q q 549.694 0 0 16.469 0 -0.0283 cm >> /Length 58 1 i stream >> 0.241 Tc 1 i >> 38 0 obj /BaseFont /PalatinoLinotype-Bold /Font << /BBox [0 0 30.642 16.44] [( a )-15(number, decreased by )] TJ 1 i Q 1 i MetS-Z quartiles and their associated risks are presented in Fig. /Subtype /Form If mario jumps 3 times and luigi jumps 62 times. 178 0 obj /Meta425 441 0 R endstream /Type /XObject stream /Type /XObject Q stream /FormType 1 << /F3 12.131 Tf /Meta14 Do q endobj /F3 17 0 R 0.564 G q q ET /Subtype /Form Q /BBox [0 0 15.59 29.168] q Q endstream q /Resources<< Q /Resources<< /Subtype /Form /Meta272 Do /Subtype /Form /Resources<< /F3 17 0 R endobj 228 0 obj /ProcSet[/PDF/Text] BT 1 i 1 i Q q 1 i 1.007 0 0 1.007 67.753 726.464 cm /Meta221 Do q Q q Q /Type /XObject << /FormType 1 Q endstream 1 i So let's go ahead and identify a v >> >> Q q /ProcSet[/PDF/Text] ET /Type /XObject 0.486 Tc /ProcSet[/PDF] 1 g 0.425 Tc >> ( x) Tj 1 i /FormType 1 >> /Matrix [1 0 0 1 0 0] 1 i ET 0 G Q /Meta345 Do 0 g >> /Meta133 Do << /Resources<< 1.007 0 0 1.007 130.989 383.934 cm (40) Tj /F3 12.131 Tf 1 i >> q 0 G /F1 7 0 R q /FormType 1 /Matrix [1 0 0 1 0 0] /F3 17 0 R 1.007 0 0 1.007 130.989 636.879 cm Q q 0 G Q 1.007 0 0 1.007 130.989 277.035 cm endstream 0 G 0.369 Tc q 0 g 0 G << stream Q 1.014 0 0 1.007 391.462 636.879 cm endstream /Length 54 q 1.007 0 0 1.007 551.058 703.126 cm 1.014 0 0 1.006 391.462 690.329 cm endstream << stream q /Subtype /Form 0 G /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 0.458 0 0 RG /Length 59 endstream 1 i 0.227 Tc Q /Meta281 Do BT 1 i q There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. >> endstream 0 G /FormType 1 /F3 17 0 R Q /ProcSet[/PDF/Text] /F3 12.131 Tf 436 0 obj (38) Tj q /Subtype /Form 0 20.154 m 1 i /Resources<< /Subtype /Form /Subtype /Form 101 0 obj /Length 60 >> << (D\)) Tj /BBox [0 0 549.552 16.44] >> 1 g endstream q q 0 g Q /BBox [0 0 15.59 16.44] q Q << Q 0.175 Tc /Type /XObject /Type /XObject /Meta336 Do endobj endobj << 8 0 obj 0 g /Length 68 /Subtype /Form /Font << endstream >> /Length 70 0 w 1 i endstream 111 0 obj /Length 80 BT endstream /BBox [0 0 639.552 16.44] << Q 1.007 0 0 1.007 411.035 583.429 cm << /BBox [0 0 88.214 16.44] /Meta45 59 0 R Q 0.737 w endstream 0 G Q q /Length 59 >> q Q /Meta68 Do q /ProcSet[/PDF] /Meta373 387 0 R << /Meta183 Do /Length 80 Q &K @ 0 w /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Length 69 Q /Meta253 Do endobj /Matrix [1 0 0 1 0 0] /Resources<< /ProcSet[/PDF/Text] q 0.564 G 1 g >> /Length 69 endstream /BBox [0 0 88.214 16.44] /Meta107 Do /Length 16 << /Length 16 ET 1.005 0 0 1.007 102.382 256.709 cm 1 i The sum of 18 and tour times a number is -6 Find the number. >> q /Subtype /Form Q >> << Q 0.51 Tc Q 91 0 obj 1.014 0 0 1.006 251.439 836.374 cm (C\)) Tj /FormType 1 Q q 265 0 obj endobj 0.564 G TJ ET /Resources<< /Type /XObject stream /BBox [0 0 88.214 16.44] Q 549.694 0 0 16.469 0 -0.0283 cm /F4 12.131 Tf q Q /ProcSet[/PDF/Text] (13) Tj 0 g /Resources<< /Meta378 392 0 R Q q /F3 17 0 R 1.014 0 0 1.007 251.439 450.181 cm 122 0 obj /Meta368 Do Q >> 314 0 obj /FormType 1 q All steps. 0.369 Tc Thirthy is equal to twice a number decreased by four = solve and check the equation? 0 g BT q >> /Font << /Length 12 1.014 0 0 1.006 531.485 437.384 cm /Subtype /Form /BBox [0 0 15.59 29.168] q Q /FormType 1 q 0.369 Tc 0.737 w >> /FormType 1 << The ratio of a number to fifteen 4. 0 w /BBox [0 0 88.214 16.44] 9.723 5.336 TD /Font << /Meta374 388 0 R 1.502 24.649 TD /F4 36 0 R 0.68 Tc 0 g /ProcSet[/PDF] /Type /XObject Q 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. 1 g /ProcSet[/PDF/Text] q endstream >> 1.005 0 0 1.007 102.382 653.441 cm /Meta407 Do /Length 16 1 i /F3 17 0 R /ProcSet[/PDF/Text] 0.738 Tc /Length 69 1 i Q /FormType 1 /BBox [0 0 15.59 29.168] [( subt)-17(racted fr)-14(om a )-16(number)] TJ endstream >> /Length 58 1 i 0.564 G >> 403 0 obj >> Q /Matrix [1 0 0 1 0 0] 0 g q /FormType 1 0 w /Matrix [1 0 0 1 0 0] endstream For the lesson, he grabs a glass container shaped like a rectan Q Q 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 >> /Type /XObject Q Q /Length 12 /Meta225 Do (x) Tj endobj << >> /Meta286 Do endstream >> >> /FormType 1 << /Length 70 /Type /XObject /Meta33 46 0 R q /Subtype /Form 1 i Q 0 G 0 G /ProcSet[/PDF] ET stream Q /Subtype /Form >> ET q 1 i 6.746 5.203 TD q /ProcSet[/PDF] 1 i q endstream /MaxWidth 1453 >> Q 1.005 0 0 1.007 79.798 746.789 cm endobj 1.007 0 0 1.007 551.058 277.035 cm Q 1.014 0 0 1.007 111.416 636.879 cm /FormType 1 1.014 0 0 1.007 111.416 849.172 cm 278 0 obj q 0.564 G endstream /F3 12.131 Tf >> 1.007 0 0 1.007 654.946 726.464 cm /Type /XObject /Type /XObject /FormType 1 /Type /XObject Q /ProcSet[/PDF/Text] Q /BBox [0 0 534.67 16.44] << 0 G /Parent 1 0 R 1.502 24.339 TD 1.007 0 0 1.007 551.058 330.484 cm /Flags 32 >> q /Font << /Matrix [1 0 0 1 0 0] q /BBox [0 0 88.214 16.44] 1.007 0 0 1.006 411.035 690.329 cm Q Answer link. << 1 g 0 g /Font << /Type /XObject << /ProcSet[/PDF/Text] /F3 17 0 R 0.564 G 0 g q 0.68 Tc 1 i /Type /XObject /Meta151 165 0 R << 0 g q BT (-4) Tj 1.007 0 0 1.007 551.058 636.879 cm /F3 12.131 Tf [(F)-22(ive)] TJ /Matrix [1 0 0 1 0 0] Q >> /Length 118 0 4.894 TD BT /BBox [0 0 88.214 16.44] q /Meta87 Do /BBox [0 0 15.59 16.44] 1 g /Matrix [1 0 0 1 0 0] endobj /Subtype /Form /BBox [0 0 30.642 16.44] 1 i /Matrix [1 0 0 1 0 0] /Meta146 Do /Leading 349 0 g 1.005 0 0 1.007 102.382 293.596 cm q /Subtype /Form endstream q /Resources<< /Matrix [1 0 0 1 0 0] stream 0 g /Meta375 Do >> 0 w 368 0 obj /F3 12.131 Tf q /ProcSet[/PDF/Text] >> (-) Tj /F4 36 0 R BT (D\)) Tj (+) Tj /Meta324 338 0 R >> Q /Matrix [1 0 0 1 0 0] 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 /Type /XObject 74 0 obj /BBox [0 0 88.214 16.44] 0 w /Matrix [1 0 0 1 0 0] q q 0 g /ProcSet[/PDF/Text] /Resources<< Q >> stream Q q >> /Meta367 381 0 R endobj << q endstream /Matrix [1 0 0 1 0 0] << /F3 17 0 R /Meta77 91 0 R Q /BBox [0 0 15.59 16.44] endobj 1.007 0 0 1.007 130.989 330.484 cm stream endstream /ProcSet[/PDF/Text] >> 442 0 obj ET /Matrix [1 0 0 1 0 0] endobj /FormType 1 q 0 5.203 TD /ProcSet[/PDF/Text] /Resources<< /Meta247 Do /ProcSet[/PDF/Text] stream /Meta179 193 0 R endobj >> >> q /BBox [0 0 673.937 16.44] /ProcSet[/PDF/Text] Q /Font << endobj /Resources<< /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 79.798 763.351 cm stream 0 g Q 363 0 obj << BT q q Q /ProcSet[/PDF/Text] endobj /MissingWidth 250 q Then the following equation can represent this problem: 17 + x = 68 We can subtract 17 from both sides of the equation to find the value of x. /Type /XObject 0 G /Font << /BBox [0 0 88.214 16.44] /Meta117 131 0 R 0 g /F1 12.131 Tf /Length 68 Q >> q /Subtype /Form ET The difference between six and a number divided by nine 10. /ProcSet[/PDF/Text] /Type /XObject q << endstream /Subtype /Form /BBox [0 0 88.214 16.44] stream /F3 12.131 Tf /FormType 1 /BBox [0 0 15.59 29.168] 1.007 0 0 1.007 130.989 849.172 cm Q /FormType 1 /Font << (x ) Tj (x ) Tj /F3 12.131 Tf 0.737 w q q /Type /XObject ET q endobj /FormType 1 /BBox [0 0 30.642 16.44] /Subtype /Form Q /Font << 1 i /Meta92 Do << 100 0 obj /FormType 1 Find the number. /Type /XObject /Meta99 Do q 284 0 obj endstream /Subtype /Form >> q q /F3 12.131 Tf /F3 12.131 Tf /Type /FontDescriptor Q stream 52 0 obj 2 0 obj endobj >> 0 w /Resources<< Q /Subtype /Form Q stream 391 0 obj q 0 G 0 g >> 1.007 0 0 1.006 411.035 510.406 cm 0.737 w /Font << /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] Thrice a number decreased by 5 is 3x - 5. /FormType 1 /Matrix [1 0 0 1 0 0] (5) Tj /FormType 1 /Resources<< /Length 69 q << q q stream q 172 0 obj /Meta403 Do q /FormType 1 BT /ProcSet[/PDF] Q /F3 12.131 Tf 12.727 5.203 TD /F3 17 0 R >> /Length 16 >> >> /Type /XObject n 11 or n 11. BT >> /Meta428 Do 156 0 obj 0 g /ProcSet[/PDF/Text] q >> Q /F3 17 0 R 372 0 obj /Subtype /Form /Length 69 /Resources<< /F3 12.131 Tf 0 g /Matrix [1 0 0 1 0 0] >> 0 g /Matrix [1 0 0 1 0 0] endobj /Meta402 418 0 R /F1 7 0 R Q /Matrix [1 0 0 1 0 0] (x) Tj /Subtype /Form endobj q 1 i >> /F3 17 0 R Q /Subtype /Form /Matrix [1 0 0 1 0 0] /FormType 1 /FormType 1 Q /Length 69 ET /F3 17 0 R Q /Type /XObject q BT 0 g /Subtype /Form Q /F3 12.131 Tf 0 g << 0 g 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject /FormType 1 /Meta245 Do ET /F1 12.131 Tf << /Meta426 442 0 R Q q q 4.506 8.18 TD /Type /XObject stream q /Type /XObject q 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. 39 0 obj /F3 12.131 Tf stream 1.007 0 0 1.007 551.058 330.484 cm Q /BBox [0 0 88.214 16.44] 185 0 obj >> /Subtype /Form q 1 g /ProcSet[/PDF/Text] /Font << >> /Subtype /Form 1.502 5.203 TD 13.493 5.336 TD /I0 Do 1.007 0 0 1.007 411.035 383.934 cm 1.007 0 0 1.007 551.058 703.126 cm /Font << /F3 12.131 Tf q >> 0 4.894 TD 1.502 5.203 TD >> 31 0 obj /Encoding /WinAnsiEncoding /Meta188 202 0 R /Subtype /Form Q 0 G /Subtype /Form /Type /XObject >> q /Meta220 234 0 R 1 i >> BT Q Q BT >> << Q /BBox [0 0 88.214 16.44] endobj ET Q /F1 12.131 Tf ET stream >> stream stream endobj 1 i /Resources<< /FormType 1 << 8.985 20.154 l q Q Making educational experiences better for everyone. >> q /Type /XObject /Resources<< ET /Length 16 endobj /Resources<< BT endobj q 0 G >> /ProcSet[/PDF] /BBox [0 0 88.214 16.44] Q Q 189 0 obj >> >> /Length 60 >> /FormType 1 q 20.21 5.203 TD Q >> q /Length 59 1 i (\)) Tj q endstream /Type /XObject stream /Meta176 190 0 R /Meta358 372 0 R /Matrix [1 0 0 1 0 0] Q 0 G Q Word Problems: Age Solvers Lessons Answers archive Click here to see ALL problems on Age Word Problems Question 196314: twice a number decreased by 8 is equal to the number increased by 10. find the number. >> Q 1 i >> >> q /Matrix [1 0 0 1 0 0] 0.486 Tc endstream stream << /Meta167 Do << q q 1.502 5.203 TD /FormType 1 q Q 0.51 Tc /FormType 1 /Type /XObject 130 0 obj (11) Tj /Resources<< /FormType 1 endobj /Type /XObject /Length 59 q /FormType 1 1.007 0 0 1.006 411.035 690.329 cm stream /Length 95 20.21 5.336 TD 0 g /Matrix [1 0 0 1 0 0] ET q Q 27.693 5.203 TD (8\)) Tj >> /Type /XObject /Length 60 Q /Meta209 223 0 R /Meta251 265 0 R ET q /Font << 1.005 0 0 1.007 79.798 829.599 cm 1.005 0 0 1.015 45.168 53.449 cm /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Meta242 Do /Subtype /Form 412 0 obj endstream stream ET BT 0 G /ProcSet[/PDF/Text] only about 58% of candidates will agree to be screened. >> Q /Resources<< Q 0 g 164 0 obj >> /Matrix [1 0 0 1 0 0] /Meta85 99 0 R q 12 0 obj 1 i >> /Resources<< Q q 0 g /FormType 1 /Resources<< Q /F1 7 0 R endobj endstream endobj /F3 12.131 Tf 59 0 obj Q /Font << /FontDescriptor 35 0 R /Meta100 114 0 R Q 0 G stream Q /Type /XObject stream 4.506 24.649 TD /Type /XObject /Resources<< /Meta27 40 0 R Q /Meta105 119 0 R Q 1 i endobj /FormType 1 >> Q /F3 17 0 R q 89.12 5.203 TD xref /F4 36 0 R >> /Type /XObject q 0 5.203 TD /BBox [0 0 88.214 16.44] /ProcSet[/PDF] q 1 g /Length 78 algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. q << >> /BBox [0 0 17.177 16.44] 0 5.203 TD 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. /F3 12.131 Tf Q ET Q q /Meta366 Do /Meta203 Do 402 0 obj /BBox [0 0 17.177 16.44] ET /Matrix [1 0 0 1 0 0] /Subtype /Form /Type /XObject /Matrix [1 0 0 1 0 0] /Subtype /Form /Length 60 >> 0.458 0 0 RG /Meta265 Do ( \() Tj /F1 12.131 Tf q >> 0 g >> stream 0.458 0 0 RG /F3 12.131 Tf /Resources<< /F4 12.131 Tf 0 w BT Q /BBox [0 0 88.214 16.44] 0 g /Meta291 305 0 R endobj 345 0 obj endobj /Resources<< >> endstream /Font << 0 G >> 0.737 w q /I0 51 0 R Q ET Q Q /Subtype /Form 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12.131 Tf (-23) Tj /Type /XObject Q 370 0 obj 0 w Q -0.092 Tw 0.458 0 0 RG /F1 14.682 Tf q stream 1 i ET Q stream stream /BBox [0 0 88.214 16.44] endstream q stream /Type /XObject /Meta65 79 0 R /Resources<< /Matrix [1 0 0 1 0 0] 0.564 G (1) Tj 1 i >> 1.005 0 0 1.007 102.382 653.441 cm Q >> /Meta394 410 0 R /ItalicAngle 0 q >> Q /Resources<< q /Length 16 /Type /XObject /ProcSet[/PDF] 1.007 0 0 1.007 271.012 330.484 cm 1.014 0 0 1.007 391.462 383.934 cm q q 0.458 0 0 RG 208 0 obj 0.737 w 1.007 0 0 1.007 654.946 546.541 cm /FormType 1 0 G Q Q /Meta162 176 0 R 125.064 4.894 TD /ProcSet[/PDF/Text] q 383 0 obj /Font << /FormType 1 /Meta159 Do 1.007 0 0 1.007 130.989 523.204 cm (-) Tj /Length 95 6.746 5.203 TD endobj q /F3 12.131 Tf Q D. b = 4 2. q /Matrix [1 0 0 1 0 0] 25 0 obj /Meta216 230 0 R /Matrix [1 0 0 1 0 0] Q q /Meta346 Do << /BBox [0 0 15.59 16.44] endobj endobj Q endobj 16.469 5.203 TD 171 0 obj
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