0 G endstream endobj 1 i Q Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /Type /XObject 0 g /F3 17 0 R BT q /FormType 1 Q /Type /XObject /ProcSet[/PDF/Text] /F3 12.131 Tf >> /F3 12.131 Tf (C\)) Tj 0 g /ProcSet[/PDF] /Length 64 Q Q (-8) Tj /Length 16 /Type /XObject /Length 69 >> /Subtype /Form endstream ET 1.007 0 0 1.007 67.753 473.519 cm /FormType 1 q /Meta222 236 0 R endstream /Meta369 383 0 R /FormType 1 162 0 obj ET /Meta337 Do ET /Resources<< /Subtype /Form q /Type /XObject >> Q /Type /XObject /F3 12.131 Tf Q /FormType 1 198 0 obj endstream /Meta138 152 0 R /BBox [0 0 15.59 16.44] stream >> /Matrix [1 0 0 1 0 0] ET S << Q -0.22 Tw BT 0 w 331 0 obj /F4 12.131 Tf 32 = 2a + 8: The quotient of fifty and five more than a number is ten. 1.007 0 0 1.007 271.012 330.484 cm /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] 360 0 obj /Font << /ProcSet[/PDF] /Type /XObject stream Q Q BT 1 i /FormType 1 /Resources<< /Resources<< >> q q stream 0 w >> endobj /Matrix [1 0 0 1 0 0] stream endobj Q /FormType 1 By the . >> /Font << /Meta145 Do /FormType 1 /Font << /F3 12.131 Tf 1 i 0 g Q /F1 14.682 Tf 1 g 0 g >> 1.014 0 0 1.007 111.416 383.934 cm stream /ProcSet[/PDF/Text] Q q >> Two times the sum of a number x and five c.) a number x times the sum of five and two d.) the sum of five times a number x and two 2.) 0 w /Subtype /Form /ProcSet[/PDF/Text] 0 G /ProcSet[/PDF] /Meta170 Do /Type /XObject 150 0 obj -0.126 Tw 1.007 0 0 1.007 551.058 636.879 cm (D\)) Tj 1 i 0.486 Tc Q /Meta159 173 0 R ET 1. /Length 65 Grad - B.S. /FormType 1 /Meta364 378 0 R /Subtype /Form Q << /ProcSet[/PDF/Text] 1 i /ProcSet[/PDF/Text] >> 0 g q /FormType 1 endobj endobj 0.458 0 0 RG >> 0 G /Subtype /Form /Type /XObject endstream << Q 0 G >> stream Next, the problem says that "x" would be equal to twice a number added by 5. q Q /FormType 1 1 i /BBox [0 0 15.59 16.44] ET stream BT /Length 69 0.458 0 0 RG /Length 12 q 2.238 5.203 TD >> 13.464 5.203 TD 1 g 1.007 0 0 1.007 67.753 546.541 cm 20.21 5.203 TD /Matrix [1 0 0 1 0 0] << 178 0 obj /F3 12.131 Tf /BBox [0 0 15.59 16.44] 1 i /Meta359 373 0 R q /ProcSet[/PDF/Text] >> Q endobj 1.007 0 0 1.007 45.168 763.351 cm /Subtype /Form /Meta312 Do q /BBox [0 0 88.214 16.44] 0.737 w endobj /Meta305 Do Q /Meta119 Do << q stream 0 G /Resources<< >> 0 w 1 i /Subtype /Form stream ET q /Type /XObject /Matrix [1 0 0 1 0 0] Advertisement Answer No one rated this answer yet why not be the first? >> 0 g /F3 17 0 R /FormType 1 /Resources<< 16 0 obj /Meta0 5 0 R Q /Meta102 Do 1.014 0 0 1.007 531.485 277.035 cm 2.238 5.203 TD /Meta125 139 0 R /Subtype /Form >> >> endstream 0 G 1.014 0 0 1.007 251.439 776.149 cm /Font << 6.746 5.203 TD Q 0 g 146 0 obj /F1 7 0 R /ProcSet[/PDF] 0.737 w 1.007 0 0 1.006 411.035 437.384 cm Q /ProcSet[/PDF/Text] /Resources<< (-9) Tj Q BT /Matrix [1 0 0 1 0 0] 0.51 Tc /F3 12.131 Tf q /Type /XObject ET /Font << /Length 69 /FormType 1 1 i /BBox [0 0 88.214 16.44] 0.737 w Q 12.727 5.203 TD 0 w BT 0 g q 1 i 1 i /Meta132 Do /Ascent 1050 /FormType 1 q q /Type /XObject 11.99 8.18 TD /Matrix [1 0 0 1 0 0] << Q endobj /Subtype /Form /Font << >> 1.005 0 0 1.007 79.798 796.475 cm endobj /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 79.798 730.228 cm /FormType 1 0.524 Tc q /ProcSet[/PDF/Text] 1.502 5.203 TD /BBox [0 0 549.552 16.44] 0 w 0 G 0.458 0 0 RG ET stream /F3 17 0 R stream stream /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) /Length 79 /Subtype /Form /F3 17 0 R Q q /Type /XObject /F3 12.131 Tf /Type /XObject /Length 16 /Matrix [1 0 0 1 0 0] << endobj 1.014 0 0 1.007 251.439 277.035 cm -y. /Type /XObject 0.458 0 0 RG >> /Subtype /Form /Subtype /Form q q 1.007 0 0 1.007 130.989 583.429 cm /F3 12.131 Tf /FormType 1 >> >> 0.458 0 0 RG /Meta130 144 0 R 549.694 0 0 16.469 0 -0.0283 cm /F3 17 0 R /F1 7 0 R /Matrix [1 0 0 1 0 0] /Meta96 Do << << /Type /XObject >> /F3 12.131 Tf Q 0 G /BBox [0 0 673.937 68.796] /FormType 1 endstream endstream /BBox [0 0 673.937 16.44] Q q endobj endstream /BBox [0 0 88.214 35.886] 1.014 0 0 1.007 391.462 383.934 cm /I0 51 0 R /Resources<< 154.289 4.894 TD /Length 69 endobj 0.737 w /Resources<< endstream /FormType 1 /BBox [0 0 17.177 16.44] ET >> >> Q q >> /BBox [0 0 15.59 16.44] >> stream /BBox [0 0 534.67 16.44] 0 g /Type /XObject /Font << Q ET Q Q /Type /XObject /BBox [0 0 15.59 29.168] endstream q /Type /XObject >> /Resources<< Q q 0.369 Tc 1.007 0 0 1.007 67.753 599.991 cm << /Resources<< /Length 118 >> /FormType 1 >> stream /ProcSet[/PDF] BT /Meta421 437 0 R 1 g /FormType 1 0.564 G stream q /Resources<< /Length 12 >> << endobj >> << /ProcSet[/PDF/Text] Q endstream q 0 5.203 TD Q /Resources<< endstream Q Q 371 0 obj /Type /XObject Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . Check out a sample Q&A here. /FormType 1 679.036 293.596 m /ProcSet[/PDF] << 0.564 G q 0.737 w /Resources<< >> Q 0.786 Tc /Length 59 >> 0.564 G 220 0 obj q 0.564 G /Length 80 endobj 1.005 0 0 1.007 102.382 799.486 cm /Subtype /Form /Resources<< 1.014 0 0 1.007 391.462 776.149 cm /Font << 0 g /Meta327 341 0 R /Font << Q << endobj q /Meta3 12 0 R 0 g /Matrix [1 0 0 1 0 0] /Type /XObject endstream q Q /Subtype /Form (+) Tj /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Length 16 /Font << 1 i /Meta108 Do Q q 0 g >> endobj << (D) Tj << >> /BBox [0 0 534.67 16.44] 368 0 obj 0.737 w q 1.014 0 0 1.007 251.439 849.172 cm /Type /XObject /Meta339 353 0 R /Length 87 /Matrix [1 0 0 1 0 0] q /BBox [0 0 88.214 35.886] 0 w endobj /F3 17 0 R endstream q Table 1. /Resources<< endobj q 381 0 obj [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ Q /FormType 1 >> 0 g /Meta201 215 0 R 1 g 0.425 Tc q endstream /Type /XObject >> >> q /F4 12.131 Tf endobj Q q 0.155 Tc q Q endstream q 1 i /Subtype /Form 0 g /Meta90 104 0 R /ProcSet[/PDF/Text] /F3 17 0 R /Subtype /Form 1.007 0 0 1.006 411.035 690.329 cm /Meta135 Do q Q 1.005 0 0 1.007 102.382 473.519 cm /FormType 1 Q /BBox [0 0 639.552 16.44] 0 g 0.738 Tc /ProcSet[/PDF] Q q /ProcSet[/PDF/Text] /FormType 1 >> q Q stream /Length 60 /Meta48 62 0 R >> q /ProcSet[/PDF] /Type /XObject 345 0 obj q 0.68 Tc q /ProcSet[/PDF] 1 i Q /Matrix [1 0 0 1 0 0] /Resources<< 1.005 0 0 1.007 102.382 347.046 cm /Resources<< /Font << /F3 17 0 R << w/Honors. [(The )-19(quotient of )] TJ /Length 16 endobj >> q /FormType 1 Q ET /ProcSet[/PDF/Text] /Length 58 endobj ET [tex]\sin (\pi -x)=\sin x[/tex]. 0.838 Tc >> /BBox [0 0 15.59 29.168] >> << /Length 59 endobj /Meta47 61 0 R /Meta228 242 0 R 1 i /Matrix [1 0 0 1 0 0] stream /F3 17 0 R >> << /BaseFont /PalatinoLinotype-Roman >> endobj /ProcSet[/PDF/Text] 1 i 1 i Q stream endstream >> Q /Resources<< /FormType 1 142 0 obj 47.933 5.203 TD S stream 0 G 1 i /ProcSet[/PDF/Text] 1.014 0 0 1.007 531.485 330.484 cm 1 i >> ET /Meta226 Do >> /BBox [0 0 88.214 16.44] 185 0 obj 0 G /Meta418 434 0 R q /Length 16 -37 VI 2. q /Matrix [1 0 0 1 0 0] q >> 0 g 3 0 obj /F3 17 0 R ET /Subtype /Form << >> endobj 0.369 Tc /ProcSet[/PDF/Text] 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. 0 g q /Resources<< 0 G stream BT /Subtype /Form q q 0 g endstream q 0 G /F3 17 0 R /Font << 0 g BT stream q q >> >> 1 g endobj >> (x) Tj Q /Matrix [1 0 0 1 0 0] 1 i endstream q Notice that we used the variable \large {d} d in our equation to stand for our unknown value. stream /F3 12.131 Tf /Length 118 Q /Subtype /Form Q >> /F3 12.131 Tf << Q stream stream BT 0 g Q stream 1.014 0 0 1.007 531.485 636.879 cm 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. >> /BBox [0 0 673.937 14.853] endstream >> /BBox [0 0 15.59 16.44] >> Q q 1 i << Q 0.737 w 1.007 0 0 1.007 271.012 383.934 cm /Meta36 Do stream /Matrix [1 0 0 1 0 0] /Length 63 /Subtype /Form >> 0 g /Subtype /Form << 0.458 0 0 RG /Length 68 q /Length 16 >> 1 g Q /BBox [0 0 88.214 16.44] << endobj q /Resources<< (2) Tj /F3 12.131 Tf >> 1.005 0 0 1.007 79.798 829.599 cm /F3 12.131 Tf /Meta284 Do /Length 68 << /Meta413 Do /F1 12.131 Tf >> 194 0 obj Q /Meta45 59 0 R endobj BT ET >> >> /Subtype /Form Q q /BBox [0 0 30.642 16.44] /Resources<< stream 0 G /F3 17 0 R /BBox [0 0 88.214 16.44] /BBox [0 0 534.67 16.44] ET >> q 252 0 obj ( \() Tj q (C\)) Tj >> /Subtype /Form >> /ProcSet[/PDF] /FormType 1 /MaxWidth 1397 /FormType 1 1.007 0 0 1.007 130.989 583.429 cm q /Resources<< Q >> endstream q 1 g 0.737 w BT Q q /Meta244 258 0 R 0 g 0 G /BBox [0 0 15.59 16.44] /F3 17 0 R q q Q Q Q /Meta280 Do q 1 i /ProcSet[/PDF/Text] /Type /XObject 1.014 0 0 1.007 531.485 523.204 cm ET /Font << 0 g 1.007 0 0 1.007 551.058 523.204 cm endobj 1.007 0 0 1.006 551.058 763.351 cm Q 1 i What is marios jumps times luigis jumps. /Type /XObject << q /ProcSet[/PDF] 0.737 w 0.564 G /Length 54 << q /F3 12.131 Tf /Length 70 >> Q 1 i >> q /Type /XObject Find an answer to your question Twice a number decreased by 8gives 58. /Length 69 Q ET q >> (C\)) Tj 1.007 0 0 1.007 271.012 450.181 cm /ProcSet[/PDF] stream Q /Resources<< ET /Length 59 /Meta157 171 0 R /Meta260 274 0 R >> /Subtype /Form 0 g /Type /XObject /BBox [0 0 88.214 16.44] endstream << q /Type /XObject << Q endobj stream /Resources<< >> Q >> Q Q /FormType 1 /Meta63 Do /Subtype /Form (-20) Tj q endobj 0.369 Tc /ID [] q ET 1.007 0 0 1.007 411.035 849.172 cm Q 0 g 0 g /Font << endstream q stream q 1 i 0 g q /Meta401 417 0 R 0.458 0 0 RG Q 16.469 5.336 TD << endstream Q /FormType 1 /Subtype /Form 1 i 25 0 obj /Matrix [1 0 0 1 0 0] >> /Resources<< q >> q >> /Length 12 Answer by Mathtut (3670) ( Show Source ): 135 0 obj /Font << /Type /XObject << endobj 0 G /Length 16 /F4 12.131 Tf BT q q /Meta268 Do endstream /Type /XObject >> 0.311 Tc 132 0 obj Q /Length 16 q ET /Meta416 432 0 R >> /Resources<< << endobj /F3 17 0 R stream >> /Font << BT /F3 17 0 R Q 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 1 i q 0 w q /Meta315 Do /Meta56 70 0 R 1 i >> >> /Length 69 q 1 i stream Q >> Q q Q q /Type /XObject 106 0 obj Q /FormType 1 0 g q /ProcSet[/PDF] Q 0.51 Tc endobj /Length 54 >> 0 g Q q /Font << /F3 17 0 R /F4 12.131 Tf q /Type /XObject << << 65 0 obj endobj /BBox [0 0 534.67 16.44] 21.713 20.154 l 1.007 0 0 1.006 551.058 437.384 cm BT q /Resources<< >> ([x ) Tj BT q /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF/Text] /FormType 1 Q endobj /Subtype /Form Q (A\)) Tj /Meta294 Do endstream /Subtype /Form 0 g ET /Resources<< endstream >> /Resources<< /I0 Do /Matrix [1 0 0 1 0 0] (x) Tj Q /FormType 1 Q 396 0 obj (x) Tj /Resources<< q stream << /FormType 1 /F3 17 0 R 3.742 5.203 TD q 1 i 1.502 5.203 TD 0 G (6\)) Tj Q endstream 0.524 Tc /Subtype /Form Q >> /Meta82 Do Q /Subtype /Form 0.737 w /FormType 1 /Resources<< ET endobj q /BBox [0 0 88.214 16.44] 0 5.203 TD 1.007 0 0 1.007 551.058 277.035 cm 1 i /Type /XObject /Font << >> >> Q /Subtype /Form 0 g 0 w Q /Type /XObject << /F1 7 0 R /BBox [0 0 88.214 16.44] Q 1.007 0 0 1.007 271.012 776.149 cm /ProcSet[/PDF/Text] /Resources<< /Meta71 Do endstream /Meta151 165 0 R /Type /XObject /ProcSet[/PDF/Text] /F3 12.131 Tf /Meta142 156 0 R q /ProcSet[/PDF/Text] 1 i 1.007 0 0 1.007 67.753 400.496 cm /FormType 1 /F1 12.131 Tf /Length 69 Q 0 g /Matrix [1 0 0 1 0 0] /Length 74 /F4 36 0 R /Matrix [1 0 0 1 0 0] 0.737 w q /Type /XObject You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 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. 1 i Find the number. 1.005 0 0 1.007 102.382 872.509 cm 0 g << /ProcSet[/PDF/Text] 0.737 w endobj /ProcSet[/PDF/Text] >> /ProcSet[/PDF/Text] (\)) Tj q /ProcSet[/PDF] /Font << /Length 69 >> 0 G >> /Type /XObject /Meta404 Do 1.005 0 0 1.007 102.382 816.048 cm Q ET Q /Resources<< /Resources<< >> << 0 5.203 TD Q /Type /XObject stream endobj BT 10.487 5.203 TD /Resources<< /Meta362 376 0 R Q 1 i [( t)-14(imes a num)-16(ber)] TJ /Meta309 Do ET << >> q Q endstream /Resources<< /Subtype /Form << (C\)) Tj 3.742 8.18 TD 10 0 obj endstream 1.007 0 0 1.007 271.012 636.879 cm /F3 12.131 Tf q Q (C\)) Tj endstream /Subtype /Form /Meta278 Do Q /F3 12.131 Tf (C) Tj /Font << /F3 17 0 R /Meta295 Do Q /ProcSet[/PDF/Text] (D\)) Tj /Meta30 Do stream 0 5.203 TD 0 w /Resources<< >> 16.469 5.203 TD 160 0 obj /Type /XObject >> 16.469 5.336 TD Q /Meta56 Do Q ET q q /Length 16 Q /Meta195 209 0 R (9\)) Tj /Subtype /Form /Font << BT /Meta52 66 0 R /Meta64 78 0 R /Subtype /Form Q 154 0 obj /ProcSet[/PDF/Text] 20.21 5.203 TD q /Meta360 Do /Subtype /Form 0 g /Resources<< /Type /XObject q 0.737 w /F3 17 0 R >> 0 w Q /FormType 1 Q /Length 69 0 w /FormType 1 q q /Matrix [1 0 0 1 0 0] /Resources<< /Matrix [1 0 0 1 0 0] 353 0 obj endstream 19.474 20.154 l 20-n c.) n+20 d.) 20+n 3.) 0.564 G (B\)) Tj 1 g >> endobj (1\)) Tj /Font << /Meta109 123 0 R /Meta263 277 0 R /Meta105 Do 35 0 obj /Font << 0.737 w 0.564 G q 1 i /Matrix [1 0 0 1 0 0] q 0.297 Tc The sum Of twice a nu4ber What is the number? Q << /Meta180 Do 280 0 obj /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 12 0 obj /ProcSet[/PDF/Text] /Type /XObject endobj (-8) Tj q q /ProcSet[/PDF/Text] 0 g Q Q Q endobj /BBox [0 0 88.214 16.44] q stream Q 0 g endobj Q /Meta10 Do /Length 16 /BBox [0 0 639.552 16.44] 1.007 0 0 1.006 551.058 763.351 cm >> /Meta212 Do /ProcSet[/PDF] /Font << /Resources<< /Length 69 >> /Type /XObject /FormType 1 /Meta31 44 0 R Q Q 52.412 5.203 TD /F3 12.131 Tf /F3 12.131 Tf /Meta372 Do /Meta226 240 0 R Q 0 w 1 i /Length 58 1.007 0 0 1.007 130.989 330.484 cm 0 g q q 1.014 0 0 1.006 251.439 437.384 cm /Resources<< /Meta148 Do >> << Q Q /Length 63 /FormType 1 Q endstream 22.478 4.894 TD /BBox [0 0 88.214 16.44] 1 i 1 i Q q endobj 0 g /ProcSet[/PDF] /Type /XObject /Type /XObject ET /ProcSet[/PDF] endobj /Meta9 Do q >> q /ProcSet[/PDF] 314 0 obj ET /Font << Q 43 0 obj /Length 59 q /Matrix [1 0 0 1 0 0] /Length 294 Q stream Q /Resources<< 1 i (\)) Tj 0 w << 1.007 0 0 1.007 551.058 277.035 cm << /Resources<< 0 g 1 g stream /ProcSet[/PDF/Text] /Meta383 397 0 R Q Q BT ET >> /Meta196 Do q /Subtype /Form >> Q /Subtype /Form 1.014 0 0 1.007 391.462 277.035 cm endstream endobj /Length 59 /F4 12.131 Tf /Subtype /Form /Meta102 116 0 R /Matrix [1 0 0 1 0 0] /Meta96 110 0 R /Meta257 271 0 R Q /Meta171 Do endobj << 1 i endstream /Meta135 149 0 R 1 i >> /Type /XObject Q BT /Font << q -0.463 Tw 0.838 Tc 38.948 5.203 TD >> /BBox [0 0 534.67 16.44] 0.524 Tc 0.68 Tc Q 1 g /Matrix [1 0 0 1 0 0] /Font << << << /F4 12.131 Tf BT /Meta11 22 0 R Q /Type /XObject /Meta304 318 0 R q q 1.007 0 0 1.007 271.012 849.172 cm >> q 79 0 obj endobj /Type /XObject BT stream /Resources<< ET << Q /ProcSet[/PDF/Text] Q /Subtype /Form 1.014 0 0 1.007 531.485 330.484 cm 1.014 0 0 1.006 111.416 763.351 cm << Q Testosterone is the primary sex hormone and anabolic steroid in males. Q Q >> q q Q q Q 1 g Q 234 0 obj 1.014 0 0 1.006 251.439 690.329 cm 0.786 Tc Q Find the number. q Q /Subtype /Form 1 i /Matrix [1 0 0 1 0 0] 94.364 5.203 TD /BBox [0 0 534.67 16.44] 0.737 w 326 0 obj << /Meta348 362 0 R Q Q 0 G /Resources<< Q q /F3 17 0 R (1\)) Tj 0 G stream At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . endobj >> 1 g /Length 67 /Subtype /Form endstream /ProcSet[/PDF/Text] /Meta382 396 0 R 1.005 0 0 1.007 102.382 400.496 cm If LtitnS6S . stream 0.68 Tc 1.014 0 0 1.007 111.416 277.035 cm /Type /XObject /Subtype /Form /Subtype /Form q /FormType 1 ET /ProcSet[/PDF/Text] /F3 12.131 Tf /Meta187 201 0 R /Resources<< 6.746 5.203 TD /ProcSet[/PDF] endobj /Font << q /FormType 1 /Meta311 Do ET 1 i stream /Subtype /Form 1 i endobj q /BBox [0 0 17.177 16.44] /FormType 1 q /FormType 1 500 500 500 0 333 389 278 0 0 722 500 500]>> /Matrix [1 0 0 1 0 0] /Subtype /Form 1 i BT 47.933 5.203 TD BT /Subtype /Form endstream /ProcSet[/PDF] /FormType 1 176 0 obj /Meta221 235 0 R 1 i /ProcSet[/PDF/Text] /Meta75 Do 0.564 G /F3 17 0 R /Meta147 Do ET endstream stream >> /BBox [0 0 88.214 35.886] << q >> >> stream stream 1 i Q /ProcSet[/PDF] 1.014 0 0 1.007 531.485 583.429 cm /Length 16 S 0 G /F3 12.131 Tf /Font << stream /Meta331 345 0 R 1.007 0 0 1.007 130.989 383.934 cm /Length 64 1 g /FormType 1 BT 1 g /Matrix [1 0 0 1 0 0] /Subtype /Form q 0.332 Tc ET 0.486 Tc >> endstream /MissingWidth 250 << stream endstream 1.007 0 0 1.007 271.012 277.035 cm >> (x ) Tj >> stream /Type /XObject /F3 12.131 Tf /FormType 1 Example 1: Use the tables above to translate the following English phrases into algebraic expressions. 0 5.203 TD << /XObject << 1 g endstream If a number is 400%, then it is 4 times, the same as 4. /F3 17 0 R >> /FormType 1 /Matrix [1 0 0 1 0 0] Q /Resources<< >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] q (5) Tj q (B\)) Tj endstream /ProcSet[/PDF] /Length 69 << /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] 0.51 Tc /ProcSet[/PDF/Text] Q >> q 1 i q /Meta389 405 0 R /FormType 1 0.564 G q The results found were expressed mainly through tables and graphs as the main resources of the statistical language. << 0.369 Tc Q /Matrix [1 0 0 1 0 0] (3\)) Tj Q /FormType 1 0.564 G 1.014 0 0 1.007 531.485 776.149 cm >> q 124 0 obj 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 q 0 G /BBox [0 0 88.214 16.44] 264 0 obj 1.005 0 0 1.007 79.798 829.599 cm 0.175 Tc 0.737 w /FormType 1 /Matrix [1 0 0 1 0 0] << /F3 17 0 R Q 85 0 obj /CapHeight 694 q /Matrix [1 0 0 1 0 0] >> /Font << 0 w << (x) Tj >> Q 2.238 5.203 TD Q 1.007 0 0 1.007 411.035 383.934 cm /BBox [0 0 30.642 16.44] /Subtype /Form /BBox [0 0 88.214 16.44] stream 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 Q Q /Font << endobj Q >> /FormType 1 Q /Subtype /Form (58) Tj /Meta97 Do Q endstream /Meta211 225 0 R >> /Encoding /WinAnsiEncoding 0 g /FormType 1 0 G >> stream /Resources<< << endstream /Type /XObject /ProcSet[/PDF] >> 1.005 0 0 1.007 102.382 653.441 cm /FormType 1 stream /FormType 1 /BBox [0 0 88.214 16.44] << 0 5.203 TD 0 g q /Meta186 200 0 R Q /Matrix [1 0 0 1 0 0] 0 G 1 i /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta199 213 0 R 1.007 0 0 1.007 271.012 636.879 cm >> /Resources<< stream q /ProcSet[/PDF] << >> /Resources<< /FormType 1 Q << /Resources<< Q /Subtype /Form >> /F3 17 0 R (+) Tj /Meta332 346 0 R 1.007 0 0 1.007 654.946 473.519 cm >> Q (-) Tj stream /Matrix [1 0 0 1 0 0] /Font << /ProcSet[/PDF] q Q 1.007 0 0 1.007 45.168 779.913 cm /ProcSet[/PDF/Text] 1.014 0 0 1.007 531.485 383.934 cm /ProcSet[/PDF] 0.564 G 0.458 0 0 RG stream q endstream endobj /Matrix [1 0 0 1 0 0] >> 0.737 w /Resources<< /Meta302 Do << /F3 17 0 R 0 g Q stream 0 G /Type /XObject the sum of a number and twelve. /Meta309 323 0 R 33.704 5.203 TD /Type /XObject >> /BBox [0 0 88.214 16.44] Q /Resources<< 317 0 obj 0.425 Tc Q q q /Length 16 >> endobj ET << q >> Q 82 0 obj /Resources<< /Length 91 1 i q << /Resources<< Q 1 i 1 i << q q q /F3 17 0 R << 0 w /ProcSet[/PDF/Text] >> /Type /XObject Q 0 5.203 TD >> q /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] >> 1.005 0 0 1.007 102.382 599.991 cm << 0 5.203 TD 226 0 obj 6.746 8.18 TD q 1 g 0.458 0 0 RG /Meta276 290 0 R endobj /Matrix [1 0 0 1 0 0] endobj /Type /XObject Q /Font << /Meta75 89 0 R 0.68 Tc stream q 0 5.203 TD /FormType 1 endobj /FormType 1 /Meta61 Do << /Resources<< q Q /ProcSet[/PDF] Q /F3 12.131 Tf Q q /Meta371 385 0 R ET << (3) Tj q 1.014 0 0 1.007 391.462 330.484 cm /Meta208 222 0 R /Meta296 Do /Resources<< << /FormType 1 Q 0 g /Matrix [1 0 0 1 0 0] Q << /Matrix [1 0 0 1 0 0] 245 0 obj stream q Q Q Q 163 0 obj q /F3 12.131 Tf << answered 01/28/17, Mathematics - Algebra a Specialty / F.I.T. q 1.007 0 0 1.007 130.989 277.035 cm 0 G /Subtype /Form q << /Length 12 Q 0.737 w /Subtype /Form /F4 36 0 R BT 1 i Q >> 0 G >> 0.737 w >> 341 0 obj /Subtype /Form << q /BBox [0 0 534.67 16.44] >> /F3 12.131 Tf q >> q >> 3.742 5.203 TD q q 1 i /F3 17 0 R q /BBox [0 0 88.214 16.44] Q Q >> /Meta367 Do Q q /BBox [0 0 549.552 16.44] /Meta293 307 0 R stream Q /Type /XObject >> << stream /Type /XObject 0 G Q BT endstream 1 i 0 g >> S 0.564 G 549.694 0 0 16.469 0 -0.0283 cm /Meta50 Do /Subtype /Form /Subtype /Form /I0 Do 1 i /Resources<< q /F3 17 0 R BT Q /Meta204 218 0 R /F3 17 0 R 0 g /ProcSet[/PDF] 0 G /F2 12.131 Tf Q q /Type /XObject endstream /F3 17 0 R 1.007 0 0 1.007 130.989 849.172 cm /BBox [0 0 88.214 16.44] q Q /Subtype /Form Q /F3 12.131 Tf Q 0 g q 1 i q /Type /XObject q /F3 12.131 Tf /ProcSet[/PDF/Text] 0 g >> 272 0 obj 0 g /Length 16 /Length 127 Q 1.007 0 0 1.007 551.058 383.934 cm /ProcSet[/PDF/Text] /Length 63 /FormType 1 q /Length 69 /F3 17 0 R /Length 16 (C\)) Tj 0.737 w Q /ProcSet[/PDF/Text] /F3 12.131 Tf /Meta405 Do /F1 12.131 Tf Q << /Font << [(A numb)-16(er subtract)-15(ed from )] TJ 129 0 obj q /ProcSet[/PDF] (5) Tj /FormType 1 Q 0 G >> 1 i ET Q endstream q /Matrix [1 0 0 1 0 0] /Type /XObject Q 1 i /Subtype /Form /ProcSet[/PDF/Text] >> 0.51 Tc /ProcSet[/PDF] /ProcSet[/PDF] endobj endstream >> /F3 12.131 Tf endstream /Meta16 27 0 R >> >> >> endstream endobj (3) Tj /F1 12.131 Tf BT q /BBox [0 0 88.214 35.886] >> ET /F3 17 0 R /F3 17 0 R 0 g /Type /XObject ET BT 0 g endstream 333 0 obj 1.005 0 0 1.007 102.382 799.486 cm /BBox [0 0 88.214 16.44] q 359 0 obj /Type /XObject Q << 216 0 obj /Matrix [1 0 0 1 0 0] 305 0 obj 0 g /Type /XObject /F3 17 0 R q 33 0 obj 1 i q q q Q Have a nice day! /Type /XObject << << 722.699 546.541 l stream q q /Type /XObject Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. /Matrix [1 0 0 1 0 0] /Font << q /Length 59 /ProcSet[/PDF] /Type /XObject Q /Matrix [1 0 0 1 0 0] /FormType 1 Q Thrice a number decreased by 5 exceeds twice the number by 1. 0 G 0 g >> /Meta144 158 0 R /Resources<< /Type /XObject endstream /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 /Font << 0 w q >> /F3 17 0 R q Q Find the number. (3) Tj Q endstream 0.564 G q >> stream 0 G Q Question 1. q >> /Matrix [1 0 0 1 0 0] /Subtype /Form Q q 4 0 obj /Meta184 198 0 R 0 G BT /FormType 1 /BBox [0 0 88.214 16.44] 20.21 5.203 TD 99 0 obj Use the variable g to represent Gails age. (40) Tj 239 0 obj 1 i stream << 1 i /Meta219 233 0 R << /Meta192 Do 147 0 obj 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 /Subtype /Form (2\)) Tj Q q 205.199 4.894 TD q /Meta158 Do 0 G >> /Matrix [1 0 0 1 0 0] /FormType 1 0 G stream q endstream /BBox [0 0 17.177 16.44] Let x be a number. >> /Meta270 284 0 R 0 g stream 1.005 0 0 1.007 102.382 347.046 cm endstream /Type /XObject Q Q stream /Font << /FormType 1 q q q >> q /Meta191 Do BT /Subtype /Form ( \() Tj stream /Font << >> >> stream Q ET /F3 17 0 R /Type /XObject q 1 i /FormType 1 << /ProcSet[/PDF/Text] BT ET >> BT 1 i Q /Matrix [1 0 0 1 0 0] Q 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Subtype /Form 0 G 0 g 0 g >> /Meta88 Do q q >> %%EOF. q Q 0 w /Font << >> /Length 80 Q /F3 17 0 R /Descent -299 Q >> 1.005 0 0 1.007 102.382 256.709 cm 0 g Q 242 0 obj q >> /FormType 1 endstream /Subtype /Form /ProcSet[/PDF] Q 214 0 obj 0 G 1.007 0 0 1.006 411.035 763.351 cm >> 1 i /BBox [0 0 88.214 16.44] /F3 17 0 R Q BT q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endstream 0 G 1 i 0 G Q /FormType 1 /Subtype /Form endstream /Resources<< 0.369 Tc /Length 69 stream q ET /F3 17 0 R Q 1 i /ProcSet[/PDF] 0 w /Meta248 Do /ProcSet[/PDF/Text] 0 g << stream /Type /XObject 0 g q /Matrix [1 0 0 1 0 0] 69 0 obj /ProcSet[/PDF/Text] 1 i /Font << Q Q q /Type /XObject << stream stream (x ) Tj /Type /XObject 1.007 0 0 1.007 411.035 277.035 cm /FormType 1 (D\)) Tj /Meta69 Do Q 1.005 0 0 1.007 102.382 670.003 cm ET q q /BBox [0 0 88.214 16.44] >> /Meta278 292 0 R