q /Type /XObject ET endobj 0.564 G Q /Font << 0 w /BBox [0 0 88.214 16.44] /Subtype /Form >> /BBox [0 0 549.552 16.44] >> /Length 69 Q /Type /XObject 1.007 0 0 1.007 271.012 776.149 cm /Font << Q 435 0 obj /Length 73 /Meta9 Do 0 5.203 TD 1.007 0 0 1.007 45.168 846.161 cm q (5) Tj /BBox [0 0 17.177 16.44] 0 g (2) Tj /FormType 1 q 2. /Meta16 Do Q 0 g /Meta57 71 0 R /BBox [0 0 534.67 16.44] /I0 Do /BBox [0 0 534.67 16.44] 0 g 179 0 obj Q endstream 1 i /Subtype /Form 0.737 w 0 G /BBox [0 0 88.214 16.44] endobj << << q /Font << Q >> 0 w << /Resources<< S >> /Meta284 Do 0 w /Font << Q Q q q 318 0 obj 0.227 Tc /Meta37 50 0 R >> /Resources<< Q q 10.487 5.203 TD >> /Length 70 1.007 0 0 1.007 551.058 636.879 cm /Subtype /Form 1.007 0 0 1.007 271.012 383.934 cm endobj 1 i >> q /Subtype /Form /Meta122 Do - 9737014. /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endobj >> endstream 349 0 obj q Find an answer to your question Twice a number decreased by 8gives 58. q /Meta15 Do endstream /Meta258 272 0 R stream 445 0 obj Q endobj << /Meta299 Do >> So let's go ahead and identify a v q 19.474 5.203 TD >> endobj -0.047 Tw 0 G 0 G /Meta191 205 0 R Q q (1\)) Tj Q /FormType 1 /Meta275 289 0 R Q 0 g stream 1 g /Resources<< q 1 i /F1 12.131 Tf /F1 7 0 R (5) Tj 0.458 0 0 RG /Meta336 Do endobj All steps. /F3 17 0 R /ProcSet[/PDF/Text] /Meta222 Do /Subtype /Form /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] q q >> /Meta269 Do /Font << q 1.007 0 0 1.007 271.012 523.204 cm /Meta320 334 0 R q 0 g endobj /Subtype /Form /Type /XObject /Matrix [1 0 0 1 0 0] 0 5.203 TD /Meta95 Do /Meta353 367 0 R /FormType 1 << >> /Meta257 Do 6.746 5.203 TD Q >> 0 G q Q 549.694 0 0 16.469 0 -0.0283 cm >> 0.458 0 0 RG /Resources<< /Length 118 the quotient of five and a number 7.) 1.007 0 0 1.007 130.989 583.429 cm 0 g >> Q /Meta46 Do 0 G (9\)) Tj Q stream endstream 0 g 1 i /Matrix [1 0 0 1 0 0] /FormType 1 /Type /XObject Q 0.369 Tc /ProcSet[/PDF] q Q 1.007 0 0 1.007 551.058 383.934 cm /FormType 1 endobj endstream 0 g (C\)) Tj BT 1 i 0 w /F1 7 0 R stream >> /Length 69 /Meta243 257 0 R /Font << q >> /Meta363 Do /Font << 0 G /Matrix [1 0 0 1 0 0] 1 i 0.425 Tc Q /Resources<< 1 i /FormType 1 ET endobj >> 0.564 G /Matrix [1 0 0 1 0 0] endstream Q >> /Length 16 << 0 w /ProcSet[/PDF/Text] endstream q /Matrix [1 0 0 1 0 0] >> 1.005 0 0 1.007 102.382 816.048 cm q endobj ET /F3 17 0 R /Meta111 125 0 R /Resources<< Q /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] endstream endobj /Type /XObject q Q >> endstream Hence, the number is 6. >> Q ET 0 g /Resources<< 0 G >> Q endobj q stream /Type /XObject /Length 69 /Meta221 Do q 1 i /Type /XObject endstream ET /BBox [0 0 534.67 16.44] endstream Q q Q /Meta311 Do /Matrix [1 0 0 1 0 0] /Resources<< stream >> /Subtype /Form 1.007 0 0 1.007 411.035 330.484 cm ET /Font << 340 0 obj 52 0 obj >> 0 g /Length 59 Q the sum of a number and twelve. /Resources<< stream Q /Subtype /Form /ProcSet[/PDF] >> q >> /F3 17 0 R q /Meta271 285 0 R /Type /XObject << q 9.723 5.336 TD q 0 g 6.746 5.203 TD << stream >> Q 0.564 G /Matrix [1 0 0 1 0 0] 88 0 obj 224 0 obj 0 G Q stream 0.838 Tc /Matrix [1 0 0 1 0 0] /Type /XObject /ProcSet[/PDF] q /Subtype /Form >> >> /Meta8 Do 0.564 G >> /Matrix [1 0 0 1 0 0] (8\)) Tj q 1.014 0 0 1.007 251.439 849.172 cm /Type /XObject q Q q q BT precision and actual right or wrong answers. /Meta289 Do /Type /XObject (13) Tj stream /Subtype /Form 381 0 obj ET /ProcSet[/PDF/Text] endstream Q q /Subtype /Form /Subtype /Form << /I0 Do >> /ProcSet[/PDF/Text] q endobj S , Prove the following >> q 0 g 1.502 5.203 TD 0.458 0 0 RG /F3 17 0 R 1.005 0 0 1.007 79.798 862.723 cm stream 0.425 Tc Q 1.007 0 0 1.007 411.035 383.934 cm /Meta415 Do q 0 g /FormType 1 >> 1 i /Matrix [1 0 0 1 0 0] stream endobj q (B\)) Tj >> 0.175 Tc q q 1 i 0 G stream 1.007 0 0 1.007 130.989 636.879 cm >> BT /Length 65 endstream /Matrix [1 0 0 1 0 0] endobj /FormType 1 q /Type /XObject endobj stream /ProcSet[/PDF/Text] 1 i endobj /Subtype /Form (x) Tj /Length 16 q << /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] ET /Meta369 Do (D\)) Tj /FormType 1 /Length 69 0 G 1 g /Matrix [1 0 0 1 0 0] /Subtype /Form ET 0 g q Q: when six times a number is decreased by 4, the result is 8. Q /Meta398 Do /Matrix [1 0 0 1 0 0] >> Q /Subtype /Form endobj /Subtype /Form << q /Subtype /Form q 32.201 5.203 TD (iii) 25 exceeds a number by 7. /Subtype /Form /Meta130 Do endobj /Meta350 Do ET q >> 1 g /Type /XObject Q 1 g /I0 51 0 R 0.458 0 0 RG >> /Type /XObject endstream q endobj q 0 g ET /Font << >> /Meta66 Do 0.458 0 0 RG BT q q q 58 decreased by twice Gails age. /Subtype /Form q ET q /XHeight 447 /BBox [0 0 15.59 16.44] /Resources<< q /Resources<< Q Q q endobj >> /ProcSet[/PDF] >> endstream >> Q /Subtype /Form Q /Matrix [1 0 0 1 0 0] >> /Subtype /Form Q /Font << /Subtype /Form 0 G /Matrix [1 0 0 1 0 0] /Type /XObject q /CapHeight 694 >> 0.269 Tc 0.241 Tc /FormType 1 0 g (x ) Tj Q /Font << Q 20.21 5.336 TD /Resources<< 19.474 20.154 l Q 1 g /Matrix [1 0 0 1 0 0] /Type /XObject /Matrix [1 0 0 1 0 0] 1 i /Meta125 Do /Meta319 333 0 R q 13.464 5.203 TD Q 1.005 0 0 1.007 102.382 546.541 cm Q /ProcSet[/PDF/Text] /Length 107 /BBox [0 0 88.214 16.44] 0 g q /Meta37 Do >> /F3 12.131 Tf /ProcSet[/PDF/Text] 1 i /Subtype /Form endobj 0 g /Length 68 0.564 G endstream /Length 58 q /Subtype /Form Q 0.564 G q Q >> 1 g << endstream BT >> /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g /FormType 1 endobj 1.007 0 0 1.007 67.753 293.596 cm stream (38) Tj 2 See answers pharry1800 pharry1800 Answer: 2n-58 Step-by-step explanation: olivbreadh olivbreadh Answer: 2x-116 or 2(x-58) Step-by-step explanation: Transalate it to numbers and operations: => 2(x-58) => 2x-116 You won't have a solid number since its not an equation. endstream q endobj /Meta140 154 0 R /F3 17 0 R stream << >> 1.005 0 0 1.007 79.798 813.037 cm /BBox [0 0 15.59 29.168] 0 g q q Q /ProcSet[/PDF] (x ) Tj 0 g endstream q stream endobj 0 w /Meta150 Do endobj /F1 12.131 Tf /Meta33 46 0 R /Length 16 /Font << 1 i >> q 1 i endobj 1.007 0 0 1.007 411.035 583.429 cm /Type /XObject /Meta386 Do Q q /FormType 1 endobj /FormType 1 1.007 0 0 1.007 130.989 636.879 cm q 1.014 0 0 1.007 251.439 383.934 cm 0.838 Tc q /F3 17 0 R 0 g /ProcSet[/PDF] 1 g BT /ProcSet[/PDF] 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 /ProcSet[/PDF] 0 G << /Meta163 Do >> << Q stream stream /Matrix [1 0 0 1 0 0] View the full answer. 0 g /Matrix [1 0 0 1 0 0] Q q /Font << q 332 0 obj << 0 g /ProcSet[/PDF/Text] 1.014 0 0 1.006 111.416 763.351 cm >> q /BBox [0 0 15.59 16.44] (4\)) Tj /BBox [0 0 30.642 16.44] (1) Tj >> 47.933 5.203 TD q /BBox [0 0 15.59 16.44] /Type /XObject q [2] Twice a number increased by four is twenty-one. /Length 65 /Font << /FormType 1 q 0.458 0 0 RG 1.007 0 0 1.007 130.989 330.484 cm 364 0 obj 1 g Q /FormType 1 q /Meta240 254 0 R Q ET Q 0 20.154 m Q endobj /Meta293 307 0 R endobj /Font << /Font << 267 0 obj 0.369 Tc Q /BBox [0 0 15.59 16.44] /FormType 1 >> 1.007 0 0 1.007 551.058 703.126 cm 1.007 0 0 1.007 130.989 849.172 cm >> /Meta224 Do q 0 g >> >> endobj /BBox [0 0 88.214 35.886] Q /ProcSet[/PDF] q /Subtype /Form endobj Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. 9.723 5.336 TD 1.007 0 0 1.007 271.012 776.149 cm /Meta151 Do 1 g /Matrix [1 0 0 1 0 0] 189 0 obj >> /Subtype /Form stream Q Q q << 0 5.203 TD Q endstream /Meta149 163 0 R >> >> /Meta25 38 0 R /Meta130 144 0 R /StemH 88 1.007 0 0 1.007 411.035 383.934 cm 1 g 0.486 Tc /Subtype /Form stream q /F3 12.131 Tf Q /ProcSet[/PDF] >> 12 0 obj /Meta44 Do /FormType 1 /F4 36 0 R Q >> << q >> q 0.737 w Q /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF/Text] 0 G >> Q >> /Length 69 /Font << Six subtracted from a number 6. q Calculate a 15% decrease from any number. [(A number )-17(divided by )] TJ /F3 17 0 R /Length 65 ET (7\)) Tj >> 1 i endstream Q /Resources<< )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] /Type /XObject /Meta302 316 0 R Q /Type /XObject BT /Font << endobj 17.234 5.203 TD q Q BT >> /Type /XObject stream >> >> /Meta404 Do 0.737 w ET 1 i 184 0 obj /Meta395 Do q /F3 12.131 Tf /ProcSet[/PDF] >> /Matrix [1 0 0 1 0 0] /BitsPerComponent 1 /Resources<< q 0 g Q /Resources<< Q Q ( x) Tj q /Subtype /Form /Meta35 48 0 R q /Meta109 123 0 R 358 0 obj Q Q BT /BBox [0 0 88.214 35.886] >> /Subtype /Form 0 g Q Q Q BT q endobj 0 g endstream 1.014 0 0 1.006 251.439 437.384 cm q stream Q /Font << /Meta143 157 0 R >> 1 i /Type /XObject q 0.564 G /ProcSet[/PDF/Text] 0 g 186 0 obj << endstream /ProcSet[/PDF] BT /F3 17 0 R /ProcSet[/PDF/Text] Q: A number increased by 5 is equivalent to twice the same number decreased by 7. You can specify conditions of storing and accessing cookies in your browser, Twice a number, decreased by 58 is less than 112, Mr. Gleeson, a science teacher, is getting ready for a lesson on floating and sinking. /Meta200 214 0 R BT 0.524 Tc /Meta245 Do /F3 12.131 Tf (C\)) Tj /Subtype /Form q /Length 54 << /FormType 1 /Resources<< /Type /XObject 0 g q /FormType 1 6.746 5.336 TD endobj /ProcSet[/PDF] 1 i BT /Matrix [1 0 0 1 0 0] Find the number. Q /ProcSet[/PDF/Text] 54 0 obj Q /Length 69 q 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 1.007 0 0 1.006 551.058 763.351 cm /Resources<< 1.007 0 0 1.007 411.035 330.484 cm 1.007 0 0 1.007 411.035 583.429 cm /Type /XObject q >> ET /Matrix [1 0 0 1 0 0] q 0.737 w 0 56.451 TD /F3 17 0 R (x ) Tj /FormType 1 stream q endobj endstream ET Q << /Matrix [1 0 0 1 0 0] endstream 1.007 0 0 1.007 67.753 726.464 cm 1 g 1.007 0 0 1.007 130.989 523.204 cm /ProcSet[/PDF/Text] BT /Font << Q endstream /Subtype /Form >> q Q q 0 w >> /Subtype /Form /Type /XObject [tex]\sin (\pi -x)=\sin x[/tex]. >> 1 i That was 1/8 of the points that he scored Q >> 1.007 0 0 1.006 551.058 836.374 cm q ( \() Tj BT endstream Q 0.564 G q endstream /Meta394 410 0 R Q /Subtype /Form q q C. Twice a number decreased by ten is at most 24. >> 0 G >> /ProcSet[/PDF/Text] /Font << /Meta171 185 0 R BT 0.564 G /FormType 1 Q >> /Type /XObject /Type /XObject /F1 7 0 R 1 i 0 g q >> stream /Matrix [1 0 0 1 0 0] stream 0 g [(1)-25(0\))] TJ Q 0 w << Q /Type /XObject stream Q 1.005 0 0 1.007 102.382 799.486 cm /ProcSet[/PDF] 201 0 obj 211 0 obj 351 0 obj >> /ProcSet[/PDF/Text] Q /Meta306 Do 0 G 1 g >> /BBox [0 0 15.59 16.44] q /Meta374 388 0 R /Matrix [1 0 0 1 0 0] /Length 118 endobj /BBox [0 0 88.214 16.44] /Font << << stream q /Resources<< endstream 1 g 0 G BT This site is using cookies under cookie policy . q (D\)) Tj q >> /BBox [0 0 88.214 35.886] << >> >> /Subtype /Form /BBox [0 0 88.214 16.44] stream /Length 59 q << /FormType 1 /Resources<< /F4 36 0 R /BBox [0 0 88.214 16.44] /LastChar 120 /Meta6 15 0 R /Subtype /Form 1 g /F3 17 0 R /Subtype /Form /F3 12.131 Tf /Meta390 406 0 R 22.478 5.336 TD /Subtype /Form 44 0 obj BT stream /F3 12.131 Tf stream /Meta136 150 0 R /Meta179 193 0 R 13.493 5.336 TD Q q >> stream q , Prove the following /Subtype /Form << >> 1.502 7.841 TD stream /Meta406 Do /BBox [0 0 639.552 16.44] >> endobj stream stream /BBox [0 0 17.177 16.44] q >> /Font << Q Q << /Matrix [1 0 0 1 0 0] endobj 0 g /ProcSet[/PDF/Text] >> Q endstream 1 i q q /Matrix [1 0 0 1 0 0] >> /FormType 1 BT stream << /Font << /Length 59 /Resources<< /Flags 32 0 G 0 G /Font << (-) Tj /Type /XObject >> 347 0 obj /Meta183 197 0 R endobj /Font << /Resources<< >> /Meta74 88 0 R 1.007 0 0 1.007 551.058 583.429 cm q Q /Meta88 102 0 R Q << 1.007 0 0 1.007 130.989 277.035 cm 0 g 1.007 0 0 1.007 551.058 636.879 cm /Length 64 (11) Tj 1 Data in this Fast Fact represent the 50 states and the District of Columbia. >> >> /Width 734 0 G 0 g 0 G /Length 69 0 g Q q 316 0 obj /Meta412 428 0 R 0.737 w [4] One half of a number decreased by fourteen is twenty-one << /Type /XObject Q 1.007 0 0 1.007 411.035 277.035 cm /ProcSet[/PDF/Text] 0 g 1.014 0 0 1.006 391.462 763.351 cm /Matrix [1 0 0 1 0 0] /Meta82 96 0 R /Matrix [1 0 0 1 0 0] /Meta316 Do 0 g There were x cookies at the beginning of a party. /FormType 1 0.51 Tc /I0 51 0 R 0 g endobj endstream /BBox [0 0 15.59 16.44] /BBox [0 0 15.59 16.44] BT Q 0 g endstream endstream /Resources<< /BBox [0 0 673.937 16.44] endobj 437 0 obj 1.007 0 0 1.007 654.946 347.046 cm /BBox [0 0 88.214 16.44] Q 0 w 0 g /Subtype /Form 0 5.203 TD /Length 16 /Type /XObject 0.458 0 0 RG Q >> /ProcSet[/PDF/Text] Q (38) Tj To find: The. Number Outcomes 1 42 2 41 3 . /F3 12.131 Tf /F1 12.131 Tf /BBox [0 0 88.214 35.886] /FormType 1 0 5.203 TD 672.261 872.509 m Q 0.458 0 0 RG q q /Matrix [1 0 0 1 0 0] ET /Subtype /Form Q 0 G q 1.007 0 0 1.007 411.035 636.879 cm /Matrix [1 0 0 1 0 0] /Length 59 /F1 12.131 Tf q >> q q q Q q 0.369 Tc Q endstream q Q 237 0 obj Q /Font << << << 0.68 Tc /F3 17 0 R 210 0 obj /Length 80 Q 1.005 0 0 1.007 102.382 653.441 cm endobj q q 389 0 obj q /Font << /ProcSet[/PDF] 102 0 obj 0 w q 1.005 0 0 1.007 79.798 829.599 cm 139 0 obj >> >> << 6.746 24.649 TD 549.694 0 0 16.469 0 -0.0283 cm q q 0.382 Tc q Q 0 w >> Q q /Length 70 /BBox [0 0 15.59 16.44] << q /BBox [0 0 88.214 16.44] /Length 70 >> 1 i /BBox [0 0 15.59 16.44] /Meta389 Do /Type /XObject q q q q 52.412 5.203 TD Q 0.369 Tc q /Resources<< endobj q /Meta342 356 0 R 1.007 0 0 1.007 551.058 523.204 cm /Subtype /Form /Flags 32 /Meta351 365 0 R /Length 16 /FormType 1 /BBox [0 0 88.214 16.44] 16.469 5.203 TD << /ProcSet[/PDF] 162 0 obj q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] >> Q 0 g (3) Tj q 1.007 0 0 1.007 551.058 383.934 cm /Meta67 81 0 R endobj stream /Matrix [1 0 0 1 0 0] Q endobj Q >> /ProcSet[/PDF] /Type /XObject /BBox [0 0 673.937 16.44] << Q endstream /ProcSet[/PDF/Text] Q endstream /ProcSet[/PDF/Text] /F1 7 0 R q << endstream /ProcSet[/PDF/Text] /Type /XObject 0 G /Subtype /Form q 1.005 0 0 1.006 45.168 879.284 cm endstream /Matrix [1 0 0 1 0 0] endstream /Resources<< Q << 1.007 0 0 1.007 271.012 330.484 cm >> Phrase : Expression : 4 times some number : 4x: twice a number : 2y : one-third of some number : the product of a number and 12 : 12w: Some examples of common phrases and corresponding . /Length 12 1 i << [(thir)17(te)15(en)] TJ Q Q endstream /Meta298 Do Q 0 5.336 TD 0.369 Tc -0.084 Tw 1 i endstream /Matrix [1 0 0 1 0 0] >> 0 g /BBox [0 0 15.59 16.44] TJ q 2.238 5.203 TD /Meta71 85 0 R Q ET /Matrix [1 0 0 1 0 0] >> /Resources<< /Resources<< 0 g /Resources<< You can also contact the clerk of court in the county you received the ticket. endstream /Meta323 Do q /Meta174 Do /Matrix [1 0 0 1 0 0] q endobj ET Q /Resources<< 1.007 0 0 1.007 551.058 383.934 cm /Subtype /Form /Font << 1.014 0 0 1.007 391.462 849.172 cm q q 0 20.154 m 1 i -0.099 Tw /Length 69 << >> endstream endobj 0.458 0 0 RG Q Q /Font << 32 = 2a + 8: The quotient of fifty and five more than a number is ten. 1.007 0 0 1.007 271.012 636.879 cm q /Resources<< /ProcSet[/PDF] << Q q /Length 66 722.699 347.046 l q /Type /XObject 0 g /F3 12.131 Tf q 1.007 0 0 1.007 45.168 713.666 cm ET /Subtype /Form /FormType 1 >> /FormType 1 1 i /Ascent 1050 q /Resources<< q 0.564 G 1.007 0 0 1.007 271.012 277.035 cm Q /Resources<< Q >> 1 i Q BT /Type /XObject endstream 0 g /BBox [0 0 534.67 16.44] >> stream endstream >> Q /ProcSet[/PDF] 1 i 0.524 Tc /Matrix [1 0 0 1 0 0] 81 0 obj Q /F3 12.131 Tf 1 i /Matrix [1 0 0 1 0 0] /Meta124 138 0 R 0.369 Tc /FormType 1 endobj 0 w endstream q /BBox [0 0 673.937 14.853] endstream 1.007 0 0 1.007 271.012 636.879 cm /F3 12.131 Tf 0.564 G 1 i BT Q /Resources<< /Subtype /Form 0 G endstream /Meta96 110 0 R Q q /Matrix [1 0 0 1 0 0] stream Q /Length 69 0.458 0 0 RG >> 0.564 G 384 0 obj /Subtype /Form /Font << endstream /Subtype /Form endobj /Count 2 /Meta250 264 0 R 77 0 obj q /Font << stream Q stream /Meta413 Do << q endstream q /Meta429 445 0 R >> 1.007 0 0 1.007 551.058 330.484 cm /Length 80 /Meta307 Do /Meta138 152 0 R 1.005 0 0 1.007 102.382 546.541 cm /Length 118 << q 0.458 0 0 RG /F4 12.131 Tf stream endobj << /Resources<< stream >> /F3 12.131 Tf S 0 w >> endstream BT /Resources<< (A\)) Tj /Resources<< endstream 0 G /FormType 1 /Subtype /Form /BBox [0 0 88.214 16.44] << /ProcSet[/PDF/Text] 1 i ( x) Tj 1 g Q /BBox [0 0 88.214 16.44] 1.007 0 0 1.006 411.035 690.329 cm 442 0 obj /Font << 0 g q 1.007 0 0 1.007 130.989 523.204 cm Q /F3 12.131 Tf /F3 17 0 R /ProcSet[/PDF] /Type /XObject ET /FormType 1 >> Q Q /F3 17 0 R /Matrix [1 0 0 1 0 0] q /Subtype /Form ET q /F3 17 0 R q /Type /XObject endobj endstream endstream /Subtype /Form 1.007 0 0 1.007 411.035 583.429 cm 0.51 Tc endobj /ProcSet[/PDF/Text] >> D. Twice a number decreased by ten is less than 24. Q endobj /Meta216 230 0 R << /Type /XObject Q >> 1 g endstream stream 1 i << Q endstream stream /Length 68 What word phrase can you use to represent 5x + 2? /BBox [0 0 15.59 16.44] 1 i /Length 54 q q /Meta111 Do 1.007 0 0 1.007 67.753 872.509 cm /Font << 20.21 5.203 TD /Type /XObject >> 1 i /Subtype /Form /Type /XObject /Meta69 Do 1.005 0 0 1.007 79.798 862.723 cm /F3 12.131 Tf 0 w >> /BBox [0 0 88.214 16.44] q /Length 16 2.238 5.203 TD /Type /XObject /XHeight 476 1 i Q 271 0 obj /Font << A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. Q /Resources<< stream ET /Type /XObject /Meta352 Do /Type /XObject << Q Q endobj q /Font << >> /Type /XObject Q /Meta292 Do 0 G /Meta255 269 0 R >> /Resources<< /Length 69 >> endstream >> /Matrix [1 0 0 1 0 0] 22 0 obj endobj 0 g /ProcSet[/PDF/Text] /Length 70 0.737 w /ProcSet[/PDF/Text] 1.005 0 0 1.007 102.382 726.464 cm 1.014 0 0 1.007 531.485 450.181 cm /BBox [0 0 88.214 16.44] /F3 17 0 R Q /Matrix [1 0 0 1 0 0] ET /Length 69 /Meta287 Do 1 i /Font << /Meta240 Do 1 i 0 g /Subtype /Form q Q /F3 17 0 R ET /Type /XObject 0 G q /Meta194 Do q q /BBox [0 0 88.214 16.44] q ET q Q >> q q [( a )-15(number, decreased by )] TJ 1 i /Resources<< Q /Meta44 58 0 R /ProcSet[/PDF/Text] /Meta280 Do BT /Meta370 384 0 R endobj 446 0 obj 3.742 5.203 TD 0.369 Tc 1 i >> q /Meta160 174 0 R 0 w /Type /XObject 400 0 obj /Length 12 Q >> /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] endobj Answer by Mathtut (3670) ( Show Source ): endstream 0.737 w /Matrix [1 0 0 1 0 0] 0 g 0.68 Tc /BBox [0 0 15.59 29.168] >> endstream 0 G q 32.939 5.203 TD /Meta330 344 0 R 1.007 0 0 1.007 67.753 347.046 cm 1.007 0 0 1.007 130.989 383.934 cm (x ) Tj /Resources<< 1 i endobj 0.737 w << BT 0.564 G /Meta290 304 0 R >> /FormType 1 stream q /Length 104 endstream /ProcSet[/PDF] 0.458 0 0 RG 330 0 obj /I0 Do endobj Q /FormType 1 /Length 59 /Resources<< 0 g /FormType 1 >> 1 i /F1 7 0 R q /FormType 1 q >> q endstream 0.458 0 0 RG /F3 12.131 Tf