/Length 59 q /Meta385 401 0 R stream /BBox [0 0 30.642 16.44] (+) Tj /Meta168 Do endobj /ProcSet[/PDF] 22.478 4.894 TD >> /BBox [0 0 30.642 16.44] q /Matrix [1 0 0 1 0 0] /Meta228 Do /F1 7 0 R /FormType 1 stream 0.458 0 0 RG 0 G q /ProcSet[/PDF/Text] 0 w stream /Subtype /Form /Subtype /Form << 1 i /Type /XObject ET << 722.699 347.046 l /Resources<< >> (A\)) Tj /Meta194 Do /Resources<< stream q /Subtype /Form /Type /XObject >> decreased by A number decreased by twelve X - 12 subtracted from Six subtracted from a number X - 6 Multiplication ( x ) times Eight times a number 8x the product of The product of fourteen and a number 14x twice; double Twice a number; double a number 2x multiplied by A number multiplied by negative six 6x endstream >> /BBox [0 0 15.59 16.44] q endobj 1.007 0 0 1.007 271.012 330.484 cm /Meta197 Do 1.014 0 0 1.006 391.462 510.406 cm /Meta135 Do endobj 1.005 0 0 1.007 45.168 889.071 cm -0.486 Tw /ProcSet[/PDF/Text] /F1 7 0 R /F3 17 0 R (5\)) Tj stream /Length 69 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/Meta177 191 0 R /F4 12.131 Tf /Length 69 0.17 Tc 1.007 0 0 1.007 45.168 846.161 cm << stream BT Q >> Q Q stream /F1 12.131 Tf /Subtype /Form endstream Q BT 0 G stream >> /FormType 1 1 i Q 380 0 obj << q stream 0.458 0 0 RG Q /Font << Q >> 1 i /Meta155 Do /Resources<< /Subtype /Form /Meta115 Do /Font << endstream Q q 1 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Meta411 Do q /FormType 1 /F4 12.131 Tf /BBox [0 0 17.177 16.44] endobj 1 i 1 g /Meta371 385 0 R >> >> q 193 0 obj q 375 0 obj /BBox [0 0 673.937 15.562] << /Subtype /Form >> >> >> >> Q endobj /F3 17 0 R /Type /XObject /Meta185 Do Q /Type /XObject q /Subtype /Form /FormType 1 /ProcSet[/PDF] 1.007 0 0 1.006 130.989 690.329 cm endstream precision and actual right or wrong answers. /Length 118 1 i q Q endstream /Type /XObject Q /Meta281 295 0 R stream /Meta90 104 0 R /Subtype /Form 0 g /Resources<< 1.005 0 0 1.007 79.798 779.913 cm >> /Font << << /Subtype /Form /Subtype /Form 0 G (A\)) Tj 207 0 obj q /FormType 1 q /Font << endobj 0 G /Meta192 206 0 R 0 g stream ET endstream Q 1.014 0 0 1.007 531.485 277.035 cm /Subtype /Form /F3 12.131 Tf q q 326 0 obj 0.738 Tc q /Subtype /Form 1 i 1 i /Matrix [1 0 0 1 0 0] Q /Font << stream /FormType 1 BT 1.007 0 0 1.007 411.035 849.172 cm /Type /XObject 16.469 5.203 TD 236 0 obj << /Length 127 1.014 0 0 1.007 251.439 583.429 cm q endobj q >> Q 0 g q Q 0 G Q BT 0.486 Tc /Type /XObject 1.007 0 0 1.007 130.989 849.172 cm 284 0 obj 373 0 obj /Resources<< 370 0 obj /Type /XObject q q Q 1 g Q /BBox [0 0 534.67 16.44] stream << >> /Type /XObject stream /ProcSet[/PDF] only about 58% of candidates will agree to be screened. /Matrix [1 0 0 1 0 0] /Resources<< /F3 12.131 Tf endstream /Resources<< /BBox [0 0 639.552 16.44] /Meta152 Do /Subtype /Form 1. /Type /XObject q >> /FormType 1 Q q BT /FormType 1 >> 434 0 obj 0.458 0 0 RG 164 0 obj /Length 64 Q 1.005 0 0 1.007 102.382 872.509 cm q /Meta264 Do q /ProcSet[/PDF] 611 556 611 611 389 444 333 611 556 833 500]>> Thrice of a number = 3x. 0 G /Subtype /Form >> >> /ProcSet[/PDF/Text] stream q >> /Descent -216 endstream /BBox [0 0 17.177 16.44] 3 0 obj stream << 185.725 5.203 TD /Meta146 Do >> 0 G 1.005 0 0 1.007 79.798 763.351 cm /Resources<< 191 0 obj 395 0 obj /Meta270 Do Twice a number decreased by 58! Q q 0 g << (40) Tj /FontName /TestGen-Regular 203 0 obj endobj 0 G /Type /XObject q q ET /Matrix [1 0 0 1 0 0] 0 g 1 i >> 1.007 0 0 1.007 130.989 583.429 cm endobj /Meta367 Do /Font << /AvgWidth 459 endobj 0 G Q BT /Length 58 << 0 w 1 g Q q 394 0 obj /Type /XObject -0.008 Tw 0 G /BBox [0 0 88.214 16.44] 0 G /Resources<< 34 0 obj 0 G /Meta208 Do Q /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF] stream /Type /XObject q /Meta209 223 0 R 500 500 500 0 333 389 278 0 0 722 500 500]>> /FormType 1 endobj /Length 59 /Subtype /Form Q << /Subtype /Form >> Q /FormType 1 2.238 5.203 TD q q Q endstream q stream 1 i 427 0 obj q 0 g ET /Matrix [1 0 0 1 0 0] q 0.369 Tc Q /Type /XObject endstream q q Q q >> -0.058 Tw Q 0 G q /Meta71 Do /Matrix [1 0 0 1 0 0] 0.564 G /Type /XObject 411 0 obj Twice a number decreased by another number: "twice a number decreased by another number" 2*(x-y), "twice a number, decreased by another number"(2*x)-y, It's possible David. endstream /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. >> 0 G 1.014 0 0 1.007 391.462 277.035 cm 0 g 0 g /Meta50 Do /Length 16 0 w /Meta52 Do A number divided by six is eight: (k / 6) = 8. endstream /Length 16 /Meta281 Do BT /F3 17 0 R BT /Meta44 Do stream Q Q 20.975 5.336 TD /Meta88 102 0 R ET Q << /Length 12 1 i 389 0 obj endobj endobj 0 G q /FormType 1 endstream ET BT Explanation: let the number be n. then we can express division in 2 ways. /ProcSet[/PDF/Text] ET 0 G 412 0 obj 0.458 0 0 RG 1.007 0 0 1.006 411.035 437.384 cm q endobj /BBox [0 0 88.214 16.44] 0 G /Subtype /Form /FormType 1 1 i /Subtype /Form 1.007 0 0 1.007 130.989 583.429 cm /ProcSet[/PDF] >> [(1)-25(0\))] TJ stream /Resources<< 0 G /Flags 32 0 g endobj /Meta376 Do /F1 12.131 Tf stream 1 i /BBox [0 0 17.177 16.44] endobj /Resources<< /FormType 1 >> /F3 17 0 R Q endstream >> /Subtype /Form >> BT Q endstream 1.014 0 0 1.006 391.462 836.374 cm Q stream 362 0 obj Q endobj Q stream /BBox [0 0 88.214 35.886] Q /BBox [0 0 88.214 16.44] /MissingWidth 252 /F3 12.131 Tf 1 g 1 g stream Q ET 9.723 5.336 TD /F3 17 0 R BT Q /Resources<< q /ProcSet[/PDF/Text] /Length 69 /Type /XObject /Meta395 411 0 R endobj Q /Type /XObject /Meta388 404 0 R /Matrix [1 0 0 1 0 0] ET /ProcSet[/PDF/Text] 0 G /Meta42 Do /Resources<< This site is using cookies under cookie policy . /FormType 1 /F3 17 0 R /F3 17 0 R 146 0 obj 0.463 Tc 1 i BT q 153 0 obj BT (-11) Tj /Type /XObject Q >> stream /ProcSet[/PDF/Text] /Meta143 157 0 R 0 G /F3 12.131 Tf Q /Type /XObject 0 G /BBox [0 0 88.214 16.44] 1 i q /F3 17 0 R q /Resources<< q << /Subtype /Form /Flags 32 /FormType 1 q << endobj 1.007 0 0 1.006 411.035 690.329 cm Q endobj /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 411.035 583.429 cm /FormType 1 >> 0.564 G Q 1.007 0 0 1.007 411.035 636.879 cm 1 g /Resources<< >> , Prove the following >> /Subtype /TrueType (-11) Tj endobj /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 35.886] endobj /Meta371 Do Q /Resources<< /Meta239 253 0 R /Meta8 Do /Font << /Subtype /Form 1.005 0 0 1.007 102.382 670.003 cm /ProcSet[/PDF] BT /Resources<< endstream /Type /XObject 0 G 0 g endobj /Length 69 /Length 79 /Subtype /Form /ProcSet[/PDF/Text] /Font << >> /Resources<< >> >> So let's go ahead and identify a v /Subtype /Form Q q endobj 0 g 6.746 5.203 TD endstream 0 g Q /Resources<< stream 1 g >> /Meta373 387 0 R /Type /XObject /FormType 1 stream This s problem could be, interpreted either way. Q 40 0 obj q /FormType 1 /Type /XObject q 0 w q /Type /XObject q 0 w Q /Subtype /Form /F3 12.131 Tf << q 0.458 0 0 RG 1 i /Subtype /Form ET /Resources<< (B\)) Tj ET /FormType 1 Q /Meta12 23 0 R << /Subtype /Form q /Font << S q Q q endobj /F3 12.131 Tf 0 G 350 0 obj stream 1 i /F3 17 0 R 106 0 obj >> << /Length 16 >> ET /F3 12.131 Tf stream /F3 12.131 Tf /Font << /Meta121 Do q q 1.007 0 0 1.007 130.989 849.172 cm Q 6.746 24.649 TD (\)) Tj >> Most questions answered within 4 hours. Q stream >> /F3 12.131 Tf 1.005 0 0 1.007 102.382 616.553 cm /Resources<< ET Q Q << Q endobj /Subtype /Form ET >> >> Q Q 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . q /Type /XObject q /Type /XObject /Length 73 /FormType 1 0 G Q BT endstream /F3 17 0 R /F3 12.131 Tf >> 0.458 0 0 RG 0.564 G (13) Tj >> 1 i 0 G endobj /Subtype /Form q Q 353 0 obj 351 0 obj /Resources<< 0.458 0 0 RG /ProcSet[/PDF/Text] 0 G 1 i /F3 12.131 Tf q endstream endobj /Font << /Type /XObject endstream >> q << Q 0 g 1.014 0 0 1.007 111.416 583.429 cm (3) Tj Q 1.007 0 0 1.007 271.012 849.172 cm S q /Matrix [1 0 0 1 0 0] 128 0 obj /ProcSet[/PDF] endobj /F3 12.131 Tf ET /Type /XObject q /Type /XObject a and b or something else.***. endobj 1 i /Resources<< 1 i Q 0 w >> >> 0.458 0 0 RG /Meta347 361 0 R q << /Font << /BBox [0 0 15.59 16.44] endstream >> /Type /XObject endobj /FormType 1 0 g /BBox [0 0 88.214 16.44] /Meta248 262 0 R 0 G New questions in Mathematics /ProcSet[/PDF/Text] /ProcSet[/PDF] Q stream /Meta415 431 0 R 92 0 obj /Meta16 Do /ProcSet[/PDF/Text] /Meta425 Do stream 0 g algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. 0 G BT >> 1 g 0.458 0 0 RG Percent Change = (Decrease First Value) x 100% 0 G Q /Meta37 Do q /Meta32 Do Q /Matrix [1 0 0 1 0 0] 1 i 0.737 w ( \() Tj Q /Type /XObject /Subtype /Form 0 4.894 TD endobj /FontDescriptor 35 0 R 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. /Resources<< Q /Meta308 Do /Type /XObject 0 G Q >> q q Q endobj 20.21 5.203 TD /BBox [0 0 15.59 16.44] /F4 12.131 Tf endstream >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 270 0 obj q /FormType 1 0 g 1 i /Meta35 Do Q /Meta25 Do endstream /BBox [0 0 30.642 16.44] q q 0 g 1 i /Matrix [1 0 0 1 0 0] 0 G 24 0 obj 0 G Q 1.005 0 0 1.007 102.382 653.441 cm endstream /Type /XObject q Q >> /Subtype /Form /Subtype /Form Q q /StemV 94 /Meta198 Do VIDEO ANSWER: in this problem were asked to solve giving, given the following information. 0 g q 1.007 0 0 1.007 551.058 703.126 cm /FormType 1 /Length 54 /FormType 1 /Meta226 Do /F3 12.131 Tf /FormType 1 /Matrix [1 0 0 1 0 0] 0.738 Tc q /Meta272 286 0 R >> /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] q /LastChar 121 endstream 1.007 0 0 1.007 130.989 523.204 cm BT /Meta429 Do Q >> /Length 69 Q Q q ( \() Tj >> 1.007 0 0 1.007 551.058 383.934 cm ET Q >> 1.008 0 0 1.007 654.946 293.596 cm /FontBBox [-90 -216 1195 800] 0 g ET 0.458 0 0 RG /Type /XObject >> 1.014 0 0 1.006 531.485 836.374 cm Q 0 G >> Q >> 2x + 3 y equal to 13 and xy = 4 find the value of x cube + 27 y cube, x/3 2/x = 1/2please solve this question. /ProcSet[/PDF] q /FormType 1 << /BBox [0 0 88.214 16.44] 0.458 0 0 RG q /ProcSet[/PDF] 1.014 0 0 1.007 251.439 277.035 cm q /Meta20 Do Q /F3 17 0 R 339 0 obj 283 0 obj << endobj /Meta385 Do Q /BBox [0 0 549.552 16.44] Get link; Facebook; Twitter; 1 i BT 0 g >> stream LAIing for a pizza and, soft drink. /Font << /BBox [0 0 88.214 16.44] 1 i /Meta236 250 0 R /Meta264 278 0 R 0.68 Tc stream /F1 12.131 Tf 142 0 obj << 1 i q << BT Q 1.007 0 0 1.006 130.989 437.384 cm 0 G 1 i endstream endstream Q >> Q Q Q 0 5.203 TD >> q 296 0 obj /Subtype /Form Q endobj /Meta35 48 0 R 1.007 0 0 1.007 551.058 523.204 cm q /BBox [0 0 15.59 16.44] 246 0 obj 0 g Q << Q q 1.007 0 0 1.007 67.753 653.441 cm -0.021 Tw Q /FormType 1 /Resources<< /Length 118 /F3 17 0 R << endobj 0.737 w /FormType 1 281 0 obj /Meta361 Do /Font << Q q /FormType 1 /Meta87 101 0 R 0.564 G 1 i /FormType 1 105 0 obj << Q >> Q 0 G >> 0.564 G 3.742 5.203 TD endobj endobj q >> ET 1 i q /Length 59 /Length 59 /BBox [0 0 534.67 16.44] q /ProcSet[/PDF/Text] q /F3 12.131 Tf Q /FormType 1 /Matrix [1 0 0 1 0 0] 0 G This site is using cookies under cookie policy . 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.056 Tw /F3 12.131 Tf q 421 0 obj Q q 0 g 0.564 G /BBox [0 0 88.214 16.44] 1.502 5.203 TD endobj /BBox [0 0 88.214 16.44] BT >> q /Subtype /Form endobj /Font << Q (\)) Tj /F3 17 0 R /ProcSet[/PDF] stream Q >> /FormType 1 In the problem above, x is a variable. /Resources<< stream >> 165 0 obj Q (x) Tj ET >> /F3 17 0 R 210 0 obj stream /Resources<< stream 0 g /Resources<< 229 0 obj stream endstream q 1 i /Resources<< >> << /Type /XObject << << endobj (40) Tj q q /FormType 1 /Flags 32 /Type /XObject >> 1.007 0 0 1.007 271.012 450.181 cm 1 i >> Answer provided by our tutors. /Resources<< 0 w 0 5.203 TD /Resources<< 0 g /F4 12.131 Tf 0 g stream 1 i 78 0 obj ET Q q ET 144 0 obj /FormType 1 Q /Subtype /Form 1.007 0 0 1.007 45.168 813.037 cm stream /ProcSet[/PDF] /Meta359 Do /Font << 0 g /Type /XObject Q /F3 12.131 Tf 188 0 obj 1 g /Resources<< Q /F3 12.131 Tf /Resources<< /Font << /ProcSet[/PDF/Text] Q /Type /XObject /FormType 1 >> Q 0 G >> /F3 17 0 R >> >> /ProcSet[/PDF] endstream Q Q >> /FormType 1 Q /ProcSet[/PDF] >> << endobj /Resources<< stream /Meta187 Do endobj >> >> q q << (7\)) Tj /BBox [0 0 30.642 16.44] stream /Subtype /Form /Resources<< S /Type /XObject >> endobj /Subtype /Form 722.699 799.486 l endobj Q 0.737 w /Subtype /Form q /Matrix [1 0 0 1 0 0] /FormType 1 /ProcSet[/PDF] BT q q /F3 12.131 Tf << /Type /XObject Q BT 253 0 obj stream /Font << Q q q . /F3 12.131 Tf /Type /XObject q /Meta16 27 0 R Q Q 1 i >> ( x) Tj /Meta121 135 0 R endstream /Type /XObject << /Resources<< /ProcSet[/PDF] 1.007 0 0 1.007 551.058 523.204 cm stream 0 G /Type /XObject /Matrix [1 0 0 1 0 0] Q q Q /Meta218 Do Q >> endobj /Font << ET >> /Length 12 /BBox [0 0 88.214 16.44] ET >> /FormType 1 q /Subtype /Form Q Q /Subtype /Form 0 5.203 TD >> ET q /Font << /I0 51 0 R /Length 59 1 i /Type /XObject 73 0 obj ET stream 0 G /F3 17 0 R /Type /XObject 0 G 420 0 obj >> endobj 0.564 G /Matrix [1 0 0 1 0 0] 0 w (C\)) Tj 0 g /Length 189 (A\)) Tj 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. 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 . /Matrix [1 0 0 1 0 0] /Resources<< 0.458 0 0 RG /FormType 1 0 G 249 0 obj /F1 7 0 R /Type /XObject 0.458 0 0 RG -y. /BBox [0 0 15.59 16.44] stream Q 1.007 0 0 1.007 45.168 779.913 cm >> /Subtype /Form 0 g 1 i 0 G q stream >> /Meta56 70 0 R q /BBox [0 0 30.642 16.44] 81 0 obj << 20/n b.) endobj 1 g 0 G /ProcSet[/PDF/Text] q 104 0 obj /Type /XObject ET stream 0 w Q /Length 65 /Resources<< << >> Q stream Twice a number decreased by 8 gives 58. /Type /XObject >> /Type /XObject Q endobj q /F1 7 0 R 0 w /StemV 94 0 G /Type /XObject endobj 0 G q /I0 51 0 R /FormType 1 (C\)) Tj /Type /XObject /Resources<< 1 i 162 0 obj /Font << 0.369 Tc /ProcSet[/PDF/Text] /F4 36 0 R /Meta376 390 0 R /Subtype /Form q /Meta38 Do 0.737 w 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. endstream ET Q Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. Q q Q q ET Q /Meta345 359 0 R /ProcSet[/PDF] 300 0 obj endobj BT 1.005 0 0 1.007 102.382 347.046 cm /F3 12.131 Tf /F3 17 0 R q /Meta384 398 0 R 0 g endobj /Meta218 232 0 R /Meta233 247 0 R 0 g A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. q q /Length 16 stream q 1.005 0 0 1.006 45.168 879.284 cm q Q 0 G SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. q q /Subtype /Form /Matrix [1 0 0 1 0 0] /Subtype /Form << /Length 12 672.261 347.046 m Q /Matrix [1 0 0 1 0 0] Q /Meta326 Do 174 0 obj ET 0 5.203 TD 0.564 G Five times the sum of a number and four 7. /F3 12.131 Tf /Subtype /Form 1.007 0 0 1.007 551.058 277.035 cm 1.007 0 0 1.007 130.989 330.484 cm /Matrix [1 0 0 1 0 0] /Type /XObject 297 0 obj /BBox [0 0 534.67 16.44] << /Resources<< q endstream >> q 1 i /F3 12.131 Tf Q endobj /F4 12.131 Tf 0 w /Meta339 353 0 R endstream /Meta150 164 0 R 60 0 obj /BBox [0 0 15.59 16.44] ET q /Matrix [1 0 0 1 0 0] /F3 17 0 R /ProcSet[/PDF/Text] 0 G /Subtype /Form Q /BBox [0 0 88.214 16.44] >> >> /BBox [0 0 88.214 35.886] 0 g 94 0 obj 1.007 0 0 1.007 67.753 872.509 cm /BBox [0 0 534.67 16.44] q q >> /Matrix [1 0 0 1 0 0] /F3 12.131 Tf endobj (B\)) Tj Q /Length 16 Q q ET /Type /XObject 1.007 0 0 1.007 67.753 726.464 cm BT endobj /Length 65 >> >> 20.21 5.203 TD /FormType 1 /FormType 1 /Length 63 endstream 722.699 726.464 l 1.007 0 0 1.007 130.989 277.035 cm >> >> >> /Subtype /Form 0 g /Subtype /Form /Meta348 362 0 R q q /ProcSet[/PDF/Text] /Subtype /Form Q /Type /XObject /Matrix [1 0 0 1 0 0] 0 g /Matrix [1 0 0 1 0 0] Q endstream /Resources<< << >> The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. /Length 79 Q /Matrix [1 0 0 1 0 0]