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

0 g /ProcSet[/PDF/Text] 0.564 G << >> 1 i Q q /Meta40 Do 0 G 0.001 Tw 187 0 obj 1.007 0 0 1.007 271.012 383.934 cm q >> /Length 70 /Length 88 /Font << >> >> 422 0 obj endobj /Subtype /Form /F3 12.131 Tf /Meta387 403 0 R /Matrix [1 0 0 1 0 0] 0.737 w >> Q q Q 1.014 0 0 1.007 251.439 330.484 cm /ProcSet[/PDF/Text] stream Q q /Resources<< /FormType 1 endobj 0 g /Type /XObject 0 G /Matrix [1 0 0 1 0 0] Q /Meta7 Do << stream Q 292 0 obj /Font << /ProcSet[/PDF] endobj q >> /Resources<< Q q >> /Length 54 << << 1.014 0 0 1.006 391.462 510.406 cm q endstream /Type /XObject q 2x - y = 6. x + 3y = -25. /Length 108 q Q << /F3 17 0 R 0.458 0 0 RG /ProcSet[/PDF/Text] (+) Tj 0 g Q 377 0 obj ET ET >> << BT /Type /XObject /ProcSet[/PDF] endobj 0.737 w ET /Type /XObject ET /Length 59 0.564 G /BBox [0 0 30.642 16.44] /Meta177 Do /FormType 1 Q /F1 7 0 R BT /Matrix [1 0 0 1 0 0] >> endobj >> /Type /XObject 0 g 722.699 293.596 l 12.727 5.203 TD q endstream Q Q >> /Meta171 Do /Meta150 164 0 R BT q << q /FormType 1 BT q /F3 17 0 R stream 0 w >> endobj 323 0 obj >> Q /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Meta262 276 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form 0.51 Tc /BBox [0 0 88.214 16.44] /Type /XObject /Resources<< /Meta326 Do stream Q /FormType 1 0 g /FormType 1 Q 0 g << >> Q ET /BBox [0 0 88.214 35.886] endobj Q /Length 59 Q BT 4.506 24.649 TD /FormType 1 q >> Q stream endstream Q ET /Type /XObject 0 g /Meta308 Do stream q 0 5.203 TD Q Q >> << BT /F3 12.131 Tf q /FormType 1 263 0 obj endobj /BBox [0 0 17.177 16.44] /Meta338 352 0 R /Resources<< 0.51 Tc 0 5.203 TD /Font << /BBox [0 0 30.642 16.44] endstream >> stream (-11) Tj >> /FormType 1 /F3 12.131 Tf BT BT /Meta44 Do >> /Type /XObject /ProcSet[/PDF] endobj Q /Meta12 Do 1 i q q 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 << q BT 1.007 0 0 1.006 411.035 763.351 cm endstream q >> q /BBox [0 0 30.642 16.44] 1.007 0 0 1.007 67.753 400.496 cm /ProcSet[/PDF] << /Meta392 408 0 R 1.007 0 0 1.007 67.753 653.441 cm /FormType 1 Q 1 i endstream /BBox [0 0 673.937 68.796] >> q New questions in Mathematics q ET 1 i Q /BBox [0 0 17.177 16.44] Q /Subtype /Form /BBox [0 0 88.214 16.44] stream /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 872.509 cm endobj /Meta138 Do q /Matrix [1 0 0 1 0 0] << Q 1 i /Meta189 Do /Type /XObject 0.737 w q q 105 0 obj 0 g endstream /Subtype /Form /Subtype /Form (-) Tj >> (B\)) Tj Q /Matrix [1 0 0 1 0 0] /Type /XObject Q endstream /Resources<< (x ) Tj /BBox [0 0 88.214 16.44] /Resources<< /Resources<< Q /Resources<< startxref /F3 17 0 R q /Font << /Subtype /Form >> 1.014 0 0 1.006 111.416 836.374 cm /BBox [0 0 88.214 16.44] /FormType 1 369 0 obj /Resources<< Q << /Matrix [1 0 0 1 0 0] /F3 17 0 R endobj 0 w 1 i endstream q 319 0 obj endstream << >> Q /F3 12.131 Tf /Resources<< /Length 16 q /Resources<< 140 0 obj Q Q >> q (5) Tj /Length 12 stream /Meta79 Do /BBox [0 0 30.642 16.44] >> /Font << >> /ProcSet[/PDF/Text] 1 i endstream Q /F1 12.131 Tf endstream << q >> 0 g /Matrix [1 0 0 1 0 0] for the season. 0.564 G /Matrix [1 0 0 1 0 0] /Resources<< /F4 12.131 Tf /ProcSet[/PDF/Text] /F3 12.131 Tf /ProcSet[/PDF] /Length 54 /BBox [0 0 15.59 16.44] /F3 12.131 Tf 0 G endobj /Meta85 99 0 R /F3 17 0 R /Meta131 Do 0.369 Tc /Resources<< Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio endobj /Type /XObject 1 g /BBox [0 0 639.552 16.44] BT /Resources<< 0.51 Tc /Meta407 Do >> /Meta125 Do q Q q /Matrix [1 0 0 1 0 0] q /Resources<< /F3 17 0 R Q /Resources<< stream /Type /XObject Q Q /Subtype /Form /Resources<< /F3 17 0 R 1.007 0 0 1.006 411.035 763.351 cm /FormType 1 0.737 w >> stream stream Q q Q /F3 12.131 Tf 54 0 obj 20.21 5.203 TD /Length 69 /Resources<< 0 g /Type /XObject /Meta127 141 0 R /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] 1 i endstream << >> >> /Matrix [1 0 0 1 0 0] Q /Meta80 Do Q /Resources<< /Length 69 0.564 G ET 1.502 7.841 TD /Subtype /Form /Subtype /Form Q BT 0 5.203 TD /Type /XObject endstream /Meta316 Do 0 w /Type /XObject stream << << 0 G 1 i /Subtype /Form << /Meta334 Do >> 0 G 1.014 0 0 1.007 251.439 383.934 cm stream /Meta221 235 0 R /Subtype /Form >> 129 0 obj q 0 g (13) Tj >> 1 i >> >> /BBox [0 0 15.59 16.44] /Subtype /Form /Resources<< 0.297 Tc q /Type /XObject /Length 69 Q >> /Matrix [1 0 0 1 0 0] Q q endobj /Type /XObject 3.742 5.203 TD /Subtype /Form /Meta177 191 0 R q 1 i /Font << /Matrix [1 0 0 1 0 0] q q 0 20.154 m /FormType 1 Question. endstream /Meta104 118 0 R , Prove the following q 0 g q /Meta240 254 0 R q /Matrix [1 0 0 1 0 0] /F3 17 0 R >> /Resources<< 0 w /ProcSet[/PDF/Text] << 17.234 5.203 TD BT Q 45 0 obj /Meta412 Do 0 g /ProcSet[/PDF/Text] Q >> 0 g /Resources<< /Leading 349 /ProcSet[/PDF] Q 0.458 0 0 RG /ProcSet[/PDF] q /Font << 209 0 obj 0 g 1 i 0 g endobj 0 w /ProcSet[/PDF/Text] ET BT (x) Tj /Meta422 Do /Length 69 0 w >> q q 0 g 0.738 Tc Number Outcomes 1 42 2 41 3 . q 1 i Find the number. Q q 1 g q Mr. Gleeson knows that 1,000 cubic centimeters is the same as 1 liter, and he wants to figure out how many liters of water will fill the container before it overflows. 0 G >> 1.007 0 0 1.007 271.012 636.879 cm Q /MaxWidth 1248 0 w Q BT endobj endobj 1.014 0 0 1.006 251.439 437.384 cm q /Matrix [1 0 0 1 0 0] /Meta172 Do stream q /ProcSet[/PDF] endobj 43.426 5.203 TD 0 G 1 i Q Q 58 0 obj 0 w << /Meta313 327 0 R 0 G 0 G 0.564 G This site is using cookies under cookie policy . q ([x ) Tj Q /Meta78 Do /Length 16 /Length 59 0.737 w >> stream 1 i /Length 139 /Meta70 84 0 R /FontBBox [-170 -292 1419 1050] endstream /Length 69 /Type /XObject endobj 1.007 0 0 1.007 271.012 636.879 cm /Meta14 25 0 R << q 1.007 0 0 1.007 130.989 636.879 cm /F3 12.131 Tf 23.952 4.894 TD q /Meta394 410 0 R /Resources<< q Q 0 G >> endstream ET Q /Meta278 Do endstream q >> /F3 12.131 Tf /Matrix [1 0 0 1 0 0] >> stream /Meta385 Do 1 i endobj /Resources<< >> /Meta354 Do >> /Meta7 18 0 R /F3 12.131 Tf /BBox [0 0 639.552 16.44] 0.458 0 0 RG 119 0 obj /Matrix [1 0 0 1 0 0] /Length 118 Q /Length 54 q 1 g q 1 g endobj /ProcSet[/PDF] q endstream 0 g 389 0 obj endobj << Q >> 1 i /Subtype /Form /F3 17 0 R /BaseFont /PalatinoLinotype-Roman /Type /XObject /Leading 253 /Type /XObject >> /Type /Pages Q >> << /Type /XObject Q >> /Font << >> /Meta426 442 0 R >> /Meta219 Do 53 0 obj /Meta274 288 0 R (3) Tj /F3 17 0 R q /Resources<< 1.007 0 0 1.007 45.168 862.723 cm /Type /XObject (C\)) Tj 0.382 Tc endobj 0 g /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] q >> 199 0 obj BT 0 w /Meta23 34 0 R endobj /Subtype /Form ET /StemH 88 /Resources<< /Type /XObject /Font << /F3 17 0 R /Matrix [1 0 0 1 0 0] Q q /Subtype /TrueType Q /Resources<< Q /Matrix [1 0 0 1 0 0] /Font << q /Resources<< /Length 16 /Font << 0 g stream /Font << << endstream /ProcSet[/PDF] 1.007 0 0 1.007 411.035 849.172 cm % /Font << (A\)) Tj /ProcSet[/PDF] 358 0 obj 0 G 0 G /Font << q /ProcSet[/PDF] q /Resources<< 1.007 0 0 1.007 551.058 703.126 cm /ProcSet[/PDF/Text] /Meta22 33 0 R q /Meta362 376 0 R stream 0.737 w stream /Subtype /Form /Meta298 Do /Matrix [1 0 0 1 0 0] (\(x ) Tj >> 0 G /FormType 1 endobj 0 G stream /Type /XObject q q /FormType 1 endstream 20.21 5.203 TD Q 1 i 1 i q q /Subtype /Form Q 39 0 obj 0 g /Meta353 367 0 R 0 G /BBox [0 0 88.214 16.44] << Q 0 g /Type /XObject 1.007 0 0 1.007 551.058 330.484 cm q /Meta267 Do /Type /XObject 33 0 obj q 0 w 0 g /Type /Font q /Matrix [1 0 0 1 0 0] /Subtype /Form See Solution. /FormType 1 /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 779.913 cm 1 i endstream 38 0 obj /ProcSet[/PDF] (8\)) Tj 1.007 0 0 1.006 411.035 510.406 cm /Meta262 Do q q /Resources<< stream 26.219 5.336 TD q /F3 17 0 R q 357 0 obj 32.201 5.203 TD /Resources<< q /FormType 1 q q Q 1 i Expert Solution. 0 g Six subtracted from a number 6. ET /Length 12 /Matrix [1 0 0 1 0 0] /Meta400 Do q 1.005 0 0 1.007 102.382 347.046 cm q /Meta388 404 0 R endstream /Matrix [1 0 0 1 0 0] >> 0.458 0 0 RG (+) Tj 416 0 obj endstream 722.699 546.541 l q >> (x) Tj 1.007 0 0 1.007 271.012 849.172 cm 1.014 0 0 1.007 531.485 703.126 cm q /BBox [0 0 88.214 16.44] stream q /F3 12.131 Tf /Length 63 /Meta179 Do /Meta84 Do >> Q /Matrix [1 0 0 1 0 0] /Meta215 Do endobj /Meta188 Do /F3 12.131 Tf /Resources<< Q 1.005 0 0 1.007 102.382 816.048 cm 0 w 1 i /FontName /TestGen-Regular 0.486 Tc Q /Length 16 1 g /Subtype /Form 0.271 Tc q Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 330.484 cm q /Subtype /Form ET q Q /FormType 1 /Subtype /Form >> q 0 g /F3 12.131 Tf q /FormType 1 /FormType 1 stream Q stream /Length 118 /ProcSet[/PDF] 0.737 w q 1.502 8.18 TD (x) Tj 36 0 obj >> /Matrix [1 0 0 1 0 0] q /Length 16 Q /Meta46 Do 0.564 G /Meta379 Do /Length 58 Q /FormType 1 /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Type /Font /Meta170 184 0 R Q q /Length 16 /F3 17 0 R /ProcSet[/PDF] q /ProcSet[/PDF] /Type /XObject q /BBox [0 0 17.177 16.44] 0 w /BBox [0 0 88.214 16.44] endstream /Subtype /Form /F3 12.131 Tf /BBox [0 0 88.214 16.44] Q q /Resources<< endobj q 0.458 0 0 RG Q Q stream /Meta178 Do /Font << Q /Type /XObject q 0.458 0 0 RG /Resources<< Q 1 i endstream 0.458 0 0 RG stream 0 w endobj endstream (6\)) Tj Solution. /Length 16 /Resources<< Q /Resources<< ET 166 0 obj 0 g Q Q /F3 12.131 Tf 0 w 1.007 0 0 1.007 130.989 636.879 cm >> 1.007 0 0 1.007 551.058 277.035 cm /Meta90 Do Q 0 g 0.51 Tc 0 w ET >> /BBox [0 0 15.59 16.44] 246 0 obj Q endstream /Meta100 Do Find the number. endstream stream 320 0 obj /ProcSet[/PDF] 80 0 obj 265 0 obj /Length 54 endstream 236 0 obj /ProcSet[/PDF] 0 G /FormType 1 q Q /Font << 1.007 0 0 1.007 130.989 776.149 cm << /F4 12.131 Tf /Meta293 307 0 R /FormType 1 /Font << /Font << /Subtype /Form q 0 G 355 0 obj q Q /BBox [0 0 30.642 16.44] << the other number. Q 0 g /Type /XObject Q /Subtype /Form Q q q 0.68 Tc /Font << >> << endstream 1 i /Subtype /Form /Subtype /Form stream endobj endobj BT Q /Matrix [1 0 0 1 0 0] >> << 0 g q endstream /Meta81 Do q 2x - 15 = -27. Q 0 G endobj q stream Q << Q << /Meta216 Do (+) Tj endstream 1.502 5.203 TD 183 0 obj /Type /XObject ET 1.014 0 0 1.007 111.416 636.879 cm /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] << >> /F1 12.131 Tf /Length 16 /Matrix [1 0 0 1 0 0] /F3 17 0 R << ET endstream ET endobj endobj stream /F3 17 0 R /Type /XObject /F3 12.131 Tf /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 w 101.849 5.203 TD Q stream q >> 0 g /FormType 1 /Meta277 Do /Length 12 Q /Meta339 353 0 R Q 1.502 5.203 TD 1 i /BBox [0 0 88.214 16.44] (13) Tj q << 0 g /Type /XObject 0 5.203 TD BT >> BT /BBox [0 0 30.642 16.44] ET /F1 7 0 R /FormType 1 /Subtype /Form endstream /FormType 1 endstream q /FormType 1 (3\)) Tj BT Q Q /BBox [0 0 15.59 16.44] 1 i q /FormType 1 Q 0 5.203 TD q /F3 17 0 R 78 0 obj << /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Subtype /Form q stream q /BBox [0 0 88.214 16.44] Q /FormType 1 /BBox [0 0 88.214 16.44] 2 0 obj 0 g 0 g /Length 59 q 145 0 obj >> stream 2. endstream Q Q /ProcSet[/PDF/Text] endstream /F3 12.131 Tf -0.486 Tw Q /FormType 1 (x ) Tj /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Meta140 154 0 R [(F)-22(ive)] TJ Q 1 i >> BT /Meta207 221 0 R Q q q /Length 54 ET 1 i Q << 1 i << 0 w /F3 17 0 R /Subtype /Form q ET 0 G /Meta66 Do << /ItalicAngle 0 /Type /XObject /Length 60 /Font << /BBox [0 0 15.59 29.168] >> >> /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] 0.68 Tc 1.014 0 0 1.006 531.485 510.406 cm (C) Tj 1.502 24.649 TD /Type /XObject endobj >> VIDEO ANSWER: in this problem were asked to solve giving, given the following information. /Meta312 326 0 R /Subtype /Form With this, we get: "3x-8". >> 188 0 obj Q 1.014 0 0 1.006 251.439 690.329 cm /BBox [0 0 88.214 35.886] 1.007 0 0 1.006 130.989 690.329 cm q /FormType 1 /Subtype /Form /F4 12.131 Tf q BT /BBox [0 0 88.214 16.44] q 0 g stream 0.737 w /FormType 1 0.737 w 1.014 0 0 1.007 111.416 277.035 cm /Resources<< q >> /Resources<< q /Matrix [1 0 0 1 0 0] /FormType 1 A rectangular garden has a width that is 8 feet less than twice the length. /F3 17 0 R /Matrix [1 0 0 1 0 0] 54.679 5.203 TD /Meta105 Do /Matrix [1 0 0 1 0 0] /FormType 1 /MediaBox [0 0 767.868 993.712] /Meta42 56 0 R ET Q >> BT /Subtype /Form /Length 69 /FormType 1 Q >> stream 0 g 250 0 obj 0 G (2) Tj Solution: 0.486 Tc /ProcSet[/PDF] Q 352 0 obj /Subtype /Form BT Q >> Q Q 1 i /F3 17 0 R 108 0 obj Q q 11.99 8.18 TD /ProcSet[/PDF] >> /BBox [0 0 15.59 16.44] /ProcSet[/PDF] /Meta350 364 0 R q 6.746 5.203 TD << Q q q /Subtype /Form 1.502 24.649 TD /FormType 1 >> /Meta140 Do /Meta218 Do q /FormType 1 q /FormType 1 /Matrix [1 0 0 1 0 0] Q Q 266 0 obj the sum of a number and twelve. stream Q /Type /XObject endstream endstream /Matrix [1 0 0 1 0 0] endstream /Length 91 stream Q /Meta423 439 0 R endobj 1 i 271 0 obj q 130 0 obj q >> ET /Length 245 0 g endobj Q 0 g 0 5.203 TD 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 . /FormType 1 q << 1 i /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /Subtype /Form /BBox [0 0 549.552 16.44] /Type /XObject << /FormType 1 >> q q /FormType 1 /F3 12.131 Tf endstream /Matrix [1 0 0 1 0 0] /F4 12.131 Tf Q 0 g /Meta111 125 0 R Q 0 G /Subtype /Form Q >> q Q endstream q /Length 67 BT /FormType 1 BT Q (x) Tj /BBox [0 0 15.59 16.44] >> /FormType 1 /Subtype /Form endstream 38.182 5.203 TD /F4 12.131 Tf /Matrix [1 0 0 1 0 0] /Length 16 Q /Resources<< stream >> endobj Q /Length 60 0 g /Length 58 /Length 69 endstream stream stream (B\)) Tj /F3 17 0 R /Resources<< >> >> Q 1 g a Question /Subtype /Form 398 0 obj /Type /XObject stream q Q >> endobj Q /BBox [0 0 17.177 16.44] q /Meta337 Do /Length 54 << 0.369 Tc /Subtype /Form /Type /XObject BT endstream 0 g /LastChar 45 Q >> Q endobj q /Length 69 12.727 24.649 TD /XHeight 471 stream q ET endobj endstream ET q Q endobj /FormType 1 /Matrix [1 0 0 1 0 0] endstream /Font << 0.369 Tc q 0 g 1.007 0 0 1.007 551.058 583.429 cm /Font << /Type /XObject 1.502 24.339 TD 1.007 0 0 1.007 551.058 703.126 cm /FormType 1 /ProcSet[/PDF/Text] /Subtype /Form Q q 0 G /Subtype /Form Thrice a number decreased by 5 exceeds twice the number by a unit. /BBox [0 0 17.177 16.44] /Resources<< 220 0 obj /Subtype /Form q /Subtype /Form /Meta273 Do q Q /Length 59 endobj 1.007 0 0 1.007 411.035 330.484 cm 0 G /Type /FontDescriptor q /Meta420 Do /Subtype /Form 0 g >> Q /ProcSet[/PDF] Q /Type /XObject /FormType 1 /Matrix [1 0 0 1 0 0] >> Q endobj 0 w endstream 421 0 obj /Length 59 1 i /Meta48 62 0 R /BBox [0 0 88.214 16.44] /ProcSet[/PDF] 1 i BT 0 G /F3 17 0 R /ProcSet[/PDF] >> /F3 17 0 R This s problem could be, interpreted either way. Q /Resources<< /Subtype /Form /Meta47 61 0 R /BBox [0 0 88.214 16.44] q endobj Q << 424 0 obj Q Q /Type /XObject BT 0.37 Tc /Length 58 /Type /XObject Q /Matrix [1 0 0 1 0 0] /Font << /ProcSet[/PDF] ET /Meta122 Do BT 1 i q /Type /XObject Q q << q Q /Length 118 ET /Type /XObject << /Meta378 Do endstream endobj 0 5.203 TD BT q endobj /Meta141 Do q q >> 1 g 1.007 0 0 1.007 130.989 636.879 cm /Meta65 Do 1.007 0 0 1.006 130.989 437.384 cm /Resources<< 0 g Q BT endobj /Resources<< >> /Matrix [1 0 0 1 0 0] /Subtype /Form Q Find the number. 1 i /Matrix [1 0 0 1 0 0] /Font << /F3 17 0 R 423 0 obj 0.737 w >> /Meta374 388 0 R nine increased by a number x. 0 g /BBox [0 0 88.214 35.886] >> /Resources<< 0 g /FormType 1 q /F3 12.131 Tf /Meta419 435 0 R (3) Tj Q /Subtype /Form << Diabetes, if left untreated, leads to many health complications. /Type /XObject Q stream 0 G 0 G /FormType 1 0 g stream /ProcSet[/PDF/Text] /Length 16 1.005 0 0 1.007 102.382 546.541 cm q 0.458 0 0 RG /Meta387 Do /F3 12.131 Tf q /Subtype /Form /Subtype /Form endobj /Length 59 /Type /XObject 0.369 Tc /Matrix [1 0 0 1 0 0] /FormType 1 Q 1.007 0 0 1.007 271.012 849.172 cm >> /Length 60 q /Font << 1 i /Length 16 0 5.203 TD 225 0 obj << 1.007 0 0 1.007 551.058 636.879 cm /FormType 1 /Meta215 229 0 R /Meta194 208 0 R /Meta232 246 0 R /F3 17 0 R Q /Length 69 q (A\)) Tj /ProcSet[/PDF] 1 i q q /Meta66 80 0 R 0 w /Meta183 197 0 R endobj << /F3 12.131 Tf Q BT stream Q endstream 0.134 Tc 1.007 0 0 1.007 271.012 383.934 cm /FormType 1 0 g q q << q /Subtype /Form /Meta56 70 0 R /Subtype /Form q stream q /Type /XObject /Matrix [1 0 0 1 0 0] >> 0 w ET /Length 68 q /ProcSet[/PDF/Text] 1.007 0 0 1.006 551.058 836.374 cm 0 G /ProcSet[/PDF] stream q /Meta306 320 0 R endstream Q Q /Resources<< endstream 0.737 w 0.838 Tc /Subtype /Form q /ProcSet[/PDF] >> Q stream Q /Length 67 /BBox [0 0 30.642 16.44] >> /Matrix [1 0 0 1 0 0] endstream /Type /XObject 0.737 w Q endstream /F3 17 0 R /F3 17 0 R /Meta400 416 0 R Q 185 0 obj 0.564 G 0.311 Tc 46 0 obj q /Resources<< /Type /XObject Q Q 0 w So let's go ahead and identify a v stream 722.699 347.046 l endstream Twice a number when decreased by 7 gives 45. >> stream Q 0 G 1 i q /Font << >> Q endobj q q /Meta355 Do 0.564 G Q << /Matrix [1 0 0 1 0 0] 0 g q << ET (A\)) Tj << 0.786 Tc /Length 59 1.502 8.18 TD /Type /XObject 0 G /Subtype /Form /Length 19882 >> q endobj >> 1 i 0 g 0 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Font << endstream >> q /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 29.168] 0 G 0 G 1.502 7.841 TD 0 g Q Q n 11 or n 11. q 0.564 G q 3.742 5.203 TD stream /Meta92 Do q << 1 i Q /Length 69 /Meta125 139 0 R q /F3 17 0 R 20.21 5.203 TD stream Twice a number decreased by . /Type /XObject >> Q /Resources<< q << q BT 1 i 0 g /Resources<< /BBox [0 0 673.937 68.796] /Subtype /Form /Length 69 Q /Resources<< 1 i q >> >> 0 G Q 1.502 5.203 TD /ProcSet[/PDF] /F3 12.131 Tf >> q 0.307 Tc /BBox [0 0 88.214 16.44] /Font << 235 0 obj >> 1 i /FormType 1 14.23 24.649 TD 0 5.203 TD /F3 12.131 Tf stream 0.737 w Q /Subtype /Form stream endstream 1.014 0 0 1.007 391.462 383.934 cm /F3 17 0 R /Type /XObject Q 0 5.203 TD >> 1 g q q >> 582 546 601 560 395 424 326 603 565 834 516 556]>> Q >> 1 i stream /Meta165 179 0 R /Resources<< /Meta251 Do /Resources<< Q /FormType 1 /Type /XObject endstream >> q Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. (x ) Tj /Meta298 312 0 R endobj /Meta299 313 0 R /Meta60 74 0 R q /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta398 414 0 R /ProcSet[/PDF/Text] Q /ProcSet[/PDF/Text] /Resources<< 0 g /Type /XObject q >> >> 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject endobj stream /ProcSet[/PDF/Text] /Type /XObject Q 0 G /Font << q /FormType 1 ET /Length 16 /Resources<< >> /BBox [0 0 639.552 16.44] 1.007 0 0 1.007 271.012 523.204 cm You can also contact the clerk of court in the county you received the ticket. /Length 59 /Type /XObject /Font << /F3 12.131 Tf q >> /Meta373 Do /ProcSet[/PDF/Text] /Meta197 Do q q endobj Q q Q /Matrix [1 0 0 1 0 0] Q /Meta395 Do 372 0 obj << /Type /XObject 1 g /ProcSet[/PDF] q /Length 64 q q -0.058 Tw 0.175 Tc q /Meta359 Do Q /Meta81 95 0 R BT 0 g 228 0 obj q 22.478 5.336 TD q q /Type /XObject Q /Matrix [1 0 0 1 0 0] 1 i q q q /Subtype /Form endobj (+) Tj << 0 g Q endobj /FormType 1 (\)) Tj /Resources<< Q 1 g 1.007 0 0 1.007 411.035 849.172 cm /F3 17 0 R /Meta155 169 0 R B. endstream Two fewer than a number doubled is the same as the number decreased by 38. >> 1.007 0 0 1.007 130.989 776.149 cm >> stream q /Matrix [1 0 0 1 0 0] (D\)) Tj /Length 59 /BBox [0 0 88.214 16.44] q (5) Tj /CapHeight 692 BT >> 0.737 w >> /Subtype /Form /BBox [0 0 88.214 16.44] /F1 7 0 R /Matrix [1 0 0 1 0 0] Q q 1 i >> >> endobj 184 0 obj << 1 i /Resources<< endobj q q 0.564 G q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 383.934 cm /F4 12.131 Tf /Length 12 0 5.203 TD Q q Now that you know the meaning of the key words you can read the problem differently. /Font << q 0.838 Tc /F3 17 0 R -0.021 Tw BT endobj Q /F3 17 0 R 0 w << endobj << 114 0 obj 0.564 G Q stream /Meta6 Do 0 g 1 i /Length 16 >> /Length 118 endobj q 1.007 0 0 1.007 551.058 383.934 cm 0 g >> 20.21 5.203 TD /Matrix [1 0 0 1 0 0] endstream endobj 0.737 w /ProcSet[/PDF/Text] stream q /Resources<< q 0.737 w q /Meta384 398 0 R (C) Tj /F3 12.131 Tf Q 0 g /FormType 1 /FormType 1 /ProcSet[/PDF/Text] /FormType 1 /ProcSet[/PDF/Text] BT /F4 12.131 Tf q S /FormType 1 /Font << /ProcSet[/PDF/Text] /Type /XObject /Meta320 334 0 R /Type /XObject /Matrix [1 0 0 1 0 0] Q 0 5.203 TD << /BBox [0 0 15.59 16.44] Phrase : Expression : 4 times some number : 4x: twice a number : 2y : one-third of some number : the product of a number and 12 : 12w: Some examples of common phrases and corresponding . q /Meta270 Do Q /Length 57 q endobj Q Q ET (4\)) Tj stream Q 30.699 5.203 TD Q /F3 17 0 R /Type /XObject (x) Tj /Matrix [1 0 0 1 0 0] ET /BBox [0 0 88.214 16.44] Q /Subtype /Form /Subtype /Form /Subtype /Form /Resources<< ( x) Tj (B\)) Tj q /Resources<< 332 0 obj endstream Q q 196 0 obj Q /Meta198 Do /Font << Q /F3 17 0 R View the full answer. /Matrix [1 0 0 1 0 0] q >> endstream /Meta163 Do ET 0 20.154 m /FormType 1 >> /Length 60 /F3 12.131 Tf /Subtype /Form endobj /FormType 1 q endobj /Subtype /Form 0 G /Subtype /Form stream /Meta124 138 0 R BT /F4 12.131 Tf /Resources<< /Resources<< 0.458 0 0 RG q /Meta353 Do q 1 g Q Q (-20) Tj >> endobj ET /Subtype /Form 0 w stream (58) Tj /Meta99 113 0 R /Subtype /Form /Length 80 >> q endstream endstream /F1 7 0 R /Resources<< Q 6.746 5.336 TD q /Type /XObject Q stream /Meta261 275 0 R endstream 1 i Q q /FormType 1 >> >> /Matrix [1 0 0 1 0 0] 126 0 obj stream stream /Matrix [1 0 0 1 0 0] /Font << /FormType 1 0 g >> /F3 12.131 Tf 0.838 Tc 0 G /Font << /Matrix [1 0 0 1 0 0] [(A number )-17(divided by )] TJ Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /F3 12.131 Tf endobj /Meta312 Do >> 722.699 400.496 l Q /ProcSet[/PDF] /Length 118 q /Meta124 Do 0 g /F3 17 0 R 353 0 obj Q >> /ProcSet[/PDF] << /Font << /Meta272 286 0 R 175 0 obj Q /Matrix [1 0 0 1 0 0] << /F3 12.131 Tf endstream 0 g endobj Q /Meta269 283 0 R /Type /XObject endobj >> 0 g /Font << Q /Meta208 Do 15.731 5.336 TD Q endobj Q >> /Meta152 Do >> endstream /MissingWidth 250 >> q /Subtype /Form >> /XObject << endobj q ET Q >> Q /FormType 1 1.005 0 0 1.007 102.382 473.519 cm q endstream Q >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] stream /F3 17 0 R Q 0 w Q Q Q SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. q 1 g ET 0.737 w /Resources<< << /Type /XObject /Resources<< q q /F3 17 0 R endobj /FormType 1 /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] (38) Tj q /Meta282 296 0 R 0 g 1 g /Length 16 /ProcSet[/PDF/Text] q /ProcSet[/PDF] q stream /Length 70 0 g 288 0 obj /ProcSet[/PDF] ] /Meta235 249 0 R 1 i 0 w /ProcSet[/PDF] 0.737 w 0.564 G Q 1 i BT Q 0.486 Tc /Length 69 0 w q Q /Subtype /Form endstream 1.014 0 0 1.007 111.416 636.879 cm Q 0.458 0 0 RG endstream ET Q /Subtype /Form /Meta429 Do /ProcSet[/PDF/Text] 21.713 20.154 l endobj 1.007 0 0 1.007 45.168 813.037 cm endobj >> /F1 7 0 R 0 G 1.007 0 0 1.006 551.058 763.351 cm 0.737 w /BBox [0 0 88.214 16.44] >> /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Subtype /Form /FormType 1 Q /ProcSet[/PDF/Text] /Meta129 Do /Type /XObject q q /Ascent 1050 /BBox [0 0 30.642 16.44] Q 1 i endobj /Font << /Meta297 Do 0 g Q ET /Type /XObject 0 20.154 m /Length 16 1 i q /Type /XObject endobj /ProcSet[/PDF] Q >> /F3 17 0 R /Meta160 174 0 R /Font << Q /Resources<< Q 1 g 0.786 Tc 248 0 obj 1 i endobj /Resources<< q Q [( a )-15(number, decreased by )] TJ endstream /Font << >> HOPE HELPS .3. /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] /Resources<< Q Q (7\)) Tj endstream >> q Q 1.014 0 0 1.006 531.485 690.329 cm stream q Q 0 G >> 180 0 obj >> 0.369 Tc stream 47.933 5.203 TD /Subtype /Form >> /Length 16 endstream q 1 i stream /Length 69 << >> q >> Q BT 0 g (1\)) Tj BT endobj q /Meta196 Do 0 G S Q 347 0 obj q >> 214 0 obj 20.21 5.203 TD /Matrix [1 0 0 1 0 0] << 19.474 5.203 TD /Subtype /Form 1.005 0 0 1.007 102.382 743.025 cm /Type /XObject 0 g >> Q 1 g /Type /XObject stream 123 0 obj /Matrix [1 0 0 1 0 0] stream Q endobj 0.68 Tc /Matrix [1 0 0 1 0 0] Q /Length 59 endstream /Type /XObject 1 i /DecodeParms [<> ] q 29 0 obj << /Subtype /Form stream /F3 17 0 R /FormType 1 -0.092 Tw 2.238 5.203 TD /Type /XObject /Type /XObject 159 0 obj /ProcSet[/PDF] 1.005 0 0 1.007 102.382 616.553 cm endstream 0 g endobj /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF/Text] Q 1 g /Font << Q /Resources<< /ProcSet[/PDF] >> /BBox [0 0 88.214 35.886] q /Matrix [1 0 0 1 0 0] BT 0 w /Length 66 (2) Tj /ProcSet[/PDF] stream 0.737 w 0 G stream << /BBox [0 0 17.177 16.44] 1.007 0 0 1.007 271.012 277.035 cm 1.007 0 0 1.007 411.035 330.484 cm Q 0 5.203 TD /Subtype /Form /Meta157 171 0 R /ProcSet[/PDF/Text] 234 0 obj 257 0 obj 1.007 0 0 1.007 130.989 277.035 cm >> /BBox [0 0 673.937 16.44] Q q /Meta320 Do 1 i /F3 12.131 Tf q Q Q 400 0 R BT S stream /BBox [0 0 15.59 16.44] 0 g 0 G 6.746 5.203 TD Q 0 G Q /FormType 1 q Q 399 0 obj endstream Q 30 0 obj q q BT Q 1 i ET q /Resources<< >> Q /ProcSet[/PDF] Q q q stream Q 0 g endstream stream /Meta209 Do q << Q >> q 1.005 0 0 1.007 102.382 653.441 cm /BBox [0 0 534.67 16.44] Q q /BBox [0 0 88.214 16.44] >> 0.737 w /Subtype /Form 0 G Q 361 0 obj >> /Type /XObject Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . 1 i q /Font << BT /Length 70 /ProcSet[/PDF/Text] /Meta271 285 0 R /ProcSet[/PDF/Text] >> Q /FormType 1 /Matrix [1 0 0 1 0 0] 1 i /Length 118 /Length 16 /Subtype /Form 1 i /BBox [0 0 88.214 16.44] /ProcSet[/PDF] Choose the correct one. 157 0 obj ET q 0 5.336 TD endstream 12.727 5.203 TD >> /Length 69 178.979 5.203 TD (x) Tj /Type /XObject << /Subtype /Form /Font << /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Resources<< endstream /F1 7 0 R /Matrix [1 0 0 1 0 0] In addition, testosterone in both sexes is involved in health and well-being . Q BT 0.458 0 0 RG /Resources<< q endstream 1.005 0 0 1.007 79.798 829.599 cm 0.838 Tc 1 i /Subtype /Form /Matrix [1 0 0 1 0 0] /Font << 1 g << 0 g stream q /F3 12.131 Tf /Type /XObject S endstream endobj stream endobj -0.008 Tw 139 0 obj /Resources<< Q >> BT /Length 65 /Length 118 140781 endstream ET 0 g >> 1 i q 1.005 0 0 1.007 79.798 746.789 cm /Subtype /Form /FormType 1 0 g q Q 85 0 obj /FormType 1 /Matrix [1 0 0 1 0 0] >> endstream 0 G endobj >> stream 397 0 obj 1.007 0 0 1.007 411.035 583.429 cm 63 0 obj q stream /Meta226 Do >> Q 0 G 1.007 0 0 1.007 411.035 330.484 cm Q Q /Matrix [1 0 0 1 0 0] 1 g 0 5.203 TD /F3 17 0 R /Length 69 /Meta330 Do << BT /Meta409 Do /Meta98 Do /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 0 g /ProcSet[/PDF] Q Q >> ET BT /Meta325 Do /Type /XObject /Matrix [1 0 0 1 0 0] /Length 69 /Meta149 163 0 R q q >> Q /Meta222 236 0 R (D) Tj /Length 69 >> Double or twice a number means 2x, and triple or thrice a number means 3x. /Font << /Meta248 Do q /Meta22 Do 1 g 1 i Q /F3 12.131 Tf /Subtype /Form Q >> q 0 g Q /Meta131 145 0 R /F3 12.131 Tf /Type /XObject Q Q >> q (x) Tj >> endobj q q << q /ProcSet[/PDF/Text] 0 G 1.007 0 0 1.007 411.035 383.934 cm /Resources<< endobj endobj stream Q /BBox [0 0 88.214 16.44] 0 20.154 m /Type /XObject 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. /Font << BT q q >> /Resources<< /Matrix [1 0 0 1 0 0] /FormType 1 q /Length 66 q /Meta253 267 0 R Twice a number decreased by another number: "twice a number decreased by another number" 2*(x-y), "twice a number, decreased by another number"(2*x)-y, It's possible David. endobj q /Type /XObject 1.007 0 0 1.007 271.012 523.204 cm >> q >> 161 0 obj Q /Font << endstream /F3 17 0 R 325 0 obj 1 g q Q Q /Length 59 /Meta37 50 0 R /BBox [0 0 30.642 16.44] /Length 69 0 g >> >> /Meta380 Do /Meta204 218 0 R << Q (8\)) Tj /Type /XObject q Q >> /F3 12.131 Tf endobj stream /Resources<< /Type /XObject endstream q /FormType 1 /Font << Q Q 0 w Q 0 G /Meta291 Do >> /ProcSet[/PDF/Text] 66 0 obj 1 i Q q 0 g /Length 69 /Meta368 Do /ProcSet[/PDF/Text] >> 1.007 0 0 1.007 551.058 277.035 cm Q /Matrix [1 0 0 1 0 0] endstream Q BT /Resources<< endobj /BBox [0 0 88.214 16.44] 0 g 1 i /F1 12.131 Tf /Matrix [1 0 0 1 0 0] /FormType 1 endstream q /Type /XObject q /I0 51 0 R << Q 1 g 1 i (B\)) Tj endobj >> BT Q /ProcSet[/PDF/Text] 52.412 5.203 TD /Resources<< /FormType 1 /BBox [0 0 88.214 16.44] Q /Matrix [1 0 0 1 0 0] Answer (1 of 8): Solution: let the number be x. Q Q endstream q /Resources<< 0 g /Subtype /Form 1 i /BBox [0 0 15.59 16.44] /Length 58 /FormType 1 /Type /XObject endobj 0 g 354 0 obj 19.474 20.154 l /Meta314 328 0 R /Meta21 Do stream (D\)) Tj ( \() Tj 1 i 160 0 obj endobj << 1 i Q 0.458 0 0 RG >> (5) Tj /F3 17 0 R q /ProcSet[/PDF/Text] /Type /XObject -0.084 Tw /Font << 672.261 799.486 m 0 g /Length 70 0 g endstream 0.458 0 0 RG /Font << stream stream Q Q /FormType 1 >> q /Meta239 253 0 R >> Q 0.458 0 0 RG 1.007 0 0 1.007 271.012 583.429 cm >> gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. Q /Resources<< /FontBBox [-90 -216 1195 800] 0.737 w Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] stream /Resources<< /Subtype /Form 0.564 G /Meta58 72 0 R 220.931 4.894 TD endobj q BT << Q /Font << BT >> Q /Resources<< >> 1 i q /Length 12 0 G Q stream /FormType 1 27 0 obj 0.737 w q q 137 0 obj /F3 12.131 Tf endstream /Matrix [1 0 0 1 0 0] 1 i /ProcSet[/PDF] /Length 16 >> 0.524 Tc stream >> 0.564 G BT endstream Q /Subtype /Form /F4 36 0 R /BBox [0 0 88.214 16.44] >> 1 i Q q /Resources<<
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