twice a number decreased by 58

>> 167 0 obj (x ) Tj stream /Meta239 253 0 R Q Q >> q /Meta405 421 0 R /Meta222 Do /Resources<< 0.524 Tc >> /F3 17 0 R /Type /XObject /Length 16 /Matrix [1 0 0 1 0 0] 1.502 24.649 TD 0.369 Tc /Resources<< /Meta34 Do >> 0 g BT >> ET Q /FormType 1 /Meta354 368 0 R /ProcSet[/PDF] /Subtype /Form /Length 69 << 549.694 0 0 16.469 0 -0.0283 cm /ProcSet[/PDF/Text] /Kids [ endobj Q /Type /XObject /Meta355 Do /ProcSet[/PDF] >> 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 418 0 obj >> /Font << /Meta9 20 0 R 117 0 obj BT stream /F3 17 0 R Q Q /Subtype /Form 21 0 obj q >> /FormType 1 0.564 G /Font << >> /Resources<< endobj /FormType 1 /Subtype /Form /BBox [0 0 534.67 16.44] stream Q endstream Q /Length 16 /ProcSet[/PDF/Text] /Meta395 Do /Type /XObject 1 i 0 w /Meta404 Do /F3 17 0 R /Subtype /Form 20.21 5.203 TD /Type /XObject 125.064 4.894 TD /Type /XObject /Length 79 /ProcSet[/PDF] /Type /XObject /Font << /Length 63 Q >> stream >> q >> /Length 12 /ProcSet[/PDF/Text] /Type /XObject << /Font << >> 218 0 obj /Font << >> 0 G 1.005 0 0 1.006 45.168 879.284 cm /BBox [0 0 15.59 29.168] q q BT /Font << /FontBBox [-170 -292 1419 1050] 0 G 442 0 obj 0 g /AvgWidth 657 >> /Subtype /Form >> 1 i /Length 12 /FormType 1 /Resources<< /Meta60 Do Q 1 i << endstream 40 0 obj Q /Meta411 Do q << Q q 0 g /Meta49 Do 1 i /BBox [0 0 15.59 16.44] q /Length 16 0 g q Q 0 G /ProcSet[/PDF] ET 141 0 obj 0 G q /Matrix [1 0 0 1 0 0] /Meta215 229 0 R /ProcSet[/PDF] /F1 7 0 R 1.007 0 0 1.007 271.012 849.172 cm /Matrix [1 0 0 1 0 0] >> /Subtype /Form 1 g endstream Q /Meta2 9 0 R 0 20.154 m Q q /Subtype /Form >> 1 i /Length 118 /Matrix [1 0 0 1 0 0] endstream Q >> /Meta232 246 0 R 0 g endstream q /F4 12.131 Tf endstream 1 i endobj q >> 0.458 0 0 RG 9.723 5.336 TD >> 297 0 obj Q /F3 17 0 R >> /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] endstream /F1 12.131 Tf /Subtype /Form Q stream /Count 2 >> /Meta151 165 0 R 110 0 obj /Type /XObject /BBox [0 0 534.67 16.44] /BBox [0 0 15.59 16.44] /Length 118 q >> 0 G 0 w /ProcSet[/PDF] /FormType 1 /Subtype /Form /ProcSet[/PDF/Text] Q (B\)) Tj 0 5.203 TD /Length 69 1 g endobj /Type /XObject 220.931 4.894 TD q >> 33.704 5.203 TD >> 1 i /Resources<< /Type /XObject /F3 12.131 Tf >> 1 i 1.007 0 0 1.006 411.035 763.351 cm q q /Type /XObject q /Meta51 65 0 R /ProcSet[/PDF] >> endstream /BBox [0 0 30.642 16.44] 94.364 5.203 TD /Type /XObject /Matrix [1 0 0 1 0 0] stream >> /Matrix [1 0 0 1 0 0] /Length 59 0 g /Resources<< 1 g /Matrix [1 0 0 1 0 0] /Pages 1 0 R /Length 63 /Matrix [1 0 0 1 0 0] /FormType 1 endstream /Meta173 187 0 R >> Q << 0 G q q (iv) A number exceeds 5 by 3. >> /Type /XObject endstream >> 1.007 0 0 1.007 130.989 776.149 cm /FormType 1 q BT /FormType 1 q << /Subtype /Form /Type /XObject BT /Resources<< 0.564 G endstream stream stream /F3 12.131 Tf endstream 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 /ProcSet[/PDF] /Resources<< 0 w stream /Length 70 >> Q Q /Meta276 Do 0 g -0.041 Tw /Subtype /Form BT /Meta183 Do /Meta272 286 0 R /Resources<< Q stream /BBox [0 0 30.642 16.44] /Meta216 Do >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] 1.005 0 0 1.007 79.798 730.228 cm << 0 g (A\)) Tj /ProcSet[/PDF/Text] endstream q 177 0 obj /FormType 1 q q /Meta333 347 0 R endstream ET /Meta191 Do /F3 12.131 Tf ET /ItalicAngle 0 /FormType 1 1.007 0 0 1.006 130.989 690.329 cm 0 g /Resources<< /Length 69 Q /BBox [0 0 88.214 16.44] << 0.486 Tc /Meta98 Do endobj >> /Meta97 111 0 R q q ET /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] q /Type /XObject Q q /ProcSet[/PDF] /Resources<< /Subtype /Form >> /ProcSet[/PDF] /Subtype /Form /Length 54 So we have twice of a mystery number decreased by three, and that is all going to be 31. Q q Q /Meta71 85 0 R Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf /ProcSet[/PDF/Text] q /F3 12.131 Tf /Meta311 325 0 R /Meta426 Do /FormType 1 >> /FormType 1 /Font << q /F3 17 0 R /Length 59 /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 531.485 450.181 cm /Type /XObject Solution: Let the number be x. << endstream q /Type /XObject /Meta258 272 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. BT S 1 i 0 G 1 i q /Subtype /Form q 443 0 obj 1.005 0 0 1.007 102.382 726.464 cm q /Resources<< >> /FormType 1 /FormType 1 stream 1.007 0 0 1.007 130.989 277.035 cm endobj /Meta363 377 0 R endobj q BT /FormType 1 6.746 5.203 TD /BBox [0 0 88.214 16.44] 28 0 obj 1.007 0 0 1.007 271.012 583.429 cm /Type /XObject /Meta295 309 0 R q ET 0.458 0 0 RG 363 0 obj /Type /XObject stream 429 0 obj /BBox [0 0 15.59 29.168] << 0 g 1 i /Subtype /Form /Font << /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 616.553 cm /F3 17 0 R -0.062 Tw /Length 58 stream 351 0 obj q Q /F1 7 0 R 19.474 20.154 l << Q 1 i 722.699 599.991 l stream Q 1 i 30.699 5.203 TD The results found were expressed mainly through tables and graphs as the main resources of the statistical language. /Meta109 123 0 R stream 0.737 w /F3 17 0 R >> 0.737 w /ProcSet[/PDF/Text] /Meta212 Do endobj /BBox [0 0 15.59 16.44] /Subtype /Form /Resources<< 0 G >> [(1)-25(0\))] TJ q Q >> q BT /Meta92 106 0 R ET 1 g 1.007 0 0 1.007 130.989 776.149 cm /BBox [0 0 88.214 16.44] 1 g stream 0.564 G 0.425 Tc /ProcSet[/PDF] /Subtype /Form q 0 G << /Type /XObject BT 286 0 obj /Subtype /Form 368 0 obj BT q BT 0.564 G /F3 12.131 Tf /F4 36 0 R >> /Length 118 stream q >> 0.564 G 1 i q Q 1.014 0 0 1.007 531.485 330.484 cm q /F3 12.131 Tf 1.007 0 0 1.007 271.012 383.934 cm << endobj >> ET /Length 69 << 52 0 obj Step 1/1. >> 0 G Q >> /FormType 1 /F3 12.131 Tf /Font << q >> BT 0 g /Type /XObject 1.007 0 0 1.007 551.058 277.035 cm << q 1.007 0 0 1.007 551.058 703.126 cm ET Q /Font << 45 0 obj q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf Q /Resources<< Q 345 0 obj /Resources<< q >> q endstream q /Type /XObject >> q << /BBox [0 0 88.214 16.44] 129 0 obj q 1 i endobj /Matrix [1 0 0 1 0 0] /Subtype /Form 0 G /Type /XObject /Type /XObject /Type /XObject Q stream << /FormType 1 0.458 0 0 RG /Resources<< (-4) Tj Q /BBox [0 0 639.552 16.44] /Resources<< BT ET stream << >> /Length 59 /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] /F3 17 0 R 1 i 11.99 24.649 TD Q Q q (7\)) Tj /Leading 349 endstream endobj /Meta25 Do /BBox [0 0 15.59 29.168] /FormType 1 q /F3 12.131 Tf Q /Meta378 Do >> Number Outcomes 1 42 2 41 3 . q Q Q 1.014 0 0 1.006 531.485 763.351 cm /F1 12.131 Tf /Meta250 264 0 R 0.37 Tc 1.005 0 0 1.007 102.382 347.046 cm /FormType 1 0 G (x) Tj endobj /F4 36 0 R 174 0 obj q << /Resources<< 56 0 obj /Meta201 215 0 R >> Q /Meta384 398 0 R 0 g /Matrix [1 0 0 1 0 0] Q >> Q stream /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 277.035 cm q << /Root 2 0 R /Length 85 BT /ProcSet[/PDF/Text] stream BT Q Most questions answered within 4 hours. /Subtype /Form << Q << >> 0.369 Tc /Matrix [1 0 0 1 0 0] 337 0 obj Q q q Select the correct mathematical statement for the following equation. q /Resources<< /Length 80 q /Meta64 Do Q /FormType 1 /Length 12 Let the 2nd number be y. endobj Q Q (A\)) Tj /BitsPerComponent 1 /Length 57 1.007 0 0 1.007 654.946 872.509 cm /Subtype /Form >> q /FormType 1 q ET /FormType 1 << /Meta130 Do stream q 0.564 G stream /Length 69 >> endstream << /Meta93 Do 0.68 Tc q /Length 19882 >> Q /Length 69 << ET >> >> Q 1 i /Length 64 0 w 25 0 obj >> ET 0 G ET Q /BBox [0 0 88.214 16.44] 384 0 obj /FormType 1 /Type /XObject /Length 12 /Meta42 Do 1.007 0 0 1.006 551.058 836.374 cm Q Q 0.458 0 0 RG ET 1.005 0 0 1.007 102.382 473.519 cm /FormType 1 /ProcSet[/PDF] /Resources<< << BT /F3 17 0 R >> 1 i /Meta235 249 0 R 0.425 Tc /Meta353 Do >> 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 G q /Meta313 327 0 R Q << /BBox [0 0 88.214 16.44] stream /ProcSet[/PDF/Text] stream /BBox [0 0 15.59 16.44] Q 0 5.336 TD /BBox [0 0 88.214 16.44] /Meta205 219 0 R /Matrix [1 0 0 1 0 0] ET /XObject << Q q /FormType 1 q /F3 17 0 R q 1 i Q /Resources<< /F4 12.131 Tf (4\)) Tj Q 0 G /Meta413 Do 376 0 obj 1 i stream Q Q /Resources<< q Q BT 0 g /ProcSet[/PDF/Text] 0 G 1.005 0 0 1.007 102.382 490.08 cm 0.564 G << >> stream 1.007 0 0 1.007 45.168 763.351 cm /Subtype /Form /Matrix [1 0 0 1 0 0] Q /Meta369 383 0 R /Subtype /Form 0 g stream Q q 15 0 obj q 0.458 0 0 RG Q >> /FormType 1 /Type /XObject << q stream >> endobj Q /Subtype /Form 178.979 5.203 TD >> Q endobj q /F1 7 0 R /FormType 1 Q endstream Q >> /Subtype /Form Q stream /FormType 1 endobj 1 i /BBox [0 0 534.67 16.44] 0.369 Tc Q BT 0 g /BaseFont /PalatinoLinotype-Bold /Meta107 Do /Meta94 108 0 R /BBox [0 0 30.642 16.44] 1 i /Font << q ET >> stream /Font << /Type /FontDescriptor 549.694 0 0 16.469 0 -0.0283 cm 0.737 w 1.007 0 0 1.007 130.989 583.429 cm /F3 12.131 Tf 1.014 0 0 1.007 391.462 636.879 cm >> /Meta182 196 0 R 0 G 0.737 w /F3 17 0 R Q 132 0 obj >> /Resources<< /Matrix [1 0 0 1 0 0] /Meta171 185 0 R q 0 g /Matrix [1 0 0 1 0 0] >> endobj << ET Q /Meta208 222 0 R >> 0 G /Subtype /Form [2] Twice a number increased by four is twenty-one. /Meta273 287 0 R endstream /Subtype /Form stream Three times a number equals fifteen 3. endobj /Meta370 Do /Subtype /Form /F3 17 0 R << >> q Q stream 0.458 0 0 RG BT Q q /Matrix [1 0 0 1 0 0] /Meta228 242 0 R q 1 i endobj q 0 G /FormType 1 q Q /FormType 1 /F3 12.131 Tf /Meta283 297 0 R /Type /XObject << 6 0 obj >> >> q q /Matrix [1 0 0 1 0 0] endobj /Font << Q endstream >> /Meta343 Do /Resources<< /ProcSet[/PDF/Text] 0.31 Tc /BBox [0 0 639.552 16.44] 0.458 0 0 RG << 1 i Q Q 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. /XObject << 1 g Q 1.007 0 0 1.007 551.058 523.204 cm ET >> q 1 i q Q 20.975 5.336 TD 1 i endstream BT /Subtype /Form Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. 1.005 0 0 1.007 102.382 400.496 cm /F3 17 0 R 0 w 1.502 5.203 TD q << Q >> At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . 0.524 Tc Q BT >> /FormType 1 0 4.894 TD /Subtype /Form endstream q >> BT /Type /XObject /Length 16 [(A numb)-16(er subtract)-15(ed from )] TJ /BBox [0 0 88.214 16.44] >> endobj /Resources<< /FormType 1 /Length 69 >> /Resources<< /BBox [0 0 88.214 35.886] NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. << endobj /Length 59 326 0 obj /Matrix [1 0 0 1 0 0] /F3 17 0 R >> ET 5.98 7.841 TD /BBox [0 0 15.59 16.44] >> ET /Font << 1 i (A\)) Tj 314 0 obj 1 g 0 g /Resources<< >> endstream /BBox [0 0 30.642 16.44] Q Q /F3 12.131 Tf Q 0 g /Meta179 Do 0 G /FormType 1 0 G Q ET /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q /Resources<< ET Q 0.737 w /FormType 1 ET >> /Subtype /Form endobj q 1 i endobj >> << q (5) Tj /FormType 1 << endobj /Length 69 0 G /BBox [0 0 15.59 29.168] 266 0 obj 1.007 0 0 1.007 271.012 636.879 cm >> 0.17 Tc /Font << q BT /Flags 32 /Meta371 385 0 R /Font << /FormType 1 ET 221 0 obj Q endobj << /Length 16 << /Matrix [1 0 0 1 0 0] /Meta265 279 0 R stream Q Q /Meta55 69 0 R q ET /ProcSet[/PDF] /F3 12.131 Tf 0 w /FontDescriptor 10 0 R /Resources<< 0 g 1.007 0 0 1.007 654.946 872.509 cm /Meta312 Do 1 g 0.737 w /Subtype /Form << Q /Length 16 q /ProcSet[/PDF/Text] /Font << stream endstream 1 i 1 i q q << /FormType 1 /Meta412 428 0 R 1 g << Q 0 g Q >> 0.134 Tc /Resources<< /Font << /Matrix [1 0 0 1 0 0] /Length 69 endstream Q 0 g q q q The result is 8 less than 10 times the number. /BBox [0 0 534.67 16.44] 0 G /FormType 1 /FormType 1 1 i /Meta115 Do 0 w q /Meta190 204 0 R >> >> endstream /F1 12.131 Tf /Subtype /Form /Resources<< Q 0 g /BBox [0 0 88.214 16.44] >> /Resources<< >> BT Q ET /Subtype /Form /BBox [0 0 30.642 16.44] /Type /XObject /F3 12.131 Tf 77 0 obj Q /Resources<< /Length 206 Q Q endstream 2. /Type /XObject q /ProcSet[/PDF] 32.201 20.154 l 0 G << 295.086 4.894 TD /F3 12.131 Tf Q 0 g 1 i /F3 17 0 R endobj 0.458 0 0 RG endstream << /Length 69 387 0 obj /F3 12.131 Tf >> /Meta267 Do >> 1 i /Meta292 306 0 R endobj /Font << /F1 7 0 R /Matrix [1 0 0 1 0 0] 173 0 obj 311 0 obj endstream >> 0 G 1 i >> /Length 59 /Length 294 >> (13) Tj /Matrix [1 0 0 1 0 0] [4] One half of a number decreased by fourteen is twenty-one q /ProcSet[/PDF/Text] /Subtype /Form q endstream 0 G endstream Q 0.458 0 0 RG endobj 0 G /Subtype /Form q << /Meta58 72 0 R /Resources<< /ProcSet[/PDF] /F3 12.131 Tf /Meta41 55 0 R /F3 12.131 Tf q q /Font << /FormType 1 Q q >> /Length 59 Q /FormType 1 /FormType 1 Q /Length 16 /Meta315 Do ET /Subtype /Form /Resources<< stream 378 0 obj stream /F3 17 0 R /Matrix [1 0 0 1 0 0] >> /Meta69 Do Q /Subtype /Form /Meta34 47 0 R >> endstream 0.307 Tc /ProcSet[/PDF] q 0 g Q 174.501 5.203 TD stream q Q /Type /XObject 1.007 0 0 1.007 551.058 636.879 cm /Resources<< stream endobj /ProcSet[/PDF/Text] ET /Subtype /Form 2. 0 g /ProcSet[/PDF/Text] 0 g /Meta287 Do q /Resources<< Answer (1 of 8): Solution: let the number be x. BT /Subtype /Form >> /Resources<< 0 g /Resources<< >> /Subtype /Form 1.007 0 0 1.007 271.012 849.172 cm 1 i /Type /XObject Twice a number is decreased by 9, and this sum is multiplied by 4. /F3 17 0 R /Meta87 Do 352 0 obj /F1 7 0 R Q /Subtype /Form Q >> stream /ProcSet[/PDF] Q BT Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 523.204 cm /Type /XObject 0 G /Matrix [1 0 0 1 0 0] 0 G /I0 Do q q 0.564 G /BBox [0 0 15.59 29.168] (vi) If 12 is subtracted from a number, the result is 24. /F3 12.131 Tf /FormType 1 /AvgWidth 459 endobj q /Meta131 Do q stream /FormType 1 q /Meta136 Do >> BT 0.564 G /Resources<< 1 i /Subtype /Form Q >> 0 5.203 TD /BBox [0 0 549.552 16.44] Q /Meta12 23 0 R q q 0 g 2.238 5.203 TD /ColorSpace [/Indexed /DeviceGray 1 ] 0 G /BBox [0 0 639.552 16.44] 1 g q >> << /Matrix [1 0 0 1 0 0] 409 0 obj /FormType 1 Q 1 g >> Q q /F3 17 0 R 1 i 1 i q /Meta111 Do endobj 0 g /ProcSet[/PDF] Q q endobj Q ET 1.007 0 0 1.007 130.989 636.879 cm 0 w BT /Meta338 Do q /Length 54 /ProcSet[/PDF/Text] 1 i endobj Q stream /Meta96 Do /Type /Font q 1.007 0 0 1.006 551.058 437.384 cm /Length 118 Find the number. q /StemV 94 /Type /XObject /Resources<< >> 0 5.203 TD /Meta69 83 0 R endstream Q /Meta331 345 0 R /Meta15 26 0 R endobj >> 1 i ET 323 0 obj /BBox [0 0 88.214 16.44] 20.975 5.336 TD 1.502 24.649 TD /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Q /Type /XObject 0.382 Tc 357 0 obj /Resources<< 0.737 w 0 w /FormType 1 /Type /XObject Q >> 1.007 0 0 1.006 130.989 690.329 cm /Subtype /Form >> /Matrix [1 0 0 1 0 0] 159 0 obj >> Q << 0 G 0 g /BBox [0 0 88.214 16.44] /Length 60 >> q >> /Type /XObject q endstream 0.564 G 240 0 obj endobj /Matrix [1 0 0 1 0 0] /FormType 1 /Type /XObject /Subtype /Form [( a )-15(number, decreased by )] TJ /Resources<< q q Q /FormType 1 >> BT >> for the season. q 1.502 5.203 TD /FormType 1 stream /Type /XObject 0 g BT 403 0 obj /Type /Catalog Q q endobj q >> 0.737 w q 1st step. /Length 16 /Matrix [1 0 0 1 0 0] 243 0 obj 1 i 0 g /Resources<< 1 g 0.486 Tc endobj << (D\)) Tj /ProcSet[/PDF] /F1 7 0 R 0.024 Tw /FormType 1 BT endobj /Type /XObject ET q endstream /Font << /ProcSet[/PDF] 176 0 obj << q Q /ProcSet[/PDF] /Meta161 Do /Matrix [1 0 0 1 0 0] 0 w /F3 12.131 Tf /Resources<< /Subtype /Form 0.737 w Q /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /Meta63 Do 0 4.894 TD 0 G /Subtype /Form /Type /XObject /ProcSet[/PDF/Text] stream 1.007 0 0 1.007 67.753 653.441 cm >> >> /ProcSet[/PDF/Text] endstream stream 0 g /FormType 1 /Type /Pages /BBox [0 0 88.214 16.44] Q /ProcSet[/PDF/Text] << 0 g /Meta117 Do q stream Q Q /FormType 1 1.007 0 0 1.007 271.012 330.484 cm /ProcSet[/PDF/Text] endstream Q /Meta346 Do /Length 66 /F3 12.131 Tf 0 w -0.486 Tw << q 13.493 5.336 TD stream /BBox [0 0 88.214 16.44] stream (2) Tj /Type /XObject >> 12.727 5.203 TD 25.454 5.203 TD stream /FormType 1 >> Q /Subtype /Form 0.425 Tc /Meta45 59 0 R 335 0 obj q /Matrix [1 0 0 1 0 0] /Subtype /Form 0 g /FormType 1 << /Resources<< /Length 16 /Matrix [1 0 0 1 0 0] ET /Length 69
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