twice a number decreased by 58pulte homes raleigh corporate office

0 g /FormType 1 /F2 11 0 R /Length 16 194 0 obj /F1 7 0 R 325 0 obj Q q Q 1 i >> >> /Subtype /TrueType Q /ProcSet[/PDF/Text] /I0 Do 1 g /FormType 1 << 1.007 0 0 1.007 271.012 383.934 cm /Matrix [1 0 0 1 0 0] /Resources<< 0.68 Tc (3) Tj 0 g >> /Length 59 /Subtype /Form 1 g q 1.007 0 0 1.007 130.989 776.149 cm Q >> q q << /BBox [0 0 534.67 16.44] /Type /XObject /Matrix [1 0 0 1 0 0] BT 0.564 G 1.007 0 0 1.007 271.012 383.934 cm /Matrix [1 0 0 1 0 0] >> 0 g /Meta349 Do (-) Tj (D\)) Tj endobj 1 i /FormType 1 Q /Subtype /Form Q >> >> endobj /FormType 1 >> /Meta334 348 0 R (iv) A number exceeds 5 by 3. q /BBox [0 0 15.59 16.44] endstream /Meta73 87 0 R 0 g /FormType 1 0 g >> << /Matrix [1 0 0 1 0 0] /Length 69 q >> /Meta277 291 0 R a Question 0 G q (-) Tj 1.005 0 0 1.007 102.382 293.596 cm q Get a free answer to a quick problem. 0 G 35 0 obj stream /Meta347 361 0 R /Resources<< endstream ET /Type /XObject Q q 0 g stream /Length 16 1.014 0 0 1.007 251.439 383.934 cm /FormType 1 >> 1 g BT /Subtype /Form Q /Meta105 Do /Subtype /Form 136 0 obj /F3 12.131 Tf >> 0 5.203 TD Q << /XObject << >> q 0 g endobj [(MULTIPLE CHOICE. q q /FormType 1 >> q >> Q 1 i Was this answer helpful? stream 0 g 0 g >> /I0 Do Q >> q Q endstream 0 5.203 TD 1.005 0 0 1.007 102.382 256.709 cm stream q /Meta90 104 0 R /ProcSet[/PDF] BT 0.271 Tc stream 1 i /Subtype /Form 157 0 obj /Meta320 334 0 R >> q /BBox [0 0 88.214 16.44] q 437 0 obj Q 0.458 0 0 RG endobj q q [( times )15(a numb)22(er and )] TJ q /Resources<< /F1 12.131 Tf /Subtype /Form >> ET /ProcSet[/PDF] stream endstream ET endobj << Q /FormType 1 /F3 12.131 Tf /F3 12.131 Tf S << 1 i /Meta198 Do ET 1 i << /Resources<< /Subtype /Form q /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] (-4) Tj endstream q Q /Type /XObject endstream /Subtype /Form /Length 68 /Subtype /Form 1.007 0 0 1.007 411.035 383.934 cm >> 1.007 0 0 1.007 67.753 726.464 cm 354 0 obj Q endobj ET /BBox [0 0 88.214 16.44] 298 0 obj ET q q q /Length 108 0 g /Resources<< (-20) Tj Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. Q 1 i /Subtype /Form /Matrix [1 0 0 1 0 0] >> 0 w >> [4] One half of a number decreased by fourteen is twenty-one /Resources<< 0 G /FormType 1 q Q endobj >> Q 1.007 0 0 1.007 271.012 703.126 cm /Length 16 0 G 0.369 Tc q BT >> /BBox [0 0 88.214 16.44] 1.007 0 0 1.006 411.035 437.384 cm Q 1 i stream /Meta8 Do 1 i /Meta68 82 0 R /Resources<< /Length 67 q >> Thrice a number decreased by 5 exceeds twice the number by 1 is . /BBox [0 0 88.214 16.44] 0.51 Tc 0.738 Tc 0 G endobj >> -0.382 Tw /BBox [0 0 534.67 16.44] /Kids [ 0 g /Matrix [1 0 0 1 0 0] 0.737 w >> stream 0 g q 1 i >> >> [tex]\sin (\pi -x)=\sin x[/tex]. q BT /Length 294 q endstream /Meta265 279 0 R 0.564 G /Meta364 Do 1 i 0 4.894 TD /Type /XObject 1.007 0 0 1.007 67.753 293.596 cm >> Q q endstream << /Resources<< endstream 1 i 0.737 w q 0.458 0 0 RG >> 1.007 0 0 1.007 67.753 347.046 cm 0.564 G 1 i >> 1.007 0 0 1.007 130.989 636.879 cm q 187 0 obj q /Matrix [1 0 0 1 0 0] (5) Tj /ProcSet[/PDF/Text] Q stream 1.007 0 0 1.007 67.753 293.596 cm endobj Q >> 0 g >> q /Subtype /Form q /F3 12.131 Tf endstream /BBox [0 0 15.59 16.44] >> /FormType 1 /Meta4 13 0 R q /ProcSet[/PDF/Text] /FormType 1 Q /Meta22 33 0 R q << /Meta375 389 0 R >> 57.656 5.203 TD q 1 i 0 g /F3 12.131 Tf q MetS-Z quartiles and their associated risks are presented in Fig. /Meta358 Do /F3 17 0 R /FormType 1 >> /Subtype /Form 0 G /Meta407 Do /Meta4 Do endobj endstream >> /FormType 1 endobj Q q BT << endobj q q 0.564 G /Subtype /Form /Type /XObject (\)) Tj /Meta58 Do /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] << BT /Subtype /Form 0.458 0 0 RG /ProcSet[/PDF] /Length 16 >> 1.007 0 0 1.007 67.753 653.441 cm 12.727 5.203 TD /Resources<< endstream Q q Q Q 1.007 0 0 1.007 45.168 763.351 cm Q ET 0.564 G Q 1.007 0 0 1.007 411.035 583.429 cm 413 0 obj ET 1 i /BBox [0 0 88.214 16.44] endstream endobj answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. >> Q /Type /XObject /Subtype /Form (v) 5 subtracted from thrice a number is 16. 0 w /LastChar 121 >> q /FormType 1 endobj endobj Q endobj /BBox [0 0 15.59 16.44] 1 g 0.458 0 0 RG 0.297 Tc endstream stream Q 1 i /Meta166 180 0 R Twice a number is decreased by 9, and this sum is multiplied by 4. endstream /Meta271 285 0 R /Meta408 424 0 R /BBox [0 0 534.67 16.44] /Type /XObject /Meta94 108 0 R /BBox [0 0 88.214 16.44] /Subtype /Form /FormType 1 0 g /Meta365 Do Q << /Meta117 Do endstream /Meta236 250 0 R endstream 419 0 obj /FormType 1 47 0 obj /Type /XObject /F3 17 0 R endobj >> /Matrix [1 0 0 1 0 0] endobj 101 0 obj endstream q >> << endstream 0.369 Tc stream 0.838 Tc /Meta335 349 0 R 0 w /Meta184 198 0 R /Subtype /Form /Resources<< /Meta345 Do /FormType 1 BT Q /F3 17 0 R /Subtype /Form /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q /Type /XObject 0 G /ProcSet[/PDF] /Length 78 1 i /F3 12.131 Tf stream /Subtype /Form 1.007 0 0 1.007 45.168 813.037 cm /F3 17 0 R 0.458 0 0 RG q << >> >> q >> Q 1 i endstream endstream /F1 12.131 Tf /Font << q << /Resources<< /Length 16 /LastChar 45 0 G /Subtype /Form /F3 17 0 R Q Q endobj /F4 12.131 Tf Q /Length 16 /Resources<< By the . 0 w 1 i q /FontBBox [-568 -307 2000 1007] >> q stream << 0 G stream << 0.458 0 0 RG q << 0 5.203 TD /Resources<< endstream >> Q /Subtype /Form /ProcSet[/PDF/Text] Q Q /Resources<< /Resources<< stream /Meta71 Do >> The value of k is: (b) 3 (d) 0 (a) 4 (c) -4 TL ing:, 1)take a graph and draw two perpendicular lines to obtain four uadrants 2)draw any object using straight line 3) write the coordinates of each point o q 1.007 0 0 1.007 45.168 746.789 cm /Subtype /Form << q q /BBox [0 0 17.177 16.44] >> /Resources<< /Meta424 Do /Resources<< /Resources<< Q ET /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] Q 1 g endobj 1 i Q q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /F3 17 0 R /Font << >> q 0 w Q (-) Tj q /Meta42 56 0 R BT stream 0 G q endobj ET Q 0 g 0 g /ItalicAngle 0 /Resources<< >> BT stream 0 5.203 TD << /Meta202 Do 6.746 8.18 TD q q 0 g 1.005 0 0 1.007 79.798 779.913 cm << /Meta132 Do /Resources<< stream Q >> stream /Matrix [1 0 0 1 0 0] /BBox [0 0 17.177 16.44] /Font << Q stream 1 i /Resources<< 1.007 0 0 1.007 130.989 583.429 cm q Q 0 G -0.021 Tw /Meta86 100 0 R 0 g 0 G BT /Meta209 Do /Font << stream stream The quotient of a seven and a number 9. 1.007 0 0 1.006 551.058 763.351 cm 0 G 0 g 424 0 obj 0.737 w 1 i stream >> q >> /Length 16 >> >> /Length 16 /Meta16 Do startxref /F2 11 0 R /Subtype /Form Q 61 0 obj 439 0 obj 0 g /I0 51 0 R >> 1 i /BBox [0 0 88.214 16.44] Q 0 g 1 i 105 0 obj >> endobj /Length 91 q Q /Font << /Type /XObject ET Q 0 G << 212 0 obj >> Q 8 0 obj stream /Subtype /Form q /F3 17 0 R << q 1 i /F3 12.131 Tf /Font << >> /Type /XObject Q Q Q ET 0 g /Font << /Type /XObject /ProcSet[/PDF] /Type /XObject q >> q q /FormType 1 /Meta339 353 0 R << /Matrix [1 0 0 1 0 0] 0 g Q 0.564 G /Length 69 /Length 59 /ProcSet[/PDF/Text] >> 32.201 5.203 TD >> /FormType 1 << 0.737 w >> endstream /Resources<< q 1.007 0 0 1.007 130.989 636.879 cm /FormType 1 q -0.486 Tw BT Q >> 0 g stream endobj endstream /FormType 1 /Matrix [1 0 0 1 0 0] /Subtype /Form q stream 0 g >> /FormType 1 q /Subtype /Form 257 0 obj /Subtype /Form On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini BT 0 G /Subtype /Form (C\)) Tj /BBox [0 0 88.214 35.886] endstream ( x) Tj /FormType 1 q 34 0 obj >> /Subtype /Form >> q 1 i (x ) Tj 0 56.451 TD 1 i >> /Length 16 262 0 obj Q q 0.68 Tc 0.737 w /Subtype /Form /BBox [0 0 534.67 16.44] << /Length 54 /ProcSet[/PDF] /Subtype /Form Q Q /F3 12.131 Tf /Length 59 1.014 0 0 1.007 531.485 277.035 cm q Q endobj Q 349 0 obj 0 5.203 TD q << q first we change the sentence in formula as the following, i think this is clear &easy to understand .i hope it helps you, This site is using cookies under cookie policy . /F1 7 0 R /Font << /F1 12.131 Tf 0.564 G endstream >> Q /BBox [0 0 15.59 16.44] Q 1.014 0 0 1.007 251.439 277.035 cm Q /Type /XObject /Meta309 323 0 R endstream /Subtype /Form >> /Font << q Q 1 i /F3 12.131 Tf >> endstream 0 g 0 5.203 TD stream stream stream 13.464 5.203 TD (C\)) Tj >> q /FormType 1 /Subtype /Form True False /Meta214 228 0 R endobj Twice a number decreased by 8 gives 58. Q 1 i /Meta294 308 0 R Patients' reasons for declining screening were not collected . << << 0 g q /Matrix [1 0 0 1 0 0] q endstream /ProcSet[/PDF/Text] << /Resources<< /Subtype /Form /F3 12.131 Tf /Matrix [1 0 0 1 0 0] >> -0.16 Tw 1.014 0 0 1.007 111.416 450.181 cm q (3) Tj Q /Subtype /Form /F3 12.131 Tf >> /XHeight 471 5.98 7.841 TD 0 G 0 g /BBox [0 0 88.214 16.44] /BBox [0 0 534.67 16.44] 0.458 0 0 RG 1 i /Subtype /Form 1.014 0 0 1.007 531.485 703.126 cm endstream 1.007 0 0 1.007 67.753 473.519 cm >> endstream /Meta238 252 0 R stream /Resources<< /Font << /Meta57 Do Q q /Resources<< /Type /XObject /Type /XObject endobj stream /ItalicAngle 0 q endstream /Matrix [1 0 0 1 0 0] stream 1 i /Type /XObject /Type /XObject /Resources<< 213 0 obj /BBox [0 0 15.59 16.44] 1.005 0 0 1.006 45.168 879.284 cm 0 g q Q >> 0.564 G Q /FormType 1 /F1 12.131 Tf /Font << q BT q /Meta38 Do /Font << /ProcSet[/PDF/Text] /Meta174 Do q /FormType 1 stream /Meta222 Do /Meta66 80 0 R q 0 G 1.007 0 0 1.006 130.989 690.329 cm >> BT 1 i >> Find the number. endstream Q /ProcSet[/PDF/Text] /ProcSet[/PDF] stream Q /BBox [0 0 88.214 16.44] >> /Type /XObject 0 G /FontDescriptor 16 0 R /Matrix [1 0 0 1 0 0] >> >> BT >> Q /Resources<< stream 0 g /Matrix [1 0 0 1 0 0] /Meta47 Do q /BBox [0 0 88.214 16.44] q 1.007 0 0 1.007 411.035 383.934 cm Q /Meta133 Do >> q 389 0 obj /F3 17 0 R Q 122 0 obj /Subtype /Form /Type /XObject >> /F1 12.131 Tf >> 0.737 w >> 0 g 0 G Q /Subtype /Form /Meta187 201 0 R 0.458 0 0 RG /ProcSet[/PDF/Text] 19.474 20.154 l << Q 1 i >> /BaseFont /PalatinoLinotype-Roman 0.297 Tc >> /Font << /Meta210 224 0 R 225 0 obj 38.182 5.203 TD /F1 12.131 Tf /Length 64 Q stream /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 130.989 523.204 cm Q q /Subtype /Form BT /Meta278 Do 0 g /Resources<< /F3 12.131 Tf (\)) Tj << /Font << /Matrix [1 0 0 1 0 0] /BBox [0 0 549.552 16.44] q 374 0 obj q /FormType 1 stream >> >> (-) Tj endobj /Resources<< /Subtype /Form /F3 17 0 R /F3 17 0 R 0.564 G endstream q Q /Length 13 /Meta275 Do endobj << Q q >> >> 0.737 w /FormType 1 >> 336 0 obj q /Meta89 Do Q >> /Subtype /Form /FormType 1 /Resources<< >> /Matrix [1 0 0 1 0 0] /Type /XObject q /Meta397 Do /BBox [0 0 88.214 16.44] In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. /FormType 1 >> 33 0 obj q 0 G /FormType 1 1 i /Font << Q /Subtype /Form 0 g /Meta390 Do /F3 12.131 Tf Q stream Q /Resources<< q /Length 16 /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] /F3 12.131 Tf Q (5) Tj >> /Length 70 32 0 obj /Matrix [1 0 0 1 0 0] q 1 g /Subtype /Form q 0 G stream >> q /ProcSet[/PDF] Q 0 w endstream endobj q 0 g Q /F3 17 0 R 0 g ET 0.458 0 0 RG Q 0 5.203 TD << /BBox [0 0 88.214 16.44] q /Length 65 1 g /FormType 1 0 G /FormType 1 1.005 0 0 1.007 102.382 546.541 cm >> Q /Meta72 Do 0.564 G /Meta368 382 0 R /F1 12.131 Tf 1.005 0 0 1.007 102.382 546.541 cm 1.007 0 0 1.007 654.946 599.991 cm 139 0 obj 1 g /FormType 1 /FormType 1 If a number is 400%, then it is 4 times, the same as 4. /ProcSet[/PDF/Text] q /Font << >> endobj /BBox [0 0 15.59 29.168] 0 w /Meta244 258 0 R /Resources<< endobj q Q /Meta87 Do 1.007 0 0 1.006 130.989 690.329 cm endstream q Q q /Matrix [1 0 0 1 0 0] 0 G /Meta117 131 0 R ET 0 g /ProcSet[/PDF/Text] /F3 17 0 R Q Q 1 g /F3 12.131 Tf 1.502 8.18 TD endobj 1.007 0 0 1.007 67.753 546.541 cm q /F3 12.131 Tf BT endstream >> /Font << 1 i q Q /Length 69 /Resources<< /Meta89 103 0 R BT (iii) 25 exceeds a number by 7. stream /BBox [0 0 15.59 29.168] 0 g q ET >> >> to represent the numbers. /Font << Q /Type /XObject /Meta248 Do Q >> 320 0 obj Q q >> q /Resources<< /Resources<< /Resources<< >> q 0.486 Tc >> Q /Subtype /Form >> q q endobj /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /BBox [0 0 88.214 16.44] stream /Subtype /Form /Meta186 200 0 R >> >> 13.493 5.203 TD 0.564 G >> 1 g << /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Length 118 q q Q Q /Meta232 246 0 R Q 20.21 5.203 TD endobj 1 i /Length 12 428 0 obj 0.564 G /FormType 1 Results: patients with type 2 diabetes mellitus predominated with 95.0 %. 201 0 obj 1.007 0 0 1.006 551.058 836.374 cm (58) Tj Q ( x) Tj /ProcSet[/PDF/Text] 0.564 G /F4 12.131 Tf 1 i /ProcSet[/PDF/Text] endobj 13.493 5.336 TD q >> /Resources<< /Subtype /Form /Meta36 Do endobj 0 G /Resources<< /Meta221 Do 0.68 Tc q /Resources<< /Matrix [1 0 0 1 0 0] 0.737 w D. Twice a number decreased by ten is less than 24. (C) Tj 196 0 obj /ProcSet[/PDF/Text] /StemH 94 /Type /XObject /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Font << >> /Resources<< endstream q /BBox [0 0 639.552 16.44] Q Q /Matrix [1 0 0 1 0 0] 1 g Q /Meta301 315 0 R Q /FormType 1 /Resources<< /Resources<< /F3 17 0 R /Subtype /Form (2) Tj 1 i Q /Type /XObject /Matrix [1 0 0 1 0 0] /Length 77 0 w /Matrix [1 0 0 1 0 0] Solution. endstream /Type /XObject /Subtype /Form /Matrix [1 0 0 1 0 0] 134 0 obj /Meta177 191 0 R /Meta353 367 0 R /Subtype /Form /FormType 1 /FormType 1 6.746 5.203 TD q 1 g >> stream /Length 69 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) Q BT /Resources<< >> << >> Q /Meta106 120 0 R /Meta175 Do /Subtype /Form 0.524 Tc /Meta252 Do 32.201 20.154 l /Matrix [1 0 0 1 0 0] q endobj endstream >> q /Length 59 endstream endstream ET >> /Meta197 211 0 R endobj /Resources<< /Meta272 286 0 R /Matrix [1 0 0 1 0 0] /F3 17 0 R 0.425 Tc >> 1 i endstream 0.178 Tc Q >> 123 0 obj q /Meta185 199 0 R /Font << /Type /XObject /Meta346 Do 1 i endobj << /Font << Q endobj /BBox [0 0 88.214 35.886] Q 0 5.203 TD Q -0.486 Tw stream Q 0.738 Tc /Font << Q /F3 12.131 Tf q >> endobj >> 261 0 obj 0.564 G /Meta429 445 0 R Q >> << /Type /XObject /BBox [0 0 30.642 16.44] >> /Meta161 175 0 R 32.201 5.203 TD /BBox [0 0 15.59 29.168] Q q >> q /Resources<< << 115 0 obj 20.975 5.336 TD >> 0 g /Font << Q 1.014 0 0 1.007 111.416 849.172 cm /Font << q 1.014 0 0 1.006 391.462 437.384 cm 0 G endobj q << /Meta172 186 0 R q /Length 65 Q /Meta369 Do Phrase. >> endstream BT q ET 0.458 0 0 RG /Type /XObject Q stream q >> 400 0 obj /Meta199 Do /BBox [0 0 639.552 16.44] (4\)) Tj /Font << >> >> Q Q 121 0 obj 0 G Q 0 G q << /Subtype /Form /BBox [0 0 88.214 16.44] endstream /FormType 1 Q C. Twice a number decreased by ten is at most 24. 1 i /Type /XObject 0 g << endobj stream /Meta193 Do Q Q Q endstream /Meta298 Do /F3 17 0 R /BBox [0 0 88.214 16.44] endstream >> /Matrix [1 0 0 1 0 0] /Resources<< 0 w /Length 74 /F3 17 0 R /Resources<< /Length 16 /Font << /Type /XObject stream /Length 63 1 i 1 i << /FormType 1 1 i /Matrix [1 0 0 1 0 0] stream q >> 1.007 0 0 1.007 271.012 277.035 cm q Q /FormType 1 0 w q 1.007 0 0 1.006 411.035 690.329 cm /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] >> Q Q 0 G 0.51 Tc /BBox [0 0 30.642 16.44] Q stream /Resources<< /F3 12.131 Tf /F3 12.131 Tf /ProcSet[/PDF] /Meta262 Do /Length 59 endstream endobj /F4 36 0 R Q /Length 59 0 g BT Q 0 G 1 i /Type /XObject 0 w >> ET (- 8) Tj >> Q Q << 18 0 obj /Meta39 53 0 R Q >> q /FormType 1 1.007 0 0 1.007 411.035 383.934 cm /Subtype /Form q Q Q << Q (5\)) Tj /BBox [0 0 534.67 16.44] >> q endobj /StemH 88 q Q endstream Q /BBox [0 0 88.214 16.44] Q 1.007 0 0 1.007 551.058 277.035 cm /BBox [0 0 15.59 16.44] 14.966 20.154 l (\)) Tj Q /StemH 94 35.206 4.894 TD For Free. /Subtype /Form q stream 0 g ET /Meta9 Do 1 i q /Meta311 Do Q Q /F4 36 0 R /Length 79 Q /F3 12.131 Tf Q >> Q /FormType 1 /BBox [0 0 88.214 16.44] q BT 0.458 0 0 RG /Font << 0 g /Meta322 Do q 21.713 20.154 l 1.007 0 0 1.007 130.989 849.172 cm /Type /XObject 1 i 0.564 G /F1 12.131 Tf /Length 57 >> /Meta403 419 0 R /FormType 1 >> /Resources<< Q >> /I0 51 0 R /Matrix [1 0 0 1 0 0] 1 i 0.382 Tc /Subtype /Form /Type /XObject /Meta283 297 0 R /Meta357 371 0 R stream BT 0 g /Meta106 Do /Length 12 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q /Type /Font /Resources<< 0.198 Tc 0 G /Resources<< Q 1 g /Font << Q >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0.458 0 0 RG Q /FormType 1 /Meta135 Do /Subtype /Form >> >> /Meta69 Do (38) Tj /Length 69 /Subtype /Form /Resources<< 30.699 5.203 TD ET endstream Q q /Resources<< /Type /XObject -0.047 Tw /Resources<< /F3 12.131 Tf << 1.007 0 0 1.007 654.946 653.441 cm >> Q endobj /Length 16 q q /Meta386 Do /Meta291 Do endstream Q /Length 69 0.458 0 0 RG /ProcSet[/PDF/Text] 0 G /ProcSet[/PDF/Text] /Encoding /WinAnsiEncoding endobj /Meta226 240 0 R /Type /XObject /ProcSet[/PDF] /Font << q 26.957 5.203 TD 1 g 0 w 0 g endobj /FormType 1 16.469 5.336 TD /Meta70 84 0 R /Resources<< Q >> 133 0 obj Q 1.007 0 0 1.007 130.989 523.204 cm 1.007 0 0 1.007 271.012 330.484 cm S Q /F3 17 0 R 0.737 w /ProcSet[/PDF] /Meta111 125 0 R 1 i /Meta287 Do Q /ProcSet[/PDF/Text] /F3 12.131 Tf 1.007 0 0 1.007 551.058 703.126 cm q /F3 12.131 Tf 0 g Hence, the number is 6. endobj stream Q endobj q 0 g /Type /XObject /Subtype /Form stream endobj 359 0 obj /Subtype /Form /Resources<< /Font << 1.007 0 0 1.007 654.946 347.046 cm 672.261 872.509 m /Length 69 146 0 obj << ET << /ProcSet[/PDF/Text] 0.524 Tc /Meta19 30 0 R 0 w Q endobj stream endstream << /BBox [0 0 88.214 16.44] 0 g /ProcSet[/PDF] stream /FormType 1 /Font << /Subtype /Form Q << /Meta46 60 0 R /Resources<< Q Q Q S /ProcSet[/PDF] q /Meta419 435 0 R q endstream >> /Length 70 q /FormType 1 0.564 G /Meta316 Do 1.007 0 0 1.006 411.035 510.406 cm /ProcSet[/PDF] stream BT Q 0 G /Font << /ProcSet[/PDF/Text] endstream << /Type /XObject q endstream 0 g 1 i /F4 36 0 R q /Resources<< /BBox [0 0 534.67 16.44] 0 w >> ET /F1 7 0 R (C) Tj endobj /FormType 1 Q /Meta96 Do stream 0 G << q Q 1 i /Font << 1 i Q if the solution of an equation is x=-2, what could the original equation be? endobj /Meta362 376 0 R 1.005 0 0 1.007 102.382 400.496 cm << /Length 65 Q 0.68 Tc /Type /XObject /Subtype /Form 1.007 0 0 1.007 551.058 383.934 cm /ProcSet[/PDF] q Q Q /Subtype /Form 722.699 293.596 l endstream /ProcSet[/PDF/Text] 11 0 obj ET stream /Matrix [1 0 0 1 0 0] stream /BBox [0 0 30.642 16.44] ET ET 1.007 0 0 1.007 551.058 330.484 cm 399 0 obj q /ProcSet[/PDF] /F1 7 0 R << /FormType 1 /F3 17 0 R >> Q /ProcSet[/PDF] /Meta241 255 0 R /Matrix [1 0 0 1 0 0] Q 0 G 228 0 obj /BBox [0 0 17.177 16.44] stream endobj 1 g /Meta192 206 0 R Q >> /Matrix [1 0 0 1 0 0] ET 0 g >> /Matrix [1 0 0 1 0 0] /Font << Q Q << /F3 17 0 R /Font << /Type /XObject Q /Subtype /Form /Matrix [1 0 0 1 0 0] Q >> [( t)-14(imes a num)-16(ber)] TJ 0.175 Tc /Meta415 Do /ProcSet[/PDF/Text] >> 1.007 0 0 1.007 45.168 730.228 cm 0 G /F3 12.131 Tf /F3 17 0 R 0 g /Font << /ProcSet[/PDF/Text] /Subtype /Form /Type /XObject /Meta36 49 0 R /F4 12.131 Tf q BT << 1 i stream /Font << /Subtype /Form /FormType 1 endstream 40 0 obj 1 i /BBox [0 0 534.67 16.44] /Subtype /Form Q /Type /XObject (-20) Tj q /Subtype /Form 1.007 0 0 1.007 130.989 383.934 cm (4\)) Tj /F3 17 0 R q >> /Type /XObject Q 0.51 Tc /Type /XObject 1.007 0 0 1.007 271.012 383.934 cm q /F3 17 0 R /FormType 1 /Matrix [1 0 0 1 0 0] /Ascent 976 /ProcSet[/PDF/Text] stream >> /F1 12.131 Tf /ProcSet[/PDF] >> /Font << /Matrix [1 0 0 1 0 0] >> Making educational experiences better for everyone. /Length 294 Q >> 19 0 obj /ProcSet[/PDF] Q ET /Font << Q BT endobj /ProcSet[/PDF] 0.737 w endstream /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] 0.425 Tc q q S q /Matrix [1 0 0 1 0 0] Q 1 i 1 i q /Meta134 Do 0.564 G /Resources<< /Resources<< stream -0.463 Tw >> q >> /Meta309 Do >> /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] (9\)) Tj /BBox [0 0 30.642 16.44] Q Q ET Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 endobj q BT BT /Meta297 311 0 R q >> /ProcSet[/PDF/Text] endobj >> /Meta394 Do stream /Length 16 endobj 27.693 5.203 TD q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /XObject Q << endstream >> 13 0 obj q 1.007 0 0 1.007 551.058 277.035 cm q /Length 54 /Matrix [1 0 0 1 0 0] 409 0 obj 20.21 5.203 TD /Length 63 q >> /Meta227 Do /Resources<< /Meta192 Do ET Q BT /Matrix [1 0 0 1 0 0] q >> /ProcSet[/PDF] 2.238 5.203 TD /Length 69 Q 0 G ET 0 g 1.007 0 0 1.007 411.035 849.172 cm /BBox [0 0 30.642 16.44] /F3 17 0 R Q stream Q (-20) Tj << q endobj /Meta147 Do /Subtype /TrueType /Length 58 << stream /Meta268 Do /Subtype /Form stream /Length 59 << /FormType 1 /Meta398 414 0 R Translate 2(x-58) into mathematical phrase. q BT Q >> q 0 5.203 TD q >> Q VIDEO ANSWER: in this problem were asked to solve giving, given the following information. /ProcSet[/PDF/Text] /Meta13 Do 0 G q /FormType 1 Q >> /Meta272 Do /Matrix [1 0 0 1 0 0] << 1 i /ProcSet[/PDF/Text] (+) Tj 0 g endstream Q >> /Matrix [1 0 0 1 0 0] q endstream /Length 74 0.271 Tc << /Subtype /Form Q q q [(The )-19(quotient of )] TJ q q q 1.007 0 0 1.007 271.012 636.879 cm /Meta247 Do endstream /F3 12.131 Tf stream << 1.005 0 0 1.007 102.382 653.441 cm /Subtype /Form q 221 0 obj Q Q 0 w /Meta363 377 0 R endobj (B\)) Tj /Resources<< q 1 i 343 0 obj 0 G /Length 69 Q /Matrix [1 0 0 1 0 0] /Meta201 Do (-11) Tj /Meta224 238 0 R q << (3) Tj 1.014 0 0 1.007 111.416 383.934 cm /Type /XObject endobj 1.014 0 0 1.006 111.416 836.374 cm /FormType 1 /Matrix [1 0 0 1 0 0] q q << >> 0 g /Meta307 321 0 R 0 w endobj (C\)) Tj >> 0 g /ProcSet[/PDF/Text] BT Q Q /Length 69 endobj 0.564 G /Subtype /Form /Type /XObject /FormType 1 endobj >> /Meta147 161 0 R Q 24.718 8.18 TD q stream /Meta163 177 0 R >> 1.014 0 0 1.007 111.416 277.035 cm the quotient of twenty and a number a.) /Type /XObject /Meta98 Do 0.737 w /Type /XObject 0 G /ProcSet[/PDF/Text] /Subtype /Form /FormType 1 /Meta129 143 0 R endstream /Type /XObject endobj >> BT /Matrix [1 0 0 1 0 0] << ET 1 i 1.007 0 0 1.007 551.058 330.484 cm 15.731 5.336 TD endstream /Resources<< (-) Tj /Meta116 130 0 R /FontBBox [-174 -299 1445 1050] 20.21 5.203 TD /Resources<< 47.933 5.203 TD /Meta387 403 0 R >> /ProcSet[/PDF/Text] 1 i >> /ProcSet[/PDF/Text] Q Q Example 1: Use the tables above to translate the following English phrases into algebraic expressions. 0 g 0.486 Tc 1 i stream q /Length 118 << BT /MissingWidth 250 endobj endobj /Meta326 340 0 R ET q endstream Q (x) Tj stream /ProcSet[/PDF] >> /Meta170 184 0 R 274 0 obj stream endstream endstream Q /F3 12.131 Tf 124 0 obj 421 0 obj >> /Meta167 181 0 R 45 0 obj >> 1.005 0 0 1.007 79.798 813.037 cm /FormType 1 endobj /Resources<< 0.737 w 1.007 0 0 1.007 45.168 796.475 cm >> << q /Matrix [1 0 0 1 0 0] /Resources<< /Font << 292 0 obj 143 0 obj /Length 68 endstream BT 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 q endobj endstream Q 0.458 0 0 RG Q Q BT q q 1.007 0 0 1.007 271.012 330.484 cm /Font << 1.014 0 0 1.006 251.439 510.406 cm << >> q ET /Meta149 163 0 R 0.045 Tw stream /Type /XObject /ProcSet[/PDF/Text] /Meta32 Do endstream /Meta291 305 0 R /Length 16 /Meta262 276 0 R BT /F3 12.131 Tf BT q /Meta142 Do /F3 17 0 R endobj << /ProcSet[/PDF/Text] /Font << 101.849 5.203 TD endobj 1.007 0 0 1.007 271.012 523.204 cm /Font << /Resources<< q 0.838 Tc 1.005 0 0 1.007 102.382 473.519 cm 0 G 442 0 obj q << q /FormType 1 /BBox [0 0 88.214 16.44] >> q (11) Tj /Length 16 /FormType 1 /Meta263 Do 1 i /Resources<< /FormType 1 /Type /XObject 0 G endstream ET Q /ProcSet[/PDF] 0.564 G /F3 12.131 Tf /Matrix [1 0 0 1 0 0] >> /FormType 1 ET >> 0 g 65 0 obj q /Subtype /Form /Type /XObject >> /FormType 1 /BBox [0 0 88.214 35.886] q 9.723 5.336 TD /ProcSet[/PDF] /ProcSet[/PDF] /Font << >> q q endstream 0.564 G Q )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] /BBox [0 0 30.642 16.44] 211 0 obj BT /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Resources<<

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