1 g Q q Q 1.007 0 0 1.007 411.035 277.035 cm endobj /I0 Do /Resources<< Q 1.007 0 0 1.007 551.058 703.126 cm /ProcSet[/PDF] /FormType 1 0 g << Q /Matrix [1 0 0 1 0 0] /Meta21 32 0 R Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. (D\)) Tj >> >> Q Q << /Type /XObject /Matrix [1 0 0 1 0 0] /Length 16 << Q Q /Subtype /Form /ProcSet[/PDF] /ProcSet[/PDF] BT /BBox [0 0 639.552 16.44] Q /FormType 1 /Contents [399 0 R] /Subtype /Form BT /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] 0 g >> /Font << 0 5.203 TD << /Length 16 /Meta105 119 0 R /Resources<< q /ProcSet[/PDF] /Resources<< /F4 36 0 R 1 i q /Subtype /Form /Meta328 342 0 R /Type /XObject /Subtype /Form /Meta397 Do >> 0 g 14.23 24.649 TD endobj /Meta237 251 0 R q /Meta317 Do 243 0 obj Q /Meta272 Do 1 i /Matrix [1 0 0 1 0 0] 438 0 obj /Meta278 Do /Meta264 Do 0 G /Type /Font Q /Meta74 Do Q S /Matrix [1 0 0 1 0 0] /Type /XObject Q >> 1.502 24.649 TD q Q q /Meta271 285 0 R /F4 12.131 Tf >> BT /Length 69 /F3 17 0 R >> endobj /Resources<< /Resources<< /ProcSet[/PDF/Text] Q /Resources<< stream /FormType 1 stream stream /Subtype /Form 1.007 0 0 1.006 551.058 763.351 cm /Meta250 Do /FormType 1 /FormType 1 1 i /Meta37 Do endobj /F4 36 0 R stream /BBox [0 0 30.642 16.44] /Meta297 Do Q >> 24 0 obj Q q /FormType 1 1 i /BBox [0 0 88.214 16.44] 0.458 0 0 RG 0.737 w >> endstream endstream /Subtype /Form 1.014 0 0 1.007 391.462 636.879 cm Q /Subtype /Form endstream /Length 16 BT This site is using cookies under cookie policy . 0 G Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. >> q Q /F1 7 0 R Q >> BT ET Q 1.007 0 0 1.007 551.058 703.126 cm /FormType 1 /Length 69 q /ProcSet[/PDF/Text] /Resources<< stream /Font << 268 0 obj /Subtype /Form 0.458 0 0 RG /Subtype /Form /Subtype /Form 1 g /Meta234 Do q Q /Length 66 stream >> q << /Matrix [1 0 0 1 0 0] q /Type /XObject << 0.564 G 333.269 5.488 TD q Q 16.469 5.203 TD /Length 69 >> q endstream endobj stream 0.458 0 0 RG /Length 59 Q q /Resources<< 0.28 Tc q BT /Length 65 /Meta177 191 0 R /F3 17 0 R endobj stream 25.454 5.203 TD /Resources<< /FormType 1 /Type /XObject /Subtype /Form BT (5\)) Tj 2.238 5.203 TD /Resources<< /Type /XObject >> /Resources<< endobj Q /Font << 0 g /FormType 1 q >> /Subtype /Form endobj << /BBox [0 0 15.59 16.44] 384 0 obj q 0 g /BBox [0 0 639.552 16.44] ( x) Tj 0 5.203 TD stream q 1 i endstream q >> endstream 0 20.154 m q endobj endobj 1.007 0 0 1.007 411.035 636.879 cm q >> endstream /Meta279 293 0 R /F3 12.131 Tf Q 0 w q (3) Tj q q /Resources<< /ProcSet[/PDF] >> q q 0 G /FormType 1 /Size 447 endstream /BBox [0 0 15.59 16.44] q endobj >> q /F3 17 0 R /F3 17 0 R >> q 0 G Q q 0 g << Q /Type /XObject q /ProcSet[/PDF/Text] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] Q 0 g q 186 0 obj q /Matrix [1 0 0 1 0 0] /Meta53 67 0 R 0.564 G /Type /XObject q 0 5.203 TD /F3 17 0 R 0 5.203 TD /F4 12.131 Tf endobj /BBox [0 0 534.67 16.44] q /Meta129 Do endobj 0.524 Tc 152 0 obj /BBox [0 0 88.214 16.44] /F3 17 0 R Q /Meta302 316 0 R Get a free answer to a quick problem. q >> endstream q /Type /XObject /Meta137 Do endstream /BBox [0 0 88.214 16.44] /Type /XObject /FormType 1 /Meta323 Do Q q 322 0 obj /Meta60 Do /Meta200 214 0 R >> /Subtype /Form /Meta57 Do >> Q Q /Meta55 Do 0.37 Tc /Subtype /Form BT 1.014 0 0 1.007 531.485 523.204 cm /BBox [0 0 88.214 35.886] /Length 16 endstream endobj 1.014 0 0 1.007 111.416 703.126 cm Q /AvgWidth 445 /Resources<< q q /Meta352 366 0 R Q 0 G 0 G (7\)) Tj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 1 i /FormType 1 22 0 obj /Subtype /Form ET stream 1.007 0 0 1.006 130.989 690.329 cm /Resources<< Twice the difference of a number and three totals twelve 8. ET q << /F1 12.131 Tf 1 i /FontBBox [-568 -307 2000 1007] << 1 i 0 g /Resources<< /Resources<< Q q Q -0.041 Tw /Resources<< /F3 17 0 R /Resources<< endobj 0.458 0 0 RG >> (-11) Tj Find the number. 0 w endobj 0 5.203 TD 1 i q /FormType 1 /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] >> /Resources<< endobj q >> 1 i /FormType 1 /FormType 1 endobj /Meta340 Do q 1 i 1 i /ProcSet[/PDF] /Meta44 Do stream /Subtype /Form 0 20.154 m /Meta152 Do /ProcSet[/PDF] endstream stream /ProcSet[/PDF] q 371 0 obj Percent Change = (Decrease First Value) x 100% 1.005 0 0 1.007 79.798 713.666 cm /Subtype /Form 0 w q stream Q /ProcSet[/PDF/Text] /F3 17 0 R Q endstream 0 g 0 G 40.45 4.894 TD /Meta308 Do endstream /Meta375 389 0 R (B\)) Tj 66 0 obj /Font << 54.679 5.203 TD /Meta100 Do /Resources<< /Resources<< 321 0 obj /Type /XObject 0 w stream /Font << Q -0.16 Tw /F3 17 0 R /Length 16 endstream endobj << 0 5.203 TD 0.564 G /FormType 1 /Font << stream /ProcSet[/PDF/Text] /F3 12.131 Tf /Resources<< /FormType 1 << BT ET q /Font << 1 i q /BBox [0 0 15.59 29.168] /Type /XObject 0.564 G /Matrix [1 0 0 1 0 0] /Resources<< (D\)) Tj /Matrix [1 0 0 1 0 0] Q /Meta2 Do /Font << Q >> Q 1.007 0 0 1.007 130.989 383.934 cm [4] One half of a number decreased by fourteen is twenty-one /Subtype /Form >> /Meta303 317 0 R stream /Subtype /Form q ET Q endobj 0.738 Tc >> /Length 58 stream /Resources<< 0 G /Subtype /Form Q /F3 17 0 R >> stream /Font << endobj >> /Length 69 << Q 201 0 obj 1 g q /ProcSet[/PDF] Q Q 0 g q 0 w << BT BT Q endobj >> >> q /FormType 1 0 G endobj Q Q endstream >> /F1 12.131 Tf /Type /XObject 0 w endobj q endstream /Subtype /Form /Matrix [1 0 0 1 0 0] 1 i >> /Matrix [1 0 0 1 0 0] Q q /Type /XObject /Type /XObject >> /Meta80 94 0 R >> Q /Subtype /Form 0.369 Tc 0.369 Tc /Resources<< Q stream endstream 1.007 0 0 1.007 411.035 277.035 cm /FormType 1 << Q 1 i /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] /Resources<< q q /FormType 1 /Type /XObject << Q 139 0 obj /ProcSet[/PDF] q >> endobj (2) Tj q /FormType 1 >> /Meta348 362 0 R 0 5.203 TD /Length 65 q q Q Q stream /F3 12.131 Tf /BBox [0 0 88.214 16.44] >> 0.241 Tc /Type /XObject stream q /Meta200 Do Double or twice a number means 2x, and triple or thrice a number means 3x. /Subtype /Form >> /ProcSet[/PDF/Text] >> >> << << Q >> 0 w /Resources<< /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] << >> Q Q Q /Matrix [1 0 0 1 0 0] /Resources<< endobj Q endstream Q /Meta412 428 0 R /Resources<< 0 g stream /ID [] q /ProcSet[/PDF/Text] >> Q endobj The difference between six and a number divided by nine 10. 1.007 0 0 1.007 130.989 636.879 cm Q /Font << stream 1 g 0 g /BBox [0 0 673.937 15.562] endobj Q q Q /Length 59 /BBox [0 0 15.59 16.44] /Font << q /Type /XObject BT << Q 87 0 obj Q endstream q 1.007 0 0 1.007 551.058 523.204 cm /Subtype /Form q Q 0 w endstream >> q stream 1.014 0 0 1.007 531.485 583.429 cm /Subtype /Form 0.737 w 0 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /FormType 1 1.007 0 0 1.007 67.753 400.496 cm Q (C) Tj /Length 16 ET 340 0 obj /Length 69 << 1.007 0 0 1.007 551.058 636.879 cm /Length 73 q Q 0 G /BBox [0 0 88.214 16.44] /Meta213 227 0 R Q S /Subtype /Form 0 G /F4 36 0 R /FormType 1 1 i << 1 i /F3 12.131 Tf /Subtype /Form stream 0 g /Type /XObject BT q /FormType 1 << /Type /XObject << /Length 60 /FormType 1 >> q >> Q /Matrix [1 0 0 1 0 0] 0 g /BBox [0 0 15.59 16.44] q Q 11.99 24.649 TD stream /Matrix [1 0 0 1 0 0] /Meta192 Do Q >> endstream 1.007 0 0 1.007 654.946 473.519 cm >> 0.51 Tc q Q /Meta208 Do /Type /XObject /Meta83 Do stream >> endstream 410 0 obj 0 g 1.007 0 0 1.007 551.058 330.484 cm /Type /XObject endobj /FormType 1 /Matrix [1 0 0 1 0 0] -0.106 Tw /Meta389 405 0 R 382 0 obj 0 g 395 0 obj endobj 6.746 5.203 TD Q ET /Resources<< >> 1.005 0 0 1.007 45.168 889.071 cm q >> /Resources<< /Font << >> /Subtype /Form Q 1. q 0 w Was this answer helpful? q 0 g << 3.742 5.203 TD /BBox [0 0 30.642 16.44] endstream /Type /XObject Q >> q Q 0 G /FormType 1 0 g endobj Q /Meta262 276 0 R Q /BBox [0 0 15.59 16.44] /FormType 1 178.979 5.203 TD 0 G >> /BBox [0 0 639.552 16.44] ET 0 G /Matrix [1 0 0 1 0 0] /FormType 1 /Subtype /Form q Q (C\)) Tj /Resources<< /Type /XObject /F3 12.131 Tf 1 i q ET endobj >> /F3 12.131 Tf ET /Type /XObject q q /Resources<< (C\)) Tj << /BBox [0 0 15.59 29.168] Q Q q stream /Meta167 181 0 R /ProcSet[/PDF/Text] 359 0 obj 161 0 obj /Type /XObject /Matrix [1 0 0 1 0 0] /F3 17 0 R q 394 0 obj /ProcSet[/PDF/Text] -0.463 Tw >> ET ET /Meta133 147 0 R /Font << /Type /XObject /Subtype /Form /Meta327 341 0 R Q /Resources<< /ProcSet[/PDF/Text] /Meta407 Do /Resources<< << >> /F3 12.131 Tf /Meta115 129 0 R 0.564 G /Subtype /Form /ProcSet[/PDF/Text] q 0.458 0 0 RG Q /F3 17 0 R q stream << Q 0 g Q BT 1.014 0 0 1.007 111.416 277.035 cm /F4 36 0 R 1 i /ProcSet[/PDF/Text] >> stream Q /Length 16 q 0 g stream /Meta48 62 0 R >> << /ProcSet[/PDF] /F3 17 0 R << q /Meta187 Do 0.737 w 82 0 obj /FormType 1 95 0 obj << /Length 16 (vi) If 12 is subtracted from a number, the result is 24. /BBox [0 0 30.642 16.44] /Meta419 435 0 R Q /BBox [0 0 534.67 16.44] >> /Length 54 /Subtype /Form /Subtype /Form /BBox [0 0 88.214 16.44] Q Q >> 1.014 0 0 1.007 531.485 583.429 cm Q Q q /Meta114 128 0 R q /Meta224 Do /Type /XObject stream q Q /ProcSet[/PDF/Text] /Type /XObject 1 i /Resources<< 1 i endstream /FormType 1 >> Q q /CapHeight 692 /Subtype /Form >> /ProcSet[/PDF/Text] /Meta341 355 0 R endstream 2 times x minus 58 C. twice the difference of a number and 5 B. twice a number decreased by 58 D. 2 times the sum of a number and 58 Answer: B. Step-by-step explanation: twice - (2) number - (x) 58-(58) Edukasyon. /Subtype /Form /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0 G Q 0.458 0 0 RG /Type /XObject q /Matrix [1 0 0 1 0 0] /Type /XObject endstream >> /Meta264 278 0 R Q 2x - 15 = -27. 291 0 obj q 19.474 20.154 l Q << Q 1 g >> 281 0 obj << BT >> endobj /Font << /Subtype /Form >> 1 i /FormType 1 0 G << /Subtype /Form /Meta75 89 0 R /Resources<< ET /FormType 1 /FormType 1 /Meta140 154 0 R /Font << /ProcSet[/PDF/Text] /Meta246 260 0 R Q endobj 38 0 obj /F3 12.131 Tf >> 0 g BT 353 0 obj 1 i /F1 12.131 Tf /Font << ET q 1 i 441 0 obj Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM << << >> >> >> q endobj q /Type /XObject q >> /Font << /Subtype /Form >> /F3 12.131 Tf 0.564 G Q /Length 16 /BBox [0 0 30.642 16.44] endobj /Subtype /Form Q /ProcSet[/PDF/Text] endobj Q endobj q /Resources<< /MissingWidth 250 Q /FormType 1 1 i endobj >> 0 g /Font << /F3 17 0 R /Resources<< /Subtype /Form /F3 12.131 Tf /F3 12.131 Tf 672.261 726.464 m Q /FormType 1 Q >> 1 i q endobj >> >> /Font << >> (C\)) Tj >> Q ET >> Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 << /ProcSet[/PDF/Text] >> (+) Tj /Type /XObject q endobj q 0.838 Tc /Font << q q /Type /XObject 0.737 w q Q /Type /XObject Q >> (D\)) Tj << /Length 59 /BBox [0 0 88.214 16.44] Q 1 i 1 i >> /Font << ( \() Tj >> /Type /FontDescriptor 1 i /Meta33 Do /ProcSet[/PDF] /Type /XObject q (\(x ) Tj >> /FormType 1 279 0 obj /Matrix [1 0 0 1 0 0] 0 G /Matrix [1 0 0 1 0 0] endstream 443 0 obj /Font << /Meta32 45 0 R 0 G 1 i Q 327 0 obj /Meta341 Do ET (-11) Tj /Meta258 272 0 R BT /Subtype /Form /Resources<< 1 i Q 0.458 0 0 RG endobj endobj (x) Tj 1 i BT Q endstream Advertisement Answer No one rated this answer yet why not be the first? 0 w /BBox [0 0 88.214 16.44] q /Length 118 q ET q /BBox [0 0 30.642 16.44] >> Twice a number decreased by . /FormType 1 ET 53 0 obj 181 0 obj Q /ProcSet[/PDF/Text] /Length 12 << 0 g /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] 0 g /Resources<< >> /FormType 1 Q (3) Tj /Font << q >> >> >> /BBox [0 0 639.552 16.44] /F1 7 0 R /Length 80 /ProcSet[/PDF/Text] TJ Q /Matrix [1 0 0 1 0 0] /Resources<< 0.737 w >> BT /Subtype /Form q /Type /XObject Q endobj BT [( and )16(a nu)26(mbe)18(r)] TJ 1.007 0 0 1.007 271.012 383.934 cm /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.007 130.989 523.204 cm endstream stream [(F)-22(ive)] TJ /FormType 1 (iv) A number exceeds 5 by 3. /BBox [0 0 534.67 16.44] 0 5.203 TD Q /Meta424 Do A number = an unknown number which can be represented by a variable, usually x. Q q /Type /XObject endstream In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. Q /Type /XObject