/Meta211 225 0 R >> q /Meta334 Do /Meta179 Do /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject /F3 12.131 Tf 1.005 0 0 1.007 102.382 653.441 cm (9\)) Tj /ProcSet[/PDF] q (1) Tj 0.524 Tc 291 0 obj /Meta56 70 0 R q 672.261 599.991 m >> /Type /XObject endobj 331 0 obj /Length 118 /Subtype /Form /Length 59 Q 1.005 0 0 1.007 102.382 546.541 cm 1.014 0 0 1.007 391.462 583.429 cm q /Font << /Meta79 93 0 R /XObject << (2\)) Tj /FormType 1 SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. /FormType 1 1 i /Type /XObject /F3 12.131 Tf /I0 51 0 R q 0 G /F3 17 0 R /BBox [0 0 15.59 16.44] << >> /F3 17 0 R >> /Resources<< /ProcSet[/PDF/Text] 132 0 obj /BBox [0 0 88.214 16.44] /Meta27 40 0 R 431 0 obj /Subtype /Form q >> endstream q /ProcSet[/PDF/Text] BT /Subtype /Form endstream q stream /Resources<< /Type /XObject 78 0 obj 0.737 w Q /Meta91 Do 0 G q 433 0 obj >> (-9) Tj 611 556 611 611 389 444 333 611 556 833 500]>> /Matrix [1 0 0 1 0 0] endobj 1.007 0 0 1.007 271.012 636.879 cm BT Q >> q >> /Matrix [1 0 0 1 0 0] Q /Meta209 223 0 R q 1.007 0 0 1.007 271.012 450.181 cm /Meta37 Do Q q /BBox [0 0 88.214 16.44] /Type /XObject q stream 0 g >> 12.727 5.203 TD /Matrix [1 0 0 1 0 0] /Meta85 Do /Meta300 Do >> /Subtype /Form BT 1.502 5.203 TD 0.564 G << Q /Type /XObject Q /Resources<< /Length 60 q /Meta393 409 0 R /FormType 1 >> Answer only. q 0.786 Tc 1 i Q 0 G /Length 59 /BBox [0 0 30.642 16.44] 0.737 w BT /Type /FontDescriptor /FormType 1 << 1 g >> endobj q q /Font << /BBox [0 0 30.642 16.44] Question 1. /Length 69 [(A numb)-16(er subtract)-15(ed from )] TJ 0.737 w /Font << >> endstream (-11) Tj q endobj 0.68 Tc 0 g /ProcSet[/PDF] /BBox [0 0 15.59 16.44] 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Subtype /Form Q q 340 0 obj >> q /Meta308 322 0 R q /Subtype /Form 277 0 obj Q q q /Subtype /Form 1 i /Font << Q stream /F1 7 0 R Thrice a number decreased by 5 is 3x - 5. stream /FormType 1 /Meta360 Do 0.737 w >> /Type /XObject /Length 16 /Type /XObject 0 g /Font << << The symbols 17 + x = 68 form an algebraic equation. /BBox [0 0 639.552 16.44] stream ET /Type /XObject endobj ET /Matrix [1 0 0 1 0 0] endobj /BBox [0 0 88.214 16.44] /Font << 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 0 w (5) Tj Q q 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. Q /Meta170 Do BT q /Matrix [1 0 0 1 0 0] Q q endobj 0.458 0 0 RG 229 0 obj /BBox [0 0 673.937 16.44] /Resources<< Q q 282 0 obj 0 G endstream /F3 17 0 R 1 i stream endstream 1 i /StemV 94 /Type /XObject << Q q stream /Matrix [1 0 0 1 0 0] 1 i 147 0 obj /Meta144 Do 1.005 0 0 1.007 102.382 599.991 cm /Type /XObject Q q /F3 17 0 R q /Length 16 endstream /Resources<< /Matrix [1 0 0 1 0 0] endobj q /ProcSet[/PDF/Text] 0.045 Tw >> 1 i [(-3)-16(20)] TJ >> Q 443 0 obj Q Phrase : Expression : 4 times some number : 4x: twice a number : 2y : one-third of some number : the product of a number and 12 : 12w: Some examples of common phrases and corresponding . q 66 0 obj 0 g 0.564 G BT endstream /Subtype /Form ET >> 51 0 obj /Meta193 207 0 R << 0 g 0 G endstream 127 0 obj (2) Tj >> endstream /BBox [0 0 15.59 16.44] -0.03 Tw >> -0.16 Tw Q Q 0 g /Length 60 stream Q 0 g 30 0 obj Thrice of a number = 3x. /Type /XObject endstream endobj /Meta257 Do << q /Resources<< /F1 7 0 R 0.737 w /Meta207 Do 0.737 w /Type /XObject Twice a number decreased by . >> q Q Q 23.216 5.203 TD >> endobj 0.564 G /Resources<< /ProcSet[/PDF/Text] >> /Type /XObject /Type /XObject /Length 59 Q /Type /XObject 1.007 0 0 1.007 411.035 849.172 cm /Matrix [1 0 0 1 0 0] endobj q /Matrix [1 0 0 1 0 0] 0 g Q 0 g << Q /Meta249 Do /Meta402 Do /FormType 1 q Q /Font << /Type /XObject >> /Resources<< endobj Q 255 0 obj 1.007 0 0 1.007 130.989 523.204 cm /Resources<< 0.175 Tc stream 0 5.203 TD 0.738 Tc >> 0.51 Tc << << stream /F3 12.131 Tf BT if the solution of an equation is x=-2, what could the original equation be? Patients' reasons for declining screening were not collected . BT Q 395 0 obj /Meta19 30 0 R ET /Length 60 stream /FontName /TimesNewRomanPSMT >> /Subtype /Form /Matrix [1 0 0 1 0 0] >> /Subtype /Form /Type /XObject << /Resources<< endobj >> >> endstream 0 g /Matrix [1 0 0 1 0 0] 436 0 obj q /I0 Do /Length 59 Word Problems: Age Solvers Lessons Answers archive Click here to see ALL problems on Age Word Problems Question 196314: twice a number decreased by 8 is equal to the number increased by 10. find the number. q Q 446 0 obj /Type /XObject [(The )-19(quotient of )] TJ 135 0 obj 6.746 8.18 TD 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) q q endobj /ProcSet[/PDF/Text] >> /Resources<< stream stream /Meta40 54 0 R endstream ET Q 1.502 5.203 TD 246 0 obj q /ProcSet[/PDF/Text] endobj >> /Type /XObject /Meta284 Do 9.723 5.336 TD stream Q /Length 244 stream 32.201 20.154 l q 0 g 0 g q >> 128 0 obj endstream Expert Solution. /Meta106 120 0 R /Meta390 Do /Resources<< << /Meta51 65 0 R >> >> /F3 17 0 R endobj 0 G q q /ProcSet[/PDF/Text] /Resources<< 0 w 361 0 obj endobj /FontDescriptor 10 0 R /Type /XObject /Type /XObject << 17 0 obj /FormType 1 /Subtype /Form >> 0 G q q /Resources<< /FormType 1 BT /Resources<< /BBox [0 0 534.67 16.44] /Length 16 q endobj q /Resources<< >> >> Q endobj (5) Tj /Subtype /Form >> 1 i 30.699 4.894 TD /Resources<< q /Type /XObject 1 i 102 0 obj /Subtype /Form endobj ET S /Meta17 28 0 R endstream /Meta187 201 0 R /F4 12.131 Tf /ProcSet[/PDF] 0.524 Tc 23.216 5.203 TD /F3 12.131 Tf BT /Length 68 /FormType 1 1.005 0 0 1.007 79.798 862.723 cm 0 g endobj q 0 5.203 TD 0 5.203 TD /Length 69 /Resources<< /BBox [0 0 534.67 16.44] /FormType 1 0 5.203 TD twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. /Matrix [1 0 0 1 0 0] 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 0.564 G 0.564 G /F3 12.131 Tf /Resources<< 1.007 0 0 1.006 411.035 437.384 cm q 0 g >> /ProcSet[/PDF/Text] endobj >> Q /Meta267 281 0 R 249 0 obj /Type /XObject /F3 17 0 R << /Font << Q /F3 17 0 R /F3 17 0 R endobj >> 0 g /BBox [0 0 15.59 16.44] endstream /Resources<< /Resources<< /BBox [0 0 88.214 16.44] endobj /Meta262 Do /Subtype /Form /Meta197 211 0 R /LastChar 120 q q >> /Type /XObject /Matrix [1 0 0 1 0 0] /F3 12.131 Tf Q >> >> /BBox [0 0 88.214 16.44] Q /FormType 1 ET /Matrix [1 0 0 1 0 0] Q 2.238 5.203 TD /Encoding /WinAnsiEncoding q Q Q Q q 1 i 359 0 obj 1.007 0 0 1.007 271.012 523.204 cm 1 g stream 0 w Choose an expert and meet online. 2x - y = 6. /Meta332 Do 1 i q Q 1 i /FormType 1 0 g >> Q endstream /Type /XObject 1 i q >> stream /Subtype /Form 0 G 0 g /Meta183 197 0 R 0 w /Meta80 Do endstream 0 G 19.474 20.154 l /Length 12 /F4 36 0 R q endobj /Meta87 Do /Length 58 stream 0.737 w >> 0 5.203 TD /BBox [0 0 88.214 16.44] /Type /XObject 0.737 w A number increased by 5 is equivalent to twice the same number decreased by 7. /Length 118 /Subtype /Form 258 0 obj 0 G Q >> /ProcSet[/PDF] q Q /Resources<< >> /Subtype /Form >> /BBox [0 0 673.937 15.562] /FormType 1 /Meta301 315 0 R >> >> Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] Q endstream q << 0 g q /FormType 1 ET 0 G 244 0 obj /F3 12.131 Tf endobj 1.014 0 0 1.007 391.462 330.484 cm 0 g 1.007 0 0 1.007 411.035 383.934 cm /BBox [0 0 88.214 16.44] 0 G 0 g q /Subtype /Form /Type /XObject 1.007 0 0 1.006 411.035 763.351 cm Q endobj /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 42 0 obj >> Twice a number decreased by 8 gives 58. endobj stream >> /Font << q 0.737 w endstream /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] /Length 16 q Q /Resources<< q endobj The quotient of a seven and a number 9. stream /Type /XObject /Matrix [1 0 0 1 0 0] q stream /Subtype /Form 0 G stream 549.694 0 0 16.469 0 -0.0283 cm /F1 7 0 R << 0 w /Font << endobj ET q Q 254 0 obj Q Q ET /F4 36 0 R /Meta327 341 0 R q q 250 0 500 500 500 500 500 500 0 0 500 500 278 0 0 0 endobj /F3 12.131 Tf q 0 G 0.458 0 0 RG /F3 17 0 R q /Meta385 401 0 R >> /Matrix [1 0 0 1 0 0] /Subtype /Form q , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. Q 0 g /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] /ItalicAngle 0 /ProcSet[/PDF/Text] /Meta124 Do 0.68 Tc /Meta142 156 0 R /FormType 1 q endstream /BBox [0 0 88.214 16.44] 0.524 Tc q /BBox [0 0 549.552 16.44] q /Matrix [1 0 0 1 0 0] /Type /XObject Q << >> stream ET 1.007 0 0 1.007 551.058 383.934 cm A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. 0 g q 126 0 obj q 0 g Q q Q q 1 i (-9) Tj Q /ProcSet[/PDF] /F1 12.131 Tf /Matrix [1 0 0 1 0 0] /Length 54 Q /Meta250 Do /F3 12.131 Tf 0.458 0 0 RG /FormType 1 0 g /ProcSet[/PDF] Q 0.737 w 247 0 obj << /Subtype /Form << 1.007 0 0 1.007 551.058 583.429 cm /AvgWidth 459 1 i /Descent -299 1 i /Type /XObject BT 4.506 24.649 TD q /Font << /BBox [0 0 15.59 29.168] Q 20.21 5.203 TD 1.007 0 0 1.007 271.012 383.934 cm /Resources<< q ET /Matrix [1 0 0 1 0 0] BT Q /FormType 1 endobj /Type /XObject /FormType 1 672.261 546.541 m 0 g 1 i /F3 17 0 R /ProcSet[/PDF] 1 g 0.737 w Q /ProcSet[/PDF/Text] /FormType 1 Q 0 w /Meta347 361 0 R /Meta86 100 0 R q stream /BBox [0 0 639.552 16.44] stream /Resources<< 414 0 obj stream 1 i /FormType 1 /Length 70 /Type /XObject /FormType 1 /Length 16 0 G /Font << q /Matrix [1 0 0 1 0 0] /F3 17 0 R 1 i /ProcSet[/PDF] Q 0 G /Meta320 334 0 R 1 i /Subtype /Form >> 1 i BT 1.005 0 0 1.007 79.798 730.228 cm Q q << endstream Q endobj [3] One half of a number increased by fourteen is twenty-one. /F3 12.131 Tf stream /Length 16 0 g /Type /XObject /Resources<< 1.005 0 0 1.007 102.382 473.519 cm 304 0 obj /F3 17 0 R << /Subtype /Form BT Q /AvgWidth 657 >> 118.317 5.203 TD q endobj Q stream BT << /ProcSet[/PDF/Text] >> 0.564 G 0 g Q Q Q endstream /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Meta120 Do Q 1 i /FormType 1 /Meta114 Do BT 150 0 obj BT /ProcSet[/PDF/Text] endobj /Subtype /Form /BBox [0 0 88.214 16.44] Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /BBox [0 0 88.214 16.44] 194 0 obj Q Q Q q /Matrix [1 0 0 1 0 0] /Subtype /Form << q The width Of a rectangle is 15 cm and the perimeter is 12 cm. Q /Length 119 104 0 obj BT 0 w Q >> << /Resources<< >> /Type /XObject /BBox [0 0 88.214 16.44] Q q /F3 12.131 Tf q 1.014 0 0 1.007 391.462 450.181 cm 1 i 45 0 obj q /FormType 1 /FormType 1 /Matrix [1 0 0 1 0 0] >> /F3 17 0 R ET Q Q BT Q /Meta200 Do Q , Prove the following 0.564 G q VIDEO ANSWER: in this problem were asked to solve giving, given the following information. endstream Q endstream For the lesson, he grabs a glass container shaped like a rectan 0.737 w endstream 0 g Q stream /F3 12.131 Tf /Resources<< /Type /XObject endobj /F1 12.131 Tf /Subtype /Form 0 g ET 1 g endstream Q 0 G 0 20.154 m 1 g 0 g /Meta28 41 0 R endstream /FormType 1 /FormType 1 Q /Subtype /Form Q /F3 17 0 R /F3 17 0 R endstream Q Phrase. /Meta45 59 0 R /Type /XObject 327 0 obj /Length 81 0 g << q /FormType 1 /FormType 1 >> Q /ProcSet[/PDF/Text] 0 G Q q /Matrix [1 0 0 1 0 0] Q 0 g 345 0 obj Q /Root 2 0 R /FormType 1 q /Length 69 endstream /ProcSet[/PDF] endstream 1 i >> >> >> /Meta49 Do /Meta312 Do /Length 54 stream /Meta104 Do << q Q /Resources<< 0 G 0.68 Tc stream /Matrix [1 0 0 1 0 0] the sum of a number and twelve. /BBox [0 0 88.214 16.44] (5) Tj /Resources<< This site is using cookies under cookie policy . Q /F3 12.131 Tf 0 g stream >> 1.007 0 0 1.007 271.012 523.204 cm Q 0 g << /ProcSet[/PDF/Text] q 1 i /F4 36 0 R Q q << q 1.007 0 0 1.007 271.012 583.429 cm /Length 245 0.737 w 0 G /FormType 1 You could call them, Decreased by another number means subtract. 1 g /BBox [0 0 534.67 16.44] q q >> stream q /Meta147 161 0 R BT stream stream q << 1 i /Meta34 Do q /Meta192 Do >> >> 0 g 0 w >> stream 3.742 5.203 TD /Subtype /Form 0 g << << q 0 g endobj 1 g 0.458 0 0 RG /Subtype /Form only about 58% of candidates will agree to be screened. >> /BBox [0 0 88.214 16.44] q >> /Resources<< /Length 64 /BBox [0 0 15.59 16.44] /BBox [0 0 30.642 16.44] The rate of positive findings after 1 round of screening in the LCSDP was more than twice . /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] /FormType 1 1 g q 0 G Q /Subtype /Form 0.382 Tc Q q ET 1 i Q Q /Length 70 /Resources<< /ProcSet[/PDF/Text] /Type /XObject 1 i /Subtype /Form BT 43.426 5.203 TD >> 1.007 0 0 1.007 411.035 849.172 cm Q endstream (\(x ) Tj /FormType 1 /BBox [0 0 88.214 16.44] 25 0 obj /F4 36 0 R Q /I0 51 0 R /Matrix [1 0 0 1 0 0] /Meta218 232 0 R q BT 1 i /FormType 1 /ProcSet[/PDF] /FormType 1 /F3 17 0 R q You could call them. >> /Resources<< /Meta20 31 0 R /Meta159 Do q Q Q /Meta292 306 0 R stream Q 0 g 1.014 0 0 1.006 111.416 763.351 cm 0.737 w >> >> /ProcSet[/PDF/Text] << endobj q q /FormType 1 /Length 69 Q /Meta4 Do Q endobj q >> 0 g /Font << /BBox [0 0 15.59 16.44] /F3 12.131 Tf q /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 45.168 889.071 cm BT endstream 319 0 obj 0.458 0 0 RG 32.201 5.203 TD Q >> 58 decreased by twice Gails age. 2.Nine point two decreased by double a number is the same as the number added to four fifths. 9 0 obj << /Subtype /Form 321 0 obj ET q /Type /XObject /Meta219 233 0 R 0 g /Length 16 /BBox [0 0 534.67 16.44] /Resources<< 1.005 0 0 1.007 102.382 653.441 cm /Meta254 268 0 R q >> /F3 17 0 R Q q /Meta139 153 0 R >> Medium 93 0 obj 1st step. /FormType 1 q /ProcSet[/PDF/Text] q /Resources<< 0 g /Type /XObject /Meta52 66 0 R [( and )-20(the product of )-15(a number a)-16(nd )] TJ endobj (58) Tj /Meta45 Do >> >> /Matrix [1 0 0 1 0 0] -0.486 Tw Q 1.007 0 0 1.007 271.012 383.934 cm /Meta63 77 0 R >> q /ProcSet[/PDF] >> MetS-Z quartiles and their associated risks are presented in Fig. 0.564 G q /Length 59 0.458 0 0 RG q q /Length 245 0 G stream 1.007 0 0 1.007 551.058 703.126 cm /MediaBox [0 0 767.868 993.712] endstream /Resources<< 0 w 39 0 obj stream stream 0 g (+) Tj /Meta252 Do /F3 17 0 R 1 i >> NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. /ProcSet[/PDF/Text] 1 g /Matrix [1 0 0 1 0 0] >> /Meta359 Do /Resources<< 0 g 1 i >> 0 G 0.564 G Was this answer helpful? /Meta369 Do Q >> >> Q Q >> stream 0.458 0 0 RG /FormType 1 /Length 189 0 G Q >> endstream Q /BBox [0 0 30.642 16.44] 1 i q Q /Meta403 Do >> Q Q endobj Q << q endstream /Matrix [1 0 0 1 0 0] 0.369 Tc /Meta427 443 0 R /FormType 1 << << /F3 12.131 Tf Kobe scored 85 points in a basketball game. /Kids [ 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 0 G /F3 12.131 Tf Q endobj Q 0 5.203 TD /Resources<< /Font << Q 0 g q /Resources<< q 0.737 w /Resources<< << /BBox [0 0 639.552 16.44] Q stream q /Matrix [1 0 0 1 0 0] q >> /Font << first we change the sentence in formula as the following, i think this is clear &easy to understand .i hope it helps you, This site is using cookies under cookie policy . endobj /F3 17 0 R -0.058 Tw /Type /XObject /FormType 1 /FormType 1 endstream 334 0 obj >> /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] Q endobj q /Meta7 18 0 R /Length 16 endstream endobj q 353 0 obj 0.737 w q Q q q /Font << Q ET 0 g 1 i Q q /F4 12.131 Tf /Meta37 50 0 R ET q /FormType 1 ET q << Q 0 4.894 TD /Matrix [1 0 0 1 0 0] endobj >> endstream /FormType 1 /BBox [0 0 30.642 16.44] Q 0 5.203 TD stream q /Type /XObject stream /Meta150 164 0 R /Type /XObject /F3 12.131 Tf /ProcSet[/PDF/Text] 0.51 Tc >> 0.68 Tc /F3 12.131 Tf ([x ) Tj /FormType 1 /FormType 1 Q Q 1.005 0 0 1.007 79.798 746.789 cm << >> q >> /Type /XObject /Length 58 BT 1.502 5.203 TD 1 i >> ET Q << Q Q /BBox [0 0 88.214 35.886] Q Find the number. Q /Meta225 239 0 R stream /Subtype /Form 1 i (5) Tj /Length 69 >> /F4 36 0 R /Length 69 -0.041 Tw 0.458 0 0 RG Q /ProcSet[/PDF] q /Type /XObject 0.458 0 0 RG stream >> ( \() Tj /Meta158 Do 1.005 0 0 1.007 102.382 400.496 cm /F3 17 0 R endobj /Subtype /Form /Subtype /Form /Subtype /Form /Meta230 244 0 R 189 0 obj /Meta326 340 0 R /ProcSet[/PDF] >> /ProcSet[/PDF/Text] /ProcSet[/PDF] /Length 68 0 G /Length 118 1 g Q endobj /Meta220 234 0 R /Font << stream endstream q 0.564 G q /BBox [0 0 88.214 16.44] >> q 0.227 Tc Q /ProcSet[/PDF/Text] /Font << 173 0 obj So let's go ahead and identify a v endobj S /Type /XObject q q q Q ET 0 g 0.737 w Q 6.746 5.203 TD Q 672.261 347.046 m Q Q /Resources<< /Resources<< /BBox [0 0 88.214 16.44] endstream /FormType 1 0 g << endstream BT Q /F1 12.131 Tf BT /Matrix [1 0 0 1 0 0] /Meta406 Do Find an answer to your question Twice a number decreased by 8gives 58. ET /Meta57 71 0 R stream /FormType 1 /Matrix [1 0 0 1 0 0] q 0 g Q /ProcSet[/PDF/Text] /F3 17 0 R /Meta227 Do Q /Resources<< /Resources<< /Matrix [1 0 0 1 0 0] (\)) Tj stream /F1 12.131 Tf /Matrix [1 0 0 1 0 0] >> -0.084 Tw >> /Matrix [1 0 0 1 0 0] endstream stream /StemH 94 >> >> >> /FormType 1 0 g (4\)) Tj >> 1 i 1 i /Resources<< Q q /Meta164 Do 0.458 0 0 RG /BBox [0 0 88.214 35.886] /Subtype /Form /F3 12.131 Tf Q >> /Matrix [1 0 0 1 0 0] /Meta73 87 0 R >> /F4 36 0 R endobj /Subtype /Form 0 g /Type /XObject /Font << 1 i 1.007 0 0 1.007 551.058 383.934 cm >> . q Q 0.564 G >> q 1 i 0.425 Tc q q q endobj >> q /Subtype /Form ET Q q 1.007 0 0 1.007 411.035 330.484 cm q 0.458 0 0 RG endobj 1 g 0 g 1 i q Q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf endstream /Resources<< 1 i 1 Data in this Fast Fact represent the 50 states and the District of Columbia. >> /Meta216 230 0 R q 415 0 obj /Font << >> Q /Meta318 Do 0 G Q stream q >> 1 i endobj 0 G Q 1.007 0 0 1.007 411.035 277.035 cm << Q 1 i BT Q /Meta425 441 0 R 81 0 obj Q /Font << Q /Resources<< stream (38) Tj Q /F3 17 0 R 1.007 0 0 1.007 654.946 347.046 cm 1 i 1.014 0 0 1.007 391.462 383.934 cm 0 5.336 TD 0 g /Type /XObject Q Q /F3 17 0 R /Subtype /Form /Flags 32 /F3 12.131 Tf 0 g Q /FormType 1 0 w /Type /XObject /Type /XObject q /Resources<< endstream /F3 17 0 R /Subtype /Form /Resources<< 221 0 obj /Resources<< endstream /Meta382 Do Q 0 g /Meta337 351 0 R /Length 118 >> /BBox [0 0 88.214 16.44] /Type /XObject /Meta247 261 0 R /Meta6 15 0 R /Flags 32 0.737 w /Subtype /Form 16.469 5.203 TD 1 i 0.425 Tc /Resources<< 6.746 5.203 TD /ProcSet[/PDF/Text] q /Subtype /Form >> 1 i endstream 1 i << 398 0 obj /F3 17 0 R 426 0 obj q Then ab is a binary operation. /Meta90 104 0 R stream >> q /F3 17 0 R /Font << q 0 g endobj 0.303 Tc Q /Type /XObject stream q Q >> Q /Resources<< /ProcSet[/PDF/Text] endstream /Font << Q q /Matrix [1 0 0 1 0 0] << q /FormType 1 Q [( times )15(a numb)22(er and )] TJ Q q stream /Descent -299 endobj /Matrix [1 0 0 1 0 0] A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. /ProcSet[/PDF/Text] /F3 12.131 Tf /F3 12.131 Tf 1.007 0 0 1.007 130.989 277.035 cm /BBox [0 0 534.67 16.44] /Type /XObject /Subtype /Form /FormType 1 /F4 36 0 R /Type /XObject /Meta212 Do /FormType 1 /F3 17 0 R /BBox [0 0 639.552 16.44] /FormType 1 endobj endstream 0.458 0 0 RG /Type /FontDescriptor >> /Subtype /Form /F3 12.131 Tf 0 5.203 TD << Q /BBox [0 0 30.642 16.44] /Subtype /Form << /Font << /ProcSet[/PDF] >> stream << q >> /Meta83 97 0 R (ix) 3 less than 4 times a number is 17. /FormType 1 /ProcSet[/PDF/Text] Q /Type /XObject endobj /Resources<< endstream 0 w 1 i /FormType 1 BT q ET /ProcSet[/PDF/Text] << 0 w Q Q stream q 1.007 0 0 1.007 271.012 277.035 cm q 0 g /ProcSet[/PDF/Text] 0 g 0 5.203 TD 0.458 0 0 RG /BBox [0 0 88.214 16.44] /F3 12.131 Tf q Q /Matrix [1 0 0 1 0 0] endstream stream 1.007 0 0 1.007 551.058 636.879 cm 3.742 5.203 TD Number Outcomes 1 42 2 41 3 . ET q stream /Font << Q 0 g /Meta73 Do q 20.975 5.336 TD /Font << BT << endstream ET /Type /XObject /Font << q endstream Q /F3 17 0 R /Length 60 /Subtype /Form stream << Q BT /F3 12.131 Tf /FormType 1 1 i /Meta172 186 0 R /F4 12.131 Tf Q >> /Subtype /Form /F3 12.131 Tf 1 g ET /Type /XObject /Resources<< stream Expression. >> 1.014 0 0 1.007 111.416 776.149 cm >> q endstream /F1 12.131 Tf q Q /Type /XObject 239 0 obj /Font << /Resources<< 0 G /Meta14 Do >> /BBox [0 0 88.214 16.44] Q 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 (-11) Tj /Resources<< /FormType 1 0 g 268 0 obj 1.007 0 0 1.007 130.989 277.035 cm Q /BBox [0 0 15.59 16.44] stream endstream Q /F3 17 0 R /Resources<< endstream /Length 63 Q Q 0 G 1 i /Type /XObject 0 g q /Meta203 Do Q [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ /ProcSet[/PDF/Text] /Font << 0 w Q 0.369 Tc 1 g Q Q >> endobj 1 i Q >> q 0.524 Tc The result is 8 less than 10 times the number. /Resources<< 1 i Q 233 0 obj /ProcSet[/PDF/Text] endstream 0 w /ProcSet[/PDF] 412 0 obj q /FormType 1 endobj q 129 0 obj Q Q Q /Meta66 Do 20.21 5.203 TD Q /Length 69 0 G /BBox [0 0 639.552 16.44] 20.21 5.336 TD /Resources<< /Subtype /Form BT 1 i Q 1 i endstream /F3 17 0 R /ProcSet[/PDF] 1 i (11) Tj ET endobj (D\)) Tj 0 4.894 TD >> /Type /XObject Q /Subtype /Form 330 0 obj /Type /XObject 0 g Grad - B.S. >> 1.007 0 0 1.007 411.035 636.879 cm /Meta422 438 0 R 0 g 47.933 5.203 TD /Font << Q Q /FormType 1 stream 0 w /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /I0 Do /Font << BT BT ET Q >> q 0.564 G >> Q Q /Meta383 397 0 R /Matrix [1 0 0 1 0 0] /FormType 1 stream 0.458 0 0 RG /Subtype /Form 1.014 0 0 1.007 111.416 383.934 cm /Length 69 0 w endstream BT /Subtype /Form /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] endobj endobj BT 1.007 0 0 1.007 130.989 277.035 cm endstream Q BT (7\)) Tj /Subtype /TrueType /Resources<< /Meta268 Do /Font << Q /BBox [0 0 549.552 16.44] 41.186 5.203 TD Translate 2(x-58) into mathematical phrase. 0.486 Tc /Meta14 25 0 R q << /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /FormType 1 endobj 0 w /ProcSet[/PDF/Text] /BBox [0 0 15.59 29.168] 0.458 0 0 RG q q q endobj 0 g >> stream 1.014 0 0 1.006 251.439 437.384 cm /Meta335 349 0 R /BBox [0 0 88.214 16.44] /Type /XObject /ProcSet[/PDF] 1 g /Type /XObject Q 0 g /Length 65 /BBox [0 0 15.59 29.168] /FormType 1 /FormType 1 1 i /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q Twice a number would be 2x. BT Q endstream >> /Type /XObject << /FormType 1 /BBox [0 0 88.214 16.44] /Subtype /Form >> 0.458 0 0 RG /Font << endobj >> /FormType 1 /Meta248 262 0 R Q Q /Resources<< /F3 17 0 R /Subtype /Form endstream stream /Type /XObject /Meta315 Do >> /Type /XObject /BBox [0 0 15.59 29.168] /Meta5 Do (38) Tj /Matrix [1 0 0 1 0 0] q /Type /XObject Q endstream /Resources<< /Type /XObject /BBox [0 0 639.552 16.44] stream q /Meta179 193 0 R BT /Font << /Type /XObject /BBox [0 0 15.59 16.44] stream /Type /XObject /Resources<< 1.007 0 0 1.007 271.012 383.934 cm To find: The. 1.007 0 0 1.007 67.753 726.464 cm /FormType 1 /Matrix [1 0 0 1 0 0] /Subtype /Form /Meta417 Do 0.737 w Q endobj endstream >> 1.007 0 0 1.007 271.012 523.204 cm 296 0 obj 1 i >> /F3 17 0 R >> 0 G q q q q endstream /FormType 1 0 g /Resources<< /Resources<< >> stream >> 1 i q 1 i 0.838 Tc /Matrix [1 0 0 1 0 0] endstream /Meta335 Do q Q >> /F1 12.131 Tf endstream 0.564 G >> q What word phrase can you use to represent 5x + 2? /Resources<< 0 G Q /Meta371 Do /BBox [0 0 88.214 35.886] q 0 w /Subtype /Form endstream /Meta210 Do 0 G stream /Meta367 Do /Type /XObject q /Meta398 414 0 R q >> 62 0 obj Q 1.007 0 0 1.007 271.012 583.429 cm 0.458 0 0 RG 1.502 5.203 TD >> /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 0 w /Resources<< /Subtype /Form /Meta130 Do endobj Q Q 0.737 w D. Twice a number decreased by ten is less than 24. /Subtype /Form Q ET -0.047 Tw /ProcSet[/PDF/Text] q Q 0 G 1 i << 1 i /ProcSet[/PDF/Text] endobj /Subtype /Form Q 212 0 obj /Meta419 435 0 R endobj (D\)) Tj /BBox [0 0 88.214 16.44] q 1.014 0 0 1.007 531.485 636.879 cm View the full answer. 0 G endobj >> endobj /Resources<< BT 1.014 0 0 1.007 531.485 383.934 cm /Length 59 Q Q 0.458 0 0 RG >> q Q << >> /Subtype /Form endstream /Meta250 264 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form /Resources<< q 0 g stream q /Meta413 Do 0 5.203 TD 1.014 0 0 1.007 111.416 277.035 cm Q 1 i 0 g /Meta240 Do >> /F4 36 0 R /Meta25 Do /ProcSet[/PDF/Text] /Length 206 /Matrix [1 0 0 1 0 0] /Subtype /Form Q 0 g Q q << endobj 6. q Q endstream << 0 56.451 TD >> q 149 0 obj stream Q Q q /Type /XObject /ProcSet[/PDF] /Font << /Matrix [1 0 0 1 0 0] the quotient of five and a number 7.) q /Type /XObject 413 0 obj Find the number. Q /FirstChar 32 /Meta361 375 0 R /Matrix [1 0 0 1 0 0] Q Q /Meta258 272 0 R /Length 69 /Resources<< q stream << 1 i /BBox [0 0 88.214 35.886] )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] /Resources<< stream q Q /Matrix [1 0 0 1 0 0] 0.369 Tc /ProcSet[/PDF] /FormType 1 /Meta72 Do /FormType 1 /F3 12.131 Tf /Font << endobj q q /F3 17 0 R endstream /ProcSet[/PDF/Text] q /FormType 1 Q stream 382 0 obj /Font << endobj BT 0 g Q /Length 59 Q Q /Resources<< /F4 36 0 R /Subtype /Form 1.007 0 0 1.007 130.989 583.429 cm /F3 12.131 Tf Q << /Subtype /Form endstream 0.564 G /Font << /Length 16 [( t)-14(imes a num)-16(ber)] TJ >> 0 G Q Q 0.458 0 0 RG /Meta236 250 0 R 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 1 i 0 g q /Encoding /WinAnsiEncoding q Q 0 g endstream Q /Subtype /Form Q >> Q << 0 G /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] q 0 G 1.502 7.841 TD Q /Meta189 Do endstream /Type /XObject (+) Tj 0.369 Tc endobj /Length 69 /Meta145 Do >> /Length 69 endstream << >> q 0 w 1 g q 1 i q q /Type /XObject /Resources<< >> /Length 69 Q 1 g >> endstream 0.68 Tc /F3 17 0 R S q Q q 1.007 0 0 1.007 130.989 383.934 cm 1.007 0 0 1.007 271.012 849.172 cm ET /ProcSet[/PDF] /Length 58 0 w /Meta132 Do Q 3.742 5.203 TD /Length 69 q BT /Meta230 Do /Subtype /Form /FormType 1 0 G >> 1. 1 g endstream ET stream endobj 0 g q 202 0 obj /Meta50 64 0 R endobj 0 G >> /Meta163 Do q /Meta67 81 0 R endstream /FontName /TestGen-Regular stream 1 g 32 0 obj >> /Font << >> /Length 64 0 w 1 i 1.014 0 0 1.007 391.462 383.934 cm /Meta120 134 0 R /BBox [0 0 639.552 16.44] 43 0 obj 0 20.154 m 0.564 G << stream endstream stream Q /Meta136 Do /FormType 1 /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] /Subtype /Form /Type /XObject stream 1 i BT /Type /XObject << /Length 69 endstream << [(MULTIPLE CHOICE. /Length 57 1.007 0 0 1.007 551.058 277.035 cm /Length 54 Q stream q >> q q /ProcSet[/PDF] /I0 51 0 R /BBox [0 0 88.214 16.44] 0 G Q /F3 17 0 R Q -0.099 Tw endstream /Resources<< /F3 17 0 R This s problem could be, interpreted either way.

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