twice a number decreased by 58

>> /Meta272 Do 1.014 0 0 1.007 391.462 330.484 cm /Meta191 205 0 R 0.458 0 0 RG >> Q LAIing for a pizza and, soft drink. Q 0.564 G endobj /F3 12.131 Tf Q 0.369 Tc >> 1 i 0.369 Tc q 0.369 Tc endobj 250 0 500 500 500 500 500 500 0 0 500 500 278 0 0 0 Q << /Meta406 Do q Q Q q endobj /Meta203 Do /Font << /Length 69 /Subtype /Form stream 1 i (7\)) Tj 422 0 obj Q /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] Q >> q q /ProcSet[/PDF] 370 0 obj /Length 59 /F3 17 0 R /FormType 1 /Matrix [1 0 0 1 0 0] >> >> ET 1 i << /FormType 1 0 G Q /ProcSet[/PDF/Text] 0.307 Tc ET stream 0.564 G BT /Type /XObject 1 i stream /F1 7 0 R 1.007 0 0 1.007 271.012 277.035 cm ET 0 g 0 w /Resources<< Q /Meta281 Do /Subtype /Form /F3 17 0 R /Length 118 stream << ET 0.297 Tc Select the correct mathematical statement for the following equation. /FormType 1 1 i /Length 80 /Type /XObject Q 3.742 5.203 TD /Type /XObject 0.737 w /Meta5 14 0 R (13) Tj >> 0 G /FormType 1 endstream q /Meta225 Do 1 i 0 G 0 g Q /Matrix [1 0 0 1 0 0] /Subtype /Form /F3 12.131 Tf Q /Matrix [1 0 0 1 0 0] 0.564 G >> >> /ProcSet[/PDF/Text] /CapHeight 476 Q /ProcSet[/PDF/Text] (D) Tj endstream q endobj 0 G 1 i q q /BBox [0 0 88.214 16.44] >> 94.364 5.203 TD 1 i q 0 5.203 TD /Subtype /Form /Resources<< << >> Q /F1 7 0 R stream /FormType 1 Q /Length 67 ET q /ProcSet[/PDF] 0.564 G 549.694 0 0 16.469 0 -0.0283 cm /Filter [/CCITTFaxDecode] ([x ) Tj 0 G 0 g 1 i /F3 12.131 Tf >> 1 i q >> (-9) Tj /FormType 1 endobj Q /Flags 32 /F1 7 0 R /F3 12.131 Tf /Meta303 317 0 R 0 G >> 385 0 obj >> Q /Resources<< /Meta405 421 0 R /Meta20 31 0 R 6.746 24.649 TD /Meta22 33 0 R >> /Meta190 Do /Meta53 Do >> /Subtype /Form endstream /BBox [0 0 88.214 16.44] BT ET 119 0 obj 1 i Q /Length 58 >> stream /Type /XObject 1 i /FormType 1 57 0 obj 0 G stream Q >> the other number. 0 G q /Length 16 endstream Q 1 Data in this Fast Fact represent the 50 states and the District of Columbia. /Type /XObject 22 0 obj /BBox [0 0 30.642 16.44] q q (\)]) Tj q << 0.564 G /Resources<< >> << /Subtype /Form 0 g /F1 12.131 Tf /Type /XObject >> Q /Meta260 274 0 R Kobe scored 85 points in a basketball game. Q (x) Tj q (9) Tj /Matrix [1 0 0 1 0 0] /Font << /Subtype /Form >> 1.014 0 0 1.007 251.439 583.429 cm /Meta24 37 0 R ET /Meta7 Do << 1 i /BBox [0 0 534.67 16.44] q >> -0.029 Tw endstream 336 0 obj endstream 1 i /Resources<< << << >> >> 0 G /Meta322 336 0 R /Resources<< q /ProcSet[/PDF] q >> /BBox [0 0 88.214 35.886] /Font << >> 0 20.154 m endstream 1.014 0 0 1.007 251.439 776.149 cm /Resources<< 188 0 obj Q /Meta219 Do Q 1 i /Length 69 1.007 0 0 1.006 130.989 437.384 cm 1 i 1 i Q /Meta246 Do /Resources<< >> /FormType 1 >> 391 0 obj 1 g 1.007 0 0 1.007 411.035 277.035 cm 104 0 obj q /Matrix [1 0 0 1 0 0] >> Q Q stream 1.007 0 0 1.006 130.989 437.384 cm /Meta49 63 0 R q /Font << 0 g 1.007 0 0 1.007 130.989 330.484 cm >> 0 G 0 g << q (D\)) Tj q /FormType 1 >> 0.458 0 0 RG /Meta342 Do /Type /XObject 1.007 0 0 1.007 45.168 862.723 cm q endobj 1 i /FormType 1 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 << endstream /Matrix [1 0 0 1 0 0] 0 g Q 1 i 0 G Q 0.738 Tc /Font << q Q S /F3 17 0 R 3.742 5.203 TD >> /F3 12.131 Tf /Meta162 176 0 R Q >> Q /Font << /Type /XObject >> 0 G >> >> endstream /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /F3 17 0 R >> /Length 69 44 0 obj 162 0 obj >> ( \() Tj 252 0 obj /Type /XObject >> 94 0 obj /Matrix [1 0 0 1 0 0] Q /Font << /BBox [0 0 88.214 16.44] /Meta327 Do stream /ProcSet[/PDF/Text] the quotient of five and a number 7.) /FormType 1 /Meta63 77 0 R q /Matrix [1 0 0 1 0 0] Q 1 i /Length 69 1 i BT Q /Meta282 Do /Resources<< /F3 12.131 Tf /ProcSet[/PDF/Text] /Type /XObject 1.007 0 0 1.007 271.012 330.484 cm << /FormType 1 0 g /ProcSet[/PDF] 0 g Q /ProcSet[/PDF/Text] Two fewer than a number doubled is the same as the number decreased by 38. stream stream /Font << stream endstream BT /Type /XObject endstream Q /FormType 1 6.746 5.203 TD 1 i /Type /XObject 1 i /Matrix [1 0 0 1 0 0] Q endstream q /ProcSet[/PDF/Text] q 1.005 0 0 1.007 79.798 730.228 cm /Meta366 380 0 R /Subtype /Form 1.007 0 0 1.007 551.058 636.879 cm /Meta120 134 0 R Question. /Subtype /Form >> 1.014 0 0 1.007 111.416 383.934 cm /ProcSet[/PDF/Text] /Length 59 Q << >> 0.564 G /FormType 1 << q /BBox [0 0 88.214 16.44] >> Q 1 i 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . 0.227 Tc << /Length 244 Q 2. >> 0 G 0 g 114 0 obj q << q endstream /Meta184 198 0 R >> ET /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] /Subtype /Form /Resources<< Q << 105 0 obj /Meta106 Do Q /Matrix [1 0 0 1 0 0] BT 0.838 Tc /Leading 349 /BBox [0 0 88.214 16.44] ET ET 0 g Q /Resources<< stream /Resources<< /Resources<< /F3 12.131 Tf -0.382 Tw 1.007 0 0 1.007 130.989 523.204 cm 0.175 Tc << q 0 5.336 TD stream That was 1/8 of the points that he scored 361 0 obj /Length 69 >> stream stream /FormType 1 0.564 G >> 6.746 5.203 TD /FormType 1 Q q >> (x) Tj /FormType 1 0 g endobj /Length 59 /Meta333 Do Q 333 0 obj 1 i Q /Resources<< Q 0 g endobj >> /F3 17 0 R /F3 12.131 Tf /Type /XObject ET BT /Meta299 Do >> /Type /XObject /Subtype /Form >> /Type /XObject 1st step. /Meta42 56 0 R 145 0 obj /Font << /Resources<< (5\)) Tj Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. /BBox [0 0 639.552 16.44] /Resources<< /Type /XObject /Resources<< /ProcSet[/PDF/Text] Q Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] ET /Font << /Resources<< ( \() Tj /Length 16 /Subtype /Form /BBox [0 0 30.642 16.44] /Resources<< /Subtype /Form q 1.005 0 0 1.007 102.382 799.486 cm the quotient of twenty and a number a.) 0 G 0 20.154 m 20.21 5.203 TD /BBox [0 0 88.214 16.44] 13.493 5.336 TD /F4 36 0 R Q /BBox [0 0 15.59 16.44] /Length 59 /Resources<< Making educational experiences better for everyone. q Q 1 i 0.486 Tc /Matrix [1 0 0 1 0 0] /Subtype /Form endstream q (iv) A number exceeds 5 by 3. -0.463 Tw stream Q endobj q q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] (13) Tj /Type /XObject ET << >> >> q 69 0 obj endobj /Matrix [1 0 0 1 0 0] Let the 2nd number be y. /BBox [0 0 88.214 16.44] /Length 69 stream /Meta341 Do 1 i 431 0 obj 0.737 w /Meta394 Do /BBox [0 0 88.214 16.44] /Type /XObject q q endobj /Subtype /Form q /Subtype /Form /Resources<< 142 0 obj /Subtype /Form 355 0 obj 0 G /Subtype /Form /LastChar 45 endstream 20.21 5.203 TD q /Subtype /Form q Q /ProcSet[/PDF] << /Font << /Resources<< 1.005 0 0 1.007 102.382 293.596 cm /ProcSet[/PDF/Text] >> 1.007 0 0 1.007 271.012 703.126 cm 200 0 obj Q /ProcSet[/PDF] /Font << q /Length 118 1 i ET stream /ProcSet[/PDF/Text] 0.458 0 0 RG /Meta131 145 0 R 0.737 w stream q /F3 17 0 R stream q q -0.486 Tw 0 G >> /F3 17 0 R /Subtype /Form Q 0 5.203 TD 1.014 0 0 1.007 251.439 636.879 cm 1 i endobj q /Length 54 Q 1 i /Resources<< >> << 0.458 0 0 RG 167 0 obj /Type /XObject Q 1.007 0 0 1.006 411.035 437.384 cm Q 1 g 21.713 20.154 l /Type /XObject /Resources<< /Resources<< /Meta166 Do /BBox [0 0 534.67 16.44] /Type /XObject for the season. /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 111.416 277.035 cm 1 i /ProcSet[/PDF/Text] (3) Tj q /Subtype /Form ET BT >> q Q 441 0 obj 0 g /Font << q Q q 722.699 347.046 l Q q 1.014 0 0 1.007 251.439 636.879 cm Q Q 1.005 0 0 1.015 45.168 53.449 cm q Q >> q Q ET 0.737 w (B\)) Tj 50 0 obj 1 i 1.007 0 0 1.007 130.989 277.035 cm 1 g /Length 54 /Type /XObject 0.737 w /F3 12.131 Tf q /Descent -299 BT 1.005 0 0 1.007 102.382 653.441 cm Q /Length 16 /Length 16 q 1 i >> 1.007 0 0 1.007 654.946 799.486 cm endobj 1 i Q q stream 1 g /ProcSet[/PDF/Text] /BBox [0 0 15.59 29.168] 1 i 0.68 Tc 282 0 obj /Meta390 406 0 R /FormType 1 0 G endobj /Meta79 93 0 R ET 1 i 0 g Q stream Q /Subtype /Form Q 3.742 5.203 TD /Subtype /Form 1.007 0 0 1.007 411.035 636.879 cm << /Resources<< q /Matrix [1 0 0 1 0 0] 1 i /FontBBox [-170 -292 1419 1050] >> q /ProcSet[/PDF/Text] /Length 16 >> >> >> 1 i ET /Subtype /Form Q /Length 59 0 5.203 TD ET /Resources<< Q q /Meta220 234 0 R /Resources<< /Meta102 Do /Type /XObject /Length 70 endstream 0 g << q /Resources<< /FormType 1 Q q >> << 0 g /Meta156 Do 1 g 1.007 0 0 1.007 551.058 636.879 cm Q /Type /XObject 285 0 obj Q /Meta57 Do 0 G /Length 69 238 0 obj Q 0 g 27 0 obj /BBox [0 0 88.214 16.44] (x ) Tj /Meta332 346 0 R >> Q /Subtype /Form /Font << q /Subtype /Form 0 G /Length 69 q 13.464 5.203 TD q Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 776.149 cm endobj 0 g 1 i endobj q /Meta390 Do q stream /Resources<< endstream ET /BBox [0 0 88.214 16.44] ET >> Q 192 0 obj Q 1 i Q Q q 1.014 0 0 1.007 111.416 583.429 cm /Resources<< /FormType 1 /Meta42 Do endobj /ProcSet[/PDF/Text] Q 0.297 Tc Q [(1)-25(0\))] TJ /Type /XObject stream /BBox [0 0 15.59 29.168] 0.564 G /ProcSet[/PDF] /Type /XObject q Q >> /Resources<< Q /MediaBox [0 0 767.868 993.712] 0 g 672.261 599.991 m Q /Resources<< /ProcSet[/PDF] << /Meta325 339 0 R /F3 17 0 R q /F4 12.131 Tf /Resources<< [(1)-25(0\))] TJ Twice a number when decreased by 7 gives 45. /BBox [0 0 88.214 16.44] endstream Q 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . 0.564 G /Length 69 << Use the variable g to represent Gails age. endstream BT >> 0.458 0 0 RG 0 G stream endobj /FontDescriptor 10 0 R stream 0 w q /Meta85 Do << /FormType 1 /Meta359 Do /Meta335 Do /Resources<< /Length 79 0 g 1.007 0 0 1.006 411.035 437.384 cm 1 i BT stream 0 G /Length 70 /F3 12.131 Tf >> >> /F4 12.131 Tf Q /Subtype /Form q 1 g Q 1.005 0 0 1.007 79.798 746.789 cm Q q >> 100 0 obj stream >> q Q Q endstream stream >> (3\)) Tj /Meta251 265 0 R 1.007 0 0 1.007 411.035 849.172 cm 346 0 obj >> 1 i /Resources<< 156 0 obj /Resources<< q endstream S q q ET /Matrix [1 0 0 1 0 0] /Resources<< 0.737 w 1 i endobj q /Meta387 Do >> /Matrix [1 0 0 1 0 0] /Meta413 Do /F3 17 0 R >> /Type /XObject endobj /BBox [0 0 88.214 35.886] /FormType 1 endobj Percent Change = (Decrease First Value) x 100% q >> Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. BT endobj /Meta213 Do 0 g 1 g (B\)) Tj q Q 1 i q /Meta44 58 0 R 0 g 417 0 obj endstream q q /ProcSet[/PDF/Text] endobj q q /F4 36 0 R 1 g /Matrix [1 0 0 1 0 0] /F3 17 0 R endstream BT ET Get a free answer to a quick problem. Q /BBox [0 0 673.937 68.796] /FormType 1 q Q Q /BBox [0 0 15.59 16.44] /Meta91 Do >> Q /FormType 1 << /Resources<< /Resources<< stream Q 0 G Q q 1 i /FormType 1 endobj /Resources<< Q endstream /F3 17 0 R endstream By the . /Meta50 Do /FormType 1 (-20) Tj >> 0 g >> 0 G /Resources<< endobj /Meta422 438 0 R /F3 12.131 Tf /Meta238 Do &K @ 237 0 obj /Resources<< Q Q 1.005 0 0 1.007 102.382 670.003 cm Q endstream q 325 0 obj Q /Length 16 Q << ET q Q Q q /FormType 1 endobj Q /Matrix [1 0 0 1 0 0] /Font << /Meta15 Do endstream Expert Answer. 0 G /Meta223 237 0 R q >> /FormType 1 0 0 Similar questions Find the number which when decreased by 8% becomes 506. 1 i q /F3 17 0 R 0 20.154 m /Matrix [1 0 0 1 0 0] /Meta178 Do endobj /Type /XObject 13.464 5.203 TD >> /Length 16 /XObject << /Subtype /Form q endstream 66 0 obj stream >> << /F3 17 0 R 136 0 obj /F4 36 0 R Q >> q /ProcSet[/PDF] 0 g /Resources<< 0.564 G /Meta426 Do BT Q 1.014 0 0 1.006 111.416 437.384 cm /FormType 1 /Matrix [1 0 0 1 0 0] /Font << Q endstream q /Resources<< /Length 69 << /Length 68 Q Q Q 0.296 Tc 20.21 5.203 TD /Meta114 128 0 R 1 i /Length 69 0.564 G Q Q q << /Subtype /Form 0.001 Tw >> 0 g /F3 17 0 R q /FormType 1 q endstream /Matrix [1 0 0 1 0 0] stream /ProcSet[/PDF/Text] /Meta310 Do /Length 58 q >> << Q /Resources<< /F3 12.131 Tf /Type /XObject << Q endobj /FormType 1 endstream Q >> 0 G endobj /Resources<< << Q /Subtype /Form q >> endstream /Meta330 Do 0 w In addition, testosterone in both sexes is involved in health and well-being . /Resources<< 214 0 obj /F3 17 0 R q /FormType 1 endstream endstream 0.737 w /FormType 1 q q /Meta320 Do >> Q Q /Length 16 /Resources<< /Length 118 /Font << 0 w >> /Resources<< /BBox [0 0 15.59 16.44] (5\)) Tj (+) Tj endobj Q /Resources<< /Matrix [1 0 0 1 0 0] 10.487 5.203 TD /FormType 1 Q 0 g >> /ProcSet[/PDF] << endobj stream endstream /Resources<< /Resources<< >> /Length 69 /Type /XObject q q /Type /XObject ET /Length 69 /BBox [0 0 88.214 35.886] endobj /Type /XObject >> << 202 0 obj 1.007 0 0 1.007 130.989 383.934 cm /ProcSet[/PDF/Text] /Resources<< /F4 12.131 Tf 0.564 G /Font << 0.175 Tc /BBox [0 0 15.59 16.44] endobj /F3 17 0 R q q Q >> 1 i /Meta131 Do /Length 69 q BT 38.948 5.203 TD >> /F3 17 0 R endstream /Subtype /Form 0.838 Tc 258 0 obj /FormType 1 /Resources<< >> /FormType 1 /Resources<< 0.458 0 0 RG >> 1.014 0 0 1.007 531.485 450.181 cm >> /FormType 1 q /ProcSet[/PDF] /Type /XObject stream (5) Tj /Matrix [1 0 0 1 0 0] stream /Matrix [1 0 0 1 0 0] /Type /XObject 0.786 Tc 1 i q Q 0 g /FormType 1 q 0.738 Tc Q 1.007 0 0 1.007 130.989 583.429 cm 177 0 obj Q 0 g Q /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 111.416 583.429 cm /Meta48 Do q q 0 G >> /BBox [0 0 88.214 16.44] /Resources<< endobj 0 g /Subtype /Form stream /Type /XObject q /FormType 1 0.458 0 0 RG 1 i /BBox [0 0 88.214 16.44] q q >> /FormType 1 endobj /Meta358 Do /FormType 1 /FormType 1 297 0 obj << stream 2.238 5.203 TD ET /Meta14 Do >> /Length 16 /Resources<< /FormType 1 << Q /Kids [ /Resources<< >> q /Meta159 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 ET 1 i /Font << (3\)) Tj q >> Q stream /F3 12.131 Tf stream Expert Solution. /Meta124 138 0 R 1 g endobj << /FormType 1 0 5.203 TD /Type /XObject 672.261 473.519 m 18.708 17.593 TD stream q 672.261 726.464 m /F1 7 0 R >> /Length 69 0.737 w /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 67.753 347.046 cm [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ endobj q 0.332 Tc /BBox [0 0 673.937 16.44] /Meta38 52 0 R << 0 w /FormType 1 /Matrix [1 0 0 1 0 0] >> /Subtype /Form Q BT BT Q endstream /Subtype /Form endobj q /Font << >> 1 g stream /Matrix [1 0 0 1 0 0] >> q /ItalicAngle 0 /Resources<< /Font << 0 g stream /BBox [0 0 549.552 16.44] /Resources<< >> /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 636.879 cm This s problem could be, interpreted either way. ET endobj /F3 12.131 Tf << /Font << >> 1.005 0 0 1.007 79.798 796.475 cm 1.007 0 0 1.007 551.058 583.429 cm /Matrix [1 0 0 1 0 0] ET 1.007 0 0 1.007 67.753 653.441 cm << Q BT /Meta352 366 0 R 278 0 obj 549.694 0 0 16.469 0 -0.0283 cm 1.007 0 0 1.007 551.058 330.484 cm /BBox [0 0 30.642 16.44] 25 0 obj endobj /Subtype /Form 1.007 0 0 1.007 411.035 277.035 cm 383 0 obj /Subtype /Form endobj /F3 12.131 Tf /Subtype /Form /BBox [0 0 88.214 16.44] (x ) Tj BT Q 1 i q Q 1.014 0 0 1.007 391.462 450.181 cm 0.786 Tc >> stream /Meta368 382 0 R endobj >> /Meta14 25 0 R >> /Type /XObject /F3 12.131 Tf /ProcSet[/PDF] /Type /XObject 0 G endstream q stream 0 g q Q 0 g q /Font << >> /Matrix [1 0 0 1 0 0] /Type /XObject q /BBox [0 0 88.214 16.44] 0 w Q q 0.458 0 0 RG endstream Q /ProcSet[/PDF] 0 g << ET Q 1 i /ProcSet[/PDF] /Meta116 130 0 R Q BT /F4 12.131 Tf /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] Q q /FormType 1 0.68 Tc Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM Q 1 i 0 g BT 178 0 obj Q Q q stream >> /Matrix [1 0 0 1 0 0] Q >> 19.474 5.203 TD /Subtype /Form /BBox [0 0 534.67 16.44] 0 G /FormType 1 Q /Meta276 290 0 R q Q /Length 59 /ProcSet[/PDF] /Length 16 endstream q /Meta80 Do Q >> q q endstream >> Twice a number decreased by 8 gives 58. q /Type /XObject /FormType 1 52.412 5.203 TD /Meta419 Do 0.738 Tc 1 i 1 i << /F4 36 0 R stream /Meta252 266 0 R endstream 10 0 obj /Subtype /Form 0.737 w /ProcSet[/PDF/Text] Q % q stream q /BBox [0 0 88.214 16.44] q 0 G Q /Resources<< Q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 0.458 0 0 RG Q 14.966 20.154 l >> 97 0 obj 0 g q ET q >> /Type /XObject >> /Type /XObject /Length 69 /FormType 1 >> a.) /Meta244 Do -0.041 Tw 1.014 0 0 1.006 531.485 763.351 cm /BBox [0 0 673.937 16.44] /F3 12.131 Tf >> >> 0 g q 1 i /Meta268 282 0 R 0 g 1.007 0 0 1.007 45.168 846.161 cm q ET /BBox [0 0 88.214 16.44] /ProcSet[/PDF] /Resources<< 1 i >> /BBox [0 0 88.214 16.44] endobj q /Meta242 256 0 R /FormType 1 /BBox [0 0 15.59 16.44] >> q endstream /FormType 1 1 i /Type /XObject /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] << q >> << Q q /Meta39 53 0 R >> /Meta61 Do /Type /XObject /Length 69 1 g /FormType 1 q >> q q 0.737 w q q /Subtype /Form >> Q /F3 12.131 Tf >> endobj stream /Subtype /Form endobj 1 i /CapHeight 662 [( t)-14(imes a num)-16(ber)] TJ q 0.737 w stream q /Matrix [1 0 0 1 0 0] endstream (+) Tj q /Meta334 Do >> /F3 17 0 R << Q /FormType 1 /FormType 1 199 0 obj q q /F4 36 0 R >> stream >> /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 111.416 776.149 cm /Length 74 /Font << /MissingWidth 250 0 G Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . /Meta393 Do >> Q Q /F3 12.131 Tf 0 w /ProcSet[/PDF] 0 g /Meta168 Do 1.014 0 0 1.007 111.416 277.035 cm Q ET endobj 89.12 5.203 TD Q endobj endobj /Meta271 285 0 R /ProcSet[/PDF] 0 w /Matrix [1 0 0 1 0 0] q /Length 58 All steps. /Font << q << 0 G >> q /Subtype /Form /Length 12 13 0 obj q 0.68 Tc q /Font << /Meta283 Do /Subtype /Form BT q 1 i /ProcSet[/PDF] /ProcSet[/PDF/Text] /Length 59 Q >> q 1 i endobj /Subtype /Form q /ProcSet[/PDF/Text] /Type /XObject /Resources<< Q /Resources<< /Meta389 405 0 R ET (1\)) Tj >> q << 49 0 obj /F3 12.131 Tf /Font << << /Type /XObject /Type /XObject 0.458 0 0 RG 1 i /Type /XObject endstream ET 147 0 obj /FormType 1 /BBox [0 0 30.642 16.44] endstream BT >> Q q BT /Type /XObject 1.007 0 0 1.007 271.012 583.429 cm /BBox [0 0 534.67 16.44] /Subtype /Form 0 G 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 1.005 0 0 1.007 45.168 916.925 cm 13.493 5.336 TD 0.369 Tc /F3 12.131 Tf q >> q endstream Q ET (x ) Tj ET ET stream 0 G /F3 12.131 Tf Q /Length 127 0 G Q /Meta204 218 0 R /Meta221 Do 0.68 Tc q /ProcSet[/PDF/Text] endstream /FormType 1 >> 1.007 0 0 1.007 551.058 383.934 cm /Subtype /Form q 22.478 5.203 TD >> 0 w Q /Encoding /WinAnsiEncoding /Length 59 0.458 0 0 RG stream 0.564 G 1 i >> 1 i 1 i 0 g 0 w 0.458 0 0 RG /ProcSet[/PDF/Text] 375 0 obj >> /Meta217 Do /Length 69 q >> q q /Length 60 endobj Q << BT Q stream BT << endstream q 22.478 4.894 TD >> /ProcSet[/PDF/Text] q /FormType 1 /Font << >> /Meta157 171 0 R /FormType 1 q 0 g 6.746 5.203 TD ET /F1 7 0 R endobj << /Font << 0 g 0 g ET >> 6.746 5.203 TD /FormType 1 /Subtype /Form 0 g /Meta80 94 0 R 0 g /Type /XObject /BBox [0 0 88.214 16.44] q /ProcSet[/PDF/Text] endobj 275 0 obj >> /ProcSet[/PDF] >> << /F1 12.131 Tf endobj Q /Font << stream /FormType 1 /Meta171 185 0 R >> /Meta315 Do Q >> 26.957 5.203 TD /Resources<< q q endstream endobj >> /BBox [0 0 88.214 16.44] 0 g 1.005 0 0 1.015 45.168 53.449 cm q q Q The rate of positive findings after 1 round of screening in the LCSDP was more than twice . >> BT stream 0 g /Resources<< 0 G /ProcSet[/PDF/Text] If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. /Meta153 Do q << /Font << /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /Type /XObject >> Answer provided by our tutors. Q endobj Q 0 g /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] BT q /Font << >> 262 0 obj /Type /XObject >> /Font << 1.005 0 0 1.007 102.382 816.048 cm /FormType 1 BT /F4 36 0 R 1.014 0 0 1.007 531.485 277.035 cm >> 31 0 obj q /Type /XObject q 722.699 400.496 l endstream 1.005 0 0 1.007 102.382 799.486 cm /Type /XObject /F3 12.131 Tf q >> /Meta384 398 0 R endobj /FormType 1 102 0 obj /F3 12.131 Tf /LastChar 121 q /Font << endobj /Length 16 q /Length 69 Q >> 0 5.203 TD 0.564 G stream /Resources<< /FormType 1 /Length 16 Q (vii) Twice a number subtracted from 19 is 11. /BBox [0 0 88.214 35.886] /ProcSet[/PDF] /ProcSet[/PDF/Text] << /ProcSet[/PDF] >> 0.68 Tc /Subtype /Form /Type /XObject /Font << 1.007 0 0 1.007 130.989 636.879 cm /FormType 1 >> /Length 69 /Matrix [1 0 0 1 0 0] /Subtype /Form q 0.271 Tc /FormType 1 /BBox [0 0 88.214 16.44] /F1 7 0 R Q /Matrix [1 0 0 1 0 0] /Meta117 Do stream (C\)) Tj 106 0 obj ET /Subtype /Form endstream >> if the solution of an equation is x=-2, what could the original equation be? /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Font << Q /Resources<< /Resources<< /Resources<< /ProcSet[/PDF/Text] Q /Matrix [1 0 0 1 0 0] /Subtype /Form [tex]\sin (\pi -x)=\sin x[/tex]. q ET >> 0 g 0 g 1.007 0 0 1.007 271.012 583.429 cm /Subtype /Form /F1 12.131 Tf Q BT (-9) Tj /Font << /F4 36 0 R q endobj 84 0 obj Q /Font << 0.564 G Twice a number when decreased by 7 gives 45. 0 G q /Font << /Font << /Meta366 Do /ProcSet[/PDF] 0.68 Tc 0 g Q endobj 0 g /Subtype /Form /F3 12.131 Tf 1.014 0 0 1.007 531.485 330.484 cm /F1 7 0 R /Length 16 0 g 1.005 0 0 1.006 45.168 879.284 cm q Twice a number decreased by ten is greater than 24. Q << BT (x) Tj >> q /Type /XObject q Q /Type /XObject /Meta351 Do 313 0 obj 12.727 5.203 TD /Resources<< (+) Tj /BBox [0 0 639.552 16.44] 1.502 8.18 TD q endstream 340 0 obj /Meta62 76 0 R endobj /ProcSet[/PDF/Text] /Meta222 236 0 R /ProcSet[/PDF/Text] q q >> BT q q ET 1.007 0 0 1.007 271.012 583.429 cm 1 i q << /Length 68 0 g endstream 1 i /F3 12.131 Tf q >> /Matrix [1 0 0 1 0 0] q /Meta46 Do /Length 16 0 g /Subtype /Form /ProcSet[/PDF/Text] /ProcSet[/PDF] endstream endstream /FormType 1 q /Meta287 Do /Font << /Resources<< >> /Type /XObject endstream 1.007 0 0 1.007 654.946 546.541 cm >> 722.699 726.464 l /Length 69 1 i /BBox [0 0 88.214 16.44] BT /ProcSet[/PDF/Text] Q /Type /XObject endstream /Meta346 360 0 R 283 0 obj Q /F3 12.131 Tf /Type /XObject endstream /F3 12.131 Tf 1.005 0 0 1.007 102.382 473.519 cm /Meta46 60 0 R q Q /Type /XObject Q /Type /XObject stream /BBox [0 0 15.59 29.168] q /Meta262 276 0 R /Length 63 /Type /XObject q << /ProcSet[/PDF/Text] /F3 17 0 R /Subtype /Form << >> /F4 36 0 R 0 G /FormType 1 << /Meta267 Do 0 w << /Meta290 304 0 R Q q /Meta32 Do BT 0 g >> >> /Meta254 Do /Font << Q /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 endstream /ProcSet[/PDF/Text] /Subtype /Form /Matrix [1 0 0 1 0 0] Q >> Q /Meta270 284 0 R /Meta135 149 0 R /ProcSet[/PDF/Text] Q /Font << 1 i /F3 17 0 R /Length 69 /F3 12.131 Tf Q stream /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Subtype /Form 1 i Q stream >> endstream >> /Length 67 0.369 Tc >> << /Meta380 394 0 R 1.005 0 0 1.007 102.382 743.025 cm 0.458 0 0 RG Q 0 G 390 0 obj q /Subtype /Form (C) Tj ET Q /Meta365 379 0 R /Matrix [1 0 0 1 0 0] /Type /XObject >> << /BBox [0 0 15.59 16.44] /Type /XObject /F3 17 0 R endstream /Matrix [1 0 0 1 0 0] 0 w Q >> >> q /FormType 1 /Matrix [1 0 0 1 0 0] /FormType 1 Q << >> endstream Q A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. 1.007 0 0 1.007 654.946 546.541 cm BT 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject /F3 12.131 Tf >> BT 0 g /StemV 77 1 i 32 0 obj Q 0 g /ProcSet[/PDF/Text] BT stream /F3 17 0 R /Matrix [1 0 0 1 0 0] /Meta79 Do 1.007 0 0 1.007 411.035 383.934 cm /BBox [0 0 88.214 16.44] 0.458 0 0 RG [(1.1)21(2 Tran)36(sla)18(tin)23(g Alge)17(b)26(raic )18(Exp)22(res)24(si)25(on 2)] TJ /Type /Pages 0 G q /F3 12.131 Tf >> 1 i BT << /Meta41 55 0 R q Solution. /Matrix [1 0 0 1 0 0] q /BBox [0 0 534.67 16.44] /Subtype /Form endstream /Resources<< Q /Resources<< q q /BBox [0 0 30.642 16.44] 1 i /Matrix [1 0 0 1 0 0] stream << stream 0.524 Tc Q >> /Meta336 350 0 R 1 i /Meta37 50 0 R /F3 12.131 Tf (C) Tj q endobj /F3 17 0 R 220.931 4.894 TD 0 g /Length 13 /Meta296 310 0 R /Type /XObject >> /Matrix [1 0 0 1 0 0] /Meta416 Do /Meta298 312 0 R /Font << Q >> /Meta412 428 0 R Q /Length 65 >> 354 0 obj << Q endstream View the full answer. /Meta51 Do endstream /ProcSet[/PDF] >> q /Font << q /Subtype /Form 1.502 5.203 TD >> /Meta16 27 0 R q >> /Matrix [1 0 0 1 0 0] 0 g 20.21 5.203 TD /BaseFont /PalatinoLinotype-Roman /Matrix [1 0 0 1 0 0] /Subtype /Form 2.238 5.203 TD endobj (3) Tj Q 384 0 obj /Matrix [1 0 0 1 0 0] /Meta421 Do >> /Type /XObject /Subtype /Form Q 0 G 11.99 24.649 TD 140781 /F3 12.131 Tf /BBox [0 0 88.214 16.44] /ProcSet[/PDF] stream /ProcSet[/PDF] /Meta285 Do 0 G Q (A\)) Tj /Meta117 131 0 R /Resources<< 0 G << >> << /Subtype /Form endobj /Length 12 endstream /Meta239 253 0 R /FormType 1 >> >> Q q Q 321 0 obj /Subtype /Form /ItalicAngle 0 /Resources<< q /Type /XObject 126 0 obj /Meta164 178 0 R 1 g BT stream /Meta182 Do >> endstream (x) Tj endstream 0.737 w 0 w /Type /XObject 1 i q 47 0 obj q [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. /F3 12.131 Tf /Meta199 Do 175 0 obj q /Length 54 /Resources<< endobj /Type /XObject Q 1.007 0 0 1.007 271.012 277.035 cm endstream q q stream /F3 17 0 R /F3 12.131 Tf Q 1 g /FormType 1 /Meta74 88 0 R 1 g /Font << 0 G /Subtype /Form /FormType 1 /FormType 1 /Meta143 157 0 R q Q 1 i /FormType 1 0 G endobj << /Subtype /Form 1.007 0 0 1.007 67.753 726.464 cm 6.746 5.203 TD endstream >> /Meta165 Do >> /Meta299 313 0 R Q BT /Type /XObject /Resources<< /BBox [0 0 15.59 29.168] /Type /XObject q 1 i 1.007 0 0 1.007 411.035 330.484 cm endobj endobj /Subtype /Form endobj Q 0.458 0 0 RG /Subtype /Form endobj 309 0 obj 239 0 obj Q Q q /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 15.59 16.44] 1 i /Meta285 299 0 R endobj /Length 12 q /BBox [0 0 15.59 16.44] 374 0 obj /F3 17 0 R 0 g q BT endobj >> 430 0 obj 0 g /ProcSet[/PDF/Text] /Font << /F4 36 0 R /Meta97 111 0 R /BBox [0 0 15.59 16.44] /Resources<< much as how 8, Last . endobj endstream /Resources<< /Resources<< q /Resources<< BT >> 444 0 obj -0.084 Tw Q 57.656 5.203 TD Q stream 0.564 G /ProcSet[/PDF/Text] >> /Length 69 /Subtype /Form 0 G 1 i >> /Length 68 endobj 0.564 G /F3 12.131 Tf /F1 7 0 R /ProcSet[/PDF] q /FormType 1 Q /Meta289 303 0 R BT /Subtype /Form endstream /Meta225 239 0 R q 20.21 5.203 TD >> /F1 12.131 Tf q /Meta122 136 0 R 0 g 0.564 G /Length 59 Q 0.271 Tc Q q 0 g stream Q << >> /Meta310 324 0 R /Subtype /Form BT q Q << /XHeight 477 /BBox [0 0 17.177 16.44] 0 w /Meta119 133 0 R >> 291 0 obj stream Q >> Q endstream Expression. /Subtype /Form 1 i Q q /ProcSet[/PDF] Q 1.007 0 0 1.006 551.058 836.374 cm 0.564 G >> /Font << /Matrix [1 0 0 1 0 0] 0.369 Tc ET /Font << /Subtype /Form 1 i Q q >> Formula - How to Calculate Percentage Decrease. 339 0 obj q /BBox [0 0 15.59 16.44] /F1 7 0 R /Resources<< 0 w /Type /XObject << /FormType 1 Q /FormType 1 219 0 obj q 0.737 w Q /Type /XObject 1 i (x ) Tj /Meta156 170 0 R 1 g 1 i Q q endobj 333.269 5.488 TD (5) Tj 1.007 0 0 1.007 271.012 776.149 cm /Matrix [1 0 0 1 0 0] endobj /Meta23 34 0 R >> /Font << q /Resources<< /Meta0 Do BT << >> /F3 17 0 R endstream 382 0 obj 1 i Q endobj << 0.311 Tc /Type /XObject /ProcSet[/PDF/Text] /ProcSet[/PDF] Calculate a 15% decrease from any number. Q Q 0 G >> q Q endobj (D) Tj stream BT /Matrix [1 0 0 1 0 0] /Font << endstream /F3 12.131 Tf /Subtype /Form Q Q Q 1 i << 405 0 obj /Type /XObject 1 i 0.737 w /Meta363 Do >> q /F3 17 0 R Q /Subtype /Form /Meta148 Do ET 418 0 obj /Subtype /Form /Matrix [1 0 0 1 0 0] q 0.68 Tc 185.725 5.203 TD q 0 5.203 TD /ProcSet[/PDF/Text] q q 293 0 obj q 0 G << stream /Type /XObject /Length 69 q endobj ET /Type /XObject >> . q 0 g ET 22.478 5.336 TD q /Meta82 96 0 R 1 i >> >> endobj /Type /XObject 253 0 obj /Type /XObject 286 0 obj /ProcSet[/PDF] q find what is x and check it 3x+2/5 - 2x-1/6 = 2/15 pa help po need ko lng po True or False: Let a & b be any elements of S, the set of natural numbers. Q Q << /Subtype /Form endstream /FormType 1 0.458 0 0 RG Q q stream endobj /Meta55 69 0 R /Meta145 159 0 R /Subtype /Form 154 0 obj Q 0 G >> /ProcSet[/PDF/Text] << /Subtype /Form -0.092 Tw Q >> /F1 7 0 R Q Q /Meta60 Do /Type /XObject 0 g Q 22.478 5.336 TD q >> q q /ProcSet[/PDF] (2) Tj /F4 12.131 Tf stream (38) Tj >> /F3 17 0 R /F3 17 0 R stream stream ET 1.007 0 0 1.007 271.012 450.181 cm /Type /XObject q /Resources<<

How To Mount Cross Stitch On Foam Board, Life Insurance Policy Search By Social Security Number Texas, Eden Funeral Home Obituaries, Wi Dnr Fishing Regulations 2022, Articles T