twice a number decreased by 58
twice a number decreased by 58
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>> BT 0 20.154 m /F4 12.131 Tf 0.369 Tc /F3 17 0 R stream 45 0 obj >> /F3 17 0 R /Length 69 1 i /FormType 1 /Resources<< /Type /XObject Q (13) Tj /Matrix [1 0 0 1 0 0] >> 0 G Q Q q /Resources<< q >> q 1.005 0 0 1.007 102.382 653.441 cm /BBox [0 0 88.214 16.44] /Font << ( x) Tj 0 G >> endstream stream q 1 i stream endobj 218 0 obj /BBox [0 0 88.214 35.886] ET 0.51 Tc /F3 12.131 Tf Q /Resources<< /Meta66 Do /Matrix [1 0 0 1 0 0] /Meta392 Do q Q 125.064 4.894 TD /Matrix [1 0 0 1 0 0] 0 g 220.931 4.894 TD q endobj Q stream ( \() Tj endstream (6\)) Tj /BBox [0 0 17.177 16.44] >> ET 1 g (-) Tj Q >> /ProcSet[/PDF] /BBox [0 0 639.552 16.44] /Meta6 15 0 R 1.007 0 0 1.006 411.035 690.329 cm /Resources<< >> 0 g >> >> q q q /Subtype /Form 0 w 0 G (11) Tj q /Pages 1 0 R >> /F3 12.131 Tf S /Resources<< /FormType 1 q 1.007 0 0 1.006 130.989 690.329 cm << /Meta105 Do /I0 Do /ProcSet[/PDF/Text] /Resources<< /BBox [0 0 534.67 16.44] /Meta12 Do [2] Twice a number increased by four is twenty-one. /FormType 1 /FormType 1 /BBox [0 0 534.67 16.44] >> /Resources<< 0 G q /F4 36 0 R >> Q 1.007 0 0 1.007 130.989 383.934 cm /Meta323 Do >> endobj [( and )-20(the product of )-15(a number a)-16(nd )] TJ /Meta339 353 0 R q /Resources<< /BBox [0 0 534.67 16.44] endstream /ProcSet[/PDF/Text] /FormType 1 q >> q /LastChar 120 /Length 69 BT >> >> Q /ProcSet[/PDF/Text] 0 g 0 g q q /ProcSet[/PDF] /FormType 1 >> /FormType 1 << stream ET /FormType 1 q endobj /Resources<< 0.737 w 1 g << BT q q endstream /Font << 0 G /BBox [0 0 534.67 16.44] q q q 385 0 obj 0.737 w q /Matrix [1 0 0 1 0 0] >> /Meta347 Do /Meta213 Do 52 0 obj q 0 g /Resources<< /Resources<< /BBox [0 0 88.214 35.886] /Type /XObject /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] << 0 g stream q q endobj [( and )16(a nu)26(mbe)18(r)] TJ endstream 0 g 328 0 obj endstream 0.737 w 0 g >> q >> /Meta155 169 0 R << /Meta122 136 0 R /FormType 1 1.007 0 0 1.006 551.058 437.384 cm /Resources<< 20.21 5.203 TD 1 i ( x) Tj Expert Solution. << /Subtype /Form q ET 26.219 5.336 TD 9.723 5.336 TD /Subtype /Form q << q /Matrix [1 0 0 1 0 0] /Resources<< Q q 57 0 obj endstream /Subtype /Form /Type /XObject /ProcSet[/PDF/Text] q /BBox [0 0 534.67 16.44] q /F3 12.131 Tf Q Q /F3 12.131 Tf /FormType 1 37 0 obj 237 0 obj Q 1 i 0 G /Meta1 8 0 R 15 0 obj /Font << BT /F3 17 0 R /Meta109 123 0 R q /Length 69 /F3 17 0 R /Meta160 174 0 R Q stream /Length 69 /F1 7 0 R 1 i Q /FormType 1 /Type /XObject ET endstream 303 0 obj endstream /ProcSet[/PDF/Text] 197 0 obj /Meta354 368 0 R Q q >> /Subtype /Form q stream Q 0 g 1.007 0 0 1.007 67.753 546.541 cm Q /Font << Q /Subtype /Form ET /Matrix [1 0 0 1 0 0] 0 G >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] q >> /Meta184 Do /Length 16 BT /Subtype /Form /Type /XObject /Meta368 Do 0 w /Font << 19.474 5.203 TD 1 g Q 1 i q Q 0 g 0.297 Tc q /Font << /Meta88 Do 1 0 obj /Meta417 433 0 R Q 1 i /BBox [0 0 30.642 16.44] /Meta400 416 0 R 0 5.203 TD << << endobj << /Meta237 Do endstream (C\)) Tj 1.502 5.203 TD Q 1.007 0 0 1.007 271.012 636.879 cm 242 0 obj /FormType 1 q Q Q q >> Q /Resources<< ET endstream 1.005 0 0 1.007 102.382 599.991 cm /F3 17 0 R 20.21 5.203 TD q >> /ProcSet[/PDF/Text] >> q 6.746 5.203 TD endstream q q 0 g 1.007 0 0 1.007 271.012 583.429 cm 183 0 obj (-8) Tj 0.737 w /Type /XObject >> 0 G endstream 0 G 0.28 Tc /Meta25 Do 29 0 obj 1.007 0 0 1.007 130.989 849.172 cm (\)]) Tj /Resources<< /F3 17 0 R Q Q q 17.234 5.203 TD /ProcSet[/PDF] True False /Meta150 Do 1 i The sum of a number and 2 is 6 less than twice that number. >> /Type /XObject 0.486 Tc /F3 17 0 R 0 g >> 1 i /FormType 1 /FormType 1 /BBox [0 0 88.214 16.44] << /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q /Length 59 1 i q Q /ProcSet[/PDF/Text] q 0.369 Tc (C) Tj q /Resources<< q /F3 17 0 R q << /Type /XObject q /Resources<< Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 1 i /Subtype /Form stream endstream /ProcSet[/PDF/Text] >> /F3 17 0 R Q Q /Matrix [1 0 0 1 0 0] Q >> >> 0.737 w << q /ProcSet[/PDF] /BBox [0 0 534.67 16.44] endstream /ProcSet[/PDF/Text] /Type /XObject /Type /XObject Q /FormType 1 /Type /XObject endstream 1.007 0 0 1.006 551.058 763.351 cm ET >> 0.68 Tc << >> q In addition, testosterone in both sexes is involved in health and well-being . BT Q /Resources<< /Type /XObject q /Meta3 Do (9\)) Tj Q q >> 1 i 1.014 0 0 1.007 391.462 277.035 cm BT /FormType 1 q /Subtype /Form 0 5.203 TD 12.727 5.203 TD stream Q 74 0 obj endobj >> Q endstream Q /Meta338 352 0 R Q All steps. /Type /XObject 1 i /Type /XObject /Subtype /Form Q << 0.737 w /Descent -277 endstream q q /ProcSet[/PDF] /Resources<< BT /Subtype /Form You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 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. << q -0.021 Tw /Subtype /Form /Flags 32 /FormType 1 ET /Subtype /Form Q 1 i endstream q /BBox [0 0 88.214 16.44] /Meta230 Do BT /Length 54 << 124 0 obj stream /Font << 1.005 0 0 1.015 45.168 53.449 cm 150 0 obj endstream >> 0.458 0 0 RG 1.007 0 0 1.007 130.989 383.934 cm /Font << 72 0 obj -0.041 Tw q /Meta200 Do /Matrix [1 0 0 1 0 0] 1 i endobj << 0.458 0 0 RG Q /Subtype /Form 19 0 obj 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/Form q Q 0 5.203 TD Q /Font << q >> 20.21 5.203 TD /Meta14 Do q q /Length 65 167 0 obj BT /Meta179 193 0 R /Type /XObject << q /Type /XObject Q endobj /Type /XObject endobj /Type /XObject /ProcSet[/PDF/Text] /Length 104 stream /F3 17 0 R /ProcSet[/PDF] 0.737 w /BBox [0 0 88.214 16.44] 20.21 5.203 TD /Type /XObject 0.458 0 0 RG 0.458 0 0 RG 0 g q stream stream 0 G 383 0 obj (-11) Tj << >> >> (13) Tj /FormType 1 q /Length 69 /MissingWidth 250 /Resources<< /FormType 1 /BBox [0 0 15.59 29.168] endstream Q /F3 17 0 R stream q >> q 338 0 obj >> /Resources<< /Meta194 208 0 R /Subtype /Form >> q 1.014 0 0 1.007 111.416 776.149 cm BT -0.056 Tw /Meta32 45 0 R Q 0 w /F1 7 0 R Q 101 0 obj /Meta290 Do >> /Filter [/CCITTFaxDecode] << /BBox [0 0 15.59 16.44] ET endstream /Length 12 /F3 12.131 Tf 1 i q q 423 0 obj /F1 12.131 Tf >> 0.564 G /Type /XObject /BBox [0 0 88.214 35.886] 6 0 obj /Meta2 Do stream Q 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 = 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>> (D\)) Tj >> 0 5.336 TD /FormType 1 << Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 Mathematics 619 solutions /Font << q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q >> /Subtype /Form >> 1.007 0 0 1.007 551.058 636.879 cm Q /ProcSet[/PDF] 1 i q >> 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 Q /Subtype /Form q stream Q On the way, we sang songs for 20 minutes which is 10% of the time we were on the road. q /FormType 1 q /Meta138 152 0 R 389 0 obj 1 i /Length 107 q /F1 7 0 R /ProcSet[/PDF/Text] 0.271 Tc 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. /F3 17 0 R stream /FormType 1 stream /Count 2 0 w /Length 16 Q Q /Resources<< /Subtype /Form /Matrix [1 0 0 1 0 0] >> 1 i /Meta169 183 0 R 0 G /F3 17 0 R ET endobj Q 0 g /Meta365 379 0 R 1 i Q << endobj q 1.005 0 0 1.007 102.382 653.441 cm /Matrix [1 0 0 1 0 0] BT Untreated or poorly treated diabetes accounts for . /Font << q >> /Length 59 endstream 0.51 Tc q 136 0 obj /F3 12.131 Tf q 0.564 G Q /Meta155 Do /Font << 295 0 obj 0.564 G q /Subtype /Form /FormType 1 q /FormType 1 Q stream q /F1 7 0 R S Q /BBox [0 0 15.59 16.44] /Length 80 Q q /F2 12.131 Tf q 185.725 5.203 TD /Type /XObject endobj 1 i /F3 17 0 R (2) Tj 350 0 obj /ProcSet[/PDF/Text] 0.458 0 0 RG 1 i stream endstream << (58) Tj Q Q /Font << /Meta167 181 0 R q /Meta250 Do 0 g /Length 59 endobj /Meta86 Do Q 0 g /Meta265 279 0 R 55 0 obj /BBox [0 0 534.67 16.44] ET /F3 17 0 R 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) /FormType 1 endstream 1.007 0 0 1.007 67.753 599.991 cm /Meta265 Do q >> /Subtype /Form q 0 g /FormType 1 /Resources<< Q 1 i 549.694 0 0 16.469 0 -0.0283 cm endstream endstream 0 g 0.486 Tc /ProcSet[/PDF] 0 g << endobj endstream >> 0 g q /Type /XObject 247 0 obj /Font << /BBox [0 0 88.214 16.44] 1 i /F1 7 0 R /Subtype /Form >> q /Subtype /Form /FormType 1 /FontName /TimesNewRomanPSMT Q Q /Meta365 Do /DecodeParms [<> ] Q /Subtype /Form << 1.007 0 0 1.007 67.753 293.596 cm 0.737 w /Type /XObject q q q /BBox [0 0 30.642 16.44] /Meta394 Do /F3 12.131 Tf >> /Meta405 421 0 R /Length 60 0 G 1 i endobj /ProcSet[/PDF/Text] Q q /ProcSet[/PDF/Text] 1.014 0 0 1.006 391.462 510.406 cm /Info 3 0 R 0 w 0 5.203 TD 0.307 Tc 0 w /F3 12.131 Tf >> /F3 17 0 R /XHeight 471 endobj The ratio of a number to fifteen 4. q /F4 12.131 Tf 0 G Q /Meta78 Do Q ET BT q >> endobj 1 i endobj Q 1.005 0 0 1.007 79.798 813.037 cm 1 g q /Meta264 Do >> /Subtype /Form >> >> Q /Meta171 Do /BBox [0 0 88.214 35.886] /Font << /Resources<< 0 G 93 0 obj /Matrix [1 0 0 1 0 0] /Meta113 Do /Resources<< >> 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. /Type /XObject endobj 0 g 0 g /Length 106 (7\)) Tj << /F3 12.131 Tf /F3 12.131 Tf 1 i q 5.98 7.841 TD endobj q Q BT >> (x) Tj 0.458 0 0 RG >> Q /Font << /F3 17 0 R /Type /XObject Q q 1 i /ProcSet[/PDF/Text] >> q endstream Q q 0 g /Matrix [1 0 0 1 0 0] /FormType 1 /F4 12.131 Tf /Meta270 284 0 R /F3 12.131 Tf /F3 17 0 R /ProcSet[/PDF/Text] /FormType 1 >> << /Subtype /Form /Meta273 287 0 R /Type /XObject /Type /XObject 1.007 0 0 1.007 130.989 523.204 cm /Meta59 Do >> Q 0 5.203 TD /Type /XObject 0.737 w >> 1 i /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] >> 1 g endobj Q /Subtype /Form q /Subtype /Form /BBox [0 0 88.214 16.44] Q /Resources<< q 0 G >> BT BT 1 g q /Meta146 160 0 R /Matrix [1 0 0 1 0 0] /Resources<< q Q /Subtype /Form 0 g stream << >> q /Resources<< Q /BBox [0 0 88.214 16.44] stream 1 g >> >> /Meta264 278 0 R Q /F3 12.131 Tf /F4 36 0 R /Matrix [1 0 0 1 0 0] ( \() Tj 106 0 obj Q Q /FormType 1 endobj 0.458 0 0 RG 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ET Q >> /F3 12.131 Tf 3.742 5.203 TD q << stream 1.502 24.649 TD /Matrix [1 0 0 1 0 0] stream /Font << 1 i 0.458 0 0 RG endstream /Matrix [1 0 0 1 0 0] Q endstream q 0 G >> 0 g /Meta149 163 0 R /ProcSet[/PDF] 1 i endobj /Type /XObject /FormType 1 ET << (-23) Tj (B\)) Tj /Length 127 /Meta15 26 0 R 0 g 1.502 5.203 TD 0 5.203 TD q Q /Resources<< q /Meta207 Do 1.007 0 0 1.007 411.035 277.035 cm endstream Q 128 0 obj stream (2) Tj >> 1 i 1.014 0 0 1.007 531.485 583.429 cm 0.369 Tc Q q /Subtype /Form q Q << /Subtype /Form /ProcSet[/PDF] 1.007 0 0 1.007 551.058 636.879 cm Q 103 0 obj 1 i Q stream /Subtype /Form /Subtype /Form 0 g /Matrix [1 0 0 1 0 0] /Meta8 Do 1 i Q BT /Font << >> /Matrix [1 0 0 1 0 0] stream >> >> /Encoding /WinAnsiEncoding << >> /Meta369 Do q q Q 1.007 0 0 1.007 130.989 636.879 cm endstream q >> endstream q 1 i If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. /F3 12.131 Tf /I0 51 0 R 0.271 Tc ET >> 0 G >> /F3 17 0 R /F3 17 0 R endobj /Length 59 q /Matrix [1 0 0 1 0 0] /Resources<< 0.17 Tc /Subtype /Form 1 i 0.458 0 0 RG endobj Q endstream stream /BBox [0 0 17.177 16.44] q /Font << endstream /FormType 1 /Subtype /Form 0 g q 1 i 0 g q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Length 69 Q >> /BBox [0 0 15.59 16.44] (-) Tj /BBox [0 0 30.642 16.44] 11.99 8.18 TD Q >> 0 g /Length 58 q /Meta55 Do 0 w endobj endobj << Q 0 5.203 TD >> /Length 16 >> << /Resources<< /FormType 1 stream >> 1.005 0 0 1.007 79.798 846.161 cm (x ) Tj Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] endobj q q Q (+) Tj 422 0 obj q >> Q q Q << /Meta161 175 0 R 179 0 obj /BBox [0 0 88.214 16.44] << >> /FormType 1 /Meta163 177 0 R Q 0.68 Tc /BBox [0 0 639.552 16.44] /FormType 1 /Length 70 >> /FormType 1 /ProcSet[/PDF/Text] >> Question. >> [(1)-25(0\))] TJ /Meta283 297 0 R /FormType 1 /Length 16 << /Length 54 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 . Q /Length 245 0 G Q /BBox [0 0 88.214 16.44] << 0 G endstream >> endstream /Meta381 Do /Matrix [1 0 0 1 0 0] 0 g 1.007 0 0 1.007 551.058 636.879 cm q >> >> Q /F3 12.131 Tf 32.201 5.203 TD /ProcSet[/PDF/Text] 384 0 obj 0 w /Meta82 96 0 R /Meta174 188 0 R stream q stream /Length 57 0.37 Tc /ProcSet[/PDF/Text] /Subtype /Form /Type /XObject q 0.241 Tc Q Q /FormType 1 << /BBox [0 0 88.214 16.44] >> /Meta180 194 0 R 1 i 0.737 w << q /Meta185 199 0 R 1 i ET endstream 0 g /Meta131 145 0 R /FormType 1 /ProcSet[/PDF/Text] endstream /Meta192 Do 1.007 0 0 1.006 411.035 763.351 cm /Type /XObject /Type /XObject The result is 8 less than 10 times the number. /F3 17 0 R Q stream >> 0 g 1 i /Font << gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. q q /BBox [0 0 534.67 16.44] Q /Matrix [1 0 0 1 0 0] 0.458 0 0 RG q >> /Type /XObject >> /BBox [0 0 15.59 16.44] /Type /XObject /Subtype /Form stream /Length 16 /Meta429 Do 0 g >> Q 0 5.203 TD /Length 73 /Meta304 318 0 R /F3 12.131 Tf q >> >> endobj /Meta25 38 0 R endobj Q 1.005 0 0 1.007 102.382 743.025 cm 1.007 0 0 1.007 130.989 523.204 cm /Meta219 Do Q stream << Q /Subtype /Form /Subtype /Form /Meta339 Do Q >> q q /ProcSet[/PDF] 1.005 0 0 1.007 102.382 799.486 cm >> endstream Q /Length 66 endobj /Meta309 323 0 R /Subtype /Form endobj q /Meta395 411 0 R Q /Subtype /Form 1 i 1.007 0 0 1.006 551.058 437.384 cm stream Q 0 w stream << 0.369 Tc /Length 69 endstream Q /Matrix [1 0 0 1 0 0] /Font << Q /Length 103 0 g ET 280 0 obj /Type /XObject /Meta203 217 0 R 1.007 0 0 1.007 130.989 776.149 cm 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 << Q BT 0 G << 0.68 Tc q /Meta154 Do << /Meta106 Do endobj /BBox [0 0 15.59 16.44] Q endstream 0.458 0 0 RG /FormType 1 /Meta301 315 0 R Q Q /Matrix [1 0 0 1 0 0] q (B) Tj Q 0 g q << endstream /Font << /BBox [0 0 88.214 16.44] /Resources<< stream q >> /Meta393 Do << 54 0 obj 1 i Q q << << Q (D\)) Tj 0 5.203 TD /FormType 1 /Subtype /Form /Meta94 108 0 R 0 g /Length 64 /Meta33 46 0 R >> Answer only. endstream endstream stream 0 g q endobj /Meta328 342 0 R Q 1.007 0 0 1.007 271.012 636.879 cm 1.014 0 0 1.007 111.416 523.204 cm /F1 7 0 R /Resources<< /Length 80 >> 1 i /BBox [0 0 88.214 16.44] 0 G /Matrix [1 0 0 1 0 0] 1 i /Length 69 q -0.463 Tw 20.21 5.203 TD stream /Meta235 Do 0 g 120 0 obj 32.201 5.203 TD 1.007 0 0 1.007 411.035 330.484 cm 0 w 0 G /F3 12.131 Tf (2\)) Tj /Type /XObject 227 0 obj 722.699 599.991 l /BBox [0 0 15.59 16.44] 0 g /Resources<< /Font << /Type /XObject Q Q /Meta279 Do /Font << q /Type /XObject /Font << ET 0.68 Tc q 295.086 4.894 TD /Type /XObject /Meta295 Do 0 5.203 TD (B\)) Tj /Subtype /Form 1.005 0 0 1.007 102.382 599.991 cm q /BBox [0 0 88.214 16.44] 1.007 0 0 1.006 411.035 763.351 cm q endstream << /ProcSet[/PDF] q Q 0 g 1.007 0 0 1.006 130.989 437.384 cm q So we have twice of a mystery number decreased by three, and that is all going to be 31. 1 g /ProcSet[/PDF/Text] 230 0 obj Q 1.007 0 0 1.007 67.753 872.509 cm /Type /XObject Q /Resources<< /Meta280 Do /Resources<< 1.007 0 0 1.007 411.035 583.429 cm << endstream q 1 g ET /F3 17 0 R /F3 17 0 R 0.564 G 1 i /Subtype /Form 1.007 0 0 1.007 130.989 383.934 cm ET /ProcSet[/PDF] 0.369 Tc 1.007 0 0 1.007 271.012 450.181 cm /Meta397 413 0 R BT q /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] endobj >> /BBox [0 0 30.642 16.44] stream /Resources<< q /Type /XObject q /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 0.786 Tc BT q 275 0 obj q /Matrix [1 0 0 1 0 0] << 373 0 obj 0.737 w /Matrix [1 0 0 1 0 0] Q >> 220 0 obj ET << /Subtype /Form q q /Resources<< /Subtype /Form /Meta123 Do /MaxWidth 1397 Q 358 0 obj 1.005 0 0 1.007 102.382 872.509 cm endobj /Contents [399 0 R] q 3.742 5.203 TD 0 G 10.487 5.203 TD << /Subtype /Form /F3 17 0 R endobj 3.742 24.649 TD 1 i /FormType 1 0.737 w /F3 17 0 R q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] >> 0 g q q 1 i /F3 17 0 R Q 0.564 G /Type /XObject << /FormType 1 (-23) Tj endobj /F3 17 0 R 0.68 Tc /Length 69 stream /Meta50 64 0 R stream stream Q >> 1 i /Meta356 Do 1 i >> In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. /ProcSet[/PDF] /ProcSet[/PDF/Text] /Subtype /Form /ProcSet[/PDF] [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /ProcSet[/PDF/Text] endstream /Meta17 Do << endstream /Matrix [1 0 0 1 0 0] /Subtype /Form endstream /Resources<< 1.007 0 0 1.007 411.035 636.879 cm 135 0 obj q /BBox [0 0 88.214 16.44] /Type /XObject endobj q /Matrix [1 0 0 1 0 0] 20.21 5.203 TD 174 0 obj /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q q endobj >> Q q 722.699 799.486 l /FormType 1 /FormType 1 stream /Length 67 /Length 68 endobj Q Q /Matrix [1 0 0 1 0 0] 0.737 w 117 0 obj 1 i /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Meta84 98 0 R 0 w 1.005 0 0 1.007 102.382 400.496 cm 1.007 0 0 1.007 411.035 277.035 cm 193 0 obj ET >> /Font << >> /Type /XObject /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject /ProcSet[/PDF/Text] endobj 1.014 0 0 1.007 391.462 636.879 cm /Font << 0 g 0.737 w 0 g Q Q ET endobj if the solution of an equation is x=-2, what could the original equation be? /F3 12.131 Tf 0.198 Tc 0 G q /FormType 1 endstream >> /Resources<< q /Subtype /Form endobj q 0.738 Tc Q Q 0 G /BBox [0 0 17.177 16.44] 257 0 obj /Type /XObject Q >> 1.005 0 0 1.007 102.382 546.541 cm Q /FormType 1 /Subtype /Form >> Q q /Meta129 Do endstream stream /Subtype /Form 1.502 5.203 TD endobj Q If LtitnS6S . /Matrix [1 0 0 1 0 0] Q /Meta333 Do 433 0 obj q /Matrix [1 0 0 1 0 0] q 0.297 Tc /Type /XObject endstream << >> /Meta117 131 0 R /Meta318 Do /F3 12.131 Tf endstream >> /FormType 1 >> /Type /XObject /Matrix [1 0 0 1 0 0] (C) Tj stream /FormType 1 0 g q endobj ET endstream 314 0 obj 1 i 57.656 5.203 TD BT BT /FormType 1 q 1 i /Matrix [1 0 0 1 0 0] /Font << /Length 16 q ET >> 1.014 0 0 1.006 391.462 763.351 cm /Meta361 Do 434 0 obj q How many points did Kobe score in the season? Q endobj >> >> ET q /Subtype /Form /ProcSet[/PDF/Text] /Length 69 /Length 64 /F3 12.131 Tf BT BT (+) Tj /Matrix [1 0 0 1 0 0] 1 i /Meta23 Do stream q q q /I0 51 0 R /Font << 0 g /F3 17 0 R q 0.564 G /Type /XObject 24.718 8.18 TD 1.005 0 0 1.007 102.382 670.003 cm /FormType 1 Q >> Q Q /Meta423 439 0 R 0.51 Tc << /Subtype /Form /F3 12.131 Tf >> Q endstream /Meta169 Do << >> /Meta310 324 0 R Q /Meta160 Do 1.007 0 0 1.007 271.012 330.484 cm /Meta62 Do >> endstream (D\)) Tj /Type /XObject q 0 5.203 TD ET 8 0 obj /FormType 1 Q q startxref 1 i 0.564 G /ProcSet[/PDF/Text] /F3 12.131 Tf << >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 1.007 0 0 1.006 551.058 763.351 cm endstream /Font << << q 3.742 5.203 TD /Meta381 395 0 R >> S /FormType 1 1 i q >> Q /Meta379 393 0 R 1.008 0 0 1.007 654.946 293.596 cm /Subtype /Form /Length 69 /Meta244 Do (-20) Tj /MissingWidth 250 0 G >> /Subtype /Form /Subtype /Form Q 274 0 obj /ProcSet[/PDF/Text] q /Length 16 /Subtype /Form 1 g B. 1 g q endobj endobj q stream stream Q /Meta120 Do /Type /XObject 62 0 obj 0 g 549.694 0 0 16.469 0 -0.0283 cm q >> /Subtype /Form Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . /F3 17 0 R 1.007 0 0 1.007 45.168 862.723 cm >> 1 i 0 g Q /Font << /Length 88 >> >> q >> /F3 17 0 R endstream 430 0 obj >> BT q ET /Type /XObject >> /BBox [0 0 15.59 16.44] /F3 17 0 R >> 204 0 obj Q q /Type /XObject stream q 0 5.203 TD /Resources<< /Subtype /Form /Matrix [1 0 0 1 0 0] /Type /XObject endstream Q /Meta219 233 0 R Q << /Length 245 Q 1 i 1.502 8.18 TD 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 0 G /BBox [0 0 15.59 16.44] /Meta348 Do << /Meta334 Do /Matrix [1 0 0 1 0 0] /Meta241 255 0 R Q >> q /Subtype /Form 1.005 0 0 1.007 102.382 726.464 cm /Meta357 371 0 R /Resources<< stream Q /Matrix [1 0 0 1 0 0] /Font << /ProcSet[/PDF/Text] /FormType 1 endobj endobj 1.014 0 0 1.006 391.462 836.374 cm ET /Font << /FormType 1 Q /Meta302 316 0 R endobj /Meta125 Do /MaxWidth 1248 << 1.007 0 0 1.007 67.753 599.991 cm /Meta362 376 0 R >> Q 1.502 5.203 TD 0.564 G q endobj endstream 0 g /ProcSet[/PDF/Text] /F3 17 0 R /BBox [0 0 15.59 16.44] >> /Meta358 372 0 R 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. Q endstream 1 i BT /F3 17 0 R 0 w endobj Q /F3 17 0 R << /Meta200 214 0 R /Matrix [1 0 0 1 0 0] q This s problem could be, interpreted either way. /ProcSet[/PDF/Text] /ProcSet[/PDF] /Meta18 29 0 R /BBox [0 0 534.67 16.44] 1.007 0 0 1.007 271.012 849.172 cm /Resources<< >> q q /ProcSet[/PDF/Text] ( \() Tj 1 g q /Meta153 167 0 R Q /Length 68 q q Q 1.014 0 0 1.007 531.485 330.484 cm Q 87 0 obj /F3 17 0 R (8\)) Tj 1.007 0 0 1.007 130.989 383.934 cm 0 w /Type /XObject >> /FormType 1 0 w /F3 17 0 R 0.564 G << >> 0.564 G ET 403 0 obj >> q << 0 g 1 g /FormType 1 /F3 17 0 R >> q /Meta358 Do Q q Making educational experiences better for everyone. 2.238 5.203 TD Q endstream /ProcSet[/PDF] << 0 g /Matrix [1 0 0 1 0 0] << ET /Length 60 /Meta281 Do q 0 g 0 g endobj Q 3.742 5.203 TD >> 77 0 obj Q q << >> /Resources<< stream 22 0 obj Q /Length 80 endobj /F3 17 0 R /FormType 1 /Length 69 /F4 36 0 R q , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. >> /Resources<< /Meta412 Do q S 0 w q 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twice a number decreased by 58