twice a number decreased by 58

twice a number decreased by 58

/FormType 1 236 0 obj ET 339 0 obj >> /Resources<< /Meta362 376 0 R /Length 60 >> << 20.21 5.203 TD /Font << **Note: You could choose any variable you want. >> stream /Meta269 Do New questions in Mathematics 0.838 Tc Q q 0.458 0 0 RG << /Resources<< /Meta177 191 0 R /Meta57 Do endobj 229 0 obj 0.737 w /Length 60 (-) Tj << << /Meta125 139 0 R 1 g /Matrix [1 0 0 1 0 0] 400 0 R endobj >> /Subtype /Form /ProcSet[/PDF/Text] 0 G q 722.699 293.596 l Q Q Q /BBox [0 0 534.67 16.44] 1 i 0.564 G q 0 g /Subtype /Form 0.737 w 0 g /F4 36 0 R Q 65.906 4.894 TD /Length 58 /Font << /Font << q 1.005 0 0 1.007 102.382 616.553 cm Q /FormType 1 /Length 69 Q /Subtype /Form >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 1 g q 0 G /Type /XObject BT Q ET /Meta246 Do Thrice a number decreased by 5 is 3x - 5. /Font << 0 g /Meta226 240 0 R q 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. q 0 G q /Type /XObject /Resources<< (+) Tj 1.007 0 0 1.007 411.035 330.484 cm Q 1 0 obj /Matrix [1 0 0 1 0 0] /Subtype /Form >> stream (-) Tj endobj /F3 17 0 R /ProcSet[/PDF] q >> Q 1.007 0 0 1.007 271.012 636.879 cm /FormType 1 /Type /XObject 0 g 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Resources<< 1 i 19 0 obj /Matrix [1 0 0 1 0 0] stream /Resources<< >> 0 G 121 0 obj endstream >> Q >> << Q 1 i endstream q q 1 g 0 g endobj /Font << /Meta110 Do /Resources<< 375 0 obj ET >> << q /Meta261 Do Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. /Subtype /Form 1 i q /Type /XObject 1 g 0 G 0 g 0 G q 16.469 5.203 TD 0.458 0 0 RG endobj q /F3 12.131 Tf /Resources<< Q 0.425 Tc Q 1.005 0 0 1.007 102.382 473.519 cm BT BT Q q Q 343 0 obj /Type /XObject >> << In other terms, 52-nx The problem is asking that you subtract twice a number from 52. q 1 i /Meta341 355 0 R << endstream >> q q << 1.007 0 0 1.007 67.753 599.991 cm q >> /Matrix [1 0 0 1 0 0] stream /Meta258 Do q /F1 12.131 Tf endstream Q /Font << stream 0 G 1 i Q endobj Q /Matrix [1 0 0 1 0 0] nine increased by a number x. /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 672.261 726.464 m Q q >> Q /Meta0 5 0 R endobj >> q /FormType 1 /Length 58 << /Type /XObject Q 1 i 1.007 0 0 1.006 411.035 690.329 cm 42 0 obj Q /BBox [0 0 15.59 16.44] /Type /XObject Q q /Resources<< q 0.524 Tc /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 /F3 12.131 Tf /Meta111 125 0 R /FormType 1 ET /ProcSet[/PDF] endstream 1.014 0 0 1.007 391.462 277.035 cm q (9\)) Tj 0.737 w No packages or subscriptions, pay only for the time you need. /Length 69 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Meta176 190 0 R 6.746 5.203 TD 0 g << >> /Resources<< /Meta183 197 0 R Q /Type /XObject /Meta78 92 0 R endobj /Meta41 55 0 R endobj Q endobj >> View the full answer. 205.199 4.894 TD 0.458 0 0 RG /Subtype /Form /ProcSet[/PDF/Text] Q q 0 g /Meta229 243 0 R q q endstream 173 0 obj /Meta183 Do /BBox [0 0 88.214 16.44] /Meta349 Do (+) Tj Q /FormType 1 << q q /Subtype /Form /Ascent 1050 /F3 12.131 Tf /Meta419 Do q 100 0 obj q /Count 2 /BBox [0 0 15.59 16.44] >> 0 g Q 0 5.336 TD /Meta404 420 0 R >> BT /F3 12.131 Tf 1 i >> /F3 17 0 R 0.737 w /FormType 1 1.005 0 0 1.007 102.382 653.441 cm endobj /Resources<< /F3 12.131 Tf >> q Q (-23) Tj /Type /Font /Type /XObject q >> /ProcSet[/PDF/Text] Q Q stream /Matrix [1 0 0 1 0 0] /Meta218 Do stream 203 0 obj /Meta377 391 0 R /Meta362 Do /Subtype /Form ( x) Tj endobj /Length 69 0 g >> /Subtype /Form q /Subtype /Form << /Subtype /Form stream q /ProcSet[/PDF/Text] /Type /XObject 0.564 G q 1 i Q Q >> endobj /Resources<< stream stream /Type /FontDescriptor endstream /Type /XObject Thrice a number decreased by 5 exceeds twice the number by 1 is . BT 260 0 obj 0 G /BBox [0 0 534.67 16.44] (6\)) Tj >> 0.737 w /Resources<< >> Q /Type /XObject Q 1.007 0 0 1.007 411.035 383.934 cm Q /Meta332 346 0 R BT endstream q q >> /FormType 1 0.198 Tc 0.738 Tc -0.106 Tw /Meta14 25 0 R stream 217 0 obj << Q /F3 17 0 R q 0.486 Tc /Subtype /Form /ProcSet[/PDF] /F3 12.131 Tf q 1.007 0 0 1.007 551.058 523.204 cm ET /Length 69 q 6.746 5.203 TD S q stream << Q 1 i /Font << q 0.463 Tc 1.007 0 0 1.007 551.058 703.126 cm /Length 116 /Subtype /Form stream 0 g >> 1 i /Length 12 ET /Subtype /Form Q >> /F3 12.131 Tf endstream Q /Meta220 234 0 R Q >> stream Q Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . /Font << /ProcSet[/PDF] q /BBox [0 0 15.59 29.168] >> 1 i 0 4.894 TD stream endobj (A\)) Tj q q Q /Meta421 Do 26.219 5.203 TD endobj /Type /XObject endstream Q >> (1\)) Tj >> /ProcSet[/PDF/Text] q endobj 0 g /Matrix [1 0 0 1 0 0] Q q 1 i /Resources<< q /Matrix [1 0 0 1 0 0] 0 g /BBox [0 0 88.214 16.44] q Q >> q 0 G endstream Q /Matrix [1 0 0 1 0 0] 0.737 w /ProcSet[/PDF/Text] q 118.317 5.203 TD q /F3 17 0 R q Q 0.564 G 338 0 obj /F3 17 0 R q 0.271 Tc >> You can specify conditions of storing and accessing cookies in your browser. /Meta7 Do /StemV 88 /ProcSet[/PDF] endobj ET q Q /F1 12.131 Tf /F3 17 0 R /Subtype /Form BT 0 w /BBox [0 0 15.59 29.168] q Q >> << /ProcSet[/PDF/Text] q endobj /Length 69 Q (x) Tj Q /Subtype /Form >> 421 0 obj 1 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. /Subtype /Form Q << /FormType 1 /FormType 1 /Type /XObject >> ET 0.369 Tc /ProcSet[/PDF/Text] q Q /Leading 253 Q 0 g q /Type /XObject 1.005 0 0 1.013 45.168 933.487 cm endstream /Meta417 433 0 R /Length 69 stream /Resources<< Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /Matrix [1 0 0 1 0 0] 0 g (2) Tj /Type /XObject /Subtype /Form q /Type /XObject /Meta215 229 0 R stream /F3 12.131 Tf /Resources<< >> /FormType 1 /Type /XObject q /Resources<< 1 i q /Meta388 404 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0 G 1.014 0 0 1.007 251.439 330.484 cm /BBox [0 0 88.214 16.44] >> /Resources<< /Meta30 Do 94 0 obj >> endobj endstream 0 g stream /Length 70 ET 0 G /Meta74 Do endstream Q 1.007 0 0 1.007 551.058 523.204 cm q (\)) Tj BT /Type /XObject /Resources<< endstream >> stream stream q q Q Q /Meta225 239 0 R /Subtype /Form Q /Matrix [1 0 0 1 0 0] Q /Subtype /Form 102 0 obj Q 1 i C. Twice a number decreased by ten is at most 24. 0 g /Subtype /Form >> /F3 17 0 R /Subtype /Form q 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. q /Meta286 300 0 R /Meta157 Do 0 g << /Length 16 q << >> /Subtype /Form /Type /XObject /Matrix [1 0 0 1 0 0] q /Meta52 66 0 R Q /ProcSet[/PDF] stream >> /Resources<< /Matrix [1 0 0 1 0 0] /Resources<< /ProcSet[/PDF/Text] << Q /FormType 1 /ProcSet[/PDF/Text] /F1 7 0 R /FormType 1 >> /FormType 1 q >> /Subtype /Form 1 i /Length 59 endobj q To find: The. /F3 12.131 Tf >> BT Q Twice a number when decreased by 7 gives 45. BT 0.564 G 0.425 Tc 0 g stream Q /BBox [0 0 88.214 16.44] /Meta47 Do q Q << (x ) Tj /Meta159 173 0 R /Matrix [1 0 0 1 0 0] /Meta181 195 0 R /F3 17 0 R /BBox [0 0 88.214 16.44] Q /FormType 1 >> (C) Tj >> /ProcSet[/PDF/Text] /FormType 1 stream << 0 g Q Find the number. q /BBox [0 0 30.642 16.44] ET /Subtype /Form << /Resources<< >> >> 0 G BT q /Meta299 313 0 R /Length 60 /Meta350 364 0 R /Resources<< q Q /ProcSet[/PDF/Text] (-8) Tj Q q /Type /XObject endstream Q /Subtype /Form 0.458 0 0 RG /Subtype /Form 13.464 5.203 TD q ET /ProcSet[/PDF] 1 i 1.007 0 0 1.007 411.035 383.934 cm >> q Q /BBox [0 0 15.59 16.44] << Q /BBox [0 0 639.552 16.44] q endobj >> Q Q /Meta60 74 0 R /FormType 1 /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] >> endstream 354 0 obj Q /F3 17 0 R ET /Meta211 225 0 R >> /Type /XObject /Meta168 182 0 R Q /BBox [0 0 549.552 16.44] >> /Type /Page >> Q /Subtype /Form /Resources<< << endobj Q Choose the correct one. If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. q Q /Type /XObject /F3 17 0 R >> >> stream q q q /Subtype /Form /FormType 1 0 5.203 TD 0.458 0 0 RG 263 0 obj >> /Meta34 Do 0.737 w 0 g q The result is 8 less than 10 times the number. Q 611 556 611 611 389 444 333 611 556 833 500]>> /BBox [0 0 673.937 14.853] q 1.007 0 0 1.007 411.035 277.035 cm 238 0 obj 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. >> Q /Meta239 253 0 R << /F3 17 0 R endobj /BBox [0 0 88.214 16.44] 0 g 0.68 Tc 0 G 0 w /F3 17 0 R /Subtype /Form stream If mario jumps 3 times and luigi jumps 62 times. Q /F3 17 0 R /Length 69 q BT 0.458 0 0 RG /Meta3 Do 0.737 w /FormType 1 (38) Tj 1.014 0 0 1.007 391.462 703.126 cm q q /Font << /Subtype /Form /BBox [0 0 30.642 16.44] >> q /Matrix [1 0 0 1 0 0] endobj S /ProcSet[/PDF] /Resources<< /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] stream 0 G Q 1 i >> >> q /Subtype /Form /Matrix [1 0 0 1 0 0] q (2) Tj 1.007 0 0 1.007 271.012 636.879 cm /Meta389 Do /BBox [0 0 15.59 16.44] endobj /BBox [0 0 88.214 16.44] endstream /BBox [0 0 30.642 16.44] )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] Q 1 i /Type /XObject /Resources<< /Subtype /Form q /F3 17 0 R /Meta305 Do 0 w 1 i 0.458 0 0 RG q /BBox [0 0 534.67 16.44] /Length 59 0 w /Subtype /Form q stream (v) 5 subtracted from thrice a number is 16. BT stream 87 0 obj << BT >> /ProcSet[/PDF/Text] Q 389 0 obj >> /Type /XObject 1.007 0 0 1.007 130.989 636.879 cm stream q 352 0 obj /Meta389 405 0 R Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. q /Meta43 Do /Length 59 Q endstream /BBox [0 0 15.59 16.44] /Subtype /Form /FormType 1 /Length 69 /FirstChar 32 1.502 5.203 TD Q q >> /BBox [0 0 88.214 16.44] 198 0 obj /Subtype /Form Q /Meta245 259 0 R 0.486 Tc /Subtype /Form 140 0 obj /Meta110 124 0 R a Question /FormType 1 Q endobj (5\)) Tj 0.564 G 1.007 0 0 1.007 67.753 872.509 cm /ProcSet[/PDF] /ProcSet[/PDF/Text] /Type /XObject q /Length 16 stream q >> q /MissingWidth 250 /Resources<< 0 g 1 i 1.007 0 0 1.007 551.058 383.934 cm 1.005 0 0 1.007 79.798 829.599 cm endobj ET /Matrix [1 0 0 1 0 0] endobj /Type /XObject /BBox [0 0 17.177 16.44] >> Q q q (B) Tj Q /F3 17 0 R >> >> /F3 12.131 Tf q /Meta426 442 0 R [4] One half of a number decreased by fourteen is twenty-one 199 0 obj /Font << 0.51 Tc 20.975 5.336 TD >> /Resources<< << /Length 68 /Subtype /Form q /Meta276 290 0 R /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] 70 0 obj 0.458 0 0 RG /Meta259 Do << /Type /XObject q 0 g stream Q Q /Matrix [1 0 0 1 0 0] /Font << Q /ProcSet[/PDF] Q Q 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 . /Subtype /Form /Meta203 Do Q >> /ProcSet[/PDF] << q /Meta46 60 0 R 0 g /Meta40 54 0 R 1 of this study. q /BaseFont /PalatinoLinotype-Bold q BT q /BBox [0 0 15.59 16.44] /Type /XObject endstream /BBox [0 0 88.214 35.886] /Type /XObject endobj >> Q q 0.37 Tc /Length 16 << Q 1 i Q Q stream stream 1.007 0 0 1.007 130.989 849.172 cm 0 g /F3 17 0 R endstream Q q /BBox [0 0 88.214 16.44] /FormType 1 Q /Length 16 /Length 69 /FormType 1 1 i Q << endstream >> q /BBox [0 0 88.214 16.44] /Resources<< q 1 g endstream >> /Resources<< Q 0 g /Matrix [1 0 0 1 0 0] 12.727 5.203 TD 0 g 0 g Q /Font << q /ProcSet[/PDF/Text] /Type /XObject -0.056 Tw /F3 12.131 Tf /Meta205 219 0 R BT /Meta275 289 0 R 0.737 w >> /Type /XObject BT >> This site is using cookies under cookie policy . Q 0 g q q >> Q /Meta182 196 0 R q 1.014 0 0 1.007 251.439 776.149 cm /Length 12 /Length 69 /ProcSet[/PDF] /Subtype /Form << /Font << stream BT /F4 36 0 R 403 0 obj q 1.007 0 0 1.007 271.012 703.126 cm >> /Meta142 Do >> endobj endobj A number = an unknown number which can be represented by a variable, usually x. endobj q All steps. 3.742 5.203 TD >> ET q 0 G (4\)) Tj /Subtype /Form 0 5.203 TD /BBox [0 0 15.59 29.168] /Resources<< >> /Resources<< q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0 g 722.699 473.519 l 0 g /BBox [0 0 88.214 16.44] endobj /Meta18 29 0 R endstream Q 0 G /Resources<< 22 0 obj 341 0 obj stream q Q q endstream >> >> /Resources<< endobj /Meta80 Do >> /MaxWidth 1453 Q /Meta378 Do 0.68 Tc >> endobj /Resources<< /F1 12.131 Tf /Resources<< /ProcSet[/PDF/Text] /F3 17 0 R endobj endobj /FormType 1 >> 0.68 Tc ET ET endobj << 1.007 0 0 1.007 411.035 277.035 cm /ProcSet[/PDF/Text] 1.007 0 0 1.007 411.035 583.429 cm (-23) Tj /BBox [0 0 673.937 16.44] 1 i Q q ET /Type /XObject 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. Q /F1 12.131 Tf Q << >> /Meta37 50 0 R Q /F3 12.131 Tf BT endstream >> 0 g /Meta192 Do q This gives us: "2x+5". q 1 g 0 g stream 2 0 obj Q 1 i << >> endstream /Type /XObject Q /FormType 1 >> >> q << Q /Meta66 80 0 R /Length 59 Q Q << /Font << /F3 17 0 R /FormType 1 [(Answe)20(r Key)] TJ 1 g stream ET endstream /Type /XObject /Subtype /Form endobj /Meta17 Do /Meta156 Do Q /Subtype /Form /F3 17 0 R >> /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 776.149 cm ET q 1.005 0 0 1.007 102.382 653.441 cm 0.564 G Q /Matrix [1 0 0 1 0 0] 1 i endobj 234 0 obj 0.838 Tc stream 0.369 Tc Q 0.458 0 0 RG BT [(th)-28(e di)-18(ffe)-14(ren)-23(ce o)-28(f )] TJ /Resources<< 0.175 Tc q /Type /XObject 1 g /F4 12.131 Tf >> >> /Type /XObject q 0.297 Tc stream 7 0 obj 0.564 G >> /Resources<< /Descent -299 /Resources<< /Meta368 382 0 R q /Type /XObject 672.261 599.991 m /BBox [0 0 88.214 16.44] BT q q Q (B\)) Tj >> Q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 130.989 583.429 cm /ProcSet[/PDF/Text] Q /Meta208 Do Q endobj ET q stream endstream endobj 0.737 w q /ProcSet[/PDF] Let x the unknown number. q BT 1 i /F3 17 0 R Q 1.007 0 0 1.007 551.058 636.879 cm >> Q /Meta185 199 0 R /Resources<< /Matrix [1 0 0 1 0 0] /Meta278 292 0 R /Type /XObject >> /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] 0.51 Tc >> >> 1 i /FormType 1 289 0 obj << /FormType 1 endobj (40) Tj 0.564 G 0.564 G >> /Matrix [1 0 0 1 0 0] 0.786 Tc q << /Meta36 49 0 R 1 i 0.458 0 0 RG 1.007 0 0 1.007 130.989 636.879 cm /Subtype /Form /BBox [0 0 534.67 16.44] Q Q << /Subtype /Form 0 G /ProcSet[/PDF] >> /F3 17 0 R Q endstream q endobj 1 i /FormType 1 0 G /Matrix [1 0 0 1 0 0] 133 0 obj Q /BBox [0 0 88.214 16.44] Q 0.425 Tc q /Matrix [1 0 0 1 0 0] q /BBox [0 0 15.59 16.44] Q q endobj 1 i /BBox [0 0 17.177 16.44] /BBox [0 0 88.214 16.44] Q 0 g q 1 g >> /Type /XObject endobj 0.564 G /FormType 1 BT Q /ProcSet[/PDF] Q Q endobj /Resources<< /Type /XObject endobj Q endstream /BBox [0 0 15.59 29.168] S q 1 i << We are asked to find the number, so, we could assign the number as "x". 1.007 0 0 1.007 130.989 383.934 cm /ProcSet[/PDF/Text] /F3 12.131 Tf /FormType 1 0 5.203 TD >> endstream endstream 0.737 w q (38) Tj Q /Meta43 57 0 R /FormType 1 0 G 257 0 obj >> 1 g A) 5 more than a number. You could call them. >> endstream 0.297 Tc /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] q /Font << /MaxWidth 1397 /Resources<< /Resources<< /ProcSet[/PDF/Text] 0 g 1 i Q /Meta381 Do /Type /XObject /FormType 1 >> Q (1\)) Tj q 1.007 0 0 1.006 130.989 437.384 cm Q >> /ProcSet[/PDF/Text] 0 5.203 TD >> 1.007 0 0 1.007 551.058 703.126 cm /Meta285 Do /Type /XObject BT >> 0 g /Length 69 0 g q (B\)) Tj /FormType 1 0.458 0 0 RG << endstream << q q 0.297 Tc /F1 12.131 Tf q By the . /ProcSet[/PDF/Text] 22.478 5.336 TD /Type /XObject >> /Font << (x) Tj 0.737 w 185.725 5.203 TD /Subtype /Form >> /Matrix [1 0 0 1 0 0] 0 g q /Type /XObject Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. /FormType 1 /Resources<< Q >> endobj q 1.005 0 0 1.007 102.382 400.496 cm /Type /XObject 0 G 52 0 obj ET /Subtype /Form 1 i /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 271.012 636.879 cm stream endstream /Meta173 187 0 R /BBox [0 0 88.214 16.44] endobj /Length 16 1 i 1 i /Matrix [1 0 0 1 0 0] BT /Matrix [1 0 0 1 0 0] /Length 70 /Meta179 193 0 R BT /Resources<< /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] [( and )-20(the product of )-15(a number a)-16(nd )] TJ /Matrix [1 0 0 1 0 0] stream /Resources<< 0.564 G 0 G 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 /Resources<< 1.007 0 0 1.007 551.058 703.126 cm 1 i Q /BBox [0 0 88.214 16.44] endstream /Resources<< Q endobj 0.524 Tc Q /Subtype /Form endstream /Meta340 Do 0.369 Tc >> /Type /XObject q 1 i /F3 17 0 R /Length 78 /Length 69 >> /Resources<< stream /BBox [0 0 17.177 16.44] 1.005 0 0 1.006 45.168 879.284 cm >> 1 i 127 0 obj /Meta24 37 0 R 0 g /Matrix [1 0 0 1 0 0] 0 w 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. Get a free answer to a quick problem. 1.007 0 0 1.007 130.989 383.934 cm 1.007 0 0 1.007 130.989 523.204 cm Q /ProcSet[/PDF/Text] 0 G /Meta40 Do q endobj q 0.737 w 0.458 0 0 RG 0 g 0.737 w 0.564 G endobj BT Q endobj /Resources<< (-) Tj 1.007 0 0 1.007 551.058 636.879 cm q /ProcSet[/PDF] /BBox [0 0 549.552 16.44] /Meta343 357 0 R /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] q /Font << 0.51 Tc q 351 0 obj /Resources<< /Length 127 0 w BT /ProcSet[/PDF/Text] q /Resources<< 1.014 0 0 1.007 111.416 330.484 cm /FormType 1 -0.382 Tw /ProcSet[/PDF] /FormType 1 stream /Meta102 116 0 R ET 0.564 G q /FormType 1 Answer link. 0.564 G /ProcSet[/PDF/Text] /Meta236 250 0 R endstream 0.737 w 0 g q 0.737 w q /Length 69 endobj /Length 119 endobj 106 0 obj << 0 g 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. /Type /XObject Q /Meta387 Do (-11) Tj 0 G 440 0 obj Q >> /Font << /Meta312 Do /Meta251 Do q q /F3 12.131 Tf /FormType 1 /Meta204 218 0 R /Font << /Meta206 220 0 R Q /FormType 1 /Length 54 /Font << stream >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Meta298 Do /Meta315 329 0 R Q q /BBox [0 0 15.59 16.44] /BBox [0 0 15.59 16.44] /Type /XObject 1 i Q q q Q /Resources<< (x) Tj endstream 1.007 0 0 1.007 551.058 523.204 cm Q /Subtype /Form /Resources<< << Q 0 g Q /Meta373 387 0 R 1 i Q /Type /XObject 1.014 0 0 1.006 391.462 437.384 cm /ProcSet[/PDF/Text] /Length 68 0 5.203 TD >> /Length 69 Q /Length 70 Q 9 0 obj << endobj 347 0 obj ET 1 i Q q >> 0 w /Matrix [1 0 0 1 0 0] /Type /XObject >> /ProcSet[/PDF] /ProcSet[/PDF] 0.737 w 1 i stream 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 >> /Type /XObject Q 0 g /Type /XObject /Meta105 Do 1 g >> 123 0 obj /Matrix [1 0 0 1 0 0] /Meta181 Do endstream BT (+) Tj q /Meta342 356 0 R 333.269 5.488 TD stream /FormType 1 >> endobj endobj ET /Meta156 170 0 R 0 G 178.979 5.203 TD stream << /FormType 1 349 0 obj /BBox [0 0 88.214 16.44] >> (x) Tj q >> /FormType 1 0.737 w /BBox [0 0 17.177 16.44] (-) Tj 1.014 0 0 1.007 111.416 277.035 cm endstream /Font << 1 g 0 w 0 w /Font << stream 1 i /ProcSet[/PDF/Text] endobj >> q >> 1 i /BBox [0 0 88.214 16.44] 0 G >> Q /F3 12.131 Tf >> /Length 69 /Matrix [1 0 0 1 0 0] /FormType 1 0.524 Tc Q q Q BT ET q Q 0.458 0 0 RG 0 g /Length 12 q stream stream /Resources<< 1.007 0 0 1.007 67.753 400.496 cm BT 0 g Q Q /Meta215 Do << /Type /XObject /Meta32 45 0 R Q /Type /XObject /Length 69 q /F4 12.131 Tf /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] stream >> q endobj /Matrix [1 0 0 1 0 0] /Font << >> endstream q /FormType 1 Q 0.458 0 0 RG /Length 118 Q q 0 g /Type /XObject /Meta284 Do >> Q /Length 68 stream Q /Length 69 /Length 54 /ProcSet[/PDF/Text] stream 0 g Q for the season. Q >> 672.261 872.509 m Q 1.007 0 0 1.007 67.753 599.991 cm Q /F1 7 0 R 1 i q /BBox [0 0 88.214 16.44] q 1 i /Meta141 Do 20.975 5.336 TD 219 0 obj /Font << >> /Meta99 Do endobj >> 1 i >> 0.786 Tc /ProcSet[/PDF/Text] /Length 294 0 g >> /BBox [0 0 88.214 16.44] >> << 192 0 obj endstream 0.737 w q >> /Font << ET /FormType 1 1 i q 306 0 obj /Matrix [1 0 0 1 0 0] q 0.564 G >> /Type /XObject 0.737 w /Size 447 1 i 1 i BT /Type /XObject 0 G endobj /ProcSet[/PDF] q /FontName /TestGen-Regular >> /F3 17 0 R >> Q 0.737 w /Font << q q /Length 12 1 i q endstream endobj /BBox [0 0 88.214 16.44] << 20.975 5.336 TD 1 i >> q Just type into the box and your calculation will happen automatically. /FormType 1 Q endobj /Subtype /TrueType endstream Q /Meta65 79 0 R q endobj /Type /XObject Q Twice a number decreased by ten is at least 24. /BBox [0 0 88.214 16.44] /FormType 1 23.952 4.894 TD 21.713 20.154 l 0.458 0 0 RG 1 i 27 0 obj stream The sum Of twice a nu4ber What is the number? q Q /ProcSet[/PDF/Text] /Resources<< q 1.007 0 0 1.007 654.946 799.486 cm BT q 1.007 0 0 1.007 271.012 776.149 cm Q >> Q /Meta140 Do >> 1 g /Matrix [1 0 0 1 0 0] 108 0 obj /Meta222 236 0 R /FormType 1 /Meta384 398 0 R /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] /Font << /Type /XObject /Meta382 396 0 R /Type /XObject /F3 17 0 R startxref /FormType 1 1.014 0 0 1.007 531.485 330.484 cm Q 0 G 9.723 5.336 TD /Resources<< /Resources<< /Meta332 Do /F3 12.131 Tf 0 g /Meta377 Do (A\)) Tj /F3 17 0 R /F3 17 0 R /Matrix [1 0 0 1 0 0] q ET endstream >> 1.008 0 0 1.007 654.946 293.596 cm BT 0 g Answer only. /Length 16 In the problem above, x is a variable. >> /ProcSet[/PDF/Text] /Type /XObject >> /Resources<< >> 0.458 0 0 RG endstream /Subtype /Form /Matrix [1 0 0 1 0 0] Q /FormType 1 /Matrix [1 0 0 1 0 0] /Type /XObject q (D\)) Tj /Length 59 Twice a number decreased by 8 gives 58 find the number Advertisement Loved by our community 24 people found it helpful Xiphodon Step-by-step explanation: 2x-8=58 2x=66 x =33 Hope it helps Please mark as brainliest Find Math textbook solutions? 413 0 obj stream stream /F3 12.131 Tf q >> >> stream q q Q /F3 17 0 R /Matrix [1 0 0 1 0 0] 1 i /F3 12.131 Tf /Type /XObject ET /Matrix [1 0 0 1 0 0] 1 i 0 G q (+) Tj /F3 12.131 Tf ET ET 291 0 obj (iv) A number exceeds 5 by 3. 0 g /ProcSet[/PDF/Text] q >> /Type /XObject /Subtype /Form Q Q /FormType 1 stream /Meta260 Do Q Percent Change = (Decrease First Value) x 100% Q 0 g 0.564 G 0 20.154 m q ET /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] /Resources<< endobj Q 200 0 obj BT /ProcSet[/PDF/Text] 0.737 w Q BT /F3 17 0 R /BBox [0 0 30.642 16.44]

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