twice a number decreased by 58

Q 0.564 G 1 i q /ProcSet[/PDF] >> >> 1.007 0 0 1.006 130.989 437.384 cm << 270 0 obj Q >> 0.425 Tc 1.005 0 0 1.006 45.168 879.284 cm << Q Q /Resources<< Q 0.737 w 1.007 0 0 1.006 411.035 690.329 cm Q /Meta190 Do Q 1 i Q Answer: 52 decreased by twice a number in algebric expression Step-by-step explanation: The problem is asking that you subtract twice a number from 52. stream 32.201 5.203 TD endstream Choose the correct one. /Subtype /Form /Meta24 37 0 R 20.21 5.203 TD q 6.746 5.203 TD 0 20.154 m >> endobj endobj 0.737 w /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.006 411.035 763.351 cm /Subtype /Form q << 0 g >> 417 0 obj 1.007 0 0 1.007 271.012 703.126 cm Q 1 i /Matrix [1 0 0 1 0 0] /F3 12.131 Tf Q /BBox [0 0 88.214 16.44] stream q q endstream Q q 1 g endobj /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 130.989 776.149 cm BT ( x) Tj 0.564 G /Type /XObject stream 0 w /Matrix [1 0 0 1 0 0] >> >> /Length 16 q endstream >> The results found were expressed mainly through tables and graphs as the main resources of the statistical language. 1.005 0 0 1.013 45.168 933.487 cm 0 G /F3 17 0 R 235 0 obj 1.502 24.339 TD >> endstream ET >> >> q /Subtype /Form /Subtype /Form /ProcSet[/PDF] /FormType 1 66 0 obj /Meta165 179 0 R /Type /XObject /Resources<< /Type /XObject 1 g endstream /Resources<< /Type /XObject >> endobj ET >> 1 i BT Q /Meta54 68 0 R /Meta207 221 0 R 1 i BT /Type /XObject >> << Q 1 i q >> /Subtype /Form /Matrix [1 0 0 1 0 0] endobj << stream (9) Tj 26.957 5.203 TD /BBox [0 0 88.214 16.44] 1 i >> << endobj >> 1 i /Meta243 257 0 R 0 g endobj /Type /XObject /F3 17 0 R ET << endobj >> 1.007 0 0 1.007 271.012 583.429 cm q Q /BBox [0 0 88.214 35.886] /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 45.168 889.071 cm >> >> endobj q Q >> q << /Length 69 /Length 63 q q /ProcSet[/PDF] 322 0 obj stream 1 g 0.369 Tc (D\)) Tj /ProcSet[/PDF/Text] Q q /BBox [0 0 15.59 16.44] q Q 0 5.203 TD 1 i endobj Q 397 0 obj /Subtype /Form << q /F4 12.131 Tf 1.007 0 0 1.007 271.012 849.172 cm 1 i endstream /Subtype /Form /Meta8 19 0 R 0 5.203 TD 0 G Q /Length 69 0 g >> 1.014 0 0 1.007 531.485 330.484 cm >> q Q endstream /ProcSet[/PDF] 1.005 0 0 1.007 102.382 653.441 cm /Subtype /Form << /Matrix [1 0 0 1 0 0] >> q 0 g /ProcSet[/PDF/Text] 0 w /Resources<< << (9\)) Tj 1.007 0 0 1.007 551.058 330.484 cm Q /Matrix [1 0 0 1 0 0] 0 g /Font << 0 g /Meta34 47 0 R >> /F3 17 0 R Q stream 1 g (x ) Tj /Meta287 Do 0 g /Meta420 436 0 R Q 0 w /Font << Q /Type /XObject /Type /XObject /Subtype /Form /Meta267 Do /Meta272 286 0 R /Length 64 >> ET /Matrix [1 0 0 1 0 0] /Meta360 Do /BBox [0 0 17.177 16.44] /ProcSet[/PDF/Text] 0.382 Tc >> q Q /Length 118 /Matrix [1 0 0 1 0 0] /F3 17 0 R 0 4.894 TD 0 5.203 TD endobj /Font << 96 0 obj endobj 1 i q 0 w Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. endstream /Meta55 69 0 R 1 i >> /F3 12.131 Tf /ItalicAngle 0 q /F3 12.131 Tf >> /Type /XObject >> /Resources<< Q << /Length 16 q 0.564 G 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 /Meta23 34 0 R endobj Q >> 1 i Q (40) Tj /Type /XObject 0.311 Tc /Subtype /Form /ProcSet[/PDF/Text] stream /BBox [0 0 534.67 16.44] >> Q << /ProcSet[/PDF/Text] 0.37 Tc S /Meta132 146 0 R By the . >> 0 w 1.007 0 0 1.007 67.753 799.486 cm /FormType 1 /Meta198 212 0 R endstream 141 0 obj /F3 17 0 R /F3 17 0 R q /F3 12.131 Tf >> << BT 0 G Q 0.458 0 0 RG q 134 0 obj q q 1 i Q /Meta252 266 0 R >> << 41 0 obj Q q /XObject << Q Q Q Q Q q /Meta61 Do /FormType 1 /FormType 1 /Matrix [1 0 0 1 0 0] 0 w endstream Q >> /ProcSet[/PDF/Text] 420 0 obj 13.493 5.336 TD q q stream 26.219 5.336 TD That was 1/8 of the points that he scored /Resources<< >> >> Q /Subtype /Form BT 1 i >> 0.369 Tc q /Matrix [1 0 0 1 0 0] 97 0 obj 1 i /Meta49 63 0 R /BBox [0 0 15.59 16.44] q /Length 16 1 g (+) Tj endstream /F3 17 0 R /F3 12.131 Tf /ProcSet[/PDF] stream q /FormType 1 >> q /FontName /TimesNewRomanPSMT 0 g endobj /BBox [0 0 88.214 16.44] 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. q /FormType 1 /ProcSet[/PDF/Text] q /FormType 1 << /Meta253 Do 1 g /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] Q >> endobj /F3 17 0 R 1 g stream q << Q 0.737 w /FormType 1 >> Q /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] /Meta29 42 0 R >> >> << /F4 36 0 R /Meta265 Do /F3 17 0 R 23.952 4.894 TD /F3 12.131 Tf 0.458 0 0 RG >> >> /Resources<< >> /ProcSet[/PDF/Text] q q >> /FormType 1 50 0 obj >> q /Meta48 62 0 R Q /FormType 1 >> /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] /Length 12 266 0 obj /Length 16 0.564 G 0.737 w ET /BBox [0 0 15.59 16.44] ET endstream /Length 54 q 0 g q 0.737 w /F3 12.131 Tf q Q 0 w endobj 0 G /Meta196 Do /FormType 1 ET Q (2) Tj >> Q /F3 17 0 R /FormType 1 /Meta298 Do stream /Meta92 106 0 R /Matrix [1 0 0 1 0 0] q /Subtype /Form /Subtype /Form /Leading 349 endobj /Length 16 >> 0 G Q >> 0 g >> /F3 17 0 R Q >> 1.007 0 0 1.007 551.058 636.879 cm 0 g endstream endobj << q /FormType 1 /Resources<< /Matrix [1 0 0 1 0 0] /Font << BT endstream BT q /Meta345 359 0 R /Font << /Type /XObject /Meta364 378 0 R 0 g BT 1 g >> 1 i q /BBox [0 0 639.552 16.44] /Length 70 q /F3 12.131 Tf 1 i << >> >> Q >> q >> q /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Meta57 Do /F3 17 0 R /Length 16 0.458 0 0 RG /ProcSet[/PDF] 0 g >> BT /Subtype /Form endobj Q /Length 54 /FormType 1 q 0.524 Tc BT (C) Tj (5\)) Tj 0 G /F3 12.131 Tf 0 g Q 1 g >> Q BT /Meta291 Do /Meta347 361 0 R /Resources<< /Length 59 /Length 16 endstream endobj /Meta40 Do Q 0.737 w Q stream /BBox [0 0 88.214 16.44] stream >> endobj >> Q endstream Q /Type /XObject 0 G >> >> >> /Length 108 0.458 0 0 RG /FormType 1 1 i stream q 1 i /Meta105 119 0 R 1 i endobj endstream >> /FormType 1 /Type /XObject 0.458 0 0 RG Q 0 G 0 g /FormType 1 Q endstream So we have twice of a mystery number decreased by three, and that is all going to be 31. 130 0 obj stream 369 0 obj /Resources<< endobj /Type /XObject Twice a number when decreased by 7 gives 45. 1.007 0 0 1.006 411.035 690.329 cm Q /Meta413 429 0 R >> /Resources<< /Length 16 /Type /XObject 0 G q 0.786 Tc 1 g /Subtype /Form /BBox [0 0 534.67 16.44] 1 i /BaseFont /PalatinoLinotype-Roman q /Resources<< 1.005 0 0 1.015 45.168 53.449 cm 1.007 0 0 1.007 130.989 383.934 cm 1 i /Type /XObject /Subtype /Form /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Length 16 /FormType 1 stream /Length 54 /Length 16 >> endobj BT /Subtype /Form 280 0 obj /Meta37 Do endobj q /Meta422 438 0 R /Type /XObject /Meta256 270 0 R 123 0 obj << /Length 16 q /Font << 52.412 5.203 TD q /Meta117 131 0 R endstream 0 w ET /Meta26 39 0 R ET >> /Meta196 210 0 R 0 g BT >> Q 0 g stream stream Q Q S /Type /XObject /FormType 1 Q 1.007 0 0 1.007 654.946 872.509 cm 0 g 20.975 5.336 TD /Type /XObject Q >> endstream /Length 2252 /Type /XObject endobj /F3 12.131 Tf /Length 54 endobj Q /Subtype /Form BT /FormType 1 /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] q 0 20.154 m >> /Meta250 264 0 R /Resources<< >> /Resources<< ET ET /Length 16 If mario jumps 3 times and luigi jumps 62 times. /ProcSet[/PDF/Text] 0 g 0.738 Tc >> /Font << 1 i /ProcSet[/PDF/Text] Q /BBox [0 0 639.552 16.44] Q /Resources<< /F4 36 0 R stream q q /Meta41 55 0 R >> Q 0 5.203 TD the sum of a number and twelve. BT nine increased by a number x. /Resources<< /Meta216 230 0 R q >> /Resources<< Q precision and actual right or wrong answers. 1 g /FormType 1 /Resources<< /Subtype /Form /BBox [0 0 30.642 16.44] Q /Meta191 205 0 R stream , Prove the following endstream 137 0 obj q /Resources<< Q q /Font << endstream q << /BBox [0 0 17.177 16.44] q 0 w /Type /XObject ET endobj >> /FormType 1 /Meta197 Do /Meta266 280 0 R Q Q BT 0 g 188 0 obj 0.564 G endobj /ProcSet[/PDF/Text] Q /Type /XObject /F1 14.682 Tf /Type /XObject BT >> endobj /F4 12.131 Tf /FormType 1 /Resources<< %%EOF. endstream /Matrix [1 0 0 1 0 0] Q /Resources<< << 0.564 G ET D. Twice a number decreased by ten is less than 24. /BBox [0 0 88.214 16.44] /Resources<< 1 i q 0 g /FormType 1 Q /Resources<< 0 G /Subtype /Form 224 0 obj 0 4.894 TD /FormType 1 /Resources<< Q /Meta79 93 0 R /ProcSet[/PDF] /Length 69 q (D\)) Tj /Meta140 Do 0 w 1.014 0 0 1.007 391.462 330.484 cm /AvgWidth 445 /Type /XObject /Meta340 354 0 R 0.369 Tc /DecodeParms [<> ] BT /Meta386 402 0 R /Meta363 377 0 R 226 0 obj /FontDescriptor 6 0 R BT 0.737 w >> Q >> q BT endobj endstream Q 0.68 Tc Q /F3 12.131 Tf /Meta9 20 0 R << 88 0 obj << /Length 59 q 0.369 Tc q stream q ET /Matrix [1 0 0 1 0 0] /Length 70 /Meta414 Do 213 0 obj Q /Resources<< /ProcSet[/PDF] Q q /FormType 1 q Q /Meta320 Do /ProcSet[/PDF/Text] /Resources<< /FormType 1 q q /Meta407 Do q 0 g Q /Type /XObject /I0 51 0 R /Font << >> 0 G Q /Meta418 Do /Length 69 /Type /XObject /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q 1.007 0 0 1.007 551.058 277.035 cm BT 1 i BT 1.007 0 0 1.007 551.058 383.934 cm 1.007 0 0 1.007 45.168 779.913 cm endobj 0 w /Font << /Length 59 /Subtype /Form << endstream q 1.007 0 0 1.007 654.946 347.046 cm q /F3 12.131 Tf /Height 22 (+) Tj 0 g 200 0 obj (5) Tj /FormType 1 Q >> 0 g /Type /XObject /Font << << 1 i /FormType 1 0 g Q /Subtype /Form -0.092 Tw stream q /ProcSet[/PDF/Text] Q 1 i 0.786 Tc /Length 60 >> 1 i 1 i q /Length 99 /Meta167 Do /Matrix [1 0 0 1 0 0] >> Q 0 g q q q q Q 0.271 Tc /Length 65 >> /FormType 1 >> q q Q >> /FontBBox [-90 -216 1195 800] q q q endobj Q endstream /Matrix [1 0 0 1 0 0] Q q 0 g 0.524 Tc 0.564 G Q /Ascent 1050 /Meta381 395 0 R endobj /Subtype /Form << stream q q /Meta201 Do /FormType 1 q /Font << >> 0 w /Meta154 168 0 R Q q Q q 0 G /Font << Q q q ET Q Q /Font << Q /Length 60 /Matrix [1 0 0 1 0 0] >> 261 0 obj endobj 1 i (+) Tj q /Meta31 44 0 R q /Matrix [1 0 0 1 0 0] Q q 1.014 0 0 1.007 111.416 523.204 cm Twice a number when decreased by 7 gives 45. /Resources<< Q >> Q << /BBox [0 0 15.59 29.168] endobj 1 i 1.502 7.841 TD 0.737 w /Meta220 Do /Meta190 204 0 R /FormType 1 /FormType 1 Q >> q twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. /Type /XObject q /Subtype /Form stream Q /Length 16 -0.03 Tw /FormType 1 /Length 65 0 g >> q /Resources<< /Meta30 43 0 R Q >> /Type /XObject /MediaBox [0 0 767.868 993.712] Q stream /Matrix [1 0 0 1 0 0] q /Meta55 Do 0 G Q /Matrix [1 0 0 1 0 0] Twice = two times, double. stream q Q 6.746 5.203 TD Q endstream /Meta258 Do << Q 0 w Q 0 g 1 i /Meta223 Do /Font << /ProcSet[/PDF] Q /Meta323 Do endobj 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 stream Q /Type /XObject >> >> /Meta384 Do endobj 0 G q endstream /Type /XObject endobj >> 282 0 obj Q Q Q Q /Length 116 /Meta81 95 0 R BT 327 0 obj << /Matrix [1 0 0 1 0 0] q q endstream 1 i /Subtype /Form Q >> << 2x - 15 = -27. Q endstream 0 g 1 i 1 i /F3 17 0 R >> q /ProcSet[/PDF/Text] /Subtype /Form q endobj /Font << q /Matrix [1 0 0 1 0 0] /Meta355 Do /FormType 1 q /Type /XObject /FormType 1 >> /F3 17 0 R stream /Meta337 351 0 R /Subtype /Form << 1 i Q /Type /XObject /Matrix [1 0 0 1 0 0] endobj /F3 17 0 R 0.369 Tc >> Q stream /Matrix [1 0 0 1 0 0] /F3 17 0 R /F4 36 0 R Q Q /Meta391 Do Q q ET >> /Matrix [1 0 0 1 0 0] endstream 0 g (58) Tj 1 i /Subtype /Form >> /Meta206 Do /Resources<< 1.502 5.203 TD /Font << stream /Resources<< 0.564 G /Subtype /Form /FormType 1 0 G q 1.014 0 0 1.007 111.416 776.149 cm q /FormType 1 /F3 12.131 Tf Q 0 w << (vi) If 12 is subtracted from a number, the result is 24. Q /Length 74 ET endobj 350 0 obj 22.478 4.894 TD 0 G q 1 i /Meta394 410 0 R q /Matrix [1 0 0 1 0 0] stream /Subtype /Form q BT 0.458 0 0 RG endobj Q /Type /XObject 316 0 obj Q q 0.564 G 0 G endobj /F3 12.131 Tf /Type /XObject /F3 17 0 R Q /Resources<< Q 1 g >> << ET stream Q /Meta359 373 0 R BT >> >> 1 i /Meta230 Do (D\)) Tj q S Q << /Subtype /Form q 215 0 obj 205.199 4.894 TD 1.007 0 0 1.007 411.035 277.035 cm /FormType 1 1 i /FormType 1 /Length 106 ET q q << stream BT /Type /XObject 0.486 Tc 0.737 w 0.838 Tc q 1 g 0.369 Tc q /F1 12.131 Tf /Type /XObject stream (-20) Tj 672.261 347.046 m (- 8) Tj q endstream q /F3 17 0 R /Font << endobj 362 0 obj Q Q /F1 12.131 Tf >> /FormType 1 0 g /Ascent 976 /ProcSet[/PDF/Text] 296 0 obj endobj /Subtype /Form q >> Q /Font << q 0.17 Tc 9.723 5.336 TD Q 6.746 5.203 TD q /Matrix [1 0 0 1 0 0] endobj /Meta307 321 0 R (+) Tj >> >> endobj 385 0 obj 1.007 0 0 1.007 45.168 829.599 cm /Length 118 /Subtype /Form 0 g /Font << /Meta151 165 0 R << q q 1.005 0 0 1.007 102.382 310.158 cm Q /Length 69 26 0 obj q << /FormType 1 BT /Matrix [1 0 0 1 0 0] Q /FormType 1 BT >> 297 0 obj >> ET q endstream /Meta83 Do /Meta315 Do /F3 12.131 Tf /Type /XObject /Meta133 147 0 R q q /Type /XObject q 319 0 obj /F3 17 0 R /ProcSet[/PDF] q /F3 12.131 Tf /Subtype /Form /F3 17 0 R << >> Q Q >> >> Q 0 20.154 m 0 g Q /FormType 1 >> Q Thrice of a number = 3x. /StemV 77 /F3 17 0 R /Meta44 Do q << /Length 54 Twice the difference of a number and three totals twelve 8. /Matrix [1 0 0 1 0 0] /Subtype /Form /Type /Page /Meta101 115 0 R /Meta283 Do 0 g q Q >> /Matrix [1 0 0 1 0 0] /Meta235 Do << endobj >> >> /Meta364 Do BT /Type /XObject BT 1 i << /Subtype /Form stream >> q stream Q 0.737 w /Meta340 Do << /ProcSet[/PDF/Text] /FormType 1 /Resources<< q << /ProcSet[/PDF/Text] Q 0 5.203 TD /Length 58 endobj q /Matrix [1 0 0 1 0 0] << 0 5.203 TD /Matrix [1 0 0 1 0 0] >> /Meta314 Do /ProcSet[/PDF/Text] Q /FormType 1 Number Outcomes 1 42 2 41 3 . << Q /FormType 1 >> >> /Meta297 311 0 R q /Meta393 Do endstream /FormType 1 /ProcSet[/PDF] Q /ProcSet[/PDF/Text] 1 i 0.458 0 0 RG q 0.564 G endstream ([x ) Tj /Length 12 Q /Meta310 Do /BBox [0 0 88.214 16.44] /F3 12.131 Tf >> /BBox [0 0 15.59 16.44] >> Q 0.425 Tc /FormType 1 /Meta19 30 0 R /Resources<< /Length 294 /Type /XObject /F3 12.131 Tf /Meta158 172 0 R 0 G Q /Matrix [1 0 0 1 0 0] << stream 1.007 0 0 1.007 411.035 383.934 cm /Type /XObject /F3 17 0 R endobj q /Resources<< /F3 12.131 Tf 0 g 1.005 0 0 1.007 102.382 616.553 cm << /BBox [0 0 88.214 16.44] 288 0 obj Q Q /Length 16 /Resources<< stream /BBox [0 0 23.896 16.44] /Matrix [1 0 0 1 0 0] /Resources<< q 0 g endobj q 1.014 0 0 1.006 531.485 690.329 cm q Q /Font << /Resources<< /F3 17 0 R /BBox [0 0 673.937 16.44] >> /Matrix [1 0 0 1 0 0] << /Subtype /Form /Meta252 Do /XHeight 471 << endstream /Resources<< /Subtype /Form Q endstream endstream >> /Subtype /Form /Resources<< /BBox [0 0 88.214 16.44] 0 g /Matrix [1 0 0 1 0 0] /Length 16 23.952 4.894 TD ET /BBox [0 0 15.59 16.44] 1 i q 217 0 obj /Resources<< /F3 12.131 Tf /Meta107 Do /F3 17 0 R >> 0 G /Type /XObject >> 1 i Q >> /Resources<< endobj /FormType 1 BT /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] 1 i endobj >> 341 0 obj /Subtype /Form Q (B\)) Tj endstream endstream Q 206 0 obj stream 0 G endstream /Font << endobj q /Length 16 /F3 17 0 R << >> BT /Descent -299 ET Q 0.564 G 1 i /ProcSet[/PDF/Text] 1.007 0 0 1.007 411.035 583.429 cm /ItalicAngle 0 q /Length 118 stream 0 g >> 0 w (40) Tj Q /BBox [0 0 88.214 16.44] endstream 0 G Mat q q 0 G << q << 0 g (B\)) Tj Q Q Q /Type /XObject /Type /XObject q /Matrix [1 0 0 1 0 0] /Meta188 Do Q /Resources<< q Q /F3 17 0 R >> 1.014 0 0 1.007 391.462 523.204 cm /Length 16 0 g /Meta289 303 0 R Q Q Q 0 g >> /Resources<< stream /Resources<< >> Q /BBox [0 0 88.214 35.886] >> 1.007 0 0 1.007 654.946 599.991 cm /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] Q /Meta28 Do /FormType 1 Q 0 g 0 g /BBox [0 0 88.214 16.44] >> q Q stream /Type /XObject Q /Subtype /Form q >> >> 413 0 obj /Font << endobj 0 g Q /Resources<< /Meta34 Do endobj endstream stream BT BT /Font << Q /Type /XObject << /Meta164 178 0 R /Subtype /Form /FormType 1 stream Q 208 0 obj 9.723 5.336 TD Q Was this answer helpful? /Type /XObject 0.564 G 1 i /BBox [0 0 88.214 16.44] >> /Resources<< endstream /FormType 1 0 w 194 0 obj /ProcSet[/PDF/Text] /Subtype /Form /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] Q 0 G /Matrix [1 0 0 1 0 0] /Resources<< (B\)) Tj /F3 12.131 Tf 1 i 442 0 obj Q /FormType 1 1 i q 0.68 Tc q /Meta271 Do /Length 68 Q /FormType 1 148 0 obj Q The observed mean MetS-Z was at inclusion 0.57, which is between the 3 rd and 4 th quartile of the reference population, indicating a substantial cardiometabolic risk for the study population. >> 1 i 0.564 G >> to represent the numbers. q q Q /Type /XObject >> /Font << /Resources<< q q >> /Subtype /Form >> /Resources<< 0.458 0 0 RG /Subtype /Form /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] /F3 12.131 Tf >> /Resources<< stream BT q >> /Matrix [1 0 0 1 0 0] /F3 17 0 R 23 0 obj ET 0.737 w 0 g Q /Subtype /Form /ProcSet[/PDF/Text] 6.746 5.203 TD /Meta113 Do /Meta274 Do 0 G BT /Resources<< q /BBox [0 0 549.552 16.44] >> 94 0 obj 0 G /Type /XObject 0 g Q Q /Subtype /Form endobj /FormType 1 /FormType 1 /Meta68 82 0 R Q endobj Question 1. /ProcSet[/PDF] Q q /BBox [0 0 15.59 16.44] << Q 1 i /Type /XObject /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 324 0 obj endobj Q 0 w 1 i endstream 422 0 obj Q 11.99 24.649 TD (x) 6 times a number is 5 more than the number. endstream 1 g >> 0.564 G endstream /Meta174 Do stream [(F)-22(ive)] TJ 1.007 0 0 1.006 411.035 437.384 cm BT Q Answer by Mathtut (3670) ( Show Source ): /FormType 1 /F1 12.131 Tf 1.005 0 0 1.007 102.382 400.496 cm 1 i 1.007 0 0 1.007 411.035 277.035 cm Q /Meta205 Do q endstream >> stream (x) Tj q 93 0 obj 0.564 G stream /Meta338 Do /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endobj q /Meta11 Do << >> /Matrix [1 0 0 1 0 0] /Type /XObject << >> << endstream (3) Tj /Meta295 Do /ProcSet[/PDF/Text] Q (38) Tj 0 g q >> /FormType 1 endobj q ET 0 g q stream /Type /XObject 0.369 Tc Find the number. 0 20.154 m /Font << -0.021 Tw /FormType 1 /Subtype /Form /Subtype /Form 1 i /Length 16 >> /Length 16 Q 1.014 0 0 1.006 391.462 763.351 cm /Length 58 ET /Meta90 Do q endstream /F3 17 0 R 1 g 0 w /Subtype /Form /Meta209 223 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form /ProcSet[/PDF/Text] stream /ProcSet[/PDF/Text] 0 w 0.369 Tc /Meta214 Do /FormType 1 q /Length 80 q endstream ET /Meta336 Do /Meta145 159 0 R Q 1 i /Matrix [1 0 0 1 0 0] Q /Font << Q stream Q 0 G /Font << /Subtype /Form n 11 or n 11. 0 G /F3 17 0 R 549.694 0 0 16.469 0 -0.0283 cm << /FormType 1 /MissingWidth 252 endstream BT Q endobj /Type /XObject 6.746 24.649 TD 1 i Q Q /Meta427 Do >> /ProcSet[/PDF/Text] q stream /Font << /Meta72 Do q endstream Q 0 g /FormType 1 /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 746.789 cm >> /Matrix [1 0 0 1 0 0] /Type /XObject 1.014 0 0 1.007 111.416 636.879 cm (5) Tj /Meta213 Do /F3 17 0 R >> /Type /XObject q << /Meta331 Do q /Subtype /Form (x) Tj 0.458 0 0 RG 0.458 0 0 RG /Length 69 1 i >> /Resources<< /FontName /PalatinoLinotype-Roman 0.68 Tc q /Matrix [1 0 0 1 0 0] ET /Meta416 432 0 R Q stream q 549.694 0 0 16.469 0 -0.0283 cm Q /Type /XObject /Meta21 32 0 R Q Q Q /Subtype /Form /Font << /Meta368 382 0 R 1.007 0 0 1.007 67.753 546.541 cm Q /BBox [0 0 88.214 35.886] q /Type /XObject Q >> >> 0 5.203 TD q BT stream >> q 0 G q /Meta71 85 0 R /BBox [0 0 88.214 16.44] /FormType 1 Q q stream >> endstream /Subtype /Form 103 0 obj 421 0 obj >> Q Q /FormType 1 2 See answers pharry1800 pharry1800 Answer: 2n-58 Step-by-step explanation: olivbreadh olivbreadh Answer: 2x-116 or 2(x-58) Step-by-step explanation: Transalate it to numbers and operations: => 2(x-58) => 2x-116 You won't have a solid number since its not an equation. 0 G 0 g /Resources<< /Length 16 /Resources<< endobj /FormType 1 /Subtype /Form q 0 G 0 g q /Type /XObject 161 0 obj >> /ProcSet[/PDF/Text] q Q q Q /Resources<< 1 i Q Q Q /Meta346 360 0 R >> /Matrix [1 0 0 1 0 0] /Subtype /Form /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 0.738 Tc 136 0 obj 37 0 obj /Meta214 228 0 R 220 0 obj q /Meta390 406 0 R stream endobj endobj /Subtype /TrueType /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 654.946 546.541 cm stream endobj >> ET Q /FormType 1 endstream Q (3) Tj 26.219 5.203 TD 0.524 Tc << /Length 12 150 0 obj /F3 17 0 R >> /Resources<< Q Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. 0 g /Subtype /Form Q >> Q BT endobj /Resources<< /Meta109 Do stream >> 5 0 obj /Subtype /Form Q >> /F3 12.131 Tf stream /Meta203 217 0 R q /Resources<< /FormType 1 q /Type /XObject /Type /XObject 63 0 obj q /BBox [0 0 88.214 16.44] BT /Length 16 >> /Meta43 Do /Subtype /Form endobj 0.737 w 0 g /Meta116 Do >> 1.005 0 0 1.007 79.798 813.037 cm << endstream >> [(-3)-16(20)] TJ 12.727 5.203 TD /ProcSet[/PDF/Text] >> /Meta120 Do (D\)) Tj /F3 12.131 Tf >> 326 0 obj Q q q Q 1.014 0 0 1.007 251.439 523.204 cm /F2 12.131 Tf /Meta99 Do << 95 0 obj /F3 12.131 Tf 1 i Q q >> 1.005 0 0 1.007 45.168 889.071 cm /Resources<< 0 w 219 0 obj /Meta282 296 0 R (38) Tj Q /Font << 1.005 0 0 1.007 45.168 916.925 cm q - 9737014. S q >> q << -0.021 Tw >> q 0.838 Tc 1 i ET q /BBox [0 0 88.214 16.44] 1 i stream 0 g 162 0 obj q q /BBox [0 0 88.214 16.44] 101 0 obj /BBox [0 0 17.177 16.44] /Meta45 59 0 R << ET /Matrix [1 0 0 1 0 0] /Resources<< 549.694 0 0 16.469 0 -0.0283 cm stream q >> BT /Length 69 (x) Tj stream /F3 17 0 R BT stream >> endstream >> endobj /FormType 1 0.737 w q << /XObject << /Resources<< endstream >> endobj 1.014 0 0 1.007 531.485 776.149 cm q /Matrix [1 0 0 1 0 0] /Font << >> Q endobj /Length 60 >> /Matrix [1 0 0 1 0 0] /Type /XObject 3.742 5.203 TD /Resources<< /Length 69 /Meta383 397 0 R

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