twice a number decreased by 58

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

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/ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> q 0 G /Length 68 stream >> endobj /Length 16 /Length 59 321 0 obj stream q /Length 70 /Matrix [1 0 0 1 0 0] Q q endobj /Meta184 Do /Length 59 /Font << 1 i >> 0 5.203 TD q /Meta168 Do Q q endstream q /ProcSet[/PDF/Text] /Meta290 304 0 R /Meta140 Do /ProcSet[/PDF] >> 20.21 5.203 TD 1.007 0 0 1.007 654.946 799.486 cm ET Q /FormType 1 /Font << /Type /XObject 0 w /Subtype /Form >> /Resources<< /ProcSet[/PDF/Text] 1.007 0 0 1.006 551.058 836.374 cm /FormType 1 /F1 12.131 Tf Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM /Meta5 Do Q q 0.737 w /Subtype /Form Q 336 0 obj 0 g Q Q 0.786 Tc /F4 36 0 R endstream /Type /XObject Q /Meta287 Do endstream /F3 17 0 R 1 i 1 i 0 5.203 TD >> endobj q << /Font << /Matrix [1 0 0 1 0 0] Q Q 0 w 328 0 obj q << << 318 0 obj /Font << >> /Resources<< << q << >> Q >> /Font << 1 i /Font << Q Q Q Q Q q /F3 12.131 Tf >> >> >> /Matrix [1 0 0 1 0 0] 103 0 obj /Length 59 q ET /Length 69 /Subtype /Form The sum of 18 and tour times a number is -6 Find the number. stream q Q q q /Meta66 Do /Meta335 Do /Resources<< stream /Length 63 endobj 54 0 obj , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. stream >> /Meta358 Do /F3 17 0 R /FormType 1 /Meta233 247 0 R /Type /XObject /ProcSet[/PDF] endobj Q /Resources<< 0.737 w /Matrix [1 0 0 1 0 0] 0.737 w Q stream BT /F3 17 0 R /FormType 1 << BT q 239 0 obj BT /Length 54 endobj q Q /Type /XObject (D) Tj endstream /Type /XObject /F3 12.131 Tf << << stream /Resources<< /Meta187 201 0 R >> endstream /ProcSet[/PDF/Text] endstream /Meta37 50 0 R /Matrix [1 0 0 1 0 0] stream /Type /XObject stream endobj stream /Resources<< /Length 78 /Leading 150 Q /FormType 1 endstream >> endobj q 1.005 0 0 1.007 102.382 546.541 cm q >> q Q /F3 17 0 R 0 g >> q /FormType 1 Step 1/1. BT /BBox [0 0 17.177 16.44] /Resources<< 1 g << Q << 1.005 0 0 1.007 79.798 746.789 cm 0 G q q Q /F3 17 0 R 1.007 0 0 1.007 551.058 583.429 cm Q stream 32 0 obj endstream 1.502 5.203 TD q Q q endstream q 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. 1 g q 0 g /Matrix [1 0 0 1 0 0] /Subtype /Form << /Type /XObject /FormType 1 q If a number is 400%, then it is 4 times, the same as 4. stream 1.007 0 0 1.006 411.035 690.329 cm /BBox [0 0 88.214 16.44] /Font << 35.206 4.894 TD /Length 69 q q /Font << endobj 0 g 0 g endstream 16.469 5.336 TD BT 0.458 0 0 RG (x) Tj /Resources<< /Resources<< /Meta251 265 0 R Q q /FormType 1 /Meta76 Do 1 i 0 4.894 TD Q /Length 69 >> << /ProcSet[/PDF/Text] << >> 672.261 653.441 m endstream >> Q /ProcSet[/PDF/Text] 252 0 obj /Resources<< << 20.21 5.336 TD /Meta31 Do /Type /XObject 0 g >> /F3 12.131 Tf That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. Q Q /Meta198 Do Q 0 5.203 TD endobj /Type /XObject /Subtype /Form q << 672.261 400.496 m BT << /Length 16 q q 79 0 obj /Resources<< >> /Resources<< q -y. >> Q Q 0 g stream q /FormType 1 Q /Resources<< /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] q (-) Tj << /Type /XObject /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] >> ET Q 0 5.203 TD (-20) Tj /Length 16 0.369 Tc 722.699 726.464 l >> BT stream /Type /XObject stream endobj /Resources<< 1.007 0 0 1.007 551.058 277.035 cm /Type /XObject 1 i q 1 i Q Q Q q Abstract: The aim of the study was to investigate the expression of miR-155 in plasma and peripheral blood mononuclear cells (PBMCs), the effects of miR-155 on the apoptosis rate /Type /XObject stream 1 i Q << Q << 112 0 obj 1.008 0 0 1.007 654.946 293.596 cm /Resources<< Q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /Meta31 44 0 R Q q /FormType 1 /Length 16 >> /BBox [0 0 88.214 16.44] << /Length 69 0 g 1 g /Font << Q 0.564 G /Subtype /Form Q Q << 582 546 601 560 395 424 326 603 565 834 516 556]>> endobj /ProcSet[/PDF/Text] /F4 36 0 R q q stream /Subtype /Form >> 0 5.203 TD >> stream 1 i Q 0 g /Subtype /Form 1 g 0.198 Tc /Length 59 1 i /F3 17 0 R /Type /XObject (B\)) Tj /FormType 1 /Meta425 441 0 R q >> /Type /XObject /Meta404 Do /ProcSet[/PDF] /ProcSet[/PDF] /Length 81 /F1 12.131 Tf stream Q /F3 12.131 Tf 0 5.203 TD 285 0 obj Q q q /Type /XObject >> /Matrix [1 0 0 1 0 0] 0 g /FormType 1 /Resources<< /Matrix [1 0 0 1 0 0] /Meta302 Do /Resources<< /Resources<< 0.458 0 0 RG /Meta159 Do q >> /Matrix [1 0 0 1 0 0] BT 1 g Q Twice a first number decreased by a second number is 6. q q , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. endobj >> 244 0 obj q /Matrix [1 0 0 1 0 0] 163 0 obj 0 G 0 g /Length 69 /Meta202 216 0 R >> q 19 0 obj 0 g /F4 36 0 R (x) Tj /BBox [0 0 88.214 16.44] endobj /F3 17 0 R /FormType 1 /F3 17 0 R endstream /Font << ET 1 i /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] stream BT 1 i Q >> /Meta292 306 0 R >> /ProcSet[/PDF] /BBox [0 0 88.214 16.44] stream BT ET >> 260 0 obj q endobj 0 w Q q Q /Resources<< Diabetes, if left untreated, leads to many health complications. /Resources<< /Meta418 Do 0.737 w /Resources<< 549.694 0 0 16.469 0 -0.0283 cm /Subtype /Form >> 0.51 Tc /Font << /Type /XObject >> Q /Length 16 293 0 obj 1.007 0 0 1.007 45.168 779.913 cm endstream /Font << 1 i /ProcSet[/PDF/Text] /Type /XObject Q /Meta281 295 0 R /Matrix [1 0 0 1 0 0] /Meta392 408 0 R >> /FormType 1 >> q 404 0 obj /F3 17 0 R /Font << /Resources<< << endstream /Resources<< /ProcSet[/PDF/Text] 1.007 0 0 1.007 67.753 872.509 cm 1.007 0 0 1.007 67.753 293.596 cm endobj stream 85 0 obj Q 341 0 obj /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Meta99 Do /Type /XObject 0 G /Resources<< >> << /Matrix [1 0 0 1 0 0] /Resources<< /Resources<< /Meta315 329 0 R 1.502 5.203 TD /Resources<< >> 0 g /ProcSet[/PDF] q endstream 1.007 0 0 1.007 271.012 849.172 cm ET /Matrix [1 0 0 1 0 0] /Resources<< >> /BBox [0 0 88.214 16.44] >> 1.014 0 0 1.006 391.462 836.374 cm /F3 12.131 Tf Calculate a 15% decrease from any number. /F3 17 0 R endobj ET /Subtype /Form Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. /Subtype /Form q q 0 g >> /F4 36 0 R 0.458 0 0 RG 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 /Font << /Resources<< /F3 17 0 R >> 1 i /Resources<< /Length 63 "49 . /Matrix [1 0 0 1 0 0] Q << /BBox [0 0 88.214 16.44] endstream /ProcSet[/PDF] You could call them, Decreased by another number means subtract. q stream Q q 1 i endstream << Q Q 1.014 0 0 1.007 531.485 703.126 cm /Resources<< q /BBox [0 0 88.214 16.44] q /Type /XObject 0.564 G /Length 116 1 i q /Meta126 140 0 R 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. 1 i /Type /XObject q endstream /Resources<< 119 0 obj /Font << /Font << Q (x) Tj >> q BT Q stream /Type /XObject /Matrix [1 0 0 1 0 0] Q /FormType 1 /Subtype /Form /Encoding /WinAnsiEncoding /Font << 0.738 Tc 0.564 G 1 g Q endobj 0 G /Type /XObject 37 0 obj q << Q << endstream q q 96 0 obj /Font << /FormType 1 Q << /Length 59 [( and )16(a nu)26(mbe)18(r)] TJ ET -0.463 Tw q The quotient of a seven and a number 9. 0 g 192 0 obj /Meta289 303 0 R /Type /XObject /Subtype /Form 0 g /BBox [0 0 534.67 16.44] /FormType 1 /Matrix [1 0 0 1 0 0] /FormType 1 /Subtype /Form Q /Subtype /Form << /Matrix [1 0 0 1 0 0] Q /Meta130 144 0 R Q /Subtype /Form q >> /Matrix [1 0 0 1 0 0] Use the variable g to represent Gails age. >> -0.067 Tw 0.458 0 0 RG stream 6.746 5.203 TD /Matrix [1 0 0 1 0 0] /Resources<< /Resources<< /Meta366 380 0 R >> /BBox [0 0 88.214 35.886] /FormType 1 /ProcSet[/PDF/Text] stream Q Q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 583.429 cm /Meta297 311 0 R 94.364 5.203 TD /Matrix [1 0 0 1 0 0] stream 0 G 343 0 obj 9.723 5.336 TD /Length 16 Q /F4 12.131 Tf 0.564 G q /Meta427 443 0 R /BBox [0 0 15.59 16.44] q /F3 17 0 R q /F3 17 0 R 1.007 0 0 1.006 551.058 763.351 cm q q 3.742 5.203 TD endobj 0 5.203 TD q Q /Meta2 Do ET /Type /XObject endstream /Meta6 Do 0 5.203 TD /F1 7 0 R /Matrix [1 0 0 1 0 0] stream 16.469 5.336 TD /Resources<< 369 0 obj 1 i /Length 12 >> endobj Q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] q /Resources<< (-8) Tj << q stream /BBox [0 0 88.214 16.44] 0.369 Tc /Type /XObject /Subtype /Form /BBox [0 0 534.67 16.44] q /Type /XObject Q /Font << endstream /Meta221 Do /Subtype /Form q q stream Q /Length 16 a and b or something else.***. /ProcSet[/PDF/Text] 1.014 0 0 1.006 251.439 690.329 cm /Matrix [1 0 0 1 0 0] BT /Matrix [1 0 0 1 0 0] << /Meta103 Do 0 5.203 TD Q /Length 69 0.786 Tc endobj /Type /XObject /Type /XObject >> (-) Tj q /Subtype /Form /ProcSet[/PDF] /F3 12.131 Tf endstream 0 g Q /BBox [0 0 15.59 16.44] Q /Subtype /Form /Type /XObject 1 i /Length 68 /F3 12.131 Tf /BBox [0 0 88.214 16.44] Q /Length 69 0 g >> endobj Q 0.37 Tc /Subtype /Form stream << /Meta204 Do endobj q Q /F3 12.131 Tf 1 i /Length 69 43.426 5.203 TD 0 g q stream 0.737 w 0 G /Matrix [1 0 0 1 0 0] /Meta216 230 0 R /Matrix [1 0 0 1 0 0] >> 0.458 0 0 RG 124 0 obj (1\)) Tj -0.126 Tw endobj 403 0 obj << 0 G 0.737 w endstream q q 0.458 0 0 RG /ProcSet[/PDF] BT /Meta415 Do 0 g /BaseFont /PalatinoLinotype-Bold ET endobj q Q /F4 12.131 Tf 1.007 0 0 1.006 551.058 437.384 cm /FormType 1 0.68 Tc endstream BT The sum of a number and 2 is 6 less than twice that number. /Subtype /Form 1 i /BBox [0 0 88.214 16.44] q q >> 259 0 obj Q /ProcSet[/PDF/Text] ET 1 i q /Matrix [1 0 0 1 0 0] Q Q /F3 17 0 R Q 0 g /Subtype /Form 0 G stream 1.005 0 0 1.007 45.168 916.925 cm /MediaBox [0 0 767.868 993.712] 0 g >> /F3 12.131 Tf Q Q >> /ProcSet[/PDF] endstream The observed mean MetS-Z was at inclusion 0.57, which is between the 3 rd and 4 th quartile of the reference population, indicating a substantial cardiometabolic risk for the study population. /F3 12.131 Tf /Subtype /Form >> /FormType 1 1 i /ProcSet[/PDF] 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 . /FormType 1 stream /Resources<< << >> /Meta286 300 0 R /Font << /I0 51 0 R Let x the unknown number. 549.694 0 0 16.469 0 -0.0283 cm endstream << Q /Resources<< /Meta83 Do 396 0 obj 1.005 0 0 1.007 102.382 293.596 cm 0.458 0 0 RG >> /Meta123 Do 0 g Q /Type /XObject q /FormType 1 -0.382 Tw /Length 54 q ET q 1 i endobj /ProcSet[/PDF] 0 G q Q 0 g /ProcSet[/PDF/Text] 1 g endobj q 240 0 obj q 0 g >> /FormType 1 0 g q stream 1 i /Meta203 217 0 R q endobj >> (B\)) Tj /BBox [0 0 30.642 16.44] >> >> 1 i /Length 59 q /Subtype /Form /Resources<< Q q >> endobj /Resources<< /Resources<< Q 1.007 0 0 1.007 551.058 703.126 cm /F3 17 0 R Q 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. /BBox [0 0 88.214 16.44] /Length 69 0 g /ProcSet[/PDF] For Free. stream /Meta278 Do q 0 w /BBox [0 0 30.642 16.44] /Meta155 Do /Subtype /Form /Type /XObject /Filter [/CCITTFaxDecode] /F3 12.131 Tf Q 1.007 0 0 1.007 271.012 523.204 cm Q /Meta292 Do 0.564 G Q /Meta165 179 0 R /Meta157 Do /Matrix [1 0 0 1 0 0] << endobj 1.007 0 0 1.007 45.168 862.723 cm stream /Type /XObject << 0 w Q 243 0 obj 1 i /BBox [0 0 549.552 16.44] /Meta363 377 0 R >> /XObject << Q q 412 0 obj /Type /FontDescriptor 1.007 0 0 1.007 411.035 330.484 cm 0 g 395 0 obj /Subtype /Form /F3 12.131 Tf 0 G /Resources<< 0.737 w /Meta149 Do q /Resources<< stream /Font << Q 359 0 obj >> q /FormType 1 /Font << endobj Q /Subtype /Form >> >> Q Q /Font << endobj /FormType 1 0 g /F3 17 0 R /ProcSet[/PDF/Text] endstream >> q 1.014 0 0 1.007 391.462 849.172 cm /ProcSet[/PDF] Q << >> 10 0 obj ET endobj gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. BT BT << Q /Length 245 Q 0.458 0 0 RG q (B) Tj q /Length 59 /Meta58 Do 436 0 obj /Subtype /Form 1 i /Font << /F3 12.131 Tf /Resources<< 27.693 5.203 TD q /Matrix [1 0 0 1 0 0] /FormType 1 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 ET /Resources<< q q Q (8\)) Tj ET q q /Resources<< q /XObject << 1.014 0 0 1.006 531.485 510.406 cm 1 i 0 g /ProcSet[/PDF/Text] 18 0 obj Q 1.005 0 0 1.007 102.382 872.509 cm q 1.007 0 0 1.007 551.058 383.934 cm Q << /F3 17 0 R q /Resources<< /Subtype /Form 0 G /Meta263 Do /Resources<< Q q 0 G 277 0 obj endobj /Length 16 Q /F3 17 0 R >> q 253 0 obj 0 G /Subtype /Form /FormType 1 /Meta26 Do << 0 g /Meta67 81 0 R /FormType 1 /FormType 1 /Meta250 264 0 R 0 G /BBox [0 0 15.59 16.44] 9.723 5.336 TD endstream /F3 17 0 R q endobj Q /FormType 1 /Font << Q q 1 g endobj 0 5.203 TD Q /Meta189 203 0 R /Type /XObject 0 g Q Q Q q q /ProcSet[/PDF/Text] /F3 12.131 Tf 0 g /Length 68 /Type /XObject ET 0 g >> endobj stream ET >> /Subtype /TrueType /FormType 1 q 0 g /BBox [0 0 30.642 16.44] /FontName /PalatinoLinotype-Roman /ProcSet[/PDF/Text] endobj /Subtype /Form /Subtype /Form q /F3 12.131 Tf q 0.369 Tc /Meta206 220 0 R Q /Length 69 /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 69 Q q ET Q /Resources<< Decreased by another number means subtract. (13) Tj /Meta191 205 0 R Q 179 0 obj Select the correct mathematical statement for the following equation. endobj << Q 0.564 G >> << 35,000 worksheets, games, and lesson plans, Spanish-English dictionary, translator, and learning, **Note: You could choose any variable you want, to represent the numbers. q /F4 12.131 Tf /Font << q q /F3 12.131 Tf 0 5.203 TD q q 21 0 obj 87 0 obj 1.007 0 0 1.007 654.946 653.441 cm >> /Meta36 Do /Meta78 Do (4) Tj /FormType 1 /Font << (11) Tj 1 i Q /Meta422 Do /FormType 1 /Meta356 370 0 R /Length 74 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /Type /XObject /ProcSet[/PDF] q /BBox [0 0 534.67 16.44] 0 g Q 1 g BT /ProcSet[/PDF] /Leading 253 ET q >> << /FormType 1 >> /FontName /PalatinoLinotype-Bold /Matrix [1 0 0 1 0 0] endstream >> >> 65 0 obj 0 w >> Q q 0.155 Tc /Meta149 163 0 R q 1 i 0 G /Type /XObject stream /F1 7 0 R ET 20.21 5.203 TD -0.16 Tw 0.564 G 1 i 0 G /BBox [0 0 88.214 16.44] /ProcSet[/PDF] endstream /FormType 1 17.234 5.203 TD endstream endobj q >> 297 0 obj Q /F1 7 0 R /F3 12.131 Tf >> /FormType 1 /Meta154 Do /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] endobj /Meta136 150 0 R 0.458 0 0 RG q /Resources<< stream 1.007 0 0 1.007 45.168 829.599 cm stream 0 G endobj q /Font << /Length 16 stream /F1 14.682 Tf /Type /XObject /Type /XObject << BT endstream /Subtype /Form /BBox [0 0 673.937 27.581] [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. (C\)) Tj q Q >> /Resources<< /Resources<< Q ET (\)) Tj /Font << /Subtype /Form /F3 17 0 R q Translate 2(x-58) into mathematical phrase. BT Q 1 i /Subtype /Form /Type /XObject [(Fiv)25(e ti)18(me)16(s)] TJ Q << Q 1 i q q (x) Tj q /Meta108 122 0 R /Resources<< (-20) Tj << You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 find the number, 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. q q Q /Meta302 316 0 R q Q /BBox [0 0 15.59 29.168] /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] Q 1 i Medium /Length 54 /ProcSet[/PDF/Text] >> q >> >> endobj /ProcSet[/PDF/Text] 0 g >> 1 i 2. Q /Type /XObject /Subtype /Form /BBox [0 0 88.214 16.44] stream 0 g /Meta268 Do 12.727 5.203 TD /Resources<< ET 1.007 0 0 1.007 271.012 776.149 cm stream /F3 17 0 R Q /F3 12.131 Tf 101.849 5.203 TD 0 G 1 i Q q 0 w /Resources<< /Resources<< 0 5.203 TD /ProcSet[/PDF/Text] Q /Matrix [1 0 0 1 0 0] /Length 69 /Length 54 stream q /F3 17 0 R endobj 0.564 G 0.458 0 0 RG /BBox [0 0 15.59 29.168] /F4 36 0 R /Font << >> >> (4\)) Tj 0 w /Resources<< >> A number = an unknown number which can be represented by a variable, usually x. ET 1.005 0 0 1.007 102.382 400.496 cm endobj /Resources<< 1 g >> /ProcSet[/PDF] /Length 12 Q 224 0 obj Q Q /Subtype /Form q Q stream endstream /Subtype /Form /Meta312 Do (iii) 25 exceeds a number by 7. /Meta297 Do q /Length 118 1 i /Length 69 0 G /Type /XObject /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] endstream Q /Font << << /Type /XObject Q stream /ProcSet[/PDF/Text] Q q /Resources<< 0.737 w q << q /FormType 1 Q /Resources<< q 0.458 0 0 RG BT /Matrix [1 0 0 1 0 0] q endstream >> 0 g /Meta57 71 0 R >> /Meta329 Do /F3 12.131 Tf /FormType 1 BT /Resources<< q Q 1 i stream endobj /FormType 1 /Matrix [1 0 0 1 0 0] /Subtype /Form q /Subtype /Form 1 i 0.458 0 0 RG /ProcSet[/PDF] /ProcSet[/PDF] 0 w 0.458 0 0 RG q /Resources<< /Meta175 189 0 R q q /Subtype /Form /BBox [0 0 15.59 16.44] 0 G >> 1 i >> /BBox [0 0 30.642 16.44] Q /Type /XObject /F3 12.131 Tf (x ) Tj Q (58) Tj q 0.297 Tc Q /Length 104 BT 1.005 0 0 1.007 102.382 653.441 cm /Meta116 Do q >> /FormType 1 30.699 5.203 TD /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] /Resources<< stream 1.005 0 0 1.006 45.168 879.284 cm /Length 70 /Subtype /Form /BBox [0 0 88.214 16.44] q 0 g Q >> /MaxWidth 1248 q 1 g /Length 69 0 g >> /Matrix [1 0 0 1 0 0] endstream 0 G Q /Meta342 356 0 R >> ET 0.737 w /BBox [0 0 88.214 16.44] BT Q /Subtype /Form /Subtype /Form q >> /Length 69 q BT 0.297 Tc /ProcSet[/PDF/Text] /Meta215 Do /BBox [0 0 534.67 16.44] q >> /BBox [0 0 15.59 29.168] ET /FormType 1 S /BBox [0 0 88.214 16.44] Q 1 i endstream 0.297 Tc q 57 0 obj /Meta42 Do /BBox [0 0 88.214 16.44] /Meta147 161 0 R /BBox [0 0 88.214 35.886] /BBox [0 0 15.59 16.44] << endstream >> 378 0 obj /ProcSet[/PDF/Text] 1 i /Meta92 Do 0 G endstream 1 i 0 g 1.008 0 0 1.007 654.946 293.596 cm 0 G q /F4 12.131 Tf q /Matrix [1 0 0 1 0 0] /Length 118 Q Q /FormType 1 ET /Meta185 Do q 1.005 0 0 1.007 102.382 473.519 cm endobj Q 0 G /Resources<< /Resources<< 303 0 obj If LtitnS6S . q q 0 g BT << /Length 59 Q /Length 69 /FormType 1 endstream BT endstream /Length 69 endobj /Length 57 BT Q 1 i /Font << /ProcSet[/PDF] Q >> q q endstream 1 i << Q q Q /Meta24 Do 0.458 0 0 RG /Meta133 147 0 R q 0.564 G >> 242 0 obj /ItalicAngle 0 /Type /XObject /Type /XObject /BBox [0 0 88.214 16.44] Q endstream /Matrix [1 0 0 1 0 0] If a number is 50%, then it is a half - the same as 0.5 or 1/2. endstream Q /Meta264 Do /FormType 1 /Length 69 1.014 0 0 1.006 391.462 437.384 cm 0 5.203 TD /BBox [0 0 534.67 16.44] BT /Meta386 402 0 R /BBox [0 0 88.214 35.886] 97 0 obj >> /Meta37 Do 0.31 Tc /Length 119 0 g Q Q /Meta124 Do endstream 0.737 w S /Type /XObject q stream 1.014 0 0 1.006 531.485 690.329 cm Q Q << 376 0 obj [( a )-15(number, decreased by )] TJ 0 5.203 TD >> 0 g 0.738 Tc ET endobj /Resources<< /Subtype /Form q 171 0 obj 0.458 0 0 RG /Type /XObject q >> 1 i 1.007 0 0 1.007 130.989 277.035 cm Q /Meta177 191 0 R Q >> (+) Tj /F3 17 0 R No packages or subscriptions, pay only for the time you need. Q 1 i /Font << Q 1 g >> /F3 12.131 Tf ET /Length 16 /Matrix [1 0 0 1 0 0] /Meta194 208 0 R Q /Meta393 409 0 R 0.737 w << >> Q Q /Meta200 214 0 R /BBox [0 0 15.59 16.44] /Meta351 Do endstream /F3 12.131 Tf /ProcSet[/PDF/Text] q /Meta240 254 0 R endstream endobj 1.005 0 0 1.007 79.798 813.037 cm endobj /Meta89 Do Next, the problem says that "x" would be equal to twice a number added by 5. /Type /XObject 1 g /F3 17 0 R /FormType 1 stream /BBox [0 0 15.59 16.44] Q 0 G /F4 36 0 R Q Find the number 1 See answer Advertisement >> 1.007 0 0 1.007 654.946 726.464 cm 0 w stream 130 0 obj Q 0 g Q endobj /Type /XObject 1 i 1.007 0 0 1.007 271.012 330.484 cm 156 0 obj >> 1.007 0 0 1.007 67.753 347.046 cm 1.007 0 0 1.007 551.058 330.484 cm stream /Matrix [1 0 0 1 0 0] /Meta286 Do endobj /F3 12.131 Tf 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject >> /Type /XObject q q /Meta239 Do 20.21 5.203 TD endstream /BBox [0 0 534.67 16.44] /Meta111 Do q endobj /Meta161 Do endobj 0.458 0 0 RG 250 0 500 500 500 500 500 500 0 0 500 500 278 0 0 0 /Meta280 Do q /Resources<< q 0 w q >> stream [(1)-25(0\))] TJ /Type /XObject /Meta46 Do 1.014 0 0 1.007 251.439 277.035 cm q q 0 g saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. Q >> /Meta29 42 0 R 1 g /FormType 1 BT /ProcSet[/PDF] BT 0.68 Tc 1 i /Type /XObject Q /BBox [0 0 639.552 16.44] q Q >> << endobj endobj /Length 69 ET /FormType 1 /ProcSet[/PDF] >> q << /FormType 1 /Type /XObject >> >> /BBox [0 0 88.214 16.44] endstream << q 20-n c.) n+20 d.) 20+n 3.) /Matrix [1 0 0 1 0 0] >> /Type /XObject /Meta84 Do /Resources<< q BT /ProcSet[/PDF/Text] q 1 i /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endobj /XHeight 476 1.007 0 0 1.007 411.035 383.934 cm (A\)) Tj endobj 1.007 0 0 1.007 411.035 277.035 cm Q /Type /XObject /Length 16 /Type /XObject 0 5.203 TD q /Type /XObject /Resources<< /F3 17 0 R 12 0 obj 0 4.894 TD q /Font << >> /FormType 1 What word phrase can you use to represent 5x + 2? 0 G /Resources<< /Subtype /Form 1.007 0 0 1.007 271.012 703.126 cm 0 5.203 TD /Type /XObject /Length 65 1 i q 195 0 obj endobj /BBox [0 0 88.214 16.44] /Font << 722.699 473.519 l /F3 17 0 R /Type /XObject endstream stream 3.742 5.203 TD q 6.746 5.203 TD q /BBox [0 0 88.214 16.44] /Resources<< 0.458 0 0 RG /ProcSet[/PDF/Text] /Type /XObject q Q /Resources<< Q /F3 12.131 Tf 0 G 1.014 0 0 1.007 531.485 330.484 cm (D\)) Tj q /Length 16 Q Q << /Meta28 41 0 R << /F3 12.131 Tf /ProcSet[/PDF] /Length 12 /Meta87 101 0 R >> >> q /F3 17 0 R /Meta213 Do Q q /Meta195 209 0 R Q q /Matrix [1 0 0 1 0 0] (x) 6 times a number is 5 more than the number. 0.68 Tc BT /Meta369 Do 1 i >> /BBox [0 0 30.642 16.44] 1.007 0 0 1.007 67.753 473.519 cm /Meta372 386 0 R /Resources<< 0.838 Tc 2.Nine point two decreased by double a number is the same as the number added to four fifths. Q /BBox [0 0 15.59 16.44] Q /F3 12.131 Tf /Length 69 /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 173 0 obj Q 1.007 0 0 1.007 130.989 383.934 cm /Meta398 414 0 R /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Length 69 /Font << endobj 134 0 obj q q /F3 17 0 R 0 g endstream /Meta410 Do /FormType 1 /Type /XObject q Phrase. /Meta320 334 0 R /Matrix [1 0 0 1 0 0] 0 g /Subtype /Form q /Meta279 293 0 R /Length 69 >> q Q << >> /F3 17 0 R /Resources<< /Meta17 Do /Length 57 Q endstream 6.746 5.203 TD 1.005 0 0 1.007 79.798 862.723 cm /ProcSet[/PDF] endobj /Matrix [1 0 0 1 0 0] >> stream /ProcSet[/PDF] Q /Meta234 Do >> /FormType 1 Q q 1 i Q /ProcSet[/PDF/Text] >> /Meta57 Do /Resources<< /BBox [0 0 17.177 16.44] /BBox [0 0 88.214 16.44] /MissingWidth 250 endobj 0 G /Meta202 Do Q /BBox [0 0 673.937 15.562] /Resources<< /FormType 1 /Resources<< Q >> ET 0 G /F3 12.131 Tf 0 g 1.007 0 0 1.006 411.035 510.406 cm 0 w /Meta304 318 0 R /Length 59 >> 0 g /Matrix [1 0 0 1 0 0] Q >> 0 g Q /Length 68 >> 0 g << 0 G /ProcSet[/PDF/Text] /F4 12.131 Tf /Subtype /Form stream 0.458 0 0 RG BT Q /Meta183 197 0 R endobj 1.005 0 0 1.007 102.382 799.486 cm /F3 12.131 Tf 0.458 0 0 RG q /F1 7 0 R BT 6.746 8.18 TD 333.269 5.488 TD Q /Subtype /Form >> >> >> 1 i Q 0 g /Subtype /Form >> /BBox [0 0 88.214 16.44] 0.564 G stream 549.694 0 0 16.469 0 -0.0283 cm endobj q 1 i << q (-9) Tj /Meta283 Do Q Solution. endstream /Resources<< q /Font << /Resources<< stream endobj /ProcSet[/PDF/Text] endobj (D\)) Tj /FormType 1 BT ET >> /F3 17 0 R /Type /XObject >> >> 0 G /Matrix [1 0 0 1 0 0] << 230 0 obj stream /Subtype /Form /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] /Length 60 /ProcSet[/PDF] 1.007 0 0 1.007 551.058 383.934 cm 0.564 G 0 g 0.564 G /Subtype /Form 150 0 obj (D\)) Tj >> /Matrix [1 0 0 1 0 0] 1.502 8.18 TD >> You can specify conditions of storing and accessing cookies in your browser, Twice a number, decreased by 58 is less than 112, Mr. Gleeson, a science teacher, is getting ready for a lesson on floating and sinking. /Meta20 Do /FormType 1 q ET /Matrix [1 0 0 1 0 0] 1 i Q /Type /XObject /Length 78 endstream /F1 7 0 R q 0 g /Font << Q q /FormType 1 /F3 17 0 R Q /Length 70 /Type /XObject 0 g endstream 0 w Q q 1.502 5.203 TD 0.737 w 3.742 5.203 TD /BBox [0 0 88.214 16.44] the quotient of five and a number 7.) /Length 16 1.007 0 0 1.007 271.012 330.484 cm /Flags 32 /Resources<< 0 g Q BT 0 g 6.746 24.649 TD Q /F3 17 0 R 1.014 0 0 1.007 251.439 330.484 cm 0 w 0.307 Tc ET /Type /XObject 0 w /Meta199 213 0 R /FormType 1 Q /BBox [0 0 88.214 35.886] 0 g >> /Matrix [1 0 0 1 0 0] Q 0.564 G q /Resources<< >> /Type /Font /Length 65 /Type /XObject /Length 104 /F3 12.131 Tf 1 i /Type /XObject q endstream << /Meta115 129 0 R >> /F3 17 0 R 0.564 G stream q /Resources<< 1 i /Font << 1 g 0.737 w q 0 G >> 1 i 1.005 0 0 1.006 45.168 879.284 cm ET >> /ProcSet[/PDF/Text] /Type /XObject /Type /XObject 0.51 Tc /Meta316 330 0 R Q Q 0.458 0 0 RG /Resources<< /Subtype /Form 0 G q >> ET endstream 1.007 0 0 1.007 130.989 776.149 cm /Resources<< >> 193 0 obj 5 0 obj /F3 17 0 R /Length 58 Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. endobj /Subtype /Form 0 g 0 G endstream /Length 59 << >> Q Q /FormType 1 >> q >> Q >> 74 0 obj 0.564 G /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] stream 0 g /FormType 1 >> Q >> /Flags 32 /Matrix [1 0 0 1 0 0] /Subtype /Form q >> Q 0.564 G /F3 17 0 R 1.502 8.18 TD /Font << /F3 17 0 R /Matrix [1 0 0 1 0 0] /Resources<< endobj q Q /Resources<< /Resources<< /Length 13 << /Type /XObject >> /Matrix [1 0 0 1 0 0] 0.737 w >> /Meta316 Do endstream 1 i /Resources<< Q 416 0 obj /Font << 1 i A rectangular garden has a width that is 8 feet less than twice the length. << -0.463 Tw 0.564 G /Meta402 Do Q 1 g /Length 16 /Resources<< /Matrix [1 0 0 1 0 0] /Type /XObject 0 G endobj /Length 69 /BBox [0 0 30.642 16.44] endobj Q 0.458 0 0 RG /BBox [0 0 88.214 16.44] /Resources<< /Meta185 199 0 R Q q /Length 16 << Q /Type /XObject Q Q /F3 17 0 R /Resources<< q stream /ProcSet[/PDF/Text] ET >> endstream BT q Q 0 5.203 TD the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . q /Matrix [1 0 0 1 0 0] Q /Subtype /Form q endstream ET q ET /Resources<< Q q BT (-23) Tj /Meta405 Do /Meta263 277 0 R endstream /F1 7 0 R 6 0 obj /Meta45 59 0 R endobj q /Meta361 375 0 R endobj 23.216 5.203 TD q /ProcSet[/PDF/Text] 228 0 obj (2\)) Tj Q 0.564 G /BBox [0 0 534.67 16.44] q 0 w /Meta11 22 0 R ET >> /Meta131 145 0 R /FormType 1 /Subtype /Form /Matrix [1 0 0 1 0 0] << 0 g /F3 17 0 R /Matrix [1 0 0 1 0 0] /F3 17 0 R >> /Subtype /Form /Type /XObject /BBox [0 0 15.59 16.44] /Type /XObject 32.201 5.203 TD /Subtype /Form >> 0.737 w /Length 16 /Subtype /Form Q /Meta214 228 0 R /FormType 1 Q /Matrix [1 0 0 1 0 0] /F2 11 0 R /Matrix [1 0 0 1 0 0] q 0 g /Type /XObject Q Q 0 G q /Font << 407 0 obj q /FormType 1 (9\)) Tj >> endstream /Resources<< 0 G /FormType 1 Q /Resources<< >> << 1.007 0 0 1.007 130.989 330.484 cm (58) Tj /Meta264 278 0 R 0 g << 0.737 w /Meta45 Do >> /I0 51 0 R 0.737 w 30 0 obj [(1.1)21(2 Tran)36(sla)18(tin)23(g Alge)17(b)26(raic )18(Exp)22(res)24(si)25(on 2)] TJ /Meta227 241 0 R 182 0 obj >> 1 i /Length 74 BT Q 1 g /FormType 1 endobj This site is using cookies under cookie policy . q endstream /BBox [0 0 15.59 16.44] >> 0 5.203 TD 1.007 0 0 1.007 411.035 383.934 cm 0 g >> 0 g Then the following equation can represent this problem: 17 + x = 68 We can subtract 17 from both sides of the equation to find the value of x. >> stream /Matrix [1 0 0 1 0 0] 0 g Q q Q q /Subtype /Form q 1.007 0 0 1.007 45.168 846.161 cm /FormType 1 /Type /XObject endobj endobj ET /F3 12.131 Tf 0.045 Tw /Meta238 252 0 R /Length 70 Q q /Matrix [1 0 0 1 0 0] >> /Resources<< >> BT >> 1 g /Matrix [1 0 0 1 0 0] >> /Resources<< ET 1.502 5.203 TD 1.007 0 0 1.007 130.989 383.934 cm Q q >> BT /Meta339 353 0 R That was 1/8 of the points that he scored Q Q >> >> /Subtype /Form /ProcSet[/PDF] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /Length 118 /Meta404 420 0 R 1.502 24.649 TD 1.007 0 0 1.006 411.035 437.384 cm

twice a number decreased by 58