/BBox [0 0 639.552 16.44] /Meta282 296 0 R 0.458 0 0 RG stream 1.007 0 0 1.007 271.012 383.934 cm q ET q q >> /Length 63 ET 1 g 1 g /Resources<< << /Subtype /Form /FormType 1 ET q q 0 w /Matrix [1 0 0 1 0 0] /I0 51 0 R /F3 17 0 R /Meta350 Do /FormType 1 0.51 Tc /Type /XObject 2.Nine point two decreased by double a number is the same as the number added to four fifths. Q /ProcSet[/PDF] /F3 17 0 R /Resources<< /Meta426 Do endobj << /Length 54 Q q Q /Length 69 q Q 1.007 0 0 1.007 654.946 653.441 cm /Subtype /Form >> /Flags 32 0 G /Meta116 Do /Meta371 Do /Length 68 Q [(F)-22(ive)] TJ /BBox [0 0 30.642 16.44] >> /Length 16 722.699 546.541 l /FormType 1 /F3 17 0 R /Font << 1.014 0 0 1.007 251.439 450.181 cm ET 0.737 w /FormType 1 1.005 0 0 1.007 79.798 862.723 cm /ProcSet[/PDF] /F3 12.131 Tf /Length 68 q endstream BT 1 g endobj >> >> /Meta197 Do /F3 17 0 R >> 0 G 1 i endstream /Resources<< /BBox [0 0 673.937 15.562] >> 406 0 obj /Matrix [1 0 0 1 0 0] /FormType 1 1.007 0 0 1.007 411.035 330.484 cm 328 0 obj Q q Q ET /Subtype /Form /Meta222 236 0 R /Resources<< endobj endstream 0.68 Tc 1.502 24.339 TD >> /Subtype /Form /Resources<< Five times a number, decreased by 58, is -23 Find the number. Q /BBox [0 0 88.214 16.44] 0 g /Meta114 Do /Meta358 372 0 R >> endstream /Meta179 193 0 R /Subtype /Form /Meta42 Do endobj /Meta205 Do q q /Type /XObject q 1.007 0 0 1.006 411.035 437.384 cm q endobj /Subtype /Form q 20.21 5.336 TD 0.737 w q /FormType 1 /BBox [0 0 30.642 16.44] endobj q /BBox [0 0 88.214 16.44] Q /F4 36 0 R 280 0 obj /BBox [0 0 88.214 16.44] >> ET /Matrix [1 0 0 1 0 0] >> Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. /Matrix [1 0 0 1 0 0] 0 G << endobj /Type /XObject >> (3) Tj /F3 17 0 R /Font << endobj >> Use the variable g to represent Gails age. 0.458 0 0 RG 0 5.203 TD q /F4 12.131 Tf 0 g endobj 397 0 obj /Resources<< /Subtype /Form /ProcSet[/PDF/Text] 0 G BT 1.007 0 0 1.007 654.946 726.464 cm /F3 17 0 R ( \() Tj Q q Q q stream q 0.737 w 0 g 6 more than twice a number: 2x+6: two less than a number: x-2: the sum of 9 and a number: 9+x: two less than three times a number: 3x-2: a number subtracted from 12 . endobj endstream << ET /BBox [0 0 15.59 16.44] Q /FormType 1 329 0 obj 1 i 6.746 5.203 TD Q /Font << 0 w /Subtype /Form endstream >> q /FormType 1 Q /Meta334 Do /Meta24 Do VIDEO ANSWER: in this problem were asked to solve giving, given the following information. endobj /Subtype /Form /Meta412 428 0 R /Matrix [1 0 0 1 0 0] stream 0 g 0.425 Tc 241 0 obj /BBox [0 0 88.214 35.886] Q 6.746 5.203 TD >> endstream 0 4.894 TD 0.458 0 0 RG /Meta398 414 0 R /BBox [0 0 30.642 16.44] stream /Subtype /Form /BBox [0 0 88.214 16.44] q 1 g 30.699 4.894 TD -0.008 Tw 1.007 0 0 1.007 130.989 636.879 cm (-20) Tj /Type /XObject /Resources<< >> ET Q /BBox [0 0 88.214 35.886] 0 w endstream /Type /XObject /Length 60 /Meta276 Do Q 320 0 obj 61 0 obj endobj /Matrix [1 0 0 1 0 0] (-) Tj << /Subtype /Form /Type /XObject /F3 12.131 Tf /Font << /F3 12.131 Tf q BT ET /Meta136 150 0 R /Meta44 58 0 R q (-9) Tj /ProcSet[/PDF/Text] (D\)) Tj /Font << 12.727 5.203 TD /Font << /Type /XObject 1 g >> ET (B\)) Tj /FormType 1 BT stream /Resources<< >> stream q (38) Tj Q /FormType 1 /Length 16 0 g >> q /Type /XObject /ProcSet[/PDF] /Resources<< /Resources<< 0 G /ProcSet[/PDF] 500 500 500 0 333 389 278 0 0 722 500 500]>> Q /FormType 1 /Meta287 301 0 R >> 15.731 5.336 TD /Matrix [1 0 0 1 0 0] q endobj endstream /ProcSet[/PDF] << 1 i >> /Meta201 215 0 R Q /Meta275 289 0 R /Font << /Length 70 0 w q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] >> 1.014 0 0 1.006 251.439 690.329 cm 0 g /F3 12.131 Tf 1 i 1 i q Q 1 i /Meta253 Do Q /Resources<< 0.37 Tc 1.007 0 0 1.007 271.012 776.149 cm You can also contact the clerk of court in the county you received the ticket. 722.699 799.486 l Q /FormType 1 q 430 0 obj Q 1.007 0 0 1.007 271.012 383.934 cm /FormType 1 >> >> q /ProcSet[/PDF/Text] 0 g q endstream /Matrix [1 0 0 1 0 0] Q /Subtype /Form /ProcSet[/PDF/Text] 0 g /F1 14.682 Tf /Meta15 26 0 R q q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] >> 277 0 obj (B\)) Tj stream /Subtype /Form stream Q Q Q 0 g stream Q /Matrix [1 0 0 1 0 0] Q Q 0 G /Length 16 endstream /Resources<< 0 g q 0.425 Tc Q 1 i /ProcSet[/PDF/Text] q (C\)) Tj q 1 i 43 0 obj BT Q /Meta137 151 0 R q Q /Font << q q 0 w /Type /XObject Q Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Length 59 /Resources<< >> 1 i /BBox [0 0 88.214 16.44] /Type /XObject 1 i /FormType 1 >> Q /Type /Page endstream /BBox [0 0 88.214 35.886] /Meta328 Do /Type /XObject 0 G Q >> q 10.487 5.203 TD Q ET /BBox [0 0 15.59 29.168] /BBox [0 0 534.67 16.44] 1.502 5.203 TD /Length 16 /F1 12.131 Tf << << /Meta407 423 0 R /Length 69 225 0 obj q /Subtype /Form q /Length 16 Q /Meta140 Do Q >> 1.005 0 0 1.007 102.382 599.991 cm 1.007 0 0 1.006 411.035 690.329 cm 0 G endobj /Type /XObject /FormType 1 stream /Meta256 270 0 R Q endstream Q /Meta139 153 0 R /BBox [0 0 534.67 16.44] /F3 12.131 Tf /Length 78 << /Subtype /Form /Resources<< Q 267 0 obj /Type /XObject /BBox [0 0 534.67 16.44] 0 G /Resources<< 1.014 0 0 1.006 531.485 510.406 cm 0 G /Type /Page (8\)) Tj 9.723 5.336 TD /Type /XObject stream >> /Matrix [1 0 0 1 0 0] /Resources<< /Type /XObject >> endstream endstream (x) Tj 1 i 1.007 0 0 1.007 654.946 347.046 cm /Resources<< /Length 59 /Type /XObject /ProcSet[/PDF/Text] >> 0 g /Meta135 149 0 R q 204 0 obj 1 i /Resources<< /ProcSet[/PDF/Text] q /Type /XObject q 233 0 obj /BBox [0 0 534.67 16.44] q q BT /Meta383 Do /Meta384 Do >> q q 0.564 G 0 G /FormType 1 /Matrix [1 0 0 1 0 0] /F4 36 0 R 0.458 0 0 RG >> [(1)-25(0\))] TJ Q /F3 17 0 R 0 g endstream 11.99 24.649 TD /Resources<< ET 1 i Q /Matrix [1 0 0 1 0 0] 214 0 obj >> /Subtype /Form Q >> Q /Subtype /Form /Type /XObject q stream /Subtype /Form 1 g >> Q endobj q Q /Matrix [1 0 0 1 0 0] q >> Q /Meta339 353 0 R /Meta180 Do >> q /Length 60 /F4 12.131 Tf /Meta304 318 0 R 119 0 obj /ProcSet[/PDF/Text] 0.786 Tc >> >> /ProcSet[/PDF/Text] /ProcSet[/PDF] endobj /Resources<< /Subtype /Form 1.014 0 0 1.006 391.462 437.384 cm 1.007 0 0 1.007 411.035 383.934 cm << /Resources<< the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . 126 0 obj /Length 12 /BBox [0 0 88.214 35.886] q Q /BBox [0 0 88.214 16.44] /Meta412 Do /Meta87 Do /FormType 1 stream /Meta209 223 0 R /F1 12.131 Tf /Matrix [1 0 0 1 0 0] /Resources<< 1 i stream << /Matrix [1 0 0 1 0 0] 216 0 obj (5) Tj 1 i /ProcSet[/PDF] 0.564 G BT q /Meta191 205 0 R ET 1 i << /Type /XObject >> >> /FormType 1 q /Length 59 << 1.014 0 0 1.006 111.416 510.406 cm 1.005 0 0 1.007 79.798 730.228 cm /F1 12.131 Tf >> /Type /XObject /Resources<< /Type /XObject 13.493 5.336 TD 30 0 obj ET q 71 0 obj Q /Subtype /Form [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. /F3 17 0 R /Length 12 /FontBBox [-90 -216 1195 800] /F3 12.131 Tf 1 i /BBox [0 0 88.214 16.44] q 0.307 Tc 0.564 G Q /Subtype /Form << /F3 12.131 Tf /FormType 1 /Meta241 Do ET 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. q endobj /Meta423 439 0 R /Length 58 /Font << /Resources<< /Meta1 8 0 R (C\)) Tj q endstream /ProcSet[/PDF/Text] /F3 17 0 R /Type /XObject >> 251 0 obj BT /Meta343 Do 1 i /ProcSet[/PDF/Text] q >> 250 0 obj 1 i ET q Q >> << /Resources<< >> q /Type /XObject /Length 16 /Meta331 345 0 R Q /Type /XObject /BBox [0 0 30.642 16.44] /Length 78 Q << (9) Tj /FontBBox [-568 -307 2000 1007] /Meta413 429 0 R /Meta279 Do Twice a number decreased by 58! /FormType 1 endobj /Subtype /Form endobj q Q stream q /Length 16 /Length 58 /Meta44 Do Q Q q /Length 67 Q q /F3 12.131 Tf 0 G 1 i /BBox [0 0 88.214 16.44] BT q q /F3 17 0 R q /ProcSet[/PDF] >> /FormType 1 0 g /Length 16 /Resources<< Q 1.007 0 0 1.007 271.012 523.204 cm Q Twice a number would be 2x. q ET /F3 17 0 R /BBox [0 0 15.59 16.44] /FormType 1 /Type /XObject /Meta0 Do endstream 1.014 0 0 1.006 531.485 836.374 cm /Meta233 247 0 R /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 0.737 w /F3 12.131 Tf Q q q /Subtype /Form /Meta318 Do /Font << /ProcSet[/PDF/Text] ET >> q q endstream /Length 16 ET /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /FormType 1 /Type /XObject /Subtype /Form q 0 g q << ET /FormType 1 << /Meta220 234 0 R stream stream /Matrix [1 0 0 1 0 0] /Meta211 225 0 R /Resources<< 1 g /Matrix [1 0 0 1 0 0] 1 g 0 g q /Font << /Meta310 324 0 R q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 0 g Q /Type /XObject >> endobj /Subtype /Form /Meta333 Do endobj Q >> << Q ET ET /Length 68 /Matrix [1 0 0 1 0 0] Q 672.261 400.496 m /ProcSet[/PDF/Text] 1 i /Font << /Meta169 183 0 R Q >> /Resources<< /Meta198 Do q 351 0 obj /Resources<< /F3 12.131 Tf >> endobj 0 G /ProcSet[/PDF/Text] >> /Resources<< 1 g /FormType 1 q endobj << << Q /BBox [0 0 30.642 16.44] q 1 i /Font << endobj /Meta262 276 0 R 0 g /Matrix [1 0 0 1 0 0] /Font << BT endstream /Length 69 >> >> Q endstream /Resources<< /ProcSet[/PDF/Text] q >> Q Q endstream q 1.007 0 0 1.007 271.012 277.035 cm q q /FormType 1 endobj /Type /XObject >> >> << 0 w /Subtype /Form /Subtype /Form /Length 54 >> Q 1 i /Meta183 197 0 R /Meta222 Do /FormType 1 /Subtype /Form /BBox [0 0 534.67 16.44] Q q /Length 59 q /Length 244 BT 0.524 Tc /Matrix [1 0 0 1 0 0] >> endstream /Meta239 253 0 R ET Q /FormType 1 (-4) Tj Q >> ET (9\)) Tj q (58) Tj /Meta298 312 0 R /Subtype /Form /I0 51 0 R >> >> q 1 i stream stream /FormType 1 0 g /Font << q q q >> Q Q 0 g /Type /XObject >> Q endstream >> Q 6.746 8.18 TD /Length 16 /Subtype /Form q In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). q /Matrix [1 0 0 1 0 0] 0 w /ProcSet[/PDF/Text] BT /ProcSet[/PDF/Text] /BBox [0 0 15.59 29.168] endstream Q >> >> /Resources<< /Type /XObject endobj /ProcSet[/PDF] /F4 12.131 Tf q q >> /Type /XObject /FormType 1 /Type /XObject BT /Font << stream >> 0 G endobj >> << /F3 12.131 Tf stream >> endobj Q << BT /Type /XObject 1 i endstream Q >> q /FormType 1 endobj /Meta300 Do stream q << stream q /Type /XObject /FormType 1 /Meta92 Do /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] Q 0.458 0 0 RG /ProcSet[/PDF/Text] q endobj BT q /FormType 1 << /Matrix [1 0 0 1 0 0] 1 i 372 0 obj /Type /XObject /Meta118 Do 1 g /ProcSet[/PDF/Text] S /Subtype /Form /ProcSet[/PDF] 0 G /FormType 1 /FormType 1 1 i 0 g /Font << >> stream /Meta151 Do 1 i /ProcSet[/PDF/Text] /F3 17 0 R ET 1.014 0 0 1.007 251.439 523.204 cm 30.699 4.894 TD 314 0 obj endstream /Type /XObject Q 6.746 5.203 TD /Meta319 Do /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /F3 12.131 Tf << /Type /XObject 2x - y = 6. x + 3y = -25. Q 0.68 Tc /ProcSet[/PDF] 0 G /F4 36 0 R /Font << S /Meta300 314 0 R endstream /Resources<< q 1 i /Length 118 Q 1 i /Length 16 0 g >> 1.014 0 0 1.007 251.439 383.934 cm Q /F3 17 0 R q endstream /FormType 1 /Resources<< >> /Type /XObject endobj /Meta261 Do /Subtype /Form 163 0 obj q /Length 19882 -0.084 Tw Q Q /Subtype /Form 1.007 0 0 1.007 551.058 523.204 cm q /ProcSet[/PDF/Text] Q 0 G /Meta321 Do endobj /Meta72 86 0 R /Length 16 Q q 346 0 obj /Type /XObject 1.005 0 0 1.007 102.382 726.464 cm 411 0 obj 0 w /Meta316 330 0 R << /Meta401 417 0 R 1 i endstream /Type /XObject /Resources<< >> /Matrix [1 0 0 1 0 0] q /Type /XObject 0 g 90 0 obj /ProcSet[/PDF/Text] 0 5.203 TD /Length 107 /F3 17 0 R 287 0 obj /Length 118 endobj /FormType 1 /FirstChar 32 /F3 17 0 R 33.704 5.203 TD /Subtype /Form /Font << 1.007 0 0 1.007 67.753 799.486 cm endstream 0.564 G >> /Meta203 Do (x) Tj << 0.524 Tc The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. /F3 17 0 R /Subtype /Form Q << endstream /Meta250 264 0 R /BBox [0 0 30.642 16.44] Q << /Meta212 Do 1 i /Length 69 Q /Resources<< /Meta235 249 0 R Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. << 0.297 Tc /BBox [0 0 17.177 16.44] 0 G /Length 79 /F1 7 0 R 1 i /ProcSet[/PDF/Text] q % /FormType 1 >> 1 i /Subtype /Form Q Q ET /FormType 1 1st step. 0 G >> /Resources<< /BBox [0 0 88.214 16.44] stream /Type /XObject ET q 407 0 obj >> q /BBox [0 0 88.214 16.44] (6\)) Tj /Matrix [1 0 0 1 0 0] endobj /Matrix [1 0 0 1 0 0] /Meta331 Do /Type /XObject /Subtype /Form 0.369 Tc 0.68 Tc 1.007 0 0 1.007 130.989 583.429 cm << >> /Meta32 Do /Resources<< /Widths [ 500 0 502]>> q Q Q Q /Resources<< BT /Length 16 >> endobj 1 g /Font << 672.261 799.486 m stream endobj 1 i Twice a number is decreased by 9, and this sum is multiplied by 4.

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