twice a number decreased by 58

0 G /Font << stream 0 G /F4 36 0 R Q q endobj /Matrix [1 0 0 1 0 0] /F3 12.131 Tf >> 1 i 0.68 Tc /ProcSet[/PDF] >> /F3 17 0 R S Q Q /F3 17 0 R 1 i endobj /ProcSet[/PDF] q Q 1.007 0 0 1.007 551.058 330.484 cm /Resources<< /Type /XObject 90 0 obj /Font << << endstream 1.007 0 0 1.007 271.012 636.879 cm /ProcSet[/PDF/Text] /Meta256 270 0 R >> stream /Meta110 124 0 R 0 g /Subtype /Form 0.51 Tc q /Matrix [1 0 0 1 0 0] Q 0 G /Type /XObject >> 0.51 Tc /Meta156 170 0 R /Meta417 433 0 R (13) Tj 146 0 obj /Matrix [1 0 0 1 0 0] 0.737 w 1 i q /Font << stream /BBox [0 0 15.59 16.44] /Type /XObject 0.458 0 0 RG /FormType 1 0 g endobj endobj endstream 347 0 obj >> q /Meta298 312 0 R /Meta35 Do endstream >> endstream Q q endobj endstream /ProcSet[/PDF] q q 12.727 24.649 TD << /Length 118 (\(x ) Tj /Resources<< /Meta185 Do /Resources<< Q stream /Resources<< >> /BBox [0 0 88.214 16.44] >> /Meta33 Do 0 g /Type /XObject q /Meta193 207 0 R /ProcSet[/PDF/Text] /Length 59 /F3 17 0 R 0.737 w /FormType 1 q /Subtype /Form This gives us: "2x+5". 0.737 w /F3 12.131 Tf /FormType 1 /F3 17 0 R /Type /XObject Q /Resources<< /Subtype /Form /Length 12 q << [( subt)-17(racted fr)-14(om a )-16(number)] TJ endobj 0 g q >> q /FormType 1 /FormType 1 346 0 obj ET >> Q /Meta404 420 0 R ( \() Tj /BBox [0 0 88.214 16.44] Q /FormType 1 Q /Font << BT /F3 12.131 Tf /F3 17 0 R endobj endobj 0 5.203 TD /Length 16 q ET >> /Meta96 110 0 R /Matrix [1 0 0 1 0 0] The sum of a number and 2 is 6 less than twice that number. 1 i Q Formula - How to Calculate Percentage Decrease. >> q BT /Resources<< q q 1.007 0 0 1.007 130.989 636.879 cm 0 5.203 TD /Meta307 Do 1.007 0 0 1.007 271.012 776.149 cm 0 G >> endobj Q /ProcSet[/PDF/Text] /Length 16 /Meta52 66 0 R endobj q 0.564 G /ProcSet[/PDF] Q Q BT 0 g /F3 12.131 Tf 1.502 5.203 TD q q stream 1 i >> 3.742 5.203 TD /Meta11 22 0 R endobj /ProcSet[/PDF/Text] 86 0 obj endobj >> /BBox [0 0 534.67 16.44] q /Meta207 Do 218 0 obj /Resources<< /ItalicAngle 0 /FormType 1 ET /F3 12.131 Tf 1.014 0 0 1.007 531.485 636.879 cm ET stream 0 G 0 g 194 0 obj Q q Q q stream ET /Matrix [1 0 0 1 0 0] /Subtype /Form /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Type /XObject endstream endstream /Type /XObject /FormType 1 1.007 0 0 1.007 654.946 400.496 cm 0 g 0.425 Tc 1.007 0 0 1.007 130.989 636.879 cm /ProcSet[/PDF/Text] 0 w stream endobj endobj (\)) Tj ET /Meta414 430 0 R endstream /Length 58 /Subtype /Form 0 g /ProcSet[/PDF/Text] 1 i q BT << 1 g /FormType 1 Q q (-20) Tj 1.007 0 0 1.007 271.012 450.181 cm /FormType 1 /Subtype /Form >> /Type /XObject /Meta44 Do /F1 7 0 R >> >> /Length 16 Q endstream endstream q /Meta184 198 0 R 359 0 obj /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 /Length 69 /Resources<< q 0 G >> 1 i Q /Subtype /Form endobj >> Q stream Q /Meta24 Do 265 0 obj >> /Resources<< 0 w /Resources<< /Meta310 324 0 R Q /Type /XObject stream >> Q 0.486 Tc q >> endstream endobj 312 0 obj q >> q (7\)) Tj 0 g 0 g q /ProcSet[/PDF/Text] (-) Tj /Resources<< Q endobj /FirstChar 43 (C\)) Tj /Meta105 119 0 R >> /ProcSet[/PDF/Text] /FormType 1 q /BBox [0 0 88.214 16.44] endstream 0 g /FormType 1 >> /FormType 1 stream >> /FormType 1 0.564 G Q endobj 0.564 G Q q 202 0 obj /Subtype /Form /F3 12.131 Tf endstream /Descent -299 /Resources<< 0.369 Tc Q Thrice a number decreased by 5 exceeds twice the number by 1 is . 0 G << BT /Length 54 /Meta103 117 0 R 1 i 1.014 0 0 1.006 111.416 836.374 cm >> << Q Q /Meta200 214 0 R 1. << Q Q /Subtype /Form Q 0.737 w 1.007 0 0 1.007 654.946 653.441 cm 0.369 Tc 186 0 obj Q Q /ProcSet[/PDF/Text] q 0 g Q stream 32.939 5.203 TD >> /Length 69 q , Prove the following /Meta269 283 0 R ET endstream 16.469 5.336 TD Q /Meta210 Do /Meta243 Do 13.493 5.336 TD 1.007 0 0 1.007 654.946 799.486 cm q /Subtype /Form /Meta28 41 0 R q 1 i stream 0.564 G Q q 1.005 0 0 1.007 102.382 310.158 cm 0 g /ProcSet[/PDF] /ProcSet[/PDF] 239 0 obj /Meta302 316 0 R /F3 12.131 Tf 0.68 Tc Q q 0.564 G q /ProcSet[/PDF/Text] /F1 7 0 R endobj Q /Resources<< 0.458 0 0 RG /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] BT endstream endstream q /Subtype /Form endstream endstream q /Type /XObject 0 g << Q >> Q 0 5.203 TD >> /BBox [0 0 30.642 16.44] /Meta418 434 0 R q endstream /F3 12.131 Tf /BBox [0 0 30.642 16.44] /Meta42 56 0 R /Meta204 218 0 R /Meta242 256 0 R Q /Type /XObject >> /BBox [0 0 15.59 16.44] q endobj /Matrix [1 0 0 1 0 0] (D\)) Tj (7\)) Tj Q Q /ProcSet[/PDF] 0.458 0 0 RG /Meta118 Do endobj >> q /FormType 1 1 g 27.693 5.203 TD /Resources<< /Type /XObject 672.261 799.486 m q /FormType 1 /Type /XObject /BBox [0 0 673.937 15.562] /Resources<< << endobj /F3 17 0 R Q /Meta422 Do /FormType 1 /Length 54 endobj stream q 1.005 0 0 1.007 79.798 796.475 cm Q /Subtype /Form Answer provided by our tutors. 1.007 0 0 1.007 67.753 799.486 cm /Meta350 Do >> /F1 12.131 Tf /StemV 94 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /FormType 1 0 G q q Q /BBox [0 0 673.937 16.44] /FormType 1 13.493 5.336 TD >> q Q << Q Q /Resources<< /F3 12.131 Tf /BBox [0 0 88.214 16.44] Q startxref Q 1 i 0 g stream q /ProcSet[/PDF] /Length 12 q /ProcSet[/PDF] q /Meta247 Do /FormType 1 0 G 0 w Q /Font << << Q 149 0 obj BT 0 5.203 TD /Length 69 q q 1 i /Resources<< >> /Meta223 Do >> Q /Resources<< 151 0 obj /Type /XObject /F1 12.131 Tf /Subtype /Form /Resources<< endstream << q 0 g endobj 0 G [( the )-24(sum of a n)-14(umber an)-14(d )] TJ 229 0 obj q 23.216 5.203 TD >> /Type /XObject /FormType 1 q /Matrix [1 0 0 1 0 0] q q 104 0 obj /BBox [0 0 549.552 16.44] /FormType 1 0 g Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . q ET This site is using cookies under cookie policy . /F1 12.131 Tf /Length 69 /Meta358 Do /Meta220 234 0 R q >> 0 G /F1 12.131 Tf Q q /Resources<< q /Resources<< endobj endobj 0 g endobj 20 0 obj /Matrix [1 0 0 1 0 0] Q >> /Type /XObject /Resources<< 0.425 Tc Q >> In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). 0.737 w 1.007 0 0 1.006 130.989 690.329 cm /Length 16 0.486 Tc /BBox [0 0 15.59 16.44] /Meta8 19 0 R Q /Meta185 199 0 R q BT 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 /Subtype /Form /F3 12.131 Tf >> >> 1 i 0.458 0 0 RG endobj /Meta54 Do Q 1.007 0 0 1.007 271.012 383.934 cm /Type /XObject 1 i BT endstream /ProcSet[/PDF] pidemiologi i Infekcionnye Bolezni. q endstream >> /ProcSet[/PDF/Text] Q q q Q q stream Q /Meta158 172 0 R endobj >> /Resources<< 1.502 5.203 TD 162 0 obj /FormType 1 >> Q q 0 g endobj /Leading 150 1.007 0 0 1.007 271.012 523.204 cm /Subtype /Form endobj (58) Tj /BBox [0 0 88.214 16.44] /Length 69 /Ascent 976 q /Resources<< 0 g 0 G 0.564 G /BBox [0 0 88.214 16.44] 1 i /Type /XObject 1.014 0 0 1.007 251.439 583.429 cm 1.007 0 0 1.007 551.058 636.879 cm Q 0 5.203 TD /Length 16 q endobj endobj /Font << ET >> /FormType 1 ( x) Tj /FormType 1 endstream stream 0.369 Tc q stream 0 g /Font << /Meta159 173 0 R 19.474 5.203 TD /BBox [0 0 88.214 16.44] /Length 59 1.007 0 0 1.007 130.989 383.934 cm /Type /XObject /ProcSet[/PDF] /Meta344 358 0 R 1 i Q >> Q 1 i >> q >> View the full answer. 1.014 0 0 1.007 111.416 583.429 cm 1 i 0.737 w /F3 17 0 R 0 G q endobj BT endstream 0.564 G 228 0 obj 192 0 obj q q /Subtype /Form /Length 59 0.175 Tc Q 0.524 Tc endstream /MaxWidth 1248 Q q /Meta24 37 0 R q >> /Font << /Matrix [1 0 0 1 0 0] Q /Meta272 286 0 R ET Q /FormType 1 1.502 5.203 TD Q /FormType 1 /Matrix [1 0 0 1 0 0] /Meta214 228 0 R (8\)) Tj 1 i 1.014 0 0 1.007 531.485 330.484 cm /FormType 1 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 . /ProcSet[/PDF/Text] /F4 12.131 Tf /ProcSet[/PDF] >> 0 G /Resources<< q << /Matrix [1 0 0 1 0 0] 0.737 w /Type /XObject >> endobj >> 0 w q endobj BT << 20.21 5.203 TD q endobj /XObject << endstream /BBox [0 0 88.214 35.886] 0.458 0 0 RG /Meta161 175 0 R /ProcSet[/PDF/Text] >> >> 367 0 obj /F3 17 0 R /Encoding /WinAnsiEncoding BT Q >> /Resources<< ET In addition, testosterone in both sexes is involved in health and well-being . /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] /Type /XObject q /FormType 1 /Type /XObject Q stream /Type /XObject 0.486 Tc stream endobj BT 1 i q >> 251 0 obj stream Two speeding tickets could increase your rate by 58% at your next renewal. Q q /Meta226 240 0 R /Meta101 115 0 R 5 0 obj /Matrix [1 0 0 1 0 0] endstream Calculate a 15% decrease from any number. 0.737 w 0 g 0 5.203 TD /Type /XObject >> Q /ProcSet[/PDF/Text] /Font << /ProcSet[/PDF/Text] Q >> /Subtype /Form >> q S /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 Q 0 g /Meta39 Do >> /Meta312 326 0 R /Matrix [1 0 0 1 0 0] q endstream 0 G 0.737 w q 1 i q /Font << q 0 g /Resources<< >> /F1 7 0 R /Type /XObject /F3 12.131 Tf q 1.014 0 0 1.007 111.416 523.204 cm /F3 12.131 Tf q << /Matrix [1 0 0 1 0 0] 549.694 0 0 16.469 0 -0.0283 cm 19.474 5.203 TD 0.458 0 0 RG /Font << >> >> >> Q Q Q /ProcSet[/PDF] /Meta263 277 0 R << /Font << -0.062 Tw << Q /ProcSet[/PDF/Text] /Type /XObject /FormType 1 /F3 17 0 R /Length 70 /Matrix [1 0 0 1 0 0] /Meta29 Do endstream /FormType 1 1 i 1.005 0 0 1.007 79.798 829.599 cm ET q endstream q 1 i endstream q >> 304 0 obj q endstream >> /Subtype /Form 138 0 obj /BBox [0 0 15.59 29.168] q Q -0.16 Tw q /BBox [0 0 88.214 16.44] >> /Font << /Font << /Font << Q q /FormType 1 q /Subtype /Form /Meta235 249 0 R << >> Q 1 i (x ) Tj 1 i Q Q /Resources<< /Subtype /Form /Type /XObject /CapHeight 476 0 g /Meta253 Do Q /FormType 1 /FormType 1 0 g Q 1 i stream endobj Q Q << /Subtype /Form /F3 17 0 R /Meta238 252 0 R /F3 17 0 R 1 g /Font << /Subtype /Form /Resources<< /Meta26 Do >> endobj /ProcSet[/PDF/Text] /F3 12.131 Tf Q >> 0.564 G << q q q 1 i stream q endobj 0 g 12.727 5.203 TD /ProcSet[/PDF/Text] endstream >> ET Q q 0 G << Q Q /BaseFont /TestGen-Regular >> 0 G /Subtype /Form 0.564 G ET endobj Q /ProcSet[/PDF] /BBox [0 0 88.214 16.44] >> endobj 0 g /Subtype /Form >> /Font << /Meta345 359 0 R /Length 16 20.21 5.203 TD 0 g ET /Font << /BBox [0 0 88.214 16.44] >> BT Q Find the length. /Resources<< ET q ET Q /FormType 1 >> /Subtype /Form 1 i 0 G /Meta78 92 0 R 0 G ET 1.007 0 0 1.007 271.012 583.429 cm /Resources<< q /FormType 1 /Length 69 1.005 0 0 1.007 102.382 400.496 cm 0.564 G << Q /F4 12.131 Tf q /F4 36 0 R /BBox [0 0 639.552 16.44] 1 i /Resources<< q 1 i ET Q 1.007 0 0 1.007 130.989 849.172 cm q 1 i (+) Tj endstream Q /ProcSet[/PDF/Text] 0 5.203 TD /Matrix [1 0 0 1 0 0] Q 1 i q << /Type /XObject /Matrix [1 0 0 1 0 0] /Subtype /Form 0.486 Tc 0.458 0 0 RG Q >> /Meta390 406 0 R /Subtype /Form q 0.332 Tc /Length 16 /F3 12.131 Tf /FormType 1 stream >> Q /ProcSet[/PDF] >> endstream Q Q 1.502 7.841 TD >> /F1 12.131 Tf endobj ET /Type /XObject /BBox [0 0 88.214 16.44] (2) Tj 0 g Q (7\)) Tj /F3 12.131 Tf 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 . >> q /ProcSet[/PDF] Q Q endobj /Subtype /Form << 0 g >> 0 w Q >> ET Q BT /Type /XObject 0 5.203 TD q >> 1 i /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Subtype /Form q endobj << q /Length 69 /F3 17 0 R 0.564 G q /Resources<< q 134 0 obj 154 0 obj 0.458 0 0 RG 1 i << 1.005 0 0 1.007 102.382 473.519 cm /Meta22 Do 45 0 obj 1 i /BBox [0 0 88.214 16.44] 1 i /F3 12.131 Tf /Descent -299 >> /Matrix [1 0 0 1 0 0] /Meta69 83 0 R Q /Font << stream Q /Subtype /Form 0 G (x) Tj >> 0.458 0 0 RG 1.005 0 0 1.007 102.382 347.046 cm /Length 118 q 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 0.737 w /Subtype /Form /Matrix [1 0 0 1 0 0] (2) Tj 0 g Q /Matrix [1 0 0 1 0 0] 1 g q Q /Info 3 0 R endobj q (1\)) Tj BT Q >> Q q /Length 16 /ProcSet[/PDF/Text] 1 i /Length 12 endstream Q 1 i /BBox [0 0 88.214 16.44] >> 2.238 5.203 TD endstream Q /Meta190 Do >> 1 i /ProcSet[/PDF] /Subtype /Form 1 g >> Q stream q >> 1 i q /Meta238 Do /Length 60 q >> /Meta49 63 0 R Q q q q 0.737 w /Subtype /Form 1.007 0 0 1.007 45.168 829.599 cm endstream 1.007 0 0 1.007 551.058 330.484 cm /Type /XObject /Length 54 /F3 17 0 R stream << 0.737 w In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. 1 i 0.463 Tc stream q << Q Double or twice a number means 2x, and triple or thrice a number means 3x. /ProcSet[/PDF] 1 g 0 G /Resources<< Q Q /Matrix [1 0 0 1 0 0] >> /Meta179 Do >> ET << << /Resources<< stream /Meta279 293 0 R 0 g /BBox [0 0 88.214 16.44] 0 G /Type /XObject q 294 0 obj >> 287 0 obj /Type /XObject /Meta194 Do stream q /Matrix [1 0 0 1 0 0] << q 0 g /FormType 1 endstream stream 1 i endstream 0.838 Tc /Length 65 /Meta264 Do endobj q q /Resources<< q /Font << << endstream /XObject << 0 w endstream /BBox [0 0 88.214 16.44] /Font << 0 g /Subtype /Form 368 0 obj endstream ET /F4 36 0 R 1 i endobj /Type /Font 0 4.894 TD Q 1 i 1 i Q 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. 143 0 obj >> /FormType 1 q Q >> /Length 16 0.134 Tc Q Notice that we used the variable \large {d} d in our equation to stand for our unknown value. q >> /Resources<< ET stream 1.007 0 0 1.007 654.946 599.991 cm q /Font << 0 g /BBox [0 0 88.214 16.44] /Type /XObject endstream 1 i Q q 1 g New questions in Mathematics q /Resources<< /MaxWidth 1397 /F3 12.131 Tf 393 0 obj /Subtype /Form /Meta182 Do /Subtype /Form 0 5.203 TD ET /FormType 1 Q 442 0 obj stream 0 w q stream /Subtype /Form 0 w Q endobj 140 0 obj 20.21 5.203 TD /Meta399 415 0 R 1 i /F1 12.131 Tf /Subtype /Form /Meta155 169 0 R endobj 0.227 Tc 400 0 R /F3 12.131 Tf q >> /Type /XObject /Meta412 428 0 R << 1 g /BBox [0 0 15.59 29.168] 273 0 obj /ProcSet[/PDF] >> 384 0 obj Q q 0.297 Tc 0 G >> 0 g /F4 12.131 Tf 0.564 G 20.21 5.203 TD >> 0 g /Resources<< 672.261 872.509 m 1.007 0 0 1.007 411.035 383.934 cm endstream q /Subtype /Form 2.Nine point two decreased by double a number is the same as the number added to four fifths. << /Resources<< 0 w /F3 12.131 Tf /Meta329 343 0 R /Type /XObject /Subtype /Form 20.21 5.203 TD 1 i /Subtype /Form q 1.007 0 0 1.007 45.168 730.228 cm /Type /XObject /Matrix [1 0 0 1 0 0] 375 0 obj >> (40) Tj /F4 12.131 Tf q /Resources<< 0.564 G /Length 63 endstream /FormType 1 /Leading 349 /ProcSet[/PDF/Text] q q /F3 12.131 Tf /BBox [0 0 30.642 16.44] >> q << /FormType 1 1.007 0 0 1.007 45.168 763.351 cm /F3 12.131 Tf << 187 0 obj Q 4.506 8.18 TD q endobj /F3 12.131 Tf /F3 12.131 Tf /Subtype /Form Q /Font << /Meta164 Do 0 g 141 0 obj >> stream << Q /Resources<< endstream /Matrix [1 0 0 1 0 0] /FormType 1 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 /Resources<< /Length 69 /BBox [0 0 15.59 16.44] << >> /Matrix [1 0 0 1 0 0] ET /Subtype /Form /Meta15 26 0 R 0 G /BBox [0 0 88.214 16.44] 0.369 Tc q /Length 59 stream >> 0 w stream q /Resources<< 1.005 0 0 1.007 102.382 546.541 cm ET << 225 0 obj /Length 70 An example of a linear inequality in one variable is A. x+y = Question: answer 1. stream /Resources<< /FormType 1 /Matrix [1 0 0 1 0 0] q 0 G 0 G 279 0 obj 0 G /LastChar 120 Q 1 i /ProcSet[/PDF/Text] Q Q /Length 81 endobj q /Meta191 205 0 R ET 0.738 Tc 0 g Q q /Type /XObject Q /BBox [0 0 15.59 16.44] /F3 17 0 R /Type /XObject 1.014 0 0 1.006 251.439 763.351 cm /ProcSet[/PDF/Text] 0 g /Meta172 Do 1 i Q Q 672.261 347.046 m Solution: Let the number be x. (\)) Tj q Q /Type /XObject (v) 5 subtracted from thrice a number is 16. Thrice a number decreased by 5 exceeds twice the number by a unit. a and b or something else.***. Q endstream /Subtype /Form q Q twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. 1.014 0 0 1.007 251.439 703.126 cm /ProcSet[/PDF] [(-3)-16(20)] TJ /FormType 1 q q 0.564 G 1.005 0 0 1.007 102.382 546.541 cm /Meta322 336 0 R >> Q 0.458 0 0 RG stream nine increased by a number x. 1 i ET ET endstream >> /F3 12.131 Tf /FormType 1 >> /Font << << 198 0 obj ET /Meta59 Do 1.005 0 0 1.007 102.382 653.441 cm endstream 0.458 0 0 RG Select the correct mathematical statement for the following equation. 0 w 1.007 0 0 1.007 271.012 583.429 cm /BBox [0 0 30.642 16.44] 6 more than twice a number: 2x+6: two less than a number: x-2: the sum of 9 and a number: 9+x: two less than three times a number: 3x-2: a number subtracted from 12 . /Meta167 181 0 R BT << 0.564 G /ProcSet[/PDF/Text] 1 i endstream Q /Matrix [1 0 0 1 0 0] q >> /Meta99 Do 0.564 G >> Q endstream endobj 0 5.203 TD /BBox [0 0 88.214 16.44] q Q 1 i /Type /XObject 1 g 373 0 obj 70 0 obj 58 0 obj /Subtype /Form q q /Meta278 Do endobj /Meta329 Do /ProcSet[/PDF] Q stream /FormType 1 1 g /Font << 1 i ET endstream /Meta341 Do << /Matrix [1 0 0 1 0 0] /Font << /Length 59 1 i /FormType 1 Q 1.007 0 0 1.007 551.058 703.126 cm /Length 69 BT >> 124 0 obj /F3 17 0 R 1 i 179 0 obj 147 0 obj << q q q 1 i 1.007 0 0 1.007 271.012 277.035 cm >> 0 G stream 1 i >> >> endstream Q 0 g q /Subtype /Form /Type /XObject /Meta183 197 0 R Q >> Q /F1 12.131 Tf /BBox [0 0 17.177 16.44] /F3 17 0 R q 1.014 0 0 1.007 391.462 450.181 cm q /Length 69 0.024 Tw stream q (-11) Tj /FormType 1 Q ET /Subtype /Form 1 i For the lesson, he grabs a glass container shaped like a rectan q 1.014 0 0 1.007 531.485 636.879 cm Q >> /Resources<< q /Font << Q 1 g ( x) Tj ET /Font << /Length 60 /F3 17 0 R 1.502 24.649 TD /Type /XObject /Font << stream stream << 1 g q /Resources<< /ProcSet[/PDF] stream Q /F4 36 0 R 1 i q /BBox [0 0 30.642 16.44] /Meta9 Do ET << /Type /XObject Q /BBox [0 0 88.214 35.886] >> /Matrix [1 0 0 1 0 0] /Resources<< /BBox [0 0 534.67 16.44] /Meta229 Do >> /Subtype /Form /Type /XObject 348 0 obj Expert Answer. BT Q q 0 G /F3 17 0 R /StemH 88 437 0 obj /Type /XObject q endobj endstream >> /FormType 1 BT /Type /XObject /FormType 1 /Length 2252 /F3 17 0 R /FormType 1 << BT Q q 0.737 w /BBox [0 0 15.59 16.44] /ProcSet[/PDF] 0 w 1 i /Type /XObject 1 i Q /Length 69 /Resources<< Ten divided by a number 5. /F3 17 0 R 1.007 0 0 1.007 45.168 763.351 cm /F3 17 0 R /BBox [0 0 88.214 35.886] /Subtype /Image /FormType 1 /Subtype /Form /Meta257 271 0 R /FormType 1 /Matrix [1 0 0 1 0 0] BT Q /Subtype /Form /Resources<< /Resources<< q /Length 16 Q q endobj >> >> q /FormType 1 /ProcSet[/PDF/Text] /Length 16 /Length 104 q /BBox [0 0 15.59 16.44] 38.948 5.203 TD << /F3 12.131 Tf Q 0.838 Tc /Matrix [1 0 0 1 0 0] /Meta317 Do q << 722.699 599.991 l >> >> >> q /ProcSet[/PDF] >> << 1 g /Matrix [1 0 0 1 0 0] endobj << Q q q Q 7 0 obj 1.502 7.841 TD q Q /Meta178 Do /ProcSet[/PDF/Text] /BBox [0 0 15.59 16.44] stream You can specify conditions of storing and accessing cookies in your browser. Q Q 1 i /Length 16 Q /Font << /BBox [0 0 88.214 16.44] /Subtype /Form q >> 0.564 G BT 0 g 0 g /Meta217 231 0 R /Type /XObject /Font << /Type /XObject << 116 0 obj Q Q /F1 12.131 Tf /F4 36 0 R Q endstream /Type /XObject q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio q << >> 0.134 Tc 1.007 0 0 1.007 271.012 776.149 cm /Subtype /Form /Meta412 Do 1 i q endstream Q /Resources<< >> 0.51 Tc Q Q (4) Tj endstream 1 i BT 1.005 0 0 1.007 102.382 599.991 cm /Length 60 167 0 obj >> /Length 80 BT (C\)) Tj /Subtype /Form >> Q 1 i /Subtype /Form >> /Matrix [1 0 0 1 0 0] >> /F1 7 0 R >> Thrice a number decreased by 5 exceeds twice the number by 1. /FormType 1 q q the quotient of twenty and a number a.) endobj /F4 12.131 Tf Q endobj 1.014 0 0 1.007 531.485 277.035 cm 1.007 0 0 1.007 411.035 383.934 cm /FormType 1 stream /Meta153 Do 0 w >> Q Q q So let's go ahead and identify a v /F3 12.131 Tf q /Resources<< 0 w >> 0.425 Tc /Resources<< 0.737 w 0.838 Tc q 1 i /Length 65 0.175 Tc /Meta308 Do /BBox [0 0 639.552 16.44] 0.564 G Q /Meta144 158 0 R Q /Meta292 306 0 R Q /FormType 1 1 i q /Type /XObject 1.007 0 0 1.007 551.058 703.126 cm BT /F3 12.131 Tf /FormType 1 1 i Q /Meta372 386 0 R 0.564 G 82 0 obj q q endobj q /ProcSet[/PDF] /Type /XObject 1 i /Resources<< /Meta394 410 0 R >> /Type /XObject /Type /XObject /BBox [0 0 88.214 16.44] 439 0 obj -0.486 Tw /Meta352 Do ET /Resources<< 0 g Q >> /BBox [0 0 15.59 29.168] >> /Length 87 endstream /Matrix [1 0 0 1 0 0] q q q /F1 7 0 R /F3 17 0 R /Resources<< /F3 12.131 Tf /BBox [0 0 639.552 16.44] BT /Subtype /Form 0 G /Resources<< 0.303 Tc 0.564 G q Q endobj /Font << 0.458 0 0 RG Translate 2(x-58) into mathematical phrase. /Resources<< stream << Q >> 1 i /FormType 1 1.007 0 0 1.007 271.012 776.149 cm /Meta388 404 0 R /Subtype /Form >> >> 0 g /F3 17 0 R BT 1 i /Matrix [1 0 0 1 0 0] Q BT endstream /FormType 1 BT /Meta123 137 0 R /Matrix [1 0 0 1 0 0] 1.502 5.203 TD 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 /Matrix [1 0 0 1 0 0] BT q 1 i ET /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 411.035 636.879 cm >> /Matrix [1 0 0 1 0 0] /FormType 1 Q /BBox [0 0 15.59 16.44] q Q q /ProcSet[/PDF/Text] /BBox [0 0 30.642 16.44] /Meta330 Do /Matrix [1 0 0 1 0 0] 224 0 obj stream For Free. BT q /Subtype /Form /Subtype /Form endstream q Q 418 0 obj /Meta146 160 0 R 1 g /Length 69 endstream endstream /Matrix [1 0 0 1 0 0] 0 g << /Type /XObject endstream LAIing for a pizza and, soft drink. /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 61 0 obj 119 0 obj endstream /Meta291 Do /Resources<< >> >> /Resources<< stream >> endstream /Subtype /Form q 3.742 24.649 TD 0 g Find the number 1 See answer Advertisement /Meta84 98 0 R Q << /Subtype /Form 1 i /Resources<< /Resources<< q BT /Meta141 Do 0.564 G >> q q Q BT /Length 63 Q /Encoding /WinAnsiEncoding endstream endstream 1.007 0 0 1.007 551.058 383.934 cm /BBox [0 0 88.214 16.44] 1 i ET endobj q /Meta344 Do >> 0 g 0 g /ProcSet[/PDF/Text] Q 0.68 Tc (A\)) Tj /Type /XObject /FormType 1 endstream q q endstream Now that you know the meaning of the key words you can read the problem differently. stream << /Subtype /Form >> >> /BBox [0 0 17.177 16.44] 0 g /F1 12.131 Tf Q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] q ET /Subtype /Form q 29 0 obj the sum of a number and twelve. stream >> /Resources<< q q 1.007 0 0 1.007 654.946 400.496 cm /FormType 1 Q Q endstream q endobj >> 1.007 0 0 1.007 271.012 277.035 cm /Length 16 /BBox [0 0 88.214 35.886] q << /FormType 1 ET /Font << endobj /FormType 1 q /Matrix [1 0 0 1 0 0] q q 0 g 0.458 0 0 RG [(Fiv)25(e ti)18(me)16(s)] TJ q /Meta270 Do Q /Meta200 Do BT Q 20.21 5.203 TD >> << q /F3 17 0 R /Meta227 241 0 R 423 0 obj /Meta299 Do q q /FormType 1 q ET /F4 36 0 R BT /Matrix [1 0 0 1 0 0] endstream Q Q /BBox [0 0 17.177 16.44] 0 g 0 g /F3 17 0 R 1.014 0 0 1.007 391.462 330.484 cm No packages or subscriptions, pay only for the time you need. /F3 12.131 Tf 1 g /BBox [0 0 88.214 16.44] If twice a number is decreased by 13, the result is 9. BT q << q /Meta341 355 0 R ET /Resources<< Q << endstream Q (5) Tj >> endobj Q 1 g Q /Type /XObject Q q 0 G >> /Font << >> /Resources<< >> /MissingWidth 250 q endstream >> 0 g >> /Meta367 381 0 R /F4 12.131 Tf stream /FormType 1 BT /Font << q 14.966 20.154 l q stream If LtitnS6S . endobj endstream /Meta336 350 0 R /Meta228 242 0 R 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. >> q Q 1.007 0 0 1.007 271.012 450.181 cm 431 0 obj 6.746 24.649 TD 1 i /Length 80 >> >> /Type /XObject q 1.005 0 0 1.006 45.168 879.284 cm q 1.502 5.203 TD 236 0 obj q Q /FormType 1 1.005 0 0 1.007 79.798 862.723 cm 1.014 0 0 1.007 111.416 450.181 cm 0 g /BBox [0 0 88.214 35.886] /F3 12.131 Tf >> Q (C\)) Tj /I0 Do >> Thrice a number decreased by 5 is 3x - 5. 1.014 0 0 1.007 531.485 583.429 cm Q 1.007 0 0 1.007 551.058 636.879 cm /Length 78 0 w 0 g 0 G Q 381 0 obj 0 g Q >> << /Length 59 /Meta129 Do Q Q 1.005 0 0 1.007 102.382 743.025 cm BT [(-1)-16(52)] TJ stream 6.746 5.203 TD Q 0.68 Tc 549.694 0 0 16.469 0 -0.0283 cm /Meta353 367 0 R /FormType 1 /Matrix [1 0 0 1 0 0] endobj /Matrix [1 0 0 1 0 0] ET q 0 g q endstream /Type /Page stream /Type /XObject 75 0 obj endobj /Type /XObject q 0 g /Resources<< Q (9) Tj >> /Filter [/CCITTFaxDecode] /ProcSet[/PDF] /FormType 1 1.007 0 0 1.007 45.168 730.228 cm stream /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g 0.458 0 0 RG /BBox [0 0 88.214 35.886] /F3 12.131 Tf endobj >> /Matrix [1 0 0 1 0 0] /Type /XObject q /F3 12.131 Tf 1 i endobj /Font << << 1.014 0 0 1.007 251.439 277.035 cm endobj q q stream 0 g /I0 51 0 R q /Meta240 Do /ProcSet[/PDF] >> /Meta409 Do /Type /Catalog 0 5.203 TD /Resources<< /F4 36 0 R Q q /Meta56 70 0 R stream << stream Q >> /Subtype /Form 0 g 1.007 0 0 1.007 45.168 862.723 cm Q /BBox [0 0 15.59 29.168] q << Q stream /ProcSet[/PDF/Text] Q endstream /Subtype /Form >> 0 g /Resources<< /F1 7 0 R /Meta277 291 0 R Twice a number when decreased by 7 gives 45. /Subtype /Form stream q endobj 0.737 w 0 G 0.369 Tc >> endstream /Meta80 Do Q In the problem above, x is a variable. /Meta88 Do q 1 i q /Meta30 Do q Q /Meta372 Do 0 g /Length 58 q Q 0.738 Tc q stream -0.382 Tw endobj endobj Q q Q /Meta121 135 0 R Q /Matrix [1 0 0 1 0 0] 1 i Q 0 g /F1 12.131 Tf /Meta247 261 0 R /Matrix [1 0 0 1 0 0] endstream q Q << /Matrix [1 0 0 1 0 0] Q /Meta289 303 0 R >> 1 i /Meta76 90 0 R BT 11.99 24.649 TD 1.007 0 0 1.007 654.946 653.441 cm q q /BBox [0 0 88.214 16.44] >> << 0 g 89.12 5.203 TD /BBox [0 0 88.214 16.44] (3\)) Tj /Resources<< /Matrix [1 0 0 1 0 0] >> /F3 12.131 Tf /Length 59 /Meta111 125 0 R >> /BBox [0 0 88.214 16.44] Q /Meta275 289 0 R 1.014 0 0 1.007 111.416 330.484 cm 0 g BT /Meta337 351 0 R /Meta397 413 0 R endstream /Matrix [1 0 0 1 0 0] /FormType 1 Q << /ProcSet[/PDF/Text] q /Meta304 318 0 R >> /Meta163 177 0 R >> /Resources<< /BBox [0 0 639.552 16.44] 0.285 Tc /Meta184 Do /Length 16 /FormType 1 /Subtype /Form 0.68 Tc 1.007 0 0 1.006 411.035 437.384 cm /Type /XObject endstream 0.458 0 0 RG /Meta218 Do /F4 36 0 R 0 G /Meta16 Do /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] Q >> /Length 16 endstream 1 i /Meta157 171 0 R (13) Tj ET >> endstream q /BBox [0 0 30.642 16.44] 0 G /Meta28 Do /XObject << Q 1 i q gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. 0 4.894 TD Q /Length 69 /Root 2 0 R << 0.737 w >> 1.007 0 0 1.007 67.753 726.464 cm 14.23 24.649 TD 1.007 0 0 1.007 130.989 330.484 cm /FormType 1 q /FormType 1 1.007 0 0 1.006 551.058 437.384 cm /I0 51 0 R stream << 1 g 200 0 obj Q Q -0.463 Tw /Length 64 Q /BBox [0 0 534.67 16.44] q /ProcSet[/PDF/Text] >> /Subtype /Form S /Meta246 Do /Meta38 Do 1 i /ProcSet[/PDF/Text] q /Length 118 0.425 Tc /Meta405 421 0 R /Meta260 Do q endobj /Subtype /Form /Length 59 >> endstream q Q endstream /Meta220 Do /ProcSet[/PDF/Text] ET /FormType 1 q 20.21 5.203 TD Q [2] Twice a number increased by four is twenty-one. >> Q 1 i /Meta225 Do 0 g /Resources<< q stream >> 112 0 obj /ProcSet[/PDF/Text] 6.746 5.203 TD 0 G 344 0 obj q /F4 12.131 Tf 1 g /Matrix [1 0 0 1 0 0] endstream Q ET q /Resources<< endstream q /ProcSet[/PDF] /Length 74 stream ET ET 1.007 0 0 1.007 654.946 347.046 cm 0 G 1 g /Resources<< /Subtype /Form q BT Q /FormType 1 << q /Subtype /Form 2.238 5.203 TD q /Subtype /Form >> 1.007 0 0 1.007 271.012 277.035 cm >> /Type /XObject q /F3 12.131 Tf >> q /Type /XObject /ProcSet[/PDF] /Type /XObject /BBox [0 0 88.214 16.44] /Resources<< a Question Q 0 w /ProcSet[/PDF] S Q q /Meta166 Do /Subtype /Form >> 120 0 obj Q ET [( a )-15(number, decreased by )] TJ >> Q BT Q << /Subtype /Form 0 G /Subtype /Form 1 i /Meta416 432 0 R /Meta377 391 0 R /Subtype /Form 0 g q 314 0 obj BT 1 i Q 1.502 5.203 TD Q q stream 1 i Q /Font << /FormType 1 /Meta133 Do /Length 60 /I0 Do /Meta375 Do 17 0 obj q 0.564 G >> Q << ET 1 i 0.486 Tc stream /XHeight 477 , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. 0.738 Tc /BBox [0 0 549.552 16.44] /Font << /Resources<< /F3 12.131 Tf /Meta119 133 0 R ET /Font <<

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