/Length 16 The sum of 18 and tour times a number is -6 Find the number. /Subtype /Form 222 0 obj 20.21 5.203 TD stream Q stream 0 g -0.047 Tw stream Q /F3 17 0 R Kobe scored 85 points in a basketball game. 0.458 0 0 RG 215 0 obj /BBox [0 0 534.67 16.44] Q In the problem above, x is a variable. q Q q q /F3 17 0 R 0 w /BBox [0 0 88.214 35.886] /Subtype /Form >> stream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Subtype /Form >> Q Q 0.369 Tc >> q /ProcSet[/PDF] /F1 7 0 R /Font << endobj 287 0 obj 0.738 Tc /ProcSet[/PDF/Text] q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 16.469 5.336 TD /Subtype /Form /Meta15 Do /FormType 1 /Meta124 Do 0 w q /Resources<< q q /Matrix [1 0 0 1 0 0] q /BBox [0 0 88.214 16.44] If mario jumps 3 times and luigi jumps 62 times. /Meta166 180 0 R /Matrix [1 0 0 1 0 0] >> 0 g -0.062 Tw Q /Type /XObject q 386 0 obj 0 g /BBox [0 0 88.214 16.44] 0.51 Tc >> stream 0.564 G [tex]\sin (\pi -x)=\sin x[/tex]. stream >> 0.458 0 0 RG endstream /Meta115 Do >> 0.369 Tc /F4 36 0 R 0 g 278 0 obj >> 0.486 Tc << q /FormType 1 /BBox [0 0 15.59 16.44] stream /Font << << /Subtype /Form /Meta271 Do 1 i 0 g /Resources<< endstream q q Q 0 G 358 0 obj ET /ProcSet[/PDF/Text] endobj 0 G 1 g 0 G /Meta196 Do /Meta1 8 0 R q /F4 12.131 Tf HOPE HELPS .3. /Meta133 147 0 R q Q q /Subtype /Form Q q (x) Tj q endstream /FormType 1 /Length 59 1.007 0 0 1.006 130.989 690.329 cm /F3 17 0 R /F3 17 0 R 549.694 0 0 16.469 0 -0.0283 cm endstream q /FormType 1 >> endobj /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Meta138 Do 0 G q q >> /ProcSet[/PDF/Text] endstream /Resources<< /ProcSet[/PDF] /FormType 1 >> stream /Font << endstream << endobj /BBox [0 0 15.59 29.168] >> stream /Meta107 121 0 R Was this answer helpful? >> /Resources<< /Length 57 >> /Resources<< [(Negativ)16(e )] TJ Q /BBox [0 0 534.67 16.44] 1 g 0 g endobj /Meta103 117 0 R endstream << 1 i 396 0 obj /F3 12.131 Tf /ProcSet[/PDF/Text] /Resources<< >> q /ProcSet[/PDF] /Meta362 376 0 R q /ProcSet[/PDF/Text] 0 G /Subtype /Form q /Meta107 Do 0 g /Length 69 0.51 Tc /Font << /ProcSet[/PDF] /Length 63 Q 1 g q BT No packages or subscriptions, pay only for the time you need. >> /Length 12 /ProcSet[/PDF] /ProcSet[/PDF/Text] ET /Meta260 Do 0.737 w Q endobj q q << 1.007 0 0 1.007 551.058 583.429 cm Q stream /ID [] 0 g S << q /FormType 1 /Resources<< 434 0 obj q (1) Tj /Meta34 47 0 R endstream /BBox [0 0 30.642 16.44] Q /Subtype /Form q stream /Meta219 233 0 R stream q 0.463 Tc >> /Resources<< /FormType 1 >> /Subtype /Form >> >> 0.524 Tc /Subtype /Form /BBox [0 0 534.67 16.44] 1 i >> Q Q 294 0 obj Q /ProcSet[/PDF] /Meta117 Do 1 g /F3 17 0 R /FormType 1 0 g 0 w >> 0 g endstream stream Q q 0 w /BBox [0 0 30.642 16.44] q << >> Q ET /Type /XObject stream /Matrix [1 0 0 1 0 0] /Font << 20.21 5.203 TD /ProcSet[/PDF/Text] /Subtype /Form >> /Resources<< >> /Length 116 /FormType 1 stream /FormType 1 Q 149 0 obj /BBox [0 0 88.214 16.44] /Subtype /Form 0.737 w q /BBox [0 0 88.214 16.44] Notice that we used the variable \large {d} d in our equation to stand for our unknown value. /ProcSet[/PDF] /F3 17 0 R 1 i 0 g /Meta241 Do In other terms, 52-nx The problem is asking that you subtract twice a number from 52. 293 0 obj 1.014 0 0 1.007 531.485 383.934 cm endstream /BBox [0 0 23.896 16.44] /ProcSet[/PDF] 1.005 0 0 1.007 102.382 293.596 cm /Meta149 163 0 R /Resources<< q (+) Tj /Meta25 Do /Meta157 171 0 R 0 g q q /Type /XObject q 0.458 0 0 RG 1 g Q >> /BBox [0 0 15.59 29.168] 1 i 0.458 0 0 RG Q q q q >> /ProcSet[/PDF/Text] (5) Tj /Subtype /Form << /ProcSet[/PDF/Text] stream /Resources<< q /F1 7 0 R /Length 294 Q >> Just type into the box and your calculation will happen automatically. 1 i Q endobj Q >> /Font << >> Q q /Meta85 99 0 R 0 g /Length 69 /Length 16 (-11) Tj Q /FormType 1 /Meta63 Do /Font << /Type /XObject stream /Matrix [1 0 0 1 0 0] q /FormType 1 /F3 12.131 Tf 0.458 0 0 RG /Type /XObject 1 g 1 i /BBox [0 0 88.214 35.886] stream Q /Subtype /Form /Matrix [1 0 0 1 0 0] Q /Resources<< 90 0 obj 65 0 obj endobj 0.51 Tc 1.014 0 0 1.007 531.485 583.429 cm /F3 12.131 Tf /Resources<< 1.014 0 0 1.007 111.416 330.484 cm q 1.007 0 0 1.007 551.058 277.035 cm Q q q << /Font << /F3 17 0 R 0 g /Resources<< S stream Making educational experiences better for everyone. /Meta260 274 0 R (D\)) Tj endstream Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. Q /F3 17 0 R << /Length 78 0 4.894 TD (11) Tj >> /Type /XObject 0 G Q q Q /FormType 1 >> >> /ProcSet[/PDF/Text] 1 i q Q /Matrix [1 0 0 1 0 0] Q 295 0 obj 1 g q 1.005 0 0 1.007 79.798 730.228 cm /F4 36 0 R /Meta232 Do endstream q >> Q endstream /F3 12.131 Tf >> Q /Matrix [1 0 0 1 0 0] 1 i /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.369 Tc /BBox [0 0 88.214 16.44] -0.099 Tw 0.564 G stream >> 1 i /BBox [0 0 15.59 16.44] BT stream stream /Type /XObject q /Meta170 Do 0 4.894 TD /F3 17 0 R /Meta52 66 0 R (C\)) Tj 1 i /Subtype /Form (5) Tj Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. >> >> endobj 0 4.894 TD /F1 7 0 R /Subtype /Form /Font << q Q >> q 380 0 obj q q q /Type /XObject stream 91 0 obj stream endobj /BBox [0 0 88.214 16.44] /Meta173 187 0 R /Subtype /Form /FormType 1 /Meta41 Do 0 w >> >> << endstream >> /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] q /Subtype /Form 0 g 1.007 0 0 1.007 130.989 277.035 cm /Subtype /Form << 0 g q /Meta96 110 0 R /Font << /Type /XObject /Length 68 1 i 3.742 8.18 TD endobj endobj Q /Matrix [1 0 0 1 0 0] /MaxWidth 1453 /Resources<< 0 w /FormType 1 ( \() Tj endobj Q Q 6.746 5.203 TD >> >> 0.37 Tc q /Meta158 Do 1 g 0 g 0.458 0 0 RG 1.007 0 0 1.007 130.989 776.149 cm /Meta148 Do /AvgWidth 401 q /Meta84 98 0 R q >> 2. 0 g ET Q 1 i /Matrix [1 0 0 1 0 0] >> Q Q BT 305 0 obj 1 g >> /ProcSet[/PDF/Text] endobj /Type /XObject endstream << ET q /Length 16 endobj /F3 12.131 Tf /FormType 1 /Subtype /Form /FormType 1 endobj Q >> /Meta385 401 0 R >> q /F3 12.131 Tf /Type /XObject 1.014 0 0 1.007 251.439 277.035 cm >> /Subtype /Form /F3 17 0 R /Length 68 q q Q q /Type /XObject /Matrix [1 0 0 1 0 0] Q stream 0.17 Tc /Meta210 224 0 R Q q 163 0 obj 722.699 293.596 l Find the number. /Resources<< BT >> /Matrix [1 0 0 1 0 0] /Type /XObject 9.723 5.336 TD 0 G /Subtype /Form endobj 0 G q ET Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM endstream 1 i /F3 17 0 R /Subtype /Form /Parent 1 0 R /BBox [0 0 30.642 16.44] Q /Meta296 Do endstream 1 g >> S /Meta101 115 0 R 0 G 0.564 G /Meta19 30 0 R 0 G 1 i /ProcSet[/PDF] q 0 w /Type /FontDescriptor q /Meta83 Do 0 G 0 G endobj /Meta163 177 0 R BT /Font << Q endobj 1 i Q >> 0.458 0 0 RG Q endstream << q endobj (+) Tj q /Matrix [1 0 0 1 0 0] Q q q 0 w 0 G 1.007 0 0 1.006 130.989 690.329 cm /I0 Do /Length 12 >> Q 3.742 5.203 TD /Matrix [1 0 0 1 0 0] endstream ET /Meta422 Do 1.007 0 0 1.007 67.753 293.596 cm /Meta314 Do /Resources<< /ProcSet[/PDF/Text] /FormType 1 /Meta37 Do /Subtype /Form q q 1 g /Length 70 /Length 69 q Twice a number decreased by another number: 0.382 Tc /Font << 1.005 0 0 1.007 45.168 889.071 cm 0 5.203 TD >> q stream -0.486 Tw /Meta120 134 0 R 0 g /BBox [0 0 15.59 16.44] Q Q /Length 16 Q /Resources<< 97 0 obj >> /Subtype /Form /F1 7 0 R /FormType 1 q Q ET /F3 12.131 Tf endstream BT >> endobj /Resources<< /BBox [0 0 30.642 16.44] /Type /XObject Q /Meta421 Do /Meta254 Do Q /Resources<< /F3 17 0 R 0.564 G 0 w << Q (6\)) Tj Q endobj 1 g 1 i q 15.731 5.336 TD stream 0 G /Meta298 Do >> endstream q 1.014 0 0 1.007 251.439 703.126 cm q Q q 0 G /BBox [0 0 88.214 16.44] << Q /I0 51 0 R >> Q stream /ProcSet[/PDF/Text] 1.007 0 0 1.007 411.035 383.934 cm /Subtype /Form /FormType 1 << /Meta167 Do /Length 68 /Length 78 ET /Type /XObject /ProcSet[/PDF] 0 g /Font << /F3 12.131 Tf /Font << /Matrix [1 0 0 1 0 0] << (3) Tj /Length 87 << 0 g /Subtype /Form q 0 g << /FormType 1 1.005 0 0 1.015 45.168 53.449 cm q Q /Resources<< stream endobj >> BT /Meta206 Do /Length 54 ET q >> q >> /Subtype /Form 1.007 0 0 1.007 130.989 583.429 cm 0 5.203 TD /ProcSet[/PDF/Text] q >> 0 g /Matrix [1 0 0 1 0 0] /Type /XObject /Type /XObject /ProcSet[/PDF/Text] Q endobj /FormType 1 313 0 obj 1.007 0 0 1.006 130.989 437.384 cm q 0 5.203 TD endobj << Q 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject 1.007 0 0 1.007 271.012 523.204 cm /Matrix [1 0 0 1 0 0] endobj /Type /XObject 1.007 0 0 1.007 271.012 330.484 cm endobj Q /Matrix [1 0 0 1 0 0] (- 4) Tj /Matrix [1 0 0 1 0 0] 0.458 0 0 RG Q /Meta307 Do endstream Q << endstream Q: when six times a number is decreased by 4, the result is 8. >> /Leading 349 endobj 183 0 obj /Length 69 q q /F4 12.131 Tf Find the number 1 See answer Advertisement 0.737 w q << << << 672.261 872.509 m 443 0 obj /Font << BT q /FormType 1 /Count 2 << q 156 0 obj -0.486 Tw /Matrix [1 0 0 1 0 0] /Meta276 Do stream Q Q >> /F1 12.131 Tf q endobj Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. 0 w if the solution of an equation is x=-2, what could the original equation be? 1 i q >> q 1 i >> 0 G q 0 g /Matrix [1 0 0 1 0 0] /Subtype /Form /Font << 0.564 G Q /XObject << >> /Length 16 >> 174.501 5.203 TD 0 G /Meta45 59 0 R Q q Q stream q a.) 0.737 w /Meta30 Do /Font << q 0 w q /Matrix [1 0 0 1 0 0] ET ET /CapHeight 476 /Meta346 Do [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. 1 i 1 g /Type /XObject << 141 0 obj /Resources<< << Q BT /Matrix [1 0 0 1 0 0] 0 G /Meta396 412 0 R /F3 17 0 R endobj - 9737014. /BBox [0 0 15.59 29.168] 1.007 0 0 1.007 551.058 330.484 cm (-8) Tj /Meta347 Do 1.005 0 0 1.007 102.382 347.046 cm 0 G 0 w q /Matrix [1 0 0 1 0 0] Q >> Q >> >> /FormType 1 /Type /XObject endobj 1 i /Length 16 BT Twice a number is decreased by 9, and this sum is multiplied by 4. stream Q Q /FormType 1 stream 1 i /Meta310 Do 0.738 Tc endobj 100 0 obj 0 G /FormType 1 Q ET /Length 16 ET /Type /XObject 1 i 0.737 w /Meta168 Do Q 1 i q q 0 g /Type /XObject /Length 69 stream endobj endobj >> /ProcSet[/PDF/Text] /F3 17 0 R endstream ET /Resources<< /Type /XObject 0.737 w endstream >> Q /Subtype /Form The difference between six and a number divided by nine 10. /Subtype /Form 23 0 obj /BBox [0 0 534.67 16.44] 722.699 347.046 l /Meta135 Do 1.007 0 0 1.007 551.058 383.934 cm 1 g /ProcSet[/PDF] Q q /F3 12.131 Tf 1.014 0 0 1.007 251.439 277.035 cm 0.458 0 0 RG q /Font << /ProcSet[/PDF/Text] /Meta50 64 0 R /ProcSet[/PDF] >> /Matrix [1 0 0 1 0 0] /Flags 32 endobj 0.369 Tc >> 0 G 27.693 5.203 TD Q /Type /XObject /BBox [0 0 534.67 16.44] /Font << /Matrix [1 0 0 1 0 0] 0.425 Tc q q TJ endstream /Font << ET 0 G Q 0 g /F3 17 0 R /BBox [0 0 30.642 16.44] q 1 i << Q 0 g q (B\)) Tj 174 0 obj Q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 776.149 cm 1 i 1 i 1.007 0 0 1.006 411.035 690.329 cm 377 0 obj /Subtype /Form (13) Tj /Meta245 Do /Subtype /Form /FormType 1 Q /Meta189 203 0 R /ProcSet[/PDF/Text] Q >> /FormType 1 79 0 obj 0 g Q >> /Meta317 331 0 R /FormType 1 1.005 0 0 1.007 102.382 417.058 cm q /Font << /Subtype /Form /Font << 0 g /Font << /Meta247 Do /Subtype /Form endstream /Matrix [1 0 0 1 0 0] 0 G Q Q /Meta218 232 0 R /ProcSet[/PDF/Text] Q /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] Q Q /Resources<< q /Meta400 Do >> /Matrix [1 0 0 1 0 0] /Type /XObject Q /Meta279 Do >> >> /ProcSet[/PDF] /BBox [0 0 88.214 16.44] (5) Tj 1 i /BBox [0 0 673.937 15.562] /Type /XObject 0.564 G q q 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 BT endobj /FormType 1 /F3 12.131 Tf q stream 1 g /Meta172 186 0 R /Subtype /Form /FormType 1 << 1 i >> /Length 69 /Subtype /Form 1 g /Font << 0 g /Meta169 183 0 R BT Q /Meta136 Do 1.005 0 0 1.007 102.382 872.509 cm << q << 1 g << /Subtype /Form >> endobj Solution. /F3 17 0 R 0.369 Tc 0.458 0 0 RG /Length 59 endstream >> endobj /Meta320 Do 0.564 G 1 i >> /Meta413 Do /Matrix [1 0 0 1 0 0] 0.737 w /Resources<< /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] << /Subtype /Form 1 i If n is "the number," which equation could be used to solve for the number? /Type /XObject /Font << 1.007 0 0 1.006 551.058 763.351 cm /Resources<< 19 0 obj endstream Q q Q /Font << 1 i >> /Meta263 277 0 R /StemH 94 /Resources<< q endstream q Q 0.458 0 0 RG /StemV 77 /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 523.204 cm 241 0 obj /Font << 315 0 obj Q 0.458 0 0 RG >> /BBox [0 0 15.59 29.168] /BBox [0 0 15.59 29.168] BT (+) Tj /Font << BT BT q ET 0 g 0 g /Meta396 Do /Length 69 << /Meta248 Do /Meta424 Do << stream (+) Tj /F1 7 0 R /BBox [0 0 88.214 16.44] /Meta367 381 0 R Q 1.005 0 0 1.007 102.382 473.519 cm stream 393 0 obj /FormType 1 1 g /FormType 1 0.564 G 0.737 w q /Font << /BBox [0 0 88.214 16.44] /Meta42 56 0 R ET 0.737 w (x ) Tj Q /F3 12.131 Tf 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 /Meta390 Do 382 0 obj << /BBox [0 0 88.214 16.44] >> /Type /XObject Q /Length 91 0 g >> endobj /Subtype /Form q endobj /Type /XObject /Meta384 Do q stream >> /Length 16 << /BBox [0 0 15.59 16.44] BT endobj q 1 g q BT Q stream Q /Length 118 stream /Length 16 S /Subtype /Form /Subtype /Form /BBox [0 0 534.67 16.44] >> 374 0 obj >> Q 0.564 G 20.21 5.336 TD q endstream 336 0 obj 0 w endstream /Resources<< q /Length 60 /Meta40 54 0 R /Meta244 258 0 R 1 i >> endstream Q (4\)) Tj q Q Q q 1 i Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. 0 G /BBox [0 0 30.642 16.44] /Meta106 Do q 0.838 Tc q >> /Length 59 /Meta426 Do /Length 59 Q /Meta197 211 0 R 1.005 0 0 1.007 79.798 779.913 cm /ProcSet[/PDF/Text] /ProcSet[/PDF] /Resources<< Find the length. Q /F3 17 0 R >> 1 i (B\)) Tj /Length 69 /Length 69 /ProcSet[/PDF/Text] << 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. /Font << S endobj 0.564 G 1 i /Subtype /Form q /FormType 1 /FormType 1 /Resources<< stream /Length 69 >> >> stream /BBox [0 0 88.214 16.44] /Meta258 272 0 R Q /Type /XObject /Resources<< 0 g q /Type /XObject /ProcSet[/PDF/Text] Q /FormType 1 0 g /Resources<< /FormType 1 /Subtype /Form q endstream /F3 12.131 Tf /Meta371 Do 0 G By the . /FormType 1 endobj 360 0 obj q Q /Length 69 1 i BT /ProcSet[/PDF] /F1 7 0 R 0 w /Font << q Q 0 G /Meta233 247 0 R 1.007 0 0 1.007 411.035 849.172 cm 0 G >> endobj q q /Resources<< 0 g Q Q endstream /Resources<< 0.369 Tc << 1 i Q 23.952 4.894 TD /Subtype /Form /Resources<< /Meta177 191 0 R /Meta4 13 0 R stream /Meta80 94 0 R /Meta401 Do endobj 89.12 5.203 TD q 0 g /F4 12.131 Tf 1 i q /Meta267 281 0 R /BBox [0 0 88.214 16.44] /FormType 1 q q 1 i /Meta329 343 0 R /F1 7 0 R /F1 14.682 Tf /Resources<< /ProcSet[/PDF/Text] /Resources<< 1 i /FormType 1 /Font << /BBox [0 0 639.552 16.44] /ProcSet[/PDF/Text] /F1 12.131 Tf Q 1 i 1.005 0 0 1.007 102.382 400.496 cm endstream q ET /F3 17 0 R /Meta95 Do 0 4.894 TD /Length 118 stream /F3 12.131 Tf /BBox [0 0 30.642 16.44] BT >> q /Type /XObject 0 g Q >> q >> /Matrix [1 0 0 1 0 0] BT BT stream q << /FormType 1 endstream Q 1 g /F3 12.131 Tf 0 w /Meta420 Do /Meta155 Do Q /Meta71 85 0 R >> q /Matrix [1 0 0 1 0 0] BT /BBox [0 0 673.937 14.853] S >> 0.564 G 20.21 5.203 TD /Subtype /Form /Type /XObject Q Q /Length 16 q /Font << /Resources<< q stream Q /Resources<< /Matrix [1 0 0 1 0 0] /Meta28 41 0 R endstream 1 i /Resources<< 406 0 obj >> /Meta70 Do /Subtype /Form Q 1 i 0.564 G ET stream stream >> 1 g >> >> stream /F3 12.131 Tf Q /Subtype /Form q 0 G /Font << q q BT q Q << /Meta152 166 0 R q << 0 g endobj /Meta338 Do /Length 58 q /BBox [0 0 88.214 35.886] /Length 67 /Subtype /TrueType q /Font << 5 0 obj 1 i q (38) Tj endstream ( \() Tj 0.564 G 1.007 0 0 1.007 411.035 383.934 cm >> q /BBox [0 0 15.59 16.44] stream q endobj 1 i Q /FormType 1 /Resources<< q q 186 0 obj BT 50 0 obj /Type /XObject q /BBox [0 0 88.214 16.44] q /Length 65 /Length 69 0 g ET 1 i endobj /Matrix [1 0 0 1 0 0] endstream /Font << /Meta137 151 0 R endstream Q endobj 1 i 0 g >> /FormType 1 /Font << Q Q 0.786 Tc /BBox [0 0 88.214 35.886] q >> endobj /F3 12.131 Tf stream /Meta108 122 0 R endobj /FormType 1 << stream 0.297 Tc endobj 88 0 obj /Subtype /Form stream 1.005 0 0 1.007 102.382 293.596 cm /FormType 1 q /Meta348 362 0 R 0 g 0 g q ET /F3 17 0 R (6\)) Tj 1 i << /Meta217 Do >> /BBox [0 0 15.59 16.44] /Type /XObject /Subtype /Form /Resources<< << /F3 17 0 R 1 i /F3 12.131 Tf Q /ProcSet[/PDF/Text] << endstream >> /Type /XObject endstream /Type /XObject 1.007 0 0 1.007 551.058 383.934 cm Q /Length 58 q 1.007 0 0 1.007 551.058 523.204 cm /BBox [0 0 30.642 16.44] 1 i /Length 16 214 0 obj /ProcSet[/PDF/Text] /MediaBox [0 0 767.868 993.712] stream /Meta43 57 0 R /Meta421 437 0 R /Subtype /Form Q q /Meta289 303 0 R q /FormType 1 endstream 1.007 0 0 1.007 271.012 776.149 cm 17.234 5.203 TD /F3 17 0 R endobj Q 276 0 obj >> /Resources<< >> 0.458 0 0 RG Q Q >> /Length 73 BT /Meta290 304 0 R q /Font << /Subtype /Form Q q << >> 0 g Q 0 g endstream 0 4.894 TD 0 g /BBox [0 0 88.214 16.44] 0 g Question 1. 0 g 217 0 obj BT >> /FormType 1 /Type /XObject 1.005 0 0 1.006 45.168 879.284 cm /BBox [0 0 15.59 16.44] Q << 0 G 1.007 0 0 1.007 130.989 383.934 cm /Font << endobj /Font << endstream >> q /Font << 0 G /Resources<< >> << Q We are asked to find the number, so, we could assign the number as "x". Q 2.238 5.203 TD /Font << endstream q endobj 0.458 0 0 RG /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] q /Meta142 156 0 R The symbols 17 + x = 68 form an algebraic equation. << >> /Meta275 289 0 R >> 0 g >> 113 0 obj /FormType 1 >> /Type /XObject /Length 68 /Meta155 169 0 R stream /BBox [0 0 17.177 16.44] q 0 g 1 i /FormType 1 /Type /XObject /StemH 88 q Q Q /Resources<< /Length 69 /Matrix [1 0 0 1 0 0] Q Find the number. q 2.238 5.203 TD BT Q 209 0 obj >> 1 i 1 i Q /Matrix [1 0 0 1 0 0] 1 i /Meta78 92 0 R >> q 0.564 G /FormType 1 0 g /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] ET endstream 0 G /FormType 1 234 0 obj >> /F3 12.131 Tf 0.564 G >> /MissingWidth 250 1.007 0 0 1.007 130.989 330.484 cm stream 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. /F3 12.131 Tf Q Q Q /Meta285 Do Andrew M. BT 0.425 Tc endstream /Meta177 Do 1.007 0 0 1.007 411.035 383.934 cm q >> q q >> q 672.261 473.519 m 0 5.203 TD /Matrix [1 0 0 1 0 0] Q 0.564 G q q << Q (4\)) Tj /FormType 1 /F3 12.131 Tf >> endstream >> 0.838 Tc q q 1.005 0 0 1.013 45.168 933.487 cm /ProcSet[/PDF/Text] /Type /XObject /ProcSet[/PDF/Text] Q (+) Tj /FormType 1 /Type /XObject 129 0 obj /Length 16 /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Length 79 /Type /XObject 25.454 5.203 TD 363 0 obj endobj 1.005 0 0 1.007 102.382 473.519 cm BT /Type /XObject /Meta36 49 0 R /Type /XObject /F3 12.131 Tf q BT >> 1.007 0 0 1.007 130.989 523.204 cm stream >> ET /FormType 1 endstream >> Twice = two times, double. /FormType 1 148 0 obj /F3 17 0 R /Length 69 /Meta311 Do /Meta262 276 0 R endobj q /Matrix [1 0 0 1 0 0] q Q /Resources<< q /Resources<< 152 0 obj endobj ET /Font << >> /Resources<< Q q /F3 17 0 R /Resources<< 1.005 0 0 1.007 79.798 862.723 cm endstream Q /Matrix [1 0 0 1 0 0] /F3 17 0 R /Meta256 Do 397 0 obj 0.458 0 0 RG Q /BBox [0 0 30.642 16.44] >> >> /Meta429 Do q q /Subtype /Form /Meta184 Do 1.007 0 0 1.007 551.058 277.035 cm << /Resources<< 0 g 0.458 0 0 RG >> 0.564 G Q ET stream q 0.564 G 1 g /BBox [0 0 639.552 16.44] 1 g /ProcSet[/PDF/Text] (x) Tj /Meta398 Do /BBox [0 0 88.214 16.44] 0.369 Tc /Font << 722.699 799.486 l >> 1 i endobj -0.22 Tw 0 5.203 TD /Type /XObject Q Q 277 0 obj 0 4.894 TD /Meta428 444 0 R /Meta285 299 0 R << Q 0 g >> 0 g 1 i q /Meta90 104 0 R /ProcSet[/PDF] q [( a )-15(number, decreased by )] TJ >> /Type /XObject stream 0 G endstream << >> /F4 12.131 Tf /BBox [0 0 88.214 16.44] >> stream q /Meta140 Do /Type /XObject endstream /Meta143 157 0 R /ProcSet[/PDF] 0 g 194 0 obj 0 G /Type /XObject /FormType 1 >> << endstream /Subtype /Form /Meta325 339 0 R (5) Tj 111 0 obj 1.502 5.203 TD /Meta154 168 0 R /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] ET >> q >> q Q >> 172 0 obj /Font << /F3 12.131 Tf /Meta248 262 0 R /Subtype /Form >> endobj q Q Q 22 0 obj 1.007 0 0 1.007 271.012 277.035 cm Q q /FormType 1 endobj << Q stream q Q >> /Matrix [1 0 0 1 0 0] >> /Type /XObject /F3 12.131 Tf /Meta111 Do 1.014 0 0 1.007 111.416 583.429 cm /F1 7 0 R endstream BT /F1 12.131 Tf 1 i 0 G 1.014 0 0 1.007 391.462 583.429 cm /Subtype /Form Q /Meta7 18 0 R /F1 12.131 Tf 0 G >> >> /Resources<< 1 i /F3 12.131 Tf /Meta110 124 0 R q >> q >> Q /Meta278 292 0 R >> 384 0 obj /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /Subtype /Form 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 << Five times a number, decreased by 58, is -23 Find the number. 1 g q q endobj /Meta5 14 0 R /Resources<< /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 251.439 636.879 cm /ProcSet[/PDF/Text] >> Q ET /Subtype /Form /Length 16 /Type /XObject /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 36 0 obj q /Matrix [1 0 0 1 0 0] >> 1.014 0 0 1.007 531.485 583.429 cm /F4 36 0 R /BBox [0 0 88.214 16.44] Q >> >> /Font << /Resources<< >> endobj /Meta340 Do endstream >> q (+) Tj (-11) Tj /Meta341 Do /Resources<< 1.007 0 0 1.007 130.989 523.204 cm /F3 17 0 R 0 g /Length 70 /Subtype /Form /BBox [0 0 673.937 15.562] /F3 17 0 R 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . /Resources<< 1.007 0 0 1.007 67.753 546.541 cm /BBox [0 0 88.214 16.44] stream /Type /XObject endstream Find an answer to your question Twice a number decreased by 8gives 58. /Matrix [1 0 0 1 0 0] 253 0 obj /Length 69 << BT >> BT Q /FormType 1 1 g Q 0 G endobj /Subtype /Form 0.564 G 274 0 obj 333 0 obj 0 g Q /Matrix [1 0 0 1 0 0] 0 g 0.458 0 0 RG Q << /Meta305 Do endstream endobj 41.186 5.203 TD /Matrix [1 0 0 1 0 0] 54.679 5.203 TD q Q /Matrix [1 0 0 1 0 0] 1 i /Length 63 Q 0 g 0 G >> BT [( and )16(a nu)26(mbe)18(r)] TJ endobj Q /Meta211 Do /Meta112 Do Q /Subtype /Form /Font << ET >> 0.458 0 0 RG Q (38) Tj /Meta240 Do Q 0 g /F4 36 0 R /Type /XObject q q >> (9) Tj /Length 58 q 1.014 0 0 1.007 111.416 583.429 cm 0.458 0 0 RG /Meta60 74 0 R /BBox [0 0 30.642 16.44] /BBox [0 0 88.214 16.44] /Meta273 Do 1.502 5.203 TD /MissingWidth 250 << Q ET /Matrix [1 0 0 1 0 0] Q /F4 36 0 R 6 0 obj /FormType 1 0.332 Tc /Subtype /Form endstream 0 g Q , Prove the following >> 0.564 G 0.737 w 1 g q << q >> q /BBox [0 0 15.59 29.168] five times the sum of a number x and two b.) /Meta5 Do stream /ProcSet[/PDF/Text] /Subtype /Form /BBox [0 0 15.59 16.44] /Meta19 Do stream stream >> 0 g 0.458 0 0 RG /Subtype /Form q /ProcSet[/PDF/Text] /Font << >> endobj q >> q /BBox [0 0 15.59 16.44] BT Q >> /F3 12.131 Tf << /Matrix [1 0 0 1 0 0] 1 i q stream q /Font << q BT 280 0 obj << endstream 1 g stream /Meta102 Do 0 g >> 267 0 obj q ET 30.699 4.894 TD endstream Formula - How to Calculate Percentage Decrease. 0.68 Tc /Length 70 1 i /Matrix [1 0 0 1 0 0] >> 263 0 obj Q 0 w >> >> /FormType 1 q /Meta308 Do /Meta300 Do 0.458 0 0 RG /FormType 1 stream