/Meta231 Do >> /FormType 1 /BBox [0 0 88.214 16.44] /F3 12.131 Tf /Matrix [1 0 0 1 0 0] endobj 722.699 872.509 l 0 g 0 g >> Q >> >> Select the correct mathematical statement for the following equation. /BaseFont /TimesNewRomanPSMT /Resources<< /BaseFont /PalatinoLinotype-Roman /Meta31 44 0 R >> 1 g /Meta35 Do /Meta289 Do << (x ) Tj /BBox [0 0 30.642 16.44] /BBox [0 0 30.642 16.44] << /Meta244 258 0 R /Type /XObject /Subtype /Form 1 i /Meta176 190 0 R BT /ProcSet[/PDF/Text] /FormType 1 Q Q endobj 267 0 obj << /Subtype /Form endstream q Q BT /Type /XObject Q 1 i q >> /Matrix [1 0 0 1 0 0] stream /Subtype /Form q /Type /XObject << q /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 0.737 w q endobj Q /Subtype /Form 0 w endstream q endstream /FormType 1 endobj 0 g >> 0.564 G 1 i 0 g 0 g Q 0 5.203 TD (58) Tj /F1 7 0 R /Type /XObject << ET q /Length 70 /Length 69 48 0 obj Q /F1 12.131 Tf 0 g endobj /Meta87 Do Q >> /ProcSet[/PDF/Text] stream /ProcSet[/PDF/Text] q ET What is marios jumps times luigis jumps. >> q (-23) Tj Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.285 Tc 0.458 0 0 RG /MissingWidth 252 /BBox [0 0 534.67 16.44] >> q /BBox [0 0 30.642 16.44] >> q /Subtype /Form endstream endstream 0.737 w Q 0 g q 0 g << /BBox [0 0 30.642 16.44] /Length 80 endstream q 0 G endobj 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 0 g 0 g 0.458 0 0 RG Q Q /Meta195 Do /Resources<< /ProcSet[/PDF/Text] >> /Type /XObject The symbols 17 + x = 68 form an algebraic equation. endobj /Meta375 389 0 R /Meta191 205 0 R /Meta145 Do /ProcSet[/PDF] endobj >> q 0 G 221 0 obj Q >> /Font << endstream 1.007 0 0 1.007 271.012 277.035 cm /Resources<< /Type /XObject >> /BBox [0 0 15.59 16.44] Q << >> /Resources<< 148 0 obj 1 i /ProcSet[/PDF] 57 0 obj /Type /XObject /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /ProcSet[/PDF/Text] Q 41 0 obj stream BT Q 1 i /Meta117 131 0 R << endstream ET >> 17.234 5.203 TD 549.694 0 0 16.469 0 -0.0283 cm BT Q 0.271 Tc /BBox [0 0 88.214 16.44] /Length 16 /Resources<< /Subtype /Form /I0 Do /Length 99 /Meta196 Do << /Meta58 72 0 R endstream endstream Q /Matrix [1 0 0 1 0 0] /Resources<< Q 1.007 0 0 1.007 551.058 277.035 cm Q 1.014 0 0 1.007 531.485 636.879 cm >> 1.502 5.203 TD /Meta112 126 0 R 0 G q /Resources<< /Resources<< Q /Subtype /Form 1 i Q 0.564 G /Type /XObject /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /Type /XObject >> /ProcSet[/PDF] 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. Q 1.007 0 0 1.007 130.989 383.934 cm Q /Resources<< Q 0.37 Tc q << 0 g ET /F3 12.131 Tf /ProcSet[/PDF] Q /F1 12.131 Tf stream 0 g q /Font << /F3 12.131 Tf << stream BT << /ProcSet[/PDF] >> Q >> BT 229 0 obj q /Type /XObject endobj /Matrix [1 0 0 1 0 0] q endobj q q q /Length 69 /Meta378 392 0 R /Meta39 Do 0 G /Meta339 Do /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /FormType 1 /Length 69 0 g /F3 17 0 R endstream /Type /XObject ET /Matrix [1 0 0 1 0 0] BT ET /Length 59 /BBox [0 0 88.214 16.44] endobj 18.708 17.593 TD << If LtitnS6S . q Q >> q << /Ascent 1050 /Meta403 Do /F3 17 0 R 1.007 0 0 1.007 67.753 473.519 cm >> /Meta112 Do BT q 0.458 0 0 RG BT /Type /XObject /FormType 1 /Subtype /Form /Type /XObject >> >> >> 0 w Q >> /Subtype /Form /FormType 1 endstream endobj >> Q 0 g q q -0.084 Tw /BBox [0 0 88.214 16.44] 0 G 0 g 0.737 w Q q /FontDescriptor 6 0 R /Length 60 q 1 i endobj /FormType 1 0 5.203 TD 1.007 0 0 1.006 411.035 690.329 cm 549.694 0 0 16.469 0 -0.0283 cm q /BBox [0 0 88.214 35.886] >> Q /Type /XObject /FormType 1 /Meta274 288 0 R q stream q endstream /BBox [0 0 15.59 16.44] /F3 12.131 Tf BT endstream /Type /XObject 1 i /BBox [0 0 88.214 16.44] 0.737 w /Resources<< 0 g stream 309 0 obj >> q >> >> q Q /Subtype /Form 1 i 403 0 obj Q /Resources<< /Matrix [1 0 0 1 0 0] stream /Type /XObject 15 0 obj 376 0 obj 0 g >> /BBox [0 0 15.59 16.44] endobj endstream /Meta270 284 0 R q /FormType 1 q /Meta32 Do stream 0 w >> endobj /Subtype /Form /Length 69 endobj q 0.369 Tc Q /Length 16 /ProcSet[/PDF/Text] 0 g Aktual'nye voprosy Vol 10, No 3 (2020) /Resources<< 406 0 obj 0 G /Type /XObject 0.564 G >> endstream 1.008 0 0 1.007 654.946 293.596 cm /BBox [0 0 639.552 16.44] 0 G /Type /XObject stream >> /Meta345 Do endobj /Meta200 214 0 R q /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] Q stream /Resources<< endstream 0 g Q /F3 17 0 R << 1.007 0 0 1.007 271.012 383.934 cm Q /Matrix [1 0 0 1 0 0] 0.68 Tc 1.005 0 0 1.013 45.168 933.487 cm >> /Matrix [1 0 0 1 0 0] /Font << /Descent -299 /Matrix [1 0 0 1 0 0] endstream 0 g endstream This gives us: "2x+5". 0.297 Tc q /Subtype /Form ( \() Tj SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. q /FormType 1 /Font << /Type /XObject /Length 59 >> 0 w /BBox [0 0 30.642 16.44] endstream 1 i << /ProcSet[/PDF/Text] q 1.005 0 0 1.007 45.168 889.071 cm /Font << Q /Meta225 239 0 R q /BBox [0 0 88.214 16.44] /ProcSet[/PDF] >> << /BBox [0 0 88.214 35.886] 0.296 Tc q /Font << BT /Length 69 /Meta23 Do 0.486 Tc /Meta313 Do 0 g q endstream /Subtype /Form /BBox [0 0 88.214 16.44] /Meta283 Do q ET /FormType 1 0 G << 0 w 30.699 4.894 TD q Q /Resources<< 1 i /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] << /FormType 1 /Matrix [1 0 0 1 0 0] ET ET /Meta322 336 0 R 333 0 obj endstream >> ET /FormType 1 Q >> /BBox [0 0 15.59 16.44] /Type /XObject 1.007 0 0 1.007 411.035 849.172 cm 292 0 obj >> BT q 290 0 obj >> 0 g /Matrix [1 0 0 1 0 0] /Meta27 40 0 R /Resources<< /ProcSet[/PDF/Text] q q ET /Resources<< 254 0 obj /Meta335 349 0 R /FormType 1 Q /Type /XObject Q 0 g endobj >> stream >> endobj 9.723 5.336 TD >> /Meta105 Do /Resources<< (x) Tj ET /F3 17 0 R /Length 54 /FormType 1 endstream /I0 51 0 R q q << Q /Meta238 252 0 R /F3 17 0 R /ProcSet[/PDF] /Resources<< 3x - 5 = 2x + 1. x = 6. /BBox [0 0 88.214 16.44] 0 g Q Q endobj /Length 70 endobj q [(The )-19(quotient of )] TJ q /ProcSet[/PDF] /FormType 1 Q >> Q 549.694 0 0 16.469 0 -0.0283 cm /Font << q /Length 67 >> >> A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. Q (\)) Tj 1.007 0 0 1.006 551.058 836.374 cm (58) Tj Q Q >> Q q /Type /XObject /BBox [0 0 639.552 16.44] /Meta167 Do /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] >> 112 0 obj /BBox [0 0 17.177 16.44] 244 0 obj Q 1 g /Font << BT Q q /Meta194 208 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form >> 1.007 0 0 1.007 45.168 763.351 cm /F3 12.131 Tf 0.564 G /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 383.934 cm << /ProcSet[/PDF] 1 g /Meta382 Do Q >> /Resources<< /Subtype /Form /Meta340 Do Q 0.68 Tc q 0 G /Type /XObject /Meta293 Do 318 0 obj q Q 6.746 5.203 TD BT << >> Q /FormType 1 273 0 obj /Meta163 Do BT endstream /Type /XObject 0 w 90 0 obj >> >> /Resources<< 0 G 1. >> /Length 16 1 i Q q /BBox [0 0 549.552 16.44] /Meta401 417 0 R Q 0.31 Tc /F3 17 0 R /FormType 1 q 0.458 0 0 RG q q q 2.238 5.203 TD >> endobj /F3 12.131 Tf 1.007 0 0 1.007 411.035 277.035 cm endobj /ProcSet[/PDF/Text] endstream Q /Matrix [1 0 0 1 0 0] stream q endstream /ItalicAngle 0 endstream /Subtype /Form q << 0.564 G 1 i /FormType 1 << Q TJ Five times a number, decreased by 58, is -23 Find the number. /Type /XObject stream /FormType 1 /I0 51 0 R >> Q 0 g /Length 69 endstream /BBox [0 0 15.59 16.44] 0 G 1 i /Type /XObject 12.727 5.203 TD >> stream /Meta302 316 0 R q >> Q >> ET /Resources<< [(E)-14(le)-23(ven)] TJ Q /Meta383 397 0 R 348 0 obj Q c Site 5 is not included in this number. /ProcSet[/PDF/Text] /Meta179 193 0 R 0 g /Meta330 Do q q /Font << /Subtype /Form endobj Q 0 g Q q /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 583.429 cm /BBox [0 0 17.177 16.44] Q stream 0 G >> 0 G stream /Resources<< /Matrix [1 0 0 1 0 0] Q /Font << q Q /Subtype /Form Q /Matrix [1 0 0 1 0 0] stream endobj endobj (x) Tj stream >> /BBox [0 0 30.642 16.44] Q /Length 59 /Resources<< Q /Font << /Type /XObject 0.564 G Q << (C\)) Tj >> /Type /XObject endobj stream 0 56.451 TD (5) Tj /Matrix [1 0 0 1 0 0] endobj << 0 G /ProcSet[/PDF/Text] 1 i >> /F3 12.131 Tf 1.007 0 0 1.007 411.035 583.429 cm /XObject << << 0 G Q /Font << /Font << /ItalicAngle 0 stream /BBox [0 0 673.937 68.796] endobj Q 1.007 0 0 1.007 67.753 473.519 cm 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 q 0 G 0 G ET 1.007 0 0 1.007 551.058 330.484 cm /Meta424 Do /F3 17 0 R /Length 58 238 0 obj /Length 12 Q /Type /XObject /ProcSet[/PDF] endstream << /FormType 1 endobj 0.297 Tc /Subtype /Form /Meta98 Do 13 0 obj Q a and b or something else.***. /Length 12 /F3 12.131 Tf endobj Want to see the full answer? Q endstream 1 i endobj /ProcSet[/PDF] to represent the numbers. 352 0 obj << endobj /ProcSet[/PDF] /Subtype /Form /Type /XObject >> 1 i /ProcSet[/PDF] 3.742 5.203 TD /Resources<< /FormType 1 /Type /XObject 0.524 Tc >> endobj -0.092 Tw ET << >> >> /Resources<< /F3 17 0 R endobj /Meta19 30 0 R q q Q Q /Type /XObject (x) Tj S q /Matrix [1 0 0 1 0 0] Q endstream 1 i 0 G Q >> /F3 12.131 Tf >> 1.005 0 0 1.007 102.382 546.541 cm 16.469 5.203 TD Q 0.738 Tc >> 549.694 0 0 16.469 0 -0.0283 cm 0 G Q Q Q /FormType 1 /Subtype /Form >> /ProcSet[/PDF] q /ProcSet[/PDF/Text] Q >> 1 i /Meta368 382 0 R /Subtype /Form /FormType 1 0.737 w 1 i ET 230 0 obj >> endstream /Length 69 7 0 obj stream /Resources<< q Q /F4 12.131 Tf stream BT 1 i 0.564 G [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /Matrix [1 0 0 1 0 0] /Type /XObject >> 0.458 0 0 RG endstream /F3 17 0 R -0.029 Tw /Meta64 Do /Matrix [1 0 0 1 0 0] Q /Type /XObject 251 0 obj 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 411.035 636.879 cm /ProcSet[/PDF] ET /Resources<< /Meta337 Do 129 0 obj Q q /Meta392 Do stream /Type /XObject /ProcSet[/PDF] BT /Subtype /Form 0 4.894 TD q /Subtype /Form endobj 0 G /BBox [0 0 88.214 35.886] [2] Twice a number increased by four is twenty-one. /Type /XObject stream 0.564 G /Font << /BBox [0 0 17.177 16.44] /Resources<< >> 1.005 0 0 1.007 102.382 653.441 cm /Meta69 Do /Length 78 /Meta245 Do >> /Resources<< q /ProcSet[/PDF/Text] /Subtype /Form /Matrix [1 0 0 1 0 0] q Q 0.564 G 0 w /Meta137 Do 0 G /Resources<< 1.007 0 0 1.007 45.168 846.161 cm /Subtype /Form 101.849 5.203 TD /ProcSet[/PDF/Text] 1.005 0 0 1.007 102.382 473.519 cm (6\)) Tj stream ET 122 0 obj 0 G 12.727 24.649 TD q /F4 12.131 Tf endstream /Meta254 268 0 R 1 g Q 1 i /Length 16 >> Q /Font << /Ascent 1050 q /Meta168 182 0 R endstream endobj Q /Matrix [1 0 0 1 0 0] stream /Meta257 271 0 R >> 0 G Q /F3 12.131 Tf Q /Type /XObject q /Length 118 0 G /BBox [0 0 30.642 16.44] q /Matrix [1 0 0 1 0 0] q 69 0 obj /Resources<< Get a free answer to a quick problem. /Matrix [1 0 0 1 0 0] q 58 0 obj >> endobj /BBox [0 0 15.59 29.168] Q /Meta105 119 0 R 1 g /Matrix [1 0 0 1 0 0] q 0 G /Resources<< q BT /Matrix [1 0 0 1 0 0] /Font << << Q 424 0 obj 0.458 0 0 RG 0 g Q 1 i << << BT /Meta353 367 0 R Q >> Q 1 i /FormType 1 /ProcSet[/PDF] << 1.007 0 0 1.007 411.035 636.879 cm q 1.007 0 0 1.007 130.989 523.204 cm 0 g >> endobj q 1 i /Meta389 Do q >> /Resources<< >> /Type /XObject q Q 3.742 24.649 TD /Type /XObject 0.786 Tc 1 i /Matrix [1 0 0 1 0 0] /Subtype /Form (x) Tj Q >> >> 0 5.203 TD q /Meta10 Do Q /Meta39 53 0 R /Subtype /Form /F3 17 0 R BT /FormType 1 q /BBox [0 0 549.552 16.44] q /Meta304 318 0 R /Resources<< q ET endstream /Length 16 /FormType 1 1.007 0 0 1.007 411.035 277.035 cm /Length 69 2x - 15 = -27. endobj 1 g /FormType 1 /MissingWidth 250 Q 22.478 5.336 TD q 0 w 0.564 G /ProcSet[/PDF/Text] Let x be a number. /Length 294 >> /F3 12.131 Tf 168 0 obj /Resources<< /F1 7 0 R Q /FormType 1 /Type /XObject q Q q /ProcSet[/PDF/Text] See Solution. q /Subtype /Form /FormType 1 0 5.203 TD ET ET /Meta248 Do Q q 0 g endstream 0.737 w 1 i /Meta210 Do q /BBox [0 0 15.59 29.168] 0 g /Type /Font /ProcSet[/PDF/Text] q /Meta113 Do /BBox [0 0 88.214 16.44] /Type /XObject /LastChar 121 q /Meta125 Do Q stream BT Q /Font << )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] >> 0.737 w /Type /XObject

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