twice a number decreased by 58twice a number decreased by 58

Q /Meta341 Do q /Meta307 321 0 R Q 0 g ET /Meta247 Do q /Type /XObject 36 0 obj /BBox [0 0 88.214 16.44] q (x) Tj q stream /Resources<< ET << A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. endstream /Resources<< /Matrix [1 0 0 1 0 0] 0 5.203 TD q Q >> ET stream /Font << 0 w /Meta21 32 0 R 0.458 0 0 RG /Length 16 Q >> /Meta304 318 0 R endstream /Resources<< /Length 59 Q /Meta10 21 0 R Q q /Matrix [1 0 0 1 0 0] /Length 16 /Subtype /Form /Meta252 Do << 1 g q 101 0 obj /Subtype /Form 1.005 0 0 1.007 102.382 599.991 cm /Meta279 293 0 R >> Q /Subtype /Form /Font << << 0.458 0 0 RG /ProcSet[/PDF] q << >> /ProcSet[/PDF/Text] 2.238 5.203 TD q 1 i /Subtype /Form >> /ItalicAngle 0 q endobj 71 0 obj /ProcSet[/PDF] /ProcSet[/PDF] >> q << /Resources<< Q Q Q /Type /XObject /Subtype /Form 1.005 0 0 1.007 79.798 779.913 cm /F3 17 0 R /Length 69 1 i /Matrix [1 0 0 1 0 0] 549.694 0 0 16.469 0 -0.0283 cm >> endobj >> 0.369 Tc Q /Subtype /Form 0 g /Subtype /Form /Subtype /Form (\)) Tj 0 G /BBox [0 0 30.642 16.44] 0 g Q /ProcSet[/PDF/Text] q >> endstream /F4 36 0 R endstream q 0 G /Length 70 Q q 0 g Q /BBox [0 0 88.214 35.886] /Meta322 336 0 R >> /Length 59 /Length 59 >> /FormType 1 /Meta241 Do Q q Q /ProcSet[/PDF/Text] /Subtype /Form (-8) Tj /Meta337 351 0 R Q >> /Matrix [1 0 0 1 0 0] 183 0 obj q Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. stream /Meta425 441 0 R /F3 17 0 R /Meta342 Do >> /ProcSet[/PDF/Text] q Q /Meta147 Do /F3 17 0 R (-) Tj Q >> /F3 17 0 R /Meta165 179 0 R /F3 12.131 Tf ET Q endobj /Subtype /Form 6.746 8.18 TD /FormType 1 Q /F3 12.131 Tf /F3 17 0 R q /Resources<< /ProcSet[/PDF/Text] 0.564 G stream >> endobj >> /ProcSet[/PDF/Text] /FormType 1 113 0 obj /Font << /Type /XObject ET Q q /Meta178 Do /StemH 94 q /Matrix [1 0 0 1 0 0] endstream BT /FormType 1 /Length 68 << /Type /XObject q 0 g /Matrix [1 0 0 1 0 0] ET /ProcSet[/PDF] stream /MediaBox [0 0 767.868 993.712] q Q /Resources<< /ProcSet[/PDF] /Type /XObject q /Type /XObject BT /MaxWidth 2000 /XObject << Q /Length 64 q Q /Matrix [1 0 0 1 0 0] Q Q 0.458 0 0 RG /Length 59 /Meta237 Do So we have twice of a mystery number decreased by three, and that is all going to be 31. /Matrix [1 0 0 1 0 0] << /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 271.012 849.172 cm Q 1 g << >> q endobj q /BBox [0 0 88.214 16.44] /Resources<< endobj Q 1.007 0 0 1.007 551.058 636.879 cm 0.001 Tw 1 i /Meta357 371 0 R /F4 36 0 R /Length 54 >> 0 g 0 g /Subtype /Form 1 i 0.486 Tc /Type /XObject << 1 i /ProcSet[/PDF/Text] /Length 63 q /FormType 1 stream /Resources<< ET /F3 12.131 Tf /Font << stream 722.699 872.509 l >> 0.68 Tc q /Meta185 199 0 R 0.51 Tc 1.005 0 0 1.007 102.382 563.103 cm /Meta8 19 0 R 0 w Q Find the number 1 See answer Advertisement 1.005 0 0 1.007 102.382 490.08 cm q q /Type /XObject /ProcSet[/PDF] 0 G q You could call them. Q 0.458 0 0 RG >> ET /Matrix [1 0 0 1 0 0] q /FormType 1 << /Matrix [1 0 0 1 0 0] Q q q Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. 1 g q >> /Font << 0 g /F3 17 0 R Q 0 g /Meta317 331 0 R >> << >> /Font << 1 g >> q /FormType 1 >> /Font << q << /Matrix [1 0 0 1 0 0] 0.369 Tc q endstream Q /Resources<< /Font << /BBox [0 0 88.214 16.44] >> /Type /XObject >> /Font << 1 i /F3 17 0 R /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 45.168 730.228 cm /Font << Q 1 i q q << q Q 1.007 0 0 1.007 271.012 277.035 cm /Resources<< >> 0 G Q /Type /XObject endobj 86 0 obj Q q 1.005 0 0 1.007 102.382 726.464 cm q /F3 17 0 R 1 i 273 0 obj 0.786 Tc /Meta84 Do >> 237 0 obj q q Q /Subtype /Form /F1 12.131 Tf /Length 59 /Resources<< 51 0 obj /Type /XObject /Type /XObject /FormType 1 q 0.51 Tc >> /Resources<< 0 G q 0 5.203 TD Q 0 g BT /Subtype /Form /Meta63 Do /Meta370 Do q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 411.035 383.934 cm /Meta209 223 0 R >> ET q endobj q /Length 19882 endobj ET /Matrix [1 0 0 1 0 0] /FormType 1 1 g endstream << /Meta168 Do Q q (2) Tj endobj q BT endobj q Q /Meta172 Do 383 0 obj Q 0 g 0 G endobj q /Subtype /Form Q q /Matrix [1 0 0 1 0 0] q ET /Subtype /Form 0.458 0 0 RG q /Matrix [1 0 0 1 0 0] /Meta123 Do 0 G q q q endobj q /Parent 1 0 R q endstream 0 G /Resources<< /ProcSet[/PDF] stream BT ET << >> 0.737 w /Subtype /Form -y. q q endstream BT q /Meta352 366 0 R /Type /XObject /F1 7 0 R q q 0 G endobj 0.285 Tc 0 w 1.007 0 0 1.007 551.058 330.484 cm Q (1) Tj q endstream /Length 16 /Meta15 Do /Resources<< /BBox [0 0 673.937 16.44] /F3 17 0 R (+) Tj endobj q ET q Q q /Length 69 1 i 1 i << /ProcSet[/PDF] q /Meta78 92 0 R If LtitnS6S . >> /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] Q << endstream /Matrix [1 0 0 1 0 0] Q q By the . >> 0 G 118.317 5.203 TD >> endstream 0 G What word phrase can you use to represent 5x + 2? endobj /Font << >> << /Meta26 39 0 R q stream (+) Tj 1 i 0.458 0 0 RG Q /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] /F1 7 0 R This site is using cookies under cookie policy . /Descent -216 0 g /CapHeight 476 /FontName /TimesNewRomanPSMT Q /Resources<< 0 g /Type /XObject 0.564 G BT >> Q endstream /Meta221 235 0 R endstream /F3 12.131 Tf Q q 1.007 0 0 1.006 411.035 510.406 cm 424 0 obj q /Meta217 231 0 R /Font << (-9) Tj >> /Meta424 440 0 R /Resources<< stream 6.746 5.203 TD << /ProcSet[/PDF/Text] Q q 1 i q (5) Tj q 1.007 0 0 1.007 130.989 277.035 cm << >> /Resources<< 1.014 0 0 1.007 391.462 383.934 cm q Q /FormType 1 >> /Length 16 /Meta34 47 0 R /F3 12.131 Tf Q /FormType 1 /BBox [0 0 30.642 16.44] 11.99 24.649 TD 9.723 5.336 TD /Meta205 Do >> << /Matrix [1 0 0 1 0 0] /Resources<< q q -0.16 Tw stream endobj >> 0 w /Type /XObject << q /ProcSet[/PDF] /Type /XObject /Type /XObject 0 g 0.737 w >> 20.21 5.203 TD endstream Q /BBox [0 0 88.214 16.44] /ProcSet[/PDF] 23.216 5.203 TD /F3 12.131 Tf /FormType 1 0 g >> 0 w 0 g Q 0 g Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . /Meta384 Do /Font << /Subtype /Form /BBox [0 0 639.552 16.44] /BBox [0 0 88.214 16.44] Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . Q 0.564 G 1.007 0 0 1.007 654.946 799.486 cm /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] >> 1 i Double or twice a number means 2x, and triple or thrice a number means 3x. Ten divided by a number 5. /Meta334 348 0 R /Length 118 ET << /BBox [0 0 88.214 16.44] 0 g q endstream /XObject << Q q /Resources<< >> 1.007 0 0 1.006 551.058 763.351 cm /Type /Page /Resources<< q /Length 118 /F3 12.131 Tf /Length 2252 0 w /Meta307 Do q /Subtype /Form endobj /Subtype /Form 0.463 Tc /BBox [0 0 30.642 16.44] 0.564 G 0.134 Tc stream 1 i /Type /XObject endstream q /Length 69 Q /Matrix [1 0 0 1 0 0] /Type /XObject Q stream /Length 16 stream << /BBox [0 0 30.642 16.44] Q /ProcSet[/PDF] >> >> Q stream 1 g 0 g 112 0 obj 0.737 w /Meta327 341 0 R Q /Matrix [1 0 0 1 0 0] >> /FormType 1 /FormType 1 /BBox [0 0 673.937 68.796] Q Q << /Subtype /Form /Font << /Subtype /Form 0 g 205.199 4.894 TD /Matrix [1 0 0 1 0 0] 0.737 w /Meta293 Do BT /Font << ET Q Q /Length 16 /F3 17 0 R /ProcSet[/PDF/Text] 0.458 0 0 RG /Font << /Type /XObject Q /I0 Do /BBox [0 0 88.214 16.44] /Meta368 382 0 R stream /Resources<< q /Length 68 BT 1.007 0 0 1.007 411.035 383.934 cm endstream 20.21 5.336 TD /FormType 1 /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] q endobj q Q q 0 g >> /Subtype /Form Q 15.731 5.336 TD 0.486 Tc Q /Meta0 5 0 R Q q 0 5.203 TD stream 149 0 obj Q 1.014 0 0 1.007 251.439 277.035 cm /Type /XObject /FormType 1 0.564 G /Length 69 ET /Meta108 Do /FormType 1 1.007 0 0 1.007 271.012 636.879 cm << >> 0 5.203 TD 0.68 Tc Q 1 i /FormType 1 /ProcSet[/PDF/Text] Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. -0.03 Tw 0 g /Type /XObject /Length 16 (6\)) Tj Q q 0 G 238 0 obj /F3 17 0 R (\)) Tj (A\)) Tj Q /Subtype /Form 0 G 1 i /Meta416 Do >> ET /Meta142 Do 312 0 obj /Length 16 q q /Type /XObject Q >> q /Subtype /Form /Subtype /Form q 52.412 5.203 TD >> q /Resources<< /Meta4 13 0 R /Resources<< ET 1 g /Subtype /Form 0.564 G /Meta275 Do /Subtype /Form endstream endstream 0 g endstream (4\)) Tj >> 1.007 0 0 1.007 411.035 330.484 cm /BBox [0 0 88.214 16.44] (x) Tj 1.014 0 0 1.007 111.416 523.204 cm /Length 65 0 G View the full answer. /Matrix [1 0 0 1 0 0] q q ET /Meta78 Do /Resources<< 33 0 obj /Length 54 endobj Q q BT /Matrix [1 0 0 1 0 0] 0.68 Tc >> /Subtype /Form BT 1.007 0 0 1.007 67.753 293.596 cm >> /Resources<< /Meta161 Do ET 0.564 G >> /Resources<< >> /ProcSet[/PDF/Text] /ProcSet[/PDF] q Q /FormType 1 endstream << stream Q Q /Meta370 384 0 R >> /Type /XObject /Meta283 297 0 R >> 1.007 0 0 1.007 551.058 636.879 cm /F3 12.131 Tf /Meta238 Do << Q /ProcSet[/PDF/Text] /Font << /Meta382 Do stream endobj >> /Meta293 307 0 R << 0 w ET /F3 12.131 Tf 290 0 obj q /Matrix [1 0 0 1 0 0] /Meta162 Do /F1 12.131 Tf << >> /XObject << /Meta158 Do Q /Length 16 /Type /XObject /Length 69 Q 0.737 w >> q /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 35.886] 32.201 20.154 l >> 0.737 w /FormType 1 /BaseFont /PalatinoLinotype-Roman /Matrix [1 0 0 1 0 0] endstream endobj /Meta238 252 0 R /Matrix [1 0 0 1 0 0] q Q /ProcSet[/PDF/Text] /Meta244 Do Q endobj >> /Resources<< 0 g /Meta164 Do 0 g /Font << >> 1.005 0 0 1.007 79.798 813.037 cm /Matrix [1 0 0 1 0 0] 0 G endstream (\)) Tj 1.007 0 0 1.007 130.989 636.879 cm 1 g << /Length 245 0.564 G /Type /XObject 0 w 255 0 obj /Meta36 Do /Meta208 Do endobj >> /Meta86 Do /FormType 1 1.007 0 0 1.007 271.012 277.035 cm >> 15.731 5.336 TD >> /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] /F3 17 0 R /FormType 1 /Subtype /Form 0.737 w 0 g Q /F3 12.131 Tf q q q stream /Length 59 q Q /Subtype /Form q 0 w 122 0 obj q /Length 54 /F3 12.131 Tf 0.458 0 0 RG /Resources<< /Type /XObject Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. Q /F1 7 0 R /Meta428 Do 110 0 obj 1 i 309 0 obj stream Q /Type /XObject Q /BBox [0 0 88.214 16.44] 0 g << q >> 0 G 169 0 obj /BBox [0 0 15.59 16.44] 1 i /Resources<< /Length 16 stream q 0 G /FormType 1 (-) Tj 0 w /Subtype /Form /FormType 1 /BBox [0 0 23.896 16.44] /Matrix [1 0 0 1 0 0] endobj 1.007 0 0 1.007 271.012 330.484 cm 1.007 0 0 1.006 411.035 437.384 cm ET /Meta81 95 0 R endobj Q /BBox [0 0 88.214 16.44] 44 0 obj /BBox [0 0 88.214 35.886] /Meta260 Do endstream /Resources<< 0 G /Matrix [1 0 0 1 0 0] q /Meta290 Do 0 g /F1 12.131 Tf /Length 16 1.007 0 0 1.007 271.012 636.879 cm >> 0 G q S ET >> Q 357 0 obj 0 5.203 TD endobj 1 i 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. /Type /XObject >> endobj /Subtype /Form 3.742 5.203 TD endstream q /Subtype /Form /Meta421 437 0 R 1 g /Meta37 50 0 R ( \() Tj 0 G /ProcSet[/PDF] Q >> /Length 80 1 g 1 g Q 164 0 obj 0 g endobj q q /Font << /ProcSet[/PDF/Text] 0 g Q 24 0 obj /Type /XObject q q >> Q /Type /XObject /Meta235 Do Q /Resources<< endstream You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. /Resources<< /FormType 1 202 0 obj stream >> If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. Q 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. 0 G /F3 12.131 Tf << Q Calculate a 15% decrease from any number. >> endobj Q 0.786 Tc q Q q /ProcSet[/PDF] Q 298 0 obj endstream /FormType 1 /BBox [0 0 88.214 16.44] /Subtype /Form 6.746 24.649 TD 0 g /Font << q 45 0 obj q /F4 12.131 Tf 0 w /Matrix [1 0 0 1 0 0] BT /Meta324 Do Q [( and )16(a nu)26(mbe)18(r)] TJ 0.737 w 1 i endobj q q endstream /Type /XObject /Length 16 Q q /Meta107 Do /Matrix [1 0 0 1 0 0] q [( times )15(a numb)22(er and )] TJ << /Subtype /Form stream /Meta79 93 0 R 1 i Q endstream endstream stream 1.502 5.203 TD 1 i /ProcSet[/PDF/Text] 1 i q /ProcSet[/PDF/Text] 1 g /XHeight 447 /Resources<< Q 98.843 5.203 TD endobj /Meta138 Do 1 i /Subtype /Form 1.014 0 0 1.007 391.462 277.035 cm Q /Matrix [1 0 0 1 0 0] 722.699 546.541 l /Count 2 q /ProcSet[/PDF/Text] /Length 70 endstream /Matrix [1 0 0 1 0 0] /AvgWidth 459 1.014 0 0 1.006 531.485 836.374 cm /Length 73 0 w << q /Resources<< /F3 12.131 Tf Q /FormType 1 BT /Length 69 (D\)) Tj /FormType 1 Q 0.68 Tc q Q /Resources<< (58) Tj endobj q /Type /XObject Q endstream 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 /Font << /Length 59 A number divided by six is eight: (k / 6) = 8. q Q 0 w endstream 1.007 0 0 1.007 271.012 636.879 cm /Subtype /Form q Q BT q /Resources<< >> /MaxWidth 1397 q /Length 59 /Type /XObject >> 1.014 0 0 1.007 251.439 450.181 cm stream >> /Length 69 /BBox [0 0 534.67 16.44] /Meta175 189 0 R /Type /XObject 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . >> stream q /Matrix [1 0 0 1 0 0] /FormType 1 1.007 0 0 1.007 271.012 450.181 cm /Type /XObject /Type /XObject (-4) Tj 1.005 0 0 1.007 102.382 653.441 cm 1.014 0 0 1.007 251.439 776.149 cm /BBox [0 0 534.67 16.44] endstream >> Q 1 i Q /Font << /Resources<< endobj q /F1 7 0 R /Subtype /Form Q /Length 59 300 0 obj /BBox [0 0 534.67 16.44] /FormType 1 1.007 0 0 1.007 45.168 779.913 cm Q 0 g >> >> 0.564 G q ( \() Tj 0.737 w << (-11) Tj >> 0 g q Q >> Q 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 1.007 0 0 1.007 271.012 776.149 cm 1.007 0 0 1.007 130.989 523.204 cm /Subtype /Form 0 G /Subtype /Form /Meta259 Do endobj Q /Matrix [1 0 0 1 0 0] endobj endobj How many points did Kobe score in the season? 0 g 0 g Q q >> q Q << Let x be a number. /FormType 1 /Subtype /Form 1 i q /Length 59 q endstream Q endobj Q >> q >> /ProcSet[/PDF/Text] q >> stream /Matrix [1 0 0 1 0 0] endstream 1 g q /ProcSet[/PDF] q /Length 69 /FormType 1 0 5.203 TD BT 280 0 obj 0 g stream endstream Q [(The )-19(quotient of )] TJ /FormType 1 /ProcSet[/PDF/Text] /Meta1 Do /Meta46 60 0 R /Subtype /Form ET /Resources<< /ProcSet[/PDF] Decreased by another number means subtract. 160 0 obj << Q Q /BBox [0 0 88.214 35.886] endobj /FormType 1 /Resources<< q << 0.524 Tc (\)]) Tj q /Descent -299 Q /Encoding /WinAnsiEncoding /F3 17 0 R [(F)-22(ive)] TJ /F3 12.131 Tf c Site 5 is not included in this number. >> /Length 59 >> /Type /XObject /F3 17 0 R endobj /FormType 1 >> stream /BBox [0 0 15.59 16.44] Q 0 g /F1 7 0 R /Matrix [1 0 0 1 0 0] [(The )-16(s)15(um )-14(of )] TJ Q q BT 1.007 0 0 1.007 271.012 277.035 cm >> /F3 12.131 Tf >> /Length 69 Q 1 i /Subtype /Form 188 0 obj Q /Resources<< (\)) Tj 1 g 314 0 obj 1.502 5.203 TD << BT 72 0 obj endstream /Meta143 Do (-) Tj 0 g /Length 16 /Meta250 Do 55 0 obj stream /Meta189 Do /Length 206 /Resources<< 0.486 Tc /Length 16 1.502 7.841 TD /Resources<< /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] 0 g >> 0.486 Tc (5) Tj >> /Type /XObject ET >> q 1 i /Type /XObject 0.68 Tc /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] ET endstream /Matrix [1 0 0 1 0 0] /FormType 1 Q /F3 12.131 Tf /Type /XObject stream 0 G /Font << 1.502 24.339 TD Q Thrice a number decreased by 5 is 3x - 5. /Font << << >> Q Q 0 g Q /Subtype /Form >> q /F3 12.131 Tf /ProcSet[/PDF/Text] /Type /XObject 1 g >> 1.014 0 0 1.007 391.462 636.879 cm /Subtype /Form Q 6.746 5.203 TD 187 0 obj >> q endstream 257 0 obj endobj 1 i 307 0 obj /FormType 1 ET ET >> q /Type /XObject q /I0 51 0 R 1 i 1.007 0 0 1.006 551.058 763.351 cm /Meta215 Do (\(x ) Tj endobj << endstream >> q >> 1.014 0 0 1.007 531.485 330.484 cm /Length 118 << [(1)-25(0\))] TJ /Length 59 Q /CreationDate (D:20140515121932-04'00') /Meta366 380 0 R BT 0 g 0.737 w 1 g /Type /XObject /Type /XObject q (v) 5 subtracted from thrice a number is 16. 1.005 0 0 1.007 102.382 799.486 cm endstream /Subtype /Form /Length 69 /Meta274 288 0 R /Meta321 335 0 R q ET endstream BT /Font << /Matrix [1 0 0 1 0 0] /Subtype /Form q q /Font << 0 w /Length 58 q 0.458 0 0 RG 0 g 20.21 5.203 TD Q /Type /XObject /BBox [0 0 88.214 16.44] /ProcSet[/PDF] >> 19.474 20.154 l stream /ProcSet[/PDF/Text] q /Resources<< ET Q /ProcSet[/PDF/Text] q q So let's go ahead and identify a v /F3 17 0 R /Subtype /Form /Meta369 383 0 R /Font << 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. /F3 12.131 Tf ET /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.006 130.989 690.329 cm >> /FormType 1 /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Type /XObject /FormType 1 /Type /XObject 1 i q /Type /XObject q >> << q 254 0 obj Q /Resources<< >> /Meta240 254 0 R 1.007 0 0 1.007 551.058 703.126 cm Thrice of a number = 3x. stream q 0 w Q Q /BBox [0 0 88.214 16.44] 30.699 5.203 TD (iii) 25 exceeds a number by 7. Q Q /Meta85 Do /Resources<< 0.51 Tc /Meta251 Do /Meta105 119 0 R q 0.737 w endobj Q 1 i /FormType 1 >> 38.948 5.203 TD Q /Type /XObject stream /FormType 1 Thirthy is equal to twice a number decreased by four = solve and check the equation? /Meta102 116 0 R /FormType 1 /Resources<< /FormType 1 >> /Length 69 stream 25.454 5.203 TD ET /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 330.484 cm /Subtype /Form ET << stream 0.271 Tc /Resources<< /Meta424 Do << 0.737 w 1.502 24.649 TD /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Subtype /Form /BBox [0 0 88.214 16.44] stream >> /FormType 1 368 0 obj /Length 69 /Matrix [1 0 0 1 0 0] << 190 0 obj Q /Meta16 Do BT /Meta277 291 0 R /Font << stream Hence, the number is 6. 78 0 obj /Type /XObject 1.007 0 0 1.007 271.012 849.172 cm /BBox [0 0 30.642 16.44] >> 41 0 obj (+) Tj ET >> 0 G 24.718 8.18 TD q 0 5.203 TD endstream 20.21 5.203 TD /Meta110 124 0 R >> Q 0 G stream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Resources<< 1 g q q Q q /Subtype /Form >> stream /Subtype /Form endstream /Resources<< Q /FormType 1 endstream 0.737 w q /Length 79 /FormType 1 /F1 12.131 Tf >> >> BT 0 5.203 TD 1.014 0 0 1.007 111.416 383.934 cm 0.838 Tc endobj 1 g /Type /XObject /Length 69 BT >> q Q endobj /Meta33 46 0 R 37 0 obj >> 0.458 0 0 RG Q 1 g 152 0 obj q 1.005 0 0 1.007 102.382 473.519 cm /Type /XObject /Length 118 /Meta86 100 0 R >> /Matrix [1 0 0 1 0 0] >> q 0 G (D\)) Tj /Meta292 Do /Type /XObject /Meta362 376 0 R 0 g 141 0 obj Q Q >> endstream /Type /XObject endstream /FormType 1 /FormType 1 endobj 0.271 Tc /Resources<< >> /Meta299 Do /Subtype /Form /Resources<< /Meta335 349 0 R Q /F3 12.131 Tf /ProcSet[/PDF] q << q /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta7 Do stream << endstream 1 i /Length 69 /Resources<< endobj (x) Tj >> 16.469 5.336 TD /Font << /ProcSet[/PDF/Text] 23.952 4.894 TD /Length 58 /Matrix [1 0 0 1 0 0] endstream BT endstream 0 G stream 41.186 5.203 TD q /Font << 243 0 obj /Meta372 386 0 R q endobj /Matrix [1 0 0 1 0 0] /FormType 1 /Length 58 endstream /Matrix [1 0 0 1 0 0] endobj /Subtype /Form /Type /XObject 1 i 0.564 G 0.369 Tc 22.478 5.336 TD Q stream /Resources<< 1.014 0 0 1.007 531.485 636.879 cm q /Type /XObject Q q 1 i 1.502 5.203 TD Q q /Type /XObject >> Q /Meta143 157 0 R /Meta263 277 0 R /Meta398 Do Q /Flags 32 /F3 12.131 Tf /Font << /Meta389 405 0 R 0.297 Tc endstream >> /FormType 1 ET /F3 17 0 R Q 154 0 obj q stream Q q q q >> stream /Length 16 q << 1.005 0 0 1.007 102.382 872.509 cm /I0 51 0 R /Matrix [1 0 0 1 0 0] stream 1 i 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 endobj 1.007 0 0 1.006 411.035 690.329 cm /F3 12.131 Tf Q q Q q >> Q /FormType 1 /Length 107 stream /BBox [0 0 17.177 16.44] /I0 51 0 R /Length 65 endobj Q /Meta83 Do 1.007 0 0 1.007 411.035 277.035 cm /BBox [0 0 639.552 16.44] ET endstream 20.21 5.203 TD /F3 17 0 R 251 0 obj /StemV 77 endobj endobj >> Q endstream /XObject << endstream stream /Meta183 197 0 R >> endstream /F3 17 0 R 0 g 0 20.154 m /FormType 1 0.564 G stream /ProcSet[/PDF/Text] /Length 59 0 w 0 G /BBox [0 0 673.937 15.562] /Type /XObject /FormType 1 >> Q /Descent -299 << 1.007 0 0 1.007 411.035 583.429 cm q /Resources<< 0 w /Type /XObject q /F1 12.131 Tf 0 g ( x) Tj Q /BBox [0 0 88.214 35.886] /Length 69 1 i /ProcSet[/PDF] /Type /XObject (1\)) Tj /Type /XObject q Q /Resources<< endstream /F2 11 0 R /Meta254 Do >> endstream /F3 12.131 Tf q /Subtype /Form endobj >> 333.269 5.488 TD << << 0 G /Type /XObject 0 w Q >> /Type /XObject q 1 i >> /Type /XObject endstream /Meta390 406 0 R 437 0 obj /FormType 1 Q Q 136 0 obj /Subtype /Form 1 g stream /F3 12.131 Tf 0 g stream q q >> /Subtype /Form 1.014 0 0 1.007 251.439 583.429 cm /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 111.416 636.879 cm /Meta161 175 0 R /Font << Q /BBox [0 0 88.214 16.44] 156 0 obj Select the correct mathematical statement for the following equation. (D\)) Tj /Type /XObject Q >> /Meta60 Do 96 0 obj 0.737 w q 0 G >> /Meta148 Do 0 w /Type /XObject BT /F3 17 0 R Q /Meta80 Do /ProcSet[/PDF] 0 g >> /BBox [0 0 88.214 16.44] 181 0 obj 355 0 obj q 0 g /Resources<< >> /FormType 1 endobj Q stream 358 0 obj q << /Matrix [1 0 0 1 0 0] q /Meta246 260 0 R 9.723 5.336 TD << /Meta219 Do stream /Font << /Length 54 >> 0.458 0 0 RG BT 1.007 0 0 1.007 271.012 383.934 cm 0 w /CapHeight 662 /BBox [0 0 88.214 16.44] ET q q /Subtype /Form /FormType 1 >> ET Twice the difference of a number and three totals twelve 8. >> q Q /Length 294 q 0 g Q Q q /FormType 1 /Meta162 176 0 R 1.007 0 0 1.007 67.753 347.046 cm
Kevin Martin Candlebox Wife, Zakrsle Zajace Na Predaj, Unity Multithreading Physics, Chichones En La Frente Que No Desaparecen, 1934 Ford 5 Window Coupe Steel Body, Articles T