>> stream Q 0.564 G 293 0 obj /Length 70 << /Subtype /Form (7\)) Tj endobj Q q /Meta142 156 0 R /F1 12.131 Tf /F3 12.131 Tf >> /ProcSet[/PDF] 672.261 653.441 m /Length 95 >> 0 g 0.175 Tc /BBox [0 0 17.177 16.44] q >> 1 i >> endobj Q /XObject << 1 i << 0 g 44 0 obj q Q Q /Meta324 Do /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 45.168 889.071 cm Q 0.838 Tc (A\)) Tj /Meta160 174 0 R BT q Q 0.458 0 0 RG /Font << /Type /XObject /F3 12.131 Tf >> Q Q 0.564 G /BBox [0 0 534.67 16.44] /F4 36 0 R 1.005 0 0 1.013 45.168 933.487 cm (-) Tj stream endobj ET /BBox [0 0 534.67 16.44] Q Q /FormType 1 q << /ProcSet[/PDF/Text] stream << 0.458 0 0 RG q /Type /XObject q /F3 17 0 R Q /Resources<< /F3 17 0 R endstream endobj endobj Q 1.005 0 0 1.007 102.382 599.991 cm stream 0.737 w endobj /Meta165 Do Q /Meta120 Do -0.486 Tw /Type /XObject >> >> 1 g Q Q q 2. 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 1.007 0 0 1.007 130.989 523.204 cm /Type /XObject endstream /Meta66 80 0 R Q /Meta38 Do /Subtype /Form 0 g << /Length 16 1 i (9) Tj 1 i 0 w /Type /XObject 0.564 G >> /Type /XObject 1 i /FormType 1 >> >> Q q >> 162 0 obj /ProcSet[/PDF/Text] Q << 1.007 0 0 1.007 271.012 330.484 cm /Font << 1 i /Meta384 398 0 R 1 i 0 G >> 429 0 obj 1 i Q >> 0.68 Tc endobj /Matrix [1 0 0 1 0 0] /Meta305 Do ET BT q /Meta264 Do 1 i /BBox [0 0 30.642 16.44] /Meta372 386 0 R 1.007 0 0 1.007 130.989 636.879 cm /Type /XObject /Meta278 Do /BBox [0 0 17.177 16.44] << Answer: 52 decreased by twice a number in algebric expression Step-by-step explanation: The problem is asking that you subtract twice a number from 52. Q ET q BT stream /Subtype /Form >> 1 i /BBox [0 0 88.214 16.44] /Length 69 << endstream /Length 16 Q 1.007 0 0 1.007 411.035 277.035 cm (D\)) Tj /Font << stream (-) Tj /Subtype /Form Q /F3 17 0 R /Meta72 Do /F3 12.131 Tf endobj 0 G /BBox [0 0 639.552 16.44] /F3 12.131 Tf q /Subtype /Form Q >> Q >> q /Font << q BT 1 i /Type /XObject /Type /XObject /Meta236 Do 0 g endstream 443 0 obj >> /F3 17 0 R >> 0 g /Subtype /Form 0 w q endobj >> endobj You could call them. endobj /Subtype /Form Q 1.014 0 0 1.006 251.439 437.384 cm >> endstream 0 g >> /FormType 1 Q /ProcSet[/PDF/Text] /F4 36 0 R Find the number. 220 0 obj /Meta247 261 0 R /Resources<< 0 20.154 m 0.425 Tc 0 g /Meta4 Do /Length 16 Q /Font << /FormType 1 /Meta286 Do /Meta129 Do Q 1 i >> >> 1.007 0 0 1.007 411.035 849.172 cm stream 1 i (x) 6 times a number is 5 more than the number. /ProcSet[/PDF] 19.474 20.154 l /ProcSet[/PDF] >> stream /Meta298 Do /ProcSet[/PDF/Text] /Subtype /Form /Type /XObject endstream stream ET >> /Resources<< /F3 17 0 R /Length 2252 /Resources<< BT q Q Q q 0 g 0 G 0 g Q >> endstream /ProcSet[/PDF/Text] 0 5.203 TD /F1 7 0 R /FormType 1 /Meta17 28 0 R /Length 69 30.699 4.894 TD q TJ B. /ProcSet[/PDF] /Meta71 Do endstream 0 g /BBox [0 0 30.642 16.44] 0.564 G q 20.21 5.203 TD /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] /FormType 1 /F3 17 0 R << /Subtype /Form /Resources<< 0.369 Tc /Resources<< >> q q /Subtype /Form 344 0 obj /Resources<< /Meta307 Do /ProcSet[/PDF] /Type /XObject /F3 17 0 R 1 g Q << endstream endobj /FormType 1 /BBox [0 0 88.214 16.44] /FormType 1 /Type /XObject endstream 1.005 0 0 1.007 102.382 743.025 cm /Length 68 549.694 0 0 16.469 0 -0.0283 cm Q stream q 0.838 Tc Q /Matrix [1 0 0 1 0 0] >> Q /Font << Q /FormType 1 Q 1.005 0 0 1.007 79.798 813.037 cm q Q q BT q /Type /XObject >> 0.458 0 0 RG /Type /XObject >> 121 0 obj /Length 69 /Subtype /Form 0.458 0 0 RG << endobj q 0 g 0 g (5) Tj /Matrix [1 0 0 1 0 0] /Type /XObject Q 0.737 w /Meta289 303 0 R /F4 12.131 Tf /F3 12.131 Tf /Matrix [1 0 0 1 0 0] >> /FormType 1 /ProcSet[/PDF/Text] 0 g /ProcSet[/PDF] /ProcSet[/PDF/Text] q /Length 73 /Subtype /Form << ET Q Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. /ProcSet[/PDF/Text] startxref /Meta377 391 0 R 0 G /BBox [0 0 30.642 16.44] /Type /XObject q << /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /ProcSet[/PDF/Text] endobj 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] q 146 0 obj endobj /Subtype /Form >> stream 0 g Q /Length 85 stream q >> /Type /XObject 0 G q Q 0 g /BBox [0 0 534.67 16.44] << endstream Q 1 g (13) Tj endobj 1 i /BBox [0 0 88.214 16.44] << /Meta406 Do /FormType 1 0 g Q << /Type /XObject /F3 17 0 R 1 i /ProcSet[/PDF/Text] << Q /F3 12.131 Tf q /Meta405 Do Q q 0.737 w a Question /Meta66 Do q 0.737 w 1.007 0 0 1.007 130.989 776.149 cm 0 G /Resources<< /F3 17 0 R 1 i 0 G /Meta341 355 0 R q /Type /XObject /Length 64 /Subtype /Form << q /Meta61 Do q /ProcSet[/PDF/Text] >> /Meta380 Do /Resources<< /Meta190 204 0 R 0 g 0 w 0.134 Tc /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] >> 178.979 5.203 TD >> stream Q /Subtype /Form >> Q 20.21 5.203 TD Q /BBox [0 0 15.59 29.168] endstream -0.106 Tw /Meta363 Do /BBox [0 0 30.642 16.44] /Type /XObject q /Meta375 389 0 R << q Q 1.007 0 0 1.007 551.058 277.035 cm q /I0 Do ET endstream 0.564 G /ProcSet[/PDF] /Font << q /ProcSet[/PDF/Text] Q q ET Q 1 g 679.036 293.596 m 0 g 0 G q -y. /F3 12.131 Tf /Meta150 Do /Meta318 Do /Meta100 114 0 R q /BBox [0 0 88.214 16.44] q >> stream << endstream 0 g /Font << /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Length 16 /Subtype /Form /FormType 1 stream q q /Meta396 412 0 R /Subtype /Form /Subtype /Image /FormType 1 /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] (6\)) Tj 1 g endstream /F3 12.131 Tf 1 i >> /Length 60 endobj /F2 11 0 R 0.369 Tc >> q /Type /XObject 0 g /Meta302 316 0 R << endobj 0 5.203 TD /BBox [0 0 88.214 16.44] Q /BBox [0 0 88.214 16.44] 0 g /Resources<< endobj /Font << 0 g ET 1 i /Resources<< 1 i Q endstream /Resources<< q stream /Type /XObject q ET << >> (\)) Tj 1 i /Matrix [1 0 0 1 0 0] /Resources<< q /ProcSet[/PDF] 722.699 293.596 l << /Type /XObject (-) Tj 1.007 0 0 1.007 551.058 636.879 cm q Let the 2nd number be y. << /ProcSet[/PDF] /Meta360 374 0 R 124 0 obj /ProcSet[/PDF] 0.564 G /F3 12.131 Tf endstream endobj 0 g 1.502 5.203 TD /Type /XObject 0 G 0 G 0 w endstream /BBox [0 0 30.642 16.44] /ProcSet[/PDF] /Meta200 Do endobj Q 0 g Q 0 g Q /Length 294 0 g q /Meta133 147 0 R q /F1 7 0 R /F3 12.131 Tf Q q >> 1 i 205.199 4.894 TD Q ET 0.458 0 0 RG 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. 254 0 obj /Meta57 Do Q /ProcSet[/PDF/Text] 363 0 obj Q /Font << Q Q 32.201 5.203 TD endobj BT /Matrix [1 0 0 1 0 0] ET >> << 314 0 obj /Font << Have a nice day! Q Q 232 0 obj 0.838 Tc 1 g 0.458 0 0 RG n 11 or n 11. /XObject << /Resources<< q 247 0 obj >> << endstream /F3 12.131 Tf /Meta14 25 0 R /Resources<< 32.201 20.154 l 722.699 599.991 l 1.007 0 0 1.007 551.058 277.035 cm /Font << endstream 0 G 1 i /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q ET >> /FormType 1 (-) Tj /Font << /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0.737 w /Meta324 338 0 R /Matrix [1 0 0 1 0 0] /Font << 1 i ET /Subtype /Form q >> ET q /Meta157 171 0 R endstream ET /BBox [0 0 17.177 16.44] Q Q >> /Length 16 /Type /XObject q [( and )16(a nu)26(mbe)18(r)] TJ >> 0 G 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) /XObject << endstream << /FormType 1 Q endobj 0.425 Tc >> /FormType 1 /Meta215 229 0 R 372 0 obj endobj << Q /FormType 1 endstream /FormType 1 /Meta41 Do /Length 54 /Meta69 83 0 R /FormType 1 16.469 5.203 TD endobj Q q /Font << q Q /FormType 1 /ProcSet[/PDF/Text] Q endstream (\)) Tj /BBox [0 0 30.642 16.44] q << 0.786 Tc 1 i /Meta174 Do 0 G /F4 36 0 R /FormType 1 Q /Font << q q On the way, we sang songs for 20 minutes which is 10% of the time we were on the road. 8.985 20.154 l /Length 69 /F3 17 0 R >> >> Q /ProcSet[/PDF/Text] 0 g >> >> /Type /XObject /Type /XObject >> endstream 0 g 0.68 Tc 1 i /Length 69 Q /Meta423 439 0 R 0 G Answer (1 of 8): Solution: let the number be x. /Meta205 219 0 R 126 0 obj /F3 17 0 R q Q stream Q Q stream stream Q 145 0 obj Q /ProcSet[/PDF] q /Type /XObject /Type /XObject >> << [(Fiv)25(e ti)18(me)16(s)] TJ q 0.564 G Q /Ascent 976 250 0 obj Q >> Three times a number equals fifteen 3. 0 g endobj Q q 0 G 0 G /Meta12 23 0 R /BBox [0 0 15.59 16.44] /F3 12.131 Tf << /Meta263 Do /Length 16 /Meta113 Do ET q >> << 1 i S q /Subtype /Form q 1 i /Matrix [1 0 0 1 0 0] /F4 12.131 Tf >> 0 g /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] Q /Matrix [1 0 0 1 0 0] Q /FormType 1 endstream Q >> endobj endstream /Type /XObject q >> /Meta401 Do /Length 65 Q /Meta82 96 0 R /Type /XObject Q /Font << q /FormType 1 stream 0.458 0 0 RG Q >> /Resources<< Q >> endobj /XHeight 447 << q << /Matrix [1 0 0 1 0 0] Q /Font << endstream /BBox [0 0 88.214 16.44] >> >> Q 1 i /Resources<< 1 i BT Q 0 g >> Q /FormType 1 0.838 Tc >> /Type /XObject (A\)) Tj >> Q 0.458 0 0 RG Q 0 g /Type /XObject BT >> /ProcSet[/PDF/Text] 0.564 G endstream /ProcSet[/PDF/Text] Q /Type /XObject ET stream Q q /ProcSet[/PDF/Text] 1.007 0 0 1.006 411.035 510.406 cm BT /Subtype /Form /ProcSet[/PDF] /FormType 1 /Length 59 /Meta270 284 0 R /Matrix [1 0 0 1 0 0] Q /Resources<< /Type /XObject /FontDescriptor 10 0 R /BBox [0 0 15.59 16.44] >> 0 g /Meta93 Do Q 0 5.203 TD endstream >> (C\)) Tj q BT 0 G >> 0 G 1 g q (5) Tj /Resources<< /Resources<< >> /Length 88 /Subtype /Form /ProcSet[/PDF] 1.014 0 0 1.007 251.439 776.149 cm q << endobj SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. /F3 12.131 Tf /Matrix [1 0 0 1 0 0] << >> /BBox [0 0 88.214 16.44] 0.786 Tc << q 0 g stream /Meta286 300 0 R /Type /XObject Q the quotient of twenty and a number a.) /Length 57 /Font << q Q /F4 36 0 R endobj (iii) 25 exceeds a number by 7. /Meta167 Do 0 g /Type /XObject /Type /XObject q /Font << >> Q /FormType 1 >> q >> /Meta293 307 0 R /Meta207 Do stream 0 w /Length 103 /Meta113 127 0 R 1.007 0 0 1.006 130.989 437.384 cm >> 1 i 25.454 5.203 TD endstream /Font << 0.269 Tc /Resources<< /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] stream << Q 0.564 G /Length 69 /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 169 0 obj q /Resources<< /FormType 1 149 0 obj /Matrix [1 0 0 1 0 0] stream (C\)) Tj Q q q q >> q Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. stream Q 0 g /ProcSet[/PDF/Text] /ProcSet[/PDF] 249 0 obj Q /Meta13 24 0 R /Resources<< 430 0 obj >> 1.007 0 0 1.007 271.012 703.126 cm /Subtype /Form /BBox [0 0 534.67 16.44] /Resources<< Q 0 G >> << << 0 g q /ProcSet[/PDF] /Meta376 390 0 R q stream /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 383.934 cm stream q q stream 0 g 0.382 Tc /Length 69 /F3 17 0 R /FormType 1 Q (+) Tj 0 G q /Type /XObject A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. /FormType 1 /F3 17 0 R 354 0 obj endstream (-) Tj endstream 4 0 R q /Meta290 304 0 R q << 0 G q /Matrix [1 0 0 1 0 0] endobj 0.564 G 133 0 obj /Meta77 91 0 R 0 g stream 0 g 1 g Q 0.737 w endstream endstream << /F3 12.131 Tf 0.369 Tc 0 w q /Type /XObject /Type /XObject /Resources<< q >> (D) Tj >> endstream /BBox [0 0 88.214 35.886] Thrice a number decreased by 5 exceeds twice the number by a unit. (D\)) Tj /Font << /F2 12.131 Tf Q /FormType 1 /Meta143 Do /Type /XObject stream /Matrix [1 0 0 1 0 0] /Meta55 Do /Meta199 213 0 R 1 i >> 0 G >> /Matrix [1 0 0 1 0 0] 0 G 0 g q endobj /Meta380 394 0 R q q /Matrix [1 0 0 1 0 0] >> /Resources<< q /F3 12.131 Tf 370 0 obj q 0.458 0 0 RG /BBox [0 0 88.214 16.44] stream /Subtype /Form This s problem could be, interpreted either way. >> q stream endstream >> /FormType 1 q << >> /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] Q << 0.458 0 0 RG 0 g 0 G 1.007 0 0 1.007 551.058 636.879 cm 0 g q Q: when six times a number is decreased by 4, the result is 8. >> /Length 59 /Font << /Meta390 Do 0 g Q q /Matrix [1 0 0 1 0 0] /Meta277 Do Q q /F3 12.131 Tf endobj /Matrix [1 0 0 1 0 0] >> endstream /F1 12.131 Tf /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] 175 0 obj For the lesson, he grabs a glass container shaped like a rectan endobj /Resources<< 0 g q >> >> /Meta284 298 0 R 1 i BT 0 4.894 TD 0.564 G /BBox [0 0 639.552 16.44] 76 0 obj >> 0 w 0 g /ProcSet[/PDF/Text] endobj /Matrix [1 0 0 1 0 0] Q 1 i /Meta283 297 0 R Q 1 i Q /Meta287 301 0 R 330 0 obj 0 G /Matrix [1 0 0 1 0 0] /Length 69 Q 0 g q q stream /Resources<< endobj /Length 69 /Length 69 1.007 0 0 1.007 411.035 277.035 cm >> stream Testosterone is the primary sex hormone and anabolic steroid in males. >> Q endobj 1 i q Q Q Q Q endstream /F1 12.131 Tf >> >> q Q stream endobj >> Step 1/1. ET endobj Q /ProcSet[/PDF/Text] endstream endstream q Q 1 i /Font << /Type /XObject /Length 67 /Resources<< 1 i endobj /Resources<< 6.746 5.203 TD >> /Type /XObject ET Q endstream 1.007 0 0 1.007 271.012 383.934 cm /Matrix [1 0 0 1 0 0] /Resources<< 1 i /Resources<< stream /ProcSet[/PDF] 128 0 obj ET 1 i /Meta389 Do << /F1 12.131 Tf endobj /FormType 1 endstream /Subtype /Form Q >> /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Type /Font /ProcSet[/PDF] /ProcSet[/PDF] Q 1 Data in this Fast Fact represent the 50 states and the District of Columbia. 216 0 obj /Subtype /Form /Matrix [1 0 0 1 0 0] q /Resources<< /Subtype /Form q q /BBox [0 0 15.59 16.44] Q /ProcSet[/PDF] /Length 67 /Subtype /Form /Type /XObject 0.737 w endstream 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 . /ProcSet[/PDF] q /FormType 1 /F3 17 0 R BT ET 0 G stream q 1.007 0 0 1.007 654.946 347.046 cm /Type /XObject q /Type /XObject /Type /XObject BT /Meta39 Do 0 g Q endobj Q q /Type /XObject /F3 12.131 Tf /StemV 77 Q /Length 16 /Font << 297 0 obj (9\)) Tj endstream >> stream /Meta382 396 0 R endobj 1 i >> /F3 12.131 Tf q Q /Width 734 endstream q 1.007 0 0 1.007 271.012 583.429 cm >> stream q q 0 G /Resources<< 0 G stream q TJ q /Subtype /Form endstream /F3 12.131 Tf endstream Q Q Q >> q 0.51 Tc 0 g /Meta30 43 0 R q /Subtype /Form >> /Resources<< /ProcSet[/PDF] (2) Tj /Meta292 306 0 R (x ) Tj >> /Meta106 120 0 R /Subtype /Form /Meta171 185 0 R /Meta190 Do /Font << << Q /F3 12.131 Tf 549.694 0 0 16.469 0 -0.0283 cm 0.564 G << endstream q endobj /BBox [0 0 534.67 16.44] /Length 12 ET /Meta201 215 0 R /Type /XObject 1 i 1.014 0 0 1.007 251.439 277.035 cm /F3 12.131 Tf /Length 59 0 G /Subtype /Form stream /Meta121 135 0 R >> q << /Length 16 >> stream endobj 722.699 726.464 l 2x + 3 y equal to 13 and xy = 4 find the value of x cube + 27 y cube, x/3 2/x = 1/2please solve this question.

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