185.725 5.203 TD Q >> q 65.906 4.894 TD Q /BBox [0 0 30.642 16.44] 1 i 1 i /Meta312 326 0 R stream /Type /XObject Q /Length 64 stream /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] q q >> 0 G q q 274 0 obj q /ProcSet[/PDF/Text] /Type /XObject Q /BBox [0 0 15.59 16.44] (40) Tj /Meta362 376 0 R endstream 0 g 1 g q Q /XHeight 476 q Abstract: The aim of the study was to investigate the expression of miR-155 in plasma and peripheral blood mononuclear cells (PBMCs), the effects of miR-155 on the apoptosis rate /DecodeParms [<> ] /Matrix [1 0 0 1 0 0] 0 G [( and )-20(the product of )-15(a number a)-16(nd )] TJ Q endobj q >> endstream /Meta63 Do /Resources<< Q 428 0 obj /FormType 1 Q 221 0 obj /Type /XObject endstream endobj /Meta280 Do q the sum of a number and twelve. /Meta285 Do 0 g endstream /Matrix [1 0 0 1 0 0] >> 18 0 obj /F3 17 0 R /Resources<< A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. /Subtype /Form 1.007 0 0 1.007 271.012 776.149 cm Q 0 g /Length 69 Q >> << 48 0 obj /Meta313 Do << << ET /BBox [0 0 639.552 16.44] /Length 78 /Matrix [1 0 0 1 0 0] /Resources<< >> q 1.007 0 0 1.007 551.058 703.126 cm >> /Subtype /Form /Resources<< << 0.458 0 0 RG 371 0 obj >> 22.478 5.336 TD (8\)) Tj endobj q 1.005 0 0 1.007 79.798 763.351 cm /Meta12 Do >> 1 i 0 g >> Q Q 1.005 0 0 1.007 79.798 763.351 cm q q Q >> /Meta389 405 0 R ET 1.007 0 0 1.007 271.012 330.484 cm endstream (-11) Tj q endobj 0.458 0 0 RG Q >> 10.487 5.203 TD Q 0 G /Length 12 endstream Q 0.737 w 1 i 1 i 1.014 0 0 1.006 531.485 690.329 cm 1.005 0 0 1.007 102.382 799.486 cm 1.014 0 0 1.007 391.462 703.126 cm >> Q answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. /BBox [0 0 17.177 16.44] << /Matrix [1 0 0 1 0 0] endobj /ProcSet[/PDF/Text] /Font << >> /Subtype /Form /Meta127 Do q /F3 12.131 Tf /Length 12 /Matrix [1 0 0 1 0 0] /Meta40 Do /F3 12.131 Tf Q 0 G endstream 1 i endobj /F1 7 0 R /Meta423 439 0 R /Type /XObject >> (\)) Tj q endstream /Resources<< /Resources<< 0 G 1.007 0 0 1.007 551.058 383.934 cm (5) Tj A number increased by 5 is equivalent to twice the same number decreased by 7. (D\)) Tj 1.005 0 0 1.007 102.382 726.464 cm /Subtype /Form endstream /FormType 1 /F4 36 0 R 0 g /Length 118 1 i 0 g stream 0 g /FormType 1 >> Q endstream (B\)) Tj /Matrix [1 0 0 1 0 0] 0 G 0.37 Tc >> /Resources<< q 1 i /Font << /Meta321 Do Q 0 G /Meta405 Do /Resources<< >> 0 G /FormType 1 672.261 726.464 m stream /I0 Do /ProcSet[/PDF/Text] /ProcSet[/PDF] Q /Resources<< 0.564 G q /Resources<< /F3 12.131 Tf /FormType 1 /FormType 1 /Resources<< /FormType 1 q (x) Tj /Font << /Length 104 stream 1.008 0 0 1.007 654.946 293.596 cm q /FormType 1 stream endstream 224 0 obj q 1.007 0 0 1.007 130.989 776.149 cm 135 0 obj >> q q /Length 65 0 g /Resources<< 0 g 268 0 obj >> >> 180 0 obj Q /ProcSet[/PDF/Text] /Type /XObject /ProcSet[/PDF/Text] /FormType 1 672.261 473.519 m stream /Subtype /Form /BBox [0 0 30.642 16.44] Q 1.007 0 0 1.007 411.035 636.879 cm 1 i /F4 36 0 R q 0 g q >> /Subtype /Form /BBox [0 0 88.214 16.44] endobj Q /BBox [0 0 17.177 16.44] /Subtype /Form 1 i Q q q VIDEO ANSWER: in this problem were asked to solve giving, given the following information. q /Length 118 /Resources<< 0 G (13) Tj /FormType 1 Q [(The )-16(s)15(um )-14(of )] TJ >> endstream /F3 12.131 Tf endstream q Q /BBox [0 0 15.59 16.44] /Resources<< BT stream /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] >> >> /Length 12 /Resources<< 1 i /Length 63 /Type /XObject q endobj /Type /XObject Q 1.007 0 0 1.007 654.946 799.486 cm /Matrix [1 0 0 1 0 0] 128 0 obj BT /FormType 1 q endobj Q endobj >> >> /Type /XObject Q 1.007 0 0 1.006 411.035 690.329 cm q 1 i 1.007 0 0 1.007 271.012 583.429 cm /FormType 1 /F3 17 0 R Q 1 i ET /Subtype /Form >> /Subtype /Form 1 g /F3 17 0 R endstream , Prove the following stream /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] Q endstream /F4 36 0 R Q 1.007 0 0 1.007 271.012 636.879 cm /F3 12.131 Tf Q Q endstream Q /BitsPerComponent 1 /F3 12.131 Tf q /Subtype /Form 0 w /F3 12.131 Tf 0 g >> Q 0 g /Resources<< 0.737 w /CreationDate (D:20140515121932-04'00') 0 5.203 TD 3.742 5.203 TD /Font << /ProcSet[/PDF] stream q /FormType 1 /FormType 1 /F3 17 0 R Q 1.014 0 0 1.006 531.485 836.374 cm 70 0 obj q /BBox [0 0 639.552 16.44] /Meta395 411 0 R /Meta170 Do << /ProcSet[/PDF] /Font << stream >> q /ProcSet[/PDF/Text] 0.737 w /ProcSet[/PDF] >> [(E)-14(le)-23(ven)] TJ stream Q 1 i >> 250 0 obj /Meta261 Do endobj 0 g 6.746 5.203 TD >> /ProcSet[/PDF] << 1.014 0 0 1.007 111.416 277.035 cm /F3 12.131 Tf Q 1.007 0 0 1.007 654.946 726.464 cm 1 i q 1.502 24.649 TD /FormType 1 1.502 5.203 TD q >> 0.737 w /F3 12.131 Tf endstream In addition, testosterone in both sexes is involved in health and well-being . 24.718 8.18 TD /Length 60 /Type /XObject Q /BBox [0 0 17.177 16.44] stream 1.005 0 0 1.007 102.382 400.496 cm 0 G q /Resources<< /Matrix [1 0 0 1 0 0] >> /Meta416 432 0 R q /ProcSet[/PDF/Text] /Type /XObject >> >> /Subtype /Form Q endstream /BBox [0 0 30.642 16.44] q 0.564 G /Resources<< << >> Q 410 0 obj Q 0.737 w endstream /Matrix [1 0 0 1 0 0] Q q (+) Tj /FormType 1 /Type /XObject 1 i /Matrix [1 0 0 1 0 0] Q Q Q endobj Q /Matrix [1 0 0 1 0 0] endstream 1.007 0 0 1.007 130.989 636.879 cm 0.458 0 0 RG 366 0 obj q q 0 g 0 5.203 TD 1.005 0 0 1.007 102.382 400.496 cm 0 g /BBox [0 0 534.67 16.44] /ProcSet[/PDF] endobj stream [(1)-25(0\))] TJ >> >> endstream 0 g Q endstream /Length 65 Q q /Subtype /Form /Length 58 q 0 g 1 i /ProcSet[/PDF] Q 0.458 0 0 RG Q q /ProcSet[/PDF/Text] BT /MediaBox [0 0 767.868 993.712] endobj /Type /Font Q /FormType 1 /Length 12 /Resources<< q stream 1.014 0 0 1.006 111.416 510.406 cm /Descent -277 /Meta265 279 0 R /FormType 1 q Q /BBox [0 0 673.937 16.44] Q /ProcSet[/PDF/Text] Q /Length 59 /Meta102 Do /Meta182 Do 0 G /F4 12.131 Tf /Type /XObject << >> /ProcSet[/PDF] 0 G Q endobj >> Q q 291 0 obj /Font << /BBox [0 0 88.214 16.44] Q /ProcSet[/PDF/Text] << endobj 0 G /Meta236 Do /Matrix [1 0 0 1 0 0] stream /F3 17 0 R endstream >> 1 g That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. endstream ET 12.727 24.649 TD /Meta328 Do /Subtype /Form /Type /XObject 273 0 obj /ProcSet[/PDF] 0.737 w /ProcSet[/PDF/Text] BT << /ProcSet[/PDF/Text] /Type /XObject /FormType 1 /Length 59 << >> Q /Font << 1 i Q 1 i BT ET /Matrix [1 0 0 1 0 0] Q /Font << BT 0.564 G /BBox [0 0 534.67 16.44] /Meta50 64 0 R << /BBox [0 0 15.59 16.44] 158 0 obj /Subtype /Form /FormType 1 >> ET /ProcSet[/PDF/Text] q 1 i /Length 59 1 i /Font << 0 g Q 0 w ET q (A\)) Tj q 232 0 obj /Font << /F3 12.131 Tf 1 g q /FormType 1 /Meta14 Do q 0 w >> Q 0 g endobj >> 312 0 obj /Meta272 Do /Meta141 Do q 0 g 14.23 24.649 TD q Q We are asked to find the number, so, we could assign the number as "x". /ProcSet[/PDF/Text] /Type /XObject >> /Meta137 151 0 R q 0 w ET /F3 17 0 R /Font << /FormType 1 /F3 12.131 Tf 182 0 obj 1.014 0 0 1.007 531.485 636.879 cm /Meta348 362 0 R /Length 79 >> Medium >> >> /Length 59 /BBox [0 0 17.177 16.44] Q 1 i q Q 0.564 G 0.564 G q q 0.458 0 0 RG /Meta37 Do /Meta354 368 0 R endobj /BBox [0 0 88.214 16.44] /BBox [0 0 17.177 16.44] Q /Resources<< /Type /XObject << /BBox [0 0 88.214 16.44] /Meta268 Do Q 0.737 w stream Q /FontBBox [-170 -292 1419 1050] /Meta228 Do /Matrix [1 0 0 1 0 0] 0 g /BBox [0 0 15.59 29.168] /FormType 1 >> /ProcSet[/PDF/Text] /Font << Q 0 G Q 0 5.336 TD /FormType 1 /F3 17 0 R >> >> /F3 12.131 Tf /Meta59 73 0 R q /Meta51 65 0 R /Matrix [1 0 0 1 0 0] >> /Meta303 317 0 R Q q /FormType 1 /Subtype /Form >> /BBox [0 0 549.552 16.44] >> 0 G (-23) Tj /F3 17 0 R q /ProcSet[/PDF/Text] /ProcSet[/PDF] Q /FormType 1 endobj >> /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 102.382 347.046 cm q /Subtype /Form >> q Ten divided by a number 5. >> /Subtype /Form /Meta163 Do /FormType 1 1 i /Meta238 Do q (x) Tj 0.332 Tc Q /Font << Q q /BBox [0 0 30.642 16.44] Q /Length 69 /FormType 1 138 0 obj 23.952 4.894 TD 0.458 0 0 RG << 0 g endobj /Meta1 8 0 R (6\)) Tj /FormType 1 endobj >> /Length 54 /Meta330 344 0 R /Flags 32 1 i 20.21 5.203 TD q q 0 w >> (\)) Tj >> stream 0 g 0 g 205 0 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stream /Font << /Length 69 >> Andrew M. endstream endobj q (+) Tj 1.014 0 0 1.007 531.485 523.204 cm q 0 G /F3 17 0 R >> /Type /XObject /Font << Q stream 0 g Q /Type /XObject 0.564 G stream /F3 17 0 R endstream /ProcSet[/PDF/Text] 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q /Font << BT q 0.737 w 1.007 0 0 1.007 67.753 799.486 cm Q /Subtype /Form endstream /ProcSet[/PDF/Text] Q Q endstream >> 283 0 obj 361 0 obj 0.227 Tc /Meta416 Do stream /Length 67 stream stream 0 G /FormType 1 /Subtype /Form If a number is 50%, then it is a half - the same as 0.5 or 1/2. q /Subtype /Form /Meta423 Do /Meta288 302 0 R stream Q q 38.182 5.203 TD 125 0 obj /Resources<< >> Q 0.737 w 0 g /BBox [0 0 88.214 16.44] (iii) 25 exceeds a number by 7. /F3 12.131 Tf endstream /BBox [0 0 88.214 16.44] endstream << 1 g >> /Type /XObject q /Subtype /Form 0.425 Tc Q q 1.007 0 0 1.006 411.035 510.406 cm Q /F3 17 0 R /Meta84 Do 0 g Q /Meta179 193 0 R /FormType 1 Q /Meta249 Do stream stream /ProcSet[/PDF/Text] /Meta164 Do >> 0.68 Tc /BBox [0 0 88.214 16.44] >> 0.369 Tc /Meta359 373 0 R 0 5.203 TD q . /Matrix [1 0 0 1 0 0] q Q q 0 g >> Q endobj q 0 w 42 0 obj Q -0.486 Tw /F4 12.131 Tf Q >> /Font << endstream /Length 2252 << 0 G /ProcSet[/PDF/Text] 0.486 Tc /F1 7 0 R endstream 0 w /Meta384 Do /Matrix [1 0 0 1 0 0] /Meta170 184 0 R (2) Tj 173 0 obj >> Q /FormType 1 q saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. q << /Type /XObject /ProcSet[/PDF] /Length 66 /Matrix [1 0 0 1 0 0] >> 10 0 obj q /BBox [0 0 30.642 16.44] >> Q /Font << /Resources<< Q >> 0.737 w q stream 1 i ET /Font << /FormType 1 << /FormType 1 /Meta310 324 0 R ET 0 G 0.311 Tc /F3 12.131 Tf 0 g BT Q /FormType 1 endstream q /Type /XObject >> endobj q q q Q Q 0 5.203 TD /Matrix [1 0 0 1 0 0] q /Type /XObject /F3 17 0 R endobj 0 w /F3 12.131 Tf /Type /XObject /ProcSet[/PDF] << /BBox [0 0 88.214 16.44] q Q stream q /Length 16 /BBox [0 0 88.214 16.44] q /Meta31 Do 73 0 obj >> /Length 12 1 i q 1 g /Matrix [1 0 0 1 0 0] ( \() Tj stream 1.014 0 0 1.007 391.462 277.035 cm (-) Tj Q /ProcSet[/PDF/Text] Q >> >> stream 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. /Length 69 Testosterone is the primary sex hormone and anabolic steroid in males. /BBox [0 0 15.59 16.44] /Meta343 Do True False ET [tex]\sin (\pi -x)=\sin x[/tex]. stream /ProcSet[/PDF/Text] >> /BBox [0 0 673.937 68.796] q Q /Meta203 217 0 R Q endstream endstream >> /FormType 1 /F3 12.131 Tf endstream /FormType 1 /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 0.737 w /Length 12 /Meta196 Do /Font << q /Meta44 Do 0 g endstream /F3 17 0 R >> stream /FormType 1 >> /Meta191 Do >> /Subtype /Form Q Q /Length 12 0 G BT Q /ProcSet[/PDF] [(Negativ)16(e )] TJ /Type /XObject /BBox [0 0 534.67 16.44] Q Q >> /Leading 150 438 0 obj /Meta304 318 0 R 115 0 obj /F3 17 0 R 1 i 0 G 0.369 Tc /Font << /Matrix [1 0 0 1 0 0] endstream /Matrix [1 0 0 1 0 0] /Meta387 403 0 R /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject 722.699 400.496 l 0.458 0 0 RG 1.007 0 0 1.007 271.012 450.181 cm /Meta33 46 0 R q /Font << A) 5 more than a number. q /Length 69 Q 162 0 obj /F3 17 0 R q /Length 16 >> Q 1 i /Font << /Meta193 Do >> /FormType 1 1 i 1 i /F3 17 0 R /Length 16 endstream /F3 17 0 R /Subtype /Form stream 20.21 5.203 TD q /Matrix [1 0 0 1 0 0] 0 w >> /Meta256 Do -0.486 Tw /BBox [0 0 534.67 16.44] Q /F3 17 0 R /F3 12.131 Tf [(thir)17(te)15(en)] TJ /Matrix [1 0 0 1 0 0] q /Type /XObject >> 362 0 obj /Subtype /Form 367 0 obj /Matrix [1 0 0 1 0 0] endobj q stream /Type /XObject Q /BBox [0 0 88.214 16.44] Twice = two times, double. << q >> stream >> Q /Length 70 /ProcSet[/PDF] << /Meta372 386 0 R >> 1.014 0 0 1.006 391.462 510.406 cm /ProcSet[/PDF] /Type /XObject Q 26.219 5.203 TD /F3 17 0 R 311 0 obj >> /Meta218 Do >> endstream >> Q q (x ) Tj 0 20.154 m Q 0.458 0 0 RG Q q 0.737 w q /Resources<< Q 0 g Q /Subtype /Form /Meta204 Do 1.007 0 0 1.007 271.012 849.172 cm /Resources<< /Meta150 164 0 R /Matrix [1 0 0 1 0 0] 47 0 obj q 25.454 5.203 TD /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 400.496 cm Q endobj << 0 g /Resources<< q Q /Resources<< << /ProcSet[/PDF] >> 1.007 0 0 1.007 551.058 583.429 cm << /FormType 1 << /Length 59 /Resources<< 0 g >> 22.478 5.336 TD 0 G >> q /Meta87 Do 1 g endobj /FormType 1 /ProcSet[/PDF/Text] /Resources<< /Resources<< (40) Tj /Resources<< 1 i In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. 0.786 Tc q 243 0 obj 1.007 0 0 1.007 551.058 703.126 cm /Matrix [1 0 0 1 0 0] Q >> 1.007 0 0 1.007 411.035 636.879 cm /FormType 1 /Matrix [1 0 0 1 0 0] S 0.458 0 0 RG /Resources<< /F3 17 0 R /Type /XObject >> 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 /F3 12.131 Tf >> q stream 0 w /Resources<< Q No packages or subscriptions, pay only for the time you need. /F3 12.131 Tf 0 G 549.694 0 0 16.469 0 -0.0283 cm /Resources<< << /Type /XObject /Resources<< 6 0 obj >> endstream 0.68 Tc q 549.694 0 0 16.469 0 -0.0283 cm q endstream q 1 g /Resources<< endstream /Type /XObject /FormType 1 endstream Q << endobj endstream 1.007 0 0 1.007 271.012 277.035 cm 1.014 0 0 1.006 251.439 690.329 cm >> endstream /Length 58 /Font << >> << /Subtype /Form endstream /BBox [0 0 15.59 16.44] 0.382 Tc Q /Matrix [1 0 0 1 0 0] 30.699 4.894 TD /F1 12.131 Tf /BBox [0 0 88.214 16.44] q /ProcSet[/PDF] ET >> /Meta292 Do >> 0.458 0 0 RG q -0.03 Tw 4 0 R 0 g q endobj >> 1 i Q /Type /XObject /FormType 1 1.005 0 0 1.007 102.382 872.509 cm This site is using cookies under cookie policy . /BBox [0 0 15.59 29.168] endstream 1.014 0 0 1.006 111.416 763.351 cm /Matrix [1 0 0 1 0 0] 1 i BT Q /Font << 0 5.203 TD q q q >> /Length 16 /Subtype /Form /Resources<< >> 328 0 obj stream /Length 16 (4\)) Tj 0.425 Tc endstream stream /Meta187 201 0 R >> endobj -0.463 Tw BT Q (D\)) Tj /F3 12.131 Tf /Font << /Meta226 Do 0.68 Tc /Flags 32 /Resources<< Calculate a 15% decrease from any number. q endstream 0.271 Tc 1.007 0 0 1.007 130.989 383.934 cm 13.493 5.336 TD /Font << /Resources<< 0 G 0 g /Length 67 1.007 0 0 1.007 551.058 277.035 cm >> Q >> 1 i 152 0 obj /BBox [0 0 88.214 16.44] /F3 12.131 Tf Q /Resources<< 223 0 obj q [(-1)-16(52)] TJ Q (D\)) Tj 0 w /F3 17 0 R S 117 0 obj /Resources<< stream /Resources<< Q /Matrix [1 0 0 1 0 0] q The sum of a number and 2 is 6 less than twice that number. (x ) Tj /Resources<< q endobj /BBox [0 0 88.214 35.886] >> 0 w endstream /Resources<< >> /Meta231 Do Get a free answer to a quick problem. BT BT /F4 12.131 Tf ET 1.007 0 0 1.006 130.989 437.384 cm /Font << /ProcSet[/PDF] q /Meta49 Do The symbols 17 + x = 68 form an algebraic equation. q /Resources<< /Length 64 >> 1 g >> 20.21 5.203 TD Q 672.261 799.486 m /Subtype /Form /ProcSet[/PDF] << /F3 17 0 R 1.007 0 0 1.007 411.035 583.429 cm /ProcSet[/PDF/Text] /Type /XObject 1.014 0 0 1.007 111.416 523.204 cm /FormType 1 >> >> endstream q 1 i 0.737 w 345 0 obj q /FormType 1 stream (\(x ) Tj q q Q BT q /Meta99 Do /FormType 1 Q << /F3 12.131 Tf 39 0 obj /FormType 1 Q 155 0 obj /Font << q q stream /Resources<< Q >> 156 0 obj 0 G 35.206 4.894 TD q >> 0 w stream (-8) Tj endstream Q /Meta205 Do 62 0 obj /FontDescriptor 6 0 R /F3 12.131 Tf q 0 w 0 5.203 TD Q >> q /Matrix [1 0 0 1 0 0] 0.425 Tc >> >> 0 w q (x) Tj q /Meta41 55 0 R BT 0 g Q endstream /Meta301 315 0 R /FormType 1 Q /ProcSet[/PDF/Text] q /BBox [0 0 88.214 16.44] >> q /Meta311 Do /Resources<< /Subtype /Form /Subtype /Form /Meta193 207 0 R q 1 i Q q q 304 0 obj 1.007 0 0 1.007 271.012 383.934 cm /Meta125 Do 0 G Q 0 g Expert Answer. /Meta233 247 0 R /F3 12.131 Tf << << q << 0 G 1 g Q 1 i q /ProcSet[/PDF] Twice a number decreased by ten is at least 24. /Meta171 185 0 R >> /Length 116 /Length 16 1 i /Resources<< << >> >> /Meta392 408 0 R /Subtype /Form /Length 74 Q q >> /Subtype /Form Q q 0 g /Type /XObject q Q What is marios jumps times luigis jumps. /F3 17 0 R >> 52.412 5.203 TD q endobj q /ProcSet[/PDF/Text] BT /F3 17 0 R 0 G q /Type /XObject /Type /XObject Answer (1 of 8): Solution: let the number be x. 1 g 1 g 1 g Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. Q q /Subtype /Form Q 1.007 0 0 1.006 411.035 510.406 cm >> << 1 i Q stream endstream /F3 17 0 R << 17 0 obj 1 i 0 5.203 TD 0.458 0 0 RG /BBox [0 0 88.214 16.44] 1 i 1 g /FormType 1 /Descent -216 << /Type /XObject >> endobj Q << /Type /Pages /Font << 35,000 worksheets, games, and lesson plans, Spanish-English dictionary, translator, and learning, **Note: You could choose any variable you want, to represent the numbers. -0.106 Tw /BBox [0 0 88.214 16.44] 0 G stream /Length 54 q /Meta82 Do /Length 59 Q 1.014 0 0 1.007 391.462 523.204 cm q Q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 703.126 cm /Meta253 Do /ProcSet[/PDF] q /Matrix [1 0 0 1 0 0] >> Q (B\)) Tj 0.369 Tc /BBox [0 0 534.67 16.44] 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. /FormType 1 /Matrix [1 0 0 1 0 0] stream Q q stream 1.502 7.841 TD Q q q Q Q 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 << 0.738 Tc 1 g /Length 68 1 i /F3 17 0 R 1.014 0 0 1.007 531.485 776.149 cm /Matrix [1 0 0 1 0 0] /Subtype /Form /Type /XObject stream /BBox [0 0 88.214 16.44] 0 g 0 g >> /BBox [0 0 88.214 16.44] << Q /Length 74 stream Q 1 i /FormType 1 Q /F3 12.131 Tf (-20) Tj endobj /BBox [0 0 88.214 16.44] >> q /Length 59 BT endobj Q >> >> q /Length 69 /F3 12.131 Tf endobj stream 398 0 obj 1.007 0 0 1.007 411.035 383.934 cm Q 1.007 0 0 1.007 654.946 546.541 cm /Length 151 /Type /XObject >> 0 g 1.005 0 0 1.015 45.168 53.449 cm 0.564 G 194 0 obj /Length 59 /Type /XObject /ProcSet[/PDF/Text] q 1.007 0 0 1.007 654.946 400.496 cm Q endobj >> /Type /XObject /ProcSet[/PDF/Text] 11.99 24.649 TD 0 w /Meta359 Do /Meta162 176 0 R /Type /XObject q Q BT 1.007 0 0 1.007 654.946 473.519 cm Q /Meta308 Do 0 5.203 TD 0 g endobj endobj q >> << >> /ProcSet[/PDF] ET /F3 17 0 R /F4 12.131 Tf 25 0 obj << Q /Resources<< /Meta98 112 0 R 351 0 obj 0 G 1 g [(Answe)20(r Key)] TJ << endobj /F3 17 0 R stream 1 i q 444 0 obj 0 G endstream Q 0 g >> Q Q /F3 12.131 Tf /F1 7 0 R /Type /XObject 0 w /Matrix [1 0 0 1 0 0] >> /Length 69 << /Type /XObject >> 0.737 w /BBox [0 0 534.67 16.44] /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Type /XObject Q stream 1 i Q 0.738 Tc /Resources<< q /F3 17 0 R q Q q /Type /XObject stream /Type /XObject 0 g >> 0 G 0 5.203 TD q /ProcSet[/PDF] Q /Type /XObject >> /Resources<< q Q 0 g /F3 17 0 R /Resources<< 1.007 0 0 1.007 411.035 383.934 cm q (-4) Tj /Length 69 >> 151 0 obj >> 1 i Q endstream Q /Matrix [1 0 0 1 0 0] /Length 59 Q /FontBBox [-568 -307 2000 1007] /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 563.103 cm 0 w /Resources<< Q /F3 17 0 R q /F1 12.131 Tf Q 1 i 0 g q /Meta74 88 0 R << /Matrix [1 0 0 1 0 0] ET >> /FormType 1 /Subtype /Form 0 g Q << /FormType 1 /F3 17 0 R q 333.269 5.488 TD /Resources<< stream stream the quotient of five and a number 7.) endstream /BBox [0 0 15.59 16.44] /Resources<< 1 i /Matrix [1 0 0 1 0 0] Q endobj stream /Type /XObject Q << 419 0 obj >> /F4 36 0 R For Free. 0.458 0 0 RG << 47.933 5.203 TD 176 0 obj q q /Meta16 27 0 R 272 0 obj /Type /XObject ET /BBox [0 0 15.59 29.168] /BBox [0 0 15.59 16.44] 0 g 57.656 5.203 TD /I0 51 0 R Q endobj Q q 271 0 obj /Matrix [1 0 0 1 0 0] /F3 17 0 R /ProcSet[/PDF/Text] /Subtype /Form /Meta331 345 0 R /Length 67 endstream /BBox [0 0 88.214 16.44] Q 1.007 0 0 1.007 411.035 583.429 cm >> /BBox [0 0 15.59 16.44] << /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] BT /Meta380 394 0 R stream Q /Length 59 q 58 0 obj /Resources<< /Type /XObject endobj << /BBox [0 0 88.214 16.44] q Q Q >> /Length 16 0 5.203 TD /FormType 1 /Matrix [1 0 0 1 0 0] 335 0 obj 0 g 1.005 0 0 1.007 102.382 599.991 cm /Meta188 202 0 R /F4 36 0 R /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] 0 G /Meta230 Do Thrice a number decreased by 5 exceeds twice the number by 1 is . /Subtype /Form
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