0.564 G Q Q Q >> /Subtype /Form /F3 17 0 R 1 i 0 g q stream Q q /FormType 1 /FormType 1 You can also contact the clerk of court in the county you received the ticket. /ProcSet[/PDF] Q endobj S endobj ET Q endstream /FormType 1 endstream /Matrix [1 0 0 1 0 0] q 209 0 obj << BT /Resources<< /BBox [0 0 534.67 16.44] stream BT stream 20.21 5.203 TD /Type /XObject /Resources<< /F3 12.131 Tf endstream Thrice a number decreased by 5 exceeds twice the number by a unit. /Length 63 /StemH 77 /Font << << BT /Matrix [1 0 0 1 0 0] /Resources<< q /Meta71 85 0 R /Matrix [1 0 0 1 0 0] q /Subtype /Form q endstream 1.005 0 0 1.007 102.382 799.486 cm /Matrix [1 0 0 1 0 0] Q 0.369 Tc 1 i (+) Tj 0 w Q BT /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q 0 g 1 i Q /Length 16 /Meta34 Do /Resources<< /Meta83 97 0 R Q << >> /Resources<< endstream << << /Meta11 Do Q /BBox [0 0 30.642 16.44] /F4 36 0 R /FormType 1 /Length 12 q /Length 70 /Filter [/CCITTFaxDecode] 89 0 obj stream q /ProcSet[/PDF/Text] /Resources<< /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] Q 0.738 Tc >> endstream BT q q /Resources<< /Length 87 1 i 0.369 Tc >> /F1 14.682 Tf Q q 0 g endobj q /Meta14 Do /F3 12.131 Tf /Meta312 Do /F3 12.131 Tf /Length 245 Q /Meta152 Do >> /Subtype /Form q /Length 16 1 i /FormType 1 549.694 0 0 16.469 0 -0.0283 cm BT >> /Meta280 Do /FormType 1 1.007 0 0 1.007 67.753 347.046 cm -0.041 Tw >> 0 g 26.219 5.336 TD /Type /XObject q >> /Length 12 >> endobj q >> /BBox [0 0 30.642 16.44] ET /Length 65 0 g Q q /Type /XObject 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 q q 0 g 1.007 0 0 1.006 130.989 437.384 cm BT endstream 1.007 0 0 1.007 271.012 330.484 cm 0 G /XObject << >> 1 i (C) Tj 0.458 0 0 RG BT /Meta405 Do /Subtype /Form 20.21 5.203 TD /F3 17 0 R Q 0.458 0 0 RG 1 i 1 i (1\)) Tj /Type /XObject 1 g BT 0.369 Tc q >> /Resources<< /Meta38 52 0 R /Length 294 0 5.203 TD /Subtype /Form /Meta98 Do /Length 16 Q q Q >> /FormType 1 /Type /XObject /F3 12.131 Tf 1.007 0 0 1.007 551.058 330.484 cm /Type /XObject endstream endstream 0.458 0 0 RG q q >> /F4 12.131 Tf 0 g /Font << q q /Subtype /Form BT 1 i /BBox [0 0 88.214 16.44] >> << /F3 17 0 R /Type /XObject BT /Meta413 429 0 R /Resources<< /ProcSet[/PDF/Text] /Subtype /Form 0 g /F3 17 0 R /Matrix [1 0 0 1 0 0] 20.21 5.203 TD q 0 g 0 G >> 0.458 0 0 RG /Type /XObject Q /ProcSet[/PDF/Text] 0.155 Tc Q /FontName /PalatinoLinotype-Roman Answer link. >> Q 191 0 obj /Meta219 Do /Resources<< /Font << Q /I0 51 0 R endobj /Meta158 Do 0 G << 1.014 0 0 1.007 531.485 277.035 cm Q >> /ProcSet[/PDF/Text] q /Subtype /Form 0 w << /Meta394 410 0 R /ProcSet[/PDF] stream 1 i Q /FormType 1 /Resources<< q BT q /Length 59 0 G endstream /F4 12.131 Tf Twice a number decreased by 58! Q /ProcSet[/PDF/Text] Q /BBox [0 0 15.59 16.44] Q /Subtype /Form /Matrix [1 0 0 1 0 0] << Q /Font << /BBox [0 0 88.214 35.886] /BBox [0 0 30.642 16.44] >> >> /FormType 1 >> /Length 59 /Matrix [1 0 0 1 0 0] q /Length 69 q ET /Meta429 Do /Font << 0 g /F3 17 0 R endstream /ProcSet[/PDF] /FormType 1 32.939 5.203 TD ET endstream /FormType 1 /FormType 1 31 0 obj 0.458 0 0 RG << endstream /Meta21 32 0 R /ProcSet[/PDF/Text] >> Q /Subtype /Form /F3 17 0 R 1.014 0 0 1.006 391.462 437.384 cm endstream endobj 1.005 0 0 1.006 45.168 879.284 cm /F1 7 0 R 125.064 4.894 TD 379 0 obj /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] 0 G 0 g /Subtype /Form 0.564 G 299 0 obj q Q /Resources<< q 411 0 obj 1 i 0 G /Matrix [1 0 0 1 0 0] /Resources<< >> Q 27.693 5.203 TD /Font << 0 4.894 TD Q q >> >> 128 0 obj endstream endobj /Resources<< >> /Subtype /Form Next, the problem says that "x" would be equal to twice a number added by 5. stream /Meta90 Do /Meta134 148 0 R q endobj 0 5.203 TD /ProcSet[/PDF] /Length 68 0 G /Matrix [1 0 0 1 0 0] endstream stream q stream >> 240 0 obj /F3 12.131 Tf << Q 0 G 1.005 0 0 1.007 102.382 473.519 cm 1 i /Meta372 386 0 R endstream 304 0 obj /Matrix [1 0 0 1 0 0] BT Q /Meta353 367 0 R endstream 1.007 0 0 1.007 654.946 872.509 cm 0 g >> 205.199 4.894 TD /Type /XObject 0.486 Tc Q /Meta105 Do endstream 0 g /Type /XObject /Length 118 1.005 0 0 1.007 79.798 796.475 cm 1 i >> q << /Font << 264 0 obj /Matrix [1 0 0 1 0 0] 0 G stream /Type /XObject -0.486 Tw /Type /XObject >> 393 0 obj Q /Length 16 /FormType 1 /BBox [0 0 15.59 16.44] >> 0 G 0.564 G /Resources<< Q endstream Q Q /Length 118 q >> Q Twice a number decreased by ten is at least 24. endobj Q endstream 1.014 0 0 1.007 391.462 776.149 cm Q Find the number. /Subtype /Form Q q /F1 7 0 R /Resources<< /FormType 1 >> /BBox [0 0 17.177 16.44] >> /BBox [0 0 673.937 15.562] >> 27 0 obj /ProcSet[/PDF] 1 i /Length 54 /Meta342 356 0 R endobj /BBox [0 0 88.214 16.44] Q /Resources<< Q >> 0 w >> /Meta176 190 0 R 121 0 obj 0 w 0 g Q Q endstream Q /Meta33 Do 0 G /Subtype /Form /Meta47 61 0 R 1 g 20.21 5.203 TD /Meta332 346 0 R endobj Q Q stream ET /ProcSet[/PDF] /F3 17 0 R >> /Length 59 << /FormType 1 /BBox [0 0 639.552 16.44] /F3 12.131 Tf >> >> 101.849 5.203 TD BT stream (4) Tj /BBox [0 0 639.552 16.44] q /Type /XObject /Meta311 Do Q /BBox [0 0 88.214 16.44] << Q q Q << /F4 36 0 R >> Q /FormType 1 >> Thirthy is equal to twice a number decreased by four = solve and check the equation? /Matrix [1 0 0 1 0 0] Q stream stream << q /Resources<< ET q >> 1.007 0 0 1.007 130.989 383.934 cm stream >> /Meta122 Do (C\)) Tj /Type /XObject /Type /XObject /Subtype /Form /FormType 1 /FormType 1 /Resources<< /Meta4 Do ET /Matrix [1 0 0 1 0 0] endobj /Resources<< /Subtype /Form stream /Resources<< endobj ET Q The result is 8 less than 10 times the number. Q (+) Tj >> (2) Tj 1.007 0 0 1.007 271.012 277.035 cm Q /Meta280 294 0 R Q /Type /XObject q >> /Subtype /Form q q The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. q /Resources<< /Length 118 /Meta37 Do /F3 12.131 Tf q q /F1 12.131 Tf Q ET [(A numb)-16(er subtract)-15(ed from )] TJ BT 57.656 5.203 TD stream /Subtype /Form stream /F3 12.131 Tf 38.182 5.203 TD endstream q endobj 1.007 0 0 1.007 551.058 636.879 cm /ProcSet[/PDF] endobj Q q /Meta141 155 0 R /Subtype /Form (A\)) Tj /Subtype /Form /FormType 1 Q /Meta28 41 0 R S /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endstream >> Q Q >> /BBox [0 0 88.214 16.44] stream endobj 220.931 4.894 TD /Resources<< 9.723 5.336 TD >> 1.014 0 0 1.006 111.416 763.351 cm BT q q 0.458 0 0 RG stream /F4 36 0 R /Meta195 209 0 R << 0 G Q /ProcSet[/PDF/Text] endstream 0.564 G Q endobj /Resources<< endobj /Type /XObject 2.238 5.203 TD Q /Length 16 /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 849.172 cm 65.906 4.894 TD >> /Matrix [1 0 0 1 0 0] endobj endstream /Length 16 The results found were expressed mainly through tables and graphs as the main resources of the statistical language. /Meta386 Do 0.737 w Notice that we used the variable \large {d} d in our equation to stand for our unknown value. /BBox [0 0 88.214 16.44] >> /ProcSet[/PDF] /Resources<< q /F3 12.131 Tf 349 0 obj /Resources<< q /F3 17 0 R 175 0 obj 1 i 0 G 0 g q 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. /Subtype /Form >> /Meta12 Do Q /Meta135 Do 161 0 obj /Meta4 13 0 R >> Q stream >> 146 0 obj 205 0 obj 0 w stream /BBox [0 0 534.67 16.44] stream << /Meta367 Do /ProcSet[/PDF/Text] q 189 0 obj /F1 7 0 R >> Q /Meta235 Do >> /Meta389 Do 0 5.203 TD Q Q 1 i /F3 17 0 R q /Type /XObject /FormType 1 /BBox [0 0 88.214 16.44] 10 0 obj q /F3 12.131 Tf /BBox [0 0 88.214 16.44] 0 g /Meta218 Do /Length 16 >> /Subtype /Form 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 << /Length 12 >> q endobj q /Type /XObject q /Meta261 Do /Meta149 163 0 R endobj 1.014 0 0 1.007 391.462 277.035 cm Q >> endstream ( x) Tj 0 g 0.564 G >> stream /Meta385 Do 0 g >> Q /Type /XObject 1 i /FormType 1 /Resources<< 1 i q 1.007 0 0 1.007 271.012 330.484 cm << /ProcSet[/PDF/Text] 1.014 0 0 1.007 531.485 330.484 cm stream 3.742 5.203 TD Q /F3 12.131 Tf q /Type /XObject 224 0 obj endobj 0.564 G /Subtype /Form BT endobj Q 284 0 obj Q q /FormType 1 /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 251.439 849.172 cm endobj endobj **Note: You could choose any variable you want. 266 0 obj /Meta48 62 0 R /Meta398 Do 1 i >> 1 i endstream 1 i ET /Resources<< q 0 4.78 TD /BBox [0 0 88.214 16.44] /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 /Subtype /Form /Meta60 74 0 R /FormType 1 /Meta270 Do /F3 12.131 Tf q >> B. Ten divided by a number 5. /Subtype /Form endobj BT /Meta410 426 0 R /BBox [0 0 88.214 16.44] q Q q q /FormType 1 0 5.203 TD /Matrix [1 0 0 1 0 0] 0.458 0 0 RG Q Q /FontDescriptor 16 0 R stream /Type /XObject 1.014 0 0 1.007 531.485 523.204 cm /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 653.441 cm 1.014 0 0 1.006 531.485 510.406 cm endstream /Meta424 Do q /Meta224 Do Q 1.007 0 0 1.007 130.989 277.035 cm >> /ProcSet[/PDF/Text] /Length 59 endobj (5\)) Tj /FormType 1 Q q stream stream 356 0 obj q 198 0 obj 1 i /Resources<< (x ) Tj >> /XHeight 476 q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 523.204 cm /ProcSet[/PDF/Text] /FormType 1 1 i Q 1.005 0 0 1.007 102.382 599.991 cm 154.289 4.894 TD 1.007 0 0 1.007 45.168 846.161 cm /ProcSet[/PDF/Text] >> q a and b or something else.***. BT /Matrix [1 0 0 1 0 0] << Q endstream /Meta141 Do /Meta139 Do 250 0 obj stream Q Q /Resources<< /Matrix [1 0 0 1 0 0] q 0 g 1 i >> >> /Matrix [1 0 0 1 0 0] endobj >> ET /Meta290 Do 0 G 1 i 1.005 0 0 1.007 102.382 293.596 cm In the problem above, x is a variable. /Resources<< endstream 0 g /FormType 1 << /BBox [0 0 88.214 16.44] 1.007 0 0 1.006 411.035 510.406 cm stream 1.007 0 0 1.007 130.989 330.484 cm /Meta243 257 0 R BT Q 1 i /F3 17 0 R /Subtype /Form >> endstream A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. >> /BBox [0 0 30.642 16.44] /FormType 1 /Meta354 368 0 R -0.486 Tw endstream /BBox [0 0 88.214 16.44] 0.458 0 0 RG << 0 g /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q /Subtype /Form endstream q BT Q /BBox [0 0 88.214 16.44] stream /Length 69 /Matrix [1 0 0 1 0 0] << Q /Type /XObject 1 i /Length 69 1 i /Meta341 355 0 R Q /Font << 1.007 0 0 1.007 271.012 849.172 cm >> stream /FormType 1 << /ProcSet[/PDF/Text] 1 i Q /Type /XObject /F1 12.131 Tf /ProcSet[/PDF] 1.014 0 0 1.006 251.439 510.406 cm /Subtype /Form /BBox [0 0 673.937 68.796] /Type /XObject /Length 16 q /BBox [0 0 88.214 16.44] 226 0 obj ET /BBox [0 0 15.59 16.44] /Subtype /Form 0 g /Meta173 Do BT Q 0 g /ProcSet[/PDF/Text] /Type /XObject q 385 0 obj /Type /XObject

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