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/Matrix [1 0 0 1 0 0] ET 0.737 w S 1 i 30.699 5.203 TD q endobj endstream /ProcSet[/PDF/Text] stream /Matrix [1 0 0 1 0 0] Formula - How to Calculate Percentage Decrease. /F3 17 0 R BT endstream 0 g /BBox [0 0 534.67 16.44] /FormType 1 endstream >> 0 G /BBox [0 0 88.214 35.886] >> q 0 5.203 TD Q /FormType 1 549.694 0 0 16.469 0 -0.0283 cm >> /Subtype /Form /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] /Length 65 /Subtype /Form /Type /XObject endobj q 0 G >> endobj /Length 69 Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. /Font << << endobj /Ascent 976 endobj /Matrix [1 0 0 1 0 0] Q /BBox [0 0 88.214 16.44] /FormType 1 q 141 0 obj endstream /Length 16 Q /FormType 1 416 0 obj 0.37 Tc Q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1 g /F3 12.131 Tf /Font << Find the number. 0 G BT /Resources<< 1 i /Subtype /Form endobj stream q 30 0 obj /Length 69 0 5.203 TD >> /Length 16 /ProcSet[/PDF] 0 G 1.007 0 0 1.007 551.058 277.035 cm Q /Meta366 380 0 R /BBox [0 0 673.937 16.44] /FormType 1 373 0 obj 111 0 obj /Resources<< Q: when six times a number is decreased by 4, the result is 8. << /Resources<< q endstream << /Length 54 stream endobj Q Q /Subtype /Form 19.474 5.203 TD Q Q q /Length 69 Q Q /ProcSet[/PDF] q >> /ProcSet[/PDF] BT /Type /XObject /Subtype /Form << /Type /XObject q /ProcSet[/PDF/Text] pidemiologi i Infekcionnye Bolezni. endobj endobj << /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 411.035 330.484 cm /Font << 0 g q >> 1.007 0 0 1.007 130.989 776.149 cm q /BBox [0 0 88.214 35.886] /Meta220 234 0 R q /BBox [0 0 534.67 16.44] q Q /Subtype /Form 1.005 0 0 1.007 102.382 743.025 cm endobj /Meta376 Do 1 i /Subtype /Form /F3 12.131 Tf LAIing for a pizza and, soft drink. stream stream q 0.297 Tc 1 g stream >> 0.564 G >> >> Q /Type /XObject endobj /XObject << 0.458 0 0 RG /Resources<< q /ItalicAngle 0 /Meta181 Do Q /Subtype /Form /Length 66 Q /BBox [0 0 88.214 16.44] /Font << << BT endstream q Q Q Q BT /Meta405 Do /Meta87 101 0 R /BBox [0 0 673.937 15.562] /Meta388 Do /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] endobj (C\)) Tj /Font << 3x - 5 = 2x + 1. x = 6. >> /Matrix [1 0 0 1 0 0] /Resources<< /Meta237 251 0 R /ProcSet[/PDF/Text] 1 i /Resources<< >> q /Meta328 342 0 R /Length 16 0 g /Resources<< 0 w Q /ProcSet[/PDF] /LastChar 121 << /Meta398 Do 0 G endstream Q /Type /XObject q Q /Resources<< 1 g /Meta309 323 0 R 0 G q /Meta269 Do %%EOF. 0.564 G /Subtype /Form Q endstream /Meta156 Do /Font << q /Subtype /Form q /BBox [0 0 15.59 29.168] /BBox [0 0 15.59 16.44] 0.524 Tc 438 0 obj Q << 1 i Number Outcomes 1 42 2 41 3 . q /Font << 1 i 1.007 0 0 1.007 654.946 726.464 cm Q >> , Prove the following 1 i /Meta12 23 0 R ET 387 0 obj Q /Font << 0 g /FormType 1 << /Meta394 Do /Type /XObject /Subtype /Form 1 i /Meta244 258 0 R /Length 244 /Type /XObject /Meta227 Do /FormType 1 /Font << "49 . 1 i BT ET /ProcSet[/PDF] >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] >> 1 i /BBox [0 0 88.214 16.44] /Meta278 292 0 R /Subtype /Form >> -y. /ProcSet[/PDF/Text] q /Meta249 Do 1.007 0 0 1.007 411.035 383.934 cm 0.369 Tc /BBox [0 0 88.214 35.886] ET stream Q stream 0 G endobj Q q /Type /XObject q 412 0 obj -0.029 Tw q /Font << /Resources<< 1 g Q (58) Tj endstream 0 g << [4] One half of a number decreased by fourteen is twenty-one Then ab is a binary operation. /FormType 1 /Matrix [1 0 0 1 0 0] 0 G q 1 i >> Q Q >> >> /BBox [0 0 88.214 35.886] q q /Resources<< 0.458 0 0 RG q q /Length 67 /Type /XObject /Meta39 53 0 R >> /Font << 0 5.203 TD Q /Resources<< /F3 12.131 Tf 1.005 0 0 1.007 102.382 490.08 cm /Resources<< (x ) Tj 0.458 0 0 RG Q New questions in Mathematics Q >> endstream >> /FormType 1 335 0 obj >> Q /BBox [0 0 88.214 35.886] endobj /Length 69 << >> BT /Resources<< 0 g /I0 Do 314 0 obj /Type /XObject Q Q q endobj 1 i >> Q /FormType 1 << 420 0 obj /BBox [0 0 88.214 16.44] q /FormType 1 << q >> /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 271.012 583.429 cm /Meta63 Do Q /FormType 1 1.007 0 0 1.007 551.058 583.429 cm 0 G /Subtype /Form /Meta260 Do If twice a number is decreased by 13, the result is 9. ET 1.007 0 0 1.007 67.753 599.991 cm Q /F3 12.131 Tf 1 i Q /FormType 1 /Type /XObject 0.737 w q /BBox [0 0 88.214 16.44] (-) Tj /Meta184 198 0 R 15 0 obj 1 g 0 g /Resources<< >> /Meta413 Do /Resources<< /ProcSet[/PDF] Twice a number decreased by 58! 95 0 obj Q >> q Q /BBox [0 0 15.59 29.168] stream /Matrix [1 0 0 1 0 0] endstream q q /Meta247 Do /Resources<< 1.005 0 0 1.007 102.382 872.509 cm q stream 0.486 Tc /Meta299 313 0 R /BBox [0 0 639.552 16.44] 259 0 obj /Meta282 296 0 R Q >> >> q 0 w /Matrix [1 0 0 1 0 0] /Meta192 Do /Type /XObject 174.501 5.203 TD /Meta202 216 0 R >> /Meta233 247 0 R /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] >> /Resources<< endobj (x) Tj /Length 58 /Meta88 102 0 R /BBox [0 0 15.59 29.168] Q /Resources<< /ProcSet[/PDF/Text] /Subtype /Form BT /Meta340 Do /Matrix [1 0 0 1 0 0] /FormType 1 endobj /Type /XObject /BBox [0 0 88.214 16.44] q stream Q /F3 12.131 Tf /Resources<< BT q endobj << (Twice) Tj Q 1.007 0 0 1.007 45.168 846.161 cm 0.564 G /FormType 1 /MaxWidth 1397 1 g /Length 78 endobj Q /F3 17 0 R /Resources<< q /FormType 1 (x) Tj >> endstream /Resources<< q 47.933 5.203 TD 1.007 0 0 1.007 130.989 523.204 cm q q /BBox [0 0 88.214 16.44] q ET 107 0 obj >> >> endobj endstream (+) Tj /Meta164 Do 0 g /Meta317 331 0 R 0 g 1 i /F3 12.131 Tf 0 g /F3 12.131 Tf << Q /Subtype /Form endstream << 0 G /FirstChar 32 Q >> /Subtype /Form /ProcSet[/PDF] /Meta190 204 0 R 333.269 5.488 TD /Matrix [1 0 0 1 0 0] << 0 g >> Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 523.204 cm Thrice a number decreased by 5 exceeds twice the number by a unit. Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] stream q /Type /XObject 1.014 0 0 1.007 111.416 330.484 cm q Q /ProcSet[/PDF] /Meta206 220 0 R /Subtype /Form /Length 54 /Length 16 1 i Q /Font << 0 g Q /ProcSet[/PDF/Text] ET 194 0 obj /Font << Q 0.458 0 0 RG q q /BBox [0 0 88.214 16.44] We are asked to find the number, so, we could assign the number as "x". q A rectangular garden has a width that is 8 feet less than twice the length. /F3 17 0 R -0.041 Tw Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] >> 0.458 0 0 RG 1 i 1 i Just type into the box and your calculation will happen automatically. /Font << Q q /Meta385 401 0 R >> 0.241 Tc q /Meta366 Do endobj Q >> q >> BT (x) Tj Q 0 G /Meta300 314 0 R /Resources<< ET /CreationDate (D:20140515121932-04'00') Q 0 w 0 g /FormType 1 endstream /ProcSet[/PDF/Text] q /BBox [0 0 88.214 16.44] q /Type /XObject /BBox [0 0 88.214 35.886] >> 0 G /Meta108 122 0 R /Meta28 Do /Resources<< /FormType 1 << /Count 2 Q /Matrix [1 0 0 1 0 0] >> 0 G [( t)-14(imes a num)-16(ber)] TJ /Resources<< 0.458 0 0 RG /BBox [0 0 30.642 16.44] 1 i Q /Resources<< ( x ) Tj /Type /XObject 16.469 5.336 TD Q >> Q /F3 17 0 R Q 1 i /Meta143 Do stream /Meta298 312 0 R /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 746.789 cm /Length 139 ET /Length 16 Q 20.21 5.203 TD /F3 17 0 R BT Q >> 11 0 obj q 0.458 0 0 RG 1 i >> q 0 g << /Subtype /Form >> endstream Q /Matrix [1 0 0 1 0 0] 1.502 5.203 TD Q stream /Matrix [1 0 0 1 0 0] >> >> ET Q >> /Font << /Resources<< /Subtype /Form 0 w 0.737 w endobj /Meta373 Do /Meta324 Do 1 i q /BBox [0 0 88.214 16.44] Q BT -0.101 Tw Q ET endobj 220.931 4.894 TD ET q 0.458 0 0 RG 1 g /FormType 1 /Length 80 >> 1 i 201 0 obj q /Resources<< 6.746 5.203 TD /Length 95 /Resources<< >> 0.838 Tc /Meta79 Do stream Q << q /Meta304 318 0 R 0 G ET /Subtype /Form Have a nice day! 51 0 obj 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 0.458 0 0 RG /Meta52 Do 0 w /BBox [0 0 15.59 29.168] /BBox [0 0 88.214 16.44] /Type /XObject Q Q >> 227 0 obj 0 G BT /Matrix [1 0 0 1 0 0] 23.216 5.203 TD endstream /BBox [0 0 88.214 16.44] /Meta240 Do [(The )-16(s)15(um )-14(of )] TJ Q << q >> q q /F3 12.131 Tf /I0 51 0 R /Type /XObject 672.261 473.519 m /F3 12.131 Tf 1 g << /Meta281 Do Q ET >> 1.007 0 0 1.007 130.989 849.172 cm endstream >> /ProcSet[/PDF/Text] >> BT Q 0 w Get link; Facebook; Twitter; /Resources<< 2.238 5.203 TD /F3 12.131 Tf endstream q Q 0 G >> /Length 59 /Matrix [1 0 0 1 0 0] >> q /Meta170 Do q /XObject << (viii) A number divided by 8 gives 7. 0 w >> Q >> Q /FormType 1 q >> /Length 69 Q Q /Meta391 Do >> Q << /BBox [0 0 88.214 35.886] 6.746 5.203 TD /Length 59 0 g stream To find: The. 1.007 0 0 1.007 551.058 383.934 cm q endobj BT 1 i q >> >> /F3 17 0 R (A\)) Tj /Matrix [1 0 0 1 0 0] Q 425 0 obj /BBox [0 0 88.214 16.44] 0 g /Meta277 Do >> 0 g /Meta357 Do /Subtype /Form >> 0 G endobj Q /Resources<< Q Q stream endobj q q Q stream /Resources<< 0 g 27 0 obj 0 G /Meta364 378 0 R endobj 0 5.203 TD Q /BBox [0 0 15.59 29.168] /Matrix [1 0 0 1 0 0] /Subtype /Form q Q 0 g (\(x ) Tj 1 i 0 g 0 G 1.005 0 0 1.007 102.382 653.441 cm 0.458 0 0 RG >> (x ) Tj BT (1) Tj 32 0 obj >> /BBox [0 0 15.59 16.44] 0 g /Resources<< /Subtype /Form /Meta194 208 0 R >> 0.458 0 0 RG 407 0 obj /Resources<< /Matrix [1 0 0 1 0 0] << 0 w 0 g q endstream /Resources<< >> BT /Meta73 87 0 R /ProcSet[/PDF] Q 1.007 0 0 1.007 551.058 277.035 cm /Meta374 Do endstream q 1 i q ET Q q q 0.68 Tc >> Q /ProcSet[/PDF/Text] >> 1 i 1 i Q q /Font << 1 i ET /I0 51 0 R << 0 g 0 G /Font << >> Q Q Hence, the number is 6. << 126 0 obj stream Q BT Q /Length 12 q Q q >> /Meta135 149 0 R 0 20.154 m /Subtype /Form >> /Descent -216 /ProcSet[/PDF] Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. 0.564 G /Subtype /Form >> /BBox [0 0 549.552 16.44] 3.742 5.203 TD q >> Q 1.014 0 0 1.007 391.462 849.172 cm BT << /Meta294 308 0 R 0 g endstream 0.737 w /Matrix [1 0 0 1 0 0] 1 g 1 i 230 0 obj 0.786 Tc 0 g [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. >> /Matrix [1 0 0 1 0 0] q q BT 6.746 5.203 TD 1 i 0 g /ProcSet[/PDF] >> 22.478 5.203 TD /BBox [0 0 534.67 16.44] 0.564 G Q /Length 59 >> a and b or something else.***. /Resources<< Q 1 i 1 i q endobj /Meta429 445 0 R /Resources<< /Type /XObject 0.564 G (1\)) Tj BT /Type /XObject 343 0 obj endstream Q >> 0.737 w >> 0 g 0 g Q Q /ProcSet[/PDF/Text] 0.524 Tc Q << In addition, testosterone in both sexes is involved in health and well-being . BT 0 g endobj /F3 12.131 Tf 0.369 Tc /Meta25 Do 0 g 1 g << q /F3 12.131 Tf /F3 17 0 R << 6.746 5.203 TD /ProcSet[/PDF/Text] q << 1.007 0 0 1.007 411.035 330.484 cm Q /BBox [0 0 88.214 16.44] endstream ET /Subtype /Form q /ProcSet[/PDF/Text] endstream 679.036 293.596 m /Meta112 Do 0.737 w /Resources<< q /FormType 1 q -0.486 Tw 1.007 0 0 1.007 130.989 277.035 cm Q 1.005 0 0 1.007 79.798 763.351 cm q (4\)) Tj /F3 17 0 R 241 0 obj 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. /ProcSet[/PDF] Q 0 g BT endstream 1.005 0 0 1.007 79.798 730.228 cm BT 1 i << 6.746 8.18 TD 1 g 0 G q (38) Tj /Length 69 >> /Type /XObject /ProcSet[/PDF] /FormType 1 endstream /F1 7 0 R /Length 69 /Subtype /Form /Length 68 /MediaBox [0 0 767.868 993.712] >> 1 i Translate 2(x-58) into mathematical phrase. /Length 69 /Matrix [1 0 0 1 0 0] Thrice of a number = 3x. BT The sum of a number and 2 is 6 less than twice that number. 32.939 5.203 TD /F3 12.131 Tf /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] endstream /Meta274 288 0 R Q /F3 12.131 Tf /Length 16 /Meta0 5 0 R >> endobj 0 G /Matrix [1 0 0 1 0 0] /Meta169 Do /F3 17 0 R BT >> 1.007 0 0 1.007 551.058 703.126 cm Q ET >> 1 i BT /Type /XObject BT /F3 12.131 Tf q /Meta4 Do q /Subtype /Form Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.425 Tc Q /Length 64 >> Q endstream /Meta361 Do 0.564 G /ProcSet[/PDF/Text] /FormType 1 Q Q stream /I0 51 0 R endobj >> 1.014 0 0 1.006 251.439 437.384 cm q BT /F3 17 0 R Q BT 286 0 obj stream 1.007 0 0 1.006 411.035 763.351 cm /Resources<< 0.838 Tc << /F3 17 0 R ET BT Q stream q Q /Meta145 159 0 R >> q 374 0 obj 0 g Q /BBox [0 0 30.642 16.44] /Length 91 0 G /FontBBox [-170 -292 1419 1050] >> /FormType 1 /Type /XObject endobj /ProcSet[/PDF] 0 g /Meta40 54 0 R /ProcSet[/PDF/Text] /F3 12.131 Tf 0 G /FormType 1 0.31 Tc q stream /Type /XObject /BBox [0 0 534.67 16.44] Q /Matrix [1 0 0 1 0 0] >> /FormType 1 >> 1 i q /Matrix [1 0 0 1 0 0] 0 G 0 g 68 0 obj 0 G >> /Font << 0 G /Length 54 1 i BT /ProcSet[/PDF] >> /Length 59 q 6.746 5.203 TD 1 i 1 i Q /Length 59 0 g q 7 0 obj (C\)) Tj << /Length 16 /F3 17 0 R q endobj >> /F1 7 0 R Q 0.564 G /Meta165 Do Q 0.737 w /F3 17 0 R /BBox [0 0 88.214 16.44] Q /Meta117 131 0 R /FormType 1 q >> /Resources<< 399 0 obj /Subtype /Form q /F3 12.131 Tf Q Q 1.014 0 0 1.007 111.416 277.035 cm q (x) Tj endstream /ProcSet[/PDF] /Font << /FormType 1 >> stream >> /Type /XObject ET 0 g /Subtype /Form 1 g q Answer link. /F3 17 0 R /ProcSet[/PDF/Text] << /Length 60 1.007 0 0 1.007 67.753 872.509 cm (1\)) Tj /Subtype /Form /Type /XObject /Resources<< 0 G Q stream 332 0 obj /Length 69 /Meta176 Do Q 423 0 obj << /Subtype /Form >> /Length 58 >> /Resources<< /Resources<< /ProcSet[/PDF] Q 0.564 G /Matrix [1 0 0 1 0 0] Q /Meta153 Do /Length 119 Six subtracted from a number 6. 2.238 5.203 TD So let's go ahead and identify a v 0.564 G /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 43.426 5.203 TD /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /FormType 1 stream /Meta276 Do ET BT /Type /XObject Q Making educational experiences better for everyone. 0.737 w >> 0.458 0 0 RG 38.948 5.203 TD 351 0 obj stream /F3 17 0 R q /Meta156 170 0 R 0 g >> Q Q Q q 1.005 0 0 1.007 79.798 796.475 cm /I0 51 0 R endobj Q >> /Resources<< endobj /Font << /Meta265 279 0 R Answer only. /FontBBox [-174 -299 1445 1050] 0 g 1.005 0 0 1.007 102.382 872.509 cm 1.007 0 0 1.006 411.035 763.351 cm /Font << /ProcSet[/PDF] 1.007 0 0 1.006 130.989 437.384 cm Q 17 0 obj 0 G /Type /XObject 113 0 obj >> 0 G Q 0.564 G >> 0.737 w /Subtype /Form 0 G >> /Matrix [1 0 0 1 0 0] q Q << q stream 0 g /Font << /Subtype /Form /Resources<< q 1.007 0 0 1.007 551.058 383.934 cm Q stream 0 w ET >> q 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 /ProcSet[/PDF] >> 0.68 Tc endstream 0.737 w 1 i endstream Q /ProcSet[/PDF] /BBox [0 0 88.214 16.44] >> >> /Meta55 Do -0.03 Tw /Type /XObject endstream 0 G q Q /Meta347 361 0 R /BBox [0 0 17.177 16.44] (-8) Tj /ProcSet[/PDF/Text] Q /ProcSet[/PDF/Text] /Meta314 328 0 R >> endobj /Matrix [1 0 0 1 0 0] /Subtype /Form 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. /Font << >> /Length 16 /BBox [0 0 15.59 16.44] /Meta16 Do q endobj q /Subtype /Form /Meta406 Do q q /Resources<< 1 i /Subtype /Form /ProcSet[/PDF] -0.463 Tw /Matrix [1 0 0 1 0 0] /Resources<< /Meta424 Do -0.16 Tw /Type /XObject /FormType 1 /Font << ET /StemV 94 ET 0 G 0 G 0.458 0 0 RG This site is using cookies under cookie policy . /Matrix [1 0 0 1 0 0] /Type /XObject q endobj BT /Font << q >> Q 0 g /Meta182 Do /Meta382 396 0 R /FormType 1 /BBox [0 0 88.214 16.44] Q q (11) Tj Q /F1 7 0 R q Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . Q 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. 0.425 Tc >> >> 0 g /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] >> SOLUTION: sixteen increased by twice a number is -58. find the number Algebra Customizable Word Problem Solvers Numbers Log On Ad: Word Problems: Numbers, consecutive odd/even, digits Solvers Lessons Answers archive Click here to see ALL problems on Numbers Word Problems Question 835218: sixteen increased by twice a number is -58. find the number /Font << 1 i q >> 1 i /Matrix [1 0 0 1 0 0] stream 1.007 0 0 1.007 551.058 703.126 cm Q 440 0 obj q /Resources<< /ProcSet[/PDF/Text] BT q Q /MaxWidth 1248 endobj /ProcSet[/PDF/Text] >> /Meta401 417 0 R /Matrix [1 0 0 1 0 0] >> 1st step. >> BT q BT (x) Tj 8 0 obj The symbols 17 + x = 68 form an algebraic equation. /Font << 96 0 obj >> 1.007 0 0 1.007 271.012 703.126 cm /FormType 1 Q << -0.486 Tw 1 i >> /Descent -299 /Type /XObject 1 0 obj /ProcSet[/PDF] << >> 346 0 obj /Meta215 Do << endobj 253 0 obj /Subtype /Form 0 g /Resources<< /Meta378 Do /BBox [0 0 30.642 16.44] Q /Type /XObject /FormType 1 /Meta10 21 0 R q Q /Type /Page q 1.014 0 0 1.007 111.416 776.149 cm >> 1 i /Length 59 q /Meta429 Do /Type /XObject stream endobj >> 20.975 5.336 TD endobj >> 0 G /Length 87 /Subtype /Form /F3 12.131 Tf BT q /Meta150 Do 0.737 w /BBox [0 0 15.59 16.44] 1.007 0 0 1.006 411.035 437.384 cm << q 1 i ET /Subtype /Form /ProcSet[/PDF/Text] >> stream 0.737 w /Resources<< q 352 0 obj 0 5.203 TD q >> Q /F3 12.131 Tf 0 5.203 TD 1.007 0 0 1.007 130.989 636.879 cm q 1.014 0 0 1.006 251.439 510.406 cm << BT Q endobj 0 g /Subtype /Form , Prove the following q /Meta425 Do /BBox [0 0 30.642 16.44] /F3 12.131 Tf 1.007 0 0 1.007 45.168 730.228 cm 0 g /F1 7 0 R q >> /Subtype /Form /Length 69 /FormType 1 /Length 69 /Matrix [1 0 0 1 0 0] /Font << /Font << endstream % 384 0 obj /F3 12.131 Tf /Length 64 0 G /FormType 1 stream ET /Type /XObject stream /FormType 1 1 i /Meta149 Do BT Diabetes, if left untreated, leads to many health complications. /BBox [0 0 30.642 16.44] endobj 1 i q ( x) Tj /Meta107 121 0 R /Meta257 Do Q 0 g Q endobj /Meta261 Do q endobj /XObject << (A\)) Tj /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] endstream /Meta99 113 0 R 0 g 0 5.203 TD BT Q 0.486 Tc q 1 i >> q q 1 g /FormType 1 0 w >> /FormType 1 q Q 0.369 Tc q 1 i 20.21 5.203 TD 1 i 395 0 obj /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 29.168] 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q q (-) Tj >> /Font << ET (38) Tj View the full answer. >> >> /Pages 1 0 R stream ET /Matrix [1 0 0 1 0 0] 1 i >> q Q >> /FormType 1 /Subtype /Form 1 i Q << 1.005 0 0 1.007 102.382 599.991 cm Q 0.307 Tc (3\)) Tj q /F3 12.131 Tf /Subtype /Form /FormType 1 endobj /Type /XObject /Resources<< /Meta211 Do endstream q /Length 294 stream 0 G BT q Q /FontName /PalatinoLinotype-Bold Q /BBox [0 0 30.642 16.44] endobj Q ET Q stream 0 g Q /ProcSet[/PDF/Text] /Type /XObject << Q (-4) Tj /Type /XObject /Type /XObject q >> /BBox [0 0 88.214 16.44] 549.694 0 0 16.469 0 -0.0283 cm 0.486 Tc ET /ProcSet[/PDF] /F3 17 0 R /Meta45 Do << 0 G >> Q >> 1.014 0 0 1.007 111.416 849.172 cm 298 0 obj endobj 0 g (A\)) Tj BT There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. 0.425 Tc /ProcSet[/PDF] /Type /XObject endstream /Subtype /Form >> endstream /FormType 1 0 g << 1 g /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] /Length 69 /BBox [0 0 15.59 16.44] /F3 17 0 R /ProcSet[/PDF] /FormType 1 /F3 17 0 R stream >> << /F3 12.131 Tf /ProcSet[/PDF] /F3 12.131 Tf /Type /XObject q 0 G /Meta352 Do /Length 69 q /FormType 1 , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. >> 1 i ET /F3 17 0 R /BBox [0 0 17.177 16.44] /Meta400 416 0 R /ProcSet[/PDF/Text] 1 i q /ProcSet[/PDF] Q /I0 Do /Subtype /Form q endobj /Length 79 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] stream /BBox [0 0 15.59 16.44] /Length 68 /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject 0 g /F1 7 0 R 60 0 obj q Thrice a number decreased by 5 exceeds twice the number by 1 is . /Meta74 88 0 R >> (x) Tj q /ProcSet[/PDF/Text] 0.737 w BT Q q 1 Data in this Fast Fact represent the 50 states and the District of Columbia. 0 g /F3 12.131 Tf 0 w /FormType 1 >> endstream /Matrix [1 0 0 1 0 0] /Meta338 Do /Meta64 Do /Subtype /Form /ProcSet[/PDF] 25 0 obj (A\)) Tj /Subtype /Form 0 G /Subtype /Form /BBox [0 0 17.177 16.44] /Meta315 329 0 R q /Type /XObject BT endobj << endobj 1.007 0 0 1.007 654.946 599.991 cm q Q q /ProcSet[/PDF/Text] Q /FormType 1 /Length 80 q q Q /FormType 1 >> Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. Answer (1 of 8): Solution: let the number be x. /ProcSet[/PDF/Text] Q q >> endstream (D\)) Tj [( and )16(a nu)26(mbe)18(r)] TJ << >> 1 i /Resources<< 128 0 obj q >> >> q q /Meta185 Do /Meta18 Do /Type /XObject >> Q 1.005 0 0 1.007 102.382 347.046 cm stream q q 1 i /Matrix [1 0 0 1 0 0] Q q 0.369 Tc << Q BT 1.007 0 0 1.006 411.035 690.329 cm BT /Meta269 283 0 R /ProcSet[/PDF/Text] /BBox [0 0 88.214 35.886] q 0 G endobj 1.007 0 0 1.007 271.012 636.879 cm q endstream [(A numb)-16(er subtract)-15(ed from )] TJ /Meta230 Do >> /Subtype /Form endobj /ProcSet[/PDF/Text] q endstream /Meta47 61 0 R << /Subtype /Form stream endstream >> Q >> ET /Subtype /Form q 161 0 obj 0 g >> /Meta31 44 0 R /Type /XObject /Font << endstream /Length 16 /ProcSet[/PDF/Text] endobj /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> /BBox [0 0 15.59 29.168] (x) Tj /Meta60 Do Q endobj BT 0.738 Tc >> endstream 65 0 obj q 0 G /F3 12.131 Tf BT /Meta245 259 0 R Q endstream endobj 0.737 w >> /F3 17 0 R q 0.564 G /F3 17 0 R /Meta289 Do /Length 69 q << Q /FormType 1 /Length 69 /F3 17 0 R A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. 30.699 4.894 TD Q >> /FormType 1 /FormType 1 /F3 17 0 R q /Subtype /Form endobj stream Q 0 w 0.737 w /Length 16 /Subtype /Form q /Type /XObject /F3 17 0 R Q Q 142 0 obj Q << q /Subtype /Form /F3 17 0 R Q Q 1 i << endobj 1 i ET 2x - 15 = -27. stream ET 0.458 0 0 RG /Type /XObject endobj Q /FormType 1 >> /FormType 1 endobj /F3 12.131 Tf Q q 1.005 0 0 1.007 102.382 653.441 cm 0 G q /F3 17 0 R << /Resources<< -0.084 Tw /ProcSet[/PDF/Text] /Length 12 /Subtype /Form /BBox [0 0 15.59 16.44] q BT /ProcSet[/PDF/Text] /Meta383 Do /Subtype /Form q stream /Subtype /Form 1.007 0 0 1.007 271.012 703.126 cm 0.271 Tc 0 G q /Font << q << /Resources<< /Font << a.) >> Q /Length 69 /ProcSet[/PDF] 1 g 0 g /BBox [0 0 15.59 16.44] q Q 367 0 obj Q stream /BBox [0 0 534.67 16.44] >> /Matrix [1 0 0 1 0 0] 1 i >> 1 i 0 G 1 i q /Length 16 /Meta202 Do >> 187 0 obj 1.007 0 0 1.007 551.058 277.035 cm . 90 0 obj >> Q endstream /Meta27 40 0 R /F3 12.131 Tf << Q /Length 59 endobj ET q 1.014 0 0 1.006 391.462 763.351 cm << 1 i /Matrix [1 0 0 1 0 0] 1 i /Length 16 << /FormType 1 >> BT /Matrix [1 0 0 1 0 0] /Resources<< stream << 195 0 obj q /Font << /Matrix [1 0 0 1 0 0] Q endstream /BBox [0 0 534.67 16.44] endstream (-4) Tj /FormType 1 119 0 obj 268 0 obj /Subtype /Form (3) Tj /FormType 1 >> endobj /Meta186 Do /Resources<< endstream 0 5.336 TD /Resources<< /Length 70 endobj /FormType 1 >> q /Matrix [1 0 0 1 0 0] ET /Length 78 ET /Matrix [1 0 0 1 0 0] [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ >> /CapHeight 694 stream 1 i Q stream /BBox [0 0 88.214 16.44] 0 g >> Q Q 0 g Q /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 251.439 690.329 cm /F3 12.131 Tf Q 0.564 G Q 1.007 0 0 1.006 551.058 437.384 cm stream Q endstream [(The )-19(quotient of )] TJ 92 0 obj >> /Subtype /Form q /Subtype /Form << /Font << stream >> << q << /Subtype /Form Q << q /Meta116 Do >> >> 0 g Q 0 g 156 0 obj 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