<< >> /Type /XObject 1.014 0 0 1.006 531.485 437.384 cm Q /FormType 1 0 G 0 G endobj Q 0 g /Meta219 233 0 R q 20.21 5.203 TD /BBox [0 0 30.642 16.44] q endobj /Subtype /Form q /F3 17 0 R >> BT 1 i Q q 0 g 1 i /F3 17 0 R /ProcSet[/PDF] << 1.007 0 0 1.007 130.989 383.934 cm 351 0 obj Most questions answered within 4 hours. (iv) A number exceeds 5 by 3. endobj /FormType 1 1.007 0 0 1.007 411.035 636.879 cm 0.564 G ET 363 0 obj 0.369 Tc Q /FormType 1 (6\)) Tj Q /Matrix [1 0 0 1 0 0] Q /Meta231 Do /FormType 1 /Meta113 127 0 R << /ProcSet[/PDF/Text] q /Type /XObject 376 0 obj >> /BBox [0 0 30.642 16.44] endobj /Type /FontDescriptor /Font << [(The )-16(s)15(um )-14(of )] TJ /Meta271 Do << ET q q 161 0 obj >> /Meta247 Do /XObject << >> q /F3 17 0 R >> /Subtype /Form << q Q ET Q Q q /Matrix [1 0 0 1 0 0] q Q >> ET /Subtype /Form /ProcSet[/PDF] /Meta320 334 0 R 0 g /Meta159 173 0 R /Meta5 14 0 R -0.486 Tw q /ProcSet[/PDF/Text] endobj /Font << 0.227 Tc /Resources<< ET /ItalicAngle 0 /Type /XObject 0 g Q /F3 12.131 Tf >> /Meta44 58 0 R q 1 i >> (13) Tj 0.463 Tc /Font << /Font << q S /Font << q 1.005 0 0 1.007 102.382 473.519 cm /F3 17 0 R /Matrix [1 0 0 1 0 0] Five times the sum of a number and four 7. /F3 17 0 R >> q q /Meta62 76 0 R q /Font << q endstream /Resources<< 1 i /Meta252 Do ET /Subtype /Form q 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-0.486 Tw /Resources<< /F3 12.131 Tf Q Q Ten divided by a number 5. << >> /Subtype /Form /Meta31 44 0 R /F3 12.131 Tf 0.564 G 2x - y = 6. x + 3y = -25. /Matrix [1 0 0 1 0 0] 0.564 G /Meta255 269 0 R (+) Tj stream stream Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. 0 w >> endobj stream /Meta70 Do 1.014 0 0 1.007 111.416 636.879 cm 1.014 0 0 1.007 391.462 703.126 cm 1 i endobj /BBox [0 0 88.214 16.44] /Font << /Subtype /Form 87 0 obj 1 i /BBox [0 0 30.642 16.44] 110 0 obj /BBox [0 0 15.59 29.168] q /ProcSet[/PDF] /Resources<< endstream /Matrix [1 0 0 1 0 0] /F3 12.131 Tf << 1 i /Matrix [1 0 0 1 0 0] Q /Length 59 /F3 12.131 Tf /Length 16 1 i /Type /XObject Q /Meta362 376 0 R 0 G q 3.742 5.203 TD /Matrix [1 0 0 1 0 0] >> 1 i /Type /XObject 1 i ET q q Q /Meta72 Do /Type /XObject /ProcSet[/PDF/Text] Q /ProcSet[/PDF/Text] endstream stream 70 0 obj << /Meta225 Do 1.007 0 0 1.007 271.012 636.879 cm 0 w 1 i 9.723 5.336 TD q 0.564 G >> /Matrix [1 0 0 1 0 0] << Q endstream q stream BT 130 0 obj BT /F1 12.131 Tf Q /Meta3 12 0 R 0.369 Tc 0.51 Tc q /Meta75 Do stream << BT /F3 17 0 R Q 346 0 obj q /Meta405 421 0 R /Matrix [1 0 0 1 0 0] 1.502 5.203 TD /BBox [0 0 534.67 16.44] Q << /Resources<< 1.007 0 0 1.007 130.989 383.934 cm q /Meta246 260 0 R Q stream /Subtype /Form 1.007 0 0 1.006 551.058 763.351 cm 219 0 obj q stream Q /Meta309 Do Q endobj q /Meta218 Do /Length 245 BT /Type /XObject /FormType 1 BT /Length 118 /FormType 1 /F3 17 0 R << q endobj /Meta235 Do 0 G >> /Font << Q q endobj endobj Q >> endstream /Font << >> 0.564 G 0 w 0.458 0 0 RG Q /Subtype /Form /Type /XObject >> q BT 0 g 0 G 0.68 Tc /Type /XObject /Resources<< q >> /F3 17 0 R 0.564 G << stream 2.238 5.203 TD 0 g /Meta374 Do 0.564 G /Font << 549.694 0 0 16.469 0 -0.0283 cm /Matrix [1 0 0 1 0 0] >> 1.007 0 0 1.007 130.989 849.172 cm /Type /XObject stream /Type /XObject 0 g >> /Matrix [1 0 0 1 0 0] stream Q q endobj Q Q /Subtype /Form /FormType 1 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/ProcSet[/PDF/Text] /Subtype /Form /FormType 1 0 5.203 TD Q endstream /F4 12.131 Tf /Type /XObject q /Meta397 413 0 R /Matrix [1 0 0 1 0 0] 1 g 0 g /Meta412 Do /Font << /Matrix [1 0 0 1 0 0] /Type /XObject /ProcSet[/PDF] q >> endobj 1 i /Type /XObject q 0 g Q q endobj Q 16.469 5.336 TD >> (D\)) Tj 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject endstream 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 . 0.285 Tc /ProcSet[/PDF] BT << Q /Resources<< /F1 12.131 Tf q 0 4.894 TD BT /ProcSet[/PDF] q /Type /XObject /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta306 320 0 R /Length 69 Q Q >> >> /Type /XObject /Matrix [1 0 0 1 0 0] 0 G 1.007 0 0 1.007 130.989 277.035 cm /Type /XObject BT "49 . 1.007 0 0 1.007 67.753 599.991 cm ET ET 1.007 0 0 1.007 271.012 849.172 cm Q << stream q /Meta387 Do /Resources<< (+) Tj q >> >> q /DecodeParms [<> ] endstream /Meta141 Do /Matrix [1 0 0 1 0 0] >> q 0 G /Font << 0 g /Matrix [1 0 0 1 0 0] >> Q q /Resources<< 0 w Q /Subtype /Form q 0.458 0 0 RG Q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 776.149 cm /Subtype /Form endobj endobj << 0.155 Tc 245 0 obj /Meta242 256 0 R 1 i >> q Q (x ) Tj endobj endobj -0.106 Tw >> >> 386 0 obj /Meta193 207 0 R 26.219 5.203 TD /Subtype /Form >> /BBox [0 0 15.59 16.44] /Meta25 38 0 R /Matrix [1 0 0 1 0 0] 1.502 5.203 TD endstream /Meta185 199 0 R 0 w 1 i 1 i 244 0 obj >> endstream BT /Matrix [1 0 0 1 0 0] Q /Type /XObject 0.458 0 0 RG /Meta333 Do 1 i Q 0 G 1 g 1.014 0 0 1.007 111.416 523.204 cm Q q stream /Font << >> /Type /XObject q >> 1.007 0 0 1.007 411.035 849.172 cm BT 1 i 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 endstream BT q /Matrix [1 0 0 1 0 0] stream q q /Meta370 Do /ProcSet[/PDF/Text] 0 5.203 TD 57.656 5.203 TD endobj (11) Tj Q /Font << q q /Subtype /Form >> >> q << >> q >> q 1 i endobj (-20) Tj /Meta13 Do endstream /Matrix [1 0 0 1 0 0] >> , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. /Type /XObject 0.524 Tc Q 0 w endstream /Length 12 1 i stream /Length 69 q /F1 14.682 Tf 0 g /Meta38 52 0 R 1 i 0.564 G Q endstream /Type /XObject 0 w 1 i /Resources<< q /ProcSet[/PDF/Text] >> q Twice a number decreased by ten is at least 24. /F3 17 0 R 0.738 Tc q 1 i 1.005 0 0 1.007 79.798 813.037 cm Q /Meta273 287 0 R /ProcSet[/PDF] /ProcSet[/PDF/Text] 331 0 obj >> 0 g 0 G q >> q endobj q >> -0.486 Tw 0 5.203 TD Q Two speeding tickets could increase your rate by 58% at your next renewal. Q stream /Matrix [1 0 0 1 0 0] /Length 69 /Type /XObject /Meta106 Do 0 5.203 TD 1 i Q q /Resources<< [(-1)-16(52)] TJ /Font << /Meta249 263 0 R /Subtype /Form /Length 68 /Type /XObject /ProcSet[/PDF] q q q ET Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. Q >> 196 0 obj /Meta308 Do Q Q Q 1.007 0 0 1.007 654.946 872.509 cm /ProcSet[/PDF] q /ProcSet[/PDF/Text] 1 i q /F3 12.131 Tf 0.737 w endstream Diabetes, if left untreated, leads to many health complications. /Font << q Q 0 G /Length 16 1 i /FormType 1 /Resources<< /Resources<< 0 g 0 g /Type /XObject /FormType 1 q (9\)) Tj Q 159 0 obj 151 0 obj /Type /XObject << /Font << /F4 36 0 R /BBox [0 0 88.214 16.44] q ET q /ProcSet[/PDF] 0 g /F4 12.131 Tf 0 g stream q Q /Meta429 445 0 R 0 g stream 0.564 G endstream >> /Length 16 BT << >> /Meta2 Do ET /Resources<< 0 g 1.005 0 0 1.006 45.168 879.284 cm q /F3 12.131 Tf 672.261 546.541 m /F3 17 0 R Q to represent the numbers. q 0 g Q /Matrix [1 0 0 1 0 0] /Meta186 200 0 R /FormType 1 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. >> 0 w q 1.005 0 0 1.007 45.168 889.071 cm 1 i /Meta86 100 0 R << stream /ProcSet[/PDF/Text] 0 G /Type /XObject /Type /XObject /F3 17 0 R 0.737 w Q endstream 0 G q /ProcSet[/PDF/Text] stream 0 G BT /BBox [0 0 88.214 16.44] /F3 12.131 Tf /Length 118 /Subtype /Form BT 1.007 0 0 1.006 411.035 763.351 cm 1.014 0 0 1.006 531.485 690.329 cm /Leading 349 BT q BT endobj /Font << /Resources<< /BBox [0 0 17.177 16.44] 0.737 w BT /Meta16 Do ET Q /Resources<< >> In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. /Meta396 412 0 R endobj /ProcSet[/PDF] 1.005 0 0 1.007 102.382 293.596 cm q /Font << /FormType 1 >> endobj Q 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. /Type /XObject /ProcSet[/PDF/Text] q 0 5.336 TD [(1)-25(0\))] TJ >> endobj Q 0 5.203 TD endstream Q 0.271 Tc /Length 68 (x ) Tj Q << 1 g >> 43.426 5.203 TD 0 g /Type /XObject /FormType 1 q /Length 16 /Matrix [1 0 0 1 0 0] << endobj 1 g stream Q /BBox [0 0 534.67 16.44] stream >> endobj /Subtype /Form q 0 g /Subtype /Form /Matrix [1 0 0 1 0 0] >> /F1 7 0 R >> >> >> /Type /XObject /F3 12.131 Tf endstream /Font << endstream Q ET << /StemH 94 /Length 54 0 g /Type /XObject stream q Untreated or poorly treated diabetes accounts for . Q >> >> Q endstream /Meta299 Do /BBox [0 0 88.214 16.44] ET Q /Font << /F3 12.131 Tf /Type /XObject Find the number. << /FormType 1 /Matrix [1 0 0 1 0 0] << /F3 12.131 Tf Q /FormType 1 0.564 G /Resources<< << >> /Font << /Meta387 403 0 R << Q BT 1 i /F3 12.131 Tf /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] ET stream /Type /XObject 0 g endstream << >> q q Q Q 1 i Q q 0.737 w Tamang sagot sa tanong: 1.) endobj >> Q 1 i 0 g 1 i endobj 1.007 0 0 1.007 67.753 546.541 cm 0.369 Tc endstream /ProcSet[/PDF/Text] << 1 i endstream /Resources<< /FormType 1 Q stream >> 1 i /Font << << /Font << /Resources<< Q 0 5.203 TD (D) Tj q endobj 0 G (9\)) Tj >> /Meta274 Do q /ProcSet[/PDF/Text] /BBox [0 0 639.552 16.44] /Meta322 336 0 R /Subtype /Form 1.502 8.18 TD << stream >> 0 g q /Resources<< 1 i q /Font << 378 0 obj Q On the way, we sang songs for 20 minutes which is 10% of the time we were on the road. /Length 16 Q /Meta85 99 0 R /Matrix [1 0 0 1 0 0] /F3 12.131 Tf BT Q 0 g /F1 12.131 Tf endobj 1.014 0 0 1.006 251.439 836.374 cm Q endstream /BBox [0 0 88.214 16.44] q endstream (5) Tj endobj << Andrew M. /F3 17 0 R endstream 72 0 obj /ProcSet[/PDF] 16.469 5.203 TD /Length 65 /BBox [0 0 30.642 16.44] endstream q q q 43.426 5.203 TD /F4 12.131 Tf /Resources<< endobj /BBox [0 0 88.214 35.886] /Length 16 1.007 0 0 1.007 551.058 383.934 cm /Meta112 126 0 R q << Q /Resources<< /Font << (3) Tj q /F3 12.131 Tf q /Meta321 335 0 R /Resources<< q /Meta200 214 0 R BT /Subtype /Form Q >> /Matrix [1 0 0 1 0 0] 1 i stream 0 g q stream /Font << endobj >> 0.737 w << Q /Matrix [1 0 0 1 0 0] >> Q endobj Q Q 216 0 obj /FontName /PalatinoLinotype-Roman q ( x) Tj << /Meta185 Do Q 1.014 0 0 1.007 251.439 776.149 cm /Font << Q /Type /XObject /Font << >> /F3 17 0 R /ProcSet[/PDF/Text] Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . /Meta396 Do /F3 17 0 R /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Length 57 q Q 0 g 1 i /Meta207 221 0 R /Type /XObject 0 5.203 TD 1 i 0 w q /F3 17 0 R startxref /Meta272 286 0 R /Subtype /Form /Resources<< 20.21 5.203 TD 0 G /ProcSet[/PDF/Text] Q 411 0 obj >> With this, we get: "3x-8". endstream /ProcSet[/PDF/Text] << 0.369 Tc Q q BT /ProcSet[/PDF/Text] /F3 17 0 R /XObject << /Matrix [1 0 0 1 0 0] q Q /Font << /Meta7 Do endstream /Matrix [1 0 0 1 0 0] Q >> /ProcSet[/PDF] endstream Q /Type /XObject /Length 12 1 i Double or twice a number means 2x, and triple or thrice a number means 3x. /Font << endstream /Type /XObject 1 i >> /F3 12.131 Tf 294 0 obj Q /BBox [0 0 30.642 16.44] /FormType 1 0 20.154 m 0 G ET Q ET 0.369 Tc (8\)) Tj endstream endobj 287 0 obj >> Q ET /FormType 1 q Q q /Meta84 Do 271 0 obj For the lesson, he grabs a glass container shaped like a rectan >> q /Resources<< /FormType 1 /Font << 337 0 obj /Matrix [1 0 0 1 0 0] 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. q q /ProcSet[/PDF] endstream 1 i /ProcSet[/PDF/Text] 1 g Q /Type /XObject << /Resources<< >> BT BT endobj Just type into the box and your calculation will happen automatically. (D\)) Tj /ProcSet[/PDF] /F3 12.131 Tf 0 g 0.369 Tc /Font << /Font << /Meta209 223 0 R 1.005 0 0 1.007 102.382 400.496 cm 295.086 4.894 TD /CapHeight 476 q 1.007 0 0 1.007 654.946 726.464 cm /Type /XObject /Type /XObject 1 g >> 136 0 obj Next, the problem says that "x" would be equal to twice a number added by 5. /FormType 1 Q 0 g /StemV 94 Q >> /Meta422 Do Q /Type /XObject /ProcSet[/PDF/Text] endstream endobj ET /Length 16 1.005 0 0 1.007 102.382 473.519 cm Q 1 i stream endstream >> /Meta199 213 0 R q Q /ProcSet[/PDF] /Meta126 140 0 R /Length 118 (C\)) Tj /XObject << Q Q SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. 26.219 5.203 TD /Type /XObject q
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