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q Q /Meta414 430 0 R Q q Q /Subtype /Form 1 i /ProcSet[/PDF] /BBox [0 0 30.642 16.44] 195 0 obj /FormType 1 /F3 17 0 R stream endobj /FormType 1 101.849 5.203 TD 0 w << 1.007 0 0 1.007 271.012 523.204 cm ET (+) Tj 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 . /Subtype /Form endstream /F3 12.131 Tf /BBox [0 0 88.214 16.44] ET /Meta82 96 0 R Q /Matrix [1 0 0 1 0 0] 0 g /FormType 1 Q q /Type /XObject /Font << /Meta326 Do >> << stream endobj 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. /BBox [0 0 88.214 16.44] endobj /BBox [0 0 673.937 16.44] /BBox [0 0 30.642 16.44] << /Matrix [1 0 0 1 0 0] 1 i >> /Resources<< 278 0 obj /Resources<< /Type /XObject /Resources<< 1.007 0 0 1.007 271.012 776.149 cm 3.742 5.203 TD /Meta66 Do >> 295 0 obj /BBox [0 0 30.642 16.44] Q 0 g stream /FormType 1 303 0 obj 0.486 Tc BT /FormType 1 1.007 0 0 1.007 45.168 713.666 cm 0 G ET q 1 i /Length 69 /Type /XObject Q >> >> /Length 79 BT /Meta322 Do 255 0 obj /Type /XObject /Meta271 Do endobj /Subtype /Form /Matrix [1 0 0 1 0 0] (3) Tj /Meta213 227 0 R >> >> ET 1.005 0 0 1.007 79.798 813.037 cm ET Q /Subtype /Form q Q BT /ProcSet[/PDF/Text] endstream /ProcSet[/PDF/Text] stream Q ET 1.014 0 0 1.007 111.416 776.149 cm /Meta393 409 0 R startxref /ProcSet[/PDF/Text] q (B) Tj BT BT /Resources<< 0.463 Tc /Type /XObject 1.014 0 0 1.006 251.439 763.351 cm stream /Length 65 0.458 0 0 RG 0 G /Length 91 /FormType 1 /BBox [0 0 88.214 16.44] 1 i /F3 12.131 Tf 412 0 obj Q endobj 1.005 0 0 1.007 79.798 813.037 cm << /Matrix [1 0 0 1 0 0] /FormType 1 0.425 Tc << q >> endstream >> /FormType 1 /FormType 1 >> /ProcSet[/PDF/Text] /BBox [0 0 15.59 16.44] 0 w /Matrix [1 0 0 1 0 0] >> 339 0 obj /FormType 1 >> Q q >> /Meta373 387 0 R /FirstChar 43 Q: when six times a number is decreased by 4, the result is 8. ET 0 5.203 TD /Length 59 Q 1 i 1 i >> stream /Meta84 Do 1 i Q << Q >> /Font << /Contents [399 0 R] Q 1.014 0 0 1.006 391.462 836.374 cm stream Q << 0.564 G << Q >> 0 G endstream 0 g /Type /Font q 23 0 obj /F3 12.131 Tf /FormType 1 (-) Tj /F4 36 0 R /Resources<< /Meta368 Do /LastChar 45 endobj /ProcSet[/PDF/Text] 82 0 obj endstream 1.007 0 0 1.007 45.168 779.913 cm 1.005 0 0 1.007 102.382 599.991 cm >> stream >> /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] 0 G 0 g q /BBox [0 0 17.177 16.44] Q 16.469 5.203 TD 0 g 1.014 0 0 1.007 251.439 330.484 cm /F1 7 0 R stream precision and actual right or wrong answers. << 1 i 441 0 obj (A\)) Tj /MissingWidth 250 stream endobj 1 i /Subtype /Form << endobj stream /F1 7 0 R Q 1 i q >> endstream Q /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 BT ET /Type /XObject 22.478 5.336 TD /Subtype /Form /BBox [0 0 639.552 16.44] Q /Meta38 52 0 R q 1 i /ProcSet[/PDF/Text] stream 1.005 0 0 1.006 45.168 879.284 cm Q 1.007 0 0 1.007 551.058 703.126 cm >> q /F3 17 0 R 1 i /Type /XObject /Type /XObject 1.007 0 0 1.007 551.058 383.934 cm /F3 17 0 R Q q /F1 7 0 R ET Q 0.369 Tc >> /F3 17 0 R Q q /F4 36 0 R /FormType 1 /Meta240 Do 1.007 0 0 1.007 411.035 583.429 cm /Meta325 339 0 R endstream 0.564 G /ProcSet[/PDF/Text] endobj q endobj endobj >> 0 g /Subtype /Form >> 1.007 0 0 1.007 654.946 599.991 cm Q Q << /F3 17 0 R q /F1 12.131 Tf Have a nice day! q /Meta409 425 0 R Q q endstream /Meta261 275 0 R 198 0 obj /BBox [0 0 88.214 35.886] /F1 7 0 R /Subtype /Form q 0.51 Tc 1.502 7.841 TD /Meta418 Do /Font << /Matrix [1 0 0 1 0 0] Q endobj q /FontDescriptor 35 0 R /Type /XObject q /Meta202 216 0 R >> (\)) Tj Q /ProcSet[/PDF] /Meta189 Do Q Q q q 1 i /FormType 1 /Resources<< 1 i /Resources<< /Meta265 Do In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). endstream /Length 57 0 g Q 1 g >> /Resources<< q /F3 12.131 Tf q /Type /XObject /BBox [0 0 88.214 16.44] Q /Meta279 Do /Subtype /Form /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 g 0.838 Tc >> [( t)-14(imes a num)-16(ber)] TJ /BBox [0 0 88.214 16.44] Q << 1.007 0 0 1.007 130.989 636.879 cm /Resources<< /Meta248 Do /F2 12.131 Tf /Length 69 Q 0.458 0 0 RG Q 1.007 0 0 1.007 45.168 730.228 cm 1.502 5.203 TD 44 0 obj /F1 12.131 Tf /Matrix [1 0 0 1 0 0] stream 0 5.203 TD q Q /Meta306 320 0 R << 381 0 obj /Resources<< Q /F3 17 0 R (x ) Tj Q << endstream /XHeight 447 /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 277.035 cm q << /Meta367 381 0 R /BBox [0 0 15.59 16.44] 1.005 0 0 1.007 79.798 730.228 cm 344 0 obj /FormType 1 q Q Q /Type /XObject stream 367 0 obj q /Subtype /Form 0 w << /Subtype /Form Q 0.524 Tc q /Meta276 Do /Resources<< 1 i Q q /Length 78 endobj /Matrix [1 0 0 1 0 0] q 6.746 5.203 TD BT >> /Meta249 263 0 R /Resources<< /Meta289 303 0 R 90 0 obj q /F3 12.131 Tf Q /F3 17 0 R << /F3 17 0 R /Meta24 Do endstream /F3 17 0 R /ProcSet[/PDF/Text] Q /Length 245 /F3 17 0 R /Meta74 88 0 R /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 849.172 cm 1 i >> /BBox [0 0 88.214 16.44] /Resources<< Making educational experiences better for everyone. (- 8) Tj << 197 0 obj /Meta64 78 0 R << /Meta111 Do 180 0 obj /Meta295 Do /Length 69 /Subtype /Form endstream /Matrix [1 0 0 1 0 0] 0.737 w >> >> /Type /XObject << BT stream /Length 16 295.086 4.894 TD /BBox [0 0 534.67 16.44] (B\)) Tj 1 i /Ascent 1050 (x) Tj [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ Q ET Q stream >> /Meta344 358 0 R 1 g /Subtype /Form endstream /Resources<< /Type /XObject >> 38.182 5.203 TD q >> 0 g /F3 17 0 R q Q Q /F3 17 0 R /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 0.369 Tc /Meta54 Do q 1 i /Length 69 0.564 G q 0 5.203 TD 62 0 obj 1.007 0 0 1.006 130.989 437.384 cm Q q q Q Q >> /Length 69 /BBox [0 0 534.67 16.44] /Subtype /Form 0 g 1 i stream endobj 0.564 G >> 0.458 0 0 RG /F3 17 0 R /Length 69 Q >> q q 0.68 Tc endstream Q /FormType 1 q 1.007 0 0 1.007 271.012 383.934 cm 0.737 w 0 g /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 1 i q BT q (1\)) Tj /Length 16 << /Meta253 Do /Meta341 Do /FormType 1 endobj >> /Subtype /Form >> stream 0 G /Font << /Type /XObject q /F3 12.131 Tf SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. Q /Meta426 442 0 R Q /Resources<< /FormType 1 << Q >> 0 G >> Q >> q BT 0 5.203 TD /Matrix [1 0 0 1 0 0] 0.737 w /Subtype /Form 276 0 obj /Type /XObject /ProcSet[/PDF] 70 0 obj /Resources<< Q q >> Q /Length 16 Q /FormType 1 >> q 1 g /Length 151 /Ascent 1050 0 G /Meta72 Do /Resources<< /Type /XObject 0 G /Meta256 Do BT /Matrix [1 0 0 1 0 0] /Type /XObject /BBox [0 0 639.552 16.44] /Meta175 189 0 R endstream /FormType 1 q /BBox [0 0 88.214 35.886] /Root 2 0 R >> stream /BBox [0 0 88.214 16.44] New questions in Mathematics 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. 0 4.894 TD stream /BBox [0 0 17.177 16.44] >> Q ET /FormType 1 /Type /XObject 1.007 0 0 1.007 654.946 347.046 cm q /BBox [0 0 88.214 16.44] BT q q /Meta178 192 0 R /Length 65 ET stream /FormType 1 Q /F1 12.131 Tf /Subtype /Form Q q q /Subtype /Form /FormType 1 Q endobj /F4 36 0 R >> >> (2) Tj >> /F3 12.131 Tf 0 G q q endobj /Meta231 245 0 R << /Resources<< q ( x) Tj >> /Font << 0 w /F3 17 0 R /Matrix [1 0 0 1 0 0] >> Q q q Number Outcomes 1 42 2 41 3 . /Length 118 q 0 g /Font << << 4.506 8.18 TD Q /Matrix [1 0 0 1 0 0] /F3 17 0 R /Resources<< /F3 17 0 R /BBox [0 0 88.214 16.44] >> >> >> /Meta392 408 0 R 26.957 5.203 TD 0.458 0 0 RG /Length 81 q << 0 g (40) Tj endstream 0.271 Tc /Matrix [1 0 0 1 0 0] Q /Resources<< /Length 12 /Length 59 /ProcSet[/PDF/Text] /BBox [0 0 30.642 16.44] ET q Q /Subtype /Form q BT endobj /BBox [0 0 534.67 16.44] S 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. endobj /FormType 1 /Type /XObject [(Answe)20(r Key)] TJ endstream /BBox [0 0 88.214 16.44] /Subtype /Form /Subtype /Form stream 1.007 0 0 1.007 654.946 726.464 cm 1.007 0 0 1.007 271.012 330.484 cm /Meta398 414 0 R Q Q /ID [] /Resources<< 0 g 0 g 1 g 0.458 0 0 RG /BBox [0 0 15.59 16.44] ET 1 i Q /FormType 1 /ProcSet[/PDF/Text] stream /FormType 1 /Matrix [1 0 0 1 0 0] q The sum Of twice a nu4ber What is the number? q >> >> BT /Resources<< /F3 17 0 R /Length 69 q endobj /Meta269 Do >> endstream /F3 17 0 R /I0 51 0 R Q Q /Meta168 Do /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 277.035 cm /BBox [0 0 88.214 16.44] q stream stream >> Q 0 G 0 w /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] 0 g /Meta179 Do 30.699 5.203 TD endstream /Meta333 Do /Length 16 q /Meta197 211 0 R 1.005 0 0 1.007 102.382 799.486 cm BT Q >> 0.486 Tc 1 g q /Matrix [1 0 0 1 0 0] >> Twice a number when decreased by 7 gives 45. 1 i 1 i 0 g /Subtype /Form Q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 0 G stream << 0.737 w /Meta274 Do /Length 16 /BBox [0 0 88.214 16.44] /Length 16 BT /Font << /FormType 1 q /F3 12.131 Tf 1.005 0 0 1.007 79.798 746.789 cm /FormType 1 ET /FormType 1 /Meta236 250 0 R >> 1 i /ProcSet[/PDF] (C) Tj << Q q q q /Matrix [1 0 0 1 0 0] /Resources<< 1 i 1.014 0 0 1.007 391.462 330.484 cm /FormType 1 0.737 w q /Meta330 344 0 R >> q endobj >> 0 g /Length 69 /ProcSet[/PDF/Text] stream 0 G 0 G << /Type /XObject BT >> /Type /XObject q /ProcSet[/PDF] 118 0 obj ET endobj 1.014 0 0 1.007 391.462 703.126 cm q 0.241 Tc /Type /XObject /Type /XObject q 0 g /Meta235 Do /Subtype /Form Q /FormType 1 endstream /Type /XObject 349 0 obj ET endstream 1.014 0 0 1.006 111.416 437.384 cm endobj 0.425 Tc endobj /Resources<< /Meta369 Do stream Q /Subtype /Form /Type /XObject 1.014 0 0 1.007 391.462 636.879 cm /F4 36 0 R Q q >> /Meta124 Do << 0.458 0 0 RG q 0 G q Q q q stream endstream 1 i q ET /Matrix [1 0 0 1 0 0] q /Font << 0 G /Meta271 285 0 R endstream /Meta44 Do >> 1 i 13 0 obj Q q /Meta207 221 0 R stream q endobj (C\)) Tj q /Length 16 We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. endstream /F3 17 0 R Q /Length 54 /F4 12.131 Tf /Matrix [1 0 0 1 0 0] 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. << BT /FormType 1 Q /Resources<< 126 0 obj /F3 17 0 R 1.007 0 0 1.007 411.035 330.484 cm endobj q /Type /XObject /F3 12.131 Tf q 159 0 obj /BBox [0 0 88.214 35.886] q q /Font << /FormType 1 q endobj q 0.737 w /Length 59 stream /Matrix [1 0 0 1 0 0] 403 0 obj >> 431 0 obj /Length 245 >> /Meta83 97 0 R q << q endstream /Meta218 232 0 R /Font << ET ET /Type /XObject endobj /Type /XObject 0.458 0 0 RG Q 335 0 obj /Meta27 40 0 R /ProcSet[/PDF] /F3 12.131 Tf /Font << << endstream /FormType 1 (2\)) Tj << ET /Resources<< /Font << /ProcSet[/PDF/Text] q q 1.005 0 0 1.007 102.382 799.486 cm ET q 104 0 obj /ProcSet[/PDF/Text] 1 i >> /Type /XObject q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] endobj ( \() Tj >> /Matrix [1 0 0 1 0 0] ET stream stream /Meta316 330 0 R 0.458 0 0 RG 1 i Thrice of a number = 3x. /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 251.439 450.181 cm stream 0 g stream /Meta228 Do 549.694 0 0 16.469 0 -0.0283 cm /Matrix [1 0 0 1 0 0] /Font << >> 298 0 obj /Length 16 /ProcSet[/PDF/Text] /Resources<< >> (C\)) Tj 1.007 0 0 1.007 551.058 703.126 cm Q 1.005 0 0 1.007 102.382 872.509 cm >> (2) Tj /Matrix [1 0 0 1 0 0] << [(-3)-16(20)] TJ /ProcSet[/PDF] 1.005 0 0 1.007 102.382 546.541 cm q Q Q q /BBox [0 0 88.214 16.44] Q /F3 17 0 R >> /Resources<< 0 G /BBox [0 0 534.67 16.44] /BBox [0 0 639.552 16.44] >> 0 G Q >> /Matrix [1 0 0 1 0 0] q 0 g q This s problem could be, interpreted either way. >> /ProcSet[/PDF] endstream /Subtype /Form Q 0 g /Meta30 43 0 R >> Q 1 i /Length 16 BT Q 1 i q 1 i >> q /Meta200 214 0 R /BBox [0 0 88.214 16.44] /Type /XObject /Subtype /Form /F3 17 0 R endobj /FormType 1 endobj 0 G 0.564 G /Type /XObject q q 0.737 w q /ProcSet[/PDF/Text] stream >> 80 0 obj q /BBox [0 0 15.59 16.44] q /Meta346 360 0 R endobj q Q 1 i /ProcSet[/PDF] << q q /Matrix [1 0 0 1 0 0] 20.21 5.203 TD endobj Q >> /Matrix [1 0 0 1 0 0] << 87 0 obj /FormType 1 /Resources<< 3.742 5.203 TD 192 0 obj (\)]) Tj /Meta9 Do Q /AvgWidth 459 q >> 1.014 0 0 1.007 111.416 277.035 cm << q q endstream /Matrix [1 0 0 1 0 0] Q 1 i /F3 12.131 Tf 0 G 1.007 0 0 1.007 411.035 636.879 cm /FormType 1 q 1.502 24.339 TD q >> /Resources<< /Count 2 0.458 0 0 RG (x) Tj 0.458 0 0 RG 1 i >> /Meta347 Do /ProcSet[/PDF] endobj >> Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. 1.007 0 0 1.007 67.753 872.509 cm /Resources<< endstream Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] endobj 0 w Q 1 0 obj Q >> 0 G 0.738 Tc 0 w 1 i , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. stream /Meta84 98 0 R /BBox [0 0 534.67 16.44] /Length 58 Q q Q >> /F1 7 0 R 0.737 w BT >> [tex]\sin (\pi -x)=\sin x[/tex]. Q /Meta327 Do 0 G 0 g /Meta2 Do q /Resources<< q 0 G /Subtype /TrueType q /Meta288 Do stream /Matrix [1 0 0 1 0 0] Q 0 g /Meta304 Do q >> /Subtype /Form (C\)) Tj /Length 67 1.014 0 0 1.007 391.462 277.035 cm 0 g 343 0 obj [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. endobj /Subtype /Form /ProcSet[/PDF] q Q >> 1.502 5.203 TD Q q A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. 0.68 Tc >> /Subtype /Form 1.014 0 0 1.007 111.416 383.934 cm /Subtype /Form 346 0 obj ET 251 0 obj 58 decreased by twice Gails age. /BBox [0 0 17.177 16.44] q endobj 125.064 4.894 TD Q endstream stream /ProcSet[/PDF/Text] 1 i /I0 Do 0.425 Tc q q /Subtype /Form << q 220.931 4.894 TD q /Resources<< /Matrix [1 0 0 1 0 0] /Meta270 284 0 R >> [tex]\sin (\pi -x)=\sin x[/tex]. /BBox [0 0 639.552 16.44] Q /FormType 1 1.005 0 0 1.007 45.168 889.071 cm /Subtype /Form >> q 1 g /Meta101 Do endstream Q q /F3 12.131 Tf endstream Q >> endobj q stream BT 0 G 1.014 0 0 1.007 111.416 330.484 cm >> 1.007 0 0 1.007 67.753 599.991 cm /Meta82 Do >> 1.007 0 0 1.006 551.058 836.374 cm q 0 w 0.737 w /Length 12 >> endstream /Subtype /Form q -0.029 Tw >> /BBox [0 0 639.552 16.44] /Type /XObject q 0 G stream /BBox [0 0 88.214 16.44] /Subtype /Form Q >> 1.007 0 0 1.007 271.012 450.181 cm Q 63 0 obj /Meta209 223 0 R /Meta229 Do stream 209 0 obj << >> Q 10 0 obj /Resources<< /Length 59 /BBox [0 0 639.552 16.44] /Type /XObject If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. Q /FormType 1 Q >> /Meta88 102 0 R /Subtype /Form /Subtype /Form /Resources<< >> /Type /XObject /Font << Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . /FormType 1 /Matrix [1 0 0 1 0 0] /F3 12.131 Tf (x) Tj /Font << /FormType 1 0 G /Resources<< q /Matrix [1 0 0 1 0 0] 0.564 G Q q endobj q /ProcSet[/PDF] ET 1.007 0 0 1.007 551.058 277.035 cm /FormType 1 /FormType 1 Q q endobj Q 1 g 178 0 obj /Length 69 Q >> Q q /Type /XObject >> q /Font << 1 i /Subtype /Form /Length 54 << << endstream /Type /XObject q endobj /Meta23 Do endstream endobj ET /Matrix [1 0 0 1 0 0] Q Q /Type /XObject /BBox [0 0 88.214 16.44] /Meta50 64 0 R 0 g endobj 0.458 0 0 RG >> ET << endobj /Resources<< endstream ] 48 0 obj /Length 69 0.425 Tc /FormType 1 /Font << /Type /XObject >> /Type /XObject /ProcSet[/PDF/Text] /F3 17 0 R >> Q Q /Matrix [1 0 0 1 0 0] stream /Meta126 140 0 R ET (x ) Tj /F3 12.131 Tf /F3 17 0 R stream Q Q: A number increased by 5 is equivalent to twice the same number decreased by 7. q << /Subtype /Form >> /FormType 1 Q /Resources<< 1.005 0 0 1.007 102.382 310.158 cm BT /Matrix [1 0 0 1 0 0] Q endstream /Matrix [1 0 0 1 0 0] stream /Meta115 Do q /FormType 1 /Meta55 69 0 R /Meta183 Do Q /BBox [0 0 549.552 16.44] endobj q >> stream q >> 340 0 obj q /Subtype /Form q /Subtype /Form /Length 16 endstream /Meta215 Do BT /Matrix [1 0 0 1 0 0] << endobj Q /ProcSet[/PDF] /Meta227 Do /Font << endstream /Length 16 /FormType 1 << Q Q endobj q Q q /Length 16 1 i >> 1.007 0 0 1.007 411.035 636.879 cm /BBox [0 0 639.552 16.44] 1.014 0 0 1.007 251.439 383.934 cm 0.737 w Q /FormType 1 /Meta5 14 0 R Q /Meta74 Do /F3 12.131 Tf Three times a number equals fifteen 3. endobj endstream endobj ET stream ET q endobj Q 3.742 24.649 TD /Meta397 Do q Q endstream /Resources<< %%EOF. Q /Length 16 246 0 obj >> 1 i 1 g /ProcSet[/PDF/Text] >> >> /BBox [0 0 673.937 68.796] 1.007 0 0 1.007 130.989 523.204 cm /BBox [0 0 88.214 16.44] /Resources<< 0 G /Meta308 Do 1 i 0 G /Ascent 891 endobj 1 g 1 g /Matrix [1 0 0 1 0 0] << twice a number decreased by 58 13 - 3x B. twice a number a divided by three = 2a / 3. five times a number x minus four = 5x - 4. thrice the sum of a number x and six = 3 (x + 6) Add your answer and earn points. 47.933 5.203 TD /FormType 1 /Resources<< Q Q >> 0 g /F3 12.131 Tf Q q stream q >> /Font << >> /F3 17 0 R /I0 51 0 R stream Q /F1 7 0 R /BBox [0 0 15.59 16.44] endstream /ProcSet[/PDF/Text] 0.227 Tc 0.564 G /Type /XObject Q /Length 69 /BBox [0 0 88.214 16.44] q 1.014 0 0 1.007 531.485 583.429 cm /F3 17 0 R /Type /XObject /BBox [0 0 88.214 16.44] 0.737 w /Subtype /Form << /ProcSet[/PDF/Text] 107 0 obj Q /BBox [0 0 88.214 16.44] BT /Matrix [1 0 0 1 0 0] /Font << << /F3 17 0 R /F3 17 0 R /Meta161 Do ET BT q stream /Type /XObject /Font << /Length 68 >> >> q /Meta75 Do /Subtype /Form 0 g Q 1 g << Q /FormType 1 >> 157 0 obj q Q /Font << /Resources<< BT BT q -0.463 Tw endstream /BBox [0 0 534.67 16.44] /BBox [0 0 17.177 16.44] /BBox [0 0 88.214 16.44] /Subtype /Form /ProcSet[/PDF/Text] 1.007 0 0 1.007 45.168 763.351 cm 322 0 obj q 0 G endobj (-) Tj 0.458 0 0 RG /FormType 1 /ProcSet[/PDF] 377 0 obj ET /Matrix [1 0 0 1 0 0] (3\)) Tj /Meta378 Do Q q 1.014 0 0 1.007 251.439 277.035 cm endobj endstream /Subtype /Form q /Subtype /Form >> Q /F3 17 0 R Select the correct mathematical statement for the following equation. /F4 12.131 Tf MetS-Z quartiles and their associated risks are presented in Fig. 0 g 1.014 0 0 1.007 391.462 523.204 cm (\)) Tj endstream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] q 1.007 0 0 1.007 130.989 383.934 cm 1.007 0 0 1.007 271.012 636.879 cm /Meta1 Do q q /Meta255 269 0 R >> /Meta245 259 0 R /Subtype /Form 0 g q Q 0 G 286 0 obj /Meta247 Do q << >> /FormType 1 /FormType 1 ET /F3 17 0 R << endobj /Meta0 5 0 R stream Q BT endobj 0.564 G endobj Q 1 i /FormType 1 q /Matrix [1 0 0 1 0 0] /Type /XObject /Matrix [1 0 0 1 0 0] q 0 w >> 0 g /Resources<< 1 i /Subtype /Form Q /FormType 1 373 0 obj /FormType 1 /Type /XObject q /Meta309 Do q q 0.564 G /BBox [0 0 30.642 16.44] q /Meta197 Do ET q endstream q 0 g 0.425 Tc 1. , Prove the following /Subtype /Form /ProcSet[/PDF/Text] Q 0.838 Tc /ProcSet[/PDF/Text] /FormType 1 /Type /XObject 1 g /F3 12.131 Tf Q /Subtype /Form q /Matrix [1 0 0 1 0 0] q 1 i 289 0 obj /F3 17 0 R /Font << Q endstream 0 g endstream /ProcSet[/PDF] >> /Subtype /Form /Type /XObject >> >> 0 G /Matrix [1 0 0 1 0 0] q Q 0.737 w q q 0 g /Resources<< /Matrix [1 0 0 1 0 0] /Length 59 /ProcSet[/PDF/Text] /Subtype /Form endobj >> endobj /Subtype /Form q /Matrix [1 0 0 1 0 0] 0 G (3\)) Tj endobj endstream q 0 g /BBox [0 0 639.552 16.44] >> /FormType 1 /ProcSet[/PDF] >> >> stream ( \() Tj (viii) A number divided by 8 gives 7. /Resources<< ET 20.21 5.203 TD (x) Tj stream q /BBox [0 0 15.59 29.168] 0 g /Meta102 116 0 R >> ET >> Next, the problem says that "x" would be equal to twice a number added by 5. /Resources<< 0 g /Meta273 287 0 R q /Meta282 296 0 R /Type /XObject 386 0 obj >> 0 G >> q /Meta233 247 0 R 0 g >> /Matrix [1 0 0 1 0 0] /F1 12.131 Tf Q /Type /XObject ET BT 1 i /Matrix [1 0 0 1 0 0] Answer by Mathtut (3670) ( Show Source ): /Meta142 Do /Resources<< 0 G /Font << S Q /Subtype /Form 14.966 20.154 l >> Q 1.007 0 0 1.007 551.058 330.484 cm BT /Matrix [1 0 0 1 0 0] /FormType 1 >> (-20) Tj /Subtype /Form 0.838 Tc /Resources<< Q -0.463 Tw /Length 12 /Meta342 Do q Q BT /Resources<< >> >> Q Q 0 g 0 g >> q endobj Q 178.979 5.203 TD 1 i 1.014 0 0 1.007 391.462 583.429 cm 3 0 obj 1.007 0 0 1.006 411.035 437.384 cm BT (C\)) Tj /Type /XObject q endstream /Length 59 endobj (-) Tj 1 i Q endobj 0 g Q 0 G stream /Meta317 Do Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. q /Resources<< /Type /XObject << q q /Font << Q /Length 118 Q /Meta128 142 0 R endobj >> >> 30.699 4.894 TD /Resources<< 47 0 obj /Resources<< /FormType 1 /Meta421 437 0 R /Type /XObject BT 1 i Q /Meta349 Do /Matrix [1 0 0 1 0 0] /Meta195 Do q BT stream Q Q >> >> /F3 12.131 Tf 0.458 0 0 RG 1 i 120 0 obj /Length 16 306 0 obj /FormType 1 Q /Matrix [1 0 0 1 0 0] q >> /Type /XObject stream /BBox [0 0 534.67 16.44] /Resources<< Q /Resources<< /BBox [0 0 88.214 35.886] 0.737 w /Length 69 BT q 357 0 obj q /Resources<< << q endstream q Q >> endobj /Meta354 368 0 R 0 G (-23) Tj /Meta33 Do 1.007 0 0 1.007 551.058 277.035 cm 239 0 obj endobj /Length 79 >> q /F3 17 0 R q >> stream /Length 69 1.005 0 0 1.007 79.798 796.475 cm /BBox [0 0 15.59 16.44] /FontDescriptor 6 0 R Q 0 g 0 G Q 0 g >> 0.737 w TJ /Matrix [1 0 0 1 0 0] 0.738 Tc Q Find the number. /F3 17 0 R /F3 17 0 R >> 0 g 416 0 obj /Matrix [1 0 0 1 0 0] /Type /XObject /FormType 1 endobj /BBox [0 0 15.59 16.44] /FormType 1 endobj /Font << 16.469 5.336 TD 0 G 0 w 1.014 0 0 1.007 251.439 583.429 cm 1 i /Type /XObject Q q /BBox [0 0 88.214 16.44] endobj /Type /XObject 1.007 0 0 1.007 130.989 277.035 cm q 213 0 obj endobj 52 0 obj endobj /Font << q /F3 12.131 Tf Q /Matrix [1 0 0 1 0 0] stream Q /FormType 1 Q 1 i ET /Length 54 endobj 0 g 0.564 G /FormType 1 >> 0.297 Tc q /Meta93 107 0 R Q VIDEO ANSWER: in this problem were asked to solve giving, given the following information. endobj /Type /FontDescriptor >> 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 /Length 16 >> /Font << The width Of a rectangle is 15 cm and the perimeter is 12 cm. q 2 0 obj Q 0 G /Type /XObject q 1.005 0 0 1.007 102.382 599.991 cm /Length 68 /Length 69 Q /Meta331 Do 0 G /F3 17 0 R /ProcSet[/PDF] >> Q /Font << /F4 36 0 R /Length 54 /Type /XObject 1 i Q /F3 17 0 R /Type /XObject BT /ProcSet[/PDF/Text] Decreased by another number means subtract. Q 0 39.216 TD q q 0 g 1 i /Subtype /Form q ET /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 67.753 473.519 cm 0 G /Font << stream 0.458 0 0 RG 0.737 w >> 1 g q q q q /Type /XObject 1 g q /StemV 88 q >> endobj ( \() Tj /Font << << 430 0 obj /Meta326 340 0 R /Length 63 >> endobj /ProcSet[/PDF/Text] 1 i << /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] >> BT Q << Q 0 G /Meta109 123 0 R Q /F3 17 0 R endstream 0.458 0 0 RG /Type /XObject >> stream [(-1)-16(52)] TJ 168 0 obj /FormType 1 q /Resources<< /Length 68 /I0 51 0 R /Resources<< /Resources<< /Meta400 416 0 R >> /F3 17 0 R /Length 59 0.564 G Twice a number, decreased by 58 is less than 112 - 18274082. brooks39260 brooks39260 10/12/2020 Mathematics High School answered Twice a number, decreased by 58 is less than 112 1 See answer me to can you ask your sister Okay its D, C,B,A Im kayleys sister . /FormType 1 q q /Resources<< q /BBox [0 0 88.214 16.44] /Resources<< /Meta405 421 0 R Q q /Meta337 Do 0 G 1.007 0 0 1.007 67.753 473.519 cm /Matrix [1 0 0 1 0 0] endobj /Subtype /Form 235 0 obj /Meta126 Do q 0.458 0 0 RG /Resources<< 1.007 0 0 1.007 45.168 796.475 cm 1.014 0 0 1.006 531.485 437.384 cm /FormType 1 >> q 0 w << Q Q >> /Length 16 Q BT >> 160 0 obj /F3 12.131 Tf /Length 67 /Font << q /ProcSet[/PDF] Q /Length 69 q /Type /XObject << >> stream 1 g (38) Tj endobj >> /Font << >> /Font << q 0 5.203 TD q /BBox [0 0 15.59 16.44] /Resources<< 0 G /Subtype /Form /ProcSet[/PDF/Text] q q /Length 64 /Meta342 356 0 R /F3 17 0 R /F3 12.131 Tf >> /ProcSet[/PDF/Text]
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twice a number decreased by 58
is complex
a process
a decision
everything
is complex