q /Font << 1.007 0 0 1.007 551.058 383.934 cm /Type /XObject 1 i 0 w BT 0 w q /Meta189 Do >> 549.694 0 0 16.469 0 -0.0283 cm << >> 0 w /Meta395 411 0 R >> /Subtype /Form q 1.007 0 0 1.007 654.946 799.486 cm Q endstream /Meta165 Do endstream q /Meta26 39 0 R /Meta66 80 0 R 0 G /Meta70 84 0 R Q q 0 g q >> 0 g /Matrix [1 0 0 1 0 0] Find the number. stream /Subtype /Form stream BT /Length 16 q /Meta357 371 0 R /Meta150 164 0 R BT endobj endstream /Type /XObject /F3 12.131 Tf Q /Meta101 Do /Subtype /Form /Length 16 endobj /BBox [0 0 639.552 16.44] 131 0 obj /ProcSet[/PDF] q /Meta173 Do ET ET q /BBox [0 0 88.214 16.44] /FormType 1 /F3 12.131 Tf q stream 0 g >> /BBox [0 0 15.59 16.44] stream endobj ET 0.564 G /F3 17 0 R Q endobj 431 0 obj /Meta3 12 0 R 0.17 Tc endobj (-23) Tj /Length 60 /F3 12.131 Tf Q /Meta195 209 0 R /Type /XObject /F3 17 0 R >> stream /Meta144 158 0 R endobj endobj endobj /Length 69 q /Subtype /Form /F1 7 0 R >> /F3 17 0 R /Resources<< q /Length 118 /ProcSet[/PDF] /FormType 1 /BBox [0 0 88.214 16.44] 1 i /BBox [0 0 17.177 16.44] /Meta189 203 0 R ET /Meta163 177 0 R 295.086 4.894 TD /Type /XObject /F3 12.131 Tf 350 0 obj 25 0 obj /Subtype /Form /F3 12.131 Tf [( times )15(a numb)22(er and )] TJ /Type /XObject 1.005 0 0 1.007 102.382 726.464 cm endstream << BT 0.564 G /FormType 1 q /Resources<< >> 0.458 0 0 RG /Resources<< BT q >> 0 w /Subtype /Form /Resources<< /Font << q /ProcSet[/PDF/Text] 0.737 w /FormType 1 0 g 0 G /BBox [0 0 30.642 16.44] >> /Meta68 82 0 R /FormType 1 1.007 0 0 1.007 130.989 330.484 cm >> 1.007 0 0 1.007 45.168 862.723 cm /Font << 0 g 1.014 0 0 1.006 531.485 690.329 cm >> /FormType 1 q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 411.035 383.934 cm 1 i 1.014 0 0 1.007 251.439 849.172 cm q Decreased by another number means subtract. 146 0 obj (-20) Tj /Meta18 29 0 R >> /FormType 1 endstream q Q >> /Meta395 Do endobj 0.369 Tc 0 w >> /Subtype /Form /Meta322 Do endobj /Subtype /Form endobj 16.469 5.336 TD /Type /XObject /Type /XObject q /Type /XObject >> Q 0.786 Tc << << A number divided by six is eight: (k / 6) = 8. /Subtype /Form endstream /Font << Q Q 0 G /Matrix [1 0 0 1 0 0] Expression. q /BBox [0 0 549.552 16.44] ET /FontName /TimesNewRomanPSMT Q (4\)) Tj stream ET >> 0.175 Tc (B\)) Tj 1 i << /Length 59 << /Subtype /Form 1 g /Type /XObject /BBox [0 0 534.67 16.44] /Resources<< 1 i /Meta375 389 0 R ET >> BT /Resources<< /Meta327 Do q Q /Length 12 /F3 12.131 Tf 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. 0.564 G Q /Matrix [1 0 0 1 0 0] Q /Subtype /Form 1.007 0 0 1.007 271.012 277.035 cm >> /Length 12 Q BT 318 0 obj /Meta339 Do 1 i 400 0 obj /FormType 1 /BBox [0 0 88.214 35.886] Twice a first number decreased by a second number is 6. (40) Tj >> Q endobj q 1 i (-) Tj 1.005 0 0 1.007 102.382 872.509 cm /F1 7 0 R 437 0 obj endstream 1.014 0 0 1.006 111.416 690.329 cm /Matrix [1 0 0 1 0 0] endstream 0 g 410 0 obj /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject Q 422 0 obj 1 i >> /Meta394 410 0 R Q endobj /FormType 1 /Subtype /Form q Q [(Negativ)16(e )] TJ >> /Type /XObject /Type /XObject /Subtype /Form /F3 17 0 R /Meta141 Do q /BBox [0 0 534.67 16.44] Q >> 1.014 0 0 1.007 111.416 330.484 cm /Font << 33 0 obj 0 G 173 0 obj 0 g (-) Tj "49 . stream endobj /Type /XObject /Matrix [1 0 0 1 0 0] >> 1 i stream 1.014 0 0 1.007 111.416 636.879 cm 0.369 Tc Q >> /Resources<< 1.007 0 0 1.007 45.168 779.913 cm stream (-) Tj /BBox [0 0 88.214 16.44] /Meta153 167 0 R Q /F1 14.682 Tf /ProcSet[/PDF/Text] endstream 1 i 0 g Q /Meta166 180 0 R endobj /ProcSet[/PDF/Text] 0 G Q 0 G 0.486 Tc Q >> /ProcSet[/PDF] 1 i /BaseFont /PalatinoLinotype-Bold << (-) Tj /FormType 1 >> << endstream /FormType 1 q 0 20.154 m 20.21 5.203 TD 1 i /Length 69 /FormType 1 q /Meta256 Do 0 56.451 TD BT /Resources<< /Resources<< q 0 5.203 TD q /Resources<< endstream Q /ProcSet[/PDF] /F4 36 0 R >> /Font << >> q Q q /FormType 1 Q 1 i -0.099 Tw 0 w 170 0 obj 0.458 0 0 RG endobj 0.564 G /Meta149 163 0 R /Subtype /Form /F3 17 0 R q /Subtype /Form 2.238 5.203 TD 0.737 w Q >> BT /F4 36 0 R 1.007 0 0 1.007 271.012 383.934 cm 0.51 Tc /F1 12.131 Tf 1.007 0 0 1.007 551.058 583.429 cm 1 i q q 0 G 0.486 Tc ET /Length 118 0 g /BBox [0 0 15.59 29.168] /Font << Q Q stream ET endobj Q /F4 36 0 R /Resources<< 1.007 0 0 1.007 551.058 636.879 cm 212 0 obj >> /F3 17 0 R >> q /Resources<< /F3 17 0 R endobj /ProcSet[/PDF] /F1 7 0 R /BBox [0 0 15.59 29.168] /Resources<< << >> /FormType 1 /Length 60 412 0 obj stream >> 64 0 obj /FormType 1 1.007 0 0 1.007 411.035 277.035 cm If n is "the number," which equation could be used to solve for the number? /ProcSet[/PDF/Text] q q Q /Resources<< /BBox [0 0 17.177 16.44] ET /FormType 1 endobj 1.014 0 0 1.007 111.416 330.484 cm BT /Matrix [1 0 0 1 0 0] q /F3 17 0 R /BBox [0 0 88.214 16.44] 1.005 0 0 1.007 79.798 796.475 cm /Meta237 Do /Subtype /Form 1.502 5.203 TD << /Meta274 Do q 25.454 5.203 TD /Encoding /WinAnsiEncoding Q /Resources<< q 1.005 0 0 1.007 102.382 293.596 cm /XObject << BT 500 500 500 0 333 389 278 0 0 722 500 500]>> 182 0 obj q ET endobj endstream endstream q << 0.737 w q /Type /XObject 3.742 5.203 TD ET >> >> /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0.458 0 0 RG 0 G /Meta115 129 0 R /F3 12.131 Tf Q >> 0.458 0 0 RG /Subtype /Form Q q Q /FormType 1 >> 0 G /FormType 1 /FormType 1 endobj Q Q /F1 12.131 Tf /Type /XObject 1 i q /FormType 1 Q You could call them. Q 1.014 0 0 1.007 251.439 383.934 cm /Matrix [1 0 0 1 0 0] 0.68 Tc Q /Type /XObject q q SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. 0 G Q /Subtype /Form 0 G 30.699 5.203 TD endstream Q /Type /XObject ET /F3 17 0 R /F3 17 0 R endstream /Type /XObject -0.041 Tw /Length 59 /Resources<< q >> /Meta289 303 0 R endobj q /BBox [0 0 549.552 16.44] /ProcSet[/PDF/Text] q BT q /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 111.416 583.429 cm >> 1.014 0 0 1.007 531.485 330.484 cm q >> q << /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1 i /Type /XObject /Meta34 47 0 R 1.007 0 0 1.007 411.035 583.429 cm Q q /Length 69 1 g >> 0.564 G /Resources<< /Resources<< Q << q Percent Change = (Decrease First Value) x 100% 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 >> 0 g q /Length 67 Q 1 i /ProcSet[/PDF/Text] q /FormType 1 /Meta219 233 0 R Q 9.723 5.336 TD Q /Meta15 26 0 R 0 w BT Hence, the number is 6. 1 g /Meta224 238 0 R 1 i BT Q Q 1.502 5.203 TD q 1.005 0 0 1.007 79.798 862.723 cm /Meta197 Do endstream /LastChar 121 0 g 389 0 obj /Type /XObject 1 i ET /Length 65 q 1 i /StemH 77 0 G /ProcSet[/PDF/Text] << 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 /BBox [0 0 30.642 16.44] /Length 16 q 0.311 Tc /F3 17 0 R 0 g ET 1.005 0 0 1.013 45.168 933.487 cm 0 g Q 166 0 obj q /Length 59 /Matrix [1 0 0 1 0 0] /Meta298 Do Q endstream /Font << endstream /Font << >> /Meta405 421 0 R stream 0.737 w 0 5.203 TD /ProcSet[/PDF/Text] /Meta164 Do /Length 294 0 g Q /F3 12.131 Tf 92 0 obj /Meta265 279 0 R /Font << 0.564 G /Type /XObject 257 0 obj /Meta175 189 0 R 41 0 obj 0 g /ProcSet[/PDF/Text] Q 1 i 0 g q Q 7 0 obj /Length 16 /Length 16 << 0.564 G Q /Font << /FormType 1 S stream /Length 12 >> /Resources<< << Q << q << /Length 78 1.007 0 0 1.007 551.058 277.035 cm endstream /ProcSet[/PDF/Text] >> Q /FormType 1 /BBox [0 0 30.642 16.44] /Subtype /Form >> Q << 0.458 0 0 RG 1.005 0 0 1.007 79.798 862.723 cm /ProcSet[/PDF/Text] /Type /XObject >> 0 w /FormType 1 /FormType 1 stream 1 i 9.723 5.336 TD /Font << endobj /Type /XObject stream 1 i /FormType 1 Twice = two times, double. 332 0 obj /F3 17 0 R Q /Meta364 378 0 R Q /Type /XObject /Meta0 5 0 R (B) Tj >> 12.727 5.203 TD /Font << q /Meta138 Do /Meta161 175 0 R 145 0 obj /BBox [0 0 30.642 16.44] << 0 g /ProcSet[/PDF/Text] 0 g /F3 12.131 Tf /Meta250 Do Q q /Font << 0 g 1 g endobj stream /Resources<< /Resources<< 1.014 0 0 1.007 111.416 636.879 cm (x ) Tj q >> << /ID [] Q Q /Matrix [1 0 0 1 0 0] /Type /XObject Q BT endobj /BBox [0 0 534.67 16.44] Q 1 i 0 g q ET /Type /XObject /F3 12.131 Tf /BBox [0 0 88.214 16.44] Q /Meta397 Do q >> /Type /XObject q /Type /XObject /Type /XObject Q endobj Q /Meta206 Do << Q 42 0 obj /Meta99 Do /Resources<< Q 0 5.203 TD /Meta418 434 0 R Ten divided by a number 5. >> ET >> BT endstream endstream /Matrix [1 0 0 1 0 0] endobj 1.007 0 0 1.007 551.058 523.204 cm /BBox [0 0 88.214 16.44] 2.238 5.203 TD /F1 7 0 R /Meta7 Do /Meta0 Do q q >> >> endstream >> >> Q 1 g Q /Type /XObject stream /Matrix [1 0 0 1 0 0] >> >> Q /BBox [0 0 88.214 16.44] /Subtype /Form /Meta267 281 0 R /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] Q stream /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Meta115 Do /Meta263 Do /Subtype /Form 47.933 5.203 TD >> /Meta231 245 0 R /Contents [446 0 R] /Meta88 Do /Meta333 Do q >> q (x) Tj q /ProcSet[/PDF] 0 G Q << 0.425 Tc Q /Meta47 61 0 R 0 G Q 267 0 obj 0.458 0 0 RG /ProcSet[/PDF/Text] /Type /XObject q >> Q Q /Type /XObject endobj /Resources<< Q >> endstream >> /Type /XObject 1.007 0 0 1.007 411.035 636.879 cm /Type /XObject /FirstChar 43 << /FormType 1 /Type /XObject Q q /Font << 1.007 0 0 1.007 45.168 829.599 cm The sum of 18 and tour times a number is -6 Find the number. Q xref >> /Resources<< 20.21 5.203 TD Q endstream >> q 1 i Q q /Meta130 144 0 R q stream /Meta29 42 0 R /Subtype /Form /Font << q Solution: Let the number be x. 1.005 0 0 1.009 45.168 905.633 cm /Font << /Meta233 247 0 R endstream 1.005 0 0 1.007 79.798 713.666 cm endobj ET 0.737 w q 1.005 0 0 1.007 102.382 653.441 cm Q << [(MULTIPLE CHOICE. 1.005 0 0 1.007 102.382 473.519 cm endobj q >> stream endobj >> /ProcSet[/PDF/Text] [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. /Subtype /Form 1.007 0 0 1.007 67.753 546.541 cm Q Q /Type /XObject stream /ProcSet[/PDF/Text] 0 w /F3 17 0 R Q /F3 17 0 R 0.68 Tc 1 i /Matrix [1 0 0 1 0 0] Q 1 i << Three times a number equals fifteen 3. Q 1 g 1.005 0 0 1.007 79.798 813.037 cm 1 i stream /Meta125 Do Q q stream stream 0 g /F3 17 0 R /I0 Do /Resources<< /F1 12.131 Tf /Matrix [1 0 0 1 0 0] /Meta273 Do /Font << q /Resources<< endobj q /Length 69 /BBox [0 0 88.214 16.44] q Q /Resources<< /FormType 1 /FormType 1 /ProcSet[/PDF/Text] ET Q Q endstream /Meta131 145 0 R /Meta39 Do /ProcSet[/PDF/Text] q /Resources<< q /FormType 1 /Matrix [1 0 0 1 0 0] 0 g /F3 17 0 R 0.737 w q q stream /Resources<< /Length 16 110 0 obj stream ET Q << Q 0 5.203 TD 6.746 5.203 TD q /Meta55 Do 52 0 obj /Subtype /Form >> /Subtype /Form /F3 17 0 R /Meta221 Do >> /Font << q /ProcSet[/PDF] >> /FormType 1 << q /Type /XObject q >> >> q 0 G q 0 g 0 w /Length 16 BT Q endstream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /FormType 1 /Resources<< ET /Resources<< 1.007 0 0 1.007 271.012 636.879 cm 1 i 0 G endobj /F3 12.131 Tf /Subtype /Form >> /Font << stream /Meta384 Do The quotient of a seven and a number 9. q >> << /Font << q >> 0.564 G /Font << /Resources<< (D\)) Tj q /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF/Text] /Meta332 Do /Length 69 /Meta16 27 0 R 0.458 0 0 RG /Length 69 /Length 54 /ProcSet[/PDF/Text] 0 G >> endstream 0.737 w >> endobj /Font << q /FormType 1 Most questions answered within 4 hours. /Resources<< /Matrix [1 0 0 1 0 0] /F3 17 0 R ET Q >> /Meta160 174 0 R 0 20.154 m q 311 0 obj q (5) Tj 1 i 1.008 0 0 1.007 654.946 293.596 cm /Subtype /Form << endobj /Meta385 Do stream /F3 17 0 R /Type /XObject /Meta105 Do /ProcSet[/PDF/Text] /Subtype /Form /F3 17 0 R >> /Type /XObject /Meta271 285 0 R 0 g 0 w Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. << Q Q /Resources<< [(The )-19(quotient of )] TJ /ProcSet[/PDF/Text] /Meta372 386 0 R 0 g Q Q 1.005 0 0 1.007 102.382 599.991 cm /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] q q >> /ProcSet[/PDF/Text] /FirstChar 32 << /BBox [0 0 88.214 16.44] /Length 16 Q /Resources<< Q /Length 16 q /Subtype /Form /ProcSet[/PDF/Text] /Meta417 Do q /Length 12 q 1.007 0 0 1.007 130.989 383.934 cm endobj q /Resources<< 0 g 65 0 obj >> 11 0 obj ET stream /Subtype /Form Q /F3 17 0 R 1 g BT q q /Meta157 171 0 R >> /ProcSet[/PDF] Q 404 0 obj (D) Tj /BBox [0 0 15.59 16.44] /FormType 1 Q Q /Subtype /Form BT 293 0 obj 18.708 17.593 TD /FormType 1 /Length 58 Q &K @ Q /Meta29 Do 1.007 0 0 1.006 130.989 437.384 cm q Q /Meta180 194 0 R endobj >> 227 0 obj 1 i /ProcSet[/PDF/Text] /Resources<< /BBox [0 0 534.67 16.44] /Length 139 /Subtype /Form >> 100 0 obj >> 0 G /Resources<< >> 75 0 obj 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 . 0.737 w 314 0 obj /Resources<< /F1 7 0 R q q /Meta202 Do 0 G q /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] /FormType 1 Q 24 0 obj /Matrix [1 0 0 1 0 0] /Subtype /Form >> q BT 0.486 Tc /ProcSet[/PDF/Text] Q /F3 17 0 R /FormType 1 /Subtype /Form stream stream 1 i q /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] /Meta105 119 0 R /Matrix [1 0 0 1 0 0] 0.425 Tc << q /Meta318 332 0 R /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0.737 w (x ) Tj /Length 59 /BBox [0 0 30.642 16.44] Q Q BT ET /FormType 1 0.564 G 69 0 obj endobj 1 g /Matrix [1 0 0 1 0 0] /Length 58 Q /Meta370 Do /FormType 1 /BBox [0 0 549.552 16.44] /Type /XObject endstream stream 0 w 0 G /Subtype /Form 1 g 1 i >> /ProcSet[/PDF] /Length 16 endobj endobj 1.007 0 0 1.007 130.989 776.149 cm /Matrix [1 0 0 1 0 0] (B\)) Tj Q stream (C\)) Tj Q /F3 12.131 Tf /Type /XObject BT /Type /XObject 0 g 1 i >> ( x) Tj q /Resources<< /Subtype /Form ET /F3 17 0 R /F3 12.131 Tf 1.007 0 0 1.007 271.012 383.934 cm stream /BBox [0 0 30.642 16.44] q << /BBox [0 0 88.214 16.44] ET stream stream 1.007 0 0 1.007 67.753 653.441 cm /ProcSet[/PDF] 0.303 Tc Q /Meta404 420 0 R Q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] << BT >> Q q 0.737 w /ProcSet[/PDF/Text] 0.737 w /FormType 1 /Subtype /Form 1.014 0 0 1.007 531.485 849.172 cm /BBox [0 0 15.59 16.44] endstream /Subtype /Form << BT 1.005 0 0 1.007 102.382 799.486 cm 1 i q q (x) Tj /Length 59 1.014 0 0 1.006 531.485 510.406 cm /Length 59 0 g /Subtype /Form /Subtype /Form /Subtype /Form << /Meta4 Do stream Q /BBox [0 0 88.214 16.44] q Q >> /Matrix [1 0 0 1 0 0] /Type /XObject 0 w q 0.564 G q q Q 0.458 0 0 RG stream >> endstream Q 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 >> 0 G /ProcSet[/PDF] Q q /Font << /Resources<< 0.271 Tc stream 1 i /Matrix [1 0 0 1 0 0] Q 1 i stream Q 0 g [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ /FormType 1 >> /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) << 0.458 0 0 RG q /BBox [0 0 30.642 16.44] endstream -0.486 Tw /Font << /CapHeight 692 /Font << 59 0 obj 0 g Q /Resources<< q /F1 12.131 Tf 1 i /Matrix [1 0 0 1 0 0] /Meta163 Do endstream Q Making educational experiences better for everyone. << ET 1 g >> endstream /FormType 1 0 5.203 TD stream /F4 12.131 Tf 0.524 Tc /ProcSet[/PDF/Text] 0.564 G /Subtype /Form /F1 12.131 Tf /Subtype /Form 1.007 0 0 1.007 45.168 746.789 cm /Resources<< Q Q >> >> >> 1.005 0 0 1.007 102.382 726.464 cm Q 1.007 0 0 1.007 411.035 383.934 cm /FormType 1 q A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. /Subtype /Form Q 1.014 0 0 1.007 251.439 583.429 cm /Meta117 131 0 R /Resources<< ET /Meta419 435 0 R 1 i Q Q 0 g >> Q /BBox [0 0 549.552 16.44] 0 g BT 1.005 0 0 1.007 79.798 746.789 cm << endobj /Subtype /Form /Resources<< /Type /XObject /Type /XObject /Matrix [1 0 0 1 0 0] >> stream q ET 35.206 4.894 TD /FormType 1 ET ET /Meta154 168 0 R /Length 16 q /Subtype /Form /Type /XObject stream 0 g 321 0 obj 1.502 5.203 TD /Meta9 Do q /Length 16 0.458 0 0 RG /Meta182 196 0 R You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. (v) 5 subtracted from thrice a number is 16. << (-8) Tj q /F3 17 0 R endstream >> >> /Type /XObject << /F3 12.131 Tf q 411 0 obj Q /Meta59 73 0 R Q Q /FormType 1 >> q Q /Resources<< << 1.007 0 0 1.007 130.989 636.879 cm q q /Meta54 Do Q q 0 g BT q Q 0 g /Meta235 249 0 R 1.005 0 0 1.007 45.168 889.071 cm /ProcSet[/PDF] /Meta414 Do /Length 69 /ProcSet[/PDF/Text] Twice the difference of a number and three totals twelve 8. endobj 1.005 0 0 1.007 102.382 816.048 cm /FormType 1 0 w endobj Q Q /BBox [0 0 673.937 16.44] endobj 0.369 Tc q >> q Q /Meta156 Do 0 g Q 120 0 obj q /Font << Q Q /Type /Page endstream /Matrix [1 0 0 1 0 0] q endstream /Font << This site is using cookies under cookie policy . 207 0 obj stream 1.007 0 0 1.006 411.035 690.329 cm >> endstream Q /F3 12.131 Tf 0.564 G q /Length 57 0 20.154 m /Resources<< /ProcSet[/PDF/Text] /MissingWidth 250 << endstream /BBox [0 0 30.642 16.44] /Font << /Meta414 430 0 R /FormType 1 << /Subtype /Form q 1 i /Subtype /Form q >> /Matrix [1 0 0 1 0 0] 0 5.203 TD /Meta369 383 0 R >> << /Matrix [1 0 0 1 0 0] << /ProcSet[/PDF] /Type /XObject /BBox [0 0 17.177 16.44] Q q 0.458 0 0 RG endstream /F3 12.131 Tf /Meta408 424 0 R /F4 12.131 Tf Q 0 g /Resources<< q q q /Length 294 1.007 0 0 1.006 411.035 690.329 cm << >> 1 i 0 G Q 0.524 Tc q Let x be a number. q endobj 3.742 5.203 TD 549.694 0 0 16.469 0 -0.0283 cm /Meta247 261 0 R 0.564 G precision and actual right or wrong answers. 3.742 5.203 TD /Subtype /Form >> /Meta365 379 0 R 1.014 0 0 1.007 531.485 583.429 cm /Font << /Meta193 Do 0 G >> >> /Subtype /Form /F3 12.131 Tf ET 12.727 5.203 TD << /Subtype /Form 1 i 1.005 0 0 1.007 102.382 563.103 cm ET endstream >> 0 G endstream 1 i >> q q << 0.737 w endobj /Meta393 409 0 R /Length 65 1 i 0 g stream stream endstream stream endobj /F3 17 0 R 0.458 0 0 RG /ProcSet[/PDF] /FormType 1 /Meta147 Do Q q /Type /XObject /FormType 1 1 i /MissingWidth 250 /BBox [0 0 88.214 16.44] /Type /XObject 0 G 438 0 obj << q >> /Length 13 Q /Meta58 72 0 R 1 i q /Length 68 endobj 1.014 0 0 1.007 531.485 523.204 cm Q 0 g >> Q 20/n b.) /Length 58 >> /Type /XObject /CapHeight 694 stream 1 i 53 0 obj Q Twice a number is decreased by 9, and this sum is multiplied by 4. q /Meta132 146 0 R /ProcSet[/PDF] 1.005 0 0 1.007 102.382 799.486 cm q q 1 i Advertisement Loved by our community 50 people found it helpful Madhvendra13 2x -8=58 2x=66 x=662 x=33 Find Math textbook solutions? (\)) Tj /Resources<< 94.364 5.203 TD /BBox [0 0 15.59 16.44] decreased by A number decreased by twelve X - 12 subtracted from Six subtracted from a number X - 6 Multiplication ( x ) times Eight times a number 8x the product of The product of fourteen and a number 14x twice; double Twice a number; double a number 2x multiplied by A number multiplied by negative six 6x 0 g 0 w 1.014 0 0 1.007 111.416 383.934 cm /Matrix [1 0 0 1 0 0] endstream >> /Subtype /Form /F3 12.131 Tf 0 G 0 5.203 TD /FormType 1 /F4 12.131 Tf /Meta291 305 0 R 0.737 w q Q 0.155 Tc /Type /XObject /Length 74 /Subtype /Form /ProcSet[/PDF] q endstream 136 0 obj 0.564 G stream >> ET 1.007 0 0 1.007 130.989 636.879 cm /Subtype /Form /Length 69 /Resources<< q 1.014 0 0 1.007 251.439 277.035 cm >> /Matrix [1 0 0 1 0 0] Q stream /Resources<< /BBox [0 0 88.214 16.44] Q endstream /Length 16 /Subtype /Form 105 0 obj 0 w /Type /XObject /Type /XObject /Subtype /Form Q 0 5.203 TD /F3 17 0 R /Resources<< Q /Meta63 Do 0 g q 418 0 obj Q 0 G /Matrix [1 0 0 1 0 0] << 1.007 0 0 1.007 130.989 776.149 cm stream endobj /Meta195 Do /Subtype /Form >> /F3 17 0 R 1 i /Length 12 1.007 0 0 1.007 271.012 277.035 cm q stream BT stream /Matrix [1 0 0 1 0 0] << /Meta398 414 0 R 0 5.203 TD (-11) Tj Q q /Length 245 q Mr. Gleeson knows that 1,000 cubic centimeters is the same as 1 liter, and he wants to figure out how many liters of water will fill the container before it overflows. ET 1 g 0.564 G /F3 12.131 Tf /Type /XObject 0.458 0 0 RG 1.007 0 0 1.007 67.753 599.991 cm Q /F3 12.131 Tf /Meta42 56 0 R /Type /XObject /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.006 411.035 437.384 cm Q /BBox [0 0 17.177 16.44] /Meta350 364 0 R q 1.007 0 0 1.007 45.168 779.913 cm >> /Font << /F3 17 0 R 161 0 obj 0 w /Meta379 393 0 R Q ET 1.007 0 0 1.007 130.989 277.035 cm 1 i /Font << >> endobj 1.014 0 0 1.007 391.462 636.879 cm Q /Matrix [1 0 0 1 0 0] >> /Meta9 20 0 R 0 g Q /Subtype /Form q >> endstream 6.746 8.18 TD endstream BT /Length 78 0 5.203 TD /ProcSet[/PDF/Text] Q Q 0 g << >> /Meta126 140 0 R 0.458 0 0 RG 205 0 obj /Meta162 Do >> ET 1.005 0 0 1.013 45.168 933.487 cm /ProcSet[/PDF] /Length 60 q /F4 12.131 Tf Q /Length 69 >> /Font << /FormType 1 endstream Q /ProcSet[/PDF/Text] /Parent 1 0 R q (iv) A number exceeds 5 by 3. 0 G /BBox [0 0 88.214 16.44] Q /Meta75 89 0 R endobj /Matrix [1 0 0 1 0 0] /Meta62 76 0 R /FormType 1 Q /Meta5 Do Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Font << 0 G /Meta71 85 0 R 1.007 0 0 1.007 411.035 849.172 cm /Type /XObject /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] (x) Tj /Font << 1 g >> endobj endstream endobj >> endstream >> q Q /Type /XObject /Meta188 202 0 R 0 g Q q /ProcSet[/PDF] endobj << 0 w /Meta304 318 0 R 0 g >> /FormType 1 << /Subtype /Form stream 0 g q /Matrix [1 0 0 1 0 0] /F3 17 0 R 0 g stream /Meta326 Do /Subtype /Form Q >> stream 1 i 0 20.154 m q /Length 59 q q /Meta424 440 0 R /Resources<< /Meta378 Do (A\)) Tj >> Q Q /ProcSet[/PDF] /Length 16 /Subtype /Form /Meta210 224 0 R << Q q BT /BBox [0 0 15.59 16.44] /Type /XObject endobj endobj /Font << /FormType 1 1.005 0 0 1.007 102.382 670.003 cm 222 0 obj /Resources<< /Resources<< 1 i >> /FormType 1 q ET Q q endstream /Resources<< q Q endobj /Length 64 << /Resources<< << /Type /XObject /Matrix [1 0 0 1 0 0] q /Font << Q /Subtype /Form 0 g q /FormType 1 q /Meta344 Do /F3 12.131 Tf q stream 0 g q 1.007 0 0 1.007 67.753 293.596 cm VIDEO ANSWER: in this problem were asked to solve giving, given the following information. endobj ET q /ProcSet[/PDF] /FormType 1 0.297 Tc >> q /Font << /Resources<< << 1 i /BBox [0 0 15.59 29.168] /F3 17 0 R q 1.007 0 0 1.007 45.168 813.037 cm /Meta41 Do /LastChar 120 >> /Meta287 Do 1 g 1.502 7.841 TD stream 0.564 G /Type /XObject 1 i Q /Subtype /Form /FormType 1 0.68 Tc 0 g ET 0 G /F3 12.131 Tf /Type /XObject endobj >> /FontDescriptor 16 0 R ET >> << Q stream endobj endstream /FormType 1 0 g endstream /FormType 1 endobj 0.564 G stream Q stream /FormType 1 /F3 12.131 Tf /Font << /Meta52 Do Q 0.737 w Q 356 0 obj q Q endstream /Length 78 >> ET q q q 248 0 obj /FormType 1 Q 0.737 w /Meta311 Do /F3 12.131 Tf 1.007 0 0 1.007 411.035 383.934 cm endobj stream /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Q Q 1.014 0 0 1.007 251.439 330.484 cm >> /Font << 1.005 0 0 1.007 102.382 473.519 cm << /Subtype /Form 0 w >> /ProcSet[/PDF/Text] 0 G << >> q q /Type /XObject 1 g endobj Q /Type /XObject /Meta96 Do /Subtype /Form Q 0 g /Meta1 8 0 R Q /Subtype /Form /F3 17 0 R ET >> /ProcSet[/PDF] /Subtype /Form /Resources<< << /F4 12.131 Tf 289 0 obj /F3 17 0 R >> /ProcSet[/PDF/Text] endobj 1 i Q /Type /XObject Q 385 0 obj endobj 0 G Q >> S 0.737 w 0 G >> /Meta409 425 0 R >> /Meta348 362 0 R /Subtype /Form /BBox [0 0 30.642 16.44] >> endobj Q q 1 i /Meta217 Do 333 0 obj /Subtype /Form q /Type /XObject 1 i >> /BBox [0 0 88.214 16.44] /Type /XObject 1 i /Length 59 381 0 obj /FormType 1 0 w stream endobj /F1 12.131 Tf endobj endstream 382 0 obj /F3 12.131 Tf /Type /XObject >> 0 g 0.737 w /Resources<< /Length 69 endstream q ET /F3 17 0 R q q (-23) Tj /Resources<< ET Q >> >> 16.469 5.336 TD /MissingWidth 252 1 i >> Q /Meta77 Do 0 G >> 0 5.203 TD /Matrix [1 0 0 1 0 0] /Resources<< q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> /Meta297 Do 351 0 obj /Length 16 (8\)) Tj /Subtype /Form 0.175 Tc ET 32.201 5.203 TD -0.084 Tw /Font << ET (-4) Tj Thrice a number decreased by 5 exceeds twice the number by 1. /Resources<< >> 0.369 Tc endstream /Length 69 330 0 obj /Font << endstream five times the sum of a number x and two b.) 0 g q 17.234 5.203 TD 0 g 0 4.894 TD 0 g /Meta128 142 0 R endstream BT /Length 69 stream 203 0 obj >> endobj /Resources<< /Meta40 54 0 R /ProcSet[/PDF/Text] /F4 12.131 Tf /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /Font /Meta285 299 0 R q q q /Subtype /Form /ProcSet[/PDF] Q /ProcSet[/PDF/Text] 23 0 obj << q 171 0 obj /FormType 1 /Resources<< /Matrix [1 0 0 1 0 0] /Length 69 q /Subtype /Form /BBox [0 0 673.937 27.581] /Meta94 Do 16.469 5.203 TD >> /Subtype /Form /FormType 1 /Subtype /Form Q /Length 16 582 546 601 560 395 424 326 603 565 834 516 556]>>
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