twice a number decreased by 58

ET 0 g /F4 36 0 R /Type /XObject 87 0 obj /FormType 1 /Subtype /Form /Meta199 213 0 R >> >> /Type /XObject endobj /ProcSet[/PDF] /ProcSet[/PDF/Text] (9\)) Tj >> 0 g /Length 69 /Subtype /Form /Resources<< q 0.564 G /F3 12.131 Tf 0.838 Tc 0 g >> (x ) Tj Q Q /BBox [0 0 673.937 15.562] 0.458 0 0 RG Q BT q /Subtype /Form >> 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 1 i /Length 16 stream BT 1.014 0 0 1.006 251.439 690.329 cm q >> 53 0 obj /Length 12 /Resources<< >> /Meta262 Do q stream stream /Meta35 48 0 R 0 g /F1 12.131 Tf /Subtype /Form >> (C\)) Tj /Type /XObject 1.007 0 0 1.007 551.058 636.879 cm << /FormType 1 >> /Subtype /Form /Meta318 332 0 R << /Subtype /Form ET << /F3 12.131 Tf q Q /F3 17 0 R /Meta92 Do /I0 Do 1.502 24.649 TD 1 i 0 w q /ProcSet[/PDF] q 0.369 Tc Q endstream Q /Type /XObject /Length 206 Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . 1 i /F3 12.131 Tf << Q /F3 12.131 Tf Q 1.007 0 0 1.007 551.058 583.429 cm endobj /Meta2 9 0 R q /Subtype /Form endobj 0.486 Tc endstream (38) Tj 0.458 0 0 RG /Length 78 stream /BBox [0 0 88.214 16.44] endobj 0.458 0 0 RG >> /Type /XObject ET q << << q /BBox [0 0 15.59 16.44] >> /Type /XObject /F1 7 0 R 0 G 0 w /BBox [0 0 88.214 16.44] /Meta20 31 0 R Find the number. Q /ProcSet[/PDF/Text] Q << /ProcSet[/PDF/Text] Q /F1 12.131 Tf >> /F3 12.131 Tf >> 1 g /Font << 0 5.203 TD 0.458 0 0 RG endobj BT 0.51 Tc endstream /BBox [0 0 673.937 14.853] 0 5.203 TD 1 i >> /Meta186 Do 1.007 0 0 1.007 411.035 277.035 cm /FormType 1 /ProcSet[/PDF] Q endstream 1 i 1.005 0 0 1.007 79.798 862.723 cm q 422 0 obj Q 1.007 0 0 1.007 130.989 583.429 cm /F3 17 0 R /Resources<< >> 0.425 Tc /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 5.203 TD /Subtype /Form >> 218 0 obj Q /FormType 1 445 0 obj /Type /XObject >> q 125 0 obj Q /F3 12.131 Tf << 0 5.203 TD q /FormType 1 /Length 16 Q /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] 582 546 601 560 395 424 326 603 565 834 516 556]>> /FormType 1 /FormType 1 /Length 16 /F1 7 0 R /FormType 1 0.458 0 0 RG Q /ProcSet[/PDF/Text] 0 w /Meta373 387 0 R >> << >> endstream BT 20.21 5.203 TD 1 i Twice a number would be 2x. /Resources<< /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] << q 1.007 0 0 1.007 67.753 293.596 cm /FormType 1 0.564 G stream 1 i /Font << >> /Matrix [1 0 0 1 0 0] 1 g /BBox [0 0 15.59 16.44] >> /Length 68 /Subtype /Form /BBox [0 0 15.59 29.168] 1.007 0 0 1.006 411.035 690.329 cm q << BT 1 g >> q [( and )16(a nu)26(mbe)18(r)] TJ >> endobj /Matrix [1 0 0 1 0 0] /Length 16 Q /BBox [0 0 88.214 16.44] Q 0 g /Matrix [1 0 0 1 0 0] Q q Q >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] q endstream 1 i q q 30.699 4.894 TD 30.699 5.203 TD 20.21 5.203 TD (38) Tj 321 0 obj q Q 1 i q /Type /XObject Two fewer than a number doubled is the same as the number decreased by 38. Q 1.007 0 0 1.007 271.012 776.149 cm stream 1 i stream 1 i /Meta47 Do 1 i /F3 17 0 R 1 i q /Meta294 308 0 R >> Q BT /Meta58 72 0 R >> /Meta323 337 0 R endobj /BBox [0 0 639.552 16.44] q << /Matrix [1 0 0 1 0 0] q Q /ProcSet[/PDF/Text] Q /FormType 1 /ProcSet[/PDF/Text] /FormType 1 q /Resources<< stream Q /I0 51 0 R /Meta396 Do Q Q Q q 0 w Q endobj Q /F3 12.131 Tf /Resources<< /Type /XObject 0 20.154 m 0.564 G /ProcSet[/PDF/Text] /Type /XObject Q /FormType 1 /Matrix [1 0 0 1 0 0] /F1 7 0 R q /BBox [0 0 88.214 16.44] /Resources<< /Type /XObject q 1 i q >> 0 G 1.502 5.203 TD 1 i endobj /Matrix [1 0 0 1 0 0] >> /Length 69 /Font << << 0.382 Tc /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 3.742 5.203 TD /XObject << stream ET 1.007 0 0 1.007 551.058 583.429 cm /ProcSet[/PDF] stream 0.311 Tc 1.007 0 0 1.007 271.012 636.879 cm stream /Type /XObject /Meta256 Do ET /Subtype /Form Q << Q /Length 60 /Type /XObject endstream >> endobj >> >> endstream q Q The ratio of a number to fifteen 4. /Length 12 /Subtype /Form /Length 67 3.742 5.203 TD >> /BBox [0 0 534.67 16.44] /Type /XObject endobj >> /Meta54 Do stream Q q Q 300 0 obj 20.21 5.203 TD 0 g /Meta409 425 0 R ET Q 0 w /BBox [0 0 88.214 35.886] endobj q A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. /Subtype /Form endstream /FormType 1 q q 318 0 obj /Meta211 225 0 R q q /Resources<< /Meta269 Do /Matrix [1 0 0 1 0 0] << q Q /BBox [0 0 30.642 16.44] /Length 118 BT /FormType 1 0.458 0 0 RG 0.297 Tc /FormType 1 0.737 w /BBox [0 0 88.214 16.44] /Length 12 /FormType 1 /Meta329 Do Q ET q 0.737 w /Subtype /Form ET 0 G /Matrix [1 0 0 1 0 0] 178.979 5.203 TD /ProcSet[/PDF] /Type /XObject ET /F3 12.131 Tf /Subtype /Form BT /Meta45 59 0 R endstream q 0.524 Tc /FormType 1 >> 1 i Q 85 0 obj /ProcSet[/PDF] /Type /XObject /F1 7 0 R endobj /Resources<< /Meta310 Do Q /Matrix [1 0 0 1 0 0] 0.458 0 0 RG Two speeding tickets could increase your rate by 58% at your next renewal. /Font << /BBox [0 0 88.214 35.886] 0 w Q 1.014 0 0 1.007 251.439 277.035 cm 549.694 0 0 16.469 0 -0.0283 cm q 0 g /F3 12.131 Tf 0 g << /FormType 1 /Meta283 297 0 R >> << q /Length 69 >> >> 57 0 obj 1 i /Meta103 Do >> /Subtype /Form /Type /XObject 427 0 obj 0 4.894 TD /Length 12 q Q q >> endstream 209 0 obj -37 VI 2. /F3 12.131 Tf /Type /XObject /ProcSet[/PDF] q /BBox [0 0 88.214 16.44] q 1.007 0 0 1.007 411.035 583.429 cm 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 >> 0 g /Meta122 Do ET Q /Matrix [1 0 0 1 0 0] 315 0 obj /Subtype /Form /Length 54 /Length 16 endobj 0 G q 1 i /Meta97 111 0 R /Type /XObject >> 0.838 Tc 1 i >> 1 i /Matrix [1 0 0 1 0 0] >> >> /Matrix [1 0 0 1 0 0] q /Type /XObject >> /Resources<< stream /Matrix [1 0 0 1 0 0] q /BBox [0 0 30.642 16.44] /BBox [0 0 88.214 35.886] /Type /XObject endstream /Type /XObject /F3 17 0 R 0 g /F3 12.131 Tf Q /Font << /Matrix [1 0 0 1 0 0] /Subtype /Form algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. /Subtype /Form 1.007 0 0 1.007 271.012 450.181 cm /Type /XObject q >> /Meta191 205 0 R Q stream >> /Matrix [1 0 0 1 0 0] 125.064 4.894 TD /Subtype /Form 0 g >> 0 g /Length 69 >> /ProcSet[/PDF] 0 g Q q << 0 g /Meta370 384 0 R >> /F3 17 0 R q >> 1 i Q Twice = two times, double. 229 0 obj endstream /ProcSet[/PDF/Text] stream Q /F3 17 0 R /BBox [0 0 15.59 16.44] 435 0 obj BT >> Q 0 w 0 5.203 TD endstream /Resources<< /Resources<< << /Meta40 54 0 R /F1 12.131 Tf endobj /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 45.168 889.071 cm 1 i >> If LtitnS6S . endobj endstream >> /Meta16 Do /FormType 1 >> endobj (40) Tj endobj << /MaxWidth 1453 Q /XObject << ET /Length 70 q /Length 80 /Resources<< 1 i >> endstream 0 5.203 TD >> Q q << q q 52.412 5.203 TD BT >> /Type /XObject /FormType 1 0 w ET q 0.564 G 0 g << 0 g /Type /XObject Q >> /Subtype /Form /Subtype /Form /Meta42 Do BT /Type /XObject q /F3 17 0 R >> Q 1.007 0 0 1.007 654.946 473.519 cm /Matrix [1 0 0 1 0 0] >> /Resources<< << << /Font << Q /F3 12.131 Tf 1. ET Q Q endobj 1 i /ProcSet[/PDF/Text] q Q 0 G /Meta368 382 0 R /Meta354 Do endobj q /Type /XObject 391 0 obj to represent the numbers. In other terms, 52-nx The problem is asking that you subtract twice a number from 52. q q q /Type /XObject /Font << Q stream << endstream Q 1.502 5.203 TD /BBox [0 0 88.214 16.44] 416 0 obj 1.007 0 0 1.006 130.989 437.384 cm Q endstream 1.007 0 0 1.007 271.012 703.126 cm /Matrix [1 0 0 1 0 0] q << /Type /XObject q 307 0 obj 0.369 Tc twice a number decreased by 58 13 - 3x B. twice a number a divided by three = 2a / 3. five times a number x minus four = 5x - 4. thrice the sum of a number x and six = 3 (x + 6) Add your answer and earn points. /Resources<< /Resources<< >> >> BT /Resources<< >> 0 g 1.007 0 0 1.006 411.035 437.384 cm 9.723 5.336 TD BT /Type /XObject 1 i 101.849 5.203 TD endobj BT You could call them, Decreased by another number means subtract. 244 0 obj 1 i /Type /XObject /FormType 1 /F3 17 0 R 1 i ET /Meta242 Do 0 5.203 TD /ProcSet[/PDF/Text] /Resources<< /F1 7 0 R /Matrix [1 0 0 1 0 0] Q /BBox [0 0 88.214 16.44] stream /ProcSet[/PDF/Text] >> ET 0 0 Similar questions Find the number which when decreased by 8% becomes 506. /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] << q >> /Subtype /Form /Type /XObject /BBox [0 0 673.937 15.562] 1.014 0 0 1.007 111.416 776.149 cm 0 g Q 1 i /F3 12.131 Tf Q endstream /Subtype /Form /FormType 1 /Subtype /Form q /Matrix [1 0 0 1 0 0] /Type /XObject /Meta6 Do Q /Meta3 Do >> /Matrix [1 0 0 1 0 0] >> /Length 69 432 0 obj q Q 0 g /Resources<< endobj q stream /Subtype /Form 0.524 Tc q /I0 Do Q /Type /XObject Q 0.737 w /F3 12.131 Tf >> stream 0 G (38) Tj /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1 i /Type /XObject << /BBox [0 0 88.214 16.44] /Meta304 318 0 R /BaseFont /PalatinoLinotype-Bold 0.271 Tc 1 i >> /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 531.485 330.484 cm >> /ProcSet[/PDF/Text] /ItalicAngle 0 Q >> /Resources<< Q /BBox [0 0 15.59 16.44] 0 5.203 TD /Meta76 90 0 R 0 g >> /F3 17 0 R Q 10.487 5.203 TD 1 g Q 326 0 obj 0 g >> q q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] endobj << q >> >> 0 G >> /Matrix [1 0 0 1 0 0] /Subtype /Form Q (7\)) Tj (B) Tj /Length 63 q 0 G endobj Three times a number equals fifteen 3. 0 g >> >> Q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf << /ProcSet[/PDF] Q 1.005 0 0 1.015 45.168 53.449 cm 0 g /Resources<< /BBox [0 0 30.642 16.44] Q q /Font << q /F3 12.131 Tf endstream /FontDescriptor 16 0 R ET /FontBBox [-568 -307 2000 1007] /ProcSet[/PDF/Text] /Meta227 241 0 R Q /F3 17 0 R /FormType 1 /Matrix [1 0 0 1 0 0] endobj /Length 58 /Resources<< endobj Q q 0 g q >> 1 g 1.007 0 0 1.007 271.012 636.879 cm /Subtype /Form 0.524 Tc >> /Meta160 Do q >> Q endstream q 19.474 5.203 TD Q (2) Tj -0.486 Tw 1.007 0 0 1.007 271.012 583.429 cm /Descent -216 /Subtype /Form >> << q 1 i /Subtype /Form >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Subtype /Form q (B\)) Tj /FormType 1 BT ET /Matrix [1 0 0 1 0 0] Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . endobj BT 1.007 0 0 1.006 551.058 437.384 cm /Length 16 endobj >> /Meta259 Do 177 0 obj >> /Root 2 0 R q /Type /XObject q Five times a number, decreased by 58, is -23 Find the number. q q 370 0 obj 0 G /Meta277 291 0 R << Q /Meta112 126 0 R /Meta352 366 0 R Q S q /BBox [0 0 88.214 16.44] stream /Meta132 146 0 R ET << >> /Matrix [1 0 0 1 0 0] Q (B) Tj Q Q t is 56: 4. 0 g endstream q /BBox [0 0 88.214 16.44] >> /Font << Q q BT Q /Subtype /Form 3.742 5.203 TD Q 0 g 1.007 0 0 1.006 551.058 763.351 cm q Q 367 0 obj /Length 118 0 G /Resources<< endobj q q /Font << /Resources<< q 0.786 Tc endstream Q << /Resources<< /F3 17 0 R 19 0 obj /Meta355 369 0 R /Meta125 139 0 R /F3 12.131 Tf endstream /Font << /MaxWidth 1248 0 G /Length 64 /FormType 1 0.737 w Q /F3 12.131 Tf q /Resources<< /Type /XObject %PDF-1.4 << Q (2) Tj /Matrix [1 0 0 1 0 0] /FormType 1 /BBox [0 0 88.214 16.44] 13.493 5.203 TD 19.474 5.203 TD /Length 58 /Length 69 /Type /XObject 6.746 5.203 TD /Meta119 133 0 R << >> /Meta156 170 0 R 0.369 Tc q >> 220.931 4.894 TD >> /ProcSet[/PDF/Text] /Meta389 Do /F3 17 0 R 1 g >> BT Q 1.007 0 0 1.007 271.012 776.149 cm /Type /XObject 0 g /ProcSet[/PDF/Text] 0.737 w q ET 0.458 0 0 RG 1 g /Meta58 Do 1 i ( x) Tj >> >> /Resources<< stream Q -0.16 Tw >> Q 13.464 5.203 TD (B\)) Tj /Type /XObject 0 G 0 g 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.458 0 0 RG 0.564 G /F4 12.131 Tf 1.007 0 0 1.007 271.012 330.484 cm Q endstream /Subtype /Form /Meta87 Do /Meta265 Do /Subtype /Form >> 0 G >> Q << q /Length 16 >> q endobj 1 i q ET /BBox [0 0 534.67 16.44] BT q /Meta147 Do /Meta117 131 0 R >> 1 i /Length 69 Q /Subtype /Form /Meta250 Do /BBox [0 0 15.59 16.44] >> q 0.737 w 679.036 293.596 m 0 g /Matrix [1 0 0 1 0 0] q >> endstream endobj Q 1.005 0 0 1.007 102.382 363.608 cm endstream /Subtype /Form 0.737 w 2.238 5.203 TD endobj /Length 69 q q q endobj >> Q /ProcSet[/PDF/Text] /Length 65 q BT q << 722.699 599.991 l stream endobj 0 G Q 1 i /F3 17 0 R stream ET /Meta243 Do 12.727 5.203 TD 0.458 0 0 RG Q Q /Subtype /Form Q /F3 12.131 Tf /BBox [0 0 88.214 16.44] /Length 59 /BBox [0 0 88.214 16.44] /ProcSet[/PDF] /Type /XObject stream endstream /Length 60 1 i endstream >> 0.458 0 0 RG /Meta266 280 0 R 0 G /Meta207 Do 127 0 obj /Length 74 /Font << endobj Q /ProcSet[/PDF/Text] Q 437 0 obj In the problem above, x is a variable. /Meta162 176 0 R 1 i /BBox [0 0 639.552 16.44] 1 Data in this Fast Fact represent the 50 states and the District of Columbia. >> /ProcSet[/PDF] /FormType 1 /F1 12.131 Tf >> /Subtype /Form stream 0 5.203 TD q 0.486 Tc /Font << Thrice of a number = 3x. Q q /Matrix [1 0 0 1 0 0] >> Q Q ET /Resources<< 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. q Q /FormType 1 << [tex]\sin (\pi -x)=\sin x[/tex]. q Six subtracted from a number 6. /Subtype /Form >> /Resources<< << Thrice a number decreased by 5 exceeds twice the number by 1 is . /F3 17 0 R /Type /XObject /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 670.003 cm Q endstream /ProcSet[/PDF/Text] 30.699 5.203 TD /FormType 1 0 4.894 TD /Matrix [1 0 0 1 0 0] ET << BT /BBox [0 0 88.214 16.44] /Font << /Length 69 /BBox [0 0 88.214 16.44] stream 0.737 w ( x) Tj << q stream q >> 1.007 0 0 1.007 411.035 330.484 cm /F3 12.131 Tf /BBox [0 0 15.59 29.168] 0 G /ProcSet[/PDF/Text] q 0.178 Tc /Font << 0 g /Subtype /Form 1 i stream /Meta324 338 0 R Q /Subtype /Form q Q /Meta424 Do >> 0 g stream 12.727 24.649 TD >> ([x ) Tj /Matrix [1 0 0 1 0 0] Q >> 0 G 1.007 0 0 1.007 551.058 330.484 cm << /Resources<< 36 0 obj Q ET /ProcSet[/PDF] /Type /XObject q q BT /F3 17 0 R << endobj q /Subtype /Form /F3 12.131 Tf >> /Meta120 Do << Q q 0.68 Tc Q endobj q /BBox [0 0 549.552 16.44] >> /Meta204 Do

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