twice a number decreased by 58
Q << endstream q 1 i 408 0 obj q /Resources<< endobj q 0 5.203 TD q << Q >> >> 72 0 obj /Resources<< >> Q [(F)-22(ive)] TJ Twice = two times, double. /FormType 1 Q /Type /XObject 0 G 1.502 24.339 TD 67 0 obj stream endobj 228 0 obj >> 219 0 obj /BBox [0 0 88.214 16.44] 0.369 Tc /Meta191 Do << q >> 672.261 347.046 m /Matrix [1 0 0 1 0 0] /Type /XObject << /ProcSet[/PDF] /Resources<< -0.056 Tw /FormType 1 stream /FormType 1 281 0 obj /FormType 1 Q: A number increased by 5 is equivalent to twice the same number decreased by 7. /Meta178 192 0 R /Meta75 89 0 R -0.486 Tw Q /ProcSet[/PDF] 0 g /Font << 0 g q /Subtype /Form >> q >> Q Percent Change = (Decrease First Value) x 100% stream 1 g /Length 54 >> Q /F3 12.131 Tf /BBox [0 0 639.552 16.44] q >> q << q 1 i 1 i (x ) Tj /Meta40 Do /Type /XObject >> endstream 144 0 obj 0.738 Tc BT /F3 17 0 R 0.458 0 0 RG /F3 12.131 Tf q /Meta42 56 0 R BT /BBox [0 0 17.177 16.44] /BBox [0 0 15.59 29.168] 327 0 obj q Q >> endobj /Meta386 402 0 R << Q << endobj 0 G /Length 69 ET q /Meta121 Do /F3 17 0 R /Subtype /Form BT 0 5.203 TD /Subtype /Form Q /Meta217 Do (C) Tj /F3 17 0 R 1.005 0 0 1.007 102.382 653.441 cm /BBox [0 0 15.59 16.44] /F1 12.131 Tf stream /Meta154 168 0 R /Resources<< /Length 54 /Meta239 Do /StemH 94 /Length 16 0.737 w Q q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q 177 0 obj endobj 1.007 0 0 1.007 130.989 330.484 cm >> Answer link. Q /Font << /Font << Q /FormType 1 /ProcSet[/PDF/Text] /MediaBox [0 0 767.868 993.712] Q 0.458 0 0 RG endobj /Subtype /Form Q 0 w /F4 12.131 Tf 1 i /Subtype /Form Q 0.564 G /FormType 1 "49 . /Resources<< /ProcSet[/PDF/Text] /Subtype /Form 1 i 1.007 0 0 1.006 411.035 763.351 cm BT Q /BBox [0 0 88.214 16.44] /Resources<< Q 0.382 Tc q >> << q /Type /XObject Q 0.564 G endobj /ProcSet[/PDF/Text] 1 i 125 0 obj Q 1.007 0 0 1.007 130.989 636.879 cm 221 0 obj /F3 17 0 R 1.007 0 0 1.007 411.035 636.879 cm /ProcSet[/PDF] 1 g /Meta176 Do q Q q /Type /XObject >> /Meta290 Do /F3 17 0 R endobj /Type /XObject 0 G 1.007 0 0 1.007 130.989 636.879 cm stream Q 1.007 0 0 1.006 130.989 437.384 cm Q q /BBox [0 0 88.214 16.44] q /Subtype /Form 421 0 obj 0.564 G /Length 64 /FormType 1 /Length 69 q Q Q /Type /XObject /Resources<< 335 0 obj /F1 7 0 R 0 G >> q stream >> 0 g [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /Matrix [1 0 0 1 0 0] The sum of a number and 2 is 6 less than twice that number. /ProcSet[/PDF/Text] Q BT /Meta266 280 0 R /F3 17 0 R endobj /FormType 1 ET /Meta272 Do Advertisement Loved by our community 50 people found it helpful Madhvendra13 2x -8=58 2x=66 x=662 x=33 Find Math textbook solutions? q 1 i >> >> 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) 1 g /Meta219 233 0 R /Font << 0 g Q /Meta15 26 0 R /Type /XObject >> 0 g << >> >> /XHeight 477 0 g q 549.694 0 0 16.469 0 -0.0283 cm /BBox [0 0 673.937 68.796] stream /BBox [0 0 534.67 16.44] /Meta134 148 0 R endobj /F4 36 0 R /Matrix [1 0 0 1 0 0] /Type /XObject 0 G >> Table 1. ET Q 1 g q 0 w 0.458 0 0 RG /Resources<< ] q Q q stream 1.502 5.203 TD 97 0 obj Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. /FormType 1 << 0 G endstream (13) Tj 330 0 obj /ProcSet[/PDF/Text] /Type /XObject /ProcSet[/PDF/Text] q 254 0 obj /Meta407 423 0 R 0.564 G Q 1 g /ProcSet[/PDF/Text] /Resources<< >> endobj << ET << 1 i ET q /ProcSet[/PDF/Text] Q /BBox [0 0 534.67 16.44] /FormType 1 q /Type /XObject /F3 17 0 R /Matrix [1 0 0 1 0 0] 0.458 0 0 RG 0 G /Type /XObject /Meta347 Do /Resources<< >> /FormType 1 /Meta22 33 0 R /Type /XObject 0 w q /Meta292 Do >> /Meta290 304 0 R 1.007 0 0 1.007 130.989 776.149 cm q /Resources<< Q Q 1.014 0 0 1.007 391.462 330.484 cm /BBox [0 0 88.214 16.44] q 0.524 Tc /Length 59 (58) Tj /Meta414 Do >> /F3 12.131 Tf q /BBox [0 0 88.214 16.44] 0 4.894 TD Q << /BBox [0 0 88.214 35.886] >> /Type /XObject /Resources<< /ProcSet[/PDF/Text] q /Length 58 stream /Type /XObject /Subtype /Form /F3 17 0 R stream /F3 17 0 R A) 5 more than a number. endobj 0 g q /Resources<< << /Subtype /Form /Subtype /Form endstream /Font << /Font << /ProcSet[/PDF/Text] (C\)) Tj 20.21 5.336 TD /Meta254 Do /Meta171 185 0 R 0 g 0.737 w ET endstream (2) Tj 0 G -0.021 Tw endobj /Meta133 147 0 R 1.014 0 0 1.007 111.416 450.181 cm Q /Subtype /Form /Matrix [1 0 0 1 0 0] /Type /XObject q /FontDescriptor 35 0 R /Meta103 117 0 R /F1 7 0 R 214 0 obj 1 i >> 0 g /Type /XObject 288 0 obj Q 0 g Q (7\)) Tj /BBox [0 0 673.937 15.562] /Matrix [1 0 0 1 0 0] >> 1 i q Q /Font << stream ET 0 g 163 0 obj 1.502 5.203 TD 0 G Q 1.014 0 0 1.007 251.439 583.429 cm /Meta227 Do /BBox [0 0 15.59 16.44] ET 8 0 obj /Matrix [1 0 0 1 0 0] BT >> Q /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] >> 1 i 68 0 obj >> q /F3 17 0 R /Subtype /Form >> endstream << endstream /BBox [0 0 88.214 16.44] /F3 17 0 R 1.007 0 0 1.007 130.989 636.879 cm Q q 0.458 0 0 RG ET /Type /XObject endstream 1 i BT ET endstream ET endobj /Meta197 Do 329 0 obj /Meta7 Do 0 g /Meta410 426 0 R BT /BBox [0 0 673.937 16.44] BT Q >> 0 w /Length 118 q 1 i /Meta242 Do << 0 g /Length 118 /Matrix [1 0 0 1 0 0] >> endstream 17.234 5.203 TD /BBox [0 0 17.177 16.44] Q If twice a number is decreased by 13, the result is 9. q q endstream BT /Font << /BBox [0 0 534.67 16.44] endstream /Font << /FormType 1 /Subtype /Form 1.007 0 0 1.007 130.989 383.934 cm /Resources<< 0 w -0.084 Tw q /Length 16 0 G /Font << (5) Tj Q /F3 17 0 R Q q >> /ProcSet[/PDF] /Meta241 255 0 R /Resources<< /F1 12.131 Tf 1.007 0 0 1.007 271.012 523.204 cm /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Resources<< Q q /Subtype /Form 1.007 0 0 1.007 271.012 383.934 cm 1. /ProcSet[/PDF] /Length 59 /ProcSet[/PDF] BT /Resources<< Find the length. >> /FormType 1 /F3 17 0 R /ProcSet[/PDF/Text] ET /FormType 1 1 i q 1 i /Font << << /Matrix [1 0 0 1 0 0] 0 g q 0 G 0 g >> 179 0 obj >> endstream q /Type /XObject 400 0 R q /Length 16 BT /Subtype /Form 0 G /Matrix [1 0 0 1 0 0] /Type /XObject q >> /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g /Type /XObject /BBox [0 0 88.214 16.44] << << Q 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 Q stream Q >> << /FormType 1 >> 1 i endobj /ProcSet[/PDF] /Meta172 Do << /F3 17 0 R 0.458 0 0 RG 0.737 w 0 g >> /BBox [0 0 15.59 16.44] endobj 328 0 obj endobj stream q Q /FormType 1 >> /Meta157 171 0 R q Q /Subtype /Form BT (A\)) Tj 1.014 0 0 1.007 391.462 383.934 cm q 0.458 0 0 RG >> /FormType 1 Q << Q stream >> (x) Tj [( subt)-17(racted fr)-14(om a )-16(number)] TJ 0 g 1.014 0 0 1.007 251.439 703.126 cm ET /Meta220 234 0 R /ProcSet[/PDF] /Meta38 Do /Meta343 Do << << 1 i /BBox [0 0 88.214 16.44] 0 w 1.007 0 0 1.007 551.058 523.204 cm q /FormType 1 /Resources<< 1.007 0 0 1.006 411.035 510.406 cm 0.458 0 0 RG 0 G /F3 12.131 Tf /F4 36 0 R /Meta393 Do Twice a number when decreased by 7 gives 45. 0 g Q /Length 58 q Q q /Font << /F4 12.131 Tf /F3 12.131 Tf q /ProcSet[/PDF] /Type /XObject /Subtype /Form /Matrix [1 0 0 1 0 0] /FormType 1 /FormType 1 /Subtype /Form stream 339 0 obj Q Q >> /F3 12.131 Tf /Length 69 Q 1.007 0 0 1.007 654.946 653.441 cm /Subtype /Image /ProcSet[/PDF/Text] /Meta415 Do /Meta312 326 0 R 1 i BT So we have twice of a mystery number decreased by three, and that is all going to be 31. /Type /XObject 0.564 G 0 G >> /Meta125 Do << Q /Matrix [1 0 0 1 0 0] q 0.458 0 0 RG 1 i 0 g >> 549.694 0 0 16.469 0 -0.0283 cm q /F3 17 0 R q 426 0 obj 0 G /Descent -216 /FormType 1 /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g >> /FormType 1 0 g 1 i 389 0 obj /ProcSet[/PDF/Text] (5) Tj /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta361 Do 65.906 4.894 TD /BBox [0 0 88.214 16.44] 0 G Q /Matrix [1 0 0 1 0 0] q 16.469 5.336 TD endstream 1.014 0 0 1.006 531.485 510.406 cm Q Q 1.007 0 0 1.007 130.989 383.934 cm ET q 0 G /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] /Length 69 0 G 1 i /Resources<< /F3 12.131 Tf 1 i /FormType 1 Q >> /FormType 1 /Matrix [1 0 0 1 0 0] /Type /XObject q endstream /Meta346 360 0 R /Type /XObject /FontDescriptor 6 0 R Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . /F3 17 0 R 9 0 obj endstream 0.564 G /Subtype /Form 1.007 0 0 1.007 411.035 636.879 cm /F1 12.131 Tf /Length 59 q /BBox [0 0 88.214 16.44] /Subtype /Form Q /Matrix [1 0 0 1 0 0] endstream Q /Font << << /Length 81 /BBox [0 0 88.214 16.44] /Length 69 0.369 Tc 95 0 obj /Meta344 Do /Length 69 1 i /ProcSet[/PDF/Text] 5.98 7.841 TD 0.51 Tc 0 g 0 g q 1.007 0 0 1.007 130.989 849.172 cm 1 i /Length 58 /Matrix [1 0 0 1 0 0] /Subtype /Form Q stream (- 4) Tj ET endobj /Type /XObject /FormType 1 Q 0 g 1.005 0 0 1.006 45.168 879.284 cm ET 0 5.203 TD >> 0.564 G /Matrix [1 0 0 1 0 0] /Type /XObject Q /Meta406 Do 1 i >> Q 0.458 0 0 RG /BBox [0 0 534.67 16.44] 0.68 Tc /BBox [0 0 88.214 16.44] q /BBox [0 0 534.67 16.44] >> BT q /F4 36 0 R /Font << (C\)) Tj endobj endstream /F3 12.131 Tf Q /Meta139 Do Q endstream << >> /Type /XObject /Resources<< Q 0 w stream >> 0 G ET >> >> >> /BBox [0 0 30.642 16.44] /Length 78 -0.058 Tw Step 1/1. /F3 17 0 R << 1 i << 1.007 0 0 1.007 271.012 849.172 cm Q 1 i 0.458 0 0 RG Q 0 G q /Type /XObject Q /F3 12.131 Tf /Resources<< ET 0 g BT 0.564 G /Type /XObject q /Font << 316 0 obj /Subtype /Form Q endobj 0 w 0 w /Meta345 Do Q /Meta372 Do stream << /F4 12.131 Tf << stream 367 0 obj 1 i /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] stream BT /Subtype /Form /Matrix [1 0 0 1 0 0] endstream 0 w /Font << /Resources<< /F1 7 0 R 0 G /BBox [0 0 534.67 16.44] q 0 g 1.014 0 0 1.007 251.439 450.181 cm 0.737 w /Meta325 339 0 R 252 0 obj /BBox [0 0 88.214 16.44] ET /Font << /ProcSet[/PDF] Q /ProcSet[/PDF/Text] /Type /XObject q q q q -0.486 Tw Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 Q /Meta151 Do 1 i /FormType 1 BT /Length 69 >> /Length 12 /Matrix [1 0 0 1 0 0] /Meta118 Do q endobj /F1 12.131 Tf q /Meta361 375 0 R /BBox [0 0 15.59 16.44] >> /ProcSet[/PDF/Text] /Type /XObject endobj 1 i << Q >> q << /BBox [0 0 88.214 16.44] /Font << /ProcSet[/PDF/Text] endstream q stream q five times the sum of a number x and two b.) /Resources<< /ColorSpace [/Indexed /DeviceGray 1 ] << /Type /XObject q /FormType 1 2. ET /BBox [0 0 15.59 29.168] >> /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 /Font << q stream << /F3 17 0 R Q 1 i /Font << >> << 0 g /Subtype /Form << endobj /ProcSet[/PDF/Text] /Meta218 Do Q /F3 12.131 Tf q /Length 118 endstream 1 i /F3 17 0 R 1 i Q 1 g q /Resources<< /Subtype /Form endstream /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 111.416 523.204 cm 0.68 Tc >> /Resources<< /F3 12.131 Tf endstream 0 w /Length 118 /BBox [0 0 534.67 16.44] /Subtype /Form /Subtype /Form ET /Type /XObject /Meta374 Do /Meta145 Do ET Q /BBox [0 0 639.552 16.44] Q /FormType 1 0 w /Type /XObject /Meta9 20 0 R Q >> 1 i stream 1.007 0 0 1.007 271.012 636.879 cm BT 1.007 0 0 1.007 411.035 277.035 cm 0 G /FormType 1 /FormType 1 /Matrix [1 0 0 1 0 0] endstream >> q /Subtype /Form (C\)) Tj /F3 12.131 Tf q << stream /FormType 1 >> Q >> q 88 0 obj >> 1 i /ProcSet[/PDF/Text] endstream /Subtype /Form >> q /F3 12.131 Tf /F3 17 0 R 0 G Q q /Matrix [1 0 0 1 0 0] 0 g /ProcSet[/PDF] 0.458 0 0 RG /Font << 0.458 0 0 RG Q ET q (\)) Tj Q /Resources<< /Matrix [1 0 0 1 0 0] endstream endstream /Length 60 /Type /XObject /Meta268 Do 1.007 0 0 1.007 67.753 799.486 cm endstream /FormType 1 Q /Length 12 371 0 obj q 1 g ET >> /Matrix [1 0 0 1 0 0] ET 0.564 G 215 0 obj 0.737 w Q 0.458 0 0 RG 446 0 obj q /Type /XObject Q >> Q endobj /Matrix [1 0 0 1 0 0] Q /Resources<< /F3 17 0 R 0.458 0 0 RG /LastChar 45 ET /BBox [0 0 88.214 35.886] /Subtype /Form 1.014 0 0 1.007 391.462 450.181 cm /Meta384 398 0 R 1 g /Type /XObject /Meta387 403 0 R Q Q q 1.007 0 0 1.007 67.753 546.541 cm /ProcSet[/PDF/Text] /ProcSet[/PDF] Q q /Length 60 endobj 0 G q /Length 12 94.364 5.203 TD stream q /BBox [0 0 88.214 35.886] 2. 1 i /Type /XObject >> stream S >> q /Meta175 189 0 R /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] q /FormType 1 /F4 12.131 Tf Q 0 w << /Font << 369 0 obj 1.005 0 0 1.007 102.382 653.441 cm 0 w >> /F1 12.131 Tf >> 1.007 0 0 1.007 271.012 636.879 cm /BBox [0 0 88.214 35.886] << 38.948 5.203 TD /Matrix [1 0 0 1 0 0] /Resources<< /BBox [0 0 88.214 16.44] stream 1 i Q 0 g 1 i Q Q 0 G >> q /Meta369 383 0 R BT /ProcSet[/PDF/Text] /F3 12.131 Tf /BBox [0 0 15.59 16.44] /Type /XObject /Subtype /Form /BBox [0 0 549.552 16.44] /Matrix [1 0 0 1 0 0] 6 more than twice a number: 2x+6: two less than a number: x-2: the sum of 9 and a number: 9+x: two less than three times a number: 3x-2: a number subtracted from 12 . 1.007 0 0 1.006 551.058 437.384 cm /Meta221 Do q /Length 69 /FormType 1 Q /F3 12.131 Tf Q >> 0 G /Type /XObject /Matrix [1 0 0 1 0 0] >> /FormType 1 /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] >> Q /Subtype /Form /Resources<< BT 0 g /F3 17 0 R BT stream stream 1.005 0 0 1.013 45.168 933.487 cm 1.007 0 0 1.007 271.012 277.035 cm >> 16.469 5.336 TD /Resources<< q 22.478 5.203 TD endobj /ProcSet[/PDF/Text] q 1 i S Q 334 0 obj q /Subtype /Form endobj ET 0.564 G Q /ProcSet[/PDF/Text] Q /Length 294 0 G /Type /XObject Q BT >> 722.699 653.441 l 30.699 5.203 TD q Q q >> 0 5.203 TD 0 g << /Leading 150 /F3 17 0 R Q [(-1)-16(52)] TJ >> 0.458 0 0 RG endobj /Font << q /Length 54 /F3 12.131 Tf q 1.007 0 0 1.007 130.989 277.035 cm >> endobj stream /Type /XObject 0.155 Tc /Width 734 1.007 0 0 1.007 130.989 277.035 cm 0 5.203 TD endstream stream 424 0 obj 1.007 0 0 1.007 551.058 330.484 cm q q 38 0 obj q /Subtype /Form << Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. /Resources<< << << /ProcSet[/PDF/Text] Q q 1 i /Meta413 Do /Resources<< 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject /Meta385 401 0 R >> 1 i Q endobj (-9) Tj 0.564 G /Length 68 endobj /Font << /Resources<< 0 G endobj 1.007 0 0 1.007 551.058 277.035 cm q stream See Solution. /Length 12 /FormType 1 0 G 301 0 obj 0.564 G q 0 g /Subtype /Form 0 G Q 1 i Q q >> /Meta119 Do q >> stream Q /Type /XObject Q 1.007 0 0 1.007 271.012 383.934 cm /Matrix [1 0 0 1 0 0] 0 g >> endstream /Matrix [1 0 0 1 0 0] /Meta132 Do /ProcSet[/PDF/Text] S /FormType 1 Q /FormType 1 /Matrix [1 0 0 1 0 0] stream stream BT Q /BBox [0 0 88.214 16.44] q q /Resources<< Q >> 1 i /Type /XObject q /Meta49 63 0 R /Matrix [1 0 0 1 0 0] 1 g q /ProcSet[/PDF] 205 0 obj /FormType 1 /Subtype /Form 0.737 w q << /BitsPerComponent 1 q ET >> >> q 32 0 obj Q /MissingWidth 250 q /Matrix [1 0 0 1 0 0] >> q /Length 69 /FormType 1 q /Type /XObject endobj 0 g /Font << Let the 2nd number be y. Q >> /Subtype /Form 0 g endstream /Meta318 332 0 R /F3 17 0 R /Matrix [1 0 0 1 0 0] 62 0 obj q 1.014 0 0 1.007 531.485 277.035 cm q >> q 1 i endstream 1.007 0 0 1.007 271.012 849.172 cm >> q << q /Resources<< >> stream >> q 0 w 0 G Q 0.458 0 0 RG /ProcSet[/PDF/Text] endobj >> q /FormType 1 >> /Length 65 0 g /Matrix [1 0 0 1 0 0] 1 0 obj q >> q Q >> 1 i /Length 70 Q << >> endobj Q /Font << 0 G /BBox [0 0 15.59 16.44] endobj /ProcSet[/PDF] /Resources<< ET /Subtype /Form q (2\)) Tj >> 0 5.203 TD endobj Q 181 0 obj << /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] Q /Meta299 313 0 R >> /Subtype /Form /Resources<< q [(MULTIPLE CHOICE. /Meta169 183 0 R Q /Matrix [1 0 0 1 0 0] /Font << 1.007 0 0 1.007 551.058 583.429 cm /Length 69 /Length 151 q << >> >> /XObject << BT /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] stream endstream >> /Meta15 Do 0.458 0 0 RG endstream /Type /XObject /Flags 32 73 0 obj /Type /XObject 0 5.203 TD stream 1 g q Q /Subtype /Form >> 1 i (\)) Tj q 0.564 G stream /Subtype /Form /XObject << q /Subtype /Form 0 G Q q q 1 i /Meta411 Do stream 11.99 24.649 TD /Meta296 310 0 R /Type /XObject Q q 48 0 obj >> 0.458 0 0 RG >> Q /Resources<< q /Meta243 Do q 1 i /Count 2 /Type /XObject Q /Type /FontDescriptor /Meta276 Do /F1 7 0 R >> 1 i /ProcSet[/PDF] /Resources<< [(The )-19(quotient of )] TJ BT q q (A\)) Tj S /ProcSet[/PDF/Text] /I0 51 0 R /ProcSet[/PDF] 266 0 obj q endobj << 318 0 obj q << /Font << Let x the unknown number. /Length 69 /Subtype /Form << 1.007 0 0 1.007 130.989 330.484 cm /Meta75 Do Q q /Type /XObject 255 0 obj 1.502 5.203 TD /Font << endstream /Resources<< /Subtype /Form endstream >> Q /Meta50 64 0 R (\)) Tj endstream 0 g Q 1 i >> 0 w endobj >> /Length 59 /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject /Resources<< endobj >> ET >> /Font << /Meta68 82 0 R 1.007 0 0 1.007 271.012 277.035 cm Q /ProcSet[/PDF] ET 0.564 G Q 0 g BT /ProcSet[/PDF] endstream 1.014 0 0 1.007 111.416 583.429 cm /Length 16 /F4 12.131 Tf q BT q (58) Tj Q endstream /Resources<< 0.458 0 0 RG 1.007 0 0 1.007 130.989 583.429 cm /Subtype /Form /FormType 1 0 g for the season. BT If n is "the number," which equation could be used to solve for the number? Q /Type /XObject endstream /Matrix [1 0 0 1 0 0] stream /Resources<< 1 i 392 0 obj endstream /Meta332 Do stream 1.005 0 0 1.007 102.382 599.991 cm 1 i stream /Font << 1.007 0 0 1.007 551.058 383.934 cm q /Meta167 181 0 R /ProcSet[/PDF/Text] >> BT /Matrix [1 0 0 1 0 0] 0.297 Tc q S ET >> 0.564 G /FormType 1 /Subtype /Form much as how 8, Last . 1.007 0 0 1.007 130.989 849.172 cm Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. >> (iv) A number exceeds 5 by 3. /Meta48 Do q /Matrix [1 0 0 1 0 0] 0.737 w 0 G /BBox [0 0 673.937 68.796] /Matrix [1 0 0 1 0 0] q 212 0 obj /F1 12.131 Tf /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Q 1.007 0 0 1.007 130.989 523.204 cm /Meta302 316 0 R Q /FirstChar 32 Q /Meta189 203 0 R 0 G 1.007 0 0 1.006 130.989 437.384 cm q /Meta158 Do /BBox [0 0 30.642 16.44] Q endstream /FormType 1 BT /FormType 1 /F3 12.131 Tf q /F3 12.131 Tf 355 0 obj Q (B) Tj q /Resources<< /Subtype /Form /FormType 1 /Meta306 Do >> 0.737 w /F1 12.131 Tf >> 1 i /Font << endstream /Meta243 257 0 R /FormType 1 q 19 0 obj endobj /Resources<< << /F1 7 0 R /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF/Text] >> 94 0 obj q /Font << /Matrix [1 0 0 1 0 0] 0 g endstream q 1 i /ProcSet[/PDF] endobj 1 g 0.17 Tc >> /Subtype /Form (D\)) Tj >> 165 0 obj /Matrix [1 0 0 1 0 0] /FormType 1 << /FormType 1 /ProcSet[/PDF] Q 0 w 9.723 5.336 TD Q /Matrix [1 0 0 1 0 0] Q >> /Length 16 /Resources<< 0 G ET /BBox [0 0 15.59 16.44] What is marios jumps times luigis jumps. 69 0 obj << /Type /XObject /Meta311 Do ET >> q 265 0 obj (4\)) Tj /Subtype /Form /BBox [0 0 88.214 16.44] 0.564 G 1 i Q /Meta333 Do 1 i /Subtype /Form /Meta405 Do >> Q /Meta41 55 0 R q /Type /XObject endstream /Subtype /Form 0 g /Type /XObject /Subtype /Form endobj 1 i /BBox [0 0 88.214 16.44] /Type /XObject 410 0 obj /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 130.989 583.429 cm /F4 36 0 R (2) Tj /FormType 1 (\)) Tj The sum of 18 and tour times a number is -6 Find the number. >> q 0.458 0 0 RG 1 g /Meta73 Do /F4 36 0 R Q >> 1 i >> /FormType 1 /ProcSet[/PDF] /Length 79 >> /Resources<< /Font << 1 i /Subtype /Form 1 i /BBox [0 0 88.214 16.44] 0.737 w /Meta210 Do the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . 1.007 0 0 1.006 411.035 690.329 cm ET /F3 17 0 R << 0 g /FormType 1 q /Subtype /Form /Matrix [1 0 0 1 0 0] -0.067 Tw Q endstream q Q >> /Type /XObject 1 i 1.007 0 0 1.006 130.989 437.384 cm >> Q , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. 0.564 G /BBox [0 0 88.214 16.44] stream /ProcSet[/PDF/Text] endstream q >> q /FormType 1 /F3 17 0 R /Font << /Subtype /Form >> /Length 59 q 348 0 obj You could call them. Q Q 0.738 Tc q 0.524 Tc 0 G /Type /XObject /ProcSet[/PDF/Text] q /ProcSet[/PDF/Text] 0 G q q /Meta276 290 0 R Q >> (9\)) Tj /FormType 1 View the full answer. >> << /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0 G /BBox [0 0 534.67 16.44] /Resources<< /Type /XObject Q /F3 12.131 Tf endobj /Meta13 24 0 R /Meta21 Do /Subtype /Form /Type /XObject q -0.486 Tw 1.007 0 0 1.007 551.058 330.484 cm Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 Q ET /Type /XObject 268 0 obj 1.007 0 0 1.007 271.012 277.035 cm 0 g /Meta88 102 0 R 3.742 24.649 TD /BBox [0 0 88.214 16.44] /Type /XObject /FormType 1 /Type /XObject Q /Matrix [1 0 0 1 0 0] 1 g stream Q /F3 12.131 Tf ET /Length 69 /Length 70 /FontBBox [-174 -299 1445 1050] << stream endstream Q 0.51 Tc q (5\)) Tj /Meta403 Do /Meta159 Do /FormType 1 q endstream 0.737 w /Matrix [1 0 0 1 0 0] q << q /Matrix [1 0 0 1 0 0] q 0.737 w stream q /BBox [0 0 534.67 16.44] Q /Resources<< /BBox [0 0 534.67 16.44] BT /Meta140 154 0 R /Subtype /Form endobj /Matrix [1 0 0 1 0 0] Q /Meta264 Do BT endstream /Matrix [1 0 0 1 0 0] /Subtype /Form 0 g /ProcSet[/PDF/Text] /F3 17 0 R 0.297 Tc >> 0 G Q q Q ET 53 0 obj [4] One half of a number decreased by fourteen is twenty-one 81 0 obj (+) Tj 1 i /FormType 1 /Meta255 Do /Subtype /Form (x ) Tj Q q >> (x) Tj /ProcSet[/PDF] /Meta27 Do BT 4 0 R 0 g /Font << endobj >> 210 0 obj 270 0 obj endstream /Subtype /Form /I0 51 0 R Q 0 w 0.737 w 294 0 obj /Meta2 9 0 R /ProcSet[/PDF/Text] 65 0 obj 1 i Q /Meta31 44 0 R 146 0 obj 1 i q 1.007 0 0 1.007 411.035 383.934 cm 0.564 G Q Q Q /FormType 1 << q 0.51 Tc /Meta138 152 0 R endobj /Meta46 60 0 R q 1 i /Meta300 Do /Matrix [1 0 0 1 0 0] >> 382 0 obj stream stream Twice a first number decreased by a second number is 6. /Font << /F3 17 0 R >> Q /BBox [0 0 88.214 35.886] >> q << Q << >> BT endstream << >> BT /Meta98 112 0 R Q /Meta81 95 0 R q Q 287 0 obj /FormType 1 >> Q /Matrix [1 0 0 1 0 0] /Subtype /Form >> q 0.737 w 0 G /Meta183 Do /Subtype /Form Q endstream /F3 12.131 Tf (+) Tj 0 5.203 TD 230 0 obj q q >> >> Q 0 w /BBox [0 0 15.59 29.168] /F3 12.131 Tf 26.219 5.336 TD /F1 7 0 R /BBox [0 0 88.214 16.44] /Meta273 Do /Subtype /Form /FormType 1 /Meta282 296 0 R 1.007 0 0 1.006 411.035 437.384 cm /Type /XObject stream q /BBox [0 0 88.214 16.44] 16.469 5.203 TD /F4 36 0 R 0 G >> Q Q /Meta354 368 0 R Q /Matrix [1 0 0 1 0 0] q Five times the sum of a number and four 7. 0.369 Tc /Meta107 121 0 R Q endobj 0 g q endstream q Ten divided by a number 5. The sum Of twice a nu4ber What is the number? 1 g /Resources<< 0.486 Tc Q /Meta236 250 0 R >> let 'x' and 'y' represent the numbers. /Meta8 Do Q 0.737 w ET /Subtype /Form stream q /Meta97 111 0 R For Free. q Q << 1 i >> >> 1.014 0 0 1.006 251.439 690.329 cm << BT /Matrix [1 0 0 1 0 0] /F3 12.131 Tf >> /BBox [0 0 30.642 16.44] 0.51 Tc stream 549.694 0 0 16.469 0 -0.0283 cm /Type /XObject 0 G 0 G /ProcSet[/PDF/Text] >> Q BT 0 g /Type /XObject /F1 7 0 R Q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1 g ET 0 20.154 m q 1.007 0 0 1.007 130.989 583.429 cm >> >> /Resources<< /F3 17 0 R /F3 17 0 R /F1 12.131 Tf q /Meta278 292 0 R /Subtype /Form stream /Length 12 << /Meta329 Do 0.463 Tc 0 w 0.564 G /Meta258 272 0 R 1 i >> q /Meta135 149 0 R 66 0 obj /F4 12.131 Tf >> endstream 1 i /BBox [0 0 17.177 16.44] endstream BT 0.68 Tc /Matrix [1 0 0 1 0 0] /Meta80 94 0 R Q /ProcSet[/PDF] /FormType 1 endobj 412 0 obj &K @ Q >> >> /Resources<< endstream 0.738 Tc endobj 0 g Q 1.502 5.203 TD Q 0 g /ProcSet[/PDF/Text] 0 G q /Type /XObject endstream 0 G Q >> Q 234 0 obj /F3 17 0 R q ET /Length 59 0 G >> /F3 12.131 Tf 0 g 1 g 0.134 Tc (-23) Tj /Meta135 Do /Resources<< q stream /FormType 1 endstream /F4 12.131 Tf /Resources<< >> /Font << >> /Length 16 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. /F3 12.131 Tf stream /Resources<< << /Length 16 0 g >> /BBox [0 0 30.642 16.44] Q q endstream >> /ProcSet[/PDF] /Meta34 47 0 R q q q >> /Meta32 45 0 R 176 0 obj 0.737 w 0 g 1 i endobj /Subtype /Form 1 i /ProcSet[/PDF] q /Subtype /TrueType endobj << 1 i 0 g /Font << q /FormType 1 0.737 w >> /Length 118 1 i /Resources<< q endstream /FormType 1 BT 422 0 obj Q /Length 59 /Resources<< /FormType 1 >> 0 g /ProcSet[/PDF/Text] >> /Meta10 21 0 R >> /Length 59 0 g stream Q >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] q /ProcSet[/PDF/Text] >> q >> /Font << /Meta203 217 0 R Q /BBox [0 0 30.642 16.44] 0 g /Meta403 419 0 R q /ProcSet[/PDF/Text] q 1 g stream /ProcSet[/PDF/Text] << /XObject << Q 86 0 obj Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM /F1 7 0 R q 0.564 G /Resources<< stream /BBox [0 0 15.59 29.168] /FormType 1 /ProcSet[/PDF/Text] Q ET /Matrix [1 0 0 1 0 0] 0.68 Tc Q ( \() Tj /F4 12.131 Tf Q << /BBox [0 0 88.214 16.44] 1 i q /Type /XObject q 0.564 G ( x) Tj 0 G (38) Tj (11) Tj >> stream 0 G 0 g >> q /Meta309 Do 0 G /Matrix [1 0 0 1 0 0] endobj 0 g 0 g q /Resources<< 1.014 0 0 1.007 391.462 583.429 cm /Type /XObject 0.458 0 0 RG /Subtype /Form q >> 0 g >> endobj /Meta87 101 0 R Q >> q /Meta378 392 0 R q q Q Q /Resources<< /FormType 1 /Resources<< 0.458 0 0 RG /Matrix [1 0 0 1 0 0] n 11 or n 11. ( x) Tj /Type /XObject q /Subtype /Form Q /Resources<< 0 G Q /BBox [0 0 88.214 35.886] /FormType 1 stream endobj BT 1.007 0 0 1.007 411.035 383.934 cm /ProcSet[/PDF/Text] q 0.564 G /Meta415 431 0 R q 1 i 1 i stream ET 0 w /ProcSet[/PDF] (-11) Tj 0 g BT Q /Meta379 Do >> /Font << /Meta12 Do q 0.737 w 409 0 obj /Matrix [1 0 0 1 0 0] q 0 g endobj << q /BBox [0 0 88.214 35.886] 0 5.203 TD /Type /XObject BT 1 i 0 g /Type /XObject >> endobj Q << /FormType 1 /Meta48 62 0 R 0 G << /BBox [0 0 88.214 16.44] q 0.458 0 0 RG /F1 7 0 R ET endobj /Matrix [1 0 0 1 0 0] (40) Tj endobj /F1 7 0 R /Matrix [1 0 0 1 0 0] 1 i /Subtype /Form /BBox [0 0 30.642 16.44] Q /Matrix [1 0 0 1 0 0] /Resources<< Kobe scored 85 points in a basketball game. 1.007 0 0 1.006 130.989 437.384 cm 203 0 obj /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] Q >> q /Resources<< ET /FormType 1 /Meta72 Do Q >> Thrice a number decreased by 5 exceeds twice the number by 1. /Type /XObject /Meta330 Do endstream /Length 16 ET /Subtype /Form Q >> BT /Font << Q /F3 17 0 R stream /F3 12.131 Tf Q endstream Q Q /Type /XObject endstream q /FormType 1 /Type /XObject 0 g 0.737 w Q 0 w 1 i stream q >> /Resources<< Q /Subtype /Form /FormType 1 /Meta19 30 0 R /F1 12.131 Tf q /ProcSet[/PDF/Text] Have a nice day! (-23) Tj q /Font << endstream 0 g Q /Length 70 >> 0 G /Matrix [1 0 0 1 0 0] /Subtype /Form 0.564 G /F3 12.131 Tf /ProcSet[/PDF/Text] << q Q /Length 70 1.014 0 0 1.006 111.416 510.406 cm >> 168 0 obj 0 g /F3 12.131 Tf /FormType 1 1.007 0 0 1.007 45.168 813.037 cm /Meta346 Do endstream q -0.058 Tw (x) Tj /ProcSet[/PDF/Text] BT 63 0 obj /BBox [0 0 30.642 16.44] q 1 i q << /Meta226 240 0 R 0 G /Meta252 266 0 R endstream q /Type /XObject /Length 59 endstream 1.014 0 0 1.007 391.462 523.204 cm 145 0 obj /Matrix [1 0 0 1 0 0] 0 G /ProcSet[/PDF] 1.008 0 0 1.007 654.946 293.596 cm Q Q /FormType 1 155 0 obj endstream If a number is 50%, then it is a half - the same as 0.5 or 1/2. /Type /XObject q 0 g 440 0 obj /ProcSet[/PDF/Text] q /Subtype /Form ET 46 0 obj endobj 208 0 obj Q 0.564 G /ProcSet[/PDF/Text] /F3 17 0 R /Matrix [1 0 0 1 0 0] twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. 346 0 obj 1 i endstream endobj Q 0.369 Tc >> Q >> BT /Type /XObject q /BBox [0 0 88.214 16.44] At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . >> Q 0.458 0 0 RG q /Resources<< /BBox [0 0 88.214 16.44] /Subtype /Form << Q 1 i 1 i /Resources<< << >> /FormType 1 /F3 17 0 R q 0 G /F3 12.131 Tf Q Thirthy is equal to twice a number decreased by four = solve and check the equation? 1 i /Meta373 387 0 R [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. q endobj 1 g /Resources<< /ProcSet[/PDF] 0 G Q /ProcSet[/PDF/Text] stream 24.718 8.18 TD endobj /Type /XObject 0 G 0.369 Tc -37 VI 2. >> >> ET /FormType 1 >> /F3 12.131 Tf /Length 70 endstream /Meta347 361 0 R /XObject << 1 g /FormType 1 BT 1.007 0 0 1.007 130.989 776.149 cm 250 0 obj 0.564 G 1 i stream /Type /XObject Q q q 1 g /FormType 1 Q /FormType 1 1.007 0 0 1.007 411.035 583.429 cm /Subtype /Form /Subtype /Form Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /Pages 1.014 0 0 1.007 391.462 636.879 cm >> Q 0 w /Resources<< /Length 80 0.564 G q 0 w /Length 116 /BBox [0 0 30.642 16.44] -0.486 Tw (D) Tj 722.699 293.596 l /Length 69 /Length 69 0.838 Tc 0 5.203 TD 1 i /Type /XObject /Resources<< /BBox [0 0 88.214 16.44] /Length 16 << stream q 1.007 0 0 1.007 130.989 383.934 cm
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twice a number decreased by 58