/BBox [0 0 639.552 16.44] endobj 1 i /F4 12.131 Tf /Length 58 Q 722.699 293.596 l q Q /Meta180 194 0 R 1.005 0 0 1.007 102.382 653.441 cm q Q endstream Q >> stream /Meta58 72 0 R endstream /F3 17 0 R stream 0.564 G /Length 118 0 G /Type /XObject /Matrix [1 0 0 1 0 0] BT /Subtype /Form /Length 59 /Type /XObject 0.51 Tc 1 g endstream /Type /XObject 1 i Q 95 0 obj Q /Font << /Font << /Length 69 1 i /F3 17 0 R endobj >> /FormType 1 q /Subtype /Form >> Q /F3 12.131 Tf ET /Subtype /Form (C) Tj 0 g 362 0 obj stream 1 i /Type /XObject Q /Resources<< endstream /ProcSet[/PDF/Text] 0.737 w 350 0 obj >> 1 i /FormType 1 stream q endstream >> Q /ProcSet[/PDF/Text] stream Q endobj 0.458 0 0 RG endstream /Meta113 127 0 R >> >> >> 94.364 5.203 TD >> /Meta57 71 0 R << Q << /Matrix [1 0 0 1 0 0] Q /Meta141 155 0 R Q /ProcSet[/PDF] /F3 17 0 R endobj 1 i endobj /ProcSet[/PDF] 6.746 5.336 TD /BBox [0 0 88.214 16.44] Q /Meta7 18 0 R Q endstream q << >> 1.007 0 0 1.007 654.946 799.486 cm endobj 0 g /Resources<< 0.737 w q BT >> /Resources<< ET /Resources<< /Meta171 185 0 R /F3 12.131 Tf 26.219 5.336 TD 0 g /Matrix [1 0 0 1 0 0] /Resources<< 0 g Q Q 1.007 0 0 1.006 130.989 437.384 cm q /Subtype /Form /Meta216 Do Q /FormType 1 Q 1.014 0 0 1.007 531.485 776.149 cm 16 0 obj stream /StemV 88 1.007 0 0 1.007 67.753 653.441 cm 1 i /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 391.462 450.181 cm >> /BBox [0 0 88.214 16.44] endstream >> /FormType 1 0.564 G 1 g 0.486 Tc q Q 1 i ( x) Tj /Subtype /Form You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. Q /ProcSet[/PDF/Text] 0.564 G endobj Q /Length 59 /FormType 1 /Length 69 0 G /F3 12.131 Tf /Meta388 Do /Type /XObject q /Type /XObject ET endobj >> /Meta181 195 0 R endobj /Length 54 >> >> /Resources<< Q /Size 447 Q 1 i Q 25 0 obj >> q BT 2x - 15 = -27. /BBox [0 0 17.177 16.44] 91 0 obj (D) Tj 306 0 obj ET q Decreased by another number means subtract. /FormType 1 stream /F3 12.131 Tf 0.458 0 0 RG 1.007 0 0 1.007 551.058 523.204 cm /Font << /Resources<< Q << 1 i >> q q >> endobj /Subtype /Form endobj << 1 g /Font << /FormType 1 0 g 0 G stream 0.369 Tc 294 0 obj 1 g 21 0 obj endobj /BBox [0 0 30.642 16.44] /ProcSet[/PDF/Text] Q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q Q Q (A\)) Tj /Length 88 Q q endobj 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 endstream /Meta188 Do BT 0.564 G Q 1 i Q q Q stream 184 0 obj Q 1.007 0 0 1.007 551.058 703.126 cm stream stream 0 G q /Resources<< This site is using cookies under cookie policy . Q 311 0 obj >> /Type /XObject 0.564 G 326 0 obj /Font << /BBox [0 0 15.59 16.44] /Length 64 /BBox [0 0 88.214 16.44] endobj q 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 BT BT /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 q /FormType 1 /F3 12.131 Tf /Subtype /Form Q /Meta80 Do 340 0 obj >> q /Matrix [1 0 0 1 0 0] 0 g Q /BBox [0 0 88.214 16.44] 0 g /F4 36 0 R 18.708 17.593 TD >> q >> >> q 1.005 0 0 1.007 79.798 746.789 cm q Q /Meta210 Do /Type /XObject /Resources<< q (4\)) Tj Q 166 0 obj /Resources<< /FormType 1 /Meta21 32 0 R q << Q 32.201 5.203 TD >> >> endstream << /Type /XObject 1 i /BBox [0 0 15.59 29.168] ( \() Tj /F3 17 0 R /Resources<< /Type /XObject /Subtype /Form << /F3 17 0 R 1 g 1 i /F1 7 0 R /Meta336 Do BT 16.469 5.336 TD Q 0.737 w Q 1 i 257 0 obj 0.564 G /BBox [0 0 15.59 16.44] q Q /Type /XObject endobj /Resources<< /Type /XObject >> /Type /XObject 1 i 187 0 obj 0 g 1.007 0 0 1.007 271.012 703.126 cm 277 0 obj endstream /Meta189 203 0 R /Length 69 /Subtype /Form /Meta242 Do /ProcSet[/PDF/Text] (40) Tj >> /Length 16 /BBox [0 0 15.59 29.168] q stream stream 1.007 0 0 1.007 271.012 523.204 cm /Meta14 Do q 0 g /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] /Type /XObject /F3 17 0 R endobj Q /ProcSet[/PDF] endobj Q 1 g /Font << /Subtype /Form Q Q >> /Type /XObject 0 w /FormType 1 q 0.51 Tc /Matrix [1 0 0 1 0 0] /Resources<< Q 0 g /Resources<< 64 0 obj << /Meta278 Do /Length 12 q >> /FormType 1 /Type /XObject >> /F3 17 0 R 1.007 0 0 1.007 67.753 726.464 cm Q 93 0 obj 363 0 obj >> 1 g << 0 g << 149 0 obj endstream /BBox [0 0 15.59 16.44] q q 205.199 4.894 TD q endobj Q /Matrix [1 0 0 1 0 0] /Meta243 257 0 R q endobj /Subtype /Form stream q (-4) Tj << endobj 1. 0 g >> 0 G ET q 0 w /BBox [0 0 88.214 16.44] 0.838 Tc endobj /Length 118 0.458 0 0 RG /FormType 1 /Length 54 /Meta271 Do /F3 17 0 R endstream 82 0 obj >> endobj /F3 17 0 R /Resources<< q q /Resources<< Q /FormType 1 (-20) Tj (C\)) Tj /Subtype /Form 2.238 5.203 TD Q 0.564 G endstream /Meta305 Do ET /Resources<< q /F4 12.131 Tf Q q 0.524 Tc 1 g (-11) Tj >> /Type /XObject >> >> BT q q << /ProcSet[/PDF/Text] /F4 36 0 R /Length 70 q /Meta135 Do q 0 g >> Q q /ProcSet[/PDF/Text] Q ET >> << 1 i 0 G /F3 12.131 Tf q 1 i /BBox [0 0 15.59 16.44] >> ET 1.014 0 0 1.007 531.485 849.172 cm endstream 1.007 0 0 1.007 130.989 636.879 cm /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 599.991 cm /Type /XObject Five times a number, decreased by 58, is -23 Find the number. BT q /Type /XObject /Matrix [1 0 0 1 0 0] /Font << 722.699 799.486 l stream 23 0 obj 0 G /Font << %PDF-1.4 /ProcSet[/PDF] q Q /FormType 1 1 i >> 5 0 obj 1.007 0 0 1.007 271.012 277.035 cm 0 G 11.99 24.649 TD endstream 0.369 Tc endobj (11) Tj /Type /XObject Q q << 1.007 0 0 1.007 130.989 583.429 cm 1.007 0 0 1.007 130.989 523.204 cm stream /I0 51 0 R 0 g 1 i If a number is 50%, then it is a half - the same as 0.5 or 1/2. q /Resources<< /Meta188 202 0 R /Meta237 251 0 R 0 g 0 g /Subtype /Form >> Q << 0 G /Type /XObject /Subtype /Form Q 0 20.154 m Q /Meta331 345 0 R /Subtype /Form 17.234 5.203 TD BT /Length 59 1.005 0 0 1.007 102.382 726.464 cm endobj >> /Matrix [1 0 0 1 0 0] endstream /FormType 1 q 1 g If twice a number is decreased by 13, the result is 9. >> /Matrix [1 0 0 1 0 0] 0 g 0.369 Tc 0 G /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] q 0.564 G 1.007 0 0 1.007 551.058 636.879 cm >> /Resources<< 1.014 0 0 1.007 391.462 523.204 cm endstream q q /Resources<< 1 i /Subtype /Form >> Q >> 1.007 0 0 1.007 654.946 400.496 cm 0 G >> 1 g q >> /Type /XObject -0.047 Tw >> 1 g 237 0 obj >> 89 0 obj endstream /Meta253 Do /Length 69 /Length 16 Q /Matrix [1 0 0 1 0 0] (-9) Tj /Subtype /Form endobj 0 g /Subtype /Form 0.564 G /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta89 Do /Meta397 Do /F3 12.131 Tf q Q 305 0 obj endstream >> Q /BBox [0 0 534.67 16.44] /Resources<< 0.458 0 0 RG Q q /Matrix [1 0 0 1 0 0] q /Resources<< /Matrix [1 0 0 1 0 0] /Meta140 Do endstream >> stream 52 0 obj /FormType 1 Q Q /Meta128 142 0 R q /Length 59 Q q 1 g (x) Tj /Length 69 /Meta336 350 0 R (x ) Tj >> Q /BBox [0 0 534.67 16.44] Q /Flags 32 /Resources<< 0 g 0.737 w ET /Meta337 351 0 R Q 0.458 0 0 RG endstream /BBox [0 0 88.214 16.44] /Meta295 309 0 R 0 G << >> stream Q /Subtype /Form 0 g /FormType 1 19.474 20.154 l /BBox [0 0 534.67 16.44] /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] stream Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. /FormType 1 Q 1.007 0 0 1.007 551.058 583.429 cm /Type /XObject q 0.296 Tc ET >> /BBox [0 0 88.214 16.44] ET 0.458 0 0 RG endstream 0.369 Tc Q stream /Meta170 Do Q Q endstream 1 i q Q >> 1.005 0 0 1.007 102.382 599.991 cm /BBox [0 0 88.214 16.44] endstream /Meta191 Do Q endstream q 1 i 8.985 20.154 l q Q q 0.001 Tw /Subtype /Form stream /Resources<< 1 i endobj q >> /ProcSet[/PDF] 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. /ProcSet[/PDF/Text] BT /FormType 1 1.007 0 0 1.007 654.946 599.991 cm /F3 12.131 Tf 1 i BT q ET /Length 59 /Font << 1 i q /Length 12 0 5.203 TD /Widths [ 500 0 502]>> q 47 0 obj /Meta374 Do /FormType 1 /Length 107 0.68 Tc << /FormType 1 >> 1 i 1 i 1 i 1 i /FormType 1 /ProcSet[/PDF] /ProcSet[/PDF] q /ProcSet[/PDF] >> endstream ([x ) Tj q 0 G stream endobj /Resources<< Q << /Meta172 Do /BBox [0 0 88.214 16.44] /Type /XObject >> 45 0 obj /Subtype /Form Q >> /Subtype /Form stream 3.742 5.203 TD q /Subtype /Form 0 5.203 TD /Info 3 0 R q /F1 12.131 Tf /Meta265 279 0 R 0 5.203 TD 1 i 1.014 0 0 1.006 251.439 763.351 cm 1 i endobj Q q /Meta387 403 0 R >> endobj q /ProcSet[/PDF] /Resources<< q /FormType 1 402 0 obj q /Type /XObject 0.737 w Q /Matrix [1 0 0 1 0 0] >> Q /Matrix [1 0 0 1 0 0] 100 0 obj endstream /BBox [0 0 88.214 16.44] >> /Font << >> 1 i /F3 12.131 Tf /Meta117 131 0 R 151 0 obj q /Type /XObject /Meta228 242 0 R Solution: Let the number be x. q 1 i 1.007 0 0 1.007 551.058 703.126 cm endstream /Type /XObject /Subtype /Form /Matrix [1 0 0 1 0 0] 1 i >> /F3 12.131 Tf /FormType 1 BT >> /F3 12.131 Tf 0.369 Tc >> stream 0.564 G /Meta158 172 0 R /Matrix [1 0 0 1 0 0] Q endstream ET q /ProcSet[/PDF/Text] ET /Length 16 Q ET 0 g endobj 0.486 Tc q q q 0 g /Subtype /Form /FormType 1 q /Type /XObject ( x) Tj (B\)) Tj /F3 17 0 R /FormType 1 0.458 0 0 RG /Meta368 382 0 R 1.007 0 0 1.007 130.989 277.035 cm Q /Matrix [1 0 0 1 0 0] Q /Font << 0 g /Font << << A number divided by six is eight: (k / 6) = 8. /Subtype /Form S q q Number Outcomes 1 42 2 41 3 . q /ProcSet[/PDF/Text] /Resources<< q (C\)) Tj /F1 12.131 Tf endstream 0.271 Tc /BBox [0 0 88.214 16.44] 0 g S q 1 i << /Length 69 >> endstream 1.007 0 0 1.007 67.753 473.519 cm 1.014 0 0 1.007 531.485 583.429 cm q /Meta257 271 0 R /Meta279 Do stream endobj q 0.564 G 0 G /F3 12.131 Tf endobj The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. 0.458 0 0 RG stream stream /FormType 1 /Resources<< q 1 g Q Q 27.693 5.203 TD q /FormType 1 0 G q /Resources<< Q /Meta40 54 0 R /Font << 112 0 obj 113 0 obj 0 g stream 0 w q /ProcSet[/PDF] /Font << endobj >> endobj /Meta406 422 0 R 0 w 65.906 4.894 TD /Matrix [1 0 0 1 0 0] /Length 16 380 0 obj 0.737 w q 1 i /ProcSet[/PDF] /BBox [0 0 88.214 16.44] /Meta285 299 0 R /Matrix [1 0 0 1 0 0] stream /Meta412 Do ET >> /ProcSet[/PDF/Text] /Length 70 Q a Question ET 0.564 G /Meta105 119 0 R Q A number increased by 5 is equivalent to twice the same number decreased by 7. q endobj endobj q 1.005 0 0 1.007 102.382 400.496 cm /F3 12.131 Tf /FormType 1 1.005 0 0 1.007 102.382 599.991 cm /Type /XObject /Subtype /Form /BBox [0 0 15.59 16.44] /Type /XObject /Meta89 103 0 R q >> 1.014 0 0 1.007 531.485 277.035 cm >> << /F3 17 0 R 1 g q >> /Meta3 Do /Meta71 Do /Subtype /Form Q >> /F3 12.131 Tf >> /FormType 1 /BBox [0 0 30.642 16.44] 0.198 Tc /FormType 1 0.155 Tc q endstream /Meta233 247 0 R >> /Resources<< -0.056 Tw 0.564 G much as how 8, Last . q /BBox [0 0 534.67 16.44] Q endobj 0 g << /Matrix [1 0 0 1 0 0] /Font << /Meta417 Do /Meta194 Do /Resources<< >> 672.261 599.991 m /Meta65 Do /F3 17 0 R >> /ProcSet[/PDF] /I0 Do /Matrix [1 0 0 1 0 0] /FormType 1 0.369 Tc /Meta138 Do << q 68 0 obj /Meta235 249 0 R 333.269 5.488 TD endstream 0 g /Encoding /WinAnsiEncoding >> << 0000000000 65535 f
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trailer /F4 36 0 R Q << << 333 0 obj /Subtype /Form Q Q /Meta276 290 0 R 1 i /BBox [0 0 88.214 16.44] /Filter [/CCITTFaxDecode] Q q Q /Length 59 2.238 5.203 TD >> /FormType 1 >> /Type /XObject /Resources<< /Meta146 160 0 R endobj >> 0.786 Tc /FontName /TestGen-Regular /BBox [0 0 88.214 16.44] 0.564 G /Meta104 118 0 R ET q 0.738 Tc stream >> /Type /XObject 0.737 w /Subtype /Form /Meta134 148 0 R /Matrix [1 0 0 1 0 0] endstream BT /Matrix [1 0 0 1 0 0] Q /Font << >> /Length 118 >> /BBox [0 0 88.214 16.44] 188 0 obj Q /Subtype /TrueType If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. 0.51 Tc BT q << /FormType 1 stream 1 g /Meta288 Do /F3 17 0 R >> Q 1 i /Meta192 Do /Type /XObject 0 g /Type /XObject /Font << endstream /FormType 1 1.005 0 0 1.007 102.382 653.441 cm Q /FormType 1 /Meta221 235 0 R /F1 12.131 Tf endobj endobj 0.486 Tc 2 times x minus 58 C. twice the difference of a number and 5 B. twice a number decreased by 58 D. 2 times the sum of a number and 58 Answer: B. Step-by-step explanation: twice - (2) number - (x) 58-(58) Edukasyon. q S 0.458 0 0 RG Q endobj 1 i /Meta128 Do 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . 0 g /Meta358 372 0 R /Length 12 /F3 17 0 R /Subtype /Form 1.014 0 0 1.007 251.439 703.126 cm endobj /Meta225 Do Q /Resources<< /F3 17 0 R /F1 12.131 Tf q /BBox [0 0 673.937 15.562] 1 i q 0 G /Length 79 /FormType 1 /Subtype /Form ( x) Tj Q 672.261 872.509 m /BBox [0 0 88.214 16.44] /Type /XObject endobj << /BBox [0 0 88.214 16.44] q BT endobj /Resources<< This gives us: "2x+5". /Font << endstream 1.007 0 0 1.006 130.989 437.384 cm endobj 426 0 obj Q /F3 12.131 Tf stream 0.458 0 0 RG q /Matrix [1 0 0 1 0 0] >> q endstream /ProcSet[/PDF/Text] Q 0 g 1 g BT /Meta41 Do q 1 i q 0 G BT 1.014 0 0 1.007 251.439 849.172 cm stream 1.007 0 0 1.007 411.035 330.484 cm endstream << /Meta204 Do 1.014 0 0 1.007 251.439 383.934 cm q q /ProcSet[/PDF/Text] 0.51 Tc stream (-) Tj ET 1 i /BBox [0 0 88.214 16.44] >> /Font << Q /ProcSet[/PDF/Text] /Meta116 Do [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ 178 0 obj 15.731 5.336 TD /Font << Q /F3 12.131 Tf 0 G q /Resources<< Q << /Resources<< >> 1 i q stream stream Q 0.564 G /Length 69 q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /FormType 1 >> 194 0 obj 1 i q endstream /Meta177 Do q /Subtype /Form /FormType 1 << 23.216 5.203 TD 1.005 0 0 1.007 102.382 473.519 cm 0 g q endobj 1 g 0 g 0 w 1.007 0 0 1.006 130.989 690.329 cm Q endobj 18 0 obj 1 g /Meta293 Do Q /FormType 1 >> 267 0 obj q /Matrix [1 0 0 1 0 0] 0.564 G Q 1 g Q endobj /Length 69 BT /Meta250 264 0 R 1.007 0 0 1.007 45.168 763.351 cm /ProcSet[/PDF] /Subtype /Form BT /FormType 1 endobj 425 0 obj 0 G -0.486 Tw /Type /XObject /Font << endobj Q Q q Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio 0 G /Type /XObject /Type /XObject q >> /F3 12.131 Tf /F4 12.131 Tf 1 i BT stream (-8) Tj 301 0 obj /FormType 1 1.502 8.18 TD /BBox [0 0 88.214 16.44] 1 g /Font << /Subtype /Form /Resources<< 0 g /Type /XObject /Meta375 389 0 R /Subtype /Form stream Q /Resources<< /Subtype /Form q 1.007 0 0 1.007 411.035 277.035 cm q Q >> stream /Resources<< /FormType 1 1.005 0 0 1.007 102.382 799.486 cm 1.502 5.203 TD 1 i /Length 16 0 g /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] (B) Tj Q /BBox [0 0 88.214 16.44] >> 445 0 obj 0.737 w 0 g 1 i /Subtype /Form Q 0 g q /Meta323 Do 1 i ET 19.474 20.154 l q 0.369 Tc q << 1 i endstream 0.737 w q q 0.458 0 0 RG q BT /Length 12 /Meta225 239 0 R /Resources<< /Subtype /Form /Matrix [1 0 0 1 0 0] >> /F1 7 0 R /Meta286 300 0 R Q endobj Q /FormType 1 /Meta67 Do Q TJ /Meta196 Do /Type /XObject /F3 17 0 R