>> /MissingWidth 250 /Length 59 Q /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] /BBox [0 0 30.642 16.44] /Length 58 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. /Font << 255 0 obj /Matrix [1 0 0 1 0 0] /F1 7 0 R /Meta369 Do >> /F3 17 0 R /Meta64 78 0 R /ProcSet[/PDF] /Font << 0 5.203 TD q stream 0.458 0 0 RG >> q You can also contact the clerk of court in the county you received the ticket. /ProcSet[/PDF] q Q /Length 16 q q >> Q q endobj << endobj /Type /XObject /Font << stream >> endstream >> 0.458 0 0 RG >> 0 g 0.458 0 0 RG >> endobj BT 0 g 1.007 0 0 1.007 45.168 746.789 cm stream q Q stream 0 G /BBox [0 0 15.59 29.168] q q endstream /Length 73 1.007 0 0 1.007 67.753 473.519 cm /Subtype /Form /Subtype /Form /Length 54 q Q 1.007 0 0 1.007 411.035 636.879 cm 300 0 obj Q /FormType 1 /Type /XObject /F3 12.131 Tf /Length 85 >> 0 G endstream /Type /XObject Q /Matrix [1 0 0 1 0 0] stream /Meta133 Do Q /Matrix [1 0 0 1 0 0] >> 0.737 w 32.201 5.203 TD >> >> 0 g /Flags 32 /F4 12.131 Tf 0 G /Resources<< q stream q 284 0 obj (+) Tj 0 g endobj /FormType 1 0.786 Tc >> /Type /XObject stream 8 0 obj /Font << /Resources<< /Matrix [1 0 0 1 0 0] /Length 16 ( x) Tj 0.369 Tc q /Length 16 >> stream 0.564 G stream /Meta91 Do 1.502 5.203 TD 1.005 0 0 1.007 102.382 799.486 cm /Type /XObject 1 i 305 0 obj << >> /Length 69 /Resources<< >> q /ProcSet[/PDF/Text] /Meta273 287 0 R Q /BBox [0 0 88.214 16.44] >> 0 g << /FormType 1 q /Subtype /Form q Q /Resources<< Q /BBox [0 0 88.214 16.44] Q >> /Matrix [1 0 0 1 0 0] 1.007 0 0 1.006 551.058 836.374 cm 406 0 obj /Meta83 97 0 R 310 0 obj Q endstream Q 1 i >> /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 271.012 330.484 cm >> /F3 17 0 R Q 0.458 0 0 RG 0.564 G 1.005 0 0 1.007 102.382 726.464 cm (5) Tj Q stream Q stream If a number is 50%, then it is a half - the same as 0.5 or 1/2. /BBox [0 0 30.642 16.44] /Meta30 Do endstream 1 g 0.458 0 0 RG ET Q 1 i 0.458 0 0 RG (B\)) Tj Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q 0.463 Tc /Font << (5) Tj /Matrix [1 0 0 1 0 0] Q >> C. Twice a number decreased by ten is at most 24. /Meta97 Do Q 3.742 5.203 TD 0.786 Tc /FormType 1 Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. Q Q >> Q >> /F3 12.131 Tf Q q q /Resources<< q 236 0 obj Q /F4 36 0 R /Subtype /Form /ProcSet[/PDF/Text] /ProcSet[/PDF] /Length 16 << 1.014 0 0 1.007 251.439 776.149 cm /Length 16 /ItalicAngle 0 /FormType 1 q << q /Matrix [1 0 0 1 0 0] 0.369 Tc endstream stream /FormType 1 /BBox [0 0 17.177 16.44] endobj /F3 17 0 R /F3 17 0 R /Matrix [1 0 0 1 0 0] /Meta143 Do [(1)-25(0\))] TJ q 0 5.203 TD q /Meta150 164 0 R /Subtype /Form /Type /XObject << /ProcSet[/PDF/Text] /ProcSet[/PDF] endobj /ProcSet[/PDF/Text] /F1 12.131 Tf Thrice of a number = 3x. stream ET A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. 0.737 w q /Meta351 Do /Subtype /Form BT 0 g 292 0 obj endobj /Subtype /Form >> endstream q /Length 16 q /Resources<< /Resources<< q 0 g BT At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . endstream /Meta245 Do 346 0 obj << /F1 12.131 Tf 0 g /Resources<< /Type /XObject Q 0.369 Tc 0 g 0.369 Tc q /Meta237 Do q 1.007 0 0 1.007 45.168 862.723 cm Q endobj 341 0 obj >> /Type /XObject q /Type /XObject >> 295.086 4.894 TD /F3 12.131 Tf /F3 12.131 Tf /FormType 1 /Resources<< >> 112 0 obj /Subtype /Form 1 i >> Q >> >> /Resources<< /FormType 1 q /Length 16 stream /Meta263 277 0 R 1 i << 293 0 obj /Type /FontDescriptor q 0 w << << 0 G endobj /ProcSet[/PDF] 0 5.203 TD /Length 54 endobj /Meta36 49 0 R 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. stream 118.317 5.203 TD Q ET q 0 G /Meta244 258 0 R /F3 12.131 Tf >> Q endobj 0.737 w q endstream 0.486 Tc endobj /ProcSet[/PDF/Text] endstream 0 g BT >> 1 i 43.426 5.203 TD 1 i q stream /F4 12.131 Tf stream /BBox [0 0 673.937 15.562] 434 0 obj 1 g BT (-11) Tj Q /Type /XObject 0 g 1 i 0 G Q /FormType 1 /Subtype /Form /BBox [0 0 88.214 16.44] Get link; Facebook; Twitter; /BBox [0 0 88.214 35.886] Five times the sum of a number and four 7. BT /Resources<< >> 0 G 1 i stream endobj /Type /XObject endobj 1 i /Font << /Meta51 65 0 R q Q >> endstream /F3 17 0 R /Type /XObject 0.297 Tc ET q /ProcSet[/PDF] 1.007 0 0 1.007 551.058 523.204 cm /ProcSet[/PDF] endobj /Type /XObject /Type /XObject << /Meta231 245 0 R 0.68 Tc 0.486 Tc Get a free answer to a quick problem. endobj Q /Resources<< /FormType 1 0.458 0 0 RG /FormType 1 0 g BT /Meta358 Do 0.737 w /Subtype /Form stream /Length 59 /Length 69 >> /Font << /Subtype /Form /FormType 1 319 0 obj /Resources<< /ProcSet[/PDF/Text] Q /BBox [0 0 639.552 16.44] /Meta129 143 0 R Q 1.014 0 0 1.007 391.462 583.429 cm /FormType 1 << 0 w /Resources<< /BBox [0 0 88.214 35.886] 0 56.451 TD 1 i 0.458 0 0 RG Q stream 57.656 5.203 TD q /Resources<< << endstream stream q Q Q q Q 0 w /ProcSet[/PDF] /Meta281 Do /F1 12.131 Tf /Resources<< q >> BT /Matrix [1 0 0 1 0 0] /Type /XObject 1 i endobj 20.21 5.203 TD /Length 16 /Matrix [1 0 0 1 0 0] Q /FormType 1 << /Matrix [1 0 0 1 0 0] /Type /XObject >> BT q >> Q 1 i Q q /ProcSet[/PDF] /FormType 1 /Meta344 Do 119 0 obj endstream /F3 12.131 Tf q 1.005 0 0 1.007 102.382 799.486 cm /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 199 0 obj 0 G >> /ProcSet[/PDF/Text] /F3 12.131 Tf 441 0 obj /Resources<< q Q /Resources<< /Resources<< BT /Length 58 /Resources<< 412 0 obj /Meta387 Do /Length 59 /Meta158 Do >> 0 w /Meta394 Do stream q endstream /Meta268 Do /Length 81 /F3 17 0 R q /Type /XObject BT 299 0 obj /Length 16 /BBox [0 0 30.642 16.44] /Resources<< /Meta93 Do q /BBox [0 0 15.59 29.168] >> /Font << 0 G /BBox [0 0 30.642 16.44] /Subtype /Form q Q /Type /XObject /Meta156 Do (D\)) Tj q /Meta63 Do /F1 12.131 Tf endobj /FormType 1 /Resources<< 184 0 obj /Type /XObject stream 0.458 0 0 RG 0.68 Tc /Length 16 >> stream /Length 68 /Meta113 127 0 R 0.369 Tc ET 1.007 0 0 1.007 271.012 636.879 cm 0.564 G if the solution of an equation is x=-2, what could the original equation be? 1 i /FormType 1 0 g 1.005 0 0 1.007 102.382 599.991 cm << 1 g >> 0 G q >> 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00000 n 0000133330 00000 n 0000133518 00000 n 0000133787 00000 n 0000133976 00000 n 0000134231 00000 n 0000134419 00000 n 0000134688 00000 n 0000134877 00000 n 0000135132 00000 n 0000135320 00000 n 0000135589 00000 n 0000135778 00000 n 0000136033 00000 n 0000136221 00000 n 0000136499 00000 n 0000136688 00000 n 0000136943 00000 n 0000137186 00000 n 0000137447 00000 n trailer Q 0 g << /Type /XObject Q BT Q 0.564 G Q /ProcSet[/PDF/Text] 333 0 obj endobj 0.838 Tc /FormType 1 1 i 0 G >> 1.007 0 0 1.007 67.753 347.046 cm /Length 16 /Resources<< << Q /StemH 88 Twice a number decreased by 8 gives 58 find the number Advertisement Loved by our community 24 people found it helpful Xiphodon Step-by-step explanation: 2x-8=58 2x=66 x =33 Hope it helps Please mark as brainliest Find Math textbook solutions? Q 0.737 w 410 0 obj 0.458 0 0 RG /ProcSet[/PDF/Text] << /F3 12.131 Tf endstream (-) Tj >> << /BBox [0 0 639.552 16.44] /Resources<< Q 0 g Q /Type /XObject /FormType 1 Solution. /BBox [0 0 88.214 16.44] q 0 g Q q q /Meta278 292 0 R Q q /BBox [0 0 30.642 16.44] /BBox [0 0 15.59 16.44] /Subtype /Form /ProcSet[/PDF/Text] 0 G /F3 17 0 R 0.564 G 0 G >> to represent the numbers. 0.311 Tc /BBox [0 0 88.214 35.886] endobj >> << endobj stream /Font << /Length 69 1 i Q Explanation: let the number be n. then we can express division in 2 ways. /ProcSet[/PDF/Text] /BBox [0 0 15.59 16.44] 0.227 Tc Q /FormType 1 Q 0.564 G >> /BBox [0 0 534.67 16.44] /F3 17 0 R /ProcSet[/PDF] /Type /XObject /Meta426 Do /Type /XObject q /Type /XObject /Resources<< >> /F3 12.131 Tf >> >> 0 w /Meta149 Do /Meta7 18 0 R >> q /F3 12.131 Tf >> BT (-) Tj >> /Meta129 Do >> /FormType 1 1 g Q 39 0 obj >> q 0 g 1.007 0 0 1.007 271.012 383.934 cm /ProcSet[/PDF/Text] q /Type /XObject /BBox [0 0 88.214 16.44] 0 G 20.21 5.203 TD Q /Font << endstream Q Q /Type /XObject >> BT q /Meta37 Do Q Q /ProcSet[/PDF] (x ) Tj Q /Meta227 241 0 R 26.219 5.203 TD Q /F3 12.131 Tf /F3 12.131 Tf 1.005 0 0 1.006 45.168 879.284 cm stream /Font << 0 g 1 g 0.458 0 0 RG /Subtype /Form /Matrix [1 0 0 1 0 0] q /FormType 1 endstream 0 g Q 1.007 0 0 1.007 551.058 703.126 cm 1 g >> 340 0 obj /Meta269 Do stream Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. endstream Q >> /Meta394 410 0 R /Resources<< /Subtype /Form q /Resources<< 185.725 5.203 TD Q Q /Matrix [1 0 0 1 0 0] the quotient of twenty and a number a.) << /ProcSet[/PDF/Text] S >> Q Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q >> Q /Length 16 0 g 140 0 obj endobj >> >> /ProcSet[/PDF] /FormType 1 0 g /Subtype /Form 0 w endobj endstream Q q /FormType 1 /Meta253 Do endobj Q 30 0 obj >> q /FormType 1 BT Q >> ET /FormType 1 endobj 0 w 0 g 0.458 0 0 RG /F3 17 0 R /BBox [0 0 30.642 16.44] BT (x) Tj << q /Subtype /Form 0.463 Tc /Font << 1 i /Matrix [1 0 0 1 0 0] 0 g >> Q q 0 g q /Matrix [1 0 0 1 0 0] 0.564 G /Length 16 /ProcSet[/PDF] 0 G stream [( the )-24(sum of a n)-14(umber an)-14(d )] TJ 0 g /Meta78 92 0 R /Meta399 415 0 R 1 i >> BT endobj 1.005 0 0 1.007 102.382 816.048 cm >> /Length 69 /LastChar 121 Q 0 g Q /F1 12.131 Tf Q endstream Q /Meta425 Do /Matrix [1 0 0 1 0 0] >> (7\)) Tj -0.092 Tw 433 0 obj The value of k is: (b) 3 (d) 0 (a) 4 (c) -4 TL ing:, 1)take a graph and draw two perpendicular lines to obtain four uadrants 2)draw any object using straight line 3) write the coordinates of each point o /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Size 447 /Length 57 0.564 G /F3 12.131 Tf /BBox [0 0 88.214 16.44] /Meta264 278 0 R /F1 12.131 Tf /Subtype /Form /F3 12.131 Tf /ProcSet[/PDF] 1 i 0.564 G /Subtype /Form ET /Font << /BBox [0 0 15.59 16.44] 1 i endobj endstream 271 0 obj 0 4.894 TD stream 1.007 0 0 1.007 551.058 383.934 cm /FormType 1 Q endobj >> /Length 78 Q 302 0 obj Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. /Font << >> Grad - B.S. /Matrix [1 0 0 1 0 0] stream Q 0 g Q BT Q /Subtype /Form (-8) Tj /Meta207 Do 75 0 obj 0 g /Subtype /Form Q 0 g q endobj q 0.458 0 0 RG /Resources<< Q /F4 36 0 R << /ProcSet[/PDF/Text] /Meta172 186 0 R stream /Length 70 Q 1.014 0 0 1.007 251.439 330.484 cm endstream 1 g q stream Q 0 g /ProcSet[/PDF/Text] To find: The. /Meta50 Do q 0.175 Tc /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Subtype /Form BT /ProcSet[/PDF] /ProcSet[/PDF] /Meta404 Do q /Meta90 Do Q /F3 12.131 Tf /F3 12.131 Tf ET q Q 60 0 obj endobj Q q 0 20.154 m /Type /XObject endstream /BBox [0 0 534.67 16.44] >> Q /Subtype /Form /Subtype /Form 0 g >> >> 26.219 5.203 TD 0 g /Meta274 Do Q 0.737 w 0.458 0 0 RG q q /Font << (C) Tj q q endobj q Q >> endstream /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Subtype /Form 1 i /Type /XObject /ProcSet[/PDF] 6.746 5.203 TD /Matrix [1 0 0 1 0 0] 216 0 obj 0.425 Tc >> /F3 17 0 R Q >> Q BT /BBox [0 0 639.552 16.44] /Subtype /Form q Q 0.458 0 0 RG ET >> 0.486 Tc endobj endstream 1 i 1.007 0 0 1.007 411.035 636.879 cm >> /Meta61 Do 0 G /Meta140 154 0 R q 0 g /Length 16 /Length 87 q 1 i q endobj /Meta149 163 0 R endstream >> 0.737 w ET /ProcSet[/PDF] Q 0 g >> q /Font << q 1 g 0 w >> /Length 74 ET 1.005 0 0 1.007 102.382 293.596 cm /FormType 1 q /FormType 1 << ET BT /Length 59 >> /BBox [0 0 15.59 16.44] 0.737 w 0 0 Similar questions Find the number which when decreased by 8% becomes 506. 0 g >> >> /Meta146 Do q /Meta336 Do /Resources<< /Resources<< q 1.007 0 0 1.007 551.058 523.204 cm

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