/Meta336 Do /FormType 1 384 0 obj /Resources<< << /Matrix [1 0 0 1 0 0] /Type /XObject Q 1 i >> endobj 1.014 0 0 1.007 111.416 703.126 cm /BBox [0 0 88.214 16.44] 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) /Matrix [1 0 0 1 0 0] Q q /Resources<< /Font << endobj 125 0 obj /Subtype /Form /Subtype /Form /Meta135 149 0 R /Meta42 56 0 R /Subtype /Form /FirstChar 43 0 G /Parent 1 0 R 230 0 obj << BT to represent the numbers. >> >> >> /Type /XObject >> /Subtype /Form /F4 12.131 Tf 1 i q /Font << /Meta380 394 0 R Q endobj /BBox [0 0 88.214 16.44] /Meta372 386 0 R >> q /Matrix [1 0 0 1 0 0] Q q BT This site is using cookies under cookie policy . >> /F3 17 0 R /ProcSet[/PDF] endstream 1 i /Length 58 endobj /ProcSet[/PDF] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] See Solution. << Q /Font << 0.51 Tc endobj Q Q /FormType 1 If you are unsure of the county, call the Administrative Office of the Court at (919) 890-1000.Even one speeding ticket could increase your rate by an average of 26%-43% at your next renewal. if the solution of an equation is x=-2, what could the original equation be? Q /Meta253 Do << >> Q >> 0 g /ProcSet[/PDF] /Resources<< q endstream Find an answer to your question Twice a number decreased by 8gives 58. q /FormType 1 /Subtype /Form 1.014 0 0 1.006 531.485 510.406 cm endobj /BBox [0 0 673.937 68.796] Kobe scored 85 points in a basketball game. 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q /Length 69 (38) Tj /Meta53 67 0 R /FormType 1 >> /Resources<< /F3 12.131 Tf Q 333 0 obj endobj The results found were expressed mainly through tables and graphs as the main resources of the statistical language. q 0 g /F4 36 0 R << 0.369 Tc /Meta129 143 0 R /Subtype /Form /Type /XObject Q stream Twice a number decreased by ten is greater than 24. /ProcSet[/PDF/Text] q q 1 i 94 0 obj /Meta2 Do /FormType 1 /Meta98 Do /Meta371 Do /Font << 20.975 5.336 TD /Meta82 96 0 R endstream /ProcSet[/PDF/Text] q /Subtype /Form 0 w BT 1 g Q The rate of positive findings after 1 round of screening in the LCSDP was more than twice . /Type /XObject /Resources<< /ProcSet[/PDF] << Q /ProcSet[/PDF/Text] endobj << Q << ET endstream 1 i << /Leading 349 Q /Length 12 /F4 12.131 Tf /F3 12.131 Tf /FormType 1 endobj /Meta300 Do Q 1.502 24.649 TD /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Meta296 310 0 R >> q /Type /XObject Q q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /BBox [0 0 30.642 16.44] Q >> (-) Tj q 1 g /Meta165 Do /F3 17 0 R q /Type /XObject endobj /BBox [0 0 30.642 16.44] /Font << 0 G Q 0 G BT Q /Meta28 Do 0 G /Subtype /Form 1.005 0 0 1.007 45.168 889.071 cm q 1.005 0 0 1.007 102.382 799.486 cm Q /Type /XObject ET Q q /Meta308 322 0 R /AvgWidth 657 124 0 obj /F3 12.131 Tf Q [(-3)-16(20)] TJ /Subtype /Form endobj New questions in Mathematics /Font << /F3 12.131 Tf /F3 12.131 Tf >> /F3 17 0 R /ProcSet[/PDF] Calculate a 15% decrease from any number. /Length 59 >> (-11) Tj Q ET << /Font << /Resources<< << /Meta328 342 0 R /F3 12.131 Tf 0.369 Tc /BBox [0 0 15.59 16.44] endstream q /Subtype /Form /Type /XObject /BBox [0 0 15.59 16.44] /Meta71 Do 1.007 0 0 1.007 67.753 653.441 cm 0 g /Length 119 /BBox [0 0 15.59 16.44] BT /Type /XObject 1 i 1.005 0 0 1.007 102.382 293.596 cm 0.458 0 0 RG endstream stream Q endstream /Meta25 Do /Resources<< ET q 0 g >> /ProcSet[/PDF/Text] /FormType 1 /FormType 1 >> q 0 4.894 TD Q stream q q Q /Resources<< q /Subtype /Form 1.007 0 0 1.007 67.753 293.596 cm >> >> >> /Subtype /Form endobj 1.005 0 0 1.007 102.382 546.541 cm q /F3 17 0 R stream /Font << /Type /XObject endobj 0 G 1.014 0 0 1.007 251.439 277.035 cm /BBox [0 0 15.59 16.44] /ProcSet[/PDF] /Length 69 q /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 q /Subtype /Form /Meta119 Do << Q 107 0 obj /BBox [0 0 30.642 16.44] [(MULTIPLE CHOICE. Q /F3 12.131 Tf 1.007 0 0 1.007 551.058 636.879 cm 0 5.203 TD Q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] Q /F3 12.131 Tf BT >> /ProcSet[/PDF] /BBox [0 0 15.59 16.44] /F3 12.131 Tf /Font << q /Resources<< Q /FormType 1 /BBox [0 0 639.552 16.44] /Subtype /Form 1.502 5.203 TD 29 0 obj -0.047 Tw Q /F1 14.682 Tf A. /Subtype /Form q 0 4.894 TD /F3 12.131 Tf /F4 12.131 Tf Q stream /Subtype /Form /Length 58 /Resources<< /BBox [0 0 673.937 16.44] 215 0 obj >> 1 i Q Q /F4 12.131 Tf /BBox [0 0 15.59 16.44] 0 g >> q 1.007 0 0 1.006 130.989 690.329 cm /Resources<< /Meta164 Do /Resources<< /BBox [0 0 88.214 16.44] /Font << /ProcSet[/PDF] 1.007 0 0 1.007 130.989 277.035 cm 85 0 obj >> /FormType 1 q q endobj endobj q stream /FormType 1 q q /F1 14.682 Tf 1 i 0 g 61 0 obj Q /Resources<< /Matrix [1 0 0 1 0 0] >> /Length 60 /Length 245 1.007 0 0 1.007 551.058 636.879 cm /FormType 1 endobj BT /F3 12.131 Tf /Meta175 189 0 R /BBox [0 0 673.937 15.562] 239 0 obj << /BBox [0 0 88.214 16.44] Thrice a number decreased by 5 exceeds twice the number by 1. Medium /BBox [0 0 15.59 16.44] /ProcSet[/PDF] /Font << /Subtype /Form stream Q 1.005 0 0 1.007 79.798 746.789 cm >> 1.014 0 0 1.007 391.462 383.934 cm 1.007 0 0 1.007 411.035 383.934 cm /BBox [0 0 88.214 16.44] Q /BBox [0 0 88.214 16.44] 1 i >> /BBox [0 0 15.59 16.44] /Subtype /Form >> 0.737 w << >> >> >> q Q 2.238 5.203 TD 0 G 0 G /ProcSet[/PDF] 0.737 w >> /XObject << q >> endobj stream /Subtype /Form 1 0 obj Q q 1.007 0 0 1.007 67.753 799.486 cm q 0.271 Tc Q 1 i /Font << >> /Meta234 248 0 R stream 1 i Q stream /Type /XObject endstream q BT endstream /FormType 1 /Type /XObject /ProcSet[/PDF/Text] /Type /XObject /Meta183 197 0 R q /Type /XObject 1 i endobj /Meta338 Do /Resources<< q Q >> 0 g /Length 69 /Matrix [1 0 0 1 0 0] /BBox [0 0 549.552 16.44] 1 i /BBox [0 0 88.214 16.44] q BT q Q /Length 70 ET /Meta140 Do /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 531.485 836.374 cm Q 549.694 0 0 16.469 0 -0.0283 cm << q << 407 0 obj Q q 1.005 0 0 1.007 102.382 653.441 cm 0 g q 0 G Q >> q 1.007 0 0 1.007 411.035 383.934 cm **Note: You could choose any variable you want. 0 w endobj /BBox [0 0 15.59 16.44] /FormType 1 /Meta19 30 0 R stream /Length 54 Q (\)) Tj 25.454 5.203 TD 0.564 G 3.742 5.203 TD /FormType 1 << /ProcSet[/PDF/Text] /ProcSet[/PDF] q 366 0 obj >> /Length 59 1.005 0 0 1.007 102.382 743.025 cm Q /BBox [0 0 88.214 16.44] /Length 79 1.005 0 0 1.007 102.382 400.496 cm Q /ProcSet[/PDF/Text] endobj q /Subtype /Form /Resources<< stream q ET /Font << 273 0 obj /Length 78 q /Length 12 q 1.007 0 0 1.006 551.058 437.384 cm ET /Subtype /Form 0.486 Tc /Matrix [1 0 0 1 0 0] q 0 G Q /Font << << q /Meta287 301 0 R >> /BBox [0 0 88.214 16.44] stream /Resources<< Q << (-9) Tj endstream 0 g /FormType 1 /Meta55 Do /Matrix [1 0 0 1 0 0] 0 g for the season. endobj /Resources<< endstream /Meta317 331 0 R endstream q (+) Tj /Resources<< >> 0.155 Tc /ProcSet[/PDF/Text] /Type /XObject /Resources<< q /Length 12 /Meta307 321 0 R Q 1 i /BBox [0 0 88.214 16.44] /Type /XObject 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. /FormType 1 /Font << /F3 17 0 R Q 20.21 5.203 TD Q stream /Meta188 202 0 R /FormType 1 /Matrix [1 0 0 1 0 0] >> /Type /XObject /Matrix [1 0 0 1 0 0] /F4 36 0 R 1.014 0 0 1.007 251.439 849.172 cm Q q -0.486 Tw stream /FormType 1 /F3 17 0 R /Subtype /Form >> /BBox [0 0 88.214 35.886] /Subtype /Form 333.269 5.488 TD 24 0 obj /FormType 1 << >> 1.007 0 0 1.007 271.012 636.879 cm Q /ProcSet[/PDF/Text] stream >> /F3 17 0 R Q >> Q q /Meta255 269 0 R /Meta257 Do /Subtype /Form Q /FormType 1 (D\)) Tj /FormType 1 q /F3 12.131 Tf 1.014 0 0 1.007 391.462 703.126 cm /Meta311 325 0 R /F3 12.131 Tf q Q >> >> endobj endstream >> 1 g endstream q [( a )-15(number, decreased by )] TJ /Length 60 /Matrix [1 0 0 1 0 0] endstream /Length 16 /Resources<< SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. /Length 69 /Length 16 1 i 316 0 obj 0.458 0 0 RG endobj endobj /Meta415 431 0 R /FormType 1 endstream endobj /Resources<< xref endstream Twice a number decreased by ten is at least 24. /Resources<< ET endobj q /Subtype /Form 1 g /Meta194 Do /BBox [0 0 88.214 35.886] /FormType 1 q /Length 63 /BBox [0 0 534.67 16.44] /Subtype /Form /BBox [0 0 30.642 16.44] endstream /Matrix [1 0 0 1 0 0] /Meta63 77 0 R Q /Length 16 /F3 12.131 Tf /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 391.462 849.172 cm /BBox [0 0 88.214 16.44] q The ratio of a number to fifteen 4. /ProcSet[/PDF/Text] /F3 17 0 R 243 0 obj 0 g ET /FormType 1 Q << /Resources<< /Meta378 392 0 R q >> /Type /XObject >> >> /Font << ET /Length 118 q /DecodeParms [<> ] /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /XObject 1 i If mario jumps 3 times and luigi jumps 62 times. stream 0 g /ProcSet[/PDF/Text] /F3 17 0 R << 1 g 1.502 5.203 TD (x) Tj /ProcSet[/PDF] /Resources<< BT /Font << /ProcSet[/PDF/Text] 0 g 0 w 1.014 0 0 1.007 251.439 776.149 cm /Length 69 /Resources<< q ET /Meta318 Do 381 0 obj << Q q /Meta290 304 0 R /Meta303 317 0 R /BBox [0 0 88.214 16.44] /Meta5 Do Q Q stream 1 i q endobj q Q /BBox [0 0 88.214 35.886] q /Type /XObject stream 1.014 0 0 1.007 531.485 776.149 cm /ProcSet[/PDF] >> (D\)) Tj /Matrix [1 0 0 1 0 0] Q (4) Tj /ProcSet[/PDF/Text] q Q endstream /Subtype /Form q /Resources<< Q -0.463 Tw /Matrix [1 0 0 1 0 0] q /Meta111 Do /Length 57 /Length 63 1 i << /Resources<< 46 0 obj Q q /Matrix [1 0 0 1 0 0] Q endobj /Resources<< /F3 12.131 Tf /Subtype /Form How many points did Kobe score in the season? 0.458 0 0 RG /F3 17 0 R Q >> 0 g >> Q /BBox [0 0 549.552 16.44] endstream >> q /Type /XObject Q << /F1 12.131 Tf ET endobj >> /Resources<< endobj /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 130.989 523.204 cm stream >> 0.369 Tc 6.746 5.336 TD 410 0 obj /Matrix [1 0 0 1 0 0] Q 1.005 0 0 1.007 79.798 796.475 cm endstream For the lesson, he grabs a glass container shaped like a rectan the quotient of twenty and a number a.) /Meta144 Do 1.007 0 0 1.007 130.989 849.172 cm stream /Meta425 Do Q BT /F3 17 0 R Q 1.005 0 0 1.007 102.382 363.608 cm 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. /Type /XObject Q /F3 12.131 Tf /Type /XObject Q >> >> /Resources<< 102 0 obj /F3 12.131 Tf endstream q >> 1 i 0 G stream << 182 0 obj ET /Type /XObject q >> Q /Subtype /Form 69 0 obj /Type /XObject /Matrix [1 0 0 1 0 0] /Length 70 endstream endobj q /Matrix [1 0 0 1 0 0] Q /F3 12.131 Tf Two speeding tickets could increase your rate by 58% at your next renewal. /Meta13 24 0 R 0.737 w q /BBox [0 0 17.177 16.44] Q 0.737 w 1.005 0 0 1.007 102.382 616.553 cm q /Subtype /Form Q /Resources<< /Matrix [1 0 0 1 0 0] 0 G endstream >> >> 0 G 0 G 0 G 57 0 obj Q /F3 17 0 R >> /ProcSet[/PDF] 1 i 1.502 5.203 TD 0 g 0 g /Meta10 Do q 0 g Q /Matrix [1 0 0 1 0 0] 0 G /Subtype /Form 0 g /Subtype /Form Q /Matrix [1 0 0 1 0 0] q /Subtype /Form q /BBox [0 0 23.896 16.44] BT endstream /Font << endobj /Subtype /Form /Matrix [1 0 0 1 0 0] endobj /Meta16 27 0 R /Meta26 Do /Resources<< /Width 734 q 0 G ET 341 0 obj 0.737 w /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Type /XObject 0.564 G Q /Subtype /Form ET q stream q /Meta144 158 0 R /Resources<< endobj >> /Meta335 Do /F3 17 0 R Q q 0 g 1.007 0 0 1.007 45.168 829.599 cm stream endstream /FormType 1 0.458 0 0 RG q Q /Subtype /Form q Q /Meta427 Do /Meta163 177 0 R stream /FormType 1 Q Q Q Answer provided by our tutors. 0 g 1 i 1 g /Length 58 /Font << endstream Q endstream /BBox [0 0 88.214 16.44] /Font << /BBox [0 0 17.177 16.44] >> /Resources<< << /Resources<< (C\)) Tj endstream /Subtype /Form q /ProcSet[/PDF/Text] << /Length 16 /Matrix [1 0 0 1 0 0] 1.014 0 0 1.006 111.416 437.384 cm /Matrix [1 0 0 1 0 0] q Q You could call them. 1 i Q Q >> /Type /XObject 0 w >> Q ET << BT 367 0 obj /Subtype /Form Q 1.007 0 0 1.007 411.035 277.035 cm ET /Meta344 358 0 R << endobj /Matrix [1 0 0 1 0 0] /F3 17 0 R /Font << >> /Subtype /Form Q /BBox [0 0 17.177 16.44] Want to see the full answer? >> endstream /Subtype /Form 1 g 1.007 0 0 1.007 45.168 862.723 cm 0 G 118.317 5.203 TD endstream /F3 12.131 Tf q ET /F3 17 0 R 1 g /Font << /Meta209 Do q endstream (58) Tj /FormType 1 /Meta298 312 0 R /Resources<< 0 5.203 TD /Meta156 Do /Type /XObject /FormType 1 /Length 69 /Resources<< /Font << 167 0 obj /Subtype /Form Q /Resources<< /Meta187 201 0 R /Font << 1.014 0 0 1.007 111.416 277.035 cm /Subtype /Form q /Meta206 220 0 R >> /Encoding /WinAnsiEncoding /Font << q 0 g stream /Meta332 Do 20.21 5.203 TD /Matrix [1 0 0 1 0 0] /FormType 1 Q >> w/Honors. stream /Subtype /Form /Type /XObject /Meta260 Do /FormType 1 ET >> 0 g 306 0 obj q 1.005 0 0 1.007 102.382 490.08 cm q Q q q /Resources<< endobj /Meta161 175 0 R Q /Subtype /TrueType Q /BaseFont /PalatinoLinotype-Bold >> /Matrix [1 0 0 1 0 0] 2. q /Subtype /Form >> /Type /XObject /Subtype /Form 428 0 obj /BBox [0 0 15.59 29.168] /Type /XObject /Resources<< /Length 69 Q 3.742 5.203 TD ET 1 i Q /Subtype /Form /Resources<< /Meta154 168 0 R 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. 154.289 4.894 TD 0.564 G ET /F3 17 0 R << stream >> /Length 16 1.007 0 0 1.007 551.058 523.204 cm >> q /F1 7 0 R endobj Q 0 g 219 0 obj /Info 3 0 R endstream Q /F3 12.131 Tf /Length 79 /BBox [0 0 88.214 35.886] /F3 17 0 R /Matrix [1 0 0 1 0 0] >> 1 i /F3 17 0 R >> /Matrix [1 0 0 1 0 0] /F3 17 0 R q /Type /XObject /Meta265 Do /ProcSet[/PDF] 58 0 obj /Matrix [1 0 0 1 0 0] /FormType 1 q /Meta167 Do ET /Resources<< /F3 17 0 R /Resources<< >> endobj Q /Meta346 Do Q Now that you know the meaning of the key words you can read the problem differently. Q endstream /StemH 88 S Q endobj q 0 g 0 g << Q /F1 7 0 R /Resources<< /Font << /Meta109 Do stream 0.458 0 0 RG 1.014 0 0 1.006 251.439 510.406 cm /Meta298 Do Q /ProcSet[/PDF/Text] stream /FormType 1 /Resources<< 0 G /FormType 1 /Type /XObject /Font << /Meta161 Do q /Meta373 Do /Matrix [1 0 0 1 0 0] q q 1.005 0 0 1.007 102.382 653.441 cm >> >> 16.469 5.203 TD 0 G /Length 64 >> /Meta39 Do 0 4.894 TD 1.007 0 0 1.007 271.012 636.879 cm /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) /F3 12.131 Tf endstream /Resources<< endobj Q 0 G /Resources<< stream >> /FormType 1 Q /Meta52 66 0 R /ProcSet[/PDF/Text] 1.007 0 0 1.007 45.168 730.228 cm q q q 7 0 obj /Subtype /Form >> 0.564 G Q stream q >> 0.68 Tc q /FormType 1 q /Length 59 722.699 799.486 l 1 i >> Q 13.464 5.203 TD Q /Length 12 /Matrix [1 0 0 1 0 0] q /Length 69 /F2 11 0 R /Meta232 246 0 R /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 156 0 obj 0 w endobj /Type /XObject /FormType 1 Q q /Meta322 Do Q 58 decreased by twice Gails age. q /Subtype /Form /FormType 1 stream 1.014 0 0 1.007 111.416 330.484 cm stream q q /FormType 1 << endobj /ProcSet[/PDF] Q 0.458 0 0 RG The difference between six and a number divided by nine 10. ET Q q /Font << BT q >> Q Q ET 0.737 w q 0 G /FormType 1 stream 0.369 Tc ET Q >> Q /Font << Q Q /Meta204 Do /Meta80 Do /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 0.564 G 212 0 obj 0.425 Tc /Resources<< /Font << 104 0 obj /Type /XObject >> Q /Meta358 Do /Type /Page << 0 g Q 283 0 obj BT /FormType 1 Q /Font << /Meta266 280 0 R 1.007 0 0 1.007 551.058 523.204 cm ET /F3 17 0 R q /Resources<< 0 g /F3 17 0 R BT endstream /ProcSet[/PDF/Text] /Meta428 Do 1 g 0 5.203 TD /F3 12.131 Tf (4\)) Tj /Subtype /Form /Length 70 0.564 G Q 1 i >> q Q ET q /Resources<< stream >> /Matrix [1 0 0 1 0 0] /BBox [0 0 17.177 16.44] Q /Subtype /Form 178 0 obj /Type /XObject /Subtype /Form (5) Tj endstream 0 5.203 TD >> 0.285 Tc Q >> >> >> Q /Length 69 0 g /FormType 1 /FormType 1 1 i >> Q q stream /Type /FontDescriptor 0.737 w 1.005 0 0 1.007 102.382 400.496 cm 1.007 0 0 1.007 45.168 763.351 cm >> /FormType 1 /Length 69 /F3 12.131 Tf 0 G /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q >> /Type /XObject /Resources<< Q /Type /XObject q >> /Type /XObject << q q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] endstream q >> >> endstream endobj q Q /F4 36 0 R endstream 309 0 obj >> Q /Length 16 q q /Meta213 227 0 R q /Resources<< /Subtype /Form /Meta421 437 0 R >> Q endobj /ProcSet[/PDF/Text] /Type /XObject >> q Q /F4 12.131 Tf 0.737 w 1 i /BBox [0 0 88.214 16.44] 0 G /Font << q Q 0 g Q q /Length 16 q /Subtype /Form Q >> 0.458 0 0 RG /FontDescriptor 35 0 R Q /BBox [0 0 88.214 35.886] >> /Font << (3) Tj /Meta415 Do >> >> /Meta139 153 0 R 0.458 0 0 RG 0 g /Font << /Length 73 20/n b.) Q 0 g /ProcSet[/PDF] endstream /Type /XObject q >> /F3 17 0 R endobj /F3 17 0 R 0 4.894 TD Q Q endobj 0 G stream endobj /ProcSet[/PDF] /F4 36 0 R Q twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. stream /F3 12.131 Tf /Matrix [1 0 0 1 0 0] >> /Resources<< 0.51 Tc Q >> 1 i q /Type /XObject /ProcSet[/PDF] /F3 17 0 R q >> Q >> /Resources<< 1.014 0 0 1.007 391.462 330.484 cm /FormType 1 q /Resources<< /ProcSet[/PDF] Q Q << endobj Q ET /F3 17 0 R /FormType 1 /Meta30 43 0 R 1.014 0 0 1.007 111.416 583.429 cm Q 0 g /ProcSet[/PDF/Text] Get a free answer to a quick problem. q /Meta241 255 0 R endobj /Type /XObject /ProcSet[/PDF/Text] ET 1 g q /F3 12.131 Tf /Meta57 Do 293 0 obj 0 g /Length 16 0.737 w 1 i q /Matrix [1 0 0 1 0 0] Q /Type /XObject /BBox [0 0 88.214 35.886] q /Length 60 /F1 12.131 Tf /FormType 1 1 i /Matrix [1 0 0 1 0 0] Q /F3 17 0 R /Resources<< 0.564 G /ProcSet[/PDF/Text] q Q q /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] /Meta337 351 0 R stream /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 713.666 cm 0 g 178.979 5.203 TD /Length 69 /BBox [0 0 88.214 16.44] Twice a number decreased by 8 gives 58. /Length 294 /Length 12 Q ET Q >> endstream 1.014 0 0 1.007 111.416 776.149 cm 1.007 0 0 1.007 67.753 473.519 cm /BBox [0 0 639.552 16.44] Q 127 0 obj << 1 i 1 i >> (+) Tj /Meta79 93 0 R q >> << endstream 1 i /BBox [0 0 15.59 16.44] q 1 i Q BT 1 i /ProcSet[/PDF/Text] endstream /FormType 1 /Subtype /Form >> 0.737 w Q 36 0 obj endobj >> Three times a number equals fifteen 3. Thirthy is equal to twice a number decreased by four = solve and check the equation? Q /Length 16 Q BT stream /Resources<< endstream >> 300 0 obj q (7\)) Tj endstream Q /Subtype /Form 0 G << /Type /XObject 0 g q /BBox [0 0 30.642 16.44] stream (-) Tj Q /ProcSet[/PDF/Text] -0.021 Tw /Resources<< q Q >> /F3 12.131 Tf /Length 16 1.502 5.203 TD /Resources<< endobj Q Q stream Q /Subtype /Form /FormType 1 q Q q q /FormType 1 endstream /Font << /BBox [0 0 15.59 16.44] endobj >> Q (-) Tj stream /Subtype /Form >> stream q 0 g >> /ProcSet[/PDF/Text] /Font << 0 G You can specify conditions of storing and accessing cookies in your browser. /Matrix [1 0 0 1 0 0] << 106 0 obj 5.98 7.841 TD Q Q /Type /XObject q stream 0 g 0.458 0 0 RG q Q /BaseFont /PalatinoLinotype-Roman /Length 69 ET Q q q q /F4 36 0 R >> /Meta88 102 0 R 1 i /Matrix [1 0 0 1 0 0] /Length 66 q 1.007 0 0 1.007 130.989 776.149 cm 0 G /ProcSet[/PDF/Text] endobj /BBox [0 0 88.214 16.44] q /Type /XObject >> Q /Matrix [1 0 0 1 0 0] q q >> /Resources<< Expression. q 52 0 obj /BBox [0 0 88.214 16.44] 285 0 obj /Length 63 0 G /Resources<< 0.786 Tc Q Q S Q q q (B\)) Tj >> q /Type /XObject /Meta234 Do 0 w /Subtype /Form /F3 17 0 R /Length 108 /Resources<< /BBox [0 0 23.896 16.44] 314 0 obj 0.564 G >> 1 i /F3 17 0 R q /ProcSet[/PDF/Text] Q 0.564 G BT Q q << << /Type /XObject /Resources<< 1 g [(The )-19(quotient of )] TJ << 1 i q 1 i Q: when six times a number is decreased by 4, the result is 8. /Matrix [1 0 0 1 0 0] 1 g /Meta216 Do q /Meta185 199 0 R /Type /XObject 0 g /Meta243 Do /F3 17 0 R stream ET /Subtype /Form >> 0 g >> >> /Type /XObject 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. /FormType 1 /F3 12.131 Tf Q Q /Length 69 endobj /FormType 1 Was this answer helpful? /BBox [0 0 88.214 16.44] /Font << 0.425 Tc /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] -0.486 Tw /Matrix [1 0 0 1 0 0] endobj (40) Tj << 1 i q 1 g /Type /XObject Q endobj q 0 G 1 i 1 i 0 G >> 0 g (B) Tj q >> /FormType 1 /Resources<< >> Q 405 0 obj 1.007 0 0 1.007 271.012 450.181 cm endstream /Meta201 Do >> BT Q /FormType 1 endstream Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q endstream stream 0 G /Subtype /Form /Meta44 58 0 R 0 g /F3 17 0 R ET /Length 58 Q /F3 17 0 R >> Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. /Font << /ProcSet[/PDF/Text] 132 0 obj << stream /FormType 1 endstream /Matrix [1 0 0 1 0 0] 0 g 0 G
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