q endstream 302 0 obj 549.694 0 0 16.469 0 -0.0283 cm /Matrix [1 0 0 1 0 0] /Subtype /Form stream 1 i /Meta320 334 0 R 388 0 obj 0.564 G q /ProcSet[/PDF/Text] Q 1 i 0 g >> 32.201 5.203 TD /Meta210 Do << /F3 12.131 Tf 0 5.203 TD endobj q stream q q 0.51 Tc endstream << /Length 70 /Length 16 q /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 330.484 cm 0 G /Type /XObject ET 213 0 obj /BBox [0 0 88.214 35.886] Q q /F1 12.131 Tf >> 1.007 0 0 1.006 551.058 763.351 cm /Matrix [1 0 0 1 0 0] 0.458 0 0 RG q 162 0 obj q Get link; Facebook; Twitter; q /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] q >> /Meta13 24 0 R endstream /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.737 w /Meta286 300 0 R Q Q q /Subtype /Form 0.737 w 0.001 Tw 0 G << Twice a number decreased by 8 gives 58. /ProcSet[/PDF/Text] endobj 0 G 220.931 4.894 TD Q 0.737 w (vii) Twice a number subtracted from 19 is 11. 0 g /Resources<< /FormType 1 /Length 69 >> /Contents [399 0 R] q 1 i Q /Font << 2.Nine point two decreased by double a number is the same as the number added to four fifths. 1.007 0 0 1.007 67.753 347.046 cm >> q >> /Length 69 q q endobj 1 g 1 i /BBox [0 0 88.214 16.44] q /Type /XObject /Type /XObject /ProcSet[/PDF] 0.564 G /BBox [0 0 30.642 16.44] /BBox [0 0 15.59 16.44] q endstream 0 5.203 TD << endstream >> 52.412 5.203 TD BT >> /Length 59 379 0 obj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.458 0 0 RG Q BT BT 1 i q q Q 403 0 obj Q /Type /XObject /Font << endobj /Meta294 308 0 R >> (B\)) Tj 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. 1 i /FormType 1 endobj /Meta250 Do endobj >> endstream Q /Length 64 /Resources<< Q /Length 59 307 0 obj q /ProcSet[/PDF] endobj endobj q endobj Q /Type /XObject /ProcSet[/PDF] 1 i 1 i 1 i q ET 0 g 1.005 0 0 1.007 102.382 599.991 cm q 0 w q >> /Meta72 86 0 R /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /F3 17 0 R stream >> 1.502 8.18 TD << Q 1.007 0 0 1.007 130.989 776.149 cm endobj 1 g q /Meta244 Do /Meta67 81 0 R 0 G /Meta196 210 0 R >> ET ET /FormType 1 q /Matrix [1 0 0 1 0 0] Q 0 G /BBox [0 0 88.214 16.44] stream q /FormType 1 stream /Meta162 176 0 R 0 w /Meta12 Do 1.502 7.841 TD 3.742 5.203 TD /Font << /Matrix [1 0 0 1 0 0] /F3 12.131 Tf endobj >> /Resources<< q /ProcSet[/PDF] /Resources<< Q /Matrix [1 0 0 1 0 0] endstream /Font << ET 17.234 5.203 TD Q /Font << << /FormType 1 /Length 16 Q BT endobj (-9) Tj 0 w /Length 59 265 0 obj >> 20.21 5.336 TD /Type /XObject endobj /Subtype /Form 0.564 G Q (x ) Tj /Subtype /Form 1 g endobj endstream q /Meta32 45 0 R /F3 17 0 R q 0 G /BBox [0 0 15.59 16.44] Q q /Meta144 Do /Font << /Font << /ProcSet[/PDF] q 1 i q Q Q /BBox [0 0 534.67 16.44] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] q Q Q 1.005 0 0 1.015 45.168 53.449 cm 0 g 1.007 0 0 1.007 130.989 277.035 cm Q 1 i /FormType 1 /FormType 1 /F3 12.131 Tf endobj Q >> /Meta30 43 0 R 1.007 0 0 1.007 45.168 763.351 cm /F3 17 0 R >> ET 1 i /BBox [0 0 534.67 16.44] /Meta163 177 0 R endobj /ProcSet[/PDF] /Resources<< a and b or something else.***. Q /Matrix [1 0 0 1 0 0] /Subtype /Form Q /Matrix [1 0 0 1 0 0] >> Q /Type /XObject 0.51 Tc /Length 64 119 0 obj >> /Font << q /Resources<< Q Q 389 0 obj 2x - y = 6. /FormType 1 /Meta142 156 0 R /Meta421 Do (4\)) Tj endobj /Size 447 Q 391 0 obj q stream >> BT >> endobj BT /Resources<< 0.737 w BT 0 G << ET q /F4 36 0 R 98 0 obj /Subtype /Form q q /Type /XObject 0 g /BBox [0 0 534.67 16.44] >> >> 0.486 Tc >> 1 i 0 g 35.206 4.894 TD 0.564 G BT q Q q On the way, we sang songs for 20 minutes which is 10% of the time we were on the road. 0 5.203 TD /FormType 1 357 0 obj /BBox [0 0 15.59 16.44] << q /ProcSet[/PDF/Text] ( x) Tj /Type /XObject q /Meta426 442 0 R BT Q 1 i /F1 12.131 Tf /Length 16 Q >> 1.014 0 0 1.007 251.439 383.934 cm q /F3 17 0 R BT /Type /XObject /ProcSet[/PDF/Text] Q Q stream q 409 0 obj 3.742 5.203 TD /BBox [0 0 88.214 16.44] 1.502 5.203 TD 0 g BT /Meta217 231 0 R /F3 17 0 R /Meta121 135 0 R 1 g /F3 17 0 R q >> 0 g /F3 17 0 R Q /Subtype /Form << Q 20.21 5.336 TD >> /Resources<< Q /Resources<< 0 g 0.458 0 0 RG Testosterone is the primary sex hormone and anabolic steroid in males. endobj q /Meta249 263 0 R /Matrix [1 0 0 1 0 0] Q endstream /ProcSet[/PDF] /BBox [0 0 88.214 16.44] twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. the quotient of five and a number 7.) endobj /ProcSet[/PDF] /FormType 1 Q Q /ProcSet[/PDF] ET /Matrix [1 0 0 1 0 0] 1 g >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 0.458 0 0 RG 0 G 1.007 0 0 1.007 67.753 473.519 cm /Resources<< >> /Resources<< endstream Q /Resources<< Q 1 i /Type /XObject /Meta267 Do q 1.007 0 0 1.007 551.058 523.204 cm /Leading 349 ET /Type /XObject 0 G 438 0 obj >> stream /Type /XObject /Type /XObject >> 0 g Q Q q /ProcSet[/PDF] q /Length 54 /Matrix [1 0 0 1 0 0] 0.28 Tc stream 0.227 Tc /Resources<< /F3 17 0 R Q The result is 8 less than 10 times the number. 1.014 0 0 1.007 391.462 583.429 cm /Meta90 Do Q /Meta332 Do ET q q /Length 67 q q q 1.005 0 0 1.007 45.168 889.071 cm q q /Meta427 443 0 R endstream /Meta54 Do 0 5.203 TD Q << 1 g endstream 0.564 G << Q 8.985 20.154 l Q 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) >> /Meta147 Do /FormType 1 << /BBox [0 0 534.67 16.44] Q /Font << q /Resources<< /ProcSet[/PDF] /FormType 1 stream Answer only. /Length 16 /Resources<< ET q /Length 65 >> ET << /Type /XObject /Font << >> 1.005 0 0 1.007 102.382 363.608 cm 0 G Q 0 G endobj 317 0 obj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 37 0 obj Q /Subtype /Form BT stream /BBox [0 0 534.67 16.44] /Subtype /Form 1.007 0 0 1.007 130.989 583.429 cm /Type /XObject /Resources<< << q /Type /XObject BT q Q stream /Type /XObject stream endstream 1.007 0 0 1.007 654.946 546.541 cm >> 351 0 obj q /Type /XObject /F3 17 0 R /BBox [0 0 639.552 16.44] Q >> 1 i endobj /Resources<< stream 0 G 0.369 Tc /Matrix [1 0 0 1 0 0] /Resources<< 1 i stream 0.738 Tc /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 111.416 437.384 cm /BBox [0 0 15.59 16.44] >> q /Subtype /Form >> 0 G /Matrix [1 0 0 1 0 0] 264 0 obj endobj 364 0 obj q Q stream 0 g /Matrix [1 0 0 1 0 0] 1.502 24.339 TD Q 0 g /Matrix [1 0 0 1 0 0] Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. Q /Resources<< endobj 104 0 obj /F3 12.131 Tf /Length 16 /F3 12.131 Tf /Font << q q 1.007 0 0 1.007 271.012 636.879 cm >> >> /ProcSet[/PDF/Text] 1.007 0 0 1.006 551.058 763.351 cm endstream q Q << 1 i /F3 12.131 Tf /Subtype /Form /BBox [0 0 639.552 16.44] /Meta258 Do /F4 12.131 Tf /Resources<< /Meta200 214 0 R /Length 69 BT /F3 17 0 R 355 0 obj 0 G endstream ET ET /Subtype /Form 59 0 obj q stream the quotient of twenty and a number a.) Q Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. 340 0 obj /ProcSet[/PDF/Text] 38.948 5.203 TD /Font << /Matrix [1 0 0 1 0 0] /Meta371 385 0 R q /ProcSet[/PDF/Text] /FormType 1 /Meta339 353 0 R /F3 12.131 Tf >> 16.469 5.203 TD /Type /XObject /ItalicAngle 0 q 1 i /Type /XObject q >> /Resources<< 0.458 0 0 RG 20.21 5.203 TD q /Type /XObject q /Type /XObject 259 0 obj /Subtype /Form 0 w Q /Subtype /Form /Subtype /Form q q /F3 12.131 Tf q /Matrix [1 0 0 1 0 0] q >> 0 G 1 g ET /Subtype /Form /Matrix [1 0 0 1 0 0] /FormType 1 Q (+) Tj /ProcSet[/PDF/Text] /Length 69 q /F3 12.131 Tf 0 G /ProcSet[/PDF/Text] 0 G endstream stream >> 93 0 obj /Length 59 /BBox [0 0 88.214 35.886] >> 1 i Q /Length 70 /Matrix [1 0 0 1 0 0] Q 1 i 344 0 obj ET 121 0 obj /ProcSet[/PDF/Text] >> /Length 118 q BT /F3 12.131 Tf q stream 0.68 Tc /Meta262 276 0 R /Meta247 Do /Resources<< /F3 12.131 Tf /F3 17 0 R /F3 17 0 R /Meta169 183 0 R /BBox [0 0 88.214 16.44] S Q /BBox [0 0 30.642 16.44] BT 1.007 0 0 1.007 551.058 703.126 cm >> >> /Length 12 << BT ET Q q Q /Matrix [1 0 0 1 0 0] 0 g Q S 0.737 w (-20) Tj /ProcSet[/PDF/Text] 1.005 0 0 1.015 45.168 53.449 cm stream Q BT 1 i /Subtype /Form (13) Tj stream /Matrix [1 0 0 1 0 0] /Font << (5) Tj 1 i /BBox [0 0 15.59 16.44] 23.216 5.203 TD Q 309 0 obj 6.746 5.203 TD /FormType 1 stream Q /Font << 1.007 0 0 1.007 654.946 473.519 cm /Type /XObject endobj /Meta425 441 0 R q 0.175 Tc endobj /Subtype /Form /Type /XObject << 242 0 obj /F3 17 0 R stream 0 G 1.007 0 0 1.007 411.035 277.035 cm 1 i q /Matrix [1 0 0 1 0 0] /Length 16 1.007 0 0 1.007 67.753 799.486 cm Q /Resources<< /FormType 1 Patients' reasons for declining screening were not collected . /FormType 1 ET /Font << /BBox [0 0 15.59 16.44] 140781 52 0 obj BT >> /FormType 1 1.007 0 0 1.007 654.946 400.496 cm /Type /XObject /Resources<< /Font << (iv) A number exceeds 5 by 3. >> /FormType 1 /BBox [0 0 15.59 29.168] Q 373 0 obj /FormType 1 endobj stream >> >> 0 w >> >> 1 i /F3 12.131 Tf 0 G << /Matrix [1 0 0 1 0 0] Q endobj /BBox [0 0 534.67 16.44] 1.502 5.203 TD Q Q 0.564 G ET 0 g << /Type /XObject >> /Subtype /Form 0.564 G Q 0 w 0.458 0 0 RG /ProcSet[/PDF/Text] 1 i /Subtype /Form /Subtype /Form stream >> Q /Resources<< 0 G q endstream /Matrix [1 0 0 1 0 0] q [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ endstream 1.014 0 0 1.006 391.462 836.374 cm q q /Resources<< << 1.007 0 0 1.007 271.012 776.149 cm (+) Tj endobj /BBox [0 0 88.214 16.44] q /F3 12.131 Tf /Subtype /Form 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. /F3 17 0 R Q q q (C\)) Tj /BBox [0 0 88.214 35.886] /F3 12.131 Tf endobj 0.458 0 0 RG 0 w << q 0 g 0.458 0 0 RG >> /BBox [0 0 88.214 16.44] >> 1.007 0 0 1.006 130.989 437.384 cm /Length 69 BT 0.68 Tc 549.694 0 0 16.469 0 -0.0283 cm Q /Length 16 1 g /Meta181 195 0 R endstream /Meta383 397 0 R ET Q 1 i 54.679 5.203 TD q << >> 30.699 4.894 TD q Q >> /Meta42 56 0 R /Type /XObject /BBox [0 0 88.214 16.44] /FormType 1 q /Matrix [1 0 0 1 0 0] 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 (7\)) Tj q 300 0 obj /Font << 1.007 0 0 1.007 654.946 599.991 cm Q /Meta139 Do /Meta281 Do /FirstChar 32 0.564 G /Matrix [1 0 0 1 0 0] /Subtype /Form /Font << stream /F3 12.131 Tf endobj >> stream 0.737 w /BBox [0 0 534.67 16.44] Q /Font << q >> 1.005 0 0 1.007 79.798 796.475 cm >> /Resources<< 1.005 0 0 1.007 45.168 916.925 cm >> q q 0 w Q /Subtype /Form q /F3 12.131 Tf 20.21 5.203 TD q /Type /XObject /ProcSet[/PDF/Text] /FormType 1 << stream /F1 7 0 R 0 g Q See Solution. << ET Q q stream /FormType 1 1 i /Resources<< 0 20.154 m << /F3 12.131 Tf 1.007 0 0 1.007 271.012 383.934 cm /ProcSet[/PDF/Text] 0 g Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . Let x the unknown number. /Meta225 239 0 R 436 0 obj 2. 0 w /ProcSet[/PDF] /ProcSet[/PDF] /FormType 1 >> stream /FormType 1 >> /BBox [0 0 30.642 16.44] 0 5.203 TD /Matrix [1 0 0 1 0 0] /FormType 1 /BBox [0 0 549.552 16.44] /BBox [0 0 88.214 16.44] /FormType 1 >> ET answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. >> Q /Meta88 Do /Descent -277 << stream 0 5.203 TD Five times a number, decreased by 58, is -23 Find the number. /Length 54 0.737 w 0 g /Length 60 273 0 obj 0 G /BBox [0 0 534.67 16.44] /Meta116 Do /CapHeight 692 >> >> << ET 0.486 Tc /Resources<< /Type /XObject BT Q /Type /XObject /BBox [0 0 88.214 16.44] << >> 404 0 obj /FormType 1 >> /Length 118
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