twice a number decreased by 58

/Meta360 374 0 R 106 0 obj /Meta52 Do /Resources<< /F4 36 0 R >> Q >> Q /Subtype /Form q 0 G BT >> 74 0 obj 0 g >> ET /Matrix [1 0 0 1 0 0] q Q /Resources<< stream endobj /Matrix [1 0 0 1 0 0] /Meta38 52 0 R /F1 7 0 R 0.737 w q /F3 17 0 R 1 i /BBox [0 0 88.214 16.44] Q q q 0 g endstream q 0 4.894 TD /ProcSet[/PDF] /FormType 1 /Subtype /Form /BBox [0 0 549.552 16.44] /BBox [0 0 88.214 16.44] endobj q << There were x cookies at the beginning of a party. >> q /Length 16 /F3 17 0 R endobj /Font << /F3 12.131 Tf 0 g /F3 17 0 R 1.007 0 0 1.007 271.012 583.429 cm /Subtype /Form BT /Type /XObject /Meta118 Do >> /Resources<< /F3 12.131 Tf /ProcSet[/PDF] /BBox [0 0 15.59 29.168] /Subtype /Form Q 84 0 obj (B\)) Tj /Meta189 203 0 R endobj A rectangular garden has a width that is 8 feet less than twice the length. /Flags 32 Q endstream 1 i stream /BBox [0 0 88.214 16.44] /F3 12.131 Tf Q /Meta159 173 0 R /F3 12.131 Tf 352 0 obj q ET /Matrix [1 0 0 1 0 0] 0 g /F3 12.131 Tf endstream Q endobj /Meta92 Do Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. /FormType 1 47.933 5.203 TD /BBox [0 0 639.552 16.44] 400 0 R /FormType 1 Q BT /BBox [0 0 15.59 16.44] /F1 12.131 Tf stream /Meta356 Do /Resources<< q 95 0 obj 0.737 w 0.458 0 0 RG 30.699 5.203 TD Q 1 i /Meta6 15 0 R >> 0 g /Resources<< 0 g >> /Meta35 Do 1.007 0 0 1.006 411.035 763.351 cm 0 g /F3 17 0 R endstream q >> BT Q /ProcSet[/PDF/Text] BT 0.564 G /Type /XObject 0 g /ProcSet[/PDF/Text] endstream 1.007 0 0 1.007 551.058 330.484 cm /Length 59 /Length 74 /Meta370 Do q >> (2) Tj /Length 16 /Matrix [1 0 0 1 0 0] /XObject << /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 391.462 383.934 cm /Resources<< >> /Length 87 /Type /XObject Q 0 g 0 G q Explanation: let the number be n. then we can express division in 2 ways. >> Q >> 0.737 w /ProcSet[/PDF] /Count 2 /FormType 1 /F3 12.131 Tf << /Length 118 /ProcSet[/PDF] 1 i >> q /BBox [0 0 15.59 16.44] endstream /Matrix [1 0 0 1 0 0] This site is using cookies under cookie policy . /Subtype /Form /Font << /ProcSet[/PDF] >> Q 0.425 Tc 92 0 obj /Meta124 Do 395 0 obj 0 g /FormType 1 Q 0 G q /FormType 1 endstream D. Twice a number decreased by ten is less than 24. 0 g The quotient of a seven and a number 9. /ProcSet[/PDF] 0 g << /Length 54 stream Q 2.238 5.203 TD /FormType 1 stream stream << BT /Matrix [1 0 0 1 0 0] /FormType 1 /Type /XObject << 1 i /Length 88 << stream /FontName /TestGen-Regular 127 0 obj ET /Meta315 Do 125.064 4.894 TD >> 445 0 obj 0 G /Meta7 Do /FormType 1 Q /Resources<< q endstream /FormType 1 /Matrix [1 0 0 1 0 0] endstream 0.458 0 0 RG 0 g >> Q 333.269 5.488 TD 1.007 0 0 1.007 551.058 636.879 cm 1 i Q >> << endstream q 1 i 1.005 0 0 1.007 102.382 363.608 cm /Resources<< /FormType 1 1.014 0 0 1.007 391.462 450.181 cm q 1.007 0 0 1.007 654.946 400.496 cm /Length 16 endstream q 70 0 obj /Meta196 210 0 R << endobj q /Meta59 Do Q Q /Resources<< 1.014 0 0 1.007 251.439 277.035 cm Q Q Q /Meta240 Do /BBox [0 0 534.67 16.44] /Subtype /Form 0 G /Meta305 Do 0 w 1.005 0 0 1.007 102.382 872.509 cm /Type /XObject Q endobj /FormType 1 /Length 69 << 1 g , Prove the following << /Meta341 Do /Meta361 Do /F3 17 0 R 246 0 obj q 0 g /Resources<< q /Meta30 Do /Resources<< 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 /Meta356 370 0 R 1 i q << 0 g /ProcSet[/PDF/Text] 0.786 Tc 1 i 1 i (+) Tj /ProcSet[/PDF] /BBox [0 0 88.214 16.44] >> 1 i Q 0 G /Length 79 >> /Type /XObject stream 144 0 obj 0 G >> q 0 G /Matrix [1 0 0 1 0 0] 1 i /Subtype /Form /Font << (D) Tj Q stream Q /Subtype /Form Q /Resources<< endobj 0.737 w 0 G Q endstream /FormType 1 /BBox [0 0 30.642 16.44] 0 g /Matrix [1 0 0 1 0 0] Q ET endstream /BBox [0 0 88.214 16.44] (x) Tj /Meta345 Do /Subtype /Form 0 G /Meta212 226 0 R 0 G Twice a number decreased by . 0 g 1 i /Font << stream BT Q 254 0 obj /Length 16 q >> Q /F1 7 0 R /F3 17 0 R /FormType 1 /F1 12.131 Tf /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] << q endobj /Length 59 q /Subtype /Form q /BBox [0 0 88.214 16.44] 0.486 Tc /Matrix [1 0 0 1 0 0] q (x) Tj 0 g q /F3 12.131 Tf BT q /ProcSet[/PDF/Text] q 0.458 0 0 RG 1.007 0 0 1.007 67.753 473.519 cm << Twice a number when decreased by 7 gives 45. /FormType 1 << /Font << 1 i q /Subtype /Form endstream 0 g ET 1.007 0 0 1.007 130.989 583.429 cm 0.738 Tc << /Resources<< 115 0 obj << Q /Matrix [1 0 0 1 0 0] five times the sum of a number x and two b.) Q /Resources<< endobj >> Q q 0 g endobj /Resources<< /ProcSet[/PDF] 170 0 obj /Type /XObject >> /BBox [0 0 15.59 16.44] /ProcSet[/PDF] 1 i endstream 1 i >> 0 g BT /Font << /Meta145 159 0 R << /Matrix [1 0 0 1 0 0] /FormType 1 q stream /Subtype /Form >> /Descent -299 endstream Q 0.564 G 0 g << /F2 11 0 R endobj Q /ProcSet[/PDF] q Q endobj /Matrix [1 0 0 1 0 0] ET /BBox [0 0 673.937 15.562] /Subtype /Form 0 g Q 1 i >> 186 0 obj /Resources<< /Type /XObject >> /Meta372 386 0 R /Font << /Matrix [1 0 0 1 0 0] ET q 14.23 24.649 TD >> q q /Meta44 Do q Q Q q /Meta137 Do endobj Q /Resources<< /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] 252 0 obj (40) Tj << /Meta274 288 0 R 1 i >> endstream 0.463 Tc 1.005 0 0 1.007 102.382 293.596 cm endstream ET Q /Meta393 Do /Resources<< q endobj << >> 0 G /Meta213 227 0 R /BBox [0 0 30.642 16.44] 1 i /Length 118 1.502 5.203 TD >> Q /BBox [0 0 88.214 16.44] /FormType 1 /FormType 1 1 i /Matrix [1 0 0 1 0 0] 173 0 obj 1.007 0 0 1.007 654.946 347.046 cm 0.564 G 382 0 obj Q >> endstream 0 0 Similar questions Find the number which when decreased by 8% becomes 506. /Meta279 293 0 R /Type /XObject 1 i /BBox [0 0 30.642 16.44] << /Type /XObject 0 w 0.425 Tc stream /ProcSet[/PDF] Q << /Type /XObject 1 i /Resources<< 1.007 0 0 1.007 271.012 636.879 cm /Resources<< endstream << stream /Font << /ProcSet[/PDF/Text] 0.737 w /Matrix [1 0 0 1 0 0] /Type /XObject (C) Tj 1 i /F4 12.131 Tf Find the number. stream /Resources<< >> /BBox [0 0 88.214 16.44] << /ProcSet[/PDF/Text] /Type /XObject /ProcSet[/PDF/Text] << /Meta313 Do /F3 17 0 R /Meta415 431 0 R >> /Font << >> /FontBBox [-568 -307 2000 1007] /Font << stream /Subtype /Form Q endobj /Type /XObject /F3 12.131 Tf /I0 Do q /FormType 1 /Subtype /Form endstream 0 G /Meta228 Do 0 g /Meta87 101 0 R 1.007 0 0 1.007 130.989 523.204 cm (B) Tj >> /Length 65 /Matrix [1 0 0 1 0 0] Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 Mathematics 619 solutions /Meta5 14 0 R >> /Matrix [1 0 0 1 0 0] 0.458 0 0 RG >> /FormType 1 Q >> /ProcSet[/PDF/Text] >> stream /Meta291 305 0 R >> /Flags 32 ET /Matrix [1 0 0 1 0 0] /Length 59 stream /Font << Q /F3 17 0 R (2\)) Tj Q Q ET q >> /Meta210 224 0 R q /F3 12.131 Tf [( and )16(a nu)26(mbe)18(r)] TJ /Resources<< Q /Meta409 425 0 R >> endobj 0 5.203 TD Patients' reasons for declining screening were not collected . /Meta238 252 0 R Answer provided by our tutors. /FormType 1 >> 349 0 obj 1 g Q q /Resources<< << 109 0 obj Just type into the box and your calculation will happen automatically. 351 0 obj /Type /XObject 1 i >> q /Resources<< 1.007 0 0 1.007 271.012 277.035 cm /Meta386 402 0 R >> Q /BBox [0 0 88.214 16.44] /F3 12.131 Tf endobj Q BT 2.238 5.203 TD BT 1 g /Length 16 Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio /FormType 1 Translate 2(x-58) into mathematical phrase. /Type /XObject << q 0 g /F1 12.131 Tf 0.737 w q /F3 17 0 R q Q >> q 0.737 w q /BBox [0 0 88.214 16.44] endstream 0.297 Tc /Type /XObject /Meta99 113 0 R /Length 54 0 G /Type /XObject /FormType 1 q 0 G **Note: You could choose any variable you want. BT >> endstream q /FormType 1 endstream 0.458 0 0 RG 2x - 15 = -27. endstream q 375 0 obj /FormType 1 stream Q Q /ProcSet[/PDF] >> /Matrix [1 0 0 1 0 0] Q /FormType 1 Q 1.007 0 0 1.007 411.035 277.035 cm 0 G Q /ProcSet[/PDF/Text] 0 g q q 1.014 0 0 1.007 531.485 583.429 cm Q In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. 1 i q Q Q /F3 17 0 R /Meta359 Do 1.014 0 0 1.007 531.485 636.879 cm Q << 22 0 obj endobj q Q 0.564 G (-) Tj /Type /XObject /F3 17 0 R Q /FormType 1 0.737 w 0.564 G 0 g q endobj Q /Resources<< 1.007 0 0 1.007 551.058 636.879 cm 1.007 0 0 1.007 271.012 703.126 cm /Meta376 390 0 R /Length 69 /Resources<< /F3 17 0 R stream /Font << >> ET 1 g /Subtype /Form /Meta186 200 0 R 205 0 obj /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /XObject 0 w << Q /Subtype /Form 0.458 0 0 RG 1 i /Type /XObject q Q q 2.238 5.203 TD 78 0 obj /Matrix [1 0 0 1 0 0] 0.458 0 0 RG >> >> 0 g /ProcSet[/PDF/Text] q endstream 1.014 0 0 1.006 391.462 437.384 cm Q /BBox [0 0 639.552 16.44] n 11 or n 11. 1.007 0 0 1.006 411.035 437.384 cm 132 0 obj /Meta37 50 0 R ET /BBox [0 0 30.642 16.44] >> 0 g 0.68 Tc /Subtype /Form 1.007 0 0 1.007 411.035 636.879 cm ET Q 0 G 0.155 Tc 1 i stream 0 w /StemV 77 BT >> 0 g endstream /Matrix [1 0 0 1 0 0] Q 1.007 0 0 1.006 130.989 437.384 cm >> /BaseFont /PalatinoLinotype-Bold 171 0 obj /Matrix [1 0 0 1 0 0] Q /Length 74 stream /F4 36 0 R /Meta86 100 0 R 0 G << stream 155 0 obj BT endobj /Resources<< q /Subtype /Form /FormType 1 /Meta398 414 0 R /FirstChar 43 /Font << /Subtype /Form Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /BBox [0 0 30.642 16.44] /Font << q Q 549.694 0 0 16.469 0 -0.0283 cm /Font << q endstream /Font << >> /Length 65 We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. endstream /FormType 1 0 G 232 0 obj >> 1.007 0 0 1.007 551.058 703.126 cm /ProcSet[/PDF/Text] /BBox [0 0 88.214 35.886] /BBox [0 0 15.59 16.44] /Subtype /Form 0.737 w 1 i 0.737 w /Encoding /WinAnsiEncoding 76.394 5.203 TD 0 5.203 TD 310 0 obj (x) Tj Q /Type /XObject >> endobj /F3 12.131 Tf >> stream >> /Matrix [1 0 0 1 0 0] 0 g Q Q q /Matrix [1 0 0 1 0 0] /Subtype /Form /Length 108 /F3 12.131 Tf [(1)-25(0\))] TJ 1.007 0 0 1.007 271.012 583.429 cm ET 0 G /FormType 1 Q 1 i << /Meta330 344 0 R /Matrix [1 0 0 1 0 0] 0.564 G >> q BT endstream << /F3 17 0 R /Type /XObject /Subtype /Form /ProcSet[/PDF/Text] /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] endobj /Meta404 Do endobj /F3 17 0 R >> /ProcSet[/PDF/Text] q 20 0 obj >> Q 0 g >> /Length 59 Q stream Q Q q /Length 16 [tex]\sin (\pi -x)=\sin x[/tex]. 1 i 0 g /Type /XObject /F3 17 0 R /Type /XObject q q >> 0.737 w q /BBox [0 0 30.642 16.44] [tex]\sin (\pi -x)=\sin x[/tex]. >> BT /FormType 1 0 G 0 g Q /Type /XObject >> /FormType 1 Q (-23) Tj << 1.502 5.203 TD Twice a number decreased by 8 gives 58. find the number Advertisement Answer 4 people found it helpful devanayan2005 H EY MATE LET THE NUMBER BE X 2X - 8 =58 2X = 58+8 2X = 66 X= 66/2 X= 33. q Q /I0 51 0 R 0 g /Resources<< q (A\)) Tj q /FontName /TimesNewRomanPSMT BT /Resources<< /Matrix [1 0 0 1 0 0] 6.746 8.18 TD Q stream stream /ProcSet[/PDF] 277 0 obj Q /Font << /Font << endstream /Type /XObject q BT stream /Meta84 98 0 R 1 i /FontDescriptor 6 0 R << 77 0 obj /BBox [0 0 17.177 16.44] stream /Subtype /Form 0 w 419 0 obj Q 722.699 546.541 l /Type /XObject endobj /Matrix [1 0 0 1 0 0] Q /Meta268 Do 0.369 Tc /Meta381 Do q Q 32.939 5.203 TD /Matrix [1 0 0 1 0 0] /Meta349 363 0 R /Type /XObject Q /F1 12.131 Tf stream /Length 16 /Matrix [1 0 0 1 0 0] >> Q /Meta121 Do >> Q Q endstream 45 0 obj Q /Resources<< /Meta388 Do /Resources<< 0.425 Tc /Type /XObject /FormType 1 /Length 68 0 w /ProcSet[/PDF] /Resources<< << 0.425 Tc 336 0 obj endstream Q 0 g 1 i /Length 59 0 g q /Resources<< Q /Subtype /Form /Meta288 302 0 R 0.564 G /Font << /Meta299 313 0 R 1 i (iv) A number exceeds 5 by 3. endstream /Font << 1.014 0 0 1.007 251.439 450.181 cm endstream /Meta323 Do << 22.478 4.894 TD /BBox [0 0 534.67 16.44] 1.007 0 0 1.007 551.058 636.879 cm Q /Subtype /Form /FormType 1 /Meta408 424 0 R Q 0 g 1 i /ProcSet[/PDF] >> /Meta185 Do /ProcSet[/PDF] << << << /Matrix [1 0 0 1 0 0] 1 i endstream >> 169 0 obj endstream /BBox [0 0 88.214 35.886] /BBox [0 0 15.59 29.168] >> 0 G /Subtype /Form /Meta134 148 0 R 114 0 obj 0 g >> 0.369 Tc /F3 17 0 R /Resources<< /Subtype /Form Q /Matrix [1 0 0 1 0 0] stream 0 g /Type /XObject >> /FormType 1 q q 1.007 0 0 1.007 271.012 330.484 cm << /BBox [0 0 88.214 16.44] q Q /FormType 1 Q endobj /Matrix [1 0 0 1 0 0] stream Q 0.458 0 0 RG 0 w /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] >> q 55 0 obj Q 0.524 Tc 0.786 Tc Q /Type /XObject Q 0.564 G endstream 1.007 0 0 1.007 45.168 813.037 cm /Length 118 1.007 0 0 1.007 45.168 730.228 cm /BBox [0 0 534.67 16.44] /Subtype /Form ET 1 i ET q /ProcSet[/PDF/Text] 672.261 653.441 m /Resources<< endstream q >> stream q q /F3 17 0 R /ProcSet[/PDF/Text] BT /Type /XObject /Resources<< 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 . 1 i 1.007 0 0 1.007 654.946 473.519 cm /Matrix [1 0 0 1 0 0] saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. q /Font << 0.68 Tc /F4 36 0 R << /F3 12.131 Tf /Matrix [1 0 0 1 0 0] Q << 0.68 Tc 328 0 obj >> /F3 17 0 R 0.524 Tc /Subtype /Form 159 0 obj 0 g stream /Subtype /Form 0.564 G 0 G >> /F3 17 0 R /Type /XObject /Resources<< q /F3 12.131 Tf stream /Meta45 Do >> stream /ProcSet[/PDF] /Length 16 << /FormType 1 q q /F1 12.131 Tf endobj /Meta33 Do 0 g /ProcSet[/PDF] >> /BBox [0 0 534.67 16.44] 0.838 Tc Q /Subtype /Form 89.12 5.203 TD 1 i /BBox [0 0 15.59 16.44] 1 i /BBox [0 0 88.214 35.886] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /FormType 1 Q >> << (5\)) Tj 0 g q 30.699 4.894 TD stream Q /Length 16 /Meta91 105 0 R q 215 0 obj /F3 12.131 Tf ET q /Matrix [1 0 0 1 0 0] 0.68 Tc >> /Meta297 311 0 R endobj stream Q stream (1\)) Tj /Font << >> /ProcSet[/PDF/Text] q endobj 1 i >> Q q >> 1 g endobj /Meta193 207 0 R /Font << q 0 G 672.261 799.486 m >> Q 1.005 0 0 1.007 102.382 400.496 cm /Subtype /Form endobj endobj Q BT endobj q 722.699 726.464 l /Type /XObject endstream (D\)) Tj q /Meta362 376 0 R Q q << endobj 1.005 0 0 1.007 102.382 743.025 cm 1 i <> /BBox [0 0 88.214 16.44] q >> endstream /Meta11 22 0 R the quotient of twenty and a number a.) BT /F3 17 0 R BT >> 1.007 0 0 1.007 130.989 277.035 cm 0.458 0 0 RG BT 1.007 0 0 1.007 551.058 277.035 cm /Subtype /Form q 0 g Q /Type /XObject /Meta181 195 0 R q q /Subtype /Form ET Q (58) Tj /Length 16 4.506 8.18 TD Q (3) Tj Q stream 23.952 4.894 TD /Type /XObject >> 1 i 1.007 0 0 1.006 411.035 690.329 cm BT >> endstream 1.014 0 0 1.007 531.485 330.484 cm endstream endobj >> /Length 59 endobj /Font << /Meta92 106 0 R /Resources<< 38 0 obj /Type /XObject /ProcSet[/PDF/Text] /Type /XObject Q /Subtype /Form 0.737 w /Matrix [1 0 0 1 0 0] >> 1 i Q 1.007 0 0 1.007 271.012 703.126 cm /ProcSet[/PDF/Text] 1 i /F4 12.131 Tf ( \() Tj Get link; Facebook; Twitter; >> stream stream /ProcSet[/PDF/Text] /Meta176 Do /Subtype /Form /Font << >> Q /Matrix [1 0 0 1 0 0] >> /F3 17 0 R /Resources<< BT stream 1.014 0 0 1.007 111.416 523.204 cm /Matrix [1 0 0 1 0 0] 333.269 5.488 TD /Font << /Meta238 Do /BBox [0 0 549.552 16.44] 1 i Q endobj /Type /XObject << >> 1 g >> 350 0 obj (x) Tj >> (3) Tj >> endobj q >> Q Q 27 0 obj /FormType 1 /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] << /BBox [0 0 88.214 16.44] >> Q /F4 12.131 Tf /BBox [0 0 88.214 16.44] q q << /Type /XObject /Meta306 320 0 R stream [(A number )-17(divided by )] TJ This gives us: "2x+5". q /Length 118 Q 1.007 0 0 1.007 271.012 523.204 cm q << /F1 7 0 R 1.502 5.203 TD Q Q /Meta335 Do /Length 69 >> Q BT endstream << 0 20.154 m << 1 i endobj >> endstream 1.005 0 0 1.007 79.798 779.913 cm /Resources<< q &K @ /Type /XObject q << Q /F1 12.131 Tf >> [(Negativ)16(e )] TJ /Subtype /Form q Q stream /Meta229 Do Q /ProcSet[/PDF/Text] q 0.269 Tc /BBox [0 0 88.214 35.886] /Type /XObject q 1 i q << 87 0 obj Q /Resources<< 0.564 G endobj /F1 7 0 R q 1 i endstream /Subtype /Form q >> >> 0 g 0 w /Length 12 q endobj /Meta351 Do /Matrix [1 0 0 1 0 0] /Meta177 Do 0 g ET 20.21 5.203 TD /Meta237 251 0 R 1.007 0 0 1.007 130.989 277.035 cm >> stream /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 713.666 cm Q >> /Resources<< 181 0 obj endstream 1 i /Subtype /Form stream Q /Type /Page 98.843 5.203 TD /Meta107 Do q /FormType 1 q /Type /XObject << Q (A\)) Tj endstream Q Q /Meta201 215 0 R >> /XHeight 477 >> endstream /Type /XObject q 287 0 obj 206 0 obj /Matrix [1 0 0 1 0 0] >> q endobj /ProcSet[/PDF] 383 0 obj /Meta39 53 0 R stream q /Meta27 Do /Meta407 Do 1 i endstream 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. /FormType 1 stream /Resources<< q /F3 17 0 R ET /F3 12.131 Tf /FormType 1 0 g >> /Font <> ET >> /Type /XObject /FormType 1 Q /Type /XObject /Type /XObject << >> 9.723 5.336 TD endobj /FormType 1 /Matrix [1 0 0 1 0 0] 0 g /Resources<< /ProcSet[/PDF] /Meta393 409 0 R q ET q q /ProcSet[/PDF/Text] >> endobj Q Q /Type /XObject 0 w Q /Length 73 >> 220.931 4.894 TD >> << q Q /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] stream Q /ProcSet[/PDF] /F4 12.131 Tf endobj /Resources<< 1.014 0 0 1.007 111.416 583.429 cm 0.737 w /FormType 1 0 g /Length 69 /ProcSet[/PDF/Text] q 0.737 w 165 0 obj 0 g /Font << /FormType 1 Q q 2x - y = 6. x + 3y = -25. /Matrix [1 0 0 1 0 0] /Meta294 Do 0 g 26.957 5.203 TD /F3 17 0 R stream >> /F3 12.131 Tf 1.007 0 0 1.007 551.058 383.934 cm /ProcSet[/PDF] q << /Type /XObject 0.564 G 0.737 w BT 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 >> << /I0 51 0 R /FormType 1 /ProcSet[/PDF] 280 0 obj BT 0.737 w endstream Q /Matrix [1 0 0 1 0 0] >> 0 g /Matrix [1 0 0 1 0 0] << q 0.68 Tc << >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Type /XObject 0.564 G q q 1st step. Q ET Q 199 0 obj endstream endobj 177 0 obj q >> Q /F4 36 0 R /FormType 1 Q Q Q ET /Matrix [1 0 0 1 0 0] >> /Encoding /WinAnsiEncoding ET q /F3 17 0 R >> 1 i /Meta72 86 0 R /FormType 1 0.737 w /Font << 0 G /Type /XObject endobj endobj /F3 17 0 R endstream >> endobj /FormType 1 /Subtype /Form BT /Meta383 397 0 R (C\)) Tj /Meta138 152 0 R >> /Matrix [1 0 0 1 0 0] stream /ProcSet[/PDF] /BBox [0 0 639.552 16.44] endstream q 257 0 obj /BBox [0 0 534.67 16.44] /Length 16 1.014 0 0 1.007 391.462 849.172 cm /Resources<< endobj Q q /Type /XObject /BBox [0 0 88.214 16.44] /FormType 1 /Subtype /Form /Meta265 279 0 R /F1 12.131 Tf /ProcSet[/PDF/Text] 0 g 1 i >> q , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. Q q /Type /XObject q /Resources<< /ProcSet[/PDF/Text] q ET << , Prove the following 1 i 0.564 G /Meta380 Do /Meta241 Do /Meta24 Do << q 0 g q >> q /Matrix [1 0 0 1 0 0] /FormType 1 >> q q >> /Subtype /Form <> 86 0 obj 1 i /Meta257 Do /BBox [0 0 88.214 16.44] >> q /Type /XObject /Meta96 Do Q endstream /Meta428 444 0 R endstream stream /ProcSet[/PDF] >> Q /BBox [0 0 88.214 35.886] endstream >> /FormType 1 /FormType 1 /Subtype /Form /Type /XObject endstream 0 g 0 g Q /Meta310 Do Q A. /ProcSet[/PDF] >> Twice = two times, double. Q ET Q /BBox [0 0 15.59 16.44] /Length 12 -0.056 Tw /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 0 w /Type /XObject q /Meta8 Do /FormType 1 endobj /BBox [0 0 30.642 16.44] Q /Type /XObject /Filter [/CCITTFaxDecode] Q endobj /Length 81 /Length 54 q >> 0.564 G /F3 17 0 R /Meta357 Do /Resources<< /F1 7 0 R /Meta10 21 0 R /Subtype /Form /Subtype /Form ET q >> /Type /XObject << >> (5) Tj /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 523.204 cm /Meta34 Do /Subtype /Form /ProcSet[/PDF/Text] q endstream 15.731 5.336 TD endobj >> /Matrix [1 0 0 1 0 0] /Length 69 >> /Subtype /Form 1 g /Font << q >> Q 0 G q >> q /FormType 1 Q /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 391.462 776.149 cm /F3 17 0 R /ProcSet[/PDF/Text] 0.458 0 0 RG /F1 12.131 Tf /Length 57 >> Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM >> >> /BBox [0 0 88.214 16.44] Q 0.369 Tc /Meta226 Do /Meta385 401 0 R /Subtype /Form 0 g q 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. 3.742 5.203 TD /FormType 1 /Meta317 331 0 R -0.382 Tw >> q /F3 12.131 Tf 1 i q endstream /BBox [0 0 88.214 16.44] >> /Resources<< Q 0 G /Subtype /Form /Length 69 /Length 12 /Subtype /Form q 0 G /Subtype /Form << Q Q 0 g /Length 16 0.737 w Q /BBox [0 0 88.214 35.886] /Meta374 388 0 R >> /ProcSet[/PDF] /Length 59 /Subtype /Form >> /Type /XObject /Resources<< /Matrix [1 0 0 1 0 0] 20.21 5.203 TD 0 g << Q (x) Tj 372 0 obj Q 1 i 1 i Q /Type /XObject /BBox [0 0 88.214 16.44] /BBox [0 0 639.552 16.44] >> >> /Meta98 Do 0.458 0 0 RG /BBox [0 0 88.214 16.44] 1 i [( subt)-17(racted fr)-14(om a )-16(number)] TJ Q /FormType 1 /Matrix [1 0 0 1 0 0] << BT >> 108 0 obj 1 g BT endobj 131 0 obj stream /Subtype /Form q /F3 12.131 Tf /BBox [0 0 88.214 16.44] << 167 0 obj << 0 w 1 i 0.737 w (-) Tj Q q /Subtype /Form q /FormType 1 Grad - B.S. 1 g 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 q /Resources<< Q endstream endobj >> /Font << /Length 64 /Meta119 Do q ET 2 times x minus 58 C. twice the difference of a number and 5 B. twice a number decreased by 58 D. 2 times the sum of a number and 58 Answer: B. Step-by-step explanation: twice - (2) number - (x) 58-(58) Edukasyon. << /BBox [0 0 23.896 16.44] << /Subtype /Form [(1)-25(0\))] TJ 1 i 174.501 5.203 TD 0 g 0 4.78 TD 1.007 0 0 1.007 130.989 636.879 cm /ProcSet[/PDF/Text] /Type /XObject /F3 17 0 R >> q stream /ProcSet[/PDF] /Length 69 0.458 0 0 RG q /FormType 1 1.007 0 0 1.007 551.058 383.934 cm >> (-) Tj q All steps. /Meta131 145 0 R endobj /BBox [0 0 15.59 16.44] /FormType 1 /Subtype /Form 0.68 Tc Q stream << /Meta333 Do /Meta146 160 0 R 0 g /Length 65 /ProcSet[/PDF/Text] << /Subtype /Form 0.838 Tc q q (-20) Tj endstream /BBox [0 0 639.552 16.44] /BBox [0 0 88.214 16.44] /Meta20 31 0 R Q q 1 g q 1 i /Type /XObject /Subtype /Form /Meta410 Do /Meta365 Do << q 0 g /ProcSet[/PDF] 0 g /F3 17 0 R /F4 12.131 Tf /F3 17 0 R 0 G BT /Font <> /Matrix [1 0 0 1 0 0] >> >> /FormType 1 /Length 244 85 0 obj 393 0 obj /Font <> /F1 7 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form /Subtype /Form ( x) Tj >> the other number. >> /Length 69 305 0 obj q 0 g Q /Matrix [1 0 0 1 0 0] 0 g >> /Subtype /Form >> /FormType 1 /Type /XObject /Meta65 Do Tamang sagot sa tanong: 1.) q BT Q q q >> endstream 0 G q 0 g /Meta419 435 0 R stream /Length 69 /Resources<< 0.425 Tc /Matrix [1 0 0 1 0 0] BT /Type /XObject /Type /XObject BT 1.007 0 0 1.007 551.058 703.126 cm /Matrix [1 0 0 1 0 0] 0 g 43.426 5.203 TD /ProcSet[/PDF/Text] Q 0 5.203 TD Q stream q endstream /Resources<< /BBox [0 0 549.552 16.44] endobj 0 g stream /F3 12.131 Tf /Resources<< /Length 119 1 i /F3 12.131 Tf ET >> >> >> /Meta204 218 0 R /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 347.046 cm 0.737 w 1 i /Matrix [1 0 0 1 0 0] endobj /Meta161 Do 266 0 obj /Type /XObject endstream 1.007 0 0 1.007 411.035 636.879 cm /Meta60 Do endstream q The sum Of twice a nu4ber What is the number? /Type /XObject << The width Of a rectangle is 15 cm and the perimeter is 12 cm. Q /Resources<< /Type /XObject /Resources<< q 0 G 225 0 obj stream endobj ET endobj /Length 59 ET Q << ET 20.21 5.203 TD >> Q stream >> /ProcSet[/PDF] q BT /ProcSet[/PDF] 1 i >> endstream Q 141 0 obj /Subtype /Form Q q >> stream 0.458 0 0 RG /Meta187 201 0 R 20.975 5.336 TD 1.014 0 0 1.006 251.439 437.384 cm 0 g q /BBox [0 0 88.214 35.886] q /Font << Q /Meta354 Do 318 0 obj Q /Meta123 137 0 R /Resources<< 130 0 obj >> Q /Meta253 Do q 188 0 obj q 418 0 obj 0 g 13.493 5.336 TD /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] 1 i 0.369 Tc 0 g /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 473.519 cm endobj Q 1 i 0 g /FormType 1 /Font << >> q /Length 69 /Resources<< (D\)) Tj /F3 17 0 R 1.007 0 0 1.007 654.946 347.046 cm /F1 7 0 R q 1.014 0 0 1.007 251.439 523.204 cm /ProcSet[/PDF/Text] >> /Meta158 Do Q Q /Resources<< Q /F3 17 0 R /Type /XObject Q 0.425 Tc 1.007 0 0 1.006 411.035 510.406 cm 89.12 5.203 TD q Q /F3 12.131 Tf /Resources<< q q Q q << 0 w q q Q q 52.412 5.203 TD stream << /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] /Font << endstream endobj /Matrix [1 0 0 1 0 0] /FormType 1 /Resources<< ( x) Tj /Type /XObject /ProcSet[/PDF/Text] Q /Type /XObject >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] ET /Subtype /Form endobj /Meta75 Do /Matrix [1 0 0 1 0 0] /F3 17 0 R q /Meta36 Do 0 g /F3 12.131 Tf /Resources<< 0 g 1.007 0 0 1.007 271.012 636.879 cm >> Q /Matrix [1 0 0 1 0 0] /FormType 1 1 i endstream q << BT stream /Length 12 /F1 7 0 R >> Q /Meta168 182 0 R endobj /Type /XObject stream <> /Subtype /Form 0.425 Tc q 0 G q /FontDescriptor 16 0 R /F3 17 0 R << /BBox [0 0 88.214 16.44] (5) Tj /Length 16 43 0 obj 1.502 8.18 TD endobj /Subtype /Form << 1.014 0 0 1.006 251.439 836.374 cm /Subtype /Form 0 5.203 TD /ProcSet[/PDF] /Resources<< 1 i /Matrix [1 0 0 1 0 0] ET 0 g /F3 17 0 R /Meta116 130 0 R ET >> /Font << /BBox [0 0 30.642 16.44] /BBox [0 0 534.67 16.44] /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] /Type /XObject Q 1.005 0 0 1.007 102.382 799.486 cm endstream /Length 70 Q Q 1.007 0 0 1.007 411.035 277.035 cm /BBox [0 0 30.642 16.44] /F3 17 0 R >> >> /Resources<< ET /FormType 1 (x) Tj >> /Resources<< /BBox [0 0 15.59 16.44] /Meta403 419 0 R q Q /Resources<< 0 g 0 w Q 65.906 4.894 TD BT /Meta248 262 0 R ET endstream << 1 i stream 1.005 0 0 1.007 102.382 400.496 cm 158 0 obj 0 5.203 TD << /Resources<< 358 0 obj >> >> /Matrix [1 0 0 1 0 0] /Resources<< 0.737 w /ProcSet[/PDF] /Type /XObject 1.014 0 0 1.007 391.462 383.934 cm ET New questions in Mathematics >> ET /Meta290 Do /Font << ET 1 i 313 0 obj Q -0.486 Tw 0 g 0 G 0.524 Tc q /Font << /Meta76 Do 136 0 obj /Meta367 Do /Length 69 /Type /XObject /FormType 1 Q 0 5.203 TD /Meta271 Do /Subtype /Form /FirstChar 32 /Type /XObject /Length 16 Q stream /Type /XObject /Resources<< /Length 59 1 i /Font << /Subtype /Form /F3 12.131 Tf (C\)) Tj /Length 59 21.713 20.154 l << >> S 0 g 1.007 0 0 1.007 411.035 636.879 cm /Resources<< BT q /Resources<< Q 1.005 0 0 1.007 45.168 889.071 cm /BBox [0 0 15.59 16.44] >> [3] One half of a number increased by fourteen is twenty-one. /Length 16 stream /F3 17 0 R /F3 17 0 R q -0.486 Tw Q >> q /Subtype /Form /Length 16 722.699 400.496 l /BBox [0 0 534.67 16.44] q << /Resources<< S stream Q 0 w /ProcSet[/PDF/Text] >> /Type /XObject endstream /I0 51 0 R >> 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. Q 0.198 Tc (D\)) Tj /Matrix [1 0 0 1 0 0] /Subtype /Form /Length 70 To find: The. /Meta99 Do Q Q q /Type /XObject Q /Type /XObject /Matrix [1 0 0 1 0 0] 446 0 obj (-11) Tj 1.007 0 0 1.006 411.035 437.384 cm stream << ET Q /Font << endstream /FormType 1 0.564 G 1.007 0 0 1.007 130.989 636.879 cm q << /FormType 1 Q /BBox [0 0 673.937 15.562] Q Q >> q ET /FormType 1 q endobj ET /BBox [0 0 15.59 16.44] /Size 447 /Type /XObject (-20) Tj 295.086 4.894 TD /FormType 1 q /Matrix [1 0 0 1 0 0] stream q 1 g Q q /FormType 1 /Length 104 /BBox [0 0 15.59 29.168] ET /F3 12.131 Tf /FormType 1 0.737 w endobj /Matrix [1 0 0 1 0 0] Q Q Q q 1.007 0 0 1.007 411.035 583.429 cm 1.007 0 0 1.007 551.058 277.035 cm /FormType 1 /Length 70 /Length 69 /Subtype /Form Q 248 0 obj endobj /Meta285 299 0 R /Subtype /Form /F3 12.131 Tf endobj /Length 16 /Subtype /Form 1.007 0 0 1.007 130.989 776.149 cm
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