/Meta271 285 0 R /BBox [0 0 88.214 16.44] ET Q q 381 0 obj endstream /FormType 1 Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. q Q Q Q /F3 17 0 R endobj (9\)) Tj /Matrix [1 0 0 1 0 0] /Meta428 444 0 R >> 0.737 w 14.966 20.154 l /Matrix [1 0 0 1 0 0] /FormType 1 /Meta42 Do /Resources<< /Meta57 71 0 R -0.084 Tw stream /Subtype /Form 1 g 1.007 0 0 1.007 411.035 583.429 cm /Font << 1 i endobj 0.458 0 0 RG /FormType 1 q (\(x ) Tj << Q /Meta197 211 0 R 0.737 w /ProcSet[/PDF/Text] 722.699 599.991 l 0 g 0 w << Q 1.007 0 0 1.007 67.753 726.464 cm /Resources<< /Meta171 Do BT q 0 G 319 0 obj 0.37 Tc [(The )-16(s)15(um )-14(of )] TJ precision and actual right or wrong answers. endstream /Length 16 endstream ET 0.737 w q 0.564 G /Subtype /Form << >> Q /Resources<< /BBox [0 0 639.552 16.44] Q 0.564 G 0 G 1.007 0 0 1.007 130.989 776.149 cm /Matrix [1 0 0 1 0 0] /Meta369 Do q endstream 1.007 0 0 1.006 411.035 690.329 cm Q << /ProcSet[/PDF] We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. Q /Resources<< q /Meta184 Do /BBox [0 0 88.214 16.44] /Meta347 361 0 R 0 G q /FormType 1 /Meta14 Do q (\)) Tj /Meta327 341 0 R Q 430 0 obj /Type /XObject endstream /Type /XObject ET q 0 g /Matrix [1 0 0 1 0 0] Q 1 i /Meta381 Do /ProcSet[/PDF/Text] /ProcSet[/PDF] /Subtype /Form q /Encoding /WinAnsiEncoding Q q /Subtype /Form 0 20.154 m /Resources<< ( \() Tj 0 g 1.007 0 0 1.007 130.989 583.429 cm 0.68 Tc << decreased by A number decreased by twelve X - 12 subtracted from Six subtracted from a number X - 6 Multiplication ( x ) times Eight times a number 8x the product of The product of fourteen and a number 14x twice; double Twice a number; double a number 2x multiplied by A number multiplied by negative six 6x endobj q ET /FormType 1 >> q 0 5.203 TD 0 w >> q /Type /XObject first we change the sentence in formula as the following, i think this is clear &easy to understand .i hope it helps you, This site is using cookies under cookie policy . >> >> endstream /Subtype /Form endobj stream >> 1.007 0 0 1.006 551.058 763.351 cm q Q 1 i /Meta72 Do >> q q << Q << 1 i 0 G /Meta135 Do q 0 g 0 G 0 G q /Type /Page /FormType 1 ET Q (5\)) Tj ET 99 0 obj /Resources<< /Meta238 252 0 R /FormType 1 1.014 0 0 1.007 531.485 277.035 cm 0 5.203 TD /Meta172 186 0 R << << 1.014 0 0 1.007 391.462 383.934 cm stream >> Q ET /Meta317 Do Q Q /F1 7 0 R << 0 g 0 G q 0 g /Resources<< /ProcSet[/PDF] /Meta272 286 0 R 0.564 G 6.746 5.203 TD /FormType 1 1 i Q << q << -0.106 Tw -0.486 Tw Q Two times the sum of a number x and five c.) a number x times the sum of five and two d.) the sum of five times a number x and two 2.) 0 G 134 0 obj endstream 0 G >> /Meta28 41 0 R 0 g /FormType 1 1.007 0 0 1.007 45.168 829.599 cm q /Length 63 /Meta260 Do Formula - How to Calculate Percentage Decrease. >> Q 59 0 obj Q /F3 17 0 R a Question q /Subtype /Form << /BBox [0 0 88.214 16.44] /Font << q Q endobj 1 0 obj /FormType 1 /Subtype /Form q Q endstream Q >> 0 g Q /F3 17 0 R /Meta316 330 0 R Q Q 0 G 1 i endstream /BBox [0 0 15.59 16.44] /Length 69 Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. /BBox [0 0 88.214 16.44] >> /Subtype /Form (-9) Tj q /Subtype /Form ( \() Tj /MaxWidth 1397 0 20.154 m 205.199 4.894 TD Q /Resources<< /FormType 1 /BBox [0 0 15.59 16.44] Q /Subtype /Form /Subtype /Form /Type /XObject /BBox [0 0 15.59 16.44] a.) (x ) Tj Q /BBox [0 0 88.214 16.44] 0 G >> 0.382 Tc /ProcSet[/PDF/Text] 0 w /Meta100 Do q /FormType 1 >> /Type /XObject /Length 118 /Resources<< /ProcSet[/PDF/Text] /F3 12.131 Tf /F3 12.131 Tf >> /Meta409 425 0 R /Subtype /Form /Meta311 Do /Font << q /Type /XObject Q q /Subtype /Form 0 g >> /FormType 1 q >> /Matrix [1 0 0 1 0 0] Q q Q Q q /FormType 1 >> 1 i /ProcSet[/PDF] /Meta132 Do /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) endobj 1.014 0 0 1.006 111.416 836.374 cm /Subtype /Form Q >> << S /Type /XObject 0.564 G /Type /XObject Q That was 1/8 of the points that he scored BT 1 i 0.737 w /Type /Page /Length 69 q >> (D\)) Tj >> Q 0 g stream /Type /FontDescriptor /Matrix [1 0 0 1 0 0] 142 0 obj 0.838 Tc Was this answer helpful? stream /BBox [0 0 534.67 16.44] /Resources<< /BBox [0 0 88.214 16.44] 0.737 w q /ProcSet[/PDF/Text] /BBox [0 0 534.67 16.44] 0.486 Tc >> 0.486 Tc /ProcSet[/PDF/Text] 66 0 obj 1.014 0 0 1.007 251.439 383.934 cm /Resources<< q >> /BBox [0 0 88.214 16.44] >> 1 i 1.007 0 0 1.007 411.035 330.484 cm /BBox [0 0 88.214 16.44] stream >> /Subtype /Form 166 0 obj /Meta313 327 0 R >> (C\)) Tj BT q endobj q Q stream stream endstream endstream /BBox [0 0 30.642 16.44] /F3 17 0 R endstream Q 188 0 obj >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] stream /FormType 1 /Resources<< /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Subtype /Form 1.014 0 0 1.007 111.416 330.484 cm 1.008 0 0 1.007 654.946 293.596 cm /BBox [0 0 15.59 16.44] /Subtype /Form endstream ET q /I0 Do 15 0 obj q stream q << >> 1 i /F3 12.131 Tf q 0 56.451 TD /ProcSet[/PDF] 0 g >> /ProcSet[/PDF] /F3 12.131 Tf >> /Length 16 /Meta62 76 0 R /Resources<< q 1.502 24.339 TD 0 w >> 0 G Q endstream The ratio of a number to fifteen 4. TJ /Length 104 Q << >> endobj q 0 G /Length 69 q Q 1.014 0 0 1.006 531.485 690.329 cm ET q q >> /Meta316 Do >> /Meta249 263 0 R Q endstream endstream >> Q /F3 17 0 R 1 i Q >> /Meta139 153 0 R 549.694 0 0 16.469 0 -0.0283 cm BT 31 0 obj 0 g q q >> BT Q (-) Tj /F1 7 0 R Q /Filter [/CCITTFaxDecode] /Meta89 Do /Meta332 Do /Meta162 Do endstream ET Q << BT /BBox [0 0 88.214 16.44] 88 0 obj 1.007 0 0 1.006 411.035 690.329 cm Q /BBox [0 0 534.67 16.44] 0 G q 20.21 5.336 TD 1 i Q /Resources<< /F3 17 0 R /Font << stream endstream 0 g q 1.005 0 0 1.007 45.168 889.071 cm >> /ProcSet[/PDF] Q << 1 i (C) Tj /Meta50 Do q /FormType 1 1 i >> 0 g endobj Q /Meta190 Do /Matrix [1 0 0 1 0 0] 177 0 obj /FormType 1 q 0 20.154 m 0.382 Tc /Widths [ 500 0 502]>> /BBox [0 0 88.214 16.44] q /Subtype /Form /FormType 1 q 1 i /Subtype /Form Q q endstream 1 g Q /Length 69 /FormType 1 32.201 5.203 TD /ProcSet[/PDF/Text] /Font << /Meta79 93 0 R q /Font << /Subtype /Form 0.564 G /Length 69 q /Meta328 Do endobj (v) 5 subtracted from thrice a number is 16. q /FormType 1 q endobj q 0 G >> stream >> 1 i << BT q endobj /Meta150 Do /Resources<< 0.737 w Q /F3 17 0 R 1.502 5.203 TD Q 0 G ET Q >> 0.458 0 0 RG endobj /Type /XObject BT Q ET >> /FormType 1 /Meta202 216 0 R /Font << << q q /F3 17 0 R q /Length 57 >> 1.007 0 0 1.007 551.058 636.879 cm /Length 65 /FormType 1 >> q q << /Length 64 q stream /Meta142 156 0 R 0 G 1.007 0 0 1.007 271.012 450.181 cm /Resources<< endobj /Matrix [1 0 0 1 0 0] 0 g stream BT q q Q endobj << 1 g endobj /ProcSet[/PDF/Text] q 1 i q ET /Subtype /Form /Font << Q /Matrix [1 0 0 1 0 0] << /BBox [0 0 88.214 16.44] q 1.005 0 0 1.007 79.798 763.351 cm /F3 17 0 R endstream /Meta214 Do /Type /XObject /Type /XObject endobj 95 0 obj 1 i /Subtype /Form BT << /Subtype /Form Q q 338 0 obj /BBox [0 0 15.59 16.44] Q Q BT Q 1.007 0 0 1.007 130.989 523.204 cm 1.007 0 0 1.007 271.012 383.934 cm /FormType 1 /Length 58 /ProcSet[/PDF/Text] Thrice a number decreased by 5 is 3x - 5. /FormType 1 q /Resources<< endobj >> BT << /FormType 1 endobj q /FormType 1 /F3 12.131 Tf q /Matrix [1 0 0 1 0 0] endobj >> 0 g /Subtype /Form 0 G ET Q q q /ProcSet[/PDF/Text] >> 248 0 obj /ProcSet[/PDF/Text] q 1 g 0 w /Subtype /Form /BBox [0 0 30.642 16.44] q /Matrix [1 0 0 1 0 0] 20.21 5.203 TD [( and )16(a nu)26(mbe)18(r)] TJ /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject /Subtype /Form /Meta207 221 0 R q (x ) Tj Q endstream Q >> (+) Tj 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 stream 1 i Q endstream 0 g Q 0 g /Subtype /Form Q Q Q /Resources<< BT /F3 12.131 Tf /Font << /Length 69 1 i q 0 g /Meta426 442 0 R /Meta378 392 0 R endstream /ProcSet[/PDF/Text] /F3 17 0 R /F3 17 0 R /Meta331 345 0 R 12.727 5.203 TD /Type /XObject stream /F3 12.131 Tf 1 i q /FormType 1 /ProcSet[/PDF/Text] /Subtype /Form q /Subtype /Form /Subtype /Form >> q /Meta33 46 0 R 0 G 0 g /Meta176 190 0 R /FormType 1 /Matrix [1 0 0 1 0 0] endstream q /Resources<< /Type /XObject endobj 1 i /BBox [0 0 15.59 16.44] q endobj q /Subtype /Form endstream Q 0 g /Matrix [1 0 0 1 0 0] q (C\)) Tj q /Resources<< /Font << /Meta33 Do Q Q >> >> 0 g /Meta282 296 0 R /Matrix [1 0 0 1 0 0] 1 i endstream stream << (58) Tj Q /ProcSet[/PDF] >> /F4 36 0 R /Meta11 Do /BBox [0 0 88.214 16.44] /Meta24 37 0 R /Matrix [1 0 0 1 0 0] /F3 12.131 Tf << >> (58) Tj Q endstream /FormType 1 stream >> /F3 12.131 Tf /Meta186 200 0 R 442 0 obj Q 0.737 w 0.458 0 0 RG >> 1.007 0 0 1.007 271.012 523.204 cm stream << 307 0 obj /Font << 0.737 w ET /Matrix [1 0 0 1 0 0] /Meta281 Do Q q /Resources<< /Length 66 /Type /XObject /Meta34 Do /Meta151 165 0 R Q /F3 17 0 R endobj 1.007 0 0 1.007 130.989 523.204 cm Q Q >> >> /Length 59 1 i q 0.564 G 1.014 0 0 1.007 391.462 277.035 cm Q 3.742 5.203 TD /Type /XObject << >> 0 G q /Type /XObject 317 0 obj /Meta114 Do /ProcSet[/PDF/Text] /Subtype /Form BT stream /Resources<< Q 0 g q Q /Meta161 Do BT << q 1.007 0 0 1.007 411.035 849.172 cm /BBox [0 0 88.214 16.44] /F1 12.131 Tf Find the number. /Meta355 369 0 R /Subtype /Form /BBox [0 0 639.552 16.44] q /Font << Q /Matrix [1 0 0 1 0 0] /Meta115 129 0 R /Resources<< /Subtype /Form Q Q >> SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. 1.014 0 0 1.007 391.462 523.204 cm /FormType 1 >> q the quotient of five and a number 7.) /Length 69 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 /Length 67 /BBox [0 0 639.552 16.44] /Meta201 215 0 R /ProcSet[/PDF] /Contents [399 0 R] /Font << 1 i 0 g 80 0 obj BT 1 g /Type /XObject q >> q /FormType 1 /Meta209 223 0 R Q /F1 12.131 Tf << /BBox [0 0 88.214 16.44] 1 i >> /Meta305 319 0 R /BBox [0 0 88.214 16.44] /Type /XObject 0 g /F3 12.131 Tf q /Type /XObject 0 w Aktual'nye voprosy Vol 10, No 3 (2020) 1.005 0 0 1.007 102.382 616.553 cm What word phrase can you use to represent 5x + 2? /FormType 1 BT 0.425 Tc 1 i q /Type /XObject /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 411.035 636.879 cm >> /Resources<< /Type /XObject q 0.458 0 0 RG Q >> 0 w ET /F3 17 0 R 27.693 5.203 TD Q 70 0 obj 0 G ( \() Tj >> Q stream 336 0 obj 155 0 obj /Length 78 /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject >> /Matrix [1 0 0 1 0 0] 0 G /ProcSet[/PDF/Text] Q Q 1.007 0 0 1.007 271.012 776.149 cm 0 G >> /BBox [0 0 88.214 16.44] q q /Meta288 302 0 R ET << Q /F3 12.131 Tf /BBox [0 0 88.214 16.44] endstream Q q /Font << 0 G ET /Subtype /Form /Length 118 /BBox [0 0 88.214 16.44] 0 g /F3 12.131 Tf 1.014 0 0 1.007 391.462 450.181 cm /Resources<< /Resources<< q /Subtype /Form /Matrix [1 0 0 1 0 0] 0.369 Tc q /Resources<< 0 G /Meta163 Do /Resources<< BT -37 VI 2. q /FormType 1 1.014 0 0 1.007 391.462 776.149 cm /Subtype /Form << >> 1.005 0 0 1.007 102.382 599.991 cm >> /Matrix [1 0 0 1 0 0] endobj /Font << 1.007 0 0 1.007 45.168 730.228 cm /Length 54 endobj 1.005 0 0 1.015 45.168 53.449 cm q 0 20.154 m Q ET Q << 0.564 G /Meta402 418 0 R q endstream 19.474 20.154 l >> /Length 69 Q Q endstream << You can specify conditions of storing and accessing cookies in your browser, Twice a number, decreased by 58 is less than 112, Mr. Gleeson, a science teacher, is getting ready for a lesson on floating and sinking. endstream /F1 7 0 R << q 1.014 0 0 1.007 251.439 383.934 cm /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] << /Type /XObject q 0 G q 0.486 Tc q >> /Type /FontDescriptor 1.007 0 0 1.006 411.035 510.406 cm /Meta277 Do q Q /Meta403 419 0 R q 1 i q q /Matrix [1 0 0 1 0 0] endstream /Resources<< /FormType 1 125.064 4.894 TD q q Q endobj 0.458 0 0 RG Q >> /BBox [0 0 17.177 16.44] >> /Meta392 Do /Meta93 Do 0 w >> q q the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . >> q /Type /XObject /BBox [0 0 673.937 15.562] /Font << /Meta281 295 0 R q endobj (+) Tj q q /Meta397 413 0 R BT 1.007 0 0 1.007 551.058 277.035 cm q stream >> stream << 1 i 0 g 0 g Q q q stream q -0.22 Tw q /Meta20 31 0 R /Meta350 364 0 R >> << **Note: You could choose any variable you want. /FormType 1 4 0 obj /ProcSet[/PDF] Q Q q /Length 67 Q 1 i q 0.458 0 0 RG 0 G q /BBox [0 0 88.214 16.44] Q 0 g Q endobj /ProcSet[/PDF/Text] Q endobj 17.234 5.203 TD Q /Type /XObject 19.474 5.203 TD 0.458 0 0 RG 0.737 w 0.458 0 0 RG Q 23.216 5.203 TD endobj endstream << /FormType 1 0.737 w q 549.694 0 0 16.469 0 -0.0283 cm Q 0 G Q /Subtype /Form 128 0 obj 0 G /F3 12.131 Tf stream endobj 0 g /Resources<< 1.007 0 0 1.007 411.035 636.879 cm 0.564 G /F1 12.131 Tf endobj (2\)) Tj q q >> /Subtype /Form /Resources<< Twice the difference of a number and three totals twelve 8. endobj 1.007 0 0 1.007 411.035 383.934 cm /Matrix [1 0 0 1 0 0] 1 i q endobj Q S /Subtype /Form 1.007 0 0 1.006 411.035 763.351 cm 1 i 0 G q 0 w >> /Meta390 406 0 R ET >> q /Length 87 >> -0.008 Tw >> /Subtype /Form 0 g (x) Tj q Answer link. /Subtype /Form /Matrix [1 0 0 1 0 0] >> 0.737 w stream 1.014 0 0 1.007 111.416 277.035 cm 1.005 0 0 1.007 102.382 293.596 cm endobj 0.737 w Q /FormType 1 0.198 Tc /Meta283 Do 0 g Q >> 0 w Q /Meta76 Do /Font << /Type /Catalog 0 G /ProcSet[/PDF/Text] /Meta75 Do >> /Type /XObject /Type /XObject 0 g /F3 17 0 R /Length 54 Q 207 0 obj /BBox [0 0 88.214 35.886] >> << /Subtype /Form q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] endobj /Meta12 23 0 R endobj /Meta310 Do q endobj /Length 74 Select the correct mathematical statement for the following equation. 1.005 0 0 1.007 79.798 713.666 cm ET /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta64 Do 0.68 Tc /Subtype /Form BT /FormType 1 /FormType 1 endstream 300 0 obj /Meta217 Do /Meta107 121 0 R 0 g /Type /XObject q endobj 1 i q /Subtype /Form 1 i 0 w >> << /Length 65 140 0 obj Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q >> 41.186 5.203 TD q >> >> /Length 69 /ProcSet[/PDF/Text] /Meta425 441 0 R /Font << Q /Meta415 431 0 R /Resources<< >> >> endobj Q /Meta27 Do /Font << 0 g BT stream Q Q /Meta245 259 0 R Q q q /Resources<< q q 0.737 w /BBox [0 0 15.59 16.44] /Font << 2. 1 i 1.007 0 0 1.007 654.946 872.509 cm /Meta424 Do 1 i /Meta147 161 0 R /FormType 1 /Subtype /Form /F3 17 0 R /Matrix [1 0 0 1 0 0] >> 0 g /BBox [0 0 30.642 16.44] q 0.564 G >> << endstream 1 i Q >> /Meta224 Do 312 0 obj >> B. /FormType 1 /F1 7 0 R Q /Matrix [1 0 0 1 0 0] ET /Meta351 Do << >> Q << >> stream /Subtype /Form q 0 g BT >> << /Meta409 Do Q /Resources<< 1.007 0 0 1.007 271.012 523.204 cm >> BT Find the number. /BBox [0 0 15.59 16.44] 120 0 obj BT endobj /Subtype /Form q q /ProcSet[/PDF/Text] q /Meta182 Do ( x) Tj /Matrix [1 0 0 1 0 0] Find the number, 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. Q: when six times a number is decreased by 4, the result is 8. >> /Font << /Meta219 233 0 R >> /Meta130 144 0 R Q /Length 68 -0.101 Tw /F1 12.131 Tf 0 g ([x ) Tj 0.369 Tc q endstream /Matrix [1 0 0 1 0 0] ET /Resources<< endobj /Length 16 >> /Length 91 0 5.203 TD Q endobj /Type /XObject 0 g >> /F3 12.131 Tf >> 0.524 Tc /F4 36 0 R /Meta418 434 0 R >> /Subtype /Form endobj /Meta46 60 0 R /Matrix [1 0 0 1 0 0] 1 i q endstream 0.737 w /FormType 1 Q endstream >> /Meta129 143 0 R Q endobj 0.524 Tc /BBox [0 0 88.214 16.44] 0 g 1 i q endobj 45 0 obj 1 i 0 g /ProcSet[/PDF] ET >> /F4 12.131 Tf endstream /F1 12.131 Tf q q /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] >> (A\)) Tj /Resources<< 0.458 0 0 RG /ProcSet[/PDF/Text] /Subtype /Form /Meta35 48 0 R 0 w /Matrix [1 0 0 1 0 0] >> /FormType 1 BT Twice a number is decreased by 9, and this sum is multiplied by 4. (8\)) Tj /Meta40 Do /Subtype /Form /ProcSet[/PDF] endobj /Matrix [1 0 0 1 0 0] Q /Subtype /Form /Length 78 /Matrix [1 0 0 1 0 0] q /BBox [0 0 639.552 16.44] /XObject << 14 0 obj >> q 295.086 4.894 TD /F3 12.131 Tf /FormType 1 >> /Resources<< endstream >> /Font << /ProcSet[/PDF/Text] 1 i endobj /F3 17 0 R endstream 0 g BT q /F4 12.131 Tf 1.005 0 0 1.007 79.798 763.351 cm q << q /Subtype /Form /BBox [0 0 88.214 16.44] /Resources<< (D\)) Tj ET /Meta195 209 0 R 419 0 obj 192 0 obj /Meta291 Do 0 5.203 TD endstream /Resources<< /F3 12.131 Tf q endobj Q endstream stream (-23) Tj Q BT (-11) Tj 1 i /Length 16 endobj We are asked to find the number, so, we could assign the number as "x". /FormType 1 /Type /XObject /F3 12.131 Tf /Length 69 endstream (B\)) Tj 0 G 13.493 5.336 TD /Meta399 415 0 R BT q Q /F3 17 0 R 0 G /Meta168 Do Q /Meta145 159 0 R 1 g q /BitsPerComponent 1 /Resources<< >> 2.238 5.203 TD 0.564 G endobj 0 G /ProcSet[/PDF/Text] Q /Resources<< /BBox [0 0 88.214 16.44] Q (-) Tj 0 g /Meta247 261 0 R /Resources<< stream 1 i stream (6\)) Tj q >> Q << /Meta406 422 0 R q /Length 59 BT (x) Tj stream >> q stream >> /F3 12.131 Tf 1 i 0 g stream /F3 17 0 R 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 /Resources<< endstream /Resources<< /F3 12.131 Tf /Meta178 Do /Resources<< /Meta340 Do /Meta365 Do 1.005 0 0 1.007 79.798 813.037 cm 0 w 1.007 0 0 1.006 551.058 836.374 cm /Meta185 Do /FormType 1 0 g endobj endstream 1.007 0 0 1.007 67.753 653.441 cm 1 g >> 0 4.894 TD ET 0.369 Tc /Font << /FormType 1 >> /Type /XObject q /Meta169 Do 1.005 0 0 1.007 102.382 546.541 cm 1.007 0 0 1.007 551.058 330.484 cm 0.458 0 0 RG q /Matrix [1 0 0 1 0 0] Q Q 1.005 0 0 1.007 102.382 799.486 cm q q BT q /Meta351 365 0 R /F3 12.131 Tf /Type /XObject Andrew M. 272 0 obj /Subtype /Form (38) Tj /BBox [0 0 88.214 16.44] /Type /XObject 1.007 0 0 1.007 130.989 383.934 cm Q 0.786 Tc endstream /FormType 1 endobj /Matrix [1 0 0 1 0 0] << q BT /Length 12 /Subtype /Form to represent the numbers. /Matrix [1 0 0 1 0 0] 361 0 obj endstream /Length 69 /Type /XObject 1.007 0 0 1.006 411.035 763.351 cm /Length 60 /BBox [0 0 15.59 16.44] /Subtype /Form /ProcSet[/PDF] (-8) Tj 1.007 0 0 1.007 654.946 653.441 cm 0 5.203 TD VIDEO ANSWER: in this problem were asked to solve giving, given the following information. BT /F3 12.131 Tf (B\)) Tj Q >> /Type /XObject Q 1 i 0.425 Tc Q ET /Font << << /Type /XObject >> /Length 16 endstream /ProcSet[/PDF/Text] q 1 i Q endobj 0.458 0 0 RG q 1 g q 1 i >> endstream /Resources<< >> /F3 17 0 R Q 0 G Q /Length 294 Q Q q q Q >> 0 G /F3 17 0 R >> 0 g << q /Meta29 Do /Matrix [1 0 0 1 0 0] q /Subtype /Form 0 447 q q /Type /XObject Q /F3 17 0 R Q (viii) A number divided by 8 gives 7. /Length 16 endobj /Resources<< Q [(E)-14(le)-23(ven)] TJ 672.261 599.991 m >> Q 1 i Q >> q /Font << /Matrix [1 0 0 1 0 0] 0 g q 1 i >> 1.005 0 0 1.007 102.382 293.596 cm /F3 17 0 R 429 0 obj /Subtype /Form 3.742 5.203 TD /Resources<< /F3 17 0 R stream Q /Type /XObject 1.014 0 0 1.007 391.462 583.429 cm Q /Resources<< /Subtype /Form q << /Meta266 Do >> Q /F3 12.131 Tf /MissingWidth 250 endstream Q /Resources<< /Meta405 421 0 R 1.007 0 0 1.007 551.058 383.934 cm /BBox [0 0 88.214 35.886] stream /Type /XObject /Resources<< 0.458 0 0 RG endobj /Length 16 q /Matrix [1 0 0 1 0 0] (-) Tj /Subtype /Form ET /ProcSet[/PDF/Text] q 1 g 0 w BT Q /F3 12.131 Tf endstream /FormType 1 /F3 17 0 R q 255 0 obj /BBox [0 0 88.214 16.44] 225 0 obj 0.68 Tc /Subtype /Form Q Q /Meta208 Do /F3 12.131 Tf 0 G /ItalicAngle 0 291 0 obj /ProcSet[/PDF/Text] /Resources<< Q Q q (-4) Tj /FormType 1 /Matrix [1 0 0 1 0 0] endstream q endstream >> BT S 0 w /Subtype /Form << >> >> /Subtype /Form /Subtype /Form q /Type /XObject BT stream 1 i /BBox [0 0 88.214 16.44] /Meta301 Do endstream 1.007 0 0 1.007 411.035 583.429 cm Q /Meta269 283 0 R Q << >> stream /Matrix [1 0 0 1 0 0] stream Q q 0.463 Tc 1.007 0 0 1.007 271.012 277.035 cm /Subtype /Form 0 g /Resources<< /Meta309 323 0 R /Meta429 445 0 R q /BBox [0 0 88.214 35.886] BT /ProcSet[/PDF] >> >> endobj /Meta223 237 0 R /BBox [0 0 88.214 16.44] Q /ProcSet[/PDF/Text] /FormType 1 endstream /Type /XObject /F3 17 0 R /FormType 1 /Resources<< q q q >> 56 0 obj 0.045 Tw Q 1 i Q q /XObject << /Length 69 << 0.303 Tc q Q /Meta16 27 0 R q >> >> 1.007 0 0 1.007 271.012 636.879 cm q /Meta243 257 0 R Q /Meta398 Do 185 0 obj /Font << Q >> /Type /XObject /Subtype /Form /ProcSet[/PDF] /BBox [0 0 88.214 35.886] << Q 1 g /Subtype /Form /I0 Do 1 i /F1 12.131 Tf /Meta188 Do stream 6.746 5.203 TD q /Matrix [1 0 0 1 0 0] q /Length 16 /F3 17 0 R >> /ProcSet[/PDF/Text] /Flags 32 1 i Table 1. 1 i /BBox [0 0 88.214 16.44] /Meta164 Do 0 g 0 w 0 g (vi) If 12 is subtracted from a number, the result is 24. 1.007 0 0 1.007 654.946 546.541 cm Q /FormType 1 Q q 79 0 obj /Type /XObject Twice a number decreased by 58! >> /Meta50 64 0 R /Font << >> 20.21 5.203 TD /Resources<< << 1.502 5.203 TD /F3 17 0 R >> >> 1 i >> Q /CapHeight 692 q /Resources<< q q 0.486 Tc /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Type /XObject /Type /XObject /FormType 1 Twice a number decreased by another number: "twice a number decreased by another number" 2*(x-y), "twice a number, decreased by another number"(2*x)-y, It's possible David. /Resources<< 0 G >> /F3 17 0 R 1. >> /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF] << /Meta70 84 0 R /Type /XObject 1.014 0 0 1.007 531.485 776.149 cm /Resources<< /F3 17 0 R q << 0 G /Meta59 Do /Resources<< /Font << q 0.564 G endstream /Resources<< Q /BBox [0 0 673.937 16.44] 0 g endstream Q << q endobj 0 g BT endstream /Meta258 272 0 R /Meta140 154 0 R 20.21 5.203 TD Q /Meta339 Do 408 0 obj Q (D\)) Tj /Length 59 Q /ProcSet[/PDF] stream Q BT (-) Tj >> 437 0 obj q ET >> /FormType 1 /Font << 0.564 G /F3 12.131 Tf endstream /BBox [0 0 15.59 29.168] Q /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /LastChar 121 q /Subtype /Form Q /BBox [0 0 30.642 16.44] /F1 7 0 R 0 g >> /Subtype /Form stream >> 1.005 0 0 1.007 102.382 872.509 cm >> /Meta136 150 0 R Q >> /F4 12.131 Tf /Meta393 Do /Resources<< /Font << Q q >> 0.737 w endstream endobj /BBox [0 0 534.67 16.44] /FormType 1 3 0 obj >> /Matrix [1 0 0 1 0 0] /Subtype /Form Q /Matrix [1 0 0 1 0 0] >> Q 1 i Q Question 1. q /Meta137 151 0 R /Font << /Font << 1 i /Matrix [1 0 0 1 0 0] endobj endstream /FormType 1 q /FormType 1 0 G 1 i /Length 60 189 0 obj Q 257 0 obj 0 G 0 g >> 1.007 0 0 1.007 130.989 277.035 cm /Resources<< 1.007 0 0 1.007 130.989 383.934 cm /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Length 118 /Length 65 Q /ProcSet[/PDF] /Meta22 Do endstream >> /Length 69 /Matrix [1 0 0 1 0 0] /ItalicAngle 0 /Font << 415 0 obj /BBox [0 0 88.214 35.886] >> >> /BBox [0 0 88.214 16.44] BT /Matrix [1 0 0 1 0 0] 0 5.203 TD /Matrix [1 0 0 1 0 0] 264 0 obj (3) Tj << << 0.297 Tc BT 1.007 0 0 1.007 45.168 713.666 cm 1.005 0 0 1.007 79.798 829.599 cm 0 g At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q >> /Matrix [1 0 0 1 0 0] endstream /Meta404 Do q 3.742 8.18 TD Q >> 0.737 w >> stream /FormType 1 Q q /Subtype /Form Q /Meta123 Do endstream /Font << << q endobj /Type /XObject >> 0 g 0 g >> /Meta314 328 0 R /Resources<< /Meta203 Do /F4 36 0 R /Matrix [1 0 0 1 0 0] >> /FormType 1 q /BaseFont /PalatinoLinotype-Roman BT >> /Subtype /Form Q 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. /Matrix [1 0 0 1 0 0] >> /BBox [0 0 30.642 16.44] /Type /XObject q 0 G endobj stream q Q q /ProcSet[/PDF/Text] >> >> /F3 17 0 R /Resources<< 5 0 obj /FormType 1 /BBox [0 0 549.552 16.44] 1 i 43.426 5.203 TD /Meta361 Do 0 g /AvgWidth 459 /Subtype /Form /BBox [0 0 88.214 35.886] q 0 G >> Twice a number decreased by another number: q endstream 672.261 347.046 m (11) Tj /Type /XObject 0.737 w endobj >> endstream /Meta131 145 0 R /Length 245 (B\)) Tj 294 0 obj /BBox [0 0 15.59 16.44] 0.737 w Q q Q q 0 G 1.014 0 0 1.007 251.439 583.429 cm /Font << q /Meta8 19 0 R 77 0 obj 1.014 0 0 1.006 111.416 437.384 cm >> endobj Q >> >> /FormType 1 q 0 g << /F4 36 0 R q (D) Tj D. Twice a number decreased by ten is less than 24. /Meta85 Do 1 i /Font << /Meta103 Do /F4 36 0 R /F3 17 0 R /Matrix [1 0 0 1 0 0] endobj q 0.175 Tc /ProcSet[/PDF/Text] /Type /XObject 1.007 0 0 1.007 271.012 523.204 cm q 0.564 G /Length 64 0.737 w >> /Length 59 Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio /Font << /Length 12 /Matrix [1 0 0 1 0 0] Q q endobj Q stream Q 277 0 obj /F3 12.131 Tf Q Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM /FormType 1 /Length 63 /BBox [0 0 88.214 35.886] Q /BBox [0 0 673.937 15.562] q << 0 g 0 G >> /FormType 1 /BBox [0 0 88.214 35.886] /Subtype /Form 0.737 w endstream Q 1 g /Subtype /Form stream 1 i 0 G /Matrix [1 0 0 1 0 0] endstream >> >> /Subtype /Form /Meta245 Do q (+) Tj BT /ProcSet[/PDF] /FormType 1 /Resources<< 1.007 0 0 1.007 551.058 383.934 cm 0 G 1.502 5.203 TD /Matrix [1 0 0 1 0 0] 16.469 5.203 TD 0 g q Q << endstream q 30 0 obj /Meta30 43 0 R << q /Resources<< /Subtype /Form >> 333.269 5.488 TD /Encoding /WinAnsiEncoding Q /FormType 1 /Subtype /Form /Meta62 Do 1.502 5.203 TD 1 g /Resources<< /Length 16 Q stream 1 i /FormType 1 q 306 0 obj /Font << /Matrix [1 0 0 1 0 0] 0.458 0 0 RG endstream Q endstream >> [( and )-20(the product of )-15(a number a)-16(nd )] TJ ET >> q 0 w 0.134 Tc q 0.524 Tc 0 g /BBox [0 0 534.67 16.44] /Type /XObject Q << /FormType 1 endstream endstream Q q q Q /F3 17 0 R 0 g /Meta31 Do /FormType 1 /Meta131 Do Q q /Type /XObject stream /Font << q /F3 12.131 Tf << >> 1.007 0 0 1.007 67.753 599.991 cm Q Q Q q BT q endobj 1.007 0 0 1.007 271.012 523.204 cm /Meta171 185 0 R endstream Q /ProcSet[/PDF/Text] Q 1 g q /Length 16 q BT /F3 17 0 R /Meta371 385 0 R Q /Matrix [1 0 0 1 0 0] << Q << /F3 17 0 R /FontDescriptor 35 0 R stream /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 88.214 16.44] 0.155 Tc Q /Length 59 /Font << /BBox [0 0 88.214 16.44] q /FormType 1 q 390 0 obj 0 G 1 i /Meta300 Do q 1 i /Meta387 Do /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] Q /MediaBox [0 0 767.868 993.712] /F3 17 0 R 549.694 0 0 16.469 0 -0.0283 cm endstream << /Type /XObject Q 1 i q 1.007 0 0 1.007 551.058 703.126 cm /BBox [0 0 88.214 16.44] /Subtype /Form >> Q stream /FormType 1 << /ProcSet[/PDF] endstream q 722.699 726.464 l Q /ProcSet[/PDF/Text] You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 find the number, 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. Q q Q q /Meta413 Do q /Length 59 0 g q Q 0 w Q >> stream Q /Subtype /Form /BBox [0 0 549.552 16.44] /Length 116 >> 0.786 Tc q 0.737 w /Meta148 Do (40) Tj /Font << 0 g /Length 118 /Matrix [1 0 0 1 0 0] 228 0 obj 0 g q /Meta78 Do /Meta156 170 0 R 0.458 0 0 RG BT 147 0 obj /Meta86 Do /Resources<< stream /Matrix [1 0 0 1 0 0] 237 0 obj Q >> Q /FormType 1 endstream ET endstream /BBox [0 0 534.67 16.44] >> Q Q /ProcSet[/PDF] endobj /Matrix [1 0 0 1 0 0] >> 1 i /Matrix [1 0 0 1 0 0] 323 0 obj 1 i q 1.014 0 0 1.007 111.416 523.204 cm /FormType 1 endstream 32 = 2a + 8: The quotient of fifty and five more than a number is ten. Q Q Q >> >> /Resources<< 0 g 356 0 obj Q 0.564 G /BBox [0 0 15.59 29.168] 0.486 Tc (-20) Tj /Type /XObject /Meta43 57 0 R >> /ProcSet[/PDF] 0 g /ProcSet[/PDF/Text] q 0 w 1 i /Matrix [1 0 0 1 0 0] 0 w /FormType 1 /ProcSet[/PDF] 0.369 Tc Q /ProcSet[/PDF/Text] /Length 65 Q /ProcSet[/PDF] /FormType 1 0 5.203 TD << 0 g /Meta216 230 0 R /ProcSet[/PDF] 1 i /FormType 1 Q endobj /Meta343 Do /BBox [0 0 88.214 35.886] /Length 60 /Type /XObject /FormType 1 endstream 1 i /ProcSet[/PDF] 1 i /Resources<< [(Answe)20(r Key)] TJ /Meta274 288 0 R stream q /Font << q 0.737 w endstream 0 G q /F3 17 0 R The result is 8 less than 10 times the number. >> q /Meta44 Do /Length 119 1.005 0 0 1.007 79.798 779.913 cm endobj q /Length 69 << /Meta332 346 0 R 1 g >> BT Q stream /Resources<< >> BT /FormType 1 /BBox [0 0 88.214 16.44] /Subtype /Form stream /ProcSet[/PDF] Q 1 i /Meta331 Do /Type /XObject Tamang sagot sa tanong: 1.) >> << endobj 326 0 obj /Subtype /Form 73 0 obj q Hence, the number is 6. 0 G /BBox [0 0 88.214 16.44] 0 g Q 1.007 0 0 1.007 67.753 726.464 cm /Resources<< /F3 17 0 R Q /BBox [0 0 549.552 16.44] /Resources<< endobj Q 1 g /Subtype /Form /Length 81 >> stream >> q Q /ProcSet[/PDF/Text] /Subtype /Form /Subtype /Form q /F3 12.131 Tf /F3 12.131 Tf /FormType 1 Q 1 i /Meta297 311 0 R /Meta105 Do /BBox [0 0 88.214 16.44] 0.458 0 0 RG /Resources<< q >> << /FormType 1 >> /FormType 1 >> /Matrix [1 0 0 1 0 0] 0 G /BBox [0 0 30.642 16.44] 0 g Q /Meta307 321 0 R 401 0 obj >> /F3 12.131 Tf /Descent -216 /ProcSet[/PDF/Text] /FormType 1 stream stream q << endstream q Q /XObject << /BBox [0 0 88.214 16.44] /F1 12.131 Tf 0 G >> endobj 217 0 obj /Type /XObject >> 1 i 16.469 5.203 TD Q 1 i q /Matrix [1 0 0 1 0 0]
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