twice a number decreased by 58

stream BT /Resources<< /Length 59 /Length 59 1 i /FormType 1 /BBox [0 0 30.642 16.44] /Type /XObject 1.502 5.203 TD q BT endstream ET 0 G /Type /XObject BT /Matrix [1 0 0 1 0 0] >> Q /BBox [0 0 88.214 16.44] 13.464 5.203 TD stream 58 decreased by twice Gails age. /BBox [0 0 534.67 16.44] /Type /XObject /FormType 1 -0.486 Tw /Meta32 Do 0.51 Tc 1 i endstream /Ascent 1050 << 414 0 obj stream /Subtype /Form /Length 69 /Subtype /Form /Matrix [1 0 0 1 0 0] /FormType 1 endstream Q >> Q >> q 0 g /Font << /ProcSet[/PDF/Text] /FormType 1 /Meta332 346 0 R 0.458 0 0 RG %%EOF. /Meta356 370 0 R 1 i >> /Meta409 Do (-) Tj q /F3 12.131 Tf 0 G -0.029 Tw << Q 0 g stream BT /F3 17 0 R /Meta147 161 0 R endstream Q /F3 12.131 Tf /Meta365 Do endobj q Q endobj /Type /XObject /BBox [0 0 30.642 16.44] >> q Q (-20) Tj endstream /Type /XObject >> endstream /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] BT /ProcSet[/PDF/Text] q << 189 0 obj >> The difference between six and a number divided by nine 10. >> /Meta367 381 0 R /ProcSet[/PDF] 0 G 0.458 0 0 RG endobj /Meta71 Do << 0 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] BT Q >> q endobj q /Resources<< /F3 17 0 R << 1.502 5.203 TD << /Subtype /Form q stream /Type /XObject /FontBBox [-90 -216 1195 800] >> 1 i /Type /XObject 0 G /Matrix [1 0 0 1 0 0] 1 i /Font << BT /FormType 1 0.458 0 0 RG /ProcSet[/PDF/Text] stream Q 406 0 obj << /Type /XObject /Meta408 424 0 R >> BT 0.458 0 0 RG /Font << That was 1/8 of the points that he scored Q /F1 12.131 Tf /Resources<< q stream Q BT >> >> endstream << ET ET endstream /Type /XObject 1.007 0 0 1.006 411.035 510.406 cm 0 g << Q If a number is 400%, then it is 4 times, the same as 4. /Resources<< 0 g endobj /Meta165 Do endobj << << q stream >> /F3 12.131 Tf Q q /StemH 94 /Meta66 Do 174 0 obj /Length 63 /ProcSet[/PDF/Text] << 0 g BT q 0.737 w /F3 17 0 R /FormType 1 /F3 17 0 R q /Type /XObject 0.51 Tc stream >> /Resources<< q 421 0 obj Q stream 206 0 obj /Resources<< >> BT /BBox [0 0 88.214 35.886] q /Type /XObject stream -0.486 Tw /StemH 88 endstream 0 G 1.007 0 0 1.007 67.753 799.486 cm 0.369 Tc Q >> /Meta57 Do 1.014 0 0 1.007 111.416 583.429 cm /Resources<< 1.005 0 0 1.007 102.382 599.991 cm 0.458 0 0 RG /Type /XObject 1 i /BBox [0 0 639.552 16.44] /F3 12.131 Tf stream endobj >> BT /Subtype /Form << /FormType 1 /ProcSet[/PDF] >> /Meta268 282 0 R Q >> /Font << 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. stream /Meta135 Do >> q 0.737 w /F3 17 0 R >> /Font << /Subtype /Form -0.021 Tw /Matrix [1 0 0 1 0 0] Q endstream endstream Q /Matrix [1 0 0 1 0 0] /Resources<< /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q /Subtype /Form >> Testosterone is the primary sex hormone and anabolic steroid in males. /ProcSet[/PDF] /Font << /Length 12 /Meta314 328 0 R >> /Resources<< 430 0 obj >> So let's go ahead and identify a v ET /Meta163 177 0 R >> stream /Meta395 Do /Subtype /Form stream q 1 i /Length 70 /Matrix [1 0 0 1 0 0] /Meta192 Do << 56 0 obj endstream 549.694 0 0 16.469 0 -0.0283 cm /Matrix [1 0 0 1 0 0] endstream ET /F3 12.131 Tf /Length 69 /Length 69 /CapHeight 662 ET 85 0 obj /ProcSet[/PDF] /FormType 1 /Length 54 (- 4) Tj endstream /Type /XObject >> /Meta136 150 0 R 0.838 Tc 1 i Q >> Q 0 G >> 0 g /F3 12.131 Tf 0 w 1.007 0 0 1.007 271.012 277.035 cm Q 0 G (13) Tj 10.487 5.203 TD 0.297 Tc /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Meta164 Do Q 1.007 0 0 1.007 654.946 599.991 cm /Resources<< Q 356 0 obj 0 G endstream ET stream Q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] endstream /Type /XObject 1 i << 0 g Q /Subtype /Form >> << q << 0.458 0 0 RG 0 w /Matrix [1 0 0 1 0 0] 246 0 obj 20.21 5.203 TD q /Font << /Type /XObject 0 g /Meta261 275 0 R ET >> /Matrix [1 0 0 1 0 0] q >> /Length 108 /Meta425 441 0 R stream /Descent -277 /Resources<< 672.261 726.464 m /Resources<< stream >> >> 1.007 0 0 1.007 551.058 523.204 cm 0 g q /ItalicAngle 0 /FormType 1 /Resources<< /FormType 1 /Font << 1 i q /FormType 1 /F3 17 0 R /Matrix [1 0 0 1 0 0] /Type /XObject q 1 i /FormType 1 ET 1 i /ProcSet[/PDF] endstream 167 0 obj Q q Q q stream << -37 VI 2. /Type /XObject /Font << /Type /XObject endobj Q /F3 17 0 R 1 i /Meta166 180 0 R /ProcSet[/PDF/Text] q /F3 17 0 R q 0 g Q q Q q the quotient of five and a number 7.) Q /Font << /Subtype /Form 0 g stream q >> (C\)) Tj /Resources<< /Meta317 Do /Subtype /Form /Meta411 Do >> 0.458 0 0 RG /Matrix [1 0 0 1 0 0] 142 0 obj Q /Type /XObject endstream q S /Font << /Meta62 Do Q q >> 1.007 0 0 1.007 271.012 523.204 cm q 0 g >> /FormType 1 /Meta60 Do Two fewer than a number doubled is the same as the number decreased by 38. 1 g 1 i >> S /Meta325 Do /F3 12.131 Tf << q stream ET q Q Q << 26.219 5.203 TD /Subtype /Form stream endobj /ProcSet[/PDF/Text] 0 w endobj /ProcSet[/PDF/Text] /Type /XObject q 1.014 0 0 1.007 531.485 849.172 cm /BBox [0 0 88.214 16.44] (x) 6 times a number is 5 more than the number. /Type /XObject >> endstream q q /Subtype /Form /FormType 1 0 g 1.005 0 0 1.007 102.382 872.509 cm /Type /XObject /Meta32 45 0 R /Font << q /Resources<< Q /FormType 1 /BBox [0 0 88.214 35.886] q ET /FormType 1 /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] /FormType 1 /Length 58 /Subtype /Form /Subtype /Form /Meta29 42 0 R BT stream << 165 0 obj q 358 0 obj Q /ProcSet[/PDF] /BBox [0 0 88.214 16.44] Q q endobj 0 g 1 i /Matrix [1 0 0 1 0 0] << Q >> endobj stream ET Q q 1 g 1.014 0 0 1.007 251.439 383.934 cm endobj /Subtype /Form << /Length 69 /Matrix [1 0 0 1 0 0] ET BT 1.007 0 0 1.007 271.012 849.172 cm /Length 70 /BBox [0 0 15.59 16.44] endobj >> /Meta385 401 0 R 1.005 0 0 1.007 79.798 862.723 cm q /Type /XObject 1 i >> 0.458 0 0 RG endstream /ProcSet[/PDF/Text] 1 i 98 0 obj /F3 17 0 R /Matrix [1 0 0 1 0 0] q >> q /ProcSet[/PDF/Text] /F3 12.131 Tf q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.006 411.035 510.406 cm 0 w /ProcSet[/PDF/Text] stream q q Q /Filter [/CCITTFaxDecode] /F3 12.131 Tf Q /Type /XObject ET 1 i /Matrix [1 0 0 1 0 0] 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 /Subtype /Form q /Type /XObject >> /Length 69 >> Q q 0 g << 0.458 0 0 RG >> 0 G 1 i /F4 36 0 R 0 w endobj >> Q /ProcSet[/PDF] 1 i /Subtype /Form /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] >> q /ProcSet[/PDF/Text] >> q Q endobj >> 0 G endstream 1 g /F3 12.131 Tf /F3 17 0 R q /F3 17 0 R /ProcSet[/PDF] /Meta412 428 0 R >> 408 0 obj We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. 1 g 1 i << /ProcSet[/PDF/Text] /Font << ET 108 0 obj stream endstream Q q /ProcSet[/PDF/Text] S endstream q >> /Meta214 Do Q endstream /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 111.416 437.384 cm /Length 12 q Q 0 g stream endobj /Matrix [1 0 0 1 0 0] endobj /Resources<< /F3 17 0 R q /Meta280 294 0 R /Meta246 260 0 R 1 i /Subtype /Form 1 g /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 251.439 437.384 cm 259 0 obj /Matrix [1 0 0 1 0 0] q 0 w /Meta264 278 0 R >> /Resources<< /Length 59 0 w 0 w 0 g Q 30.699 5.203 TD /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] /Type /XObject /Subtype /Form /ProcSet[/PDF/Text] stream 445 0 obj BT 143 0 obj Q /Length 69 q >> /BBox [0 0 88.214 16.44] 549.694 0 0 16.469 0 -0.0283 cm /F3 17 0 R BT << >> BT /Length 245 /Type /XObject /Font << q 0 5.203 TD 1.007 0 0 1.007 411.035 636.879 cm /Resources<< >> endobj (x) Tj /Resources<< << q endobj /MaxWidth 1453 /Meta277 Do 0 G /FormType 1 [(MULTIPLE CHOICE. /Font << stream 1.014 0 0 1.007 111.416 277.035 cm 1.005 0 0 1.007 102.382 799.486 cm /Font << stream 0.68 Tc (B\)) Tj q q /FormType 1 /ProcSet[/PDF/Text] 672.261 799.486 m >> 0 g 1 i 1.014 0 0 1.006 251.439 690.329 cm /Subtype /Form 1 i /Resources<< /FormType 1 << /ProcSet[/PDF/Text] >> (40) Tj << /Length 16 q /Matrix [1 0 0 1 0 0] Andrew M. /FormType 1 endobj /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 45.168 763.351 cm endstream /Meta126 140 0 R /F3 12.131 Tf /F1 12.131 Tf >> stream endstream /F3 12.131 Tf /Meta253 267 0 R /FormType 1 /Subtype /Form (D) Tj 134 0 obj 0.564 G q /Subtype /Form /F3 12.131 Tf 0 g 1 i /Type /XObject /F3 12.131 Tf /Type /XObject /F3 17 0 R 0.382 Tc >> 0.37 Tc (38) Tj /Matrix [1 0 0 1 0 0] 301 0 obj 0 g Q /Meta390 406 0 R stream << 1.014 0 0 1.007 531.485 383.934 cm 13.464 5.203 TD Q stream /FormType 1 19.474 20.154 l 0.458 0 0 RG >> /BBox [0 0 88.214 16.44] /Descent -216 ET 1 i q /Type /XObject q /Meta226 240 0 R /FormType 1 /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 130.989 636.879 cm ET /ProcSet[/PDF/Text] /F3 17 0 R endstream 0 G >> /Subtype /Form BT q >> /Length 54 /Subtype /Form Q /Matrix [1 0 0 1 0 0] q Q 0 G /Matrix [1 0 0 1 0 0] >> >> /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 q Q /Length 70 >> << /Matrix [1 0 0 1 0 0] BT /ProcSet[/PDF/Text] /F4 36 0 R /Meta1 Do 0 g /ProcSet[/PDF] /ProcSet[/PDF/Text] /Meta310 Do 0.564 G /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Meta295 Do endobj >> /Type /Page /Meta414 430 0 R >> q q q /Matrix [1 0 0 1 0 0] << /Subtype /Form /Matrix [1 0 0 1 0 0] /Resources<< 0 g Q 0.737 w >> /Resources<< >> BT /Length 69 /F3 17 0 R /F3 12.131 Tf /F3 17 0 R endobj 0.486 Tc /Length 69 0.458 0 0 RG /FormType 1 /ProcSet[/PDF] Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. /F3 17 0 R << /F3 17 0 R Q /Matrix [1 0 0 1 0 0] /Font << /F3 17 0 R /Resources<< 21 0 obj /Type /XObject q /ProcSet[/PDF] /Length 69 q /F3 12.131 Tf /Type /XObject /FormType 1 /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] Q >> >> /Meta428 444 0 R /Resources<< /Font << /BBox [0 0 534.67 16.44] 1.014 0 0 1.007 391.462 330.484 cm /FormType 1 endstream 0 w /Meta242 Do stream /Subtype /Form Q q /Length 69 /F1 14.682 Tf 0 w /FormType 1 [(The )-16(s)15(um )-14(of )] TJ 0 g /FormType 1 /Parent 1 0 R 0 g Q /BBox [0 0 88.214 16.44] 1 i q endstream /XObject << /F3 12.131 Tf /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] 1 i 0 g 90 0 obj >> >> q /Type /XObject /Subtype /Form /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF/Text] 0.737 w If mario jumps 3 times and luigi jumps 62 times. /Font << 0.425 Tc 0 G 1 i /StemV 77 0 g (3) Tj /F3 17 0 R /Font << 0 G >> Q >> /FormType 1 0 G /FormType 1 /Meta85 99 0 R /Meta51 Do 0 g >> q endobj /Font << /Length 79 0 w ET Q Q /BBox [0 0 673.937 16.44] << << /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 636.879 cm Q Q q Q 1 i /Resources<< 0 g Q /Resources<< 0.737 w /Meta238 Do 1 g endstream >> q Let x the unknown number. << /FormType 1 /BBox [0 0 88.214 16.44] /Subtype /Form (x ) Tj q /Resources<< >> /FormType 1 1.007 0 0 1.007 411.035 849.172 cm << Q >> /F1 12.131 Tf /Length 59 /Meta145 Do 1 i /FormType 1 0 g /Resources<< /Meta238 252 0 R /Length 70 q 0 G << /Subtype /Form 0 g BT endobj 11 0 obj >> 40.45 4.894 TD /Resources<< 1 i endstream /Meta263 277 0 R /ProcSet[/PDF/Text] /F3 17 0 R q >> << stream q /Resources<< /Subtype /Form /F3 12.131 Tf 1.007 0 0 1.007 67.753 599.991 cm /Type /XObject << 391 0 obj q /BBox [0 0 534.67 16.44] q 0 g ET 0 g (9\)) Tj answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. stream /Font << 0 g << endstream /F3 17 0 R Q q Q /Subtype /Form /F3 12.131 Tf /Subtype /Form << /F3 17 0 R Q 0 g 0.737 w >> /Matrix [1 0 0 1 0 0] /Meta97 Do 1 i /Subtype /Form 0 w endobj /Meta45 Do endstream Q /F3 12.131 Tf 0.564 G /Type /XObject 1 i q 1.005 0 0 1.007 79.798 813.037 cm /Type /XObject >> /Font << q Q Q 20.21 5.203 TD q q endstream BT 0 g /Meta285 299 0 R /F1 12.131 Tf >> q /Meta95 109 0 R >> BT /F3 17 0 R -0.021 Tw >> /Meta427 Do stream >> /Meta206 Do /Subtype /Form BT q /Subtype /Form stream 0 G Notice that we used the variable \large {d} d in our equation to stand for our unknown value. /Subtype /Form /Type /XObject Q -0.106 Tw 76.394 5.203 TD Q << q /Meta130 144 0 R Q q stream ET ET Q /FormType 1 BT 89.12 5.203 TD /Length 63 stream /BBox [0 0 639.552 16.44] /ProcSet[/PDF/Text] q /Type /XObject /Matrix [1 0 0 1 0 0] q ET endobj /Subtype /Form (B\)) Tj /Meta393 409 0 R q >> endstream >> 0.458 0 0 RG 393 0 obj q /Length 95 /BBox [0 0 15.59 29.168] /F4 36 0 R /F1 7 0 R >> /Length 12 Q /Length 54 BT /Meta210 224 0 R >> 1.007 0 0 1.007 130.989 330.484 cm q 1.007 0 0 1.007 45.168 796.475 cm 269 0 obj 0 G 1 i 1 g >> stream /Matrix [1 0 0 1 0 0] Q (C\)) Tj Q endobj /FormType 1 endstream 1.014 0 0 1.007 531.485 583.429 cm /F3 17 0 R 0 g >> Q /Matrix [1 0 0 1 0 0] endobj 1.007 0 0 1.007 411.035 330.484 cm /FormType 1 311 0 obj /Font << Q /Subtype /Form The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. 0 g 0 w /BBox [0 0 88.214 16.44] Was this answer helpful? q /Length 91 (2) Tj /Subtype /Form 0 g >> q q /BBox [0 0 88.214 16.44] 6.746 5.203 TD /Meta112 Do >> 0.737 w Q >> /Matrix [1 0 0 1 0 0] /FormType 1 0.564 G >> Q >> /BBox [0 0 30.642 16.44] Q Q /F1 7 0 R Q q >> 1 i Q /BBox [0 0 534.67 16.44] /Resources<< /Meta376 Do /BBox [0 0 88.214 16.44] 1 i Q /Meta314 Do /ProcSet[/PDF] Q endstream /Meta313 Do 0.68 Tc /Subtype /Form BT 251 0 obj 1.005 0 0 1.007 79.798 730.228 cm 0.369 Tc endstream q 1 i /Meta6 15 0 R 0 G 722.699 599.991 l /FormType 1 /F4 36 0 R q 30.699 4.894 TD 0 G 1.014 0 0 1.007 251.439 583.429 cm q /I0 Do /Meta16 27 0 R 1.005 0 0 1.007 102.382 490.08 cm Q /Meta223 237 0 R >> /Subtype /Form 0 G ET 0 G 0.564 G BT >> >> 63 0 obj /Meta151 165 0 R /Length 118 1.014 0 0 1.007 391.462 383.934 cm q /Subtype /Form q /BBox [0 0 30.642 16.44] endstream stream 0.68 Tc (6\)) Tj >> stream 1 i /ProcSet[/PDF/Text] << /Length 59 Q Q endstream 1.005 0 0 1.007 102.382 653.441 cm endobj >> /Font << /Root 2 0 R ET 2.238 5.203 TD q q q /F4 12.131 Tf /Meta371 385 0 R /Meta131 Do D. b = 4 2. endobj BT Q 0 g >> 0.458 0 0 RG Q 0.369 Tc /BBox [0 0 88.214 16.44] >> 0.425 Tc 0.737 w q q << /Resources<< BT 1 i /Subtype /Form /Meta8 Do 382 0 obj q /ProcSet[/PDF] /Meta261 Do /BBox [0 0 639.552 16.44] /F3 12.131 Tf /BBox [0 0 88.214 16.44] Q 0 g q BT Q Q << /Matrix [1 0 0 1 0 0] Diabetes, if left untreated, leads to many health complications. << /Meta378 392 0 R q Q /Subtype /Form /Resources<< /ProcSet[/PDF] /Resources<< Q q Q q /Meta52 Do q 0.458 0 0 RG BT << (-11) Tj /Meta212 226 0 R 146 0 obj /F3 17 0 R endstream Q /Font << q 1 i stream (58) Tj /Type /XObject 0 39.216 TD 1.014 0 0 1.007 391.462 583.429 cm /Resources<< q /ProcSet[/PDF/Text] /BBox [0 0 15.59 16.44] 0.737 w q /Subtype /Form q 181 0 obj Q /Matrix [1 0 0 1 0 0] /Subtype /Form /Meta254 Do q /Meta272 Do >> /FormType 1 /ProcSet[/PDF/Text] /FormType 1 0 5.203 TD /F3 12.131 Tf /Meta256 Do /Subtype /Form q /F3 12.131 Tf 0 G << << Q Q /Type /XObject << 1 i /FormType 1 /Length 58 1 i /ProcSet[/PDF/Text] Q /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 391.462 776.149 cm >> 1.007 0 0 1.007 45.168 713.666 cm 255 0 obj Solution: 272 0 obj 0 G /Pages 1 0 R 1 i /Length 12 Q Q 1.007 0 0 1.007 411.035 277.035 cm endstream /F3 12.131 Tf /FormType 1 /Meta281 295 0 R q /ProcSet[/PDF] BT Q /BBox [0 0 88.214 16.44] /Resources<< 1.007 0 0 1.006 130.989 437.384 cm endstream Q 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0 0 RG Q stream /FirstChar 32 /Subtype /Form Q 500 500 500 0 333 389 278 0 0 722 500 500]>> BT /Length 59 q Q >> 1.005 0 0 1.007 102.382 400.496 cm /BBox [0 0 30.642 16.44] BT << ET 400 0 obj q q >> /FormType 1 5.98 7.841 TD 1 i /Type /XObject /Meta88 102 0 R 22 0 obj Q /FormType 1 /Meta35 Do endobj /Subtype /Form q 116 0 obj 0 G 0.737 w 1 i /Length 16 /Meta373 Do << 0.564 G 0 G endobj /BBox [0 0 88.214 16.44] 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q q /F3 12.131 Tf BT /ProcSet[/PDF/Text] q /BBox [0 0 15.59 16.44] ET BT /F3 17 0 R 0.564 G /Subtype /Form 39 0 obj /BBox [0 0 534.67 16.44] << /ProcSet[/PDF/Text] /Type /XObject /BBox [0 0 88.214 16.44] q endobj /Length 99 /F3 17 0 R /Meta31 44 0 R /ProcSet[/PDF/Text] endobj /XObject << 1.014 0 0 1.007 251.439 523.204 cm << >> /Length 54 endstream /Resources<< q /Type /XObject /Resources<< 1.007 0 0 1.007 271.012 636.879 cm >> 125 0 obj Q /Meta133 147 0 R /Matrix [1 0 0 1 0 0] /FormType 1 >> /F3 17 0 R /F3 17 0 R /BBox [0 0 534.67 16.44] 1 i /Type /XObject endobj 0 G >> Q 0 g /Matrix [1 0 0 1 0 0] /Length 68 /FormType 1 0 g /Matrix [1 0 0 1 0 0] NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. endobj << Q /Resources<< 1.007 0 0 1.007 130.989 583.429 cm stream stream 0 g Q /Length 59 1 i 1.005 0 0 1.007 102.382 473.519 cm /Meta217 231 0 R stream >> /FormType 1 >> endstream endobj >> /Subtype /Form /BBox [0 0 15.59 29.168] /Meta161 175 0 R /Type /XObject 0.564 G 16 0 obj /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /ProcSet[/PDF] 315 0 obj /Type /XObject /Meta369 Do 1.014 0 0 1.007 531.485 330.484 cm /Meta280 Do /FormType 1 /Matrix [1 0 0 1 0 0] Q Q q /Resources<< stream endstream /Length 69 1 i >> Q /Meta363 377 0 R stream 1 i 0 G /Length 78 Q /Subtype /Form << /Meta15 Do /FormType 1 /FormType 1 0 G Q 1 g endstream >> /Resources<< /BBox [0 0 88.214 35.886] Q 0.458 0 0 RG BT much as how 8, Last . 0 G 0 g ET /Type /XObject /Type /XObject /Length 69 Q /Subtype /Form 38 0 obj q 101 0 obj 0.737 w /Meta167 Do /BBox [0 0 88.214 16.44] Q /Meta44 Do ET >> BT ET >> /Font << << /Subtype /Form >> /FormType 1 Q Q /BBox [0 0 88.214 16.44] /Meta391 407 0 R /F3 12.131 Tf 1 i q /Length 59 endobj ET >> q endstream -0.463 Tw Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. /Font << /F1 12.131 Tf BT Q q /Matrix [1 0 0 1 0 0] Expression. q /BBox [0 0 15.59 16.44] /Meta268 Do 0.458 0 0 RG q Q 22.478 4.894 TD << /Font << /Font << /Subtype /Form /Font << ET q endstream >> Q -0.126 Tw xref /FormType 1 /FormType 1 0.737 w Q /F3 12.131 Tf /Meta206 220 0 R >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Subtype /Form 0.458 0 0 RG /Subtype /Form 0 g /BBox [0 0 15.59 16.44] /Length 16 /Subtype /Form BT q ( x) Tj >> endobj 0 g 3.742 5.203 TD Q q /Resources<< endobj stream /FormType 1 q /Resources<< /Type /XObject BT endobj /Meta75 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