0 g /Meta190 204 0 R stream 0.241 Tc /BBox [0 0 30.642 16.44] 0 g /Length 16 /BBox [0 0 88.214 35.886] stream /Type /XObject >> 1.007 0 0 1.007 411.035 277.035 cm Q 0 g << /Matrix [1 0 0 1 0 0] /Subtype /Form q q /Meta314 Do q 1 i 0.486 Tc 1 i Answer provided by our tutors. /Font << ET /F4 36 0 R q 1.014 0 0 1.007 531.485 330.484 cm /BBox [0 0 88.214 16.44] Q /ProcSet[/PDF] << /BBox [0 0 88.214 16.44] >> stream /BBox [0 0 88.214 35.886] /Subtype /Form /Resources<< endstream Diabetes, if left untreated, leads to many health complications. /Subtype /Form 25.454 5.203 TD 0.369 Tc endstream Q Q Q endstream 1 i q /Subtype /Form q stream 1 g /Meta109 123 0 R endstream 434 0 obj /Meta289 303 0 R /Meta88 102 0 R BT /Type /XObject /FormType 1 0.737 w 0 G Q q /ProcSet[/PDF/Text] q ET 0.737 w >> 1.014 0 0 1.007 391.462 583.429 cm Q /Type /XObject 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. >> 48 0 obj /Meta337 Do /Meta171 Do endstream (x) Tj 51 0 obj 0 g >> Q /Font << /Matrix [1 0 0 1 0 0] q ET /Type /XObject /Meta186 Do q Q /Meta226 240 0 R /Meta130 144 0 R >> /Type /XObject 96 0 obj >> /ProcSet[/PDF/Text] Q 0.458 0 0 RG q q /FormType 1 /Font << 0 w Q Q 402 0 obj 396 0 obj Q /StemH 94 << /Length 58 /F3 17 0 R >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endstream Q /Meta118 Do endstream /Font << 0 5.203 TD q 1 i 1.005 0 0 1.007 102.382 363.608 cm << 0 G 0.786 Tc /Meta422 438 0 R Q q /Subtype /Form /F3 17 0 R 9.723 5.336 TD endobj /Meta207 221 0 R /Font << q Q /FormType 1 /Matrix [1 0 0 1 0 0] endstream 1 i Q 125 0 obj Q Q BT q /FormType 1 endstream - 9737014. 1 i BT >> /Meta68 Do /Matrix [1 0 0 1 0 0] >> /Length 69 0 5.203 TD 0.564 G endobj /FormType 1 Q Q q 8.985 20.154 l q -0.486 Tw q 20.21 5.203 TD >> /FormType 1 /Meta385 401 0 R Q /Type /XObject >> /Meta369 383 0 R /ProcSet[/PDF/Text] 65 0 obj 1 g /Meta177 191 0 R 0 G /Font << /Matrix [1 0 0 1 0 0] stream 0 g 0.564 G endobj 1.007 0 0 1.007 130.989 523.204 cm /ID [] 0.155 Tc /Type /XObject 24 0 obj endstream 0 G /Subtype /Form Q /FormType 1 << << [(The )-19(quotient of )] TJ /Meta368 Do /Subtype /Form /Matrix [1 0 0 1 0 0] Q Q Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Subtype /Form /ProcSet[/PDF/Text] /Length 58 387 0 obj /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] 112 0 obj << 1.014 0 0 1.007 251.439 383.934 cm /Meta425 Do Q 1.007 0 0 1.007 271.012 849.172 cm /Matrix [1 0 0 1 0 0] /Type /XObject 0 w /ProcSet[/PDF] BT /Type /FontDescriptor q /Resources<< q /FormType 1 /Resources<< -0.056 Tw /BBox [0 0 88.214 16.44] 0 g /Resources<< /Meta372 386 0 R Q Q >> q endstream Q /Meta71 Do 0.458 0 0 RG Q /F3 12.131 Tf Q /Type /XObject /Meta425 441 0 R q q /Subtype /Form 1.014 0 0 1.006 531.485 690.329 cm BT /F3 12.131 Tf /F3 12.131 Tf /FormType 1 /Meta269 Do q /Matrix [1 0 0 1 0 0] /F3 17 0 R /F3 17 0 R endobj /Subtype /Form /Type /XObject /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] BT q Q Tamang sagot sa tanong: 1.) q >> Q 0 G 0.425 Tc 0 5.203 TD /Meta198 212 0 R Q endobj 1.502 7.841 TD Q /BBox [0 0 88.214 16.44] BT 1.005 0 0 1.015 45.168 53.449 cm 0.458 0 0 RG /FormType 1 for the season. 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 /Meta206 220 0 R endstream >> Q q >> 32.201 20.154 l /ProcSet[/PDF] ET 0 g 1.014 0 0 1.007 251.439 636.879 cm 0 G /BBox [0 0 88.214 35.886] /Font << endobj Q /BBox [0 0 88.214 16.44] >> << /Meta311 325 0 R 1.007 0 0 1.007 654.946 599.991 cm /Meta261 275 0 R Q /ProcSet[/PDF/Text] 0 g q 1 i /Type /XObject /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 411.035 636.879 cm endobj q >> << 0 G q /Matrix [1 0 0 1 0 0] /F3 17 0 R If twice a number is decreased by 13, the result is 9. 0 G >> 1 i 0.458 0 0 RG /ProcSet[/PDF/Text] >> /Meta267 281 0 R BT Q BT q << /ProcSet[/PDF] q 0 G /Type /XObject >> /Meta208 Do 0 20.154 m /ProcSet[/PDF/Text] ET /BBox [0 0 17.177 16.44] q 1.502 24.649 TD << q 0 g 1 i /AvgWidth 401 /Leading 349 0 G 1.007 0 0 1.007 271.012 330.484 cm /F3 12.131 Tf Q /BBox [0 0 88.214 16.44] >> /BBox [0 0 15.59 16.44] /F4 12.131 Tf >> /Type /XObject /F1 12.131 Tf /Matrix [1 0 0 1 0 0] endstream 1 i stream /Font << /Type /XObject Q stream 0.564 G 1.005 0 0 1.007 102.382 599.991 cm BT /Subtype /Form stream /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Length 13 /Meta189 203 0 R endstream /Subtype /Form 0 G >> /Type /Page /Matrix [1 0 0 1 0 0] endstream /Type /XObject q /BBox [0 0 534.67 16.44] /Meta351 365 0 R q q Q 0.564 G Q /BBox [0 0 88.214 16.44] stream /Length 118 0.737 w /Matrix [1 0 0 1 0 0] /F3 17 0 R /F3 17 0 R /F3 12.131 Tf 0 g >> /FormType 1 /FormType 1 >> BT /Matrix [1 0 0 1 0 0] >> /F3 12.131 Tf Q stream /F3 17 0 R q /Font << q /Length 58 endobj /F3 17 0 R S endstream << endstream << BT stream /FormType 1 3.742 5.203 TD q q /BBox [0 0 30.642 16.44] 360 0 obj >> /ProcSet[/PDF/Text] q if the solution of an equation is x=-2, what could the original equation be? q Q << 0.564 G /Meta79 Do 1.007 0 0 1.007 271.012 703.126 cm Q endobj /Length 12 (9\)) Tj /FormType 1 135 0 obj /Type /XObject q 0.786 Tc /Meta196 Do /F3 17 0 R /Matrix [1 0 0 1 0 0] << stream endobj /FormType 1 >> << stream /BBox [0 0 88.214 16.44] /Meta350 364 0 R 1.007 0 0 1.007 67.753 347.046 cm 0.51 Tc stream 0 G /Widths [ 500 0 502]>> /F4 36 0 R 150 0 obj << 0 g 174 0 obj 1 i BT /ProcSet[/PDF/Text] 16.469 5.336 TD Q q Q Q /F3 17 0 R /Subtype /Form /Type /XObject /Font << >> /ProcSet[/PDF] /BBox [0 0 15.59 16.44] 0 g stream /F3 12.131 Tf stream >> 1.007 0 0 1.007 45.168 829.599 cm stream endobj q /F3 17 0 R /F1 7 0 R q Q << /F3 12.131 Tf q /Meta122 136 0 R /BBox [0 0 549.552 16.44] Find the length. /Type /XObject /Length 69 /Length 16 /Matrix [1 0 0 1 0 0] 0 g 0 g 0.458 0 0 RG endobj >> /FormType 1 /Meta175 Do /Meta170 Do /FormType 1 /ProcSet[/PDF/Text] 111 0 obj 1 i /Meta110 124 0 R endstream q >> /Matrix [1 0 0 1 0 0] endstream /Font << q /FormType 1 >> q Q /Length 54 194 0 obj So let's go ahead and identify a v Q /Meta67 81 0 R q >> /Matrix [1 0 0 1 0 0] Q /Type /FontDescriptor 0 g >> 289 0 obj 0.458 0 0 RG stream /Font << /Type /XObject /Type /XObject 1 g >> /FormType 1 >> stream 1 i 293 0 obj Q /Resources<< /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] Q /F3 17 0 R /Meta388 Do (x) Tj Q Q Q /Subtype /Form >> Q /Type /XObject 672.261 347.046 m BT >> /Meta344 Do /F3 17 0 R ET /Type /XObject stream 1.007 0 0 1.007 551.058 330.484 cm 0.425 Tc >> Q q /Length 12 /Matrix [1 0 0 1 0 0] endstream stream endobj << << >> /BBox [0 0 88.214 16.44] q /Length 118 /FormType 1 q Q 0.737 w /Meta332 Do /Font << 0.369 Tc Q /Meta116 Do stream Q Q /ProcSet[/PDF/Text] >> 258 0 obj Q /Subtype /Form 0.738 Tc Q 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 1 i ET /Resources<< /FormType 1 /BBox [0 0 15.59 29.168] 0 w 0.458 0 0 RG /Meta95 109 0 R 1 i Q /Length 54 /Resources<< /Subtype /Form Q q q /ProcSet[/PDF] /Font << << 1 i 0.458 0 0 RG endstream /Subtype /Form /Length 16 Q endobj /Meta159 Do /Resources<< /Meta209 223 0 R q 1.007 0 0 1.007 67.753 726.464 cm 0 g Q q >> Q /Meta139 Do /Type /XObject /Length 70 /Meta383 397 0 R Q 0 5.203 TD 1.007 0 0 1.007 271.012 277.035 cm /BBox [0 0 88.214 16.44] 0 G 0 w /Resources<< 1.007 0 0 1.006 551.058 836.374 cm endstream stream Q 1.005 0 0 1.007 102.382 653.441 cm /FormType 1 /F1 7 0 R >> 1 i /Meta21 Do 0 g Twice a number when decreased by 7 gives 45. /F3 12.131 Tf << endstream 134 0 obj /FormType 1 endstream /F1 7 0 R q endstream 424 0 obj /Length 59 0.458 0 0 RG 1 g BT q /Resources<< Q Q /Meta57 Do 14.23 24.649 TD >> /Meta266 Do >> 0.838 Tc 1 i endobj /FormType 1 q q >> 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 Q 1.007 0 0 1.007 551.058 523.204 cm /Type /XObject endobj << 549.694 0 0 16.469 0 -0.0283 cm /Meta350 Do ET /ProcSet[/PDF] Q q >> 1 i endstream /FormType 1 /Resources<< 1.007 0 0 1.007 271.012 776.149 cm 1.007 0 0 1.007 45.168 813.037 cm /Meta151 Do << /Font << Twice a number decreased by 8 gives 58. 0.737 w The width Of a rectangle is 15 cm and the perimeter is 12 cm. >> If a number is 400%, then it is 4 times, the same as 4. /Font << /F3 17 0 R /FormType 1 0 g Q /Meta61 75 0 R q 256 0 obj /Meta105 119 0 R stream /Length 69 /ProcSet[/PDF/Text] << /Meta99 113 0 R /Meta26 Do q 32.201 5.203 TD q /FontBBox [-568 -307 2000 1007] 672.261 653.441 m q >> endobj q /Length 69 q Q /BBox [0 0 30.642 16.44] >> Q stream /FormType 1 << q >> stream /F3 17 0 R << 436 0 obj /Meta307 Do /Meta97 111 0 R Q q /Type /XObject >> /F3 12.131 Tf /Length 63 Q /Resources<< 1.007 0 0 1.007 67.753 473.519 cm 19.474 20.154 l 0.269 Tc /Type /XObject 1.005 0 0 1.007 102.382 616.553 cm /BBox [0 0 639.552 16.44] ET >> /Font << q /F1 12.131 Tf q 0 G q endstream stream >> endstream 1 i Q /Type /XObject /Meta37 50 0 R /Length 54 177 0 obj /Length 69 Q >> /Resources<< /Subtype /Form q BT >> 1.007 0 0 1.007 411.035 583.429 cm /FormType 1 Q >> >> /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 251.439 277.035 cm /F3 12.131 Tf /ProcSet[/PDF] endobj 376 0 obj BT /ProcSet[/PDF] 0 G Q Q /Length 65 << /Subtype /Form q >> Q 1.007 0 0 1.007 411.035 383.934 cm >> /Meta68 82 0 R 395 0 obj q /FormType 1 q /ProcSet[/PDF/Text] stream (+) Tj /Resources<< >> /Meta390 406 0 R stream 1.007 0 0 1.007 551.058 277.035 cm /Matrix [1 0 0 1 0 0] /Length 16 Q /F3 17 0 R Q saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. endobj /Length 54 BT Q /BBox [0 0 88.214 16.44] 189 0 obj /StemV 94 Q /Resources<< /F3 12.131 Tf stream /FormType 1 >> /FormType 1 /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] BT >> stream 1 i 1 i 1.007 0 0 1.007 551.058 636.879 cm 1 i endstream 1.005 0 0 1.007 79.798 746.789 cm /Length 16 /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] BT /Meta78 92 0 R /Meta127 Do endstream Q /Length 16 /Type /XObject 0.738 Tc Q 1 i 13.464 5.203 TD endobj >> 1 g 181 0 obj /Resources<< /Meta19 30 0 R /ProcSet[/PDF/Text] Six subtracted from a number 6. ET << endobj /Length 65 (\)) Tj 0.458 0 0 RG /Subtype /Form q Q 1.005 0 0 1.007 102.382 473.519 cm /Meta107 Do >> /Resources<< endobj << /Subtype /Form >> Q 0.524 Tc Q /FormType 1 1.014 0 0 1.006 251.439 763.351 cm 56 0 obj Q /Matrix [1 0 0 1 0 0] 1.502 8.18 TD << 364 0 obj 1.007 0 0 1.007 130.989 583.429 cm Q Q /ProcSet[/PDF] Q << Q Q (iv) A number exceeds 5 by 3. << 1 i Q endobj q 0.37 Tc Q /ProcSet[/PDF/Text] 0 g 188 0 obj 1 i q q endstream (\)]) Tj /Length 69 q 1.007 0 0 1.007 411.035 277.035 cm endobj Ten divided by a number 5. /Resources<< q q /Length 70 << /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] Q BT /F3 12.131 Tf endobj /Resources<< 1 i 1 i /BBox [0 0 17.177 16.44] endobj 0.458 0 0 RG endobj 0.175 Tc 1 i BT Q /BBox [0 0 15.59 16.44] 0.737 w /ProcSet[/PDF] /Subtype /Form /F3 12.131 Tf 1 i Q 0 G /ProcSet[/PDF/Text] 0.458 0 0 RG >> /Meta158 Do 0 g 27 0 obj /Font << 0 g >> /Matrix [1 0 0 1 0 0] 0 w /F3 12.131 Tf Q /Type /XObject Q /Meta342 Do >> 1 i q >> /Matrix [1 0 0 1 0 0] Q Q (x ) Tj /Type /XObject /Length 16 Q /Resources<< 0 g /StemH 88 << ET Q << /Length 16 0 g 20.21 5.203 TD endobj << q 0 G /Meta377 391 0 R /Subtype /Form /Resources<< 0 g Q /Subtype /Form /ProcSet[/PDF/Text] 0.297 Tc BT /BBox [0 0 30.642 16.44] << /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] w/Honors. 0 g /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 654.946 872.509 cm /Type /Page /Subtype /Form /Meta168 182 0 R /Meta155 169 0 R 20.21 5.203 TD stream /Length 64 q (9) Tj 9.723 5.336 TD 0 g 404 0 obj Q /Meta404 Do ET 1.005 0 0 1.007 102.382 799.486 cm /Matrix [1 0 0 1 0 0] Q /Resources<< 0.458 0 0 RG BT q /Type /XObject q stream >> << Q q /BBox [0 0 17.177 16.44] >> stream q /Font << 2x - y = 6. x + 3y = -25. ET 236 0 obj stream 0.458 0 0 RG ( x) Tj 1.008 0 0 1.007 654.946 293.596 cm endstream endstream /Matrix [1 0 0 1 0 0] 0 5.336 TD 0.564 G >> q << (2\)) Tj 0.564 G /ProcSet[/PDF/Text] << stream /Matrix [1 0 0 1 0 0] /Subtype /Form endobj >> endobj /Matrix [1 0 0 1 0 0] Q endstream /F3 17 0 R /Length 54 1.005 0 0 1.007 79.798 713.666 cm ET /FormType 1 /Matrix [1 0 0 1 0 0] 0.369 Tc A number increased by 5 is equivalent to twice the same number decreased by 7. /Meta13 24 0 R Q q Q stream /Meta141 Do /Length 57 q /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] /Type /XObject << Q /Length 69 /Meta102 116 0 R [(A number )-17(divided by )] TJ /Subtype /Form /Meta19 Do /F3 17 0 R << /MaxWidth 1248 1 i 175 0 obj BT q BT q stream >> endstream Q endobj /Meta329 Do Q S (6\)) Tj /F3 17 0 R Q stream endobj /Meta342 356 0 R stream endobj S /Meta414 430 0 R /FormType 1 /Subtype /Form /F1 12.131 Tf q 0.458 0 0 RG >> /F3 12.131 Tf 1 i 0.297 Tc /Subtype /Form /Resources<< /F4 12.131 Tf << q 0.737 w 1 i 0.458 0 0 RG /Resources<< 6.746 5.203 TD /Matrix [1 0 0 1 0 0] ET /Meta180 194 0 R << Let x the unknown number. >> endobj Q 0.564 G 0 20.154 m >> >> /FormType 1 endstream /Length 59 Q BT 0.737 w endstream stream &K @ 0 G q /Font << 0 G Q Q /FormType 1 q Q /BBox [0 0 534.67 16.44] ET q endobj q << 0.227 Tc stream endstream /FormType 1 Q 0 G /Type /XObject 0 g Q 14.966 20.154 l q << 226 0 obj /Resources<< << endobj /Font << /Subtype /Form 0 w Q << q stream /Matrix [1 0 0 1 0 0] stream stream ET /Resources<< 1 i >> q 0 g endobj /F3 12.131 Tf 0 w Question 1. BT BT /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] Q stream >> /Meta45 59 0 R 0.458 0 0 RG 266 0 obj 1.014 0 0 1.006 391.462 437.384 cm /Length 69 /F1 7 0 R endstream q q /Resources<< >> >> >> /Type /XObject (A\)) Tj If n is "the number," which equation could be used to solve for the number? 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 /BBox [0 0 15.59 29.168] /Length 16 /Resources<< Q /Font << Q Q stream Q /Length 16 Q /Meta401 417 0 R /Length 80 q << ET ET /F3 12.131 Tf /Type /XObject /Font << endobj /F1 12.131 Tf q endobj 1 i 1 i BT >> /FormType 1 332 0 obj /Meta287 301 0 R << endstream 140 0 obj /FormType 1 >> q /Type /XObject >> 0 g ET q q << A. 0 g q q endstream Q /Meta29 Do Q 13.493 5.203 TD q S /Resources<< /F3 12.131 Tf /F1 7 0 R Q endobj 1 g << Use the variable g to represent Gails age. /Type /XObject 1 i stream /Meta91 105 0 R Q stream Q /ProcSet[/PDF] >> Q 1.005 0 0 1.007 79.798 779.913 cm 1 i 0.458 0 0 RG 0 g q 0 g /Meta335 349 0 R 1.005 0 0 1.007 102.382 293.596 cm >> /Meta298 312 0 R /Length 16 /ProcSet[/PDF] If you are unsure of the county, call the Administrative Office of the Court at (919) 890-1000.Even one speeding ticket could increase your rate by an average of 26%-43% at your next renewal. Q q /Meta154 168 0 R endstream stream Untreated or poorly treated diabetes accounts for . /Meta409 Do Q /Font << q 0 g /Subtype /Form Q /Resources<< q /Matrix [1 0 0 1 0 0] /Resources<< Q (-9) Tj /Resources<< Q /Meta140 Do Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. Q stream /ProcSet[/PDF/Text] << endobj q BT /Resources<< BT ET Q /ProcSet[/PDF/Text] /Subtype /Form /FormType 1 Q /Resources<< Q >> q q Two speeding tickets could increase your rate by 58% at your next renewal. q >> q 123 0 obj /Matrix [1 0 0 1 0 0] /Resources<< >> >> BT /BBox [0 0 15.59 29.168] << 0 w >> q q /Meta427 443 0 R 722.699 293.596 l >> Q q >> >> /Type /XObject >> q /Matrix [1 0 0 1 0 0] ET /Matrix [1 0 0 1 0 0] q >> 1 g q /Meta231 245 0 R /Type /XObject /Meta388 404 0 R 0.369 Tc endstream 20/n b.) >> /Matrix [1 0 0 1 0 0] /Resources<< << 164 0 obj /ProcSet[/PDF/Text] /Subtype /Form ( \() Tj >> endstream endobj >> /Meta237 Do /Meta392 Do 0 g /Subtype /Form 1.007 0 0 1.007 130.989 330.484 cm /ProcSet[/PDF] 6. 54.679 5.203 TD /Type /XObject 722.699 653.441 l >> /ItalicAngle 0 /Meta428 Do q 0.564 G 1 i q /Font << >> >> >> q ET /BBox [0 0 88.214 16.44] /Length 70 0 w 24.718 8.18 TD 0 G /Font << >> >> 0 w >> q 1.007 0 0 1.007 130.989 383.934 cm 0 G /ProcSet[/PDF] endstream /ProcSet[/PDF] 0 39.216 TD >> >> /Type /XObject (5\)) Tj (C\)) Tj Q /Length 16 0.838 Tc 1.007 0 0 1.007 67.753 726.464 cm 0 g /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 45.168 779.913 cm << 1.007 0 0 1.007 130.989 849.172 cm Q 1 i endobj /Resources<< /Matrix [1 0 0 1 0 0] /Subtype /Form /XObject << Q /Meta357 371 0 R q /Resources<< stream Q >> /Meta121 135 0 R /F3 17 0 R Q 1.007 0 0 1.007 271.012 450.181 cm endobj << Q 1.005 0 0 1.007 102.382 473.519 cm q ET MetS-Z quartiles and their associated risks are presented in Fig. /FormType 1 >> Q /BBox [0 0 88.214 16.44] /F3 17 0 R /Font << There were x cookies at the beginning of a party. q /Meta99 Do /Type /XObject 0.369 Tc /Meta116 130 0 R >> 0 g /BBox [0 0 88.214 16.44] /Type /XObject 0 5.203 TD Q endobj 1 i /FormType 1 /Meta333 347 0 R >> /ProcSet[/PDF/Text] 0 g 0 g /F3 12.131 Tf >> stream Q Q 1 i /Meta165 179 0 R q /Font << /Matrix [1 0 0 1 0 0] 380 0 obj endobj ET BT 0 g /FormType 1 >> 0 G 0.564 G 0 G 1.007 0 0 1.006 411.035 437.384 cm 0 g 0.458 0 0 RG /Type /XObject 1 i Q q /Meta328 342 0 R /Meta200 Do >> /Matrix [1 0 0 1 0 0] 299 0 obj 1 i 61 0 obj q /Meta8 19 0 R >> q 11.99 8.18 TD >> /BBox [0 0 673.937 68.796] endobj 16.469 5.336 TD /Font << 1 i /Meta21 32 0 R Q q 19.474 5.203 TD q /Meta375 389 0 R q BT /Subtype /Form 1 i 1 i endstream /Type /XObject 1 i 0.68 Tc 400 0 obj /Resources<< /Meta83 97 0 R /Length 65 /Meta0 5 0 R /FormType 1 /Subtype /Image A number = an unknown number which can be represented by a variable, usually x. /XObject << Twice a number decreased by . >> /Matrix [1 0 0 1 0 0] 0 g >> /BBox [0 0 88.214 16.44] << /Type /XObject q /ProcSet[/PDF/Text] 1 g Q 0.737 w /FormType 1 Q 1 i endstream 1.007 0 0 1.007 67.753 872.509 cm 0 G Q 1 i 0.564 G [( and )16(a nu)26(mbe)18(r)] TJ /F3 12.131 Tf 1 i Q 0.458 0 0 RG >> /BBox [0 0 534.67 16.44] endobj /Meta319 333 0 R 1 i 0.297 Tc /Meta410 Do /Resources<< Q /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 391.462 763.351 cm q >> >> >> 400 0 R >> /Matrix [1 0 0 1 0 0] 0 g stream 0 g /BBox [0 0 639.552 16.44] >> /Meta256 Do /Resources<< In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. q /Subtype /Form q 1.007 0 0 1.007 130.989 776.149 cm Q >> 0 G 93 0 obj /Subtype /Form Q q /Font << /Meta67 Do (8\)) Tj q /ProcSet[/PDF/Text] /Font << /ProcSet[/PDF] 1 g /Meta281 Do << /ProcSet[/PDF] endobj Q 0 w 32.201 5.203 TD >> endobj q /Font << /XHeight 476 endstream 0 G /F3 12.131 Tf /Subtype /Form Q /F3 12.131 Tf /Meta77 91 0 R /Meta239 253 0 R /ProcSet[/PDF/Text] << 1 g /Resources<< /Matrix [1 0 0 1 0 0] 321 0 obj Q Q >> stream /Type /XObject /F3 17 0 R 78 0 obj endstream Q /Meta326 Do endstream /Type /XObject /F3 17 0 R /Resources<< /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Type /XObject A. /Matrix [1 0 0 1 0 0] Q endobj /ProcSet[/PDF/Text] /Subtype /Form /FormType 1 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 = 1/2please solve this question. 249 0 obj Q Q >> 1.005 0 0 1.007 79.798 829.599 cm 1 i /F3 17 0 R Q /Subtype /Form q /Font << q Q /Subtype /Form ET /F3 17 0 R (D\)) Tj /Resources<< 0 g 0 g q Q 1 i /Meta379 393 0 R /F4 12.131 Tf ET BT q 0.737 w /FormType 1 /Meta45 Do << q q q 1 g endobj /Meta396 412 0 R >> /Font << /ProcSet[/PDF] /Font << Q >> 0.564 G Q 0 G >> endobj 82 0 obj /F4 36 0 R endobj /Matrix [1 0 0 1 0 0] 0 g /ProcSet[/PDF] Q /F3 12.131 Tf 438 0 obj 1.007 0 0 1.006 130.989 437.384 cm /BBox [0 0 88.214 16.44] 0.564 G Q q Q /Subtype /Form 0 G /Matrix [1 0 0 1 0 0] 1 i /Font << /I0 Do 1 i Q /Length 69
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