>> >> endobj /Length 16 1.007 0 0 1.007 271.012 849.172 cm ET 0.369 Tc /Matrix [1 0 0 1 0 0] BT 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q /Meta211 225 0 R q /Meta252 266 0 R 0.458 0 0 RG endobj endobj -0.486 Tw 1 i endobj /Type /XObject >> q /FormType 1 0.227 Tc 1 g endstream Answer by Mathtut (3670) ( Show Source ): >> >> /Resources<< 21 0 obj >> 1 g 0.458 0 0 RG >> 1.007 0 0 1.007 45.168 829.599 cm Q /Type /XObject /Matrix [1 0 0 1 0 0] q 0.369 Tc stream 1.007 0 0 1.007 271.012 776.149 cm /ProcSet[/PDF] q << /BBox [0 0 88.214 35.886] q endstream /Resources<< endobj ET /FormType 1 /Meta36 49 0 R 672.261 400.496 m - 9737014. /Meta232 Do /Resources<< 0.737 w /Meta32 Do /F3 17 0 R stream q /ProcSet[/PDF] Q /Resources<< 0 g /Resources<< >> Q << endstream q stream 0 G /FormType 1 >> BT Q q 0 G 0.241 Tc 0 w /Meta396 Do stream endstream /XHeight 476 0 g Q q 1 i 0 g 0 w >> endobj Q Q Q q In other terms, 52-nx The problem is asking that you subtract twice a number from 52. 1.014 0 0 1.006 391.462 836.374 cm /Length 69 /BBox [0 0 88.214 16.44] 0 G /Font << stream 0 g a.) /Type /XObject >> Let the 2nd number be y. /Meta52 Do /Meta371 385 0 R 0 g endstream >> /Type /XObject Q On the way, we sang songs for 20 minutes which is 10% of the time we were on the road. /Meta245 Do Q /ProcSet[/PDF] endobj 398 0 obj endobj 0.425 Tc /F3 12.131 Tf << q /Subtype /Form 1 i q /Type /XObject /Subtype /Form 1 i endstream 0.458 0 0 RG 0.524 Tc /Subtype /Form /Subtype /Form endstream 0 G Q 0 g 1 g endstream /Matrix [1 0 0 1 0 0] /F3 17 0 R >> q /Matrix [1 0 0 1 0 0] /Resources<< 1 i q /F4 36 0 R stream /Meta208 222 0 R ET q /Matrix [1 0 0 1 0 0] 0 5.203 TD (5) Tj >> endstream ET endobj /ProcSet[/PDF] /Type /XObject q /Meta366 Do q /F3 12.131 Tf /ProcSet[/PDF/Text] >> 0.737 w /Meta400 416 0 R >> stream Q /Type /XObject 0 g >> 1.005 0 0 1.007 102.382 799.486 cm 549.694 0 0 16.469 0 -0.0283 cm 0.458 0 0 RG 0 g Q /FormType 1 1 i /Resources<< /Meta376 Do /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /F4 12.131 Tf /Matrix [1 0 0 1 0 0] /FormType 1 /F3 12.131 Tf q /Type /XObject 0 g q endobj ET /FormType 1 Q /F3 17 0 R /BBox [0 0 15.59 29.168] /FormType 1 >> (D\)) Tj 0 g endstream endobj 0.737 w BT 1 g /Matrix [1 0 0 1 0 0] Q /F3 12.131 Tf /Meta288 Do /ProcSet[/PDF] q 0 w q /FormType 1 /F3 17 0 R (B\)) Tj /Meta53 Do /Meta251 265 0 R 1.007 0 0 1.007 67.753 546.541 cm 0 G 0 g 20.21 5.336 TD stream Q /Type /XObject endstream a Question /FormType 1 1 i >> Q endobj 0.786 Tc 0 g Q 1.007 0 0 1.006 130.989 437.384 cm stream endobj /BBox [0 0 88.214 16.44] /Subtype /Form /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.006 411.035 437.384 cm >> /Meta408 424 0 R BT endstream 1.008 0 0 1.007 654.946 293.596 cm /Subtype /Form >> 1 i >> BT q 1 i >> /Type /XObject /BBox [0 0 15.59 29.168] >> >> BT q Q endobj S Q /Type /XObject ET stream Q (-9) Tj endstream 0 g >> q /Meta342 356 0 R >> >> BT 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. This s problem could be, interpreted either way. /Meta155 Do q >> >> /F3 12.131 Tf Q 0 5.203 TD ET /Matrix [1 0 0 1 0 0] /Type /XObject /Font << >> 0 g /ProcSet[/PDF/Text] /Subtype /Form Q /Matrix [1 0 0 1 0 0] /Font << /Length 69 endobj stream >> /F3 12.131 Tf /ProcSet[/PDF] ET /FormType 1 174 0 obj /BBox [0 0 15.59 16.44] >> /Meta165 179 0 R q >> Q twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. /BBox [0 0 673.937 14.853] >> /Meta193 207 0 R 1 i 1 i Q stream 2.238 5.203 TD /ProcSet[/PDF/Text] >> /Resources<< stream >> q ET 1 i /F3 12.131 Tf q Q /Type /XObject >> /Type /XObject stream q /Meta321 Do >> /ProcSet[/PDF/Text] /Meta65 Do ET endstream 0.737 w q << /FormType 1 endstream /Meta95 Do q endobj Q S endstream /ProcSet[/PDF/Text] endobj 19.474 20.154 l q >> stream Q >> endstream /Type /XObject 0.564 G /Resources<< BT /I0 Do stream 1.007 0 0 1.007 130.989 636.879 cm Q q q /BBox [0 0 88.214 16.44] Q /Type /XObject endobj /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] >> endstream >> 0 G /Type /XObject /Meta47 61 0 R q /Length 69 << /Meta313 327 0 R 0 g Q /Subtype /Form /BBox [0 0 549.552 16.44] endstream 236 0 obj q Q Q 0 g >> q /Length 69 /ProcSet[/PDF] >> Q >> 1.007 0 0 1.007 411.035 277.035 cm 3 0 obj /Meta9 Do /Meta359 Do << 1 i /ProcSet[/PDF] q /Resources<< Q endstream /FormType 1 >> stream /F4 36 0 R /F3 12.131 Tf /Font << >> (x) Tj /ProcSet[/PDF] endobj 383 0 obj >> /ProcSet[/PDF/Text] Twice a number decreased by 8 gives 58. q /F3 12.131 Tf 1.005 0 0 1.007 102.382 473.519 cm Q /Width 734 /BBox [0 0 17.177 16.44] /BBox [0 0 549.552 16.44] q S q /ProcSet[/PDF/Text] 0.458 0 0 RG /ProcSet[/PDF/Text] 17.234 5.203 TD q /Resources<< /Matrix [1 0 0 1 0 0] /Subtype /Form 0 g endobj Q /BBox [0 0 88.214 16.44] endstream /Subtype /Form 219 0 obj /Meta144 158 0 R 16.469 5.336 TD Q /Resources<< >> 0 g /Font << 164 0 obj to represent the numbers. 1.007 0 0 1.007 271.012 277.035 cm /Matrix [1 0 0 1 0 0] >> 1 i 1 g q << ET /Meta216 Do /Font << /FormType 1 /FormType 1 Q q stream 1.005 0 0 1.007 102.382 653.441 cm 233 0 obj << 0.178 Tc /Length 77 << 1.007 0 0 1.007 45.168 829.599 cm q /BBox [0 0 15.59 16.44] /FormType 1 q /F1 12.131 Tf Q /Meta328 342 0 R /FormType 1 /Type /Pages BT /Meta360 Do /Length 74 q 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 q /Type /XObject /Meta300 Do Then the following equation can represent this problem: 17 + x = 68 We can subtract 17 from both sides of the equation to find the value of x. /ColorSpace [/Indexed /DeviceGray 1 ] 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.458 0 0 RG /Matrix [1 0 0 1 0 0] /Length 118 1 i /Matrix [1 0 0 1 0 0] Thrice of a number = 3x. >> << 1.502 24.649 TD 1 i 0 5.203 TD BT << Twice a number, decreased by 58 is less than 112 - 18274082. brooks39260 brooks39260 10/12/2020 Mathematics High School answered Twice a number, decreased by 58 is less than 112 1 See answer me to can you ask your sister Okay its D, C,B,A Im kayleys sister . 1 g endstream /Length 69 0 g /F4 36 0 R endstream 0 G 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. Was this answer helpful? 378 0 obj /Length 60 0.564 G /Matrix [1 0 0 1 0 0] endobj ET /FormType 1 stream /ProcSet[/PDF] q 0 G endstream << /Resources<< endstream /ProcSet[/PDF/Text] /ProcSet[/PDF] q Now that you know the meaning of the key words you can read the problem differently. ET /ProcSet[/PDF] /Type /XObject endstream 0.458 0 0 RG endstream endobj Q q q 0.68 Tc Q 0.458 0 0 RG /FormType 1 stream Q /F3 12.131 Tf /FormType 1 << 1 i q 43.426 5.203 TD 0 g /Matrix [1 0 0 1 0 0] 1 i endstream /Meta18 29 0 R >> /Meta117 131 0 R q 1 i /Meta235 Do /F3 12.131 Tf 1.014 0 0 1.007 391.462 383.934 cm Q endobj q /Length 79 0 G >> ET endobj /FormType 1 /FormType 1 /Type /XObject >> /F3 17 0 R endobj /Type /XObject /Meta80 Do endstream /Length 70 stream 0 g /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q 189 0 obj >> 1 g >> /ProcSet[/PDF/Text] q /Font << /Matrix [1 0 0 1 0 0] /Length 69 0 g /Resources<< 179 0 obj /Matrix [1 0 0 1 0 0] q /Meta263 277 0 R >> /FormType 1 >> endstream /FormType 1 endobj stream /Descent -299 1.005 0 0 1.007 102.382 799.486 cm /FormType 1 /Resources<< 0 g /Length 80 Q 1.014 0 0 1.007 111.416 277.035 cm q /F3 17 0 R /ProcSet[/PDF/Text] /Subtype /Form >> /BBox [0 0 88.214 16.44] q q /ProcSet[/PDF] /F3 17 0 R endobj /FormType 1 /ProcSet[/PDF/Text] << /Descent -277 >> >> stream 25 0 obj >> Q stream /Length 16 /Matrix [1 0 0 1 0 0] q q << /Subtype /Form q endstream Q /Resources<< Q ET q >> 1.007 0 0 1.007 130.989 583.429 cm /Resources<< endobj >> /Meta136 Do q /Matrix [1 0 0 1 0 0] 0 g Q /Matrix [1 0 0 1 0 0] ET /F3 17 0 R /Meta31 44 0 R Q q /Resources<< q 19.474 5.203 TD 48 0 obj >> stream q Q endobj /BBox [0 0 88.214 16.44] /Font << q /F3 12.131 Tf /Type /XObject ET Q stream >> /Resources<< /ProcSet[/PDF/Text] /Meta192 Do >> q /Resources<< 0 G q >> /F3 12.131 Tf endstream /Meta29 42 0 R /F1 7 0 R /Meta166 Do /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 4.894 TD q /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 391.462 690.329 cm q >> (B\)) Tj 0 g 1 i q endstream >> (+) Tj 1.007 0 0 1.007 551.058 277.035 cm What is marios jumps times luigis jumps. q q /Type /XObject >> /ProcSet[/PDF/Text] /Length 73 /Kids [ /F1 7 0 R 0.737 w /ProcSet[/PDF] q Q endstream 0 g Calculate a 15% decrease from any number. /Font << 0.369 Tc /Type /XObject q >> endstream 1.007 0 0 1.007 130.989 523.204 cm Q 0 g Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] We are asked to find the number, so, we could assign the number as "x". endobj q >> q /BBox [0 0 30.642 16.44] q 1.007 0 0 1.007 551.058 636.879 cm >> /Resources<< q /ProcSet[/PDF/Text] >> 0 g ET 311 0 obj 0.458 0 0 RG 1 i 1.014 0 0 1.007 531.485 523.204 cm 13.493 5.336 TD /Meta115 129 0 R /Meta136 150 0 R >> q 0 g /BBox [0 0 88.214 16.44] Q 0 G Q Q 0.737 w /Type /XObject /Subtype /Form (D\)) Tj Q 1 i /Meta229 Do 32.201 5.203 TD 1.007 0 0 1.007 45.168 713.666 cm BT /Resources<< Solution: Let the number be x. /F3 17 0 R Q q -0.029 Tw Q /Meta356 Do /BBox [0 0 88.214 16.44] >> 0.458 0 0 RG 440 0 obj 1 i 417 0 obj ET /Resources<< >> /Meta86 100 0 R /F3 17 0 R Q stream << >> q /BBox [0 0 88.214 35.886] 1 i stream endstream 1 i /BBox [0 0 30.642 16.44] BT (x) Tj endobj stream q 0 G q /Meta35 48 0 R /F3 17 0 R /StemV 77 1.005 0 0 1.007 102.382 670.003 cm /ProcSet[/PDF/Text] q /FormType 1 << /Info 3 0 R /Subtype /Form w/Honors. /Subtype /Form << /ProcSet[/PDF] /BBox [0 0 17.177 16.44] Q /Font << Twice a number when decreased by 7 gives 45. /Meta415 431 0 R >> 0 G /Meta227 Do stream /FormType 1 >> >> To find: The. >> 0.486 Tc q 213 0 obj endobj /Meta281 Do endstream BT 314 0 obj Q >> (+) Tj Q Q Q /ProcSet[/PDF] 1 g /BBox [0 0 88.214 16.44] >> /Type /XObject >> << 25.454 5.203 TD /Type /FontDescriptor /BBox [0 0 673.937 68.796] /Meta209 223 0 R /Meta189 203 0 R /BBox [0 0 534.67 16.44] Q >> /BBox [0 0 15.59 16.44] Q endstream /Meta324 338 0 R Q q /BBox [0 0 88.214 16.44] endobj << /FormType 1 /Meta356 370 0 R BT /Matrix [1 0 0 1 0 0] q BT 549.694 0 0 16.469 0 -0.0283 cm << stream /Length 2252 Q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 1 g /F3 17 0 R 1 g ET q /Meta80 94 0 R q /Meta284 298 0 R 1.007 0 0 1.007 411.035 583.429 cm 0 g BT endobj 1 i >> q q >> q /Type /XObject Q /Type /XObject stream Q q q /Resources<< /Meta423 Do 0.564 G q Q << 0 g q q /Font << 85 0 obj /FormType 1 q 0 w Q << endstream >> /Length 16 /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] (- 8) Tj Q /BBox [0 0 88.214 16.44] q 1 g endobj /ProcSet[/PDF/Text] 0.297 Tc stream 140 0 obj /Meta16 Do >> 0.737 w /BBox [0 0 639.552 16.44] [( t)-14(imes a num)-16(ber)] TJ 0 G /Font << /BBox [0 0 534.67 16.44] /Meta190 204 0 R << q >> << (C\)) Tj 0 g Q q >> /F1 12.131 Tf /Subtype /Form /Font << /BBox [0 0 534.67 16.44] << stream /BBox [0 0 15.59 16.44] (x) Tj q q /ProcSet[/PDF/Text] q 0 w Q endstream ET /Length 57 /ProcSet[/PDF/Text] /Resources<< 0.564 G 0.486 Tc /Length 59 You can specify conditions of storing and accessing cookies in your browser. q /Meta425 441 0 R BT endobj /Font << 0 g 0 g ET 1.014 0 0 1.007 391.462 277.035 cm /F3 17 0 R /BBox [0 0 17.177 16.44] /Meta393 Do q 0 w >> endobj /Font << /Meta331 345 0 R stream q /Subtype /Form BT >> Q Q Q Q endstream Q /Meta8 19 0 R /F3 17 0 R >> 0 G (B\)) Tj /Matrix [1 0 0 1 0 0] Two speeding tickets could increase your rate by 58% at your next renewal. /ProcSet[/PDF/Text] endobj /Matrix [1 0 0 1 0 0] 0 G stream /Length 54 Q << stream 1 g /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /Meta233 Do 582 546 601 560 395 424 326 603 565 834 516 556]>> Q /F3 12.131 Tf /Length 69 /Resources<< endobj 0 w /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 45.168 889.071 cm /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 4.506 24.649 TD (40) Tj endobj 165 0 obj Twice a first number decreased by a second number is 6. 0.458 0 0 RG /Type /XObject 0 G << 1.007 0 0 1.007 271.012 636.879 cm Q 0 G endstream Q Q 1.007 0 0 1.007 130.989 330.484 cm >> /Meta406 Do >> (x) Tj /F1 12.131 Tf q << BT >> /Matrix [1 0 0 1 0 0] 0.486 Tc >> /Subtype /Form 0.458 0 0 RG 1 of this study. stream >> /Length 16 >> 394 0 obj 0.564 G stream 0 g /Subtype /Form >> stream /ProcSet[/PDF/Text] 0 g /Matrix [1 0 0 1 0 0] /F1 7 0 R /Length 70 /Meta375 Do 1.005 0 0 1.015 45.168 53.449 cm 133 0 obj /BBox [0 0 17.177 16.44] /F3 17 0 R q /Length 69 Q q q 1.007 0 0 1.007 654.946 726.464 cm /Length 59 ET >> /Meta361 375 0 R /Meta357 371 0 R >> Q /Length 16 /Length 60 q >> 9.723 5.336 TD 369 0 obj Find the number. endstream Q 0 5.203 TD /F4 36 0 R q Q /Font << endstream q BT 0 g stream /BBox [0 0 15.59 16.44] Q /Resources<< 0.564 G ET << q q /Type /XObject << /BBox [0 0 534.67 16.44] Q q q 0.737 w << Q /Resources<< /Type /XObject Q >> /BBox [0 0 88.214 35.886] /F3 12.131 Tf q endobj -0.463 Tw 2.238 5.203 TD /Length 16 Q >> /FormType 1 endobj /FormType 1 /F3 17 0 R ET stream q Q /F3 12.131 Tf /ProcSet[/PDF/Text] /F3 17 0 R q /FormType 1 endobj /Type /Font /Resources<< 1.014 0 0 1.007 531.485 703.126 cm q /Subtype /Form Q /FormType 1 Q stream 98 0 obj >> endstream q BT Q /Meta422 438 0 R 0.486 Tc Q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] stream q q /Leading 150 >> /Type /XObject q 0 g stream /Font << /FormType 1 1 g 3x - 5 = 2x + 1. x = 6. q /ProcSet[/PDF/Text] 24 0 obj /BBox [0 0 88.214 16.44] Q q 1.007 0 0 1.007 411.035 636.879 cm >> -0.486 Tw BT /Matrix [1 0 0 1 0 0] >> stream Q /FormType 1 0.786 Tc 1.007 0 0 1.006 551.058 437.384 cm /Meta91 Do /Font << stream /Subtype /Form >> 276 0 obj 1.007 0 0 1.007 551.058 383.934 cm 1 i << /Subtype /Form 0.175 Tc Q 549.694 0 0 16.469 0 -0.0283 cm << BT /Resources<< 1 i /ProcSet[/PDF/Text] 0 g /Meta298 Do 32.201 5.203 TD q q q >> /Meta214 Do ET 0.311 Tc /Font << 442 0 obj stream q 0 g 0 G /FormType 1 (Twice) Tj 105 0 obj Q << Q << /Subtype /Form /Matrix [1 0 0 1 0 0] << Twice a number when decreased by 7 gives 45. << 0 G << 1 i << Q /BBox [0 0 88.214 16.44] >> /FormType 1 /ProcSet[/PDF] /FontName /PalatinoLinotype-Bold /Length 104 0 G 0 g /Resources<< Q /Matrix [1 0 0 1 0 0] /FormType 1 ET >> 1.005 0 0 1.015 45.168 53.449 cm >> /Subtype /Form /Length 67 1 i /Meta198 212 0 R Q q /Length 16 /Type /XObject /F3 17 0 R q q 0 G /ProcSet[/PDF] **Note: You could choose any variable you want. endobj /F4 12.131 Tf stream /Meta26 Do Q /Meta384 398 0 R /ProcSet[/PDF/Text] stream /Length 12 /Length 54 1.005 0 0 1.007 102.382 743.025 cm >> /Type /XObject /FormType 1 >> /Font << ET Q /Meta2 Do (-) Tj BT 0.564 G /Meta205 219 0 R /Resources<< 1 g /Meta351 Do /Meta253 267 0 R 20/n b.) 3.742 24.649 TD endobj /Meta4 Do endobj /Resources<< q /Subtype /Form /ProcSet[/PDF] 0 g /ProcSet[/PDF/Text] Q endobj << q >> >> /Resources<< only about 58% of candidates will agree to be screened. /ProcSet[/PDF/Text] >> /F3 12.131 Tf /MissingWidth 250 /Font << /Meta47 Do 23.216 5.203 TD /Type /XObject 1 g /Resources<< /Matrix [1 0 0 1 0 0] 59 0 obj 100 0 obj q << 0 G q q q Q 163 0 obj Q q 1.014 0 0 1.006 391.462 437.384 cm Q (3) Tj 0.155 Tc /F3 17 0 R /Resources<< 722.699 872.509 l BT ET Table 1. q /Meta285 Do >> /Resources<< q endobj /Length 16 Q Q /Meta256 Do /FormType 1 >> stream /F3 12.131 Tf 1.005 0 0 1.007 79.798 829.599 cm /Resources<< 1.014 0 0 1.007 111.416 776.149 cm /BBox [0 0 88.214 16.44] 1 i /F1 7 0 R /Length 12 0.297 Tc >> 1 g /Meta418 434 0 R << Q /F3 17 0 R << Q /FormType 1 endobj (D) Tj /Subtype /Form /Length 16 180 0 obj q q 1.007 0 0 1.007 130.989 277.035 cm q q /Length 63 q /BBox [0 0 88.214 16.44] /ProcSet[/PDF] 1.007 0 0 1.007 45.168 763.351 cm 0 G /BBox [0 0 88.214 16.44] 0.227 Tc 722.699 799.486 l 4 0 R /Length 70 /ProcSet[/PDF] /Resources<< Q >> >> q >> Q q /Matrix [1 0 0 1 0 0] >> q 1.007 0 0 1.007 654.946 653.441 cm endstream ET /Length 16 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 1 i ET 0.369 Tc /F3 17 0 R (6\)) Tj 1 i 1.007 0 0 1.007 130.989 277.035 cm >> /Font << 0.838 Tc >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q /Type /XObject /F3 17 0 R /Subtype /Form 0.737 w >> >> << /Meta64 Do endstream Q 0 G /Meta279 293 0 R 1.014 0 0 1.007 531.485 849.172 cm 1 g q 0 G BT Q ET endobj >> endstream 1 i 1 i q 1 g >> /Type /XObject /Length 67 1 i /Subtype /Form /Matrix [1 0 0 1 0 0] q 98.843 5.203 TD /Meta377 Do Q /F3 12.131 Tf endobj /Resources<< q >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] (iv) A number exceeds 5 by 3. /Type /XObject 187 0 obj >> /Resources<< 1 i q /F3 17 0 R q 400 0 obj 0.564 G Q q Q /FormType 1 Q /Matrix [1 0 0 1 0 0] /Length 16 /Meta192 206 0 R 0 G q q q /Subtype /Form /Subtype /Form /Type /XObject /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] (B) Tj 295.086 4.894 TD 0.564 G endstream 1 i /Meta129 143 0 R >> /Resources<< 0 g q endobj /Resources<< BT (\)) Tj q -0.463 Tw /Resources<< /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 277.035 cm << 0 g Q 1.014 0 0 1.007 111.416 383.934 cm /Type /XObject /Meta70 84 0 R /BBox [0 0 88.214 35.886] /Resources<< BT << /Resources<< /Type /XObject q BT q Q Q /Subtype /Form >> Q stream >> /F4 36 0 R q endobj q In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). /Meta338 352 0 R << 1 i stream /F3 17 0 R /Meta21 32 0 R q Q Q /Font << stream /BBox [0 0 15.59 29.168] /Meta120 Do /Meta341 355 0 R /F3 12.131 Tf (x ) Tj 358 0 obj 0 56.451 TD q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Length 69 0 G /Meta230 244 0 R q 1.014 0 0 1.006 111.416 437.384 cm /F3 12.131 Tf /Meta100 114 0 R stream 181 0 obj /Resources<< BT >> Q q q endstream /F3 17 0 R /Font << 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 /F3 17 0 R q Q /Resources<< >> 0.486 Tc /Subtype /Form Q q Q saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. /Length 69 endobj 1 i /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.007 271.012 330.484 cm 0.737 w stream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 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. >> stream Q /FormType 1 /Matrix [1 0 0 1 0 0] /Meta381 395 0 R 205.199 4.894 TD /Font << q 0.737 w Q 22.478 5.203 TD /Matrix [1 0 0 1 0 0] /BBox [0 0 17.177 16.44] >> ET (+) Tj 0 w /FontBBox [-568 -307 2000 1007] << /ProcSet[/PDF/Text] 1. stream q >> /ProcSet[/PDF/Text] q >> /F3 12.131 Tf >> 1 i q endobj endobj 1.007 0 0 1.007 271.012 277.035 cm If LtitnS6S . >> /F3 17 0 R /Meta13 Do Q /ProcSet[/PDF/Text] 1 g 0.458 0 0 RG >> /BBox [0 0 15.59 16.44] Q 1 g stream /BBox [0 0 88.214 16.44] ET /Matrix [1 0 0 1 0 0] /BBox [0 0 23.896 16.44] Q >> >> 0 G Q 425 0 obj /Subtype /Form /BBox [0 0 88.214 16.44] Q 1.502 5.203 TD /BBox [0 0 88.214 16.44] >> ET q >> Q /Type /XObject << /FormType 1 Example 1: Use the tables above to translate the following English phrases into algebraic expressions. 0.564 G 0.51 Tc q /F3 12.131 Tf 1.007 0 0 1.007 130.989 277.035 cm endstream endobj 197 0 obj Q q /Type /XObject q /Length 206 /Meta373 Do /F3 12.131 Tf 1 i >> << endobj 1 i /Subtype /Form /Meta197 211 0 R 356 0 obj /Meta46 Do 1.007 0 0 1.007 67.753 653.441 cm Q /FormType 1 << << q /BBox [0 0 88.214 16.44] /Subtype /Form endobj 16.469 5.336 TD /Resources<< Q 1.007 0 0 1.007 271.012 277.035 cm /Length 54 /FormType 1 /FormType 1 q q /Subtype /Form << /F3 12.131 Tf /Meta247 261 0 R Q Q 1.007 0 0 1.007 130.989 583.429 cm /Resources<< /Subtype /Form endobj /F3 17 0 R /ProcSet[/PDF/Text] /Font << Q 411 0 obj stream (x) Tj /Length 73 /Font << >> 0.564 G (x) Tj << /F3 12.131 Tf /Meta278 292 0 R endstream 0 G Q << endobj stream stream
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