2 times x minus 58 C. twice the difference of a number and 5 B. twice a number decreased by 58 D. 2 times the sum of a number and 58 Answer: B. Step-by-step explanation: twice - (2) number - (x) 58-(58) Edukasyon. 6 more than twice a number: 2x+6: two less than a number: x-2: the sum of 9 and a number: 9+x: two less than three times a number: 3x-2: a number subtracted from 12 . 1 g Q << /ProcSet[/PDF/Text] /FormType 1 1 i /Resources<< /Type /XObject q /Length 54 /Meta309 323 0 R /Resources<< /Meta334 348 0 R q Q << 1 i /Font << (D\)) Tj BT /Subtype /Form ( x) Tj /Meta127 Do 101.849 5.203 TD 0.458 0 0 RG Q << << /F3 17 0 R Q 2.238 5.203 TD 0 g Q ET q q /ProcSet[/PDF/Text] /Subtype /Form /F3 17 0 R 0 g >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> q 0 g ET /Font << endstream q 0 G /BBox [0 0 15.59 16.44] /FormType 1 Q >> /Length 70 /ItalicAngle 0 0 g q q Q 0 g 1.007 0 0 1.007 411.035 849.172 cm 0 G >> q 1.502 7.841 TD /Meta327 Do /Length 60 ET 1 i /F3 12.131 Tf /Meta74 Do >> Check out a sample Q&A here. endstream >> 1 i endstream >> endobj Q /Type /XObject Q 445 0 obj q stream 1.007 0 0 1.007 654.946 400.496 cm 0 G 405 0 obj /Resources<< twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. /Meta86 Do /Length 54 q /ProcSet[/PDF/Text] << >> 1 g endobj /FormType 1 1 g /FormType 1 << 0.369 Tc 1 i /FormType 1 /Subtype /Form (9) Tj (A\)) Tj 1 i >> /ProcSet[/PDF] 0 G 1.007 0 0 1.007 130.989 776.149 cm << BT >> << 1.014 0 0 1.006 531.485 763.351 cm 0.369 Tc /Meta161 175 0 R 1.007 0 0 1.007 271.012 383.934 cm >> (5) Tj q 0.458 0 0 RG 229 0 obj Q 1 i stream q /Meta157 171 0 R << Q /ProcSet[/PDF] /Font << /Length 59 Q /Meta28 41 0 R Q /Meta68 Do << 1 i /FormType 1 /ProcSet[/PDF/Text] 1 i /FormType 1 0 G q stream 0 G /Subtype /Form 0 g q 0 G /Meta342 356 0 R Q 1 i Q /Subtype /Form endobj q q q /BBox [0 0 549.552 16.44] 1 g 1 i Q endobj Q 1.007 0 0 1.007 411.035 636.879 cm q /Meta129 143 0 R q (\)]) Tj /Resources<< /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 531.485 510.406 cm /F1 12.131 Tf /F3 12.131 Tf endstream q 0 G endobj /Meta111 Do /Type /XObject /Resources<< endstream q Q 431 0 obj /BBox [0 0 15.59 16.44] /Type /XObject /FormType 1 q Q /F3 17 0 R 0.68 Tc /BBox [0 0 30.642 16.44] Q /Subtype /Form >> /Matrix [1 0 0 1 0 0] /FormType 1 stream 1.007 0 0 1.007 271.012 523.204 cm q 0.001 Tw /Length 64 Five times a number, decreased by 58, is -23 Find the number. 0.786 Tc 1.502 5.203 TD Q Q /Length 69 Q stream Q Q 0.737 w /Subtype /Form Q endstream q Q endstream >> /Meta348 362 0 R 0.458 0 0 RG /Meta58 72 0 R endstream Q /Meta387 Do On the way, we sang songs for 20 minutes which is 10% of the time we were on the road. << 0 w << BT /FormType 1 Q Q /Meta276 290 0 R q stream >> /Subtype /Form ET /ProcSet[/PDF/Text] q q << /Matrix [1 0 0 1 0 0] /Meta385 Do ET BT Q /F3 12.131 Tf q /Type /XObject q /Meta198 Do >> >> >> BT /Font << (iii) 25 exceeds a number by 7. 0 g >> /Type /XObject << 1 i /Subtype /Form /Font << Q 0 G /BBox [0 0 88.214 35.886] >> stream /Font << Q (40) Tj 1 g (9\)) Tj q /Meta336 Do Q 0.458 0 0 RG /Length 16 /Resources<< >> >> /Resources<< endobj >> stream Q 0 g q /Length 69 Q /Meta322 336 0 R endobj q 0 g endobj endobj /Font << 0 w 1.007 0 0 1.006 551.058 437.384 cm >> 1.014 0 0 1.007 531.485 636.879 cm q q /F3 12.131 Tf (+) Tj endobj /F3 17 0 R q /BBox [0 0 88.214 16.44] /Resources<< 0 G 0 g >> >> Q 174 0 obj 359 0 obj stream << 0 g >> 0 G /Matrix [1 0 0 1 0 0] /Type /XObject 0 G 0 5.203 TD endstream 0 g /Matrix [1 0 0 1 0 0] /Resources<< The rate of positive findings after 1 round of screening in the LCSDP was more than twice . /Type /XObject 1 i BT /Meta98 112 0 R /Meta226 240 0 R stream 0 w /FormType 1 /Resources<< >> /Meta368 382 0 R 1 i /Resources<< q 0.271 Tc /Meta91 Do 0 G 1.007 0 0 1.007 411.035 277.035 cm /ProcSet[/PDF/Text] >> /ProcSet[/PDF/Text] 0 g q q q /Subtype /Form /Meta202 Do 1 i 0 G /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 370 0 obj >> 0 20.154 m /FormType 1 >> /F1 12.131 Tf /F1 12.131 Tf /F3 12.131 Tf >> /ProcSet[/PDF] /Font << /Length 64 /Resources<< endstream stream /Meta134 148 0 R >> /BBox [0 0 88.214 16.44] >> Q Q 0 w (2) Tj /FormType 1 << q >> 14.966 20.154 l >> /Meta180 194 0 R /Subtype /Image 1.005 0 0 1.007 79.798 713.666 cm 0.458 0 0 RG 0 g 0 5.203 TD >> 0 G Q /Type /XObject endobj endstream /FormType 1 Q BT 0 g /Meta263 Do /FormType 1 722.699 799.486 l q q /Subtype /Form /ProcSet[/PDF] /F3 12.131 Tf /FormType 1 /Subtype /Form /Font << q (13) Tj >> q /BBox [0 0 88.214 16.44] Q 1 g /Meta424 Do [tex]\sin (\pi -x)=\sin x[/tex]. ( \() Tj q /Type /XObject /BBox [0 0 30.642 16.44] /Subtype /Form /BBox [0 0 673.937 15.562] 0.458 0 0 RG Q /F3 12.131 Tf q /Meta173 Do 1.005 0 0 1.007 102.382 347.046 cm endstream /Meta349 Do /F1 7 0 R /Meta235 249 0 R (7\)) Tj stream q Q /FormType 1 << /Meta357 Do 277 0 obj q D. b = 4 2. 1 i endobj (2\)) Tj Q /BBox [0 0 15.59 29.168] /Meta121 Do 0 G /Meta24 Do >> >> /FormType 1 Q /Matrix [1 0 0 1 0 0] q BT endstream /BBox [0 0 88.214 16.44] 402 0 obj 1.007 0 0 1.007 654.946 400.496 cm >> /Resources<< q >> Q 1 i /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Resources<< >> >> 1.014 0 0 1.007 531.485 636.879 cm q Q BT /Subtype /Form /F4 36 0 R 0 g Q /Subtype /Form >> 0.51 Tc ET /Resources<< 0 g q /Resources<< stream /Matrix [1 0 0 1 0 0] q /Type /XObject Q endstream /Resources<< /Meta0 5 0 R ET 150 0 obj >> >> >> (5\)) Tj >> q /Matrix [1 0 0 1 0 0] /Subtype /Form BT q /Subtype /Form /Matrix [1 0 0 1 0 0] >> >> Q Q >> /ProcSet[/PDF] (4\)) Tj /Resources<< 99 0 obj /Resources<< /Length 54 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 << /Length 16 Q Q A link to the app was sent to your phone. /Resources<< /BBox [0 0 88.214 16.44] ET >> 0.737 w Q q q 1 i (4\)) Tj /Subtype /Form /BBox [0 0 15.59 16.44] /Type /XObject Advertisement Loved by our community 50 people found it helpful Madhvendra13 2x -8=58 2x=66 x=662 x=33 Find Math textbook solutions? /Meta228 242 0 R 0 G /Subtype /Form Q << q /Meta10 21 0 R /ProcSet[/PDF] >> Q >> 0.564 G 1.007 0 0 1.007 551.058 636.879 cm q 1 i 1 i /Subtype /Form endobj Twice a number decreased by another number: /Type /XObject /Resources<< >> /Length 12 /Meta61 Do >> 0 G /Type /XObject q /Subtype /Form >> q 80 0 obj ET (-23) Tj /Resources<< q >> endobj 0 g Q /Matrix [1 0 0 1 0 0] >> Q 0.369 Tc /Subtype /Form endobj /Resources<< /Subtype /Form 0.458 0 0 RG q q q q /Meta29 42 0 R /BBox [0 0 15.59 16.44] 20.21 5.203 TD endobj 0 G /FormType 1 /Length 106 287 0 obj q /Matrix [1 0 0 1 0 0] 0.369 Tc stream >> /BBox [0 0 88.214 35.886] /Subtype /Form /BBox [0 0 30.642 16.44] 1 g 60 0 obj 1 i >> Q /Font << Q Q 0 G >> /BBox [0 0 88.214 35.886] /Meta363 Do q 0 g /Meta94 Do Q ( x) Tj q 0.564 G 1.007 0 0 1.007 130.989 277.035 cm /Font << Q The sum Of twice a nu4ber What is the number? /Subtype /Form /Meta311 325 0 R /ProcSet[/PDF] /Meta319 333 0 R >> Q stream endobj << q /Meta6 15 0 R /Meta196 Do ET 1.014 0 0 1.007 531.485 330.484 cm /Type /XObject 0.564 G /Subtype /Form 1 i Q /F3 12.131 Tf 1 i /F4 12.131 Tf 0.458 0 0 RG << 247 0 obj /Subtype /Form BT >> /Type /XObject /Subtype /Form 1 i /Meta27 40 0 R What is marios jumps times luigis jumps. Q Q /F3 12.131 Tf [(Negativ)16(e )] TJ /ProcSet[/PDF] Q /Resources<< /BBox [0 0 88.214 16.44] ET (D\)) Tj /Matrix [1 0 0 1 0 0] >> BT ET q /BBox [0 0 30.642 16.44] << /Type /XObject /Length 118 0.241 Tc /Meta415 431 0 R 1.007 0 0 1.006 130.989 437.384 cm /Length 16 38.948 5.203 TD endstream /Font << /Type /XObject 0 G Q /Font << q Q q /FormType 1 /Matrix [1 0 0 1 0 0] For the lesson, he grabs a glass container shaped like a rectan /Type /XObject 0 w A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. endstream endstream /Meta310 Do /BBox [0 0 15.59 16.44] /MaxWidth 1453 endstream << Q 1.005 0 0 1.007 102.382 653.441 cm endstream 1 i /F3 12.131 Tf /Meta405 421 0 R 6. BT ET << /Meta144 158 0 R 0 g 1 0 obj endstream (C\)) Tj Q endobj /F4 12.131 Tf 1 i q A number divided by six is eight: (k / 6) = 8. Q q endobj 0 g endstream Q /Matrix [1 0 0 1 0 0] endstream BT /BBox [0 0 88.214 16.44] /FormType 1 1.007 0 0 1.006 411.035 690.329 cm q >> Q 13.493 5.336 TD 1.005 0 0 1.007 102.382 473.519 cm 0 g Q endstream 0 G -0.101 Tw Q /BBox [0 0 88.214 16.44] /Encoding /WinAnsiEncoding endstream 204 0 obj Q q 0 g q Q 0 5.203 TD ( \() Tj >> /Meta354 368 0 R >> (C) Tj /Matrix [1 0 0 1 0 0] >> (-8) Tj /BBox [0 0 88.214 16.44] >> 1 i /Subtype /Form 0.564 G 1 i endobj Q /Type /XObject 1.007 0 0 1.007 271.012 383.934 cm >> << /BBox [0 0 639.552 16.44] q >> /Font << /BBox [0 0 88.214 16.44] endobj A number = an unknown number which can be represented by a variable, usually x. BT 2. stream >> 0 G 0 w >> Q endstream /Resources<< Q Q q >> /Type /XObject ET ET /Resources<< 9.723 5.336 TD Q /Resources<< /BBox [0 0 534.67 16.44] 0 g 0 5.203 TD /Meta299 Do /Length 16 << /Length 59 /Resources<< Negative thirteen decreased by 3 times a number x. /Meta197 211 0 R /F3 12.131 Tf 6.746 24.649 TD 1 i /Resources<< q Q /Meta254 268 0 R (40) Tj /F3 12.131 Tf 0 G /ProcSet[/PDF] q 0 w 1.007 0 0 1.007 271.012 849.172 cm 0 5.203 TD Q q >> 1 g (B) Tj Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM 0 w /BBox [0 0 88.214 16.44] Q /Subtype /Form /Resources<< /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] /Meta50 64 0 R -0.092 Tw >> /Font << endobj /F3 17 0 R BT BT 1.007 0 0 1.007 551.058 383.934 cm endobj /ProcSet[/PDF/Text] >> /Resources<< /ProcSet[/PDF] ET ET >> /Meta198 212 0 R /Subtype /Form 0.524 Tc endobj /I0 51 0 R /F3 17 0 R 0 g q >> /Resources<< /F3 17 0 R q >> q stream /Font << endobj 1 i 3.742 8.18 TD 1 i /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /FormType 1 271 0 obj ET 0.382 Tc >> 0 w 1.005 0 0 1.007 79.798 713.666 cm ET /F3 12.131 Tf 434 0 obj q << [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ /BBox [0 0 673.937 16.44] /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 88.214 16.44] Q /ProcSet[/PDF] So we have twice of a mystery number decreased by three, and that is all going to be 31. 0 g q /ProcSet[/PDF/Text] 0 G 1.007 0 0 1.006 411.035 510.406 cm << /Meta170 Do 164 0 obj 1.502 5.203 TD endstream /FormType 1 /Matrix [1 0 0 1 0 0] /Meta321 Do >> endstream /FormType 1 /Length 69 0.369 Tc /LastChar 121 /F3 17 0 R ET q 0.737 w /Meta233 247 0 R /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] >> /F3 17 0 R /Meta423 439 0 R /Resources<< q Q /Font << /Length 69 1 i /Subtype /Form /F3 17 0 R Q endobj /FormType 1 q /Meta245 Do /Resources<< << /ProcSet[/PDF] /FormType 1 /ProcSet[/PDF/Text] Q Q /Meta415 Do 84 0 obj 1 g 1 g >> >> 396 0 obj >> << BT 1 i 0.737 w /Meta416 432 0 R /Type /XObject 1.007 0 0 1.007 411.035 330.484 cm /F1 7 0 R /Matrix [1 0 0 1 0 0] /Type /XObject /FormType 1 stream Q 0 w stream 1.005 0 0 1.007 102.382 653.441 cm /Meta346 Do stream Q 1 g Q /Resources<< ET q /BBox [0 0 30.642 16.44] /FormType 1 1.014 0 0 1.006 111.416 690.329 cm /Meta386 402 0 R stream /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject Thrice a number decreased by 5 is 3x - 5. q /Resources<< 18 0 obj 0.425 Tc 0 w Q ET >> 0 G /Meta134 Do 0.458 0 0 RG endobj 0 G /Font << 0.737 w /Font << Q 30.699 4.894 TD stream 1.007 0 0 1.007 551.058 636.879 cm << >> >> q endstream /ProcSet[/PDF/Text] ET >> 0.564 G Q /F3 12.131 Tf Q -0.008 Tw /Meta283 Do /Meta24 37 0 R /Meta264 Do q /I0 Do >> Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. q endobj ( \() Tj 0 4.894 TD /Resources<< stream >> BT 0.737 w endstream 1.014 0 0 1.007 531.485 523.204 cm 1.014 0 0 1.007 251.439 330.484 cm 154.289 4.894 TD 94.364 5.203 TD endobj /Type /XObject /FormType 1 Q 1 i endstream /F3 12.131 Tf /Meta379 Do 0 4.78 TD q /Subtype /Form << stream q /Subtype /Form /Resources<< 0.838 Tc q 0 g 0 w >> q 1.005 0 0 1.007 102.382 670.003 cm -0.047 Tw 0 g 0.458 0 0 RG q (C\)) Tj 1 i endobj 1 g 0.458 0 0 RG stream /Length 68 q >> /Resources<< Q 0 G /Length 16 16.469 5.336 TD endstream Q 0.458 0 0 RG (58) Tj endstream endobj Q 207 0 obj >> 1 g >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Subtype /Form Q BT /Subtype /Form Q Q /FormType 1 /ProcSet[/PDF/Text] 16 0 obj q /F1 7 0 R stream Q Q BT 1 g q 0.838 Tc /Meta25 Do stream << /Subtype /Form 0 G /Resources<< Q q endstream 1 i /Resources<< [( subt)-17(racted fr)-14(om a )-16(number)] TJ Q Q /Resources<< 16.469 5.203 TD endobj Q BT Q /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf >> BT Q /Resources<< Q /Subtype /Form q /BBox [0 0 15.59 29.168] stream (2\)) Tj 1.014 0 0 1.006 251.439 763.351 cm >> 1 g 1 i Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 Mathematics 619 solutions >> Q 1.007 0 0 1.007 45.168 746.789 cm /FormType 1 q /Subtype /Form (x) Tj 0 g 139 0 obj /F3 17 0 R /Font << /BBox [0 0 88.214 16.44] /Meta239 Do Q stream /Meta143 157 0 R 1.007 0 0 1.007 654.946 473.519 cm Q >> >> Q If twice a number is decreased by 13, the result is 9. /Font << Andrew M. q q 0.458 0 0 RG q >> q /BBox [0 0 15.59 29.168] 410 0 obj 1 i 0.738 Tc q 218 0 obj /Subtype /Form Q /BBox [0 0 15.59 16.44] 360 0 obj /F3 17 0 R 1 g q >> %PDF-1.4 stream /FormType 1 /Meta140 154 0 R BT A rectangular garden has a width that is 8 feet less than twice the length. /F1 7 0 R >> Q 2.238 5.203 TD stream >> /BaseFont /PalatinoLinotype-Roman 1 i << endobj 0.737 w /F3 17 0 R (x) Tj endstream Q /Resources<< >> 1 g /Meta296 Do /Font << /Type /XObject 0.737 w 0 g Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . BT >> q /Resources<< q >> >> 0 g Q Q /Type /XObject Q: A number increased by 5 is equivalent to twice the same number decreased by 7. /Meta186 Do 1 i q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] endstream /F3 17 0 R /Font << 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q Q /FormType 1 q stream /Matrix [1 0 0 1 0 0] 20.21 5.203 TD >> /Type /XObject /BBox [0 0 88.214 35.886] /BBox [0 0 534.67 16.44] /Meta314 Do Q 13.464 5.203 TD ET 12.727 5.203 TD 0.458 0 0 RG /Type /XObject /Meta80 Do 0.524 Tc q -0.463 Tw Q /Subtype /Form -0.062 Tw endstream /ProcSet[/PDF] endstream >> /Matrix [1 0 0 1 0 0] q /Font << q stream -0.106 Tw /FormType 1 /F3 17 0 R >> 0.737 w endstream 113 0 obj q 0 g 0 G 0.369 Tc 1.014 0 0 1.006 251.439 510.406 cm 352 0 obj BT << 1.007 0 0 1.006 411.035 763.351 cm /Type /XObject /Type /XObject q 1.007 0 0 1.007 271.012 523.204 cm /Matrix [1 0 0 1 0 0] 0.738 Tc 0.458 0 0 RG /Resources<< endobj -0.029 Tw [(-3)-16(20)] TJ /Type /XObject /Resources<< q Q q stream Thirthy is equal to twice a number decreased by four = solve and check the equation? q endobj /Font << endstream Q >> stream /Resources<< << 354 0 obj Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0 g /FormType 1 Q /Length 16 >> /F1 12.131 Tf 382 0 obj /FormType 1 Q Let x the unknown number. /Meta219 233 0 R Q 722.699 653.441 l /Type /XObject 0 g endobj Q endobj 1.502 5.203 TD << /F1 12.131 Tf q endstream /FormType 1 q << /Matrix [1 0 0 1 0 0] (-20) Tj 249 0 obj Then the following equation can represent this problem: 17 + x = 68 We can subtract 17 from both sides of the equation to find the value of x. Q /Resources<< 1 i >>