Twice a number decreased by 58! 0.458 0 0 RG (\)) Tj /FormType 1 20.975 5.336 TD Q 0.737 w << q Q << /Type /XObject stream q q /F3 12.131 Tf 1 g Q /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject stream 0.458 0 0 RG endobj endobj 1 i 1 g (2) Tj /Matrix [1 0 0 1 0 0] 722.699 546.541 l Q Q /Type /XObject << 0 G Q (-23) Tj >> stream /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] >> stream At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . >> Q 1 i (-23) Tj /Type /XObject /Resources<< /Type /XObject /Meta222 Do BT >> 1 i /Matrix [1 0 0 1 0 0] /Length 54 Q << << stream /Width 734 /FormType 1 >> answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. Choose an expert and meet online. (A\)) Tj /Meta367 381 0 R /Length 16 q /ProcSet[/PDF/Text] q /Type /XObject q /Subtype /Form ET stream /Subtype /Form q endstream 1.005 0 0 1.007 102.382 670.003 cm >> /F1 12.131 Tf BT 377 0 obj << 0 g /Font << /BBox [0 0 88.214 16.44] /Meta229 243 0 R BT 1 i >> 0 5.203 TD /Length 59 /F1 7 0 R /Meta256 Do stream /Matrix [1 0 0 1 0 0] << /Length 16 endstream BT 16.469 5.203 TD BT 0 g endobj /Font << /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /Font << << stream 1 i /Meta225 239 0 R /ProcSet[/PDF/Text] stream 0 g 12 0 obj /F3 17 0 R << q Q /F3 12.131 Tf 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 /BBox [0 0 30.642 16.44] /Meta250 Do /Length 59 q << << Q Q 145 0 obj 0 g q 1 i Q 1.005 0 0 1.015 45.168 53.449 cm /Resources<< /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] ET /Length 69 endstream /Type /XObject 0.458 0 0 RG >> Question. Q /Type /XObject >> endobj /Length 118 >> q /Meta156 Do /Meta8 19 0 R /FormType 1 ET 1 g 0.564 G 1 i << << Q /Type /FontDescriptor /Font << (-9) Tj -0.486 Tw BT Negative thirteen decreased by 3 times a number x. /Meta241 255 0 R endstream /ProcSet[/PDF] 0 g q /MediaBox [0 0 767.868 993.712] q Q stream /Meta314 328 0 R Q >> >> /Font << a and b or something else.***. Formula - How to Calculate Percentage Decrease. >> 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. 1.014 0 0 1.007 391.462 330.484 cm /Font << /BBox [0 0 534.67 16.44] q /Type /XObject << 1.014 0 0 1.006 391.462 763.351 cm 1 g << 445 0 obj /Font << /Matrix [1 0 0 1 0 0] If a number is 50%, then it is a half - the same as 0.5 or 1/2. Q endobj /Resources<< /F4 12.131 Tf q >> 150 0 obj /BBox [0 0 15.59 16.44] /Meta390 406 0 R /ProcSet[/PDF/Text] ET Q q [(F)-22(ive)] TJ 0.178 Tc /Resources<< /BBox [0 0 88.214 16.44] q 270 0 obj 1 i /FormType 1 Q /BBox [0 0 534.67 16.44] /FormType 1 0 G 0 G /Font << 1 i /Resources<< /Subtype /Form /Meta88 Do q /Meta192 206 0 R /Type /XObject stream /ProcSet[/PDF/Text] 281 0 obj 0 G /BBox [0 0 88.214 35.886] >> /BBox [0 0 30.642 16.44] /Resources<< q /Length 69 /F3 12.131 Tf Q /Resources<< 271 0 obj endstream >> endstream 1.005 0 0 1.007 102.382 799.486 cm stream 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. ET "49 . endstream (38) Tj /Meta16 27 0 R /Flags 32 >> endstream 118.317 5.203 TD /Meta331 Do >> 0.524 Tc /Meta251 Do 0 G 1 g /Matrix [1 0 0 1 0 0] q 0 g 1.007 0 0 1.007 411.035 330.484 cm /Font << 1.007 0 0 1.007 271.012 383.934 cm Table 1. 0 G 0 w Q 260 0 obj 0 g endstream >> 0.458 0 0 RG >> 0.737 w /BBox [0 0 15.59 16.44] q >> BT /FormType 1 0.738 Tc /Length 16 /Font << q Q q Q q << /Resources<< /Matrix [1 0 0 1 0 0] BT /Resources<< Q /F3 12.131 Tf 0 g q >> q 1 i /Font << 1.014 0 0 1.006 111.416 690.329 cm /Matrix [1 0 0 1 0 0] /Meta361 375 0 R q 1 i /Meta378 Do >> endstream /Meta282 Do Q 0.458 0 0 RG stream 1.014 0 0 1.006 111.416 763.351 cm Q 1 g /Length 54 /Matrix [1 0 0 1 0 0] /Type /XObject /Subtype /Form << Q >> Q q The sum Of twice a nu4ber What is the number? q A) 5 more than a number. endobj q q /Meta399 Do /Subtype /Form >> Q 0 G 0 g /Meta239 Do Q >> /Type /XObject /F3 12.131 Tf 1 i Q q Q /Resources<< Q /Subtype /Form >> ET << /Meta273 Do /Meta368 382 0 R /F3 12.131 Tf endstream Q 195 0 obj endstream endobj endobj << Q q Q /Font << endstream << (A\)) Tj 1 i 1.007 0 0 1.007 45.168 796.475 cm /ProcSet[/PDF] q q Q /Font << /Type /XObject /ProcSet[/PDF] Q endobj endobj Q 1 i 549.694 0 0 16.469 0 -0.0283 cm /Resources<< stream >> 1.005 0 0 1.007 102.382 872.509 cm >> q /FormType 1 /Subtype /Form /Length 118 /Length 59 q ET /F3 12.131 Tf /Meta4 Do /BBox [0 0 88.214 16.44] 0.564 G /F1 7 0 R /Type /XObject stream 0 g 1.007 0 0 1.007 271.012 636.879 cm (1) Tj << BT q /Subtype /Form Q 0 G q /F3 12.131 Tf /Length 59 Q 100 0 obj 1 i /FormType 1 >> 1.007 0 0 1.007 411.035 636.879 cm endobj /F4 36 0 R /Resources<< 0 G /F3 12.131 Tf 0 G /Resources<< /Meta139 153 0 R 16.469 5.203 TD (x) Tj stream >> /Type /XObject 0 g Q /ProcSet[/PDF/Text] Q /ProcSet[/PDF/Text] Q 336 0 obj /Meta330 Do 6.746 5.203 TD /Resources<< /ProcSet[/PDF/Text] q Q 1 i << q >> >> /Resources<< 0 g /Length 57 43.426 5.203 TD /Resources<< Q /ProcSet[/PDF/Text] /Subtype /Form q Q 1 g q /FormType 1 >> /Resources<< 1 i endstream /Meta291 Do >> 1 i /F4 36 0 R /BBox [0 0 88.214 16.44] << /Matrix [1 0 0 1 0 0] q /FormType 1 1 i 0.564 G 0.564 G stream endobj /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g q 0 G 0 g 0.425 Tc >> /Subtype /Form /Meta343 Do 1 i Q /Subtype /Form >> Thrice a number decreased by 5 exceeds twice the number by a unit. >> q q >> 1.007 0 0 1.007 411.035 636.879 cm 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 /Font << 0.458 0 0 RG /F3 17 0 R /Meta123 137 0 R /Subtype /Form ET (D\)) Tj /ProcSet[/PDF] endstream 0 g 206 0 obj ( x) Tj /Matrix [1 0 0 1 0 0] /Length 294 (x ) Tj /BBox [0 0 15.59 29.168] (3\)) Tj Q << q q 0 g /Resources<< q /F3 12.131 Tf << /Meta358 372 0 R /Meta105 Do With this, we get: "3x-8". q /Font << endstream BT q >> q BT q >> >> 0 g << 0 g 1.007 0 0 1.007 271.012 583.429 cm /Length 69 306 0 obj /F3 12.131 Tf /F3 17 0 R /BBox [0 0 534.67 16.44] 1 i q Twice the difference of a number and three totals twelve 8. q /BBox [0 0 88.214 16.44] 1.014 0 0 1.006 111.416 510.406 cm /BBox [0 0 30.642 16.44] q /ProcSet[/PDF/Text] Q 0 G /FormType 1 stream 1.007 0 0 1.007 551.058 703.126 cm endobj /Meta395 411 0 R >> /Meta169 183 0 R BT Q /FormType 1 Q >> endstream Q >> (-) Tj /F3 17 0 R 32.201 5.203 TD /ProcSet[/PDF/Text] 293 0 obj /BBox [0 0 88.214 16.44] q 672.261 599.991 m 0 g /BBox [0 0 15.59 29.168] >> /Meta227 Do >> ET /Subtype /Form << 126 0 obj q q << Q 1 g endobj Q q 1.005 0 0 1.007 102.382 653.441 cm /BBox [0 0 30.642 16.44] >> 1 i /Meta287 301 0 R q /Length 85 0 5.203 TD 1 i Q q q 0 5.203 TD >> /Font << /BBox [0 0 30.642 16.44] endstream 1 i /Meta116 130 0 R 1 g 0 g Q >> /Length 69 Q /F3 17 0 R ET /XHeight 476 Q q endstream /Meta372 Do q q 0.838 Tc 286 0 obj /Type /XObject >> q 0 g Q /F3 17 0 R /Type /XObject 0 G 0 G /Type /XObject /Font << /FormType 1 0.564 G 1.007 0 0 1.007 67.753 293.596 cm endobj Q 1.005 0 0 1.007 102.382 400.496 cm 112 0 obj /BBox [0 0 88.214 16.44] Q 163 0 obj 0 G /Length 67 BT /Meta116 Do Q 1.007 0 0 1.006 411.035 763.351 cm /XHeight 471 1 i /Meta404 420 0 R 1.005 0 0 1.007 102.382 363.608 cm q /Matrix [1 0 0 1 0 0] 0 g /BBox [0 0 88.214 16.44] 0 g q -0.486 Tw 0 G /BBox [0 0 30.642 16.44] endstream 1.005 0 0 1.007 79.798 829.599 cm 0 w /Length 16 Q /Meta204 218 0 R /Type /XObject 1 g /Resources<< 1 i /Resources<< 0 G /F1 12.131 Tf /ProcSet[/PDF/Text] 0.458 0 0 RG 0 G Q /ProcSet[/PDF] 1 i /BBox [0 0 88.214 16.44] /Meta186 200 0 R /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] q (B\)) Tj q 0 g >> q >> 0.737 w ET 1 i /Meta263 Do 1.007 0 0 1.006 411.035 510.406 cm 0 g >> /Font << endstream 1 i /MediaBox [0 0 767.868 993.712] /BBox [0 0 88.214 16.44] /Font << 0 G /Type /XObject q /FormType 1 1 i 0 g /Length 64 q Q Q BT q /Length 69 endstream >> >> 0.297 Tc 0 G >> q q /Length 67 >> Q >> /F3 12.131 Tf Q /Subtype /Form Q endstream q /Resources<< Q -0.463 Tw (x) Tj 361 0 obj /Meta49 63 0 R q /Subtype /Form /Font << /Meta138 152 0 R endobj /Subtype /TrueType 0 g On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini 278 0 obj endstream c Site 5 is not included in this number. 0 g /Matrix [1 0 0 1 0 0] /Font << << /Length 60 >> Q Q /ProcSet[/PDF/Text] /Meta233 Do /Type /XObject 0 g /F3 17 0 R /BBox [0 0 88.214 16.44] Q (3) Tj Q 1 i >> /Meta23 34 0 R q /Font << q Q q >> q 0 5.203 TD Q: when six times a number is decreased by 4, the result is 8. /Type /XObject stream Q 1.014 0 0 1.007 391.462 450.181 cm endstream We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. Now that you know the meaning of the key words you can read the problem differently. Q 0 G q >> /Length 139 /Subtype /Form 334 0 obj q q endobj Five times a number, decreased by 58, is -23 Find the number. >> q Q Q /Type /XObject >> /Subtype /Form /BBox [0 0 30.642 16.44] /Type /XObject << endobj /Resources<< /F3 17 0 R Q 0 G Q q stream Q Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] Find the number. /FormType 1 /Resources<< Q /Resources<< endobj >> /FormType 1 >> /F1 12.131 Tf /Contents [399 0 R] >> /FormType 1 /Meta293 307 0 R Q 0.51 Tc /ProcSet[/PDF] /Subtype /Form q Q /Resources<< q 0 G /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] endstream >> endstream q 0.838 Tc /Meta7 18 0 R ET /F3 17 0 R >> q << /BBox [0 0 88.214 16.44] 0.786 Tc /Resources<< /I0 51 0 R 0.68 Tc /BBox [0 0 88.214 16.44] stream >> ET q /Meta285 299 0 R >> q q q Q q /Matrix [1 0 0 1 0 0] 59 0 obj /BBox [0 0 30.642 16.44] ET << Q 1 i /Meta402 Do /Subtype /Form q To find: The. 1.014 0 0 1.006 531.485 836.374 cm /FormType 1 1 i /FormType 1 /BBox [0 0 88.214 16.44] endobj 326 0 obj /BBox [0 0 88.214 16.44] /Resources<< /FormType 1 ET << 1.007 0 0 1.007 551.058 383.934 cm /Resources<< BT 121 0 obj 0 g /Meta305 Do 1 g /F3 12.131 Tf q 0.564 G 1 i 292 0 obj /ProcSet[/PDF/Text] 10.487 5.203 TD >> /F4 12.131 Tf << /Meta39 53 0 R 1 i Q Q /ProcSet[/PDF/Text] endstream 20.21 5.203 TD /Length 69 >> /Matrix [1 0 0 1 0 0] stream endobj q /ProcSet[/PDF/Text] 1 i /Matrix [1 0 0 1 0 0] 0 g Q ET 21.713 20.154 l stream 43 0 obj /Meta158 Do /BBox [0 0 88.214 16.44] endobj /BBox [0 0 673.937 68.796] /Meta25 Do 0.425 Tc ET ET Q Q endobj 1.007 0 0 1.007 411.035 277.035 cm stream /BBox [0 0 15.59 16.44] endobj /Matrix [1 0 0 1 0 0] Abstract: The aim of the study was to investigate the expression of miR-155 in plasma and peripheral blood mononuclear cells (PBMCs), the effects of miR-155 on the apoptosis rate /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) BT /F3 12.131 Tf /Subtype /Form >> /Subtype /Form 1 i /Meta38 52 0 R 20.21 5.203 TD 0.458 0 0 RG /ProcSet[/PDF] /Subtype /Form 1 g /FormType 1 BT endstream stream /Matrix [1 0 0 1 0 0] (11) Tj q /Meta209 Do 1 i /Meta165 179 0 R /Matrix [1 0 0 1 0 0] Q ET q /Meta289 Do (-23) Tj 0.458 0 0 RG Q Q /BBox [0 0 15.59 16.44] [( and )16(a nu)26(mbe)18(r)] TJ stream << /ProcSet[/PDF/Text] /Subtype /Form 672.261 799.486 m /BBox [0 0 17.177 16.44] 0.458 0 0 RG endstream /Meta177 191 0 R /Subtype /Form /Length 59 -0.463 Tw Q /BBox [0 0 88.214 16.44] /Font << stream Q /Type /XObject 0.68 Tc /Type /XObject 0 g /Meta301 315 0 R Testosterone is the primary sex hormone and anabolic steroid in males. /Meta187 201 0 R [(The )-19(quotient of )] TJ /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] >> 0 w BT 22 0 obj stream /Resources<< /Resources<< /ProcSet[/PDF] q /F1 12.131 Tf /Resources<< >> Q stream >> /Matrix [1 0 0 1 0 0] Q >> 0 G Q BT /Meta124 138 0 R stream /FormType 1 /Resources<< q endstream 0.369 Tc /Resources<< /Resources<< q q /BBox [0 0 88.214 16.44] 1.502 5.203 TD 1.005 0 0 1.007 102.382 799.486 cm /Matrix [1 0 0 1 0 0] 1 i >> /Meta428 Do /Subtype /Form << << 0 g 0 5.336 TD /Matrix [1 0 0 1 0 0] 0 g endstream 0 w /Matrix [1 0 0 1 0 0] Q /Meta99 Do /Subtype /Form >> << /Matrix [1 0 0 1 0 0] 47 0 obj q 174.501 5.203 TD /Meta127 Do [(-3)-16(20)] TJ /Resources<< 1 i /FormType 1 1.007 0 0 1.007 551.058 703.126 cm Q /F1 12.131 Tf (-20) Tj /Resources<< Q /Subtype /Form Q BT Q << Find the number. >> 1.007 0 0 1.007 551.058 523.204 cm (+) Tj /Meta23 Do << endstream /F3 17 0 R 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 q /F4 12.131 Tf stream /BBox [0 0 15.59 16.44] q /Length 16 ET /CapHeight 476 >> q q Q Q /Type /XObject Q 1 i endstream /I0 51 0 R 1 i BT endobj /Meta266 280 0 R ET /F4 36 0 R 250 0 obj /Type /XObject 0 5.203 TD << q /Meta362 Do 0 G q /Meta124 Do /ProcSet[/PDF/Text] Q endobj q 0.738 Tc >> /Matrix [1 0 0 1 0 0] /Subtype /Form q /Resources<< /Type /XObject /F3 17 0 R q q /FormType 1 /Type /XObject 0 4.894 TD /Resources<< /Subtype /Form >> 1.007 0 0 1.007 411.035 330.484 cm A number = an unknown number which can be represented by a variable, usually x. BT >> Q >> Q q q /Resources<< (8\)) Tj /Type /XObject (B\)) Tj 0.564 G q Q >> q /Matrix [1 0 0 1 0 0] q 86 0 obj << /Matrix [1 0 0 1 0 0] 0 g 0 G << /Length 19882 >> 0.51 Tc /F3 12.131 Tf 0.737 w q 1 i 1 g (x) Tj 0.369 Tc 0.564 G << /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 1 i /Matrix [1 0 0 1 0 0] Double or twice a number means 2x, and triple or thrice a number means 3x. 1.014 0 0 1.007 111.416 330.484 cm /Subtype /Form (2\)) Tj 22.478 5.336 TD BT 1.007 0 0 1.007 551.058 703.126 cm /F3 17 0 R >> >> endobj >> /ProcSet[/PDF/Text] /Subtype /Form >> 1 i stream 1.005 0 0 1.007 79.798 862.723 cm /FormType 1 /F3 12.131 Tf 0 g 1 i /ProcSet[/PDF/Text] stream 224 0 obj /Meta280 294 0 R /Length 69 /ProcSet[/PDF] stream /Length 70 /XObject << /Matrix [1 0 0 1 0 0] (2) Tj /Font << Q 108 0 obj 1.007 0 0 1.007 271.012 636.879 cm endstream /Subtype /Form 443 0 obj /BBox [0 0 88.214 16.44] Q endobj 0 g >> /ProcSet[/PDF/Text] /BBox [0 0 15.59 16.44] q Q >> MetS-Z quartiles and their associated risks are presented in Fig. /Subtype /Form q q 6.746 5.336 TD endstream q /ProcSet[/PDF/Text] /F3 17 0 R q /Meta39 Do 0 G 181 0 obj 0 g /FormType 1 q (1\)) Tj /Length 67 endstream Q /Type /XObject /Subtype /Form q q /Resources<< Q >> 1 i q -37 VI 2. /Length 59 ET 213 0 obj Q Q q Q q >> Let x the unknown number. (-8) Tj << /Matrix [1 0 0 1 0 0] q /Meta8 Do >> /BBox [0 0 88.214 16.44] endobj /Length 16 1.007 0 0 1.007 654.946 599.991 cm 238 0 obj 153 0 obj q endstream 0.737 w 0.51 Tc Q 0 g /ProcSet[/PDF/Text] /F3 17 0 R q /ProcSet[/PDF/Text] 0.564 G 0.738 Tc 0 G /Length 12 /Subtype /Form 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. >> /ProcSet[/PDF/Text] 234 0 obj 0.458 0 0 RG >> /Length 12 /Matrix [1 0 0 1 0 0] q stream endstream >> 1.014 0 0 1.007 531.485 450.181 cm /FormType 1 /Matrix [1 0 0 1 0 0] 0 g /Resources<< 0.458 0 0 RG >> /Type /XObject 0 g /F3 17 0 R /Type /XObject Q /Meta338 Do endobj q /Meta10 Do 672.261 473.519 m /Meta273 287 0 R /F1 12.131 Tf Q Q q /F3 12.131 Tf endobj /Type /XObject 0 G /F1 12.131 Tf stream stream 125 0 obj Q Q /Type /XObject Q /F1 12.131 Tf /Length 70 endobj >> /BBox [0 0 88.214 16.44] /Meta331 345 0 R BT /Meta414 430 0 R /Filter [/CCITTFaxDecode] /Resources<< BT 1 i /Resources<< 0.564 G q Q endstream 1.005 0 0 1.007 79.798 796.475 cm q ET /Matrix [1 0 0 1 0 0] (x) Tj /Meta137 151 0 R 0 g /F1 12.131 Tf 0 g q endstream 20.21 5.203 TD ET /Matrix [1 0 0 1 0 0] q 0.369 Tc endstream /Meta173 Do 1 g /Matrix [1 0 0 1 0 0] q q 0 4.894 TD /F1 7 0 R 437 0 obj 0.737 w /Subtype /Image /ProcSet[/PDF] /ProcSet[/PDF/Text] q >> endobj Q BT >> /Subtype /Form 1 i >> /Matrix [1 0 0 1 0 0] /FormType 1 /Meta228 Do ET 0 G 1 i 0 g Q BT /Font << Q /Matrix [1 0 0 1 0 0] q /Resources<< /Meta355 369 0 R 0 G /F3 12.131 Tf startxref >> 0.564 G << endobj /Resources<< Q q 0 5.203 TD 303 0 obj stream >> -0.126 Tw /Subtype /Form /ProcSet[/PDF/Text] BT 202 0 obj q /F3 17 0 R /Matrix [1 0 0 1 0 0] /Type /XObject 0 G stream stream 0 g << q /Type /XObject (x) Tj /FormType 1 0 G /ProcSet[/PDF] q /Font << /FormType 1 /Type /XObject 0.458 0 0 RG /Type /XObject 0 G 0 w 1.007 0 0 1.007 271.012 450.181 cm 1.502 5.203 TD >> Q >> /ProcSet[/PDF/Text] endstream /Matrix [1 0 0 1 0 0] /Font << Q 261 0 obj /Font << q BT /ProcSet[/PDF/Text] /Resources<< /Length 16 1 i q /ProcSet[/PDF/Text] /Subtype /Form 1.014 0 0 1.007 251.439 583.429 cm 0.332 Tc /Length 60 /Subtype /Form >> 0 g Q /Font << 182 0 obj stream 0 g /Font << 0 G /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] stream BT /ProcSet[/PDF/Text] Q 0.738 Tc endstream /BBox [0 0 15.59 29.168] /F3 17 0 R [3] One half of a number increased by fourteen is twenty-one. /Font << >> /BBox [0 0 673.937 14.853] >> /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 35.886] /FormType 1 q << << >> stream << Q /Meta236 Do /Type /XObject Q /Length 16 q Q Q (5) Tj q /Type /XObject /F3 12.131 Tf ET Q << Q 176 0 obj Q >> ET /Meta308 322 0 R 1.007 0 0 1.007 130.989 523.204 cm 1 i stream stream /Resources<< 425 0 obj << /ProcSet[/PDF/Text] >> endobj /ProcSet[/PDF/Text] Q 0.458 0 0 RG 0.369 Tc /Meta221 235 0 R /Resources<< /Length 245 2. q ET q 0 G B. ET /Matrix [1 0 0 1 0 0] >> q /FormType 1 (D\)) Tj Q endobj Q 0.68 Tc /Meta223 237 0 R 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 /Matrix [1 0 0 1 0 0] /Length 12 /Subtype /Form /BBox [0 0 30.642 16.44] /ProcSet[/PDF/Text] /Subtype /Form << 0.458 0 0 RG stream endobj endobj 1 i << Q /FormType 1 /Resources<< /Matrix [1 0 0 1 0 0] stream Q q Q >> 1.502 5.203 TD /ProcSet[/PDF] /Meta219 233 0 R /BBox [0 0 534.67 16.44] ET /Length 16 1.007 0 0 1.007 130.989 330.484 cm 99 0 obj /Subtype /Form /Resources<< 0.737 w endobj /Resources<< /Meta59 73 0 R (+) Tj 12.727 24.649 TD /Meta66 Do stream /Type /XObject /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] ET stream Q Q /FormType 1 1 Data in this Fast Fact represent the 50 states and the District of Columbia. >> endobj 0.51 Tc >> /ProcSet[/PDF/Text] stream /Meta425 Do q 1.007 0 0 1.007 551.058 277.035 cm Solution: Let the number be x. /ProcSet[/PDF] << 1 i /FormType 1 1 i >> /ProcSet[/PDF/Text] endobj /FormType 1 Q Q Q q q q 0 G q /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 636.879 cm Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. /Meta410 426 0 R (C\)) Tj BT >> (58) Tj /Resources<< /Resources<< /Meta187 Do endstream /Meta401 Do /XObject << 0.458 0 0 RG >> /Resources<< /Matrix [1 0 0 1 0 0] Q /BBox [0 0 15.59 16.44] /FormType 1 /Type /XObject q q Q >> q /Type /XObject endobj >> Q /F3 12.131 Tf 12.727 5.203 TD /Meta103 117 0 R Q >> 0 g 1 i 0 g 9.723 5.336 TD q /F3 17 0 R 1 i /Type /XObject 1.014 0 0 1.007 391.462 383.934 cm >> endobj endobj /ProcSet[/PDF/Text] q Q /F3 12.131 Tf 311 0 obj 549.694 0 0 16.469 0 -0.0283 cm 19.474 20.154 l /ProcSet[/PDF] Q q >> >> stream ET >> /Meta74 88 0 R 384 0 obj << Q stream << 76 0 obj /Length 58 /ProcSet[/PDF/Text] 1 i /F3 17 0 R 1 i q q stream q /ProcSet[/PDF/Text] 0 G /Resources<< /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0 w 33 0 obj /FormType 1 Q Q q q 315 0 obj /F3 12.131 Tf Q q >> /Matrix [1 0 0 1 0 0] /I0 51 0 R >> /Meta361 Do /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Q 0 w 8.985 20.154 l /F3 17 0 R /Font << /Length 69 (viii) A number divided by 8 gives 7. Q >> BT /Type /XObject Q /ProcSet[/PDF] endobj 89.12 5.203 TD 0.369 Tc Q >> endobj Q /BBox [0 0 88.214 35.886] /F3 12.131 Tf /Meta324 Do << 32.201 5.203 TD >> /Font << 0.369 Tc /Length 69 0.486 Tc /Meta204 Do >> endobj /F3 12.131 Tf Q 1.014 0 0 1.007 391.462 383.934 cm /Matrix [1 0 0 1 0 0] /Resources<< /Length 54 /BBox [0 0 88.214 16.44] 54 0 obj /Type /XObject 0 g /Type /XObject /Matrix [1 0 0 1 0 0] >> /Length 60 /Meta225 Do /Subtype /Form /FormType 1 >> ( x) Tj >> q (x ) Tj endobj /Font << /Font << Q endstream 0.737 w /Subtype /Form /Type /XObject 1 i /Resources<< 0 g q Number Outcomes 1 42 2 41 3 . 0 g /Resources<< 198 0 obj endstream /Subtype /Form 1.502 5.203 TD /FormType 1 Q stream /Font << %%EOF. /ProcSet[/PDF/Text] stream >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0 G >> 434 0 obj /Matrix [1 0 0 1 0 0] /Subtype /Form 1.005 0 0 1.007 102.382 653.441 cm 340 0 obj endobj 279 0 obj /F4 12.131 Tf >> endstream /Font << /Length 16 1 i /Length 65 0 g 366 0 obj /BBox [0 0 88.214 35.886] /BBox [0 0 15.59 16.44] endobj /FormType 1 stream q 0.458 0 0 RG /FormType 1 /F3 17 0 R 358 0 obj BT endstream /Meta7 Do 1 i /ProcSet[/PDF/Text] /Meta5 14 0 R (-) Tj stream Q /ProcSet[/PDF] /FormType 1 /F4 12.131 Tf Q /F3 12.131 Tf /FormType 1 >> BT >> /BBox [0 0 17.177 16.44] /Meta160 Do q (iii) 25 exceeds a number by 7. /Matrix [1 0 0 1 0 0] q BT >> 0 g 1.007 0 0 1.007 130.989 776.149 cm >> /F3 17 0 R q >> >> 268 0 obj 335 0 obj stream q /Font << /Length 16 q /Meta395 Do 190 0 obj /Matrix [1 0 0 1 0 0] -0.16 Tw endobj (B\)) Tj 0 g 0 w /Subtype /Form q /XHeight 447 0 g >> /ProcSet[/PDF] Q /Font << 95 0 obj 0.463 Tc >> /Meta379 393 0 R 0 g Q 0.51 Tc /Meta366 380 0 R 0 w /F4 36 0 R >> BT 0 4.78 TD Q 0 g stream Hence, the number is 6. q 0 G q stream >> 0 g /Subtype /Form ET >> >> /Subtype /Form q q BT ET >> q 38.948 5.203 TD /Encoding /WinAnsiEncoding /Type /XObject q /Font << Q /Font << >> ET /Subtype /Form /F3 12.131 Tf ET q /FormType 1 << /Resources<< Twice a number decreased by 8 gives 58. >> 1 i /Length 63 /Type /XObject q /Type /XObject 421 0 obj BT Q stream Q 1 i /ProcSet[/PDF] q /Subtype /Form 1.005 0 0 1.007 102.382 256.709 cm (x) Tj 0 g 404 0 obj Q /FormType 1 /Matrix [1 0 0 1 0 0] stream >> /F3 12.131 Tf /FormType 1 stream /FormType 1 S 1 i << q 1 i Q >> /Meta144 Do /Meta35 Do 0 g Q Q >> Q Q /Length 12 /Resources<< 1.007 0 0 1.007 271.012 450.181 cm << /Length 65 /Meta262 276 0 R /BBox [0 0 88.214 16.44] /BBox [0 0 639.552 16.44] Q /Resources<< /Subtype /Form q >> /Matrix [1 0 0 1 0 0] q 0 g ET /Font << /Font << Q q A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6.
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