BT 0.458 0 0 RG 1 i 0 g stream /ProcSet[/PDF] /Type /XObject q Q /Meta128 142 0 R endobj stream /F4 12.131 Tf /F3 17 0 R Aktual'nye voprosy Vol 10, No 3 (2020) endstream << << /ProcSet[/PDF/Text] q 1 i >> /F3 17 0 R /Subtype /Form >> /Type /XObject 0 g /Length 70 /Subtype /Form (x) Tj /Meta371 385 0 R << Q /BBox [0 0 15.59 16.44] Q 27.693 5.203 TD /FormType 1 0 G If n is "the number," which equation could be used to solve for the number? /Subtype /Form q Q Q 1.007 0 0 1.007 654.946 546.541 cm Q 1 i 202 0 obj 1 i /Meta392 408 0 R q 0.564 G endobj q >> 0 g Q 1 g << endobj /Subtype /Form /Length 118 Q 1.005 0 0 1.007 45.168 889.071 cm >> 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 . stream /Matrix [1 0 0 1 0 0] /Length 78 >> >> >> Q 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. q 1 g /Subtype /Form q /Matrix [1 0 0 1 0 0] 0 5.203 TD ET -0.058 Tw /Matrix [1 0 0 1 0 0] endobj 101.849 5.203 TD Q /Type /XObject /Subtype /Form 0 G /Meta422 Do 1 g Q q /Meta141 155 0 R 0.786 Tc 1 i 0 w >> >> /Resources<< >> /F4 12.131 Tf 1 i [tex]\sin (\pi -x)=\sin x[/tex]. 0 G q /Resources<< >> /Type /FontDescriptor 1.007 0 0 1.007 411.035 849.172 cm << /Subtype /Form /Subtype /Form 293 0 obj ET << /Subtype /Form /Meta400 Do q q q Q /BBox [0 0 15.59 16.44] Q /Resources<< 0 G /ProcSet[/PDF/Text] 0.564 G If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. 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 endstream Q 0 G 722.699 599.991 l /Type /XObject Q Q (58) Tj The width Of a rectangle is 15 cm and the perimeter is 12 cm. endstream endstream << Q q BT /F3 12.131 Tf /Font << endstream /Subtype /Form 0 g /Subtype /Form q /Length 69 /Type /XObject >> 0 g BT /ProcSet[/PDF] /BBox [0 0 30.642 16.44] q << Q q Q endobj /Type /XObject /BBox [0 0 88.214 16.44] q /FormType 1 q 0.458 0 0 RG /Meta225 239 0 R q 0.564 G stream 0.241 Tc /F3 17 0 R endobj /Meta50 Do /Type /XObject /Type /XObject /FormType 1 /Meta131 Do q Q /Subtype /Form stream /Type /XObject q Q /Resources<< endstream /Matrix [1 0 0 1 0 0] >> Q /Resources<< Then ab is a binary operation. Q /Type /XObject /F3 12.131 Tf q 1.007 0 0 1.007 411.035 383.934 cm Q 38 0 obj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 1 i 0.458 0 0 RG /Length 59 stream /F3 12.131 Tf 1 i stream stream /F3 17 0 R >> >> /Matrix [1 0 0 1 0 0] 0.737 w endstream /ProcSet[/PDF] /BBox [0 0 88.214 16.44] BT /Length 12 (6\)) Tj stream 206 0 obj Q /BBox [0 0 88.214 16.44] stream /Meta132 Do q Q /BBox [0 0 17.177 16.44] 0.486 Tc q /Meta362 Do /FormType 1 >> 272 0 obj >> >> BT 0.271 Tc /Resources<< 1 g q q /Resources<< endobj Q 0.524 Tc /BBox [0 0 15.59 16.44] /Meta378 392 0 R >> Q stream q /BBox [0 0 549.552 16.44] q /Font << 32.201 5.203 TD /Resources<< 0 G /F3 12.131 Tf 0 w 0 G /Type /XObject endobj >> Q (-11) Tj /Meta187 Do Q q 20.21 5.203 TD stream /ProcSet[/PDF/Text] /Resources<< 1 i /Meta310 Do /Type /XObject endobj /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] /Font << /Matrix [1 0 0 1 0 0] 0.297 Tc stream The ratio of a number to fifteen 4. << /BBox [0 0 88.214 16.44] /Resources<< q Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.737 w 1 i /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Meta19 30 0 R /Length 58 1 i Q /Resources<< /FormType 1 /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /Resources<< 0 g 0 G /Meta150 Do 582 546 601 560 395 424 326 603 565 834 516 556]>> /Resources<< >> /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] BT /FormType 1 /Meta372 386 0 R 387 0 obj /Resources<< >> 315 0 obj >> /ItalicAngle 0 0.564 G 0 5.203 TD Q /F3 17 0 R /Meta307 321 0 R BT /F3 12.131 Tf -0.486 Tw q /Length 16 /Resources<< /ProcSet[/PDF/Text] 1 i /Meta22 33 0 R ET /BBox [0 0 639.552 16.44] /F1 7 0 R 0.271 Tc 124 0 obj q >> (-9) Tj >> Q /Length 245 ET Q endstream stream q 1.005 0 0 1.007 79.798 713.666 cm q stream /BBox [0 0 534.67 16.44] q (x) Tj /FormType 1 /F4 12.131 Tf Q q stream << /Type /XObject << /Type /XObject /Type /XObject q /I0 51 0 R endobj 57 0 obj (-) Tj Q endstream /Type /XObject q Q 722.699 473.519 l 1 i 1 i ET << >> /Type /XObject q 0.458 0 0 RG stream /Font << /Type /XObject << >> (B\)) Tj 0.737 w /BBox [0 0 88.214 16.44] /Resources<< /FormType 1 /Resources<< /FormType 1 q /BBox [0 0 23.896 16.44] Q 1.005 0 0 1.007 102.382 347.046 cm q Q 0 g /ProcSet[/PDF/Text] << /Resources<< 19.474 20.154 l q 0 5.203 TD /Meta407 423 0 R Q /Resources<< 0 G 0.564 G /BBox [0 0 88.214 16.44] /Meta284 Do endstream stream 26.219 5.336 TD /Font << /Type /XObject /BBox [0 0 673.937 14.853] (4\)) Tj /Resources<< /Length 16 /Resources<< 20.21 5.336 TD 0 w 1 i 1 g 1 g stream stream /Matrix [1 0 0 1 0 0] ET << 0 w >> endobj /ProcSet[/PDF] (D\)) Tj stream /Length 69 (-4) Tj 0 G /ProcSet[/PDF/Text] Q Q << >> >> /Length 69 >> >> 18 0 obj Q Q /ProcSet[/PDF/Text] 0.737 w 278 0 obj Q /Matrix [1 0 0 1 0 0] Q Q BT 1.007 0 0 1.007 411.035 383.934 cm Q /Meta6 Do /ProcSet[/PDF/Text] stream BT stream /Length 69 /BBox [0 0 549.552 16.44] >> >> 38.948 5.203 TD >> /F3 12.131 Tf q 549.694 0 0 16.469 0 -0.0283 cm 0 w /Matrix [1 0 0 1 0 0] 0 g /Subtype /Form Q Q endstream 0 w /Type /XObject q >> /ProcSet[/PDF] 0 g /ProcSet[/PDF] endstream 1 g q (D) Tj endobj 1 i endstream q q 1 i 0 g /Length 69 233 0 obj 0.68 Tc 1.007 0 0 1.007 130.989 277.035 cm 0 g 1.014 0 0 1.007 531.485 523.204 cm Q endstream Q /Meta342 Do 0.458 0 0 RG endstream ET /Height 22 << Q /Type /XObject 350 0 obj /BBox [0 0 88.214 16.44] Q q 0.564 G /BBox [0 0 534.67 16.44] /F3 12.131 Tf q q /Matrix [1 0 0 1 0 0] 1 g /Type /XObject /FormType 1 Q Find the length. (+) Tj /Resources<< (B) Tj /Meta32 Do /F3 17 0 R /F1 12.131 Tf /Type /XObject Q endobj Q 0.786 Tc BT q q /Meta206 Do /Resources<< endobj /Subtype /Form << 1 i /Length 99 371 0 obj Q Double or twice a number means 2x, and triple or thrice a number means 3x. stream q 335 0 obj endstream Q /Length 69 Q Q q ET /Subtype /Form Q /F3 17 0 R Q >> /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 383.934 cm ET /Length 16 /Subtype /TrueType /ProcSet[/PDF/Text] >> /Subtype /Form /Matrix [1 0 0 1 0 0] >> /Type /XObject >> /Subtype /Form /Meta150 164 0 R stream 1.007 0 0 1.007 45.168 813.037 cm 1.007 0 0 1.006 551.058 763.351 cm /Type /XObject 30 0 obj ET >> Q Q >> /Resources<< /FormType 1 1 g 0.458 0 0 RG 89.12 5.203 TD >> 4.506 24.649 TD /FormType 1 12.727 5.203 TD 0 G endobj /FormType 1 Q 0 G /Font << Q /Meta343 357 0 R BT q Q >> Q >> (-) Tj stream stream q /Resources<< /Resources<< /F3 12.131 Tf Q /Ascent 1050 >> q /Matrix [1 0 0 1 0 0] endobj /Meta93 Do q endstream /Font << 85 0 obj << ET Q /FormType 1 /Meta197 Do /Meta79 Do /Matrix [1 0 0 1 0 0] >> endstream >> for the season. >> /FormType 1 q q /Resources<< ET /Type /XObject endobj 0.51 Tc stream /ProcSet[/PDF] endstream /F3 12.131 Tf endstream You could call them, Decreased by another number means subtract. /Type /FontDescriptor endobj /Type /XObject /Type /XObject 380 0 obj Q /Type /XObject /Subtype /Form stream /Type /XObject /Meta205 219 0 R << /Meta74 Do q /ProcSet[/PDF/Text] q 15.731 5.336 TD q /F3 12.131 Tf Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. /Matrix [1 0 0 1 0 0] >> /BBox [0 0 88.214 16.44] /Type /XObject /F1 12.131 Tf 0 G 0 g /ProcSet[/PDF] /F3 12.131 Tf /Meta289 Do Q: A number increased by 5 is equivalent to twice the same number decreased by 7. 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. /Descent -277 Q 1 i >> /Meta7 Do 1.007 0 0 1.007 45.168 713.666 cm /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 417.058 cm << endobj ET Q >> /BBox [0 0 534.67 16.44] /F3 17 0 R /Type /XObject /ProcSet[/PDF] 212 0 obj 0 5.203 TD q /F4 12.131 Tf q ( x) Tj 1 g /Meta257 271 0 R /F3 17 0 R 125.064 4.894 TD Q /Type /Page q >> q endobj /Subtype /Form /Length 67 BT 0 w /FormType 1 Q 0 G /Subtype /Form q /ProcSet[/PDF/Text] 0 w q /Type /XObject 1 i q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /Subtype /Form 0 g /BBox [0 0 88.214 16.44] stream >> 1.005 0 0 1.006 45.168 879.284 cm 0.564 G 1.014 0 0 1.006 391.462 763.351 cm ET /ProcSet[/PDF/Text] Q /Font << ET q In other terms, 52-nx The problem is asking that you subtract twice a number from 52. /Matrix [1 0 0 1 0 0] << /Meta276 290 0 R Q /BBox [0 0 88.214 16.44] /Length 54 0 G endstream /Meta321 Do 0.738 Tc endobj >> 0 5.203 TD q /BBox [0 0 534.67 16.44] 1.007 0 0 1.007 130.989 636.879 cm /ProcSet[/PDF] /FormType 1 10.487 5.203 TD /F3 12.131 Tf /Meta384 398 0 R stream q q /Font << 672.261 653.441 m /FormType 1 ET /FormType 1 /Length 95 >> /Resources<< /FormType 1 1 g q >> 1 g ET /Font << /Type /XObject endstream q /Font << /Subtype /Form << /FormType 1 /Matrix [1 0 0 1 0 0] Q Q stream /Meta167 Do endobj /ProcSet[/PDF/Text] 1 i [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /Meta102 116 0 R 0.37 Tc /Meta291 Do 1 i >> Q << q twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. 1.014 0 0 1.007 111.416 849.172 cm 1 i /Type /XObject 0.458 0 0 RG /Meta406 422 0 R 20 0 obj 1 i [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. 0.564 G >> /Matrix [1 0 0 1 0 0] 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 Q /Subtype /Form q /Meta71 Do /Subtype /Form << 0.458 0 0 RG /BBox [0 0 88.214 16.44] Q /Type /XObject /Type /XObject /Font << << Expert Answer. Q /Meta332 346 0 R /FormType 1 Q /Resources<< 0.738 Tc endstream endobj /Matrix [1 0 0 1 0 0] 0 G Q q /ProcSet[/PDF] endobj q /Meta194 208 0 R /Resources<< Q 0.564 G 0 G q >> , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. >> "49 . endobj 0 G /Resources<< q 1 i /F3 12.131 Tf >> /ProcSet[/PDF] /Type /XObject /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> /Meta414 430 0 R 0.458 0 0 RG << 0 g Q >> 672.261 546.541 m 323 0 obj /Resources<< 1 i Q /BBox [0 0 639.552 16.44] /FormType 1 q /Subtype /Form Q endstream /BBox [0 0 88.214 16.44] q q << /Meta40 Do Q /Meta272 286 0 R ET q endobj /Font << Q /BBox [0 0 15.59 16.44] Q >> /Resources<< /Matrix [1 0 0 1 0 0] >> BT /FormType 1 /I0 51 0 R >> /Meta8 Do >> /FormType 1 Q 1.007 0 0 1.007 271.012 703.126 cm endstream /Meta232 246 0 R stream /ProcSet[/PDF/Text] Q 52.412 5.203 TD 0 G endobj endstream ET /Resources<< /Type /XObject /F3 17 0 R q 9 0 obj 0.564 G /BBox [0 0 88.214 16.44] /Meta223 237 0 R (13) Tj /FormType 1 ET Q Q /Meta43 57 0 R 1.007 0 0 1.007 67.753 473.519 cm >> Q ET /FormType 1 /Length 16 >> endstream 0 g Q /Meta148 Do 1.007 0 0 1.007 45.168 730.228 cm q q (-) Tj /Meta283 Do /BBox [0 0 17.177 16.44] Q endobj >> /ProcSet[/PDF/Text] /FormType 1 Q Q Find the number. Q q 0 g /Length 63 /Font << /Subtype /Form Q /Type /XObject /Length 16 q 1.007 0 0 1.007 271.012 849.172 cm /Meta299 Do /Meta57 Do q Q /F1 12.131 Tf stream Q Q /Matrix [1 0 0 1 0 0] /Length 69 q 0 g endstream 1.007 0 0 1.007 67.753 599.991 cm q 0.564 G endobj /Subtype /Form 1 i endobj /Meta157 171 0 R twice a number decreased by 58 13 - 3x B. twice a number a divided by three = 2a / 3. five times a number x minus four = 5x - 4. thrice the sum of a number x and six = 3 (x + 6) Add your answer and earn points. << q << q In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). /F3 12.131 Tf /Length 65 Q 0.564 G >> 30.699 5.203 TD endstream /Font << /F3 17 0 R /BBox [0 0 534.67 16.44] 1 i 0.564 G BT 2. 1.005 0 0 1.007 102.382 293.596 cm /Length 16 q ET 0 g ET 0 5.203 TD /Resources<< endobj if the solution of an equation is x=-2, what could the original equation be? 1.007 0 0 1.007 551.058 523.204 cm 0 w 1.005 0 0 1.007 102.382 400.496 cm ET 0.737 w /Font << BT 1 i /F3 12.131 Tf 1 i /F3 17 0 R 0.369 Tc q 0.737 w endobj endstream Q endobj q 1 i BT /ProcSet[/PDF] /FormType 1 /Meta38 Do /Subtype /Form ET /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] ET /FormType 1 Q 1 g /Resources<< >> 0 g /Type /XObject 0.737 w q /Font << Q 1 i /ProcSet[/PDF] 0.458 0 0 RG q /Meta227 Do 0 g /FormType 1 Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. /Meta409 425 0 R /FormType 1 /Resources<< 0.738 Tc >> /ProcSet[/PDF/Text] /Subtype /Form 0.458 0 0 RG This site is using cookies under cookie policy . endstream Q /Meta353 Do 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 endstream That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. /F4 36 0 R q 1 i >> Q endstream /ProcSet[/PDF/Text] 0.51 Tc 1.502 7.841 TD /F1 7 0 R 1.007 0 0 1.007 271.012 450.181 cm /CapHeight 662 /Meta68 82 0 R >> >> 352 0 obj q >> Q 0 w Q ET 0 G /Font << /BBox [0 0 534.67 16.44] ET /BBox [0 0 30.642 16.44] 0 g 181 0 obj /Meta426 442 0 R If a number is 400%, then it is 4 times, the same as 4. 0 G q q /F1 7 0 R 20.21 5.203 TD /FormType 1 /BBox [0 0 15.59 16.44] 0 g Q BT endobj /FormType 1 Q 0 20.154 m /BBox [0 0 88.214 16.44] q stream /ProcSet[/PDF/Text] >> q >> << /F3 17 0 R 290 0 obj /Type /XObject >> (x) Tj /FormType 1 >> /Type /XObject endstream /FormType 1 Q q /F1 7 0 R Q q 1 i ET /Type /XObject /Subtype /Form q /Length 70 /Meta101 Do q Q >> 30.699 4.894 TD 0 g /Meta55 Do 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 /BBox [0 0 88.214 16.44] /Font << q /Type /XObject /Subtype /Form /Type /XObject 1 i 20.21 5.203 TD /BBox [0 0 15.59 16.44] /Meta13 24 0 R 1.502 5.203 TD endstream 0 g q /Length 69 << 24 0 obj /Meta251 Do Q endobj endobj 1.005 0 0 1.007 102.382 653.441 cm /F3 17 0 R /Meta296 310 0 R Q /Length 54 /Matrix [1 0 0 1 0 0] endobj 0 g 0 G >> stream New questions in Mathematics endobj Q /F3 12.131 Tf /Length 69 /F3 17 0 R /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Meta314 328 0 R ET >> 0.564 G >> 1.005 0 0 1.007 102.382 256.709 cm /Resources<< /F1 12.131 Tf /Type /XObject << /FormType 1 >> Q >> ET 0 g /Type /XObject 0.737 w >> /Length 59 Q /BBox [0 0 15.59 16.44] /Meta405 421 0 R 401 0 obj /Matrix [1 0 0 1 0 0] 0 g /Type /XObject /Length 60 endstream Q Q Q BT endobj /Meta390 406 0 R 0.564 G BT /F4 36 0 R 365 0 obj Q endstream endstream A. 0.458 0 0 RG /ProcSet[/PDF] q /Type /XObject /Resources<< 1 g q 1 i /Subtype /Form Q /Matrix [1 0 0 1 0 0] >> /BBox [0 0 88.214 16.44] >> /BBox [0 0 15.59 29.168] Q /FormType 1 On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini Q /Length 64 In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. /Meta229 Do >> Q /Meta334 Do /Subtype /Form /StemV 77 stream >> /Matrix [1 0 0 1 0 0] >> /Meta66 Do q /Length 69 << Q /Subtype /Form 1.007 0 0 1.007 654.946 799.486 cm 19 0 obj 26.957 5.203 TD q Q stream 0 g >> Q /Meta104 118 0 R (\)) Tj /Subtype /Form /Length 59 0.458 0 0 RG /Font << /Type /XObject stream /ProcSet[/PDF/Text] >> /ProcSet[/PDF/Text] q q q << Q q q /Meta72 Do 1 i q /Font << 1.014 0 0 1.007 531.485 277.035 cm 1 g q /Meta272 Do /Type /XObject endobj /Resources<< 0 g 0 g /Length 16 ET /Meta408 Do /I0 Do ET /Matrix [1 0 0 1 0 0] Q /Length 16 q /Meta126 Do (11) Tj Q /Font << >> /Meta238 Do /Matrix [1 0 0 1 0 0] /Meta196 Do /ProcSet[/PDF/Text] ET endstream /Font << /Meta170 Do 1 i Q q 0 g /Meta3 12 0 R /Resources<< /BBox [0 0 30.642 16.44] /Resources<< ( decreased by ) Tj Q Q /ProcSet[/PDF/Text] Q 1 i /ProcSet[/PDF] 2x - 15 = -27. /Resources<< q /Length 16 >> /Subtype /Form /ProcSet[/PDF/Text] Q q >> /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 1 i /Resources<< /F3 17 0 R /Type /FontDescriptor /Subtype /Form << /Resources<< /Matrix [1 0 0 1 0 0] ET /Resources<< 1 i << the other number. q << << q /Matrix [1 0 0 1 0 0] /Meta234 Do stream 0 g 1.005 0 0 1.007 79.798 779.913 cm stream q BT [( times )15(a numb)22(er and )] TJ 0 g << Q /BBox [0 0 30.642 16.44] q /Type /XObject endobj /Length 12 /Meta26 39 0 R 3.742 24.649 TD 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 0 G (13) Tj 0 g /Matrix [1 0 0 1 0 0] >> q 1.014 0 0 1.006 531.485 763.351 cm /Type /XObject /ProcSet[/PDF] q 1 i q [(1)-25(0\))] TJ 1.007 0 0 1.007 271.012 849.172 cm So let's go ahead and identify a v 1.005 0 0 1.007 102.382 400.496 cm q (9\)) Tj << >> q /F1 12.131 Tf q endobj 182 0 obj /Encoding /WinAnsiEncoding /Matrix [1 0 0 1 0 0] endobj q BT q endstream 1 i 1 g /Subtype /Form >> /F3 12.131 Tf /Subtype /Form q Answer link. /Resources<< /ProcSet[/PDF/Text] /BBox [0 0 17.177 16.44] 0.738 Tc stream << Q endstream >> /FormType 1 /F4 36 0 R endstream /ProcSet[/PDF/Text] 0.458 0 0 RG 0 g Q Q /FormType 1 Q Q /Type /XObject >> 1.502 5.203 TD 1.007 0 0 1.007 551.058 277.035 cm 0.737 w stream endobj /Subtype /Form /Length 16 /Length 70 >> /F3 17 0 R 0.737 w A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. /Meta278 Do 0 g >> BT /F3 12.131 Tf /Meta172 186 0 R ET A number = an unknown number which can be represented by a variable, usually x. /Resources<< /Meta140 154 0 R endstream /FormType 1 endobj Q Q Q endobj q BT BT endobj /FirstChar 32 /Matrix [1 0 0 1 0 0] Q 0.486 Tc >> /F3 17 0 R /Length 68 1.007 0 0 1.007 271.012 383.934 cm /Meta72 86 0 R BT Q /Type /XObject /F1 7 0 R /F4 36 0 R /Meta429 Do /Length 16 /Matrix [1 0 0 1 0 0] /Length 69 /Size 447 /ProcSet[/PDF] q q 1.007 0 0 1.007 271.012 776.149 cm q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Resources<< /Subtype /Form 0.458 0 0 RG /Meta41 Do >> /Type /XObject endstream q 0.737 w /Meta354 Do endstream << q 411 0 obj ET q Q q endstream /BBox [0 0 88.214 16.44] /Font << /FormType 1 0.458 0 0 RG Q 0.737 w /FormType 1 0.737 w ET Q >> /Subtype /Form 1 i /Resources<< >> ET 1.005 0 0 1.007 102.382 599.991 cm >> q /Matrix [1 0 0 1 0 0] /Font << 1.007 0 0 1.007 130.989 583.429 cm /F4 12.131 Tf Notice that we used the variable \large {d} d in our equation to stand for our unknown value. stream >> q 1 i 1.007 0 0 1.007 45.168 862.723 cm q >> endobj q Q /Meta88 Do 0 G >> 370 0 obj stream /FormType 1 q /FormType 1 >> Q /Subtype /Form Q /Subtype /Form /ProcSet[/PDF] 1.007 0 0 1.007 551.058 383.934 cm >> /FormType 1 /Meta377 Do 0 g 0.564 G /Meta38 52 0 R /Meta144 158 0 R q /ProcSet[/PDF/Text] 0 w /Resources<< ET << endstream q Q Q endstream BT /F4 12.131 Tf /Subtype /Form Q endstream q /Matrix [1 0 0 1 0 0] << /Font << /BBox [0 0 549.552 16.44] endobj 1.014 0 0 1.007 391.462 636.879 cm 1.005 0 0 1.007 79.798 796.475 cm q (x) Tj q stream /Subtype /Form >> q q /Subtype /Form q ET /BBox [0 0 30.642 16.44] /FormType 1 0 G endstream /Matrix [1 0 0 1 0 0] 227 0 obj /ProcSet[/PDF] 14.23 24.649 TD saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. BT 0.458 0 0 RG /Meta322 336 0 R 1 i /Meta184 198 0 R /Matrix [1 0 0 1 0 0] stream stream q Q >> q 0 w q /Length 118 0 20.154 m /F3 17 0 R Q Q /FormType 1 << /Length 69 /Meta286 Do endstream q /Leading 150 0.51 Tc /Length 58 endstream << /BBox [0 0 88.214 16.44] -0.058 Tw /F3 12.131 Tf 0.68 Tc endobj 1.007 0 0 1.007 411.035 383.934 cm >> q >> 1 i /Font << /Subtype /Image endobj ET ET 549.694 0 0 16.469 0 -0.0283 cm /F3 12.131 Tf endstream /Length 16 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 Q 16.469 5.203 TD /Meta397 Do 0 w Q Q Q 0 G stream 99 0 obj /ProcSet[/PDF] /Length 70 0 G /Meta255 269 0 R ET q >> /BBox [0 0 88.214 16.44] /Meta133 147 0 R 318 0 obj /FormType 1 /ProcSet[/PDF/Text] Q /F3 17 0 R /ProcSet[/PDF/Text] >> /Subtype /Form q q BT 1.007 0 0 1.007 271.012 636.879 cm >> >> q << << q << q 1.007 0 0 1.006 411.035 437.384 cm 0.564 G Q /F3 12.131 Tf q 1 g /ProcSet[/PDF/Text] /FormType 1 endobj q << 1 i 0.524 Tc 0 g Q /ProcSet[/PDF/Text] q Q << 1 i /Resources<< endobj << 16.469 5.203 TD /F3 12.131 Tf << q Q /Matrix [1 0 0 1 0 0] /Subtype /Form >> ( x) Tj /Subtype /Form /ProcSet[/PDF] /BBox [0 0 17.177 16.44] q endstream /Meta120 Do endobj /Subtype /Form BT 344 0 obj Q /Subtype /Form 0 g q /Resources<< /Type /XObject Answer only. 0 g /Matrix [1 0 0 1 0 0] q Q q 0.786 Tc /Meta298 312 0 R /Type /XObject /Subtype /Form endobj Q q stream /BBox [0 0 673.937 15.562] /Matrix [1 0 0 1 0 0] q /Subtype /Form endobj /CapHeight 476 0 G 0 g q /Meta348 362 0 R q >> /Meta404 Do endobj /Subtype /Form 0 G q Q /Meta247 Do 26.957 5.203 TD 1 i /Length 81 stream >> /F3 17 0 R /Type /XObject /BBox [0 0 534.67 16.44] >> endstream /F3 17 0 R q >> /Meta313 327 0 R 96 0 obj q /Meta10 Do 1.007 0 0 1.007 67.753 599.991 cm 0 g endstream >> 0 g 0 G q /F3 12.131 Tf (40) Tj 264 0 obj /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] >> 0 w endobj /ProcSet[/PDF] endobj q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf endstream /Meta151 165 0 R q /Meta98 Do 0.737 w BT << q 0.838 Tc Q >> /Meta343 Do q /Length 65 /Subtype /Form /Subtype /Form /Matrix [1 0 0 1 0 0] >> << Q /ProcSet[/PDF] endobj q endstream /Meta321 335 0 R 0.564 G endobj /Resources<< /Length 69 /BBox [0 0 17.177 16.44] ET q q q Q 1.014 0 0 1.007 531.485 383.934 cm endstream 31 0 obj 2.238 5.203 TD 0.198 Tc Q /Resources<< /Meta347 Do /Meta409 Do /Meta231 245 0 R q >> 1 i /Length 69 /Subtype /Form >> >> /F3 17 0 R >> /Type /XObject 0.31 Tc /Matrix [1 0 0 1 0 0] /Meta398 Do >> 11.99 24.649 TD >> Q >> Q /FormType 1 /Font << >> >> endobj q 1 i /Subtype /Form 258 0 obj 435 0 obj /Meta407 Do /Resources<< 20.21 5.203 TD /Type /XObject 1 i /BBox [0 0 88.214 35.886] /Type /XObject 0.369 Tc endstream q /Meta232 Do q /Subtype /Form /ProcSet[/PDF/Text] Q q stream /Font << (B\)) Tj >> 164 0 obj 0.737 w q stream /FormType 1 /Meta418 434 0 R q /Type /XObject /Matrix [1 0 0 1 0 0] /Meta349 Do (x) 6 times a number is 5 more than the number. Q /FormType 1 /FormType 1 1.014 0 0 1.007 391.462 450.181 cm BT /Meta135 Do /Meta87 Do Q ET >> Q >> 0 g q endobj /Meta244 Do q Q endobj q endobj >> >> /Matrix [1 0 0 1 0 0] Q /MissingWidth 252 /Matrix [1 0 0 1 0 0] /Resources<< /Type /XObject /FormType 1 /Resources<< /Font << q >> 6.746 5.203 TD Two fewer than a number doubled is the same as the number decreased by 38. /Subtype /Form /Subtype /Form endobj 0.458 0 0 RG 0 5.203 TD 1 i ET ET >> /LastChar 45 /Length 59 endobj (C\)) Tj endobj 1 i 0 20.154 m 1 g /Meta425 441 0 R /Type /XObject Q 6.746 5.203 TD /Type /XObject -y. 1 i 0 G << /Meta199 213 0 R /ProcSet[/PDF/Text] /Meta155 169 0 R >> /FormType 1 /F3 17 0 R 0.297 Tc /Meta80 94 0 R 1 g >> /Meta245 259 0 R 1 i /F3 12.131 Tf >> /Resources<< stream 1 i endstream /BBox [0 0 88.214 35.886] /F1 7 0 R /Matrix [1 0 0 1 0 0] Q /FormType 1 endobj S >> /Length 294 stream 0 w 0 w 439 0 obj /BBox [0 0 30.642 16.44] /Font << >> /Type /XObject Q endobj endobj /Length 59 /Matrix [1 0 0 1 0 0] q /FormType 1 /BBox [0 0 534.67 16.44] 442 0 obj stream /FormType 1 /Matrix [1 0 0 1 0 0] /Type /XObject 0.458 0 0 RG Q << BT /Font << 98.843 5.203 TD /ProcSet[/PDF] 1 i endobj 0.564 G >> /F1 12.131 Tf /Subtype /Form >> 1 i /Subtype /Form /XObject << /Length 69 /Length 69 0.737 w /BBox [0 0 30.642 16.44] << /Meta412 Do /F4 36 0 R endobj BT /Font << Q /Length 69 /Meta294 Do >> 0 w /Length 12 q q /Length 59 /Resources<< 549.694 0 0 16.469 0 -0.0283 cm Q /Length 16 /F3 17 0 R endobj /FormType 1 stream /Subtype /Form 1.014 0 0 1.007 111.416 636.879 cm q There were x cookies at the beginning of a party. /Meta186 200 0 R 1 i /Meta75 Do Q ET >> q Q (3\)) Tj /ProcSet[/PDF] >> stream /F3 12.131 Tf /Font << Q /Meta185 199 0 R 0 G q q >> /Matrix [1 0 0 1 0 0] /MissingWidth 250 0.738 Tc q /Font << /BBox [0 0 639.552 16.44] /Meta168 182 0 R endstream /Type /XObject endstream /Subtype /Form /BBox [0 0 88.214 16.44] 1 i /F3 12.131 Tf /ProcSet[/PDF/Text] q 1 g 1 i 1 i /MediaBox [0 0 767.868 993.712] /Subtype /Form /FormType 1 Q q >> endstream /ProcSet[/PDF/Text] /Meta422 438 0 R /Subtype /Form 337 0 obj /Meta334 348 0 R /Matrix [1 0 0 1 0 0] 333.269 5.488 TD /Subtype /Form 1 i /Resources<< stream << /Length 16 q q /Length 107 /ProcSet[/PDF] /Font << /Type /XObject endstream q /Subtype /Form 92 0 obj /F3 12.131 Tf >> 0 g /Meta342 356 0 R >> /Subtype /Form (x) Tj BT /Meta364 Do /Matrix [1 0 0 1 0 0] /Resources<< 1.007 0 0 1.007 551.058 523.204 cm /BBox [0 0 88.214 16.44] 0.737 w stream endobj The rate of positive findings after 1 round of screening in the LCSDP was more than twice . endobj /FormType 1 /Length 59 0 g q q >> 1 g 0 g stream /FormType 1 /FormType 1 Q q q /Matrix [1 0 0 1 0 0] >> BT /F4 36 0 R 0.37 Tc >> /BBox [0 0 88.214 35.886] >> 0.564 G >> /F3 12.131 Tf q /Length 12 /Length 16 /Meta96 Do (11) Tj >> 0 4.894 TD /Matrix [1 0 0 1 0 0] >> /Type /XObject /Matrix [1 0 0 1 0 0] q 317 0 obj /Type /XObject q >> /Font << Q stream q >> Q >> /ProcSet[/PDF/Text] /Length 118 /Meta375 389 0 R ET /Type /Font q q Q /Resources<< /Meta160 174 0 R This site is using cookies under cookie policy . Q 0.297 Tc q (5) Tj >> /FormType 1 1 i q /ProcSet[/PDF/Text] q /Resources<< 1 i 0 g 103 0 obj /ProcSet[/PDF/Text] endobj stream ET 399 0 obj /Meta277 291 0 R /ProcSet[/PDF] /F3 17 0 R /Length 118 Q 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 q /Type /XObject stream 0.738 Tc /Subtype /Form S /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /Font << 1 i 0.369 Tc Q /Subtype /Form /Matrix [1 0 0 1 0 0] stream q stream /F3 12.131 Tf 1.007 0 0 1.007 411.035 330.484 cm stream /F1 12.131 Tf Q 0 5.203 TD /BBox [0 0 88.214 16.44] q /F3 17 0 R q q Q /Font << Q 361 0 obj /Meta47 Do /Subtype /Form 0 G q 1 i /FormType 1 /Resources<< 0.564 G q Q 0 g 139 0 obj << 1 i q /FormType 1 /ProcSet[/PDF/Text] /Length 59 /ProcSet[/PDF/Text] /F3 17 0 R q >> << 0 G 69 0 obj Q /BBox [0 0 534.67 16.44] q Q /Type /XObject 1 i q 0 g /Matrix [1 0 0 1 0 0] /FormType 1 0.458 0 0 RG /Meta411 Do << >> 1.014 0 0 1.007 111.416 636.879 cm Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. endstream 0 g /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Type /XObject 1 i /Meta266 280 0 R /Meta246 Do Q Q q q /Meta335 349 0 R stream Q stream /Subtype /Form -0.126 Tw 1 g /F3 17 0 R q >> 1 i /Meta153 Do endobj 0 w /I0 51 0 R >> << >> Q q /FormType 1 /F3 17 0 R 0 G q /Resources<< /BBox [0 0 639.552 16.44] (x) Tj 0 w /Matrix [1 0 0 1 0 0] Q 81 0 obj /ProcSet[/PDF/Text] Q 0 G /Meta63 77 0 R 62 0 obj << find what is x and check it 3x+2/5 - 2x-1/6 = 2/15 pa help po need ko lng po True or False: Let a & b be any elements of S, the set of natural numbers. ET /Type /XObject /Meta206 220 0 R /FormType 1 Q >> stream 0.134 Tc 61 0 obj 0 g /Resources<< /Resources<< 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. /Resources<< stream Q << q Q /BBox [0 0 88.214 16.44] q 163 0 obj /FormType 1 Q /Resources<< /BBox [0 0 30.642 16.44] 0 g -0.101 Tw [(F)-22(ive)] TJ >> Question. /FormType 1 /Meta323 Do /Subtype /Form stream q stream Q 1 i q q Q 0 g endobj /Subtype /Form q endstream /F3 17 0 R /ProcSet[/PDF/Text] 102 0 obj /F3 17 0 R 0.564 G (-20) Tj /Meta405 Do << (-) Tj q Q Q Let x the unknown number. /ProcSet[/PDF/Text] /BBox [0 0 673.937 16.44] >> 1.014 0 0 1.007 391.462 776.149 cm q Q Q q 0 G /Length 12 0 w stream /F3 17 0 R 186 0 obj q /Type /XObject /Type /XObject endstream /Matrix [1 0 0 1 0 0] q /F4 36 0 R /Resources<< q q /ProcSet[/PDF/Text] >> q /Meta141 Do << << Q q stream 0 g q /F3 12.131 Tf 0 g 1 i 0 w >> /Matrix [1 0 0 1 0 0] BT 1 i q endobj /Subtype /Form endobj /FormType 1 stream 118.317 5.203 TD stream /Type /XObject >> Medium /BBox [0 0 88.214 35.886] 32.201 5.203 TD /ProcSet[/PDF/Text] 168 0 obj /Meta93 107 0 R /Resources<< << /ProcSet[/PDF] /Font << 0.369 Tc /StemH 94 /F3 17 0 R /Meta230 Do endstream 414 0 obj /F3 12.131 Tf 0 g 1.014 0 0 1.007 111.416 523.204 cm Q /Meta191 Do 1.007 0 0 1.007 654.946 653.441 cm 1.007 0 0 1.007 271.012 277.035 cm /Resources<< /Meta60 74 0 R 1.007 0 0 1.007 67.753 347.046 cm /BBox [0 0 15.59 16.44] >> /Meta244 258 0 R /Resources<< 1 i q 1.007 0 0 1.007 551.058 383.934 cm /Font << >> Q 0.737 w Q 1 i >> Q /Resources<< /FormType 1 /BBox [0 0 88.214 16.44] endstream endobj 0.458 0 0 RG ET /Length 16 0 w /Subtype /Form /BBox [0 0 15.59 16.44] 1 i BT Q /F2 12.131 Tf /BBox [0 0 88.214 16.44] 0.564 G << /Length 67 Q BT q endstream q endstream ET /Length 69 >> << endstream /FormType 1 Q 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. 0 g /Meta158 172 0 R 308 0 obj q /F3 17 0 R

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