/Length 16 endobj << 1.014 0 0 1.006 111.416 763.351 cm 94.364 5.203 TD q /F1 12.131 Tf /Resources<< /Subtype /Form endobj 0.737 w /Length 69 1 g 0 g 361 0 obj /ProcSet[/PDF] (6\)) Tj Q 261 0 obj endstream 304 0 obj 0 G /FormType 1 240 0 obj 1.007 0 0 1.007 654.946 872.509 cm 156 0 obj q q /Subtype /Form endobj 186 0 obj 1.502 5.203 TD q Q /Meta68 Do /FormType 1 >> q /Matrix [1 0 0 1 0 0] Q stream BT /Resources<< BT /Type /XObject 38 0 obj /F3 12.131 Tf 1.007 0 0 1.007 271.012 776.149 cm 0.737 w >> Q /Meta244 Do >> 1 g >> 408 0 obj Q 0 g (2) Tj endobj Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. BT /Length 12 /BBox [0 0 88.214 35.886] /Subtype /Form endobj 0.458 0 0 RG Q /Length 59 ET 1.005 0 0 1.007 102.382 872.509 cm Q 1 i /Meta153 167 0 R << /FormType 1 1.007 0 0 1.007 45.168 779.913 cm /Meta208 222 0 R Q endstream >> /FormType 1 /Resources<< q The quotient of a seven and a number 9. ( x ) Tj >> /Length 67 Q 0 g 278 0 obj endstream >> /Meta271 285 0 R /Type /XObject >> << endobj q 1.014 0 0 1.007 391.462 330.484 cm 1 i q << 1 i << q /F3 17 0 R Q >> q << >> 0 w ET >> [(Fiv)25(e ti)18(me)16(s)] TJ /FormType 1 0 g 1 i (A\)) Tj 324 0 obj 1.007 0 0 1.007 271.012 776.149 cm /FormType 1 /BBox [0 0 534.67 16.44] 0 w Q /StemV 94 (x) Tj 0 5.203 TD 0 5.203 TD q q -0.058 Tw 0 G /Matrix [1 0 0 1 0 0] /FormType 1 /Resources<< q Q Q << 1 i Let x be a number. >> Q >> /FormType 1 /Matrix [1 0 0 1 0 0] BT >> /FormType 1 /BBox [0 0 88.214 16.44] /Meta102 116 0 R /F4 36 0 R /Meta199 Do /Length 245 stream /Matrix [1 0 0 1 0 0] Q 1 i /F4 36 0 R endobj q Q q /ProcSet[/PDF/Text] /Meta216 230 0 R 91 0 obj q 20.21 5.203 TD /Meta191 205 0 R /ProcSet[/PDF] /ProcSet[/PDF/Text] /Type /XObject q Q Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. ET /Type /XObject q Q /BBox [0 0 88.214 16.44] << 1.502 5.203 TD q Q 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject /Matrix [1 0 0 1 0 0] q /Length 16 >> /F1 7 0 R q 1.007 0 0 1.007 45.168 730.228 cm /Type /XObject ET 117 0 obj Q 0.458 0 0 RG q Q /Matrix [1 0 0 1 0 0] endobj endobj >> Q /Meta372 Do /Length 54 stream /F3 17 0 R /Font << stream 39 0 obj >> 1 i q 0 g q 367 0 obj /Type /XObject /Matrix [1 0 0 1 0 0] Q << /Length 60 There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. Q 0.458 0 0 RG /Meta47 Do /Font << /Length 118 /F4 36 0 R Q 9.723 5.336 TD >> /Type /XObject ET 0 G 0 g /FontDescriptor 10 0 R /Resources<< /Meta115 Do 0 w 1 i (3) Tj /Resources<< 0 G BT << /Subtype /Form /Type /XObject /Font << /Subtype /Form 1.007 0 0 1.007 551.058 523.204 cm /Resources<< 0 g 0 G << << >> >> /Meta383 Do /Meta16 27 0 R /ProcSet[/PDF/Text] >> 111 0 obj /BBox [0 0 17.177 16.44] /Subtype /Form /Type /XObject /Meta392 408 0 R /Type /XObject /Resources<< /FormType 1 /Meta313 327 0 R /Length 59 /FormType 1 /Resources<< /Length 69 /Resources<< /Resources<< Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 703.126 cm q /Subtype /Form 1 i /F3 17 0 R stream Q Q /Type /XObject /FormType 1 >> >> /FormType 1 1 i 420 0 obj 322 0 obj Q /Meta117 131 0 R 20.21 5.203 TD /Type /XObject /Meta77 Do >> /Subtype /Form /F3 12.131 Tf That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. /Meta189 Do BT 145 0 obj q 215 0 obj /FormType 1 0.737 w endobj << ET /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Subtype /Form ( \() Tj /Meta5 Do 80 0 obj /Subtype /Form endobj q /Type /XObject >> endstream 0 w 1.007 0 0 1.007 654.946 400.496 cm >> 0 g Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. 0 g 0 g /Matrix [1 0 0 1 0 0] 0 w 1.005 0 0 1.007 79.798 779.913 cm /BBox [0 0 534.67 16.44] 6.746 5.203 TD Q /Meta296 310 0 R /Resources<< 1.014 0 0 1.007 111.416 450.181 cm Q 0.285 Tc /FormType 1 << q 331 0 obj endstream /Matrix [1 0 0 1 0 0] endstream 0.737 w >> 1.007 0 0 1.006 551.058 763.351 cm ET ET /Meta42 Do Q Q q 1.007 0 0 1.007 411.035 330.484 cm >> /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Subtype /Form Q /Matrix [1 0 0 1 0 0] stream A link to the app was sent to your phone. q q 1 i 0 g /Meta308 322 0 R >> /FontBBox [-568 -307 2000 1007] >> /BBox [0 0 88.214 16.44] endstream 1.014 0 0 1.006 531.485 836.374 cm /Resources<< 1.014 0 0 1.007 391.462 277.035 cm /BBox [0 0 30.642 16.44] /Resources<< /Font << 0 w /Resources<< >> q 368 0 obj /Resources<< /F3 12.131 Tf /F3 12.131 Tf ( decreased by ) Tj 0 G /Type /Font This site is using cookies under cookie policy . /BBox [0 0 673.937 27.581] 57 0 obj /Meta54 68 0 R >> << Q q stream ET /Meta269 Do 1.005 0 0 1.007 102.382 726.464 cm 722.699 400.496 l << /Meta15 Do /ProcSet[/PDF/Text] Q q stream /Meta202 216 0 R endstream Q 1.005 0 0 1.007 102.382 473.519 cm Twice a number decreased by 8 gives 58. /Subtype /Form endobj 1 i /Resources<< Q 396 0 obj /BBox [0 0 88.214 16.44] BT 0 g /Resources<< ET (D\)) Tj /Resources<< /Resources<< /BBox [0 0 639.552 16.44] endstream Q /Length 69 >> endstream endstream 82 0 obj /Meta344 Do /Type /XObject /Length 16 /Meta350 Do 1 i ET 0.297 Tc Q << 1.007 0 0 1.007 551.058 383.934 cm endobj 293 0 obj endobj Q Q -0.463 Tw /Type /XObject 0 g 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 >> /Resources<< /Meta236 Do Q >> stream 41 0 obj /Font << 0 g /BBox [0 0 88.214 16.44] Q /FormType 1 /Matrix [1 0 0 1 0 0] >> /BBox [0 0 88.214 16.44] /Length 294 /Meta246 260 0 R q stream 0 g ET stream << /FormType 1 /ProcSet[/PDF] No packages or subscriptions, pay only for the time you need. stream >> 1.014 0 0 1.007 251.439 450.181 cm /Resources<< >> >> 1.007 0 0 1.007 551.058 277.035 cm 235 0 obj q /BBox [0 0 88.214 16.44] >> q /Meta337 Do /F3 17 0 R q ET /F3 12.131 Tf 1.007 0 0 1.007 551.058 330.484 cm stream /Type /XObject /F3 17 0 R Q q Q >> 364 0 obj /Type /XObject /Meta180 Do (-) Tj Q /Matrix [1 0 0 1 0 0] 0.737 w /Resources<< /Type /XObject stream q >> stream 1 g 0.737 w /Meta97 111 0 R >> 0 g 0 5.203 TD Q /F3 17 0 R 19.474 5.203 TD >> 20.21 5.203 TD >> 218 0 obj >> Q endstream Q q 1 i Q BT ( x) Tj /Descent -216 >> q /Font << q >> >> /F3 17 0 R /Type /XObject 1 i Q /FormType 1 endstream /Matrix [1 0 0 1 0 0] >> 0 g << 6.746 24.649 TD [3] One half of a number increased by fourteen is twenty-one. 1 g [( and )-20(the product of )-15(a number a)-16(nd )] TJ stream /FormType 1 ET /FormType 1 ET 0 g stream q /Meta0 Do /Matrix [1 0 0 1 0 0] 0.737 w Q /BBox [0 0 15.59 16.44] 0 G (38) Tj 0 w /Resources<< 0.737 w Q Solution. >> ET /Font << BT /Length 69 endobj /Matrix [1 0 0 1 0 0] Q >> /FormType 1 /ProcSet[/PDF/Text] >> Q /FormType 1 1 i /Length 65 Q /Matrix [1 0 0 1 0 0] >> /Type /XObject q /Meta117 Do Q 1.007 0 0 1.007 411.035 849.172 cm /Resources<< q >> /FormType 1 ET Q endobj stream 0 w >> >> /F3 12.131 Tf >> /Length 16 /MaxWidth 1248 << 0 g /Subtype /Form Q 77 0 obj (-8) Tj ( \() Tj q 0.51 Tc endobj BT q /Matrix [1 0 0 1 0 0] endstream ET >> q q 1 g endobj 1.007 0 0 1.007 411.035 583.429 cm q /F4 12.131 Tf 87 0 obj Q >> endstream /Matrix [1 0 0 1 0 0] Q 1 i /Meta156 170 0 R Q endobj >> >> /Subtype /Form 0 w /Meta339 Do /Meta363 377 0 R /Type /XObject , Prove the following /F3 17 0 R 185 0 obj 1 g q /Font << endobj /BBox [0 0 30.642 16.44] 1.014 0 0 1.006 531.485 510.406 cm >> endstream >> BT 1 i q (7\)) Tj q stream 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 1.007 0 0 1.007 411.035 330.484 cm /Matrix [1 0 0 1 0 0] endobj (B\)) Tj 13.464 5.203 TD stream /Resources<< ET /FormType 1 1.014 0 0 1.007 531.485 330.484 cm Q ET /ProcSet[/PDF/Text] q /Resources<< /Resources<< Q /Resources<< q /Type /XObject Q -0.463 Tw 0.458 0 0 RG 0 g /Length 69 /ProcSet[/PDF] q /ProcSet[/PDF/Text] /Meta94 Do /Matrix [1 0 0 1 0 0] Q 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. /BBox [0 0 88.214 16.44] q /Type /XObject >> Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endobj /F3 12.131 Tf << 1 i endobj >> 1 i /Resources<< endstream /FormType 1 >> /Font << 1 i 16.469 5.336 TD Q 0 w q Q /F4 36 0 R Q /Type /XObject 229 0 obj q << >> endobj Q >> 1 i q Find the number 1 See answer Advertisement q /Length 16 Q 0 20.154 m /Type /XObject /FormType 1 >> /Type /XObject 0 G Q Six subtracted from a number 6. Q endstream ET Q stream BT /F1 7 0 R q /ProcSet[/PDF/Text] endobj /Subtype /Form /F3 17 0 R q >> /Meta14 25 0 R /ProcSet[/PDF] >> /F3 12.131 Tf /Subtype /Form Q /Type /XObject (+) Tj Q stream /ProcSet[/PDF/Text] /Meta381 395 0 R q >> /Font << >> Q << /Meta359 Do q endobj Q >> 0.227 Tc /MissingWidth 250 1 i Q << /FormType 1 /FormType 1 Q C. Twice a number decreased by ten is at most 24. 1 i endstream /Meta352 Do /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 /F3 12.131 Tf q /Encoding /WinAnsiEncoding /Type /XObject 1.502 5.203 TD /Meta111 125 0 R 0 g << /F3 17 0 R /Length 59 Q /ProcSet[/PDF] /F4 36 0 R stream Q << 270 0 obj Get link; Facebook; Twitter; Q /BBox [0 0 549.552 16.44] Q BT /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] >> Q q >> stream 0.737 w Q 1 i /Length 294 Q q /FormType 1 endstream 0 g BT /Font << endstream Q /Length 16 >> /Length 16 0.737 w /Subtype /TrueType /Font << /FormType 1 Q ET /Resources<< << >> endobj 1 i >> Twice a number decreased by 8 gives 58. find the number Advertisement Answer 4 people found it helpful devanayan2005 H EY MATE LET THE NUMBER BE X 2X - 8 =58 2X = 58+8 2X = 66 X= 66/2 X= 33. Q /Matrix [1 0 0 1 0 0] Q /Font << q -0.486 Tw 1.007 0 0 1.007 551.058 383.934 cm stream 0.737 w /Type /XObject endstream /F3 12.131 Tf q << q stream endobj /F3 12.131 Tf Q Was this answer helpful? endstream [(Negativ)16(e )] TJ /FormType 1 /Meta339 353 0 R Q q /Subtype /Form q q stream /Type /XObject /Meta420 436 0 R 1 i BT /F3 17 0 R 1 i 1.005 0 0 1.007 102.382 293.596 cm >> /Resources<< /FontName /PalatinoLinotype-Bold Double or twice a number means 2x, and triple or thrice a number means 3x. /Meta185 Do >> BT 351 0 obj 0 G >> >> /Type /XObject /Matrix [1 0 0 1 0 0] /Meta195 Do 416 0 obj /Meta66 Do /FormType 1 1 i Q /Length 12 Q /BBox [0 0 30.642 16.44] endobj >> /BBox [0 0 30.642 16.44] /F4 36 0 R stream /Meta341 355 0 R /BBox [0 0 15.59 16.44] q 133 0 obj /Subtype /Form q (x) Tj /ProcSet[/PDF/Text] /Font << /Subtype /Form stream 0 w /Length 69 /F1 7 0 R 1 i /Matrix [1 0 0 1 0 0] /Length 69 16.469 5.203 TD Q >> << 0 G 0.564 G endobj /Subtype /Form /FormType 1 3.742 5.203 TD /Length 118 /ProcSet[/PDF/Text] Q q >> ET Q Q endstream 1.007 0 0 1.007 551.058 703.126 cm /Resources<< Q << Q /Meta315 Do /Subtype /Form /F3 12.131 Tf /ProcSet[/PDF] q /Type /XObject 1 i /Subtype /Form /Type /XObject /Resources<< /Meta31 Do /Resources<< 0 g q /BBox [0 0 88.214 16.44] /F3 12.131 Tf q 1.005 0 0 1.007 102.382 256.709 cm 1 i endstream 0.296 Tc 103 0 obj /Font << /Resources<< q BT >> /Subtype /Form endstream /F3 17 0 R Q << /Type /XObject /Matrix [1 0 0 1 0 0] /Length 69 0.564 G 0.564 G /FormType 1 q /Matrix [1 0 0 1 0 0] 1 i Q /Length 16 1 i 1 i /FormType 1 1 i 0 w 0 g /Length 64 q /Type /XObject 0.458 0 0 RG endstream 0.737 w 0 g /F4 36 0 R Answer (1 of 8): Solution: let the number be x. stream q q << stream 1.007 0 0 1.006 130.989 690.329 cm /Type /XObject stream Q << /Meta351 365 0 R 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. >> /Resources<< q Q /F3 17 0 R Q ([x ) Tj 1.007 0 0 1.007 271.012 330.484 cm q 0 w 0 G /Font << /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] ET stream 1.014 0 0 1.007 391.462 523.204 cm >> Q 0 G 205 0 obj /FormType 1 17.234 5.203 TD /Length 12 /Resources<< Q q Q /FormType 1 q /Meta100 Do q /Length 59 q /Meta240 254 0 R endobj /ProcSet[/PDF] /Subtype /Form ET /Resources<< 0 G 722.699 347.046 l 0 G S 1 i 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 : . /Font << /Meta340 354 0 R endobj 0.458 0 0 RG 125 0 obj >> Q /BBox [0 0 15.59 16.44] Q /F3 12.131 Tf >> 0 w /Matrix [1 0 0 1 0 0] q >> 407 0 obj Q Q endobj 1 i /Length 64 381 0 obj /Subtype /Form /Matrix [1 0 0 1 0 0] Q /Font << Q 1.005 0 0 1.007 102.382 400.496 cm (A\)) Tj BT /Font << /F4 36 0 R 1.005 0 0 1.007 102.382 799.486 cm Q /Meta210 Do << ET 722.699 473.519 l 0 G /Type /XObject /Meta95 109 0 R /Subtype /Form /F3 17 0 R 1.007 0 0 1.007 271.012 523.204 cm /Resources<< endstream 1 i Q << >> Q /F3 12.131 Tf >> 0 g /Matrix [1 0 0 1 0 0] q 0 g q 1.007 0 0 1.006 411.035 437.384 cm endobj Q 0 g >> stream >> << 0.369 Tc endobj endstream 0.458 0 0 RG 0 G Q 0 g /Subtype /Form /Type /XObject 0 G /FormType 1 1.005 0 0 1.007 102.382 599.991 cm BT endobj Q Q q 1.014 0 0 1.007 391.462 450.181 cm /Font << /FormType 1 /FormType 1 /Resources<< /Length 68 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 q /Matrix [1 0 0 1 0 0] 0.17 Tc /F3 17 0 R endobj stream q >> /Subtype /Form Q 346 0 obj q Answer link. endstream q q Q endobj >> >> Q 1.005 0 0 1.007 79.798 763.351 cm q /Meta328 342 0 R q /FormType 1 /Matrix [1 0 0 1 0 0] stream 0 w << 1 i /Font << q Q >> /Meta25 38 0 R 236 0 obj /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 1 i Q 1 i q 0 G q S (x) Tj Q /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] << q >> stream /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] >> /Font << 299 0 obj 0 G 0 g /Resources<< Q endstream q /Resources<< /FormType 1 1.007 0 0 1.007 130.989 636.879 cm q >> q endstream /BBox [0 0 534.67 16.44] q /BBox [0 0 15.59 29.168] /Length 95 endobj /Meta72 86 0 R q /Meta86 Do /BBox [0 0 534.67 16.44] endobj /Matrix [1 0 0 1 0 0] endobj Q 112 0 obj endobj /ProcSet[/PDF/Text] q /Meta312 Do 549.694 0 0 16.469 0 -0.0283 cm [2] Twice a number increased by four is twenty-one. << /ProcSet[/PDF/Text] Q 1 i /AvgWidth 445 endobj Mr. Gleeson knows that 1,000 cubic centimeters is the same as 1 liter, and he wants to figure out how many liters of water will fill the container before it overflows. 1 i >> stream 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 /BBox [0 0 88.214 16.44] Q Two speeding tickets could increase your rate by 58% at your next renewal. /ProcSet[/PDF/Text] Q 0 g /Meta7 Do q q q endstream endobj /Meta334 Do Q q Q /Subtype /Form endobj /Resources<< 0.297 Tc Q >> /Encoding /WinAnsiEncoding Q /BBox [0 0 15.59 16.44] >> 0 G >> >> >> q 0.564 G endstream stream q BT >> 0 g (B\)) Tj /Meta220 Do /ProcSet[/PDF] 1.007 0 0 1.007 271.012 636.879 cm /ProcSet[/PDF] /Length 74 q endobj q 0 g q << /BBox [0 0 30.642 16.44] 245 0 obj
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