twice a number decreased by 58

twice a number decreased by 58twice a number decreased by 58

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0.564 G q 32.201 5.203 TD endobj >> /Font << /FormType 1 /F3 17 0 R (-) Tj /ProcSet[/PDF/Text] stream q /BBox [0 0 88.214 35.886] << 1 i /Meta382 396 0 R /Subtype /Form /Meta373 387 0 R 1.005 0 0 1.007 79.798 829.599 cm endstream 0.37 Tc 0 G /Resources<< Q /Subtype /Form ET Q 1 i stream /Type /XObject endobj stream 0 G /ProcSet[/PDF/Text] Q << q /Font << q 1 i ET >> /Subtype /Form q 1 i /F4 36 0 R /Meta427 Do /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /FormType 1 32 0 obj /Resources<< endstream 0 g Q 1.007 0 0 1.007 271.012 849.172 cm /FormType 1 /F3 17 0 R /Length 64 /Meta191 205 0 R Q /F3 12.131 Tf 0.737 w /FormType 1 >> endobj >> q /Length 59 0 4.894 TD 1 i /ProcSet[/PDF] 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. /Length 69 /F3 17 0 R /Length 69 /BBox [0 0 88.214 16.44] Q q 1 i 1.007 0 0 1.007 654.946 400.496 cm endobj q Q q q /Meta4 13 0 R /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] ( x) Tj /Subtype /Form endobj << 0.737 w 0.458 0 0 RG 128 0 obj 1 i >> 1.007 0 0 1.007 45.168 829.599 cm /Matrix [1 0 0 1 0 0] >> 0 G 0 G -0.058 Tw /Matrix [1 0 0 1 0 0] /Subtype /Form /Meta210 224 0 R stream q /Length 78 Q /Font << /Type /XObject /FormType 1 ET 0.737 w /Meta3 Do >> /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Length 68 endstream 0.458 0 0 RG Let x be a number. << /Type /XObject 0 g Q /Meta228 Do Q /BBox [0 0 88.214 16.44] 0 G << << Q endobj Twice a number when decreased by 7 gives 45. /FormType 1 >> /BBox [0 0 17.177 16.44] /Length 16 q (x) Tj An example of a linear inequality in one variable is A. x+y = Question: answer 1. Q 0 g q /Type /XObject endobj 1.005 0 0 1.007 102.382 363.608 cm q Q >> /Meta113 Do /F3 17 0 R endobj q stream (B\)) Tj /Subtype /Form 1.005 0 0 1.007 102.382 816.048 cm Q >> >> q q Q /Resources<< /BBox [0 0 673.937 16.44] /Subtype /Form See Solution. /Meta129 Do /FormType 1 /Font << q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Length 12 /Meta24 Do 1.007 0 0 1.006 411.035 763.351 cm /Meta148 162 0 R /BBox [0 0 15.59 16.44] ET q 0.458 0 0 RG endobj /Type /XObject /Meta66 Do /F3 17 0 R 1 i >> << 1 i >> /FormType 1 q /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] Q /FormType 1 0.458 0 0 RG Q endobj /Subtype /Form [(The )-16(s)15(um )-14(of )] TJ q /Resources<< 150 0 obj Q 1.007 0 0 1.007 130.989 383.934 cm /Subtype /Form 0.564 G /Length 16 (iv) A number exceeds 5 by 3. << /Meta284 298 0 R /BBox [0 0 88.214 16.44] endstream Q /BBox [0 0 15.59 29.168] q ET /Meta60 Do 0.458 0 0 RG /FormType 1 stream /ProcSet[/PDF/Text] /FormType 1 endobj Q 1 g /FormType 1 Q 1.502 24.339 TD (3\)) Tj q Q 1.007 0 0 1.007 551.058 330.484 cm /Meta33 Do Q Q Q /F3 12.131 Tf BT /Type /XObject /Font << >> endobj stream Q 0.737 w The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. /ProcSet[/PDF] 0.045 Tw Q >> >> 38.948 5.203 TD 1 i 179 0 obj /FormType 1 1 i << endobj Q >> 1.007 0 0 1.007 654.946 347.046 cm 226 0 obj >> >> ET /FormType 1 /CapHeight 694 /FormType 1 151 0 obj /Meta188 Do Q Q 1.007 0 0 1.007 130.989 277.035 cm q /Type /XObject q /Meta286 Do 399 0 obj stream Q >> 0 g ET q ET /Font << (B\)) Tj /Matrix [1 0 0 1 0 0] /F3 12.131 Tf The sum of a number and 2 is 6 less than twice that number. stream 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 . (5\)) Tj /FormType 1 BT /F3 17 0 R << 0 w (A\)) Tj /Matrix [1 0 0 1 0 0] 0.786 Tc q q q stream q q /Length 69 1.005 0 0 1.007 102.382 670.003 cm 172 0 obj 1.007 0 0 1.007 271.012 383.934 cm >> stream 243 0 obj /Resources<< Q A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. 0 g /Font << << 14.966 20.154 l q q 419 0 obj stream &K @ /FormType 1 endstream 406 0 obj /Meta180 Do 1 i stream 1.014 0 0 1.007 391.462 383.934 cm /Meta12 Do 0 G /BBox [0 0 88.214 16.44] /F4 36 0 R /ProcSet[/PDF/Text] This site is using cookies under cookie policy . stream endobj endobj >> 0 g /ProcSet[/PDF/Text] /F3 17 0 R /F3 12.131 Tf /F3 17 0 R ET q << endobj q << Q /F3 17 0 R q Q Q 0 g /F3 12.131 Tf << /ProcSet[/PDF] stream endobj q q /Subtype /Form 0 w 0 G stream Q 1.014 0 0 1.007 391.462 776.149 cm /Meta173 187 0 R 549.694 0 0 16.469 0 -0.0283 cm 1 g BT /BBox [0 0 534.67 16.44] >> BT endobj 1.014 0 0 1.007 111.416 277.035 cm >> Q Q /F3 12.131 Tf Q /Matrix [1 0 0 1 0 0] /F3 17 0 R 1.007 0 0 1.007 130.989 330.484 cm Q /ProcSet[/PDF/Text] 1.005 0 0 1.007 79.798 796.475 cm /Length 16 Q /FormType 1 /ProcSet[/PDF/Text] -0.058 Tw ET Q /ProcSet[/PDF/Text] 0.564 G >> 415 0 obj /Resources<< Q /Meta2 9 0 R /Matrix [1 0 0 1 0 0] Q Q q 1 i To find: The. /Meta113 127 0 R 1 i 198 0 obj /Length 69 Q Q q /Subtype /Form Q Q /Meta302 316 0 R 6.746 8.18 TD (x) Tj Q /BBox [0 0 88.214 16.44] stream BT Q q /F4 36 0 R Q BT /Resources<< /FormType 1 q 0 5.336 TD /Resources<< Q Q >> q 0 g endstream /Meta300 Do /FormType 1 q q the other number. /AvgWidth 401 >> << /Type /XObject Twice a number would be 2x. (38) Tj 0 g Q /Length 68 1 i /Matrix [1 0 0 1 0 0] 0 G /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] Q /BBox [0 0 673.937 16.44] << Example 1: Use the tables above to translate the following English phrases into algebraic expressions. endstream >> >> 0.369 Tc /Meta308 322 0 R q (A\)) Tj /Meta54 68 0 R /Matrix [1 0 0 1 0 0] endstream /Length 60 q (C\)) Tj q 22.478 5.336 TD >> -0.382 Tw 0.369 Tc /Length 59 /BBox [0 0 88.214 16.44] endstream 0 G /Length 69 0 G /Meta44 Do /Font << /Meta349 Do Q /Type /XObject << << /FormType 1 endstream >> /Resources<< (+) Tj q (2) Tj q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] ET /FormType 1 /BBox [0 0 23.896 16.44] Q >> q 1.005 0 0 1.007 102.382 473.519 cm /Matrix [1 0 0 1 0 0] Q 1.005 0 0 1.007 102.382 473.519 cm 0.458 0 0 RG /ProcSet[/PDF] endobj /Matrix [1 0 0 1 0 0] /Length 69 Q Q -37 VI 2. /F1 12.131 Tf q endobj 0 g 320 0 obj /Meta209 Do /Font << 20 0 obj endobj q [(The )-19(quotient of )] TJ /FormType 1 1 i Q /Meta123 Do /Type /XObject Q /Subtype /Form endstream /ProcSet[/PDF] ] << Q endobj /Matrix [1 0 0 1 0 0] q << q /Meta158 172 0 R /FormType 1 /FormType 1 /Leading 253 /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 743.025 cm >> >> 0 g ET /Subtype /Form /Resources<< 0 G /Length 68 /BBox [0 0 15.59 16.44] 1 i Q q 1.005 0 0 1.007 102.382 347.046 cm q /Subtype /Form /Resources<< 1 g Q /Type /XObject /Subtype /Form Q q /ProcSet[/PDF/Text] ( \() Tj /Matrix [1 0 0 1 0 0] /Meta156 170 0 R 306 0 obj /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /FormType 1 q >> /Meta402 418 0 R /Length 69 1.014 0 0 1.007 111.416 583.429 cm 1.007 0 0 1.007 67.753 347.046 cm Q 0 w /Subtype /Form q 1 g q ET /Type /XObject >> Question 1. /F1 12.131 Tf /ProcSet[/PDF/Text] 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 Q Q %%EOF. 1.007 0 0 1.007 130.989 583.429 cm << q 672.261 872.509 m Patients' reasons for declining screening were not collected . /F4 36 0 R 1 i stream stream q >> endobj /Length 12 /Meta139 Do >> Q /F3 17 0 R /FormType 1 /Resources<< BT 1.007 0 0 1.007 130.989 523.204 cm BT Q /Meta371 385 0 R 0 g q /Subtype /Form endobj /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] /Length 127 /FormType 1 q 10 0 obj << /ProcSet[/PDF] 0.369 Tc >> >> 1 i q /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 58 << /Meta77 Do /Subtype /Form endobj 305 0 obj 20.975 5.336 TD -0.486 Tw Q Q Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. 0 w >> q q endobj endobj Medium /BBox [0 0 88.214 16.44] /FormType 1 q /ProcSet[/PDF/Text] 0 g Q ET /Matrix [1 0 0 1 0 0] q << /Length 60 /FormType 1 endobj q /FormType 1 0 G endstream /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 130.989 776.149 cm Q /Type /XObject Q /Type /XObject /Length 69 S Q /ProcSet[/PDF] Q q Q << q Q >> /F3 17 0 R << 0.271 Tc Q /Matrix [1 0 0 1 0 0] /Subtype /Form endobj q ET (x) Tj /Length 12 416 0 obj (C\)) Tj /Font << 0 g /F4 12.131 Tf 1.007 0 0 1.006 411.035 763.351 cm /ProcSet[/PDF/Text] /Resources<< 1 i >> /Meta68 Do /Meta304 318 0 R 1 i (6\)) Tj /Meta271 285 0 R BT ET Thrice of a number = 3x. /Meta9 20 0 R Q q Q endobj BT 0.458 0 0 RG /F3 12.131 Tf Q endobj stream 1.014 0 0 1.007 531.485 776.149 cm /Type /XObject /ProcSet[/PDF] endstream q /Type /XObject /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /F4 12.131 Tf Q Q Q /FormType 1 1 g Q << Q >> /Meta244 258 0 R q /FormType 1 1.014 0 0 1.007 391.462 277.035 cm 0.458 0 0 RG 0 g /Matrix [1 0 0 1 0 0] Q /Meta336 Do /Font << ( x) Tj View the full answer. q /Subtype /Form 722.699 546.541 l 0 g /BBox [0 0 88.214 16.44] /Font << endstream >> 1.014 0 0 1.007 391.462 583.429 cm stream /Meta426 442 0 R /XHeight 477 stream /Font << /Meta206 Do Q /Subtype /Form endstream BT stream 1.007 0 0 1.007 654.946 400.496 cm /ProcSet[/PDF/Text] >> /Type /XObject Q Q stream 1.005 0 0 1.007 102.382 799.486 cm 0.425 Tc 57 0 obj q /F3 17 0 R /F3 17 0 R 1 i >> /FormType 1 1.502 5.203 TD /Matrix [1 0 0 1 0 0] /Resources<< Q /F3 17 0 R 0.737 w 0 g /FormType 1 0.564 G >> endstream /Meta299 Do << BT 1 g Q 0 g endobj >> stream q Q D. b = 4 2. endstream << /Resources<< << /Type /XObject /Subtype /Form 500 500 500 0 333 389 278 0 0 722 500 500]>> Q q Q endobj stream Q /Meta27 40 0 R /F1 12.131 Tf /Resources<< BT -0.486 Tw << 1 i /BBox [0 0 30.642 16.44] /Font << q (x) Tj /FormType 1 0 g /Font << 1 i 1 g 1 i BT Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. /F1 12.131 Tf /Length 59 157 0 obj q /FormType 1 722.699 400.496 l Q Q 0 G /F3 12.131 Tf >> BT /Matrix [1 0 0 1 0 0] /F1 7 0 R In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. 30.699 5.203 TD /BBox [0 0 639.552 16.44] q /FormType 1 << q /BBox [0 0 88.214 35.886] ( x) Tj >> >> q /FormType 1 << /Meta288 302 0 R /Meta165 179 0 R /Length 69 Answer: 52 decreased by twice a number in algebric expression Step-by-step explanation: The problem is asking that you subtract twice a number from 52. /Meta227 241 0 R Q Q /Length 64 1 g 0 g Q endstream Q /Font << 0.369 Tc Q >> /Matrix [1 0 0 1 0 0] algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. endobj Q Q q /ProcSet[/PDF/Text] /Subtype /Form stream endstream endstream /Type /XObject >> Q /Type /XObject 1 i /Font << endstream Q 0 g 0 w 3.742 5.203 TD Q /F3 17 0 R /ProcSet[/PDF] 33.704 5.203 TD /Subtype /Form 1 g /Length 59 endobj /FormType 1 Q endstream /Type /XObject 1.005 0 0 1.007 102.382 310.158 cm endstream Q Q /Meta207 221 0 R /Length 16 /Type /XObject q /FormType 1 1 i >> 0 G 1.014 0 0 1.007 251.439 450.181 cm 0 G Phrase. /Type /XObject q /FormType 1 0.737 w >> /Font << q Q /Resources<< /Meta285 299 0 R q /F3 17 0 R 0.737 w /Subtype /Form ET endobj >> 98 0 obj 0 G /Length 118 0 G >> 1.014 0 0 1.007 111.416 703.126 cm q /FormType 1 endstream 1.502 7.841 TD Q S 1 i 333.269 5.488 TD >> /Length 78 Q Q /Subtype /Form >> Q ET (8\)) Tj 1.007 0 0 1.007 67.753 653.441 cm endstream /ProcSet[/PDF/Text] >> 0 w >> 99 0 obj /Resources<< 1.007 0 0 1.007 654.946 546.541 cm q /Subtype /Form ET /BBox [0 0 30.642 16.44] 0.564 G ET /Meta88 102 0 R /I0 Do Q /Type /XObject endstream /BBox [0 0 88.214 16.44] /Resources<< /ProcSet[/PDF] 1 i /F4 36 0 R 1 0 obj q /F3 17 0 R /F3 12.131 Tf /Subtype /Form q >> BT 1.005 0 0 1.007 79.798 829.599 cm /FormType 1 q >> ET >> /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] /Meta189 203 0 R 1.007 0 0 1.006 130.989 437.384 cm q /F3 12.131 Tf q 0.737 w /Subtype /Form /Subtype /Form /Meta211 225 0 R stream /F3 12.131 Tf stream /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] 0 G /Length 118 0 w /Subtype /Form endobj Q /Type /XObject /Meta71 85 0 R /Matrix [1 0 0 1 0 0] >> 1 i /Type /XObject Q << 0 G Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] << Q /Length 16 q BT 383 0 obj /Meta345 Do Q 0.458 0 0 RG q /Flags 32 /BBox [0 0 534.67 16.44] 358 0 obj ET endobj q /BBox [0 0 15.59 29.168] 1.007 0 0 1.006 411.035 437.384 cm /Meta288 Do BT /Resources<< 0 g Q endstream /F3 12.131 Tf endobj /F3 17 0 R q >> 0 g -0.021 Tw << /FontName /PalatinoLinotype-Bold /Length 16 402 0 obj 20.21 5.203 TD >> /Meta78 Do q >> >> ET q /Font << /FormType 1 /Matrix [1 0 0 1 0 0] /BBox [0 0 549.552 16.44] >> /Resources<< /Subtype /Form /FormType 1 endobj >> /BBox [0 0 88.214 16.44] 1.502 5.203 TD >> /ProcSet[/PDF] 1.007 0 0 1.007 271.012 523.204 cm /Font << Q Q q /FormType 1 /Resources<< In other terms, 52-nx The problem is asking that you subtract twice a number from 52. endstream /Length 69 stream /BBox [0 0 15.59 16.44] << /FormType 1 /Font << /Type /XObject endobj Use the variable g to represent Gails age. /Resources<< >> Q 1 i BT /Type /XObject Q 0 G stream /BBox [0 0 15.59 16.44] /Subtype /Form Q 0.369 Tc /Type /XObject Q /BBox [0 0 88.214 16.44] q endobj endobj 0 g /Meta334 Do Q 1.007 0 0 1.007 45.168 779.913 cm 294 0 obj q 396 0 obj Q q 223 0 obj /Type /XObject 307 0 obj /FormType 1 /Type /XObject 1.007 0 0 1.006 130.989 437.384 cm >> Q BT << Q q BT q /Font << >> /Meta412 Do 0 w Q /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF] Q Q q /Resources<< >> /Font << q 267 0 obj 0 w (3\)) Tj q /Matrix [1 0 0 1 0 0] 183 0 obj q /Subtype /Form /Font << -0.22 Tw q ET q 0 G /ProcSet[/PDF/Text] 0 G /Meta170 184 0 R /FormType 1 ET q q q >> /FormType 1 endstream 0 G 39 0 obj 1 i /ProcSet[/PDF/Text] ET Q q Q S q q Q /ProcSet[/PDF/Text] 1 i With this, we get: "3x-8". << endstream Q /Meta174 Do /Font << q /F4 12.131 Tf (x ) Tj /BBox [0 0 639.552 16.44] >> endstream >> /Matrix [1 0 0 1 0 0] >> Q /FontDescriptor 10 0 R /Font << /ProcSet[/PDF] BT /Matrix [1 0 0 1 0 0] >> /Meta343 357 0 R (\(x ) Tj /BBox [0 0 549.552 16.44] ET BT >> 295 0 obj 19.474 5.203 TD /Type /XObject /Subtype /Form q /Length 16 /Subtype /Image >> Q 0 g /FormType 1 q stream ET 1 i endstream /Subtype /Form Q 0 5.203 TD 0.564 G endstream 0.458 0 0 RG q << BT q endobj q /Meta157 Do 0 w Q Q endstream /FormType 1 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 q endobj endstream Q 1 i Q Q /F3 17 0 R >> ET /FormType 1 20.21 5.203 TD /Font << /Type /XObject << q /Ascent 891 1.005 0 0 1.007 102.382 293.596 cm q /Font << Q 0 G q stream (-20) Tj /Meta423 439 0 R >> (2) Tj /Font << 0.564 G << /Meta333 347 0 R BT 0 g /Meta409 425 0 R Q q /F3 17 0 R stream >> /ProcSet[/PDF/Text] /Meta64 78 0 R /Resources<< /FormType 1 0 g /FormType 1 endobj q /BBox [0 0 88.214 16.44] /Meta268 Do q Q /Meta348 Do /Subtype /Form /Type /XObject /BBox [0 0 15.59 16.44] 26.219 5.203 TD (-) Tj 0.737 w endobj >> >> Q /Meta133 147 0 R BT stream /Type /XObject /Length 16 >> /Type /XObject BT 8 0 obj >> stream >> q 0 g 196 0 obj /Matrix [1 0 0 1 0 0] For the lesson, he grabs a glass container shaped like a rectan /FormType 1 q /ProcSet[/PDF/Text] /ProcSet[/PDF] /BBox [0 0 30.642 16.44] /Font << /FormType 1 /Length 58 /Resources<< Grad - B.S. >> /Font << /Subtype /Form ET /Meta137 151 0 R /Font << /FormType 1 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. q Q << 0.68 Tc 363 0 obj Q >> stream /Font << /Subtype /Form /Meta114 Do endobj 1.007 0 0 1.007 551.058 583.429 cm 35.206 4.894 TD stream q 1 g /Length 54 /Font << q stream endstream /Resources<< >> 0 g /Matrix [1 0 0 1 0 0] 1 i endstream /Resources<< /Matrix [1 0 0 1 0 0] BT Q Q Q /Meta15 26 0 R /Meta311 Do /Length 12 Q 47.933 5.203 TD /Resources<< /ProcSet[/PDF] /FontDescriptor 35 0 R Q /BBox [0 0 17.177 16.44] /Type /XObject endstream /ProcSet[/PDF/Text] /Meta215 Do endobj /FormType 1 ET endstream q 0 5.203 TD /Resources<< /Meta350 Do q /Type /XObject /F3 17 0 R Q >> /F1 7 0 R /ProcSet[/PDF/Text] 0 G Q Q /Length 16 >> Tamang sagot sa tanong: 1.) q stream /F1 12.131 Tf stream /Resources<< >> /ProcSet[/PDF/Text] 1 i 0 G /F3 17 0 R << /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] Q 1 i /ProcSet[/PDF] Negative thirteen decreased by 3 times a number x. Q >> q 1 i 0 G >> 0.458 0 0 RG /Meta101 Do Q q >> 1 i >> q >> /Font << /Resources<< 1.007 0 0 1.007 271.012 450.181 cm 1 g 68 0 obj /Meta131 Do q /Type /XObject /Type /XObject Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Font << << 1 i 361 0 obj /Length 74 /F3 12.131 Tf 1.007 0 0 1.007 271.012 583.429 cm /F3 12.131 Tf 2. /Resources<< >> (C) Tj /BBox [0 0 88.214 16.44] >> 0 G /Resources<< Q /Resources<< << [( a )-15(number, decreased by )] TJ q /Meta135 Do 370 0 obj Q endstream /Font << . /F3 17 0 R /FormType 1 /Length 59 0 G /BBox [0 0 88.214 35.886] 1 i (viii) A number divided by 8 gives 7. /Meta47 61 0 R 347 0 obj stream /Resources<< ET q /Meta8 19 0 R 0.369 Tc >> /Length 16 /Meta229 Do stream 0 w q /Type /XObject /Subtype /Form 0.458 0 0 RG q 1.007 0 0 1.007 411.035 583.429 cm 227 0 obj >> 1 i Q Q /F3 12.131 Tf 0.31 Tc /Subtype /Form 1.007 0 0 1.007 271.012 330.484 cm q BT Q stream /FormType 1 /Type /XObject /Meta352 Do /Matrix [1 0 0 1 0 0] << 0.737 w stream (-23) Tj q /Type /XObject ET >> (-8) Tj /Matrix [1 0 0 1 0 0] 1 i q q 0.737 w 0.382 Tc /Meta286 300 0 R endobj Q /FormType 1 >> /Matrix [1 0 0 1 0 0] q /Font << endobj 1 i /FormType 1 q 353 0 obj q /F3 17 0 R Next, the problem says that "x" would be equal to twice a number added by 5. /Resources<< q endobj 1 i C. Twice a number decreased by ten is at most 24. 0 4.894 TD /Meta256 Do /ProcSet[/PDF/Text] << >> 0.68 Tc /Meta182 Do /FormType 1 q ET /Meta153 Do /ProcSet[/PDF/Text] q /Font << /Type /XObject 1 i Q 0 G >> >> /Meta328 Do q /Matrix [1 0 0 1 0 0] Q Q Q Q Q q 1 i /Font << /F4 36 0 R /FormType 1 /ProcSet[/PDF] 1.007 0 0 1.007 271.012 523.204 cm 0 G /Type /XObject >> /Meta376 Do Q /Length 12 1 i /FormType 1 /Type /XObject /Length 57 q (-) Tj 0.564 G 0.68 Tc /Meta89 103 0 R Q Q /Length 58 >> /Type /XObject /BBox [0 0 15.59 16.44] Q 0 G /Meta55 69 0 R /Subtype /Form 312 0 obj /FormType 1 (-4) Tj The result is 8 less than 10 times the number. >> 132 0 obj 0.458 0 0 RG endobj >> << 1.007 0 0 1.007 551.058 277.035 cm >> /Length 69 /ProcSet[/PDF/Text] endobj /F3 17 0 R q BT endobj Q >> >> q Q q 0 g /FormType 1 /Type /XObject q << q /Meta16 Do Q BT 0 g 1 i endstream q q 0 g << Q /Meta375 389 0 R /Type /XObject /Matrix [1 0 0 1 0 0] >> If n is "the number," which equation could be used to solve for the number? ( \() Tj << /Length 118 >> 1.007 0 0 1.007 130.989 523.204 cm Q /Type /XObject /Meta196 Do /F3 17 0 R >> << /ProcSet[/PDF] Q 0 G /Resources<< << 0.458 0 0 RG /Meta289 303 0 R q /FormType 1 1 g /Meta177 191 0 R /Subtype /Form 1.007 0 0 1.007 130.989 583.429 cm endstream >> 239 0 obj stream /Matrix [1 0 0 1 0 0] 374 0 obj Q /Type /XObject >> 1.007 0 0 1.007 130.989 849.172 cm stream /Resources<< >> /Meta9 Do /F3 12.131 Tf 1 i BT BT /Meta422 438 0 R /Meta393 Do /Type /FontDescriptor 117 0 obj Q /FormType 1 BT Q /BBox [0 0 30.642 16.44] 0.737 w 0.369 Tc stream /Meta7 18 0 R /BBox [0 0 534.67 16.44] >> /ProcSet[/PDF/Text] Answer by Mathtut (3670) ( Show Source ): Q << endobj 245 0 obj q 0.458 0 0 RG /Subtype /Form /ProcSet[/PDF/Text] q /F3 17 0 R /MediaBox [0 0 767.868 993.712] /Meta134 Do 722.699 653.441 l /BBox [0 0 30.642 16.44] stream /Matrix [1 0 0 1 0 0] >> 0.369 Tc 1.007 0 0 1.007 45.168 846.161 cm 0 g Q >> 1 i << 0.458 0 0 RG q [( and )16(a nu)26(mbe)18(r)] TJ /Subtype /Form endobj 0 G << endstream >> /FormType 1 176 0 obj endstream endobj BT << q >> /F4 36 0 R /Type /XObject /Meta410 Do 0 20.154 m q << >> /Length 60 q /Length 118 /Meta277 Do 0.564 G 1.007 0 0 1.007 130.989 523.204 cm /BBox [0 0 88.214 16.44] stream >> 1 i /Length 59 /Matrix [1 0 0 1 0 0] 6.746 5.203 TD 1 g 0.458 0 0 RG q /FormType 1 /Type /XObject The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. Q >> Q 0.486 Tc /BBox [0 0 88.214 16.44] /Meta333 Do /Meta377 391 0 R Q 61 0 obj q /Resources<< S q q /ProcSet[/PDF/Text] 318 0 obj /Subtype /Form /Subtype /Form 1.007 0 0 1.007 271.012 450.181 cm Q q endstream q 1 i 253 0 obj 129 0 obj Thrice a number decreased by 5 is 3x - 5. endobj /Matrix [1 0 0 1 0 0] Q 0.737 w q 35 0 obj 0.564 G >> >> /Matrix [1 0 0 1 0 0] /F1 7 0 R /FormType 1 q q Q << endobj Q q /Flags 32 /Length 16 >> (58) Tj /Meta128 142 0 R /Matrix [1 0 0 1 0 0] /Resources<< >> q /Resources<< >> 0.458 0 0 RG endobj 1.007 0 0 1.007 130.989 523.204 cm /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 779.913 cm

twice a number decreased by 58