twice a number decreased by 58
/Length 16 endstream /Length 65 /BBox [0 0 88.214 16.44] stream 0 5.203 TD 1.005 0 0 1.015 45.168 53.449 cm /F3 12.131 Tf Q 0.737 w q Q /Matrix [1 0 0 1 0 0] >> endstream /Meta6 15 0 R >> /Length 16 /Resources<< 1 g >> /Subtype /Form /Resources<< Q /FormType 1 0 w /Meta219 Do ET Q 1 i 0 g /BBox [0 0 639.552 16.44] q << /Matrix [1 0 0 1 0 0] endstream 1.014 0 0 1.007 531.485 776.149 cm endstream stream /Font << q /Meta19 Do Q stream /Root 2 0 R /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] >> << /XHeight 477 /FormType 1 /Type /XObject /Resources<< /Subtype /Form Q /F3 17 0 R 1.007 0 0 1.007 411.035 277.035 cm /BBox [0 0 549.552 16.44] /Matrix [1 0 0 1 0 0] stream /Subtype /Form 0 4.894 TD /ProcSet[/PDF] q /Matrix [1 0 0 1 0 0] /FormType 1 0 g /Meta97 Do /Matrix [1 0 0 1 0 0] stream /Length 16 q << endstream Q /Font << 0.564 G stream /Type /XObject /BBox [0 0 673.937 16.44] q /Meta390 406 0 R ET /Length 118 Q 1 i 1 g Q 1 g /Resources<< Q /FormType 1 /F3 12.131 Tf endstream /Type /XObject q /Type /XObject 0.458 0 0 RG endobj /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] 1 i >> q /Resources<< /Meta387 Do /Meta246 260 0 R << /F3 12.131 Tf /AvgWidth 459 q /Meta157 171 0 R /Type /XObject /F1 7 0 R /Meta298 Do 0 g /Resources<< >> stream /Resources<< 95 0 obj /Meta309 323 0 R /BBox [0 0 88.214 16.44] q /Meta7 Do /Font << /Resources<< /Resources<< BT 0 w >> /FormType 1 /Type /XObject /Type /XObject 1 i >> ET stream 0.175 Tc >> /StemV 94 /BBox [0 0 88.214 16.44] /Type /XObject /Font << endobj >> >> << /Type /XObject << /Meta226 240 0 R /FormType 1 q Q 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. 296 0 obj /Meta44 58 0 R /F3 17 0 R /Subtype /Form /Subtype /Form >> /Length 16 1 i 327 0 obj /Type /XObject B. (13) Tj Q /FontName /TimesNewRomanPSMT /FormType 1 stream Q /Length 59 /Meta252 Do >> Q stream >> Q >> BT 1 g 0 5.203 TD stream >> 1.007 0 0 1.006 411.035 437.384 cm << /Meta257 271 0 R q /FormType 1 q /BBox [0 0 88.214 35.886] /FormType 1 /Matrix [1 0 0 1 0 0] /F3 17 0 R >> >> q 312 0 obj q ET /BBox [0 0 534.67 16.44] >> Q /Matrix [1 0 0 1 0 0] 0.564 G >> 16.469 5.203 TD q /Matrix [1 0 0 1 0 0] 0 g q 6.746 5.336 TD Q q Q /Resources<< >> /F3 17 0 R /F3 17 0 R /Meta150 164 0 R << /Resources<< >> q 1 i 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 /Font << /BBox [0 0 88.214 16.44] 1.502 24.339 TD /FormType 1 /I0 51 0 R >> /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] Q endstream >> endstream /Resources<< >> ET >> Q BT Q /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] >> Ten divided by a number 5. Twice a number decreased by 8 gives 58. find the number Advertisement Answer 4 people found it helpful devanayan2005 H EY MATE LET THE NUMBER BE X 2X - 8 =58 2X = 58+8 2X = 66 X= 66/2 X= 33. >> endobj 47.933 5.203 TD /Resources<< stream 1 i Q /Meta315 329 0 R /Subtype /Form /Meta391 407 0 R 0 g Q 0 g ET /Meta198 Do 0.458 0 0 RG endobj endstream >> /Type /XObject stream q >> >> Q endobj (\)) Tj /Meta265 279 0 R Q /BBox [0 0 88.214 16.44] Q endobj 0 g 0 w /F1 12.131 Tf /FormType 1 /Matrix [1 0 0 1 0 0] q >> (13) Tj endstream /Resources<< /FormType 1 Q Q /ProcSet[/PDF] /Type /XObject >> 1 i Q endobj /Meta12 23 0 R q [(-3)-16(20)] TJ Q /FormType 1 0.738 Tc >> 0 g /F3 12.131 Tf /F3 12.131 Tf endstream /Resources<< >> 0 G /Subtype /Form q /Matrix [1 0 0 1 0 0] /Font << /FormType 1 >> /ProcSet[/PDF] >> /Resources<< 1.007 0 0 1.007 411.035 636.879 cm /Length 69 /BBox [0 0 30.642 16.44] /Meta106 Do (x) Tj >> 0 w /Meta263 Do Q 293 0 obj endobj /Meta186 Do q 1.005 0 0 1.007 79.798 796.475 cm 32.201 5.203 TD 0 G 0.369 Tc >> /Type /XObject >> a and b or something else.***. q Next, the problem says that "x" would be equal to twice a number added by 5. /Font << /BBox [0 0 88.214 16.44] q %PDF-1.4 Q endobj /Length 108 >> /ProcSet[/PDF] /Length 16 Q Q (-) Tj /Length 79 /Meta326 Do q Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. /FormType 1 0.463 Tc q /Font << q 0.458 0 0 RG << /Subtype /Form q /Subtype /Form /BBox [0 0 549.552 16.44] Q 0.737 w 0 5.203 TD (\)) Tj Q q >> 6.746 5.203 TD 0 w /Matrix [1 0 0 1 0 0] /Resources<< /BaseFont /TimesNewRomanPSMT /Meta155 Do /F3 17 0 R BT >> 183 0 obj /XObject << q /Subtype /Form Q (5) Tj /Length 69 1 i q /Subtype /Form >> Q /FormType 1 q [( and )16(a nu)26(mbe)18(r)] TJ /Length 70 /Meta59 Do Q /Matrix [1 0 0 1 0 0] ( x) Tj >> 3 0 obj 19.474 20.154 l (x ) Tj /Type /XObject 1 i /Length 58 ET Q endstream 0.737 w q 1 i endstream 0 w /F3 12.131 Tf q 0 w >> 1.007 0 0 1.007 551.058 583.429 cm /F3 12.131 Tf q Q >> /Type /XObject /Font << endobj q Q >> /Meta372 Do /Type /XObject /Type /XObject 1.007 0 0 1.007 271.012 450.181 cm endstream << /F3 12.131 Tf >> 0.486 Tc /Meta253 Do Q 1 i /BBox [0 0 88.214 16.44] >> /Font << Q q /Font << /Meta248 Do >> endstream /ProcSet[/PDF/Text] /FormType 1 /Resources<< Q q /Font << /Type /XObject Q q /Meta173 Do /Subtype /Form q endstream >> /Meta410 Do 1 i /ProcSet[/PDF] << /Matrix [1 0 0 1 0 0] endstream (\)) Tj ET /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 726.464 cm endobj /Subtype /Form /Length 16 BT >> 13.464 5.203 TD /Meta278 292 0 R The result is 8 less than 10 times the number. 26.957 5.203 TD /Subtype /Form 0.564 G q /Meta151 Do /F3 12.131 Tf /Type /XObject 371 0 obj Q /Subtype /Form 1.007 0 0 1.007 271.012 383.934 cm >> endstream /Meta0 Do 0 G 1.007 0 0 1.007 45.168 829.599 cm /Subtype /Form 1.005 0 0 1.007 102.382 872.509 cm q /BBox [0 0 88.214 16.44] /Font << /FormType 1 /Meta98 Do /Resources<< /Meta286 300 0 R endstream /Meta389 Do 1 i >> 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. ET /F3 12.131 Tf ET /F1 12.131 Tf /Subtype /Form Q Q >> 1.005 0 0 1.007 102.382 417.058 cm 225 0 obj /Subtype /Form q 1.007 0 0 1.006 551.058 763.351 cm /F3 17 0 R >> endobj /Subtype /Form BT /Type /XObject 0 g >> endobj /FormType 1 endstream Q stream ET /Font << /F3 12.131 Tf /F3 17 0 R 1.005 0 0 1.007 102.382 799.486 cm Q /Resources<< >> Q endobj ET stream q BT /Resources<< /ProcSet[/PDF/Text] 0 39.216 TD /Length 67 1 g 0 G /Meta282 Do /F3 17 0 R 0.564 G << Q Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. ET Q stream /F3 17 0 R 0.458 0 0 RG /F3 12.131 Tf /BBox [0 0 673.937 16.44] /Subtype /Form 1 i 1 i endstream >> /Resources<< q 0 G (-9) Tj Q /Meta266 280 0 R /Subtype /Form >> q Q q ET /Subtype /Form 1 i /Font << << /Type /XObject /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 271.012 450.181 cm BT /FormType 1 >> /Resources<< /Type /XObject 1.502 5.203 TD /BBox [0 0 88.214 16.44] BT q 0.458 0 0 RG /Resources<< q /Length 54 /Meta87 101 0 R 277 0 obj 1.502 5.203 TD << BT >> 0.458 0 0 RG /Resources<< Q /ProcSet[/PDF/Text] >> 1 i >> /F3 17 0 R /Resources<< 0 g 1 i /BBox [0 0 15.59 16.44] [(Fiv)25(e ti)18(me)16(s)] TJ ( decreased by ) Tj 194 0 obj /Meta325 Do >> 1.007 0 0 1.007 130.989 330.484 cm /Meta206 Do Q endobj endobj /F3 12.131 Tf /Length 59 q /ProcSet[/PDF/Text] 1 i /F3 12.131 Tf 1 i /Resources<< >> 0.564 G 1.007 0 0 1.007 271.012 583.429 cm /F3 12.131 Tf /Meta179 Do q >> /Meta307 321 0 R 0.227 Tc /XObject << /Meta94 108 0 R /F4 12.131 Tf 0 G 0.564 G /BBox [0 0 30.642 16.44] 1.007 0 0 1.007 411.035 383.934 cm 234 0 obj endobj q /Type /XObject /ProcSet[/PDF/Text] Six subtracted from a number 6. /F3 17 0 R 1 g Q /Length 16 Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. q Q /F1 12.131 Tf 0 G /Meta25 38 0 R 0.369 Tc /Matrix [1 0 0 1 0 0] 335 0 obj 0.458 0 0 RG 722.699 599.991 l Q /F4 12.131 Tf >> /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf SOLUTION: sixteen increased by twice a number is -58. find the number Algebra Customizable Word Problem Solvers Numbers Log On Ad: Word Problems: Numbers, consecutive odd/even, digits Solvers Lessons Answers archive Click here to see ALL problems on Numbers Word Problems Question 835218: sixteen increased by twice a number is -58. find the number 202 0 obj >> << 0 G /Resources<< q 0 G BT << /F4 36 0 R 338 0 obj /F3 17 0 R /Font << stream /BBox [0 0 15.59 16.44] /BBox [0 0 15.59 16.44] q 0.425 Tc 1 i /Matrix [1 0 0 1 0 0] stream q 0 g 1 i q >> >> << 283 0 obj /Meta37 50 0 R /F3 17 0 R endobj endstream /Font << /Meta70 Do 0.564 G ET 0.737 w 0 g 0 g 0 g 0.369 Tc /Meta400 Do /BitsPerComponent 1 0 G 200 0 obj Q >> /BBox [0 0 17.177 16.44] 181 0 obj >> Q 0 g stream 1.014 0 0 1.007 531.485 450.181 cm >> Q /Subtype /Form /Length 16 274 0 obj 1.007 0 0 1.007 411.035 383.934 cm 0 w 1.007 0 0 1.007 67.753 400.496 cm stream >> stream >> >> stream >> /Type /XObject /Length 69 /Resources<< 0 G /Font << /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf >> Q q 43.426 5.203 TD Q endobj >> 0 G /Matrix [1 0 0 1 0 0] q ET /BBox [0 0 88.214 16.44] >> Q 0.68 Tc Q >> endobj /ProcSet[/PDF/Text] Q endstream /Subtype /Form 1 g Q 211 0 obj /F3 17 0 R /Resources<< /Subtype /Form /Resources<< Q /Type /XObject endobj /Type /XObject q endstream << q (A\)) Tj >> q 1.014 0 0 1.007 251.439 636.879 cm 0 g /Subtype /Form Q 1 g /Meta92 Do 0 g q /BBox [0 0 88.214 16.44] 0 g /Type /XObject endstream /Resources<< endobj /Meta100 114 0 R q /FormType 1 >> 0.369 Tc Q Q q q 117 0 obj /Meta91 105 0 R endobj /Subtype /Form /Meta335 Do Q q >> 1.007 0 0 1.007 67.753 347.046 cm 0.737 w /Meta343 357 0 R q /Resources<< q 1 g /ProcSet[/PDF] Q 1.007 0 0 1.007 271.012 776.149 cm Q 0 g BT 1.007 0 0 1.007 551.058 636.879 cm /BBox [0 0 88.214 16.44] 0 g << /Font << /Length 60 /FormType 1 >> /Matrix [1 0 0 1 0 0] BT 1.007 0 0 1.007 130.989 383.934 cm Q ET ET /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /F4 36 0 R q /Meta120 Do 76.394 5.203 TD A number increased by 5 is equivalent to twice the same number decreased by 7. >> >> Q 1 i >> /Resources<< /BBox [0 0 15.59 16.44] /BBox [0 0 30.642 16.44] 0 G endstream /Meta240 Do stream /F3 12.131 Tf q /ProcSet[/PDF/Text] << Q 329 0 obj /Matrix [1 0 0 1 0 0] /Meta121 Do << /Font << >> >> >> 1 g /Font << /Type /XObject /BBox [0 0 88.214 16.44] stream >> 0.458 0 0 RG 0.297 Tc BT >> stream >> ET BT /Type /XObject /Subtype /Form BT /XObject << /Meta331 345 0 R /Meta66 Do Q /FormType 1 /F3 12.131 Tf q /ProcSet[/PDF/Text] /Type /XObject 175 0 obj /BBox [0 0 534.67 16.44] endstream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /ProcSet[/PDF] Q Q >> 78 0 obj 0.737 w >> /F3 12.131 Tf q /BBox [0 0 88.214 16.44] Q endobj /Subtype /Form A rectangular garden has a width that is 8 feet less than twice the length. endstream >> 0.737 w endobj /F4 12.131 Tf /Subtype /Form 1 i /F3 17 0 R q /F3 17 0 R 0 G /Length 58 endobj /Resources<< >> /Subtype /Form endobj 1 i Q Q /Meta33 Do q endstream Q << stream ET 269 0 obj 216 0 obj stream /Resources<< 0.458 0 0 RG /Meta67 81 0 R q (40) Tj endstream /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] q endstream /FormType 1 /Type /XObject 0 G q /BBox [0 0 88.214 16.44] endstream << 0 G q endstream 0 g << /Meta4 Do /ProcSet[/PDF/Text] /BBox [0 0 639.552 16.44] >> /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Meta38 52 0 R 250 0 obj >> 0 G /F1 12.131 Tf Q << q /Subtype /Form << >> stream >> << q 0.458 0 0 RG 57.656 5.203 TD >> >> 0 g /XObject << q **Note: You could choose any variable you want. << 0 G q /Type /XObject /Resources<< /Subtype /Form /Subtype /Form /F3 17 0 R /Matrix [1 0 0 1 0 0] Q /BBox [0 0 88.214 35.886] /Length 16 q /BBox [0 0 88.214 16.44] << /BBox [0 0 673.937 27.581] /Meta169 Do q Q 20.21 5.203 TD 0 G /ProcSet[/PDF] /F3 17 0 R /Type /XObject >> 0 G /F3 12.131 Tf /FormType 1 /Font << >> /Meta104 Do A number = an unknown number which can be represented by a variable, usually x. endstream /Resources<< /Matrix [1 0 0 1 0 0] /Length 59 /Type /XObject /Font << /ProcSet[/PDF] /Meta329 Do 1.005 0 0 1.007 102.382 653.441 cm /Subtype /Form /FormType 1 0 G /Length 12 << stream 73 0 obj /F4 12.131 Tf Q /Type /XObject /Matrix [1 0 0 1 0 0] endstream /FormType 1 >> q endobj /ProcSet[/PDF/Text] /FormType 1 /Matrix [1 0 0 1 0 0] endobj /FormType 1 /Font << Q q ET endobj q ET 1 g 198 0 obj q >> /Type /XObject /Subtype /Form q 0.737 w q 0 g q Q /Matrix [1 0 0 1 0 0] /Subtype /Form 1.014 0 0 1.007 111.416 583.429 cm >> stream q 0.737 w 0.369 Tc endstream /Length 69 0 g /Font << 0 w /ProcSet[/PDF/Text] /FormType 1 /BBox [0 0 88.214 16.44] /Meta388 Do /Length 118 /Subtype /Form 0.486 Tc 1 i 0 G 0 g q Q 156 0 obj 0 g q >> /Meta360 Do ET 48 0 obj Q ( \() Tj q Q /FormType 1 ET /Meta59 73 0 R >> /ProcSet[/PDF/Text] 0.786 Tc Q /Font << << /F3 12.131 Tf 1 i /Resources<< Q /Length 69 /Subtype /Form >> q /ProcSet[/PDF/Text] /Resources<< BT endstream /Resources<< endstream 1 g /BBox [0 0 15.59 16.44] Q (3\)) Tj /F3 17 0 R 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 1 i endstream Q /FormType 1 >> /Matrix [1 0 0 1 0 0] >> << /Length 294 /FormType 1 203 0 obj 1.007 0 0 1.007 654.946 599.991 cm 1.007 0 0 1.007 551.058 277.035 cm Q Q q Q >> 1.007 0 0 1.007 271.012 583.429 cm /Meta93 107 0 R >> q q /Type /XObject >> 0 w q stream >> << 333 0 obj /BBox [0 0 88.214 16.44] Q /Subtype /Form >> /BBox [0 0 88.214 16.44] endstream q 213 0 obj q 1 i << (B\)) Tj /Subtype /Form /Matrix [1 0 0 1 0 0] q /Type /XObject ET stream /ProcSet[/PDF] [tex]\sin (\pi -x)=\sin x[/tex]. /Length 58 << /ProcSet[/PDF/Text] << q Q << endobj /ProcSet[/PDF/Text] ET >> 1.007 0 0 1.007 67.753 799.486 cm 1.007 0 0 1.007 45.168 779.913 cm >> /ProcSet[/PDF] Q Q Q ET Q << BT Q /Meta260 274 0 R endobj /ProcSet[/PDF/Text] /Type /XObject /Length 294 0.564 G 1 i q >> /Meta57 Do >> 0 w q 0.271 Tc 1 i /Font << /F3 17 0 R 1 i stream q >> 0.564 G 0 w 0 g stream >> Q endstream endobj /Length 60 endstream 393 0 obj endstream /ProcSet[/PDF/Text] q endobj >> << stream << endobj q stream Q /Font << /F3 17 0 R 1 g >> /Size 447 /Meta214 228 0 R /BBox [0 0 88.214 35.886] >> q /Length 54 << BT q /Meta365 Do 1 g endobj 0 g /FormType 1 0.737 w Q q /Type /XObject q BT /Subtype /Form >> Q 1.502 24.649 TD /Font << /Meta82 96 0 R /Subtype /Form stream /Type /XObject q /F3 17 0 R << 0 g Q (-) Tj << /BBox [0 0 30.642 16.44] >> /Length 69 << BT ET /Length 99 /FormType 1 Choose the correct one. q q 0.458 0 0 RG q 0 g q >> S /Meta51 Do /Length 245 0 w endobj /Meta428 444 0 R /Meta187 Do 116 0 obj >> 0.737 w Q >> BT Q /Meta203 Do << /F3 17 0 R /ProcSet[/PDF/Text] Q An example of a linear inequality in one variable is A. x+y = Question: answer 1. 94 0 obj >> /F3 17 0 R /Meta85 99 0 R 253 0 obj twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. stream /FormType 1 << 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. Q /Subtype /Form /Resources<< Q 0 g Q 0.737 w /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Meta187 201 0 R /BBox [0 0 15.59 16.44] q /Meta119 133 0 R /Meta160 Do 1 i 1.502 5.203 TD stream endstream 21.713 20.154 l S >> Q /ProcSet[/PDF] stream Twice = two times, double. 1 i Q /BBox [0 0 88.214 16.44] q /FormType 1 /F1 12.131 Tf 172 0 obj >> q 0 5.203 TD /Meta395 411 0 R 0.737 w /Matrix [1 0 0 1 0 0] stream Q q 20.21 5.203 TD Expression. /FormType 1 stream Q endstream << endobj /Meta29 42 0 R >> >> /Resources<< >> 1.007 0 0 1.007 271.012 636.879 cm (5\)) Tj endobj /Length 16 Q /Meta376 Do Q /ProcSet[/PDF/Text] /Meta383 Do q /Font << q endstream Q /Matrix [1 0 0 1 0 0] 178.979 5.203 TD /Font << Q q Q /F3 17 0 R >> /Meta102 Do 0.737 w Q Q q >> >> Q /Resources<< 1.007 0 0 1.006 130.989 437.384 cm << Q 1.502 5.203 TD 1.007 0 0 1.007 130.989 849.172 cm 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. ET 1.007 0 0 1.007 130.989 776.149 cm endobj q Percent Change = (Decrease First Value) x 100% 372 0 obj /Type /XObject q >> 0 w 52.412 5.203 TD >> /Resources<< 1 i >> >> /Subtype /Form q /BBox [0 0 23.896 16.44] /Matrix [1 0 0 1 0 0] q 1 i stream /Font << /Font << endstream 1 i /FormType 1 6. q ET 1 i /Meta347 Do 0.68 Tc /Type /XObject endstream 0 g endstream ET /Meta319 Do endobj /Resources<< /F3 17 0 R /FormType 1 0 g That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] /Length 59 Q /FormType 1 endobj 142 0 obj q /Subtype /Form Q /BBox [0 0 15.59 16.44] Q 0 g /FirstChar 43 /F3 12.131 Tf 0.564 G q /Resources<< /FormType 1 /Type /XObject Q /Subtype /Form q 0.737 w /BBox [0 0 88.214 16.44] endobj /Length 16 /ProcSet[/PDF/Text] q endstream 0.737 w 1 i /Font << /XObject << /XObject << /BBox [0 0 15.59 29.168] /FormType 1 Q /Length 67 >> Q /F1 12.131 Tf 98 0 obj 9.723 5.336 TD >> 1 i /FormType 1 0 20.154 m 1 i ET 1.007 0 0 1.007 551.058 383.934 cm /Meta222 236 0 R /Resources<< 315 0 obj /Matrix [1 0 0 1 0 0] /Meta262 276 0 R >> q 0 w if the solution of an equation is x=-2, what could the original equation be? endobj /Length 59 Q Q q Q 0 g Q Q /F3 17 0 R >> q /Meta394 410 0 R Q 1 i /Type /XObject /Meta18 29 0 R /Type /XObject >> endstream Q /ProcSet[/PDF/Text] q 549.694 0 0 16.469 0 -0.0283 cm 1 i >> q Q 0.564 G q /Font << << /Meta135 Do /Resources<< 65 0 obj q /F1 12.131 Tf 1 i q >> q BT q /Type /XObject Q << /Resources<< 358 0 obj /Subtype /Form >> >> endobj endstream q stream << /FormType 1 endstream /FormType 1 endobj 3.742 5.203 TD /ProcSet[/PDF/Text] q 0 g /Font << /Matrix [1 0 0 1 0 0] >> Q 0 g 1 g 425 0 obj ET >> /Length 69 /F3 12.131 Tf Q /F4 36 0 R << /Matrix [1 0 0 1 0 0] endstream Q endstream q 0 g q /Resources<< /F3 17 0 R 1.007 0 0 1.007 67.753 599.991 cm q 0.458 0 0 RG 0.737 w That was 1/8 of the points that he scored /Resources<< >> endstream 1 g 1.007 0 0 1.006 551.058 836.374 cm << stream q 0 G /F1 7 0 R /ProcSet[/PDF/Text] >> /Meta391 Do q /Type /XObject Q 0 g Q LAIing for a pizza and, soft drink. q >> /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] Q /FormType 1 q /FormType 1 /FormType 1 Q 0 G /Meta355 369 0 R q /Subtype /Form endobj Q 1.005 0 0 1.007 102.382 546.541 cm /Font << endstream /Subtype /Form stream q Q >> 0.524 Tc 1.005 0 0 1.007 102.382 726.464 cm 0 g 300 0 obj /Meta359 373 0 R 1 i gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. /F3 17 0 R Q /Meta10 Do endobj Q /ProcSet[/PDF] 1 i BT 178 0 obj /Font << /Matrix [1 0 0 1 0 0] /FormType 1 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 Q /F1 12.131 Tf /Meta396 Do 1 i 0 g /Matrix [1 0 0 1 0 0] /Meta81 95 0 R ET /Matrix [1 0 0 1 0 0] endstream 0.524 Tc stream 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. /F3 12.131 Tf Q 321 0 obj 1.007 0 0 1.006 411.035 510.406 cm ET Q /Type /XObject 1 i 1 i 1 i -0.486 Tw Two speeding tickets could increase your rate by 58% at your next renewal. /Meta274 Do 0 G /Matrix [1 0 0 1 0 0] endstream q 20-n c.) n+20 d.) 20+n 3.) Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio Q << endobj /Meta335 349 0 R >> Q << ET /Resources<< stream 1 i endstream /Type /XObject Q /Matrix [1 0 0 1 0 0] >> stream Q /FormType 1 /Matrix [1 0 0 1 0 0] 0 g /F3 12.131 Tf Q /Meta348 Do only about 58% of candidates will agree to be screened. /Matrix [1 0 0 1 0 0] q >> >> /Type /XObject 1 i endstream BT /Subtype /Form endobj 0.458 0 0 RG /Font << 1.007 0 0 1.007 271.012 776.149 cm endobj q stream >> ET 0.486 Tc /F3 12.131 Tf /Resources<< 0 g endstream /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 67.753 293.596 cm 210 0 obj /F3 12.131 Tf /Meta55 69 0 R 0 5.203 TD /DecodeParms [<> ] 1 i (B) Tj 1.007 0 0 1.007 271.012 636.879 cm /BBox [0 0 88.214 16.44] /Meta238 252 0 R 0 G (+) Tj /Font << >> q /Matrix [1 0 0 1 0 0] Q BT >> /Meta300 Do >> /Meta336 350 0 R 0 G 0.458 0 0 RG 0.564 G Q 1.007 0 0 1.007 271.012 636.879 cm /F3 17 0 R stream Q 1 i /Font << 0 5.203 TD q Q /F3 12.131 Tf endobj /Length 91 [(1)-25(0\))] TJ /Resources<< 1.005 0 0 1.007 102.382 653.441 cm /Font << q /Resources<< endobj 1.007 0 0 1.007 130.989 383.934 cm /FormType 1 q /FirstChar 32 << /ProcSet[/PDF/Text] /Meta96 Do /Meta53 Do /Matrix [1 0 0 1 0 0] 0.425 Tc /F3 17 0 R /F3 12.131 Tf 0 g 1 i q >> q /Meta197 Do /F1 7 0 R endstream /Font << q /Meta276 Do /F3 17 0 R endstream /Type /XObject 1 g q 1 i /ProcSet[/PDF/Text] (6\)) Tj 2.Nine point two decreased by double a number is the same as the number added to four fifths. ET endobj q endstream endobj 1.007 0 0 1.007 130.989 523.204 cm 1 g 0 g /Meta71 85 0 R q stream Q Q /FormType 1 /Matrix [1 0 0 1 0 0] stream /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 862.723 cm /F3 12.131 Tf 0.486 Tc /Flags 32 Q Q Q Q 1 i >> View the full answer. /StemV 88 Q /Ascent 1050 /Resources<< q stream 125 0 obj /Matrix [1 0 0 1 0 0] >> /F3 17 0 R >> << /FormType 1 /Subtype /Form /Length 54 0.311 Tc /FormType 1 /Type /XObject Q stream /ProcSet[/PDF/Text] >> Q Diabetes, if left untreated, leads to many health complications. 0 G /F3 12.131 Tf ET Q q In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). 1 i q Q /Subtype /Form Q , Prove the following 0.737 w /Resources<< /Type /XObject stream /BBox [0 0 15.59 29.168] 0 G 1 i 0.486 Tc /Font << /Resources<< q 1 g Q >> stream /FormType 1 [(The )-19(quotient of )] TJ /BBox [0 0 30.642 16.44] /Resources<< endstream 0.564 G (D\)) Tj 1 i 235 0 obj /F3 12.131 Tf 0 G q Q endstream 1 i endstream /ProcSet[/PDF] Q Q 1.005 0 0 1.007 102.382 653.441 cm Q 1.007 0 0 1.007 67.753 872.509 cm 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. >> Q 0 G /Matrix [1 0 0 1 0 0] stream 298 0 obj >> >> >> 0.738 Tc /Resources<< stream endobj /Resources<< 0 G q 109 0 obj /Matrix [1 0 0 1 0 0] /F3 12.131 Tf ET Q Q BT /F3 17 0 R Q 1 i /Meta382 396 0 R << 27.693 5.203 TD >> q /Resources<< /F3 17 0 R 0.564 G /Meta355 Do Q 1.014 0 0 1.007 111.416 383.934 cm endobj /Matrix [1 0 0 1 0 0] >> /Length 118 /Font << /Font << endobj Q 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. 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