twice a number decreased by 58twice a number decreased by 58
>> /Meta221 Do Q /Meta42 Do /FormType 1 /ProcSet[/PDF] 434 0 obj 1.005 0 0 1.007 102.382 743.025 cm 0 20.154 m 0 G /F3 12.131 Tf 1.007 0 0 1.007 271.012 277.035 cm /Matrix [1 0 0 1 0 0] Q Q /Resources<< 2.238 5.203 TD q Q q /BBox [0 0 88.214 16.44] stream stream endstream /Resources<< 1.005 0 0 1.007 102.382 363.608 cm /FormType 1 0 g 0.458 0 0 RG /ProcSet[/PDF/Text] endstream q 0 G q q endstream 140 0 obj /ProcSet[/PDF/Text] >> << /BBox [0 0 639.552 16.44] /F3 12.131 Tf /BBox [0 0 88.214 16.44] algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. 1.005 0 0 1.007 102.382 670.003 cm endstream >> BT /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endstream >> /Meta135 149 0 R /ProcSet[/PDF/Text] 1.005 0 0 1.007 79.798 796.475 cm /FormType 1 q << 1 i 1.014 0 0 1.007 251.439 636.879 cm ET 1 i Q /ProcSet[/PDF] 1 g 0.737 w /F3 12.131 Tf Q endobj /Subtype /Form 0 g q >> /ProcSet[/PDF/Text] >> Q /BBox [0 0 17.177 16.44] q 0 g q /Type /XObject q /FormType 1 Q: when six times a number is decreased by 4, the result is 8. >> /F1 12.131 Tf 6.746 5.203 TD 256 0 obj q Q (-20) Tj Q Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM 1 i BT BT 0.564 G q 1 g 271 0 obj q /Font << /Subtype /Form >> /F3 12.131 Tf >> endstream 1 i )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] /ProcSet[/PDF/Text] >> 1 i /FormType 1 /FormType 1 /Type /XObject q 1.007 0 0 1.007 551.058 383.934 cm q endobj 0 4.894 TD 223 0 obj 1. Expert Solution. endobj 0.458 0 0 RG endstream /Meta38 Do /Font << endobj BT Q Q endobj >> /Meta266 Do /F3 17 0 R [(thir)17(te)15(en)] TJ Q /Length 58 0 g 430 0 obj endstream >> 0.241 Tc BT q /Matrix [1 0 0 1 0 0] /Font << Q /Meta306 320 0 R /Font << /ProcSet[/PDF] Q >> q stream q 1.005 0 0 1.007 102.382 400.496 cm /Matrix [1 0 0 1 0 0] /Font << endobj q /Length 118 1.014 0 0 1.007 531.485 277.035 cm /FormType 1 294 0 obj Q >> 0.564 G 0 g endobj q /BBox [0 0 30.642 16.44] Q /Matrix [1 0 0 1 0 0] /Meta128 Do 0 w >> /Subtype /Form << /Resources<< 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 (D\)) Tj /Matrix [1 0 0 1 0 0] 0.486 Tc q /BBox [0 0 673.937 68.796] /Length 16 /Meta336 Do >> 140781 endstream /Font << /Resources<< 86 0 obj /ProcSet[/PDF/Text] /F3 17 0 R endstream Q q ET << 1 i Q /Meta52 Do endobj 291 0 obj /ProcSet[/PDF] /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 /Subtype /Form 148 0 obj 1.007 0 0 1.007 271.012 636.879 cm q /Matrix [1 0 0 1 0 0] q /Type /XObject >> /Length 59 q /Meta321 Do (2\)) Tj /ProcSet[/PDF/Text] /Subtype /Form /Length 65 >> /ProcSet[/PDF] q /Length 54 stream 0 w Q /Type /XObject /Length 69 /Meta161 175 0 R /Resources<< /Matrix [1 0 0 1 0 0] the quotient of twenty and a number a.) endstream Hence, the number is 6. Q << 2x - 15 = -27. >> stream /Meta361 375 0 R /Meta306 Do /F3 12.131 Tf Q q Q >> q /Type /XObject ET endstream 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 0 g >> q BT [(Answe)20(r Key)] TJ Q /Meta149 163 0 R /F3 12.131 Tf /Type /XObject 1 i 15.731 5.336 TD 0 5.203 TD /FormType 1 Q 0.564 G (B\)) Tj << ET /Meta364 Do endobj Q /F1 7 0 R /Matrix [1 0 0 1 0 0] >> /Meta112 Do /Meta316 Do /Subtype /Form q /F4 36 0 R /Font << 0 w /Matrix [1 0 0 1 0 0] Twice a number decreased by 8 gives 58. q /Meta146 Do endstream /Resources<< 0.458 0 0 RG endstream Q Q /ProcSet[/PDF/Text] q precision and actual right or wrong answers. Q endobj stream 354 0 obj /Meta68 Do q BT 1 i 0 w /FormType 1 q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] /Font << (B\)) Tj 1 i /ProcSet[/PDF] /BBox [0 0 30.642 16.44] /Meta39 Do 1.005 0 0 1.013 45.168 933.487 cm /FormType 1 /Subtype /TrueType (58) Tj /Matrix [1 0 0 1 0 0] Q 0 5.203 TD Q /Meta385 401 0 R 0 g << 1 i 132 0 obj /Resources<< q /FormType 1 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. /Matrix [1 0 0 1 0 0] /Resources<< /Subtype /Form q q /Subtype /Form The quotient of a seven and a number 9. Answer link. /Meta268 Do /Meta120 134 0 R q >> 13.464 5.203 TD Q << >> 1 i /F4 12.131 Tf /Meta394 410 0 R q /BBox [0 0 88.214 16.44] 0 g endobj Q >> >> /Type /XObject 0 g /Meta187 Do /Meta150 164 0 R >> /F1 12.131 Tf /Resources<< Q 0 g stream Q >> Q 1.007 0 0 1.007 271.012 849.172 cm /Subtype /Form /Subtype /Form The result is 8 less than 10 times the number. endobj /FormType 1 /Type /XObject /F3 12.131 Tf /Length 69 1 g q /ProcSet[/PDF] /Meta155 Do stream /Matrix [1 0 0 1 0 0] /Type /XObject /Resources<< endstream Q q /Meta155 169 0 R q /ProcSet[/PDF] /Leading 349 1 i /Subtype /Form 0.524 Tc BT q /ProcSet[/PDF/Text] 49 0 obj Find the number. /F3 12.131 Tf /BBox [0 0 88.214 16.44] /BBox [0 0 30.642 16.44] endstream q BT >> 128 0 obj >> q Q 0.458 0 0 RG /FormType 1 1.007 0 0 1.007 551.058 703.126 cm Q Q stream 1 i /ProcSet[/PDF/Text] 1 i /Resources<< << Q q /Resources<< q q 1.007 0 0 1.007 271.012 277.035 cm /I0 Do endstream 0 G endobj /BBox [0 0 15.59 16.44] /FormType 1 /BBox [0 0 17.177 16.44] /Resources<< >> 0 g /FormType 1 23.216 5.203 TD /FormType 1 /BBox [0 0 88.214 16.44] << (vii) Twice a number subtracted from 19 is 11. 0.737 w >> 3.742 5.203 TD << /Contents [399 0 R] q /Meta308 Do D. Twice a number decreased by ten is less than 24. /Font << /Resources<< 0 5.203 TD >> /Meta242 Do /Meta176 Do ET Q q 672.261 726.464 m stream 365 0 obj >> /Meta386 402 0 R Q /Subtype /Form 0 g /Font << 0 w Q 1 g /ProcSet[/PDF] /Resources<< Q /Length 16 << /BBox [0 0 88.214 16.44] stream /Length 68 0 g (5\)) Tj /Meta244 258 0 R For Free. Q Q /Meta238 252 0 R /Resources<< BT Q >> /Meta6 15 0 R /Type /XObject endstream /FormType 1 stream /Type /XObject stream 13.493 5.203 TD /Length 54 Q /Meta390 Do /BBox [0 0 88.214 16.44] q 162 0 obj /Type /XObject /ProcSet[/PDF/Text] 405 0 obj /Matrix [1 0 0 1 0 0] q (5\)) Tj >> endstream /F3 12.131 Tf (3) Tj q << (-) Tj Translate 2(x-58) into mathematical phrase. 0 g q << 0.458 0 0 RG /Length 16 >> 0 w /BBox [0 0 88.214 16.44] 0.838 Tc /FormType 1 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q /F4 36 0 R Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 0.458 0 0 RG 1.007 0 0 1.007 271.012 277.035 cm 1 i q /Subtype /Form /F3 17 0 R Q 3.742 5.203 TD 1 i Q q /FormType 1 BT 1.005 0 0 1.009 45.168 905.633 cm 0.369 Tc 1.007 0 0 1.007 551.058 523.204 cm /Resources<< >> 241 0 obj /Matrix [1 0 0 1 0 0] /BBox [0 0 673.937 15.562] q /ProcSet[/PDF] >> /Meta319 333 0 R /Meta10 Do stream << Mr. Gleeson knows that 1,000 cubic centimeters is the same as 1 liter, and he wants to figure out how many liters of water will fill the container before it overflows. /ProcSet[/PDF] Q If a number is 50%, then it is a half - the same as 0.5 or 1/2. >> 392 0 obj 1 i 57 0 obj endobj 0 g q 250 0 500 500 500 500 500 500 0 0 500 500 278 0 0 0 0 G /Resources<< >> /Font << /Subtype /Form /Meta347 361 0 R >> >> 299 0 obj 0 G 1.005 0 0 1.007 102.382 400.496 cm >> /ProcSet[/PDF/Text] 1 i Q let 'x' and 'y' represent the numbers. Q 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. /FormType 1 q /Meta80 Do /Subtype /Form /Type /XObject /Length 79 229 0 obj << /FormType 1 /Meta13 24 0 R ] /Type /XObject 0 5.203 TD BT /Meta128 142 0 R Q 0.458 0 0 RG /F3 12.131 Tf 0 g q 2. 1 i /Type /XObject 154.289 4.894 TD 1 g /Subtype /Form /Length 118 >> endstream 13.493 5.336 TD /F3 17 0 R Q ET 0 G q Q 0.564 G << stream Q << ET /FormType 1 /I0 51 0 R /BBox [0 0 88.214 35.886] /Meta350 364 0 R Q stream /Matrix [1 0 0 1 0 0] endstream 1 i 1.007 0 0 1.007 271.012 330.484 cm Q Q stream /F3 17 0 R /F3 17 0 R >> q Q >> 260 0 obj << /ProcSet[/PDF/Text] 0 g stream /Subtype /Form /BBox [0 0 549.552 16.44] << q endobj ET endobj /Meta358 372 0 R 0 g 0.458 0 0 RG 1.007 0 0 1.006 411.035 437.384 cm 0 G 1.014 0 0 1.007 111.416 849.172 cm stream 125.064 4.894 TD /ProcSet[/PDF] /Length 16 << << endobj 0 g /ProcSet[/PDF/Text] (9\)) Tj q 1 g >> /F3 17 0 R Q 0 g q /ProcSet[/PDF] >> 30.699 5.203 TD 0 g 0 g q /Matrix [1 0 0 1 0 0] /F3 17 0 R (D\)) Tj q /Resources<< /Resources<< /F3 17 0 R q /F3 17 0 R /Type /XObject 1.005 0 0 1.007 102.382 490.08 cm Q endstream q Q /Meta264 278 0 R /FormType 1 /Meta297 Do BT /Length 12 q /F3 12.131 Tf 1.005 0 0 1.007 102.382 799.486 cm 1 i [tex]\sin (\pi -x)=\sin x[/tex]. stream Q /ProcSet[/PDF/Text] /Type /XObject 0 G 0 g 1 i /Font << stream Q /ProcSet[/PDF] endobj Q endstream /Type /XObject /Subtype /Form /FormType 1 /FormType 1 endstream 0 g << 51 0 obj /BBox [0 0 534.67 16.44] 1.502 8.18 TD endobj /I0 51 0 R q 0 w 1 i /Font << /Subtype /Form 0 G /FormType 1 (x ) Tj q /Length 68 /Meta145 159 0 R /ProcSet[/PDF/Text] q S q >> 1 g /FormType 1 12.727 5.203 TD 1.005 0 0 1.006 45.168 879.284 cm /BBox [0 0 88.214 16.44] /Meta203 Do /Encoding /WinAnsiEncoding endstream endstream /Font << q q Q >> /Length 59 >> /Meta308 322 0 R /Resources<< endstream 70 0 obj >> (A\)) Tj 105 0 obj ET /Subtype /Form 0.564 G >> >> Q /Type /XObject Q Q /Length 67 20.21 5.203 TD 1.014 0 0 1.007 531.485 636.879 cm Q 165 0 obj q 0 g /Length 67 0 G 227 0 obj Q /Type /XObject gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. /Length 69 q q /Meta227 241 0 R /Resources<< Q q q 277 0 obj /Subtype /Form /Meta362 Do /FormType 1 266 0 obj /F3 17 0 R 330 0 obj q 0 G endobj /FormType 1 /Length 244 ET /Meta271 285 0 R Q 1 i q << Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . Get a free answer to a quick problem. endobj Q 0 G q /Font << >> -0.092 Tw /Matrix [1 0 0 1 0 0] /FormType 1 /ProcSet[/PDF] 16.469 5.203 TD /Meta65 Do /BBox [0 0 30.642 16.44] 0 G /Resources<< 1.007 0 0 1.007 411.035 383.934 cm 1 i (\)) Tj /Meta157 Do 1 i q /Subtype /Form 1 i Q /Font << /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Resources<< 1 g (C\)) Tj /Matrix [1 0 0 1 0 0] /F3 12.131 Tf >> stream /Type /XObject 0 G /Font << /Type /XObject >> /Length 16 << endstream /F1 7 0 R << /FormType 1 /Subtype /Form 0 G /Matrix [1 0 0 1 0 0] BT Q >> 6.746 5.203 TD endobj Q 1.007 0 0 1.006 411.035 437.384 cm endobj /Length 69 /Matrix [1 0 0 1 0 0] q q 65.906 4.894 TD /Type /XObject q >> -0.058 Tw /BBox [0 0 88.214 16.44] q q q /Meta208 222 0 R /FormType 1 /Subtype /Form Q 0 G ET endobj 549.694 0 0 16.469 0 -0.0283 cm endstream /Meta20 31 0 R 1.005 0 0 1.007 102.382 400.496 cm /Root 2 0 R Q 312 0 obj /Resources<< /Meta179 Do /Font << 0 g endobj 0 g Q 0.458 0 0 RG << q endobj /Type /XObject /FormType 1 >> Q Twice a number decreased by another number: stream endstream Q 0 g Q 1.007 0 0 1.007 551.058 636.879 cm << 0.737 w Q << q 1.007 0 0 1.007 45.168 746.789 cm /Meta410 426 0 R ET << q ET 0 w >> >> /Length 16 Q endstream Q /Subtype /Form /Meta119 133 0 R Q q /BBox [0 0 15.59 29.168] Thrice a number decreased by 5 exceeds twice the number by 1. endobj /Subtype /Form 1 i >> 0 G /Meta195 209 0 R /Meta218 Do endstream /Type /Pages q /F3 12.131 Tf There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. (x ) Tj /ProcSet[/PDF] Q /ProcSet[/PDF] 0 w /Type /XObject 0 G 433 0 obj Q Q 0 g 307 0 obj 1 i Twice a number when decreased by 7 gives 45. BT endstream /F1 7 0 R q BT >> /Meta180 194 0 R >> , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. q q Q 0 G q 0 w Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> /Font << /BBox [0 0 88.214 16.44] /F3 12.131 Tf >> 0.738 Tc >> /BBox [0 0 15.59 16.44] /Type /XObject /Type /XObject /Meta405 Do /Meta292 306 0 R /Type /XObject 6.746 5.203 TD << BT Q Q q /Meta32 Do endobj 1.502 24.339 TD /Length 68 0 g Q Q Q /Subtype /Form endobj q 1.007 0 0 1.007 130.989 277.035 cm Q /Resources<< >> 1 i /Matrix [1 0 0 1 0 0] Q /Meta57 71 0 R /F1 7 0 R 0 g /Type /XObject 0 G << stream >> endobj >> >> (6\)) Tj << << (A\)) Tj endstream 315 0 obj (9\)) Tj /Matrix [1 0 0 1 0 0] /Meta126 Do 0 G (7\)) Tj /F3 12.131 Tf endobj /ProcSet[/PDF] q /Leading 253 >> /Matrix [1 0 0 1 0 0] Q endstream stream 345 0 obj stream Q stream (4\)) Tj 0 G ET /Meta249 263 0 R 0 G /Length 65 q 0 G /I0 51 0 R /ProcSet[/PDF] /Length 119 >> /FormType 1 0.737 w Q BT endstream /BBox [0 0 88.214 35.886] /Type /XObject /FormType 1 1.007 0 0 1.007 411.035 636.879 cm /ProcSet[/PDF/Text] 0 G endobj 33 0 obj 0 5.203 TD BT 1.007 0 0 1.007 67.753 293.596 cm << q twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. q q /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] endstream stream Q endstream /Subtype /Form /Type /XObject 0.68 Tc /Type /XObject 1.005 0 0 1.007 79.798 779.913 cm >> 1.014 0 0 1.007 391.462 583.429 cm /Resources<< /F3 12.131 Tf /ProcSet[/PDF/Text] 1.007 0 0 1.007 411.035 383.934 cm endstream 1 i Now that you know the meaning of the key words you can read the problem differently. endobj stream /ProcSet[/PDF/Text] >> 89.12 5.203 TD 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. 0.024 Tw (-23) Tj 0 5.336 TD 174 0 obj 0 5.203 TD endstream 156 0 obj q stream Q /F4 36 0 R /Length 60 << Q Q 0 G << << Q q 95 0 obj >> 0 G /F3 12.131 Tf /F1 12.131 Tf >> Q /Resources<< q >> Q BT >> /Font << Q endobj q /ProcSet[/PDF/Text] Q /Meta373 Do 0 G q Q 1.007 0 0 1.007 411.035 636.879 cm /FormType 1 q /FormType 1 235 0 obj (40) Tj 0 g 1 i endobj q Q /Matrix [1 0 0 1 0 0] /Resources<< q /Meta255 Do 1.007 0 0 1.007 271.012 636.879 cm /Meta341 355 0 R stream Q /Meta105 119 0 R 203 0 obj Q 0 w /Length 54 /Font << >> q Q >> endobj /Resources<< stream 0.425 Tc 0 w /Meta95 Do q /ProcSet[/PDF/Text] /BBox [0 0 88.214 35.886] /Length 16 /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] 186 0 obj >> << /Meta45 Do Q /F3 12.131 Tf Q stream /Subtype /Form BT /Type /XObject 1 i 123 0 obj /Subtype /Form endobj /FormType 1 /Resources<< 0 g endstream /FormType 1 << 0 g q stream endstream [( a )-15(number, decreased by )] TJ /Subtype /Form /Resources<< endobj stream 0 G /Meta18 Do /ProcSet[/PDF/Text] endobj 1.014 0 0 1.006 391.462 510.406 cm /F3 17 0 R 248 0 obj >> 0.564 G (x) Tj /Type /XObject 0.838 Tc q (-23) Tj q >> q /Matrix [1 0 0 1 0 0] 0.564 G /FormType 1 >> Q q 1 i /Font << 1.014 0 0 1.007 251.439 703.126 cm /FormType 1 /FormType 1 >> 1.014 0 0 1.007 251.439 849.172 cm endstream Q q /Subtype /Form << /Length 16 /ProcSet[/PDF] endobj stream endobj stream /Length 59 446 0 obj 1.014 0 0 1.006 251.439 836.374 cm >> 0 447 (1) Tj [(Fiv)25(e ti)18(me)16(s)] TJ << /F1 12.131 Tf >> endstream /F3 17 0 R /Font << /Subtype /Form /F1 7 0 R , Prove the following q 1.007 0 0 1.007 130.989 636.879 cm Q Q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] << Q endobj /Matrix [1 0 0 1 0 0] /Length 16 How many points did Kobe score in the season? >> /Resources<< 0 g /Subtype /Form /Type /XObject /Subtype /Form /Resources<< /Meta60 74 0 R >> Q /Length 91 /Meta29 42 0 R Q /FormType 1 BT >> /Subtype /Form /Meta412 Do /BBox [0 0 88.214 35.886] << /FormType 1 /Resources<< 0.737 w /F3 12.131 Tf Q 0 G 0.737 w /Font << Q 391 0 obj endstream /Type /XObject q (\(x ) Tj >> Q 327 0 obj 0 g 0.001 Tw stream /FormType 1 /Subtype /Form 1 i 0 G /F3 17 0 R Q 0.737 w (\)) Tj Q 1.007 0 0 1.006 551.058 836.374 cm /Length 118 1 i /Meta252 266 0 R /FormType 1 0.458 0 0 RG /BBox [0 0 88.214 35.886] Q On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini /FormType 1 /Matrix [1 0 0 1 0 0] /FormType 1 /Subtype /Form /Type /XObject /Meta84 Do Q 0.838 Tc 244 0 obj Formula - How to Calculate Percentage Decrease. [(1)-25(0\))] TJ 1.007 0 0 1.006 551.058 763.351 cm endobj q q /Type /XObject /BBox [0 0 15.59 16.44] q /FormType 1 q << endstream Q /F4 12.131 Tf /Length 69 /Resources<< q /Matrix [1 0 0 1 0 0] >> /Subtype /Form 1 i endobj /F3 12.131 Tf /FormType 1 >> >> /Resources<< 0 G 0 g Q 26 0 obj << /FormType 1 BT /BBox [0 0 534.67 16.44] >> /Subtype /Form /Length 118 q q q >> /Length 16 0 G >> (38) Tj 1.007 0 0 1.007 551.058 636.879 cm /F3 17 0 R q /BBox [0 0 15.59 16.44] 22 0 obj BT Q endstream /Meta380 394 0 R endobj q /Resources<< q stream /Meta243 257 0 R 349 0 obj Advertisement Loved by our community 50 people found it helpful Madhvendra13 2x -8=58 2x=66 x=662 x=33 Find Math textbook solutions? -0.041 Tw /Subtype /Form 9.723 5.336 TD /BBox [0 0 88.214 16.44] -0.067 Tw ET /Length 16 q >> /I0 51 0 R 175 0 obj /Type /XObject /Matrix [1 0 0 1 0 0] Q (-20) Tj /ProcSet[/PDF] /Type /XObject q /Meta222 Do q /F3 12.131 Tf 0 g /ProcSet[/PDF/Text] Q >> 0.564 G /Subtype /Form 0 g /Matrix [1 0 0 1 0 0] /Type /XObject q /ProcSet[/PDF/Text] 0.564 G /Type /XObject BT /F3 17 0 R 185.725 5.203 TD /Meta164 Do Q q /FormType 1 /Matrix [1 0 0 1 0 0] 1 i 0 G q /Resources<< /Resources<< /Length 294 >> Q /Subtype /Form 0 g /FormType 1 /ProcSet[/PDF] endobj q 0 g Q << /F3 17 0 R >> /FormType 1 Q Q /BBox [0 0 639.552 16.44] >> endobj >> /ProcSet[/PDF/Text] >> endstream /F3 17 0 R /F3 17 0 R /BBox [0 0 88.214 16.44] Decreased by another number means subtract. /Subtype /Form /Length 139 1.502 5.203 TD 0.458 0 0 RG q Q 0 G 1 i endstream /BBox [0 0 88.214 16.44] >> 0 g >> q /F1 7 0 R q q /BBox [0 0 88.214 16.44] >> >> /Matrix [1 0 0 1 0 0] >> /Meta47 61 0 R stream /Meta379 393 0 R << /Font << 1 i BT /Meta365 Do << 1 i 0 w 9.723 5.336 TD /Font << 3 0 obj /ProcSet[/PDF/Text] /Length 64 1.005 0 0 1.006 45.168 879.284 cm /ProcSet[/PDF/Text] 1 i 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. 242 0 obj /Matrix [1 0 0 1 0 0] 0 G /Matrix [1 0 0 1 0 0] q 0 G /Type /XObject 283 0 obj q /ProcSet[/PDF/Text] /Meta366 Do /Subtype /Form Q q endstream /Matrix [1 0 0 1 0 0] Q Q 0.737 w >> /F3 17 0 R 1.005 0 0 1.007 102.382 653.441 cm /Resources<< Q BT ET 1 g /Subtype /Form 0 g Q 1.005 0 0 1.007 102.382 599.991 cm q >> q q >> endobj /Matrix [1 0 0 1 0 0] /Meta335 349 0 R Q ET << stream BT /Meta4 Do >> 0.564 G >> 0 4.78 TD /Meta424 440 0 R /BBox [0 0 88.214 16.44] /Meta135 Do >> >> 1.007 0 0 1.007 130.989 383.934 cm 1.007 0 0 1.007 551.058 383.934 cm endstream q Q endstream q Q ET /FormType 1 Q /BBox [0 0 88.214 35.886] q Q >> /Matrix [1 0 0 1 0 0] Q >> Q /BBox [0 0 88.214 16.44] /Subtype /Form Q endobj >> endstream /Matrix [1 0 0 1 0 0] endobj /F3 17 0 R endstream BT Q /Resources<< /ProcSet[/PDF] (58) Tj 0.51 Tc 297 0 obj /Font << >> q Q /Meta46 Do stream 0.564 G q Q /Matrix [1 0 0 1 0 0] /Meta405 421 0 R Q >> 0.564 G 0 g 1 i 6.746 5.203 TD /Subtype /Form q /Length 59 /Meta138 Do endstream /F4 36 0 R /Resources<< q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 271.012 277.035 cm /Resources<< >> 290 0 obj Q 0.458 0 0 RG 1 i 0 g Q 0 G >> /Font << 0.458 0 0 RG /F3 17 0 R /ProcSet[/PDF/Text] 16.469 5.336 TD 1 i >> /Resources<< 0 g >> /F3 17 0 R 16 0 obj 1.007 0 0 1.007 271.012 330.484 cm /Subtype /Form /Meta173 187 0 R >> 1 g q q 0 G Q 0 G stream Q >> /FormType 1 0.737 w q Q >> /Matrix [1 0 0 1 0 0] 0 5.203 TD /ProcSet[/PDF] ET /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> /Resources<< /Matrix [1 0 0 1 0 0] q /Type /XObject 1 g 0 g /BitsPerComponent 1 1 i (x) Tj 1.014 0 0 1.007 111.416 450.181 cm q /Matrix [1 0 0 1 0 0] Q 1.007 0 0 1.006 411.035 690.329 cm stream % << /Meta169 Do /F3 12.131 Tf 230 0 obj stream Q 1 i /Length 16 Twice a number would be 2x. /Matrix [1 0 0 1 0 0] ET 1 g q 238 0 obj 1.007 0 0 1.007 130.989 776.149 cm q 0.271 Tc >> stream /Meta329 Do q /Resources<< << 1 i /Meta131 Do >> Q A link to the app was sent to your phone. Q 1 i stream endstream /Subtype /Form Q 310 0 obj (5) Tj /Matrix [1 0 0 1 0 0] BT q >> >> 1 i S 1.007 0 0 1.007 411.035 277.035 cm Q /F4 36 0 R 1 i 208 0 obj >> endstream /FormType 1 /BBox [0 0 534.67 16.44] endstream /F3 12.131 Tf /Matrix [1 0 0 1 0 0] Q 0.369 Tc /Meta345 359 0 R 1 g 0 g Q /Matrix [1 0 0 1 0 0] >> /Encoding /WinAnsiEncoding a Question Q 1 g 549.694 0 0 16.469 0 -0.0283 cm [(A number )-17(divided by )] TJ >> 0 g /Resources<< 1 i q 0000000000 65535 f
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Harvey Watkins Jr Car Accident, Articles T
Harvey Watkins Jr Car Accident, Articles T