length 69 Q 1.007 0 0 1.007 271.012 523.204 cm /Length 99 /Font << 0 g (C\)) Tj stream endstream << /Font << q stream Q 1.007 0 0 1.007 45.168 779.913 cm 0.737 w 417 0 obj /Resources<< /Meta284 298 0 R /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] Q 1.007 0 0 1.007 271.012 383.934 cm endobj /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] 0.51 Tc 0 g /Type /XObject endobj Q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf << /FormType 1 >> Q BT /Matrix [1 0 0 1 0 0] /Type /Font /Resources<< /Resources<< Q /Type /XObject Q Q /F4 36 0 R q << /Length 74 Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. 6 0 obj 1 i Q endobj answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. q 0.737 w /Font << 1 i endobj Q 0 g 0.564 G 0.564 G 0 g q 0 4.894 TD << >> /Length 69 1.014 0 0 1.007 111.416 330.484 cm /Resources<< /FormType 1 q /Matrix [1 0 0 1 0 0] 80 0 obj /Matrix [1 0 0 1 0 0] /Meta401 Do /Encoding /WinAnsiEncoding 0.458 0 0 RG endobj q endstream 228 0 obj >> /Subtype /Form /Length 245 /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /Resources<< >> /Meta48 Do endstream /Resources<< q 6.746 5.203 TD /Length 16 Q /BBox [0 0 15.59 16.44] 1 i endstream << /FormType 1 0.786 Tc Q 0 G 0.838 Tc -0.463 Tw >> /FormType 1 stream 1.007 0 0 1.007 411.035 277.035 cm BT 0 g 13 0 obj Q S /Resources<< /Resources<< endstream /Length 64 endobj endobj q >> Q /Length 69 /Pages 1 0 R Q BT endstream Q q << 1 i /Meta363 Do stream Q 0.838 Tc /Font << << /BBox [0 0 534.67 16.44] stream ET /Meta331 Do endstream /BBox [0 0 88.214 16.44] 0 g (38) Tj /ProcSet[/PDF] stream /Meta217 231 0 R /Subtype /TrueType 1.007 0 0 1.007 654.946 546.541 cm Q (5\)) Tj ET Q /Resources<< /Resources<< /Resources<< 54.679 5.203 TD q BT >> 0 g Q /BBox [0 0 15.59 16.44] /Meta349 363 0 R Q BT /FormType 1 stream q /Meta363 377 0 R 0 w << /Meta380 394 0 R >> >> /BBox [0 0 88.214 35.886] q /BBox [0 0 88.214 16.44] endobj /F1 12.131 Tf >> /Resources<< Q Q << 0 g endobj << q (4\)) Tj endobj /Meta367 Do Q 0.458 0 0 RG stream /Meta360 374 0 R 0 g /BaseFont /PalatinoLinotype-Roman endobj /Subtype /Form /BBox [0 0 549.552 16.44] /F3 12.131 Tf Advertisement Loved by our community 50 people found it helpful Madhvendra13 2x -8=58 2x=66 x=662 x=33 Find Math textbook solutions? 78 0 obj q 1.007 0 0 1.007 130.989 383.934 cm >> BT 1.007 0 0 1.007 551.058 636.879 cm q << /ProcSet[/PDF/Text] q endobj /ProcSet[/PDF] >> /Length 118 /Matrix [1 0 0 1 0 0] stream >> /Type /XObject >> 1.005 0 0 1.007 102.382 400.496 cm >> /Length 59 275 0 obj endobj q [( the )-24(sum of a n)-14(umber an)-14(d )] TJ In the problem above, x is a variable. /Meta196 Do >> 0 G 52 0 obj /Type /XObject /Meta277 Do >> /Meta299 Do << 1 i 0 G >> Q q /Type /XObject /ProcSet[/PDF] /ProcSet[/PDF] S ET /ItalicAngle 0 /Subtype /Form /Type /XObject Q 1.007 0 0 1.006 551.058 836.374 cm A link to the app was sent to your phone. 1 i 186 0 obj 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject q Calculate a 15% decrease from any number. /Subtype /Form 396 0 obj 184 0 obj >> 1 i /Subtype /Form 0.458 0 0 RG /ProcSet[/PDF/Text] << /I0 Do >> ET 312 0 obj << Q /F3 17 0 R >> q /Subtype /Form /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /F3 17 0 R >> >> /Meta39 Do /ProcSet[/PDF/Text] 1 i q q q q (1\)) Tj /Length 54 >> >> 0.564 G q /Matrix [1 0 0 1 0 0] 0 g /Matrix [1 0 0 1 0 0] 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . 370 0 obj /Meta392 Do endobj Q /Length 68 BT Q /BBox [0 0 534.67 16.44] ET >> /Meta219 Do >> Q ET 1 i /Meta130 144 0 R /Ascent 1050 0 g >> /FormType 1 /Length 69 Q stream endobj /Matrix [1 0 0 1 0 0] /Type /XObject Q Q << /FormType 1 0 G /Meta301 315 0 R ET 0 g /Length 68 >> [tex]\sin (\pi -x)=\sin x[/tex]. /Subtype /Form 0 g (5) Tj /Length 69 BT 0 g /Type /XObject /F1 12.131 Tf 14.23 24.649 TD q /F1 7 0 R /F3 12.131 Tf BT << 1 g /ProcSet[/PDF] Q /Font << >> /FormType 1 /Matrix [1 0 0 1 0 0] /Resources<< q /FormType 1 Q /ProcSet[/PDF] Q 322 0 obj /Resources<< /Resources<< 1.005 0 0 1.007 102.382 347.046 cm q /FormType 1 q /Resources<< endstream Q 1 i stream 1.007 0 0 1.007 271.012 849.172 cm /Matrix [1 0 0 1 0 0] /Meta116 130 0 R q /F1 12.131 Tf /Meta332 Do /FormType 1 1 i /I0 51 0 R q stream Q /Matrix [1 0 0 1 0 0] /FormType 1 /Subtype /Form endstream /Meta213 227 0 R stream 1.007 0 0 1.007 45.168 746.789 cm endobj q q /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) << 0 G /FormType 1 /F3 12.131 Tf /Resources<< Q /Meta7 18 0 R /Resources<< /Meta355 Do /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 130.989 277.035 cm >> 1.502 7.841 TD >> stream 1 i /Length 118 1 i /ProcSet[/PDF] q /BBox [0 0 88.214 16.44] 82 0 obj /FormType 1 1 i 1 i 1 i 0 G 0 g 343 0 obj q 672.261 653.441 m 0.369 Tc q Q << /Meta404 Do /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] 0 g endobj /Meta194 Do /Resources<< q 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 . 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 /Matrix [1 0 0 1 0 0] >> /Length 16 /Meta150 Do q /Length 16 /StemV 94 /Subtype /Form /Subtype /Form q /Subtype /Form 49 0 obj /F1 7 0 R q /FormType 1 Q Q endobj 549.694 0 0 16.469 0 -0.0283 cm /Subtype /Form Q endstream 0 w >> /ProcSet[/PDF/Text] 14.966 20.154 l Decreased by another number means subtract. stream /F1 7 0 R q /F3 12.131 Tf q Q /BBox [0 0 17.177 16.44] >> 1.007 0 0 1.007 130.989 277.035 cm endstream /Meta389 Do >> 0.134 Tc Q /Font << /Matrix [1 0 0 1 0 0] << >> 9.723 5.336 TD /Length 69 1 i << 1 i /FontBBox [-90 -216 1195 800] Q /F3 12.131 Tf 1.007 0 0 1.007 654.946 546.541 cm /Resources<< >> q 0 g Q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] (5) Tj /F3 17 0 R /BBox [0 0 30.642 16.44] /FormType 1 >> q /Resources<< endobj (C\)) Tj q 1 i (58) Tj /Resources<< /Type /XObject /BBox [0 0 15.59 16.44] endobj /BBox [0 0 673.937 16.44] /Type /XObject /ProcSet[/PDF/Text] q << /Type /XObject q /Resources<< >> 271 0 obj endobj Q /FormType 1 q /Resources<< /ProcSet[/PDF/Text] Q /Font << /F1 7 0 R q Q /FormType 1 ET 1.005 0 0 1.013 45.168 933.487 cm ET /Meta151 165 0 R 0 G 0 g /MaxWidth 1248 endobj /Matrix [1 0 0 1 0 0] 10.487 5.203 TD endstream endstream << BT endobj /ProcSet[/PDF/Text] << /F1 12.131 Tf q /Meta210 Do /Font << 0.458 0 0 RG 1.007 0 0 1.007 551.058 523.204 cm BT /FormType 1 1.005 0 0 1.007 79.798 746.789 cm Q /Resources<< 1.007 0 0 1.007 654.946 653.441 cm q /F2 11 0 R endstream >> stream 1.005 0 0 1.007 102.382 417.058 cm Q &K @ 43.426 5.203 TD endobj stream /BBox [0 0 88.214 16.44] Q /Subtype /Form 0.458 0 0 RG Q >> >> endobj /Type /XObject ET /Meta232 Do /FormType 1 /F1 12.131 Tf /Type /XObject 16.469 5.203 TD q /Subtype /Form /ProcSet[/PDF/Text] >> /BBox [0 0 15.59 16.44] /Meta347 361 0 R Q /Length 69 stream /ProcSet[/PDF/Text] 0 w /F3 17 0 R /CapHeight 694 Q 0.564 G 0.175 Tc /Length 69 /Type /XObject q 0 G 1.007 0 0 1.007 654.946 400.496 cm /ProcSet[/PDF] 0.737 w 0 5.203 TD 0.737 w /FormType 1 /F3 17 0 R /ProcSet[/PDF] /Subtype /Form /BBox [0 0 639.552 16.44] Q /Meta210 224 0 R /Meta390 406 0 R Q >> /F3 12.131 Tf /FormType 1 /Subtype /Form /Matrix [1 0 0 1 0 0] /Type /XObject /Length 59 >> Q /Subtype /Form 0 G >> /ProcSet[/PDF/Text] stream Q 0.564 G /Matrix [1 0 0 1 0 0] /Meta417 Do /Meta209 223 0 R Q >> >> q Patients' reasons for declining screening were not collected . /Meta295 309 0 R /Type /XObject 0 G q q Q /Matrix [1 0 0 1 0 0] /Font << 1.007 0 0 1.007 654.946 400.496 cm << q /Type /XObject endstream /Subtype /Form q /Resources<< Q q q /Meta317 Do 1.007 0 0 1.007 411.035 849.172 cm 0 g >> 0 g >> stream Q >> /BBox [0 0 30.642 16.44] /Subtype /Form stream /Matrix [1 0 0 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