twice a number decreased by 58

/Matrix [1 0 0 1 0 0] The symbols 17 + x = 68 form an algebraic equation. /BBox [0 0 88.214 16.44] /ProcSet[/PDF] A. BT >> ET /Subtype /Form Q Q /BBox [0 0 88.214 16.44] /BBox [0 0 549.552 16.44] 0 g /Font << endobj q BT q 1 g /Matrix [1 0 0 1 0 0] q /Meta185 199 0 R Q >> endobj >> q /Length 85 /Resources<< ET 121 0 obj [(Fiv)25(e ti)18(me)16(s)] TJ endobj /Meta336 350 0 R 1.007 0 0 1.007 411.035 383.934 cm /Type /XObject endstream 0 g /Length 118 q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Meta245 259 0 R Q Q << 1.502 5.203 TD q ET Q /F3 17 0 R /Resources<< << endobj /F3 12.131 Tf q /Meta10 Do endobj /Length 151 0 5.203 TD >> endobj 1 i /Resources<< /Meta134 Do 285 0 obj stream (D) Tj /FormType 1 /ProcSet[/PDF/Text] /Type /XObject q /Font << Q << 243 0 obj /F3 12.131 Tf Q Q 1.005 0 0 1.007 102.382 293.596 cm >> 1.005 0 0 1.007 102.382 726.464 cm 0 G /Font << /Length 16 >> /ProcSet[/PDF] /Meta149 163 0 R 287 0 obj /Type /XObject 0.737 w 0.458 0 0 RG /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] endstream Q /FormType 1 1.007 0 0 1.007 551.058 703.126 cm 441 0 obj 1 i stream Q /Meta108 122 0 R Q /Resources<< Q >> /Length 78 >> /Meta221 Do Q /F1 12.131 Tf 93 0 obj q 176 0 obj /Meta278 292 0 R /F3 12.131 Tf 722.699 473.519 l Q (x ) Tj /Meta65 79 0 R 0.564 G 0.737 w >> q Q << 1 g /ProcSet[/PDF] >> 0.68 Tc Q /F3 12.131 Tf >> /Resources<< /Subtype /Form << /Subtype /Form /F3 12.131 Tf Q 0.369 Tc stream endobj 0 w /Resources<< q q [tex]\sin (\pi -x)=\sin x[/tex]. 1.005 0 0 1.007 102.382 400.496 cm q 20.21 5.203 TD endobj 9.723 5.336 TD 0 5.203 TD /F4 36 0 R /Subtype /Form /BBox [0 0 15.59 16.44] /F1 14.682 Tf /Resources<< (7\)) Tj /Type /Page /Subtype /Form 1 i << /Matrix [1 0 0 1 0 0] /F1 7 0 R 331 0 obj Q q Q >> /Matrix [1 0 0 1 0 0] Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . /Subtype /Form /Meta91 105 0 R /Matrix [1 0 0 1 0 0] 0.737 w Q Q 0 w BT /Meta330 344 0 R /Type /XObject /Type /XObject 110 0 obj 0.458 0 0 RG /Length 60 /Meta182 Do >> 0 g /FormType 1 1.007 0 0 1.006 551.058 836.374 cm Q q /FormType 1 /Resources<< /Matrix [1 0 0 1 0 0] xref /Subtype /Form /Font << /ProcSet[/PDF/Text] 1 i << endstream >> /Type /XObject stream /Meta92 Do stream q Q q 0 G 1.007 0 0 1.006 411.035 763.351 cm q << Q 0 G /F3 17 0 R endstream /Matrix [1 0 0 1 0 0] Q BT Q 1.014 0 0 1.006 251.439 437.384 cm >> /BBox [0 0 30.642 16.44] /Meta70 Do Q Q /ProcSet[/PDF] 1 g 113 0 obj 294 0 obj /Subtype /Form >> Q BT /Length 95 107 0 obj /Meta174 Do 1.007 0 0 1.007 271.012 277.035 cm 1.007 0 0 1.007 271.012 636.879 cm /Type /XObject 26.219 5.203 TD q /Matrix [1 0 0 1 0 0] /Subtype /Form 0 g 429 0 obj /Font << stream /F1 12.131 Tf /Subtype /TrueType /Encoding /WinAnsiEncoding endobj /ProcSet[/PDF/Text] 0 G /FormType 1 0.297 Tc Q Q /Subtype /Form 0 g /Subtype /Form 0.564 G /Subtype /Form endstream /BBox [0 0 534.67 16.44] << /Resources<< endobj /Font << ET /F1 12.131 Tf 0 G 0.68 Tc stream 0 G /Meta279 Do /F4 12.131 Tf /Matrix [1 0 0 1 0 0] Q Q /FormType 1 q endstream 0.458 0 0 RG 1 i /FormType 1 /Type /XObject /Subtype /Form Q << q q q /FormType 1 /ProcSet[/PDF/Text] 0 g /Type /XObject /MediaBox [0 0 767.868 993.712] Q stream /Type /XObject q 1 i /Matrix [1 0 0 1 0 0] Q 3.742 5.203 TD q << << /Meta30 43 0 R /Font << /Meta158 172 0 R /Subtype /Form /Type /XObject 286 0 obj << q /Font << /F3 17 0 R BT Q q /Matrix [1 0 0 1 0 0] >> >> /Type /XObject 0.524 Tc 0 g 0 G 0 g >> 0 G 0 g >> 1.007 0 0 1.006 551.058 763.351 cm >> Q 1 i q /FormType 1 /Type /XObject >> q Q >> /Length 16 Q /Meta174 188 0 R q endstream 0 g 0 g /Flags 32 /Meta233 247 0 R 0.737 w 0 w Q 1 i BT /Resources<< /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.007 551.058 636.879 cm /Meta300 Do /Meta85 99 0 R q 1.007 0 0 1.006 130.989 437.384 cm q << 0 g BT 722.699 347.046 l /Type /XObject /Meta106 Do /BBox [0 0 15.59 29.168] Q /Meta218 232 0 R q 0.737 w Q /Type /XObject 1 i Q q /Meta75 89 0 R endstream /BBox [0 0 88.214 16.44] [(Negativ)16(e )] TJ >> q endobj 1 i q /FormType 1 /Type /XObject << 26.219 5.203 TD endobj endobj /BBox [0 0 15.59 16.44] ET q >> 0.737 w q q /Length 68 1 i 0 g 0 G >> /ProcSet[/PDF/Text] /Subtype /Form /Length 54 1 i Q q 0 G /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] 0.271 Tc q /Meta130 Do stream /Meta268 Do 1 i >> q /Matrix [1 0 0 1 0 0] >> BT /F3 17 0 R /F3 17 0 R q q q 1.007 0 0 1.007 271.012 636.879 cm 0 g /Matrix [1 0 0 1 0 0] /Type /XObject q 1 i /F1 12.131 Tf If n is "the number," which equation could be used to solve for the number? ET /F3 12.131 Tf q q q 278 0 obj stream 0.738 Tc >> 0.51 Tc 0 G /ProcSet[/PDF] /Meta339 353 0 R Q /F3 17 0 R BT q /Font << 192 0 obj /Length 69 << Q /ProcSet[/PDF/Text] /Subtype /Form >> /Subtype /Form Q /Meta321 335 0 R << 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 Expression. /Length 59 0.737 w /BBox [0 0 673.937 16.44] /FormType 1 /Subtype /Form 1.007 0 0 1.007 551.058 277.035 cm /FormType 1 140 0 obj /Subtype /Form << q (x) Tj 1.014 0 0 1.007 391.462 450.181 cm >> >> q endobj /FormType 1 /Meta186 Do /Subtype /Form 1 i 0 G 0 g /Type /XObject /F3 17 0 R 2 0 obj BT 1 g q << 1 i 722.699 726.464 l 0 G 1.014 0 0 1.006 531.485 836.374 cm q endobj >> /XObject << /F3 12.131 Tf /Meta293 Do /Meta113 127 0 R endstream 0 g /ProcSet[/PDF] q 1.007 0 0 1.007 411.035 277.035 cm q /Type /XObject /Type /XObject stream q /Subtype /Form 1 i 0.458 0 0 RG /F3 17 0 R /ProcSet[/PDF/Text] /Meta312 Do D. b = 4 2. << /Resources<< stream Q endstream << endstream /Font << /Type /XObject ET 0.51 Tc q Q Q << /F4 12.131 Tf /FontName /TestGen-Regular Q endobj 432 0 obj 1.007 0 0 1.007 551.058 583.429 cm Q Q /Subtype /Form ET q >> /ProcSet[/PDF/Text] endstream /F3 17 0 R /FormType 1 400 0 obj /Type /XObject BT 0 g /BBox [0 0 673.937 15.562] /ProcSet[/PDF] >> /Matrix [1 0 0 1 0 0] ET 177 0 obj 1.007 0 0 1.006 551.058 437.384 cm 1 i stream BT 0 G /F3 12.131 Tf Q Q /Font << endobj 0 g Q >> stream endobj endstream q Q /Matrix [1 0 0 1 0 0] /Length 59 q q BT /Type /XObject /Matrix [1 0 0 1 0 0] 1 g 0 G /Meta36 Do /Length 69 >> /Meta296 Do q 184 0 obj /F3 12.131 Tf /Meta39 Do /Subtype /Form /Meta100 Do << /Matrix [1 0 0 1 0 0] endstream Q Q /Meta421 Do /Resources<< (B\)) Tj /F3 17 0 R ET >> << 0.369 Tc "49 . /ProcSet[/PDF] /FormType 1 ET /Subtype /Form /Type /XObject 204 0 obj /F3 17 0 R >> Q /F3 12.131 Tf /BBox [0 0 88.214 35.886] 0 G /Resources<< /Type /XObject >> q ET >> /Length 60 /Length 16 0.737 w >> >> Q the sum of a number and twelve. /ProcSet[/PDF] /F3 12.131 Tf /Matrix [1 0 0 1 0 0] (B\)) Tj q >> Q q >> q 1.007 0 0 1.007 411.035 383.934 cm 1.007 0 0 1.007 551.058 330.484 cm 1 i /Length 59 Let the 2nd number be y. >> Q /Font << >> q >> /Meta362 376 0 R /ProcSet[/PDF/Text] Q /Resources<< 1.007 0 0 1.007 67.753 347.046 cm Q 1 i << << Q Q /Matrix [1 0 0 1 0 0] 0.51 Tc 0 w q << 322 0 obj q Q 1 i Q /Length 245 /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject Q Q /Meta414 430 0 R 1 g /ProcSet[/PDF] /Resources<< Q Q Q << 0.369 Tc << /Subtype /Form /FormType 1 /Meta7 18 0 R 3.742 24.649 TD q 0 G /Font << /FormType 1 /Resources<< /FormType 1 Q Notice that we used the variable \large {d} d in our equation to stand for our unknown value. q >> /Resources<< stream Q q 43 0 obj /FormType 1 /FormType 1 << Q 0 G Q /BBox [0 0 639.552 16.44] /Length 16 /Meta162 Do /ProcSet[/PDF] endstream /XObject << 0.564 G q << Q 0.458 0 0 RG 0 g >> 1.007 0 0 1.007 271.012 849.172 cm stream /F1 7 0 R /Type /XObject Q endobj /Meta50 Do /Meta13 Do (B) Tj 0 G >> 215 0 obj 98.843 5.203 TD Q No packages or subscriptions, pay only for the time you need. /ProcSet[/PDF/Text] /Meta133 147 0 R 1 i >> /FormType 1 endstream 0 g /Matrix [1 0 0 1 0 0] q Q /F3 12.131 Tf /F3 17 0 R /Font << stream /Type /XObject ( \() Tj >> >> /F3 12.131 Tf 0 5.203 TD /Length 59 q If twice a number is decreased by 13, the result is 9. /Matrix [1 0 0 1 0 0] << Q 0 g /Meta167 181 0 R >> S /Matrix [1 0 0 1 0 0] /Meta277 291 0 R ET q 0.738 Tc 1 i /Subtype /Form 103 0 obj ET /Resources<< 439 0 obj 239 0 obj 1.007 0 0 1.007 130.989 849.172 cm Q 0 5.203 TD /Subtype /Form ET q /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] Q Q Q /Subtype /Form /Font << ET >> /Subtype /Form 1.007 0 0 1.007 551.058 523.204 cm Results: patients with type 2 diabetes mellitus predominated with 95.0 %. >> /Font << 276 0 obj q /Resources<< /F3 17 0 R /Meta362 Do /F2 12.131 Tf 0.134 Tc Q Q 89.12 5.203 TD /Length 64 /Subtype /Form /Meta187 201 0 R 1 i /FormType 1 stream /Font << stream /FormType 1 1.014 0 0 1.007 251.439 636.879 cm q /Resources<< BT /FormType 1 ET endobj q endstream If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. Explanation: let the number be n. then we can express division in 2 ways. stream 164 0 obj /Meta388 404 0 R 1.007 0 0 1.007 551.058 583.429 cm >> /ProcSet[/PDF/Text] endobj (-9) Tj Q /Length 69 0 w 0 G /Meta119 Do q endstream 0.369 Tc 0 g q (2) Tj Table 1. q 0 g >> /Matrix [1 0 0 1 0 0] >> >> /BBox [0 0 534.67 16.44] /Type /XObject Q q Q endobj << /Meta59 Do BT >> Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] q /Length 16 /Resources<< q /Font << Q endobj 0 g << /Meta30 Do 0 g BT ET 0 w Q Q /Length 54 /Subtype /Form 1 i Q /ProcSet[/PDF] << >> 1.007 0 0 1.007 551.058 383.934 cm /Meta83 Do 23.952 4.894 TD /Font << /Length 54 endstream /BBox [0 0 639.552 16.44] /Type /XObject /Resources<< /ProcSet[/PDF/Text] 1 i 410 0 obj 0 G /Matrix [1 0 0 1 0 0] /F1 12.131 Tf 92 0 obj stream Q Q >> ET /Subtype /Form /FormType 1 /Font << /Matrix [1 0 0 1 0 0] /Resources<< BT /BBox [0 0 673.937 16.44] /Subtype /Form q >> q 2.238 5.203 TD 0 G 0 G endstream /Meta273 287 0 R /Length 118 /Type /XObject 0.738 Tc Q 0.458 0 0 RG 1.502 5.203 TD /ProcSet[/PDF] 379 0 obj q >> /Resources<< Q /Matrix [1 0 0 1 0 0] /F4 36 0 R /ProcSet[/PDF/Text] q /Resources<< /FormType 1 >> 1.007 0 0 1.007 411.035 636.879 cm 1 g /Subtype /Form /FormType 1 1.007 0 0 1.007 67.753 473.519 cm Q 1.005 0 0 1.007 45.168 889.071 cm (x) Tj 1.007 0 0 1.007 130.989 636.879 cm /F3 17 0 R 21.713 20.154 l /FormType 1 /FormType 1 1 i 1.007 0 0 1.006 551.058 836.374 cm Thrice a number decreased by 5 is 3x - 5. /Meta425 441 0 R 0 w /ProcSet[/PDF] /BBox [0 0 15.59 16.44] endstream 0.564 G /Resources<< /Resources<< /Meta6 15 0 R /Type /XObject 1 i Q /Meta3 12 0 R >> q endobj >> endobj Q /F3 12.131 Tf /F3 17 0 R 0.68 Tc >> 0 g 1 i ET /F3 17 0 R stream /Resources<< << >> /ProcSet[/PDF] (+) Tj SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. endstream 0 G 0.369 Tc ET Q /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 endstream (-11) Tj 0 G /Meta145 159 0 R Q 52.412 5.203 TD >> 0.737 w Q BT << /Meta405 Do 202 0 obj q Q 1 g /FormType 1 /FormType 1 q /BBox [0 0 88.214 16.44] q << /F3 12.131 Tf /Type /XObject /Type /XObject 1 i Q Q /Resources<< /BBox [0 0 534.67 16.44] q endstream q Q q 0 20.154 m 1.005 0 0 1.007 79.798 763.351 cm q /Subtype /Form /F3 12.131 Tf (-) Tj /ProcSet[/PDF/Text] ET >> 0 g 0 G 0 w /Matrix [1 0 0 1 0 0] /Meta425 Do q /Meta167 Do 20.975 5.336 TD >> /FormType 1 /FormType 1 (B) Tj endstream Q Choose an expert and meet online. /BBox [0 0 88.214 16.44] endstream /Resources<< Q /ProcSet[/PDF] 0.737 w 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 . /Font << BT >> /Meta70 84 0 R /Resources<< q /Matrix [1 0 0 1 0 0] Q << >> 0 g /F3 12.131 Tf Q /Meta67 81 0 R << endobj Q Q q /F3 17 0 R Q 1.007 0 0 1.007 654.946 347.046 cm q q /F3 12.131 Tf /Meta82 96 0 R 0.737 w endobj Q q /BBox [0 0 88.214 16.44] /Resources<< 0.786 Tc 0 g Q /Type /XObject >> << 1.007 0 0 1.007 551.058 703.126 cm q 1 i 1.502 5.203 TD /Type /XObject Q >> (C\)) Tj q /F3 12.131 Tf /BBox [0 0 30.642 16.44] q 425 0 obj Q /BBox [0 0 30.642 16.44] 16.469 5.336 TD 0.311 Tc endobj /Font << >> /ProcSet[/PDF] >> 0.458 0 0 RG >> Q /F3 12.131 Tf Q /Resources<< 0 G BT stream Decreased by another number means subtract. 0 w >> /Type /XObject Q 0.369 Tc /Meta124 138 0 R (C\)) Tj /Meta381 395 0 R q /BBox [0 0 15.59 16.44] endstream endstream /Font << /Resources<< Q Q Q /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] /Length 16 0 g /Type /XObject BT Q >> BT stream Q /Resources<< /Encoding /WinAnsiEncoding << /Meta20 31 0 R 1.007 0 0 1.007 67.753 293.596 cm /Meta336 Do /F3 17 0 R /F3 12.131 Tf /Type /XObject q /Meta146 Do 1 i 1 i 0 g /FormType 1 /Type /XObject 0 G /Resources<< q 3.742 5.203 TD 0 w 0 g 0 g >> Q 0.68 Tc /ProcSet[/PDF] /Meta259 273 0 R /F3 12.131 Tf /Meta251 Do >> /FormType 1 /Subtype /Form /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. q /Resources<< >> >> stream 1.007 0 0 1.007 551.058 703.126 cm endstream /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q /Meta395 411 0 R stream q Making educational experiences better for everyone. Q /Resources<< BT 1.007 0 0 1.007 271.012 450.181 cm << /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 670.003 cm 1 i /F3 17 0 R << /MaxWidth 1453 q /BBox [0 0 639.552 16.44] endstream 1 i << /Length 19882 Q /Type /XObject q /Subtype /Form /Length 60 >> /F3 17 0 R /Meta249 Do /F3 12.131 Tf endstream Q Q /Matrix [1 0 0 1 0 0] q q That was 1/8 of the points that he scored q /Matrix [1 0 0 1 0 0] /FormType 1 Q /Matrix [1 0 0 1 0 0] endobj /Matrix [1 0 0 1 0 0] /Meta410 Do /Type /XObject endobj 0 5.203 TD q /BBox [0 0 639.552 16.44] /Subtype /Form /Length 69 /Meta37 Do /BBox [0 0 15.59 16.44] >> /ProcSet[/PDF/Text] >> endstream Q There were x cookies at the beginning of a party. /ProcSet[/PDF] Q /Type /XObject >> -0.126 Tw /Subtype /Form stream endobj /Resources<< 126 0 obj /Type /XObject q /F4 36 0 R /ProcSet[/PDF/Text] 0 g endstream /Type /XObject /F3 12.131 Tf q (B\)) Tj 0.564 G q Q q Q /BBox [0 0 88.214 16.44] Two fewer than a number doubled is the same as the number decreased by 38. 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/Meta207 Do /F3 17 0 R Q A link to the app was sent to your phone. Q q >> /BBox [0 0 88.214 16.44] 1 i /F3 17 0 R 1 i TJ q /Type /XObject (8\)) Tj Get a free answer to a quick problem. q Q q /Meta176 190 0 R 0.786 Tc /Resources<< /FormType 1 endobj /Meta101 115 0 R Q 1 i 0 w Q Q Q /Type /XObject /Meta292 306 0 R /Resources<< endobj /Type /XObject /Matrix [1 0 0 1 0 0] /Subtype /Form View the full answer. q 0 g << Q 0 w /Meta77 Do 0.458 0 0 RG ET /Meta133 Do /Matrix [1 0 0 1 0 0] /I0 Do q /Resources<< q /FormType 1 endobj 155 0 obj /Font << 275 0 obj /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 Q q Twice a number decreased by ten is at least 24. Q /FormType 1 stream (7\)) Tj 1 i /Meta345 359 0 R /Meta351 365 0 R Q 1 i << Q >> /Type /XObject q << ET 1 g /BBox [0 0 534.67 16.44] 0 g 1.007 0 0 1.007 45.168 846.161 cm (-4) Tj /Length 16 0 g 1.007 0 0 1.006 411.035 510.406 cm q q stream endstream /Meta278 Do /FormType 1 >> /Matrix [1 0 0 1 0 0] /Font << /Meta206 220 0 R /Matrix [1 0 0 1 0 0] Q Q /Ascent 1050 >> endstream q /Meta81 95 0 R 1 i q q /Type /XObject /F3 12.131 Tf /Type /XObject << stream 1.014 0 0 1.007 391.462 703.126 cm Q /FormType 1 endstream /Subtype /Form /Meta259 Do 0 G /F3 17 0 R endobj Q 382 0 obj /Resources<< endobj >> This site is using cookies under cookie policy . /FormType 1 Q /Font << endstream /FormType 1 /Meta136 Do ET << << /FormType 1 /F3 17 0 R stream /Resources<< /Resources<< q /Font << /FormType 1 >> /Subtype /Form Q >> /Matrix [1 0 0 1 0 0] /FormType 1 /ProcSet[/PDF/Text] 3.742 5.203 TD Q q /Resources<< << /Length 16 >> >> 1.007 0 0 1.006 130.989 437.384 cm endstream /ProcSet[/PDF/Text] q 1 i /FormType 1 -0.008 Tw 0 g q >> >> /Length 16 1 i 1.005 0 0 1.007 102.382 293.596 cm endstream q /Meta272 Do /Length 69 q q >> 0.369 Tc /MissingWidth 252 /Subtype /Form << Q 1 i /Matrix [1 0 0 1 0 0] /Length 16 /Resources<< /Meta147 161 0 R << /ProcSet[/PDF/Text] >> /Font << 1 i /Resources<< 0 G BT /Resources<< /Subtype /Form Q endstream << 1 i 1 i /BBox [0 0 88.214 16.44] /Length 66 q q Q >> endstream Q q 431 0 obj /FormType 1 Q /FormType 1 q /FormType 1 /Subtype /Form >> 134 0 obj BT << /Matrix [1 0 0 1 0 0] 217 0 obj 259 0 obj q /Resources<< stream /Type /XObject BT /Subtype /Form Q q 0 G stream 0.737 w stream >> endstream /Resources<< q q Q q endstream /ProcSet[/PDF] /Type /XObject 0 w >> << stream /F3 17 0 R q 0.737 w endstream /I0 51 0 R /Font << /ProcSet[/PDF/Text] /FirstChar 32 0.564 G /ProcSet[/PDF/Text] /BBox [0 0 30.642 16.44] 1 i 1 i 189 0 obj /Meta150 164 0 R q stream >> Q q Q Q /Meta148 Do stream 32.201 20.154 l 0 g Q /Matrix [1 0 0 1 0 0] Q Q /Meta154 Do /Matrix [1 0 0 1 0 0] endstream /Meta169 183 0 R endstream Twice a number decreased by 8 gives 58. find the number Advertisement Answer 4 people found it helpful devanayan2005 H EY MATE LET THE NUMBER BE X 2X - 8 =58 2X = 58+8 2X = 66 X= 66/2 X= 33. 80 0 obj stream /Type /XObject 0 G 0 G /Length 65 >> /Length 16 ET >> /Resources<< endstream /Meta225 Do 0 g endobj /Length 59 endstream Q q Get link; Facebook; Twitter; >> BT /Length 118 >> endstream /Meta206 Do >> << 20 0 obj /Length 69 /Subtype /Form /Subtype /Form /Matrix [1 0 0 1 0 0] >> /Subtype /Form 0.737 w 187 0 obj /F3 17 0 R /FormType 1 endstream 1 i /F3 12.131 Tf Thirthy is equal to twice a number decreased by four = solve and check the equation? 1 i >> /ProcSet[/PDF/Text] Q /Resources<< Q /Subtype /Form /Font << 0 G 72 0 obj Q /Resources<< 1 i ET Q /Matrix [1 0 0 1 0 0] /Meta44 Do 1 i 1 i 26.219 5.336 TD /Matrix [1 0 0 1 0 0] >> 0.737 w /Subtype /Form 0 g Q /BBox [0 0 88.214 16.44] q 1 i 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] endobj << stream 0 g 0 g /Resources<< 0 20.154 m endobj /Resources<< >> /Resources<< /Meta207 221 0 R Q endobj 0.564 G stream << /ProcSet[/PDF] Q stream /Resources<< /FormType 1 Q ET /Subtype /Form Q ET << q 0.458 0 0 RG q 1 g /Subtype /Form Q 0.369 Tc 0 G /ProcSet[/PDF] Q 94.364 5.203 TD /ProcSet[/PDF] >> /Type /XObject endobj Q Q /Resources<< /F3 17 0 R << /F3 17 0 R Q 0 g q -0.03 Tw (3\)) Tj /Encoding /WinAnsiEncoding >> 1.005 0 0 1.007 102.382 563.103 cm << 1.005 0 0 1.007 102.382 293.596 cm /Font << /ProcSet[/PDF/Text] /Length 16 /Length 16 0 w stream /Meta81 Do endstream q Q 0.68 Tc 1.007 0 0 1.007 271.012 776.149 cm /Type /XObject /BBox [0 0 30.642 16.44] /Length 69 stream 1 i /F3 12.131 Tf /Meta190 Do Q /FormType 1 1 i >> Q 0 G >> /BBox [0 0 15.59 16.44] >> /BBox [0 0 30.642 16.44] >> /Meta380 Do q 0 G q >> 0 g )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] 1 i endstream [tex]\sin (\pi -x)=\sin x[/tex]. q /Matrix [1 0 0 1 0 0] Q /Meta102 116 0 R 427 0 obj /F3 12.131 Tf /Type /XObject Q endstream /Subtype /Form Q endstream /Meta337 351 0 R /FormType 1 q endobj q /F3 17 0 R 0 g >> Q Solution. 1 i HOPE HELPS .3. 1.014 0 0 1.007 111.416 703.126 cm >> << Q Q 20.21 5.203 TD /Matrix [1 0 0 1 0 0] 6.746 5.336 TD q >> q /Length 54 >> /Length 69 /Type /XObject 353 0 obj /Subtype /Form /BBox [0 0 15.59 16.44] endobj Q 17 0 obj q /Length 16 endstream 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. q /Length 59 /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.486 Tc /Type /XObject 1 i 1 i 0 g /Subtype /Form /Meta284 298 0 R q q /Length 16 /Type /XObject q /Type /XObject 1 g /Meta223 Do 27.693 5.203 TD >> 0.564 G /Type /XObject /ProcSet[/PDF/Text] stream q Q >> >> Q stream 0.458 0 0 RG ET 0.227 Tc /Type /XObject q 1.005 0 0 1.007 79.798 730.228 cm ET /Type /XObject BT >> >> /Matrix [1 0 0 1 0 0] Q endobj BT >> /Resources<< 1.005 0 0 1.015 45.168 53.449 cm (A\)) Tj /Matrix [1 0 0 1 0 0] endstream /Length 16 6.746 5.203 TD << /Resources<< /Font << Q 279 0 obj /Meta46 Do /Matrix [1 0 0 1 0 0] 1 i much as how 8, Last . /BBox [0 0 88.214 16.44] /BBox [0 0 30.642 16.44] /F3 17 0 R ET /Font << endstream (+) Tj 1.007 0 0 1.006 411.035 763.351 cm stream /Resources<< /Encoding /WinAnsiEncoding ET /FormType 1 q Q /Length 69 1.007 0 0 1.007 551.058 636.879 cm << 111 0 obj /Length 67 1 i 0 w Q /Length 16 /Meta213 Do 1 g << ET >> Q endobj /Length 68 stream /F3 12.131 Tf ET >> /BBox [0 0 88.214 16.44] The sum of a number and 2 is 6 less than twice that number. q >> 0.564 G 0.737 w /Length 67 Q /Meta274 288 0 R 0 g << /F3 17 0 R /F3 17 0 R >> /ProcSet[/PDF] q Q << Q >> ET 0.51 Tc /ProcSet[/PDF] stream /Meta385 Do endstream ET Q BT << /Meta402 Do /Matrix [1 0 0 1 0 0] stream /Length 58 /Matrix [1 0 0 1 0 0] /Meta241 255 0 R Q /Resources<< endobj 0.737 w

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