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endstream /ProcSet[/PDF] 185.725 5.203 TD A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. /FontBBox [-90 -216 1195 800] /F2 12.131 Tf 0.737 w /F3 17 0 R /FormType 1 0 g 0 G 23.216 5.203 TD q 324 0 obj >> 1 i 549.694 0 0 16.469 0 -0.0283 cm Q /Meta381 395 0 R /Font << stream endobj << BT Q /Subtype /Form (11) Tj Q /Meta170 184 0 R q 0 5.203 TD Q /Subtype /TrueType /Resources<< /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Subtype /Form Q /Type /XObject 0.458 0 0 RG 0.369 Tc >> /Font << 410 0 obj 1 i /Length 16 Q endstream /F1 12.131 Tf 339 0 obj /Resources<< ET q /Type /XObject A number divided by six is eight: (k / 6) = 8. 0 G /FormType 1 /Meta125 139 0 R (B\)) Tj q 1 i 1.005 0 0 1.007 102.382 616.553 cm Q /Meta225 Do << << q /Resources<< /Type /XObject 1 g /Font << >> 0 g endobj q /Length 69 1 i /Matrix [1 0 0 1 0 0] BT stream , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. /FormType 1 /F1 7 0 R >> /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] endstream /Matrix [1 0 0 1 0 0] endstream /Meta417 Do /Meta253 Do 1.005 0 0 1.007 102.382 653.441 cm 0.737 w q q /Resources<< 128 0 obj /FormType 1 1.007 0 0 1.007 271.012 703.126 cm Q >> /Subtype /Form 1 g q 93 0 obj Q >> Q 0.51 Tc ET 0.737 w >> Q /ProcSet[/PDF/Text] /F3 12.131 Tf 0 G /Resources<< Q 0 G /Type /XObject the other number. stream << 0 5.203 TD /Length 107 Q /F3 12.131 Tf >> Q /FormType 1 174 0 obj /Resources<< /F3 12.131 Tf /Length 16 /Meta76 Do 1.007 0 0 1.007 411.035 330.484 cm 1.005 0 0 1.007 79.798 796.475 cm 427 0 obj /Meta114 Do << /BBox [0 0 639.552 16.44] 1.007 0 0 1.007 271.012 383.934 cm endobj BT /Meta281 295 0 R /Resources<< stream endobj q endobj Q /ProcSet[/PDF] /ProcSet[/PDF] >> 0 g /Meta215 Do 0.564 G q /Meta227 Do /Resources<< stream /Matrix [1 0 0 1 0 0] /Length 99 /Type /XObject /Length 60 /Type /XObject endobj endstream >> stream q 135 0 obj 1 g 65.906 4.894 TD /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] 0 g q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 67.753 473.519 cm 0 G 0 w q 0.458 0 0 RG /Meta256 Do >> q stream /ProcSet[/PDF/Text] /FormType 1 /I0 Do 0.564 G 1 i Q Q /Subtype /Form >> /F3 12.131 Tf /Matrix [1 0 0 1 0 0] endobj q 334 0 obj Q /F3 17 0 R q q 1 i >> >> 175 0 obj 0.564 G 1.007 0 0 1.007 271.012 450.181 cm ET /Subtype /Form /Length 245 Q /Meta83 Do 0 w /Font << Q q The width Of a rectangle is 15 cm and the perimeter is 12 cm. q 1 g A rectangular garden has a width that is 8 feet less than twice the length. /Meta171 Do /FormType 1 /Meta271 285 0 R /Meta69 83 0 R 26.219 5.203 TD >> 1 i 1.005 0 0 1.007 79.798 862.723 cm /Matrix [1 0 0 1 0 0] Q >> 0.564 G /BBox [0 0 88.214 16.44] 1 i /Meta263 277 0 R /FormType 1 /Font << BT 1.007 0 0 1.007 654.946 726.464 cm Q q /Length 245 endobj endobj Q 0 G q /Meta201 Do 271 0 obj /Type /XObject /Font << 35,000 worksheets, games, and lesson plans, Spanish-English dictionary, translator, and learning, **Note: You could choose any variable you want, to represent the 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Q >> /Meta400 416 0 R >> 0 G /Length 68 q 0.458 0 0 RG Thrice a number decreased by 5 exceeds twice the number by 1. /Matrix [1 0 0 1 0 0] Q endstream q q /Font << 1.502 5.203 TD 1 g >> five times the sum of a number x and two b.) >> Q 1.005 0 0 1.007 102.382 293.596 cm /Resources<< >> >> 0.564 G Q >> /Meta262 276 0 R (-) Tj /Meta282 296 0 R >> >> algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. /BBox [0 0 88.214 35.886] 344 0 obj 1 g q /Resources<< >> 94.364 5.203 TD [(F)-22(ive)] TJ 0.564 G /Contents [399 0 R] Q /Subtype /Form Q >> /Type /XObject /ProcSet[/PDF] /F3 17 0 R 83 0 obj /F3 12.131 Tf q >> 0 g Q 0.838 Tc ET /Length 65 Q /Type /XObject 0 g >> Q /ProcSet[/PDF/Text] 0 g /F4 12.131 Tf 0 G 38.182 5.203 TD /Resources<< >> /Type /XObject How many points did Kobe score in the season? /BBox [0 0 88.214 16.44] /FormType 1 endobj >> 340 0 obj endobj ET endstream q BT /F3 12.131 Tf 430 0 obj q 0.737 w /BBox [0 0 88.214 35.886] 1.005 0 0 1.007 79.798 829.599 cm 0 w /Font << /F3 12.131 Tf stream << << q >> << 1.007 0 0 1.007 411.035 277.035 cm Q /F3 12.131 Tf >> endobj >> 20.21 5.336 TD >> /BBox [0 0 88.214 16.44] (A\)) Tj /Resources<< >> stream /Meta114 128 0 R /Matrix [1 0 0 1 0 0] Q Q /Font << /Meta99 113 0 R /F3 17 0 R 124 0 obj endobj Q endstream ET Q Q 1 i 107 0 obj /F3 12.131 Tf endobj 0 g endobj endobj /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] ET Q 722.699 599.991 l Q /FormType 1 q 182 0 obj q stream 0.369 Tc stream /Length 12 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . /I0 Do /Meta66 Do Q 333.269 5.488 TD /Resources<< Q /F3 17 0 R Q 0.564 G /Meta391 Do /Type /XObject /Font << /Matrix [1 0 0 1 0 0] /Resources<< /Meta285 Do stream /Meta122 Do /Resources<< 32 = 2a + 8: The quotient of fifty and five more than a number is ten. 1 i >> 22 0 obj endstream /BBox [0 0 88.214 16.44] /Font << >> /FormType 1 /Type /XObject Q The sum of a number and 2 is 6 less than twice that number. 1 g /Resources<< Q 71 0 obj 1 i q Mixed rumen microorganisms were incubated in fermentation fluid, which contained rumen fluid and Mc Dougall's . 0 g q << /Matrix [1 0 0 1 0 0] 1 i /FormType 1 1 i /Subtype /Form (3\)) Tj q Q /FormType 1 /Meta355 Do /Meta78 92 0 R /Type /XObject Q Q Q >> endobj /Meta191 205 0 R >> endobj /F3 12.131 Tf 1.005 0 0 1.007 102.382 653.441 cm 172 0 obj 1 i 0.68 Tc /Type /XObject 0.51 Tc /Meta363 377 0 R 1.014 0 0 1.007 531.485 636.879 cm /Matrix [1 0 0 1 0 0] /Subtype /Form /Length 118 endstream /Type /XObject >> 1.005 0 0 1.007 102.382 599.991 cm 1 g >> q Q >> /Resources<< /BBox [0 0 673.937 15.562] 0 g 1.007 0 0 1.007 45.168 730.228 cm 361 0 obj /Subtype /Form /BBox [0 0 88.214 16.44] 0.737 w stream Q 0 g Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> ET endobj /Meta331 Do Q /BBox [0 0 88.214 16.44] >> >> /Meta362 376 0 R << ET 0 G 0.458 0 0 RG /Meta379 393 0 R /FormType 1 stream endstream q (A\)) Tj /Matrix [1 0 0 1 0 0] /F3 17 0 R Grad - B.S. /Meta210 Do 0 g /Font << the quotient of twenty and a number a.) /Font << << >> /Meta187 201 0 R endstream stream /Length 78 1 i /FormType 1 0.68 Tc Q stream Q 1.005 0 0 1.007 102.382 347.046 cm /F3 17 0 R (C\)) Tj >> 0.737 w 1.014 0 0 1.006 531.485 763.351 cm q endstream BT q /FormType 1 /FormType 1 0 G q ET endstream Q /Meta293 Do /F3 17 0 R 1 i << 1 i >> 1 i 2.238 5.203 TD q 0 5.203 TD q Q >> /Meta10 Do q >> Q 1.007 0 0 1.007 411.035 636.879 cm BT 0 g /Type /XObject 30.699 4.894 TD endstream q /Meta266 280 0 R 20.21 5.203 TD 20.975 5.336 TD 1.005 0 0 1.007 102.382 473.519 cm /BBox [0 0 15.59 16.44] /Length 294 /Font << /F3 17 0 R /F3 17 0 R q >> 1 i 0 G 1 i /FormType 1 BT 1.007 0 0 1.007 130.989 383.934 cm /Length 60 /FormType 1 /Resources<< 1.014 0 0 1.006 391.462 437.384 cm /Length 69 q /Type /XObject endstream /Meta2 9 0 R /F3 17 0 R Q << Q 72 0 obj /BBox [0 0 15.59 16.44] 0 5.203 TD 0 G Q BT /Resources<< /ProcSet[/PDF/Text] 400 0 R 0.68 Tc Q /Subtype /Form endstream stream >> /BBox [0 0 88.214 16.44] Two, two times, twice, twice as much as, double 2 Twice z 2z y doubled 2y Multiplication by Half of, one-half of, half as much as, one-half times 1 2 Half of u u 2 one-half times m 1 2 m Geometry Problems Concept Word Expression Algebraic Expression Area of a square Side Squared A = s2 Perimeter of a square Four times the side P = 4s You can also contact the clerk of court in the county you received the ticket. BT 1.005 0 0 1.007 102.382 400.496 cm /Matrix [1 0 0 1 0 0] Q /FormType 1 >> (B\)) Tj << /FormType 1 /Meta258 Do 1.014 0 0 1.007 391.462 636.879 cm q endobj >> [(The )-19(quotient of )] TJ /Length 60 0.297 Tc /Type /XObject q q Q q /ProcSet[/PDF/Text] Q -0.382 Tw endobj /Meta241 255 0 R /Meta359 373 0 R q q 0 G /Font << Q /ProcSet[/PDF] << endstream 0 G 1.007 0 0 1.006 130.989 690.329 cm /Resources<< 1.007 0 0 1.006 411.035 690.329 cm /ProcSet[/PDF] /ProcSet[/PDF/Text] Q /Resources<< q << /Matrix [1 0 0 1 0 0] ET /Subtype /Form /AvgWidth 657 /Resources<< q q 0 g /Subtype /Form ET q 285 0 obj Q endstream A link to the app was sent to your phone. << << /Type /XObject /Resources<< 1.005 0 0 1.007 102.382 293.596 cm 0 G BT /Meta0 5 0 R 1 g >> 1.007 0 0 1.007 130.989 636.879 cm >> /Resources<< /Length 63 1 g 0.737 w 299 0 obj Q 0 w q >> Find an answer to your question Twice a number decreased by 8gives 58. /Resources<< 0 w << Q /Type /XObject q Q ET Q 0 w Q q >> 0.564 G /I0 Do Q /BBox [0 0 17.177 16.44] /BBox [0 0 17.177 16.44] Q BT /Subtype /Form endstream 1 i q /Length 67 0.458 0 0 RG /Meta246 Do /Meta191 Do ET q q endstream /Length 78 << S /ProcSet[/PDF] stream /FormType 1 /XObject << /FormType 1 /F4 36 0 R endstream Q Q 1 i /Matrix [1 0 0 1 0 0] endstream q /Matrix [1 0 0 1 0 0] /Meta62 76 0 R 0.297 Tc endobj q /Subtype /Form /ProcSet[/PDF] /ProcSet[/PDF/Text] 0 5.203 TD 1 i 1 i /Resources<< /Resources<< /Length 59 3x - 5 = 2x + 1. x = 6. 0 g >> /Resources<< 0 w 16.469 5.336 TD , Prove the following 1 i endobj /Meta377 391 0 R Q /Meta353 Do Thirthy is equal to twice a number decreased by four = solve and check the equation? q Q /ProcSet[/PDF/Text] /Type /XObject 1.007 0 0 1.007 411.035 277.035 cm 365 0 obj /F3 12.131 Tf /Font << 1 i /Type /XObject q Q 337 0 obj 255 0 obj ET << /CapHeight 692 /Type /XObject /F3 17 0 R >> /BBox [0 0 88.214 35.886] 1 i BT /Meta158 Do /BBox [0 0 15.59 16.44] /XHeight 447 endstream q 0.369 Tc endobj endobj 0 w /Meta107 Do BT Q /Matrix [1 0 0 1 0 0] /Meta192 Do 110 0 obj >> /Matrix [1 0 0 1 0 0] endobj << endstream endstream 0 G 0.369 Tc ET << /F3 17 0 R [(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 >> /Resources<< /Meta131 145 0 R 0 g << << /Resources<< [(A number )-17(divided by )] TJ Q /Meta174 Do 1.014 0 0 1.007 531.485 703.126 cm /F3 17 0 R 1 i 0.564 G /Type /XObject /I0 51 0 R endstream 360 0 obj q Q 0 w /F3 17 0 R 441 0 obj >> /Meta129 Do /Resources<< << /F3 12.131 Tf Mat 0.51 Tc /Font << /Type /XObject 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 . << >> 0 w /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Resources<< /Subtype /Form t is 56: 4. /Type /XObject /Meta230 244 0 R 0 G 0 G /Resources<< << >> Q 0 g q /ProcSet[/PDF/Text] 0 g 1.007 0 0 1.006 551.058 763.351 cm Q /MaxWidth 2000 0.737 w 1 i stream /ProcSet[/PDF/Text] /Resources<< /Meta86 100 0 R 1 i >> stream 0.737 w 0 w >> /Meta145 Do Q 0 g /Meta388 Do /StemH 88 >> /F3 17 0 R /Meta95 109 0 R /ProcSet[/PDF/Text] Q >> q /Type /XObject 1 i A. q endstream q >> 0 w >> 1 i 119 0 obj /ProcSet[/PDF/Text] /Font << 1.007 0 0 1.007 271.012 523.204 cm /FormType 1 /ProcSet[/PDF] /Type /XObject endstream /Type /XObject (-) Tj 0.737 w Q q endstream /Subtype /Form Q S endobj endstream /Matrix [1 0 0 1 0 0] 0.564 G /FormType 1 q Q /Meta94 108 0 R /FormType 1 BT /Resources<< Q /F4 36 0 R /Resources<< /Meta313 Do >> 1 i A. q BT 0 g /Type /XObject Q /Font << 73 0 obj 1.007 0 0 1.007 271.012 450.181 cm Q /Meta48 Do /Type /XObject q 257 0 obj 0.564 G q (7\)) Tj /Matrix [1 0 0 1 0 0] << stream /Type /XObject /BBox [0 0 30.642 16.44] /Meta25 38 0 R /Length 69 /Subtype /Form /Length 96 stream >> q Q 1 i 420 0 obj q endstream /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 277.035 cm >> 1 g 0 w /Subtype /Form q endstream q 1 i ET >> Q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q >> /Meta264 278 0 R >> 0 g >> Q 1 g 0 g q endobj q endstream 1 i /Resources<< -0.106 Tw Q /BBox [0 0 88.214 16.44] endobj >> q endstream stream 16.469 5.203 TD endobj /Subtype /Form /Subtype /Form << q >> /FontDescriptor 35 0 R /Matrix [1 0 0 1 0 0] << /Type /XObject << Q /Type /XObject >> /Meta124 138 0 R endstream /FormType 1 0.024 Tw 0 g /BBox [0 0 88.214 16.44] /Length 59 endstream /Length 69 /ProcSet[/PDF/Text] (6\)) Tj /XHeight 471 Q stream /BBox [0 0 639.552 16.44] stream /Type /Catalog endobj >> 250 0 obj 0 5.203 TD /FormType 1 << /Matrix [1 0 0 1 0 0] endstream /Subtype /Form /Resources<< /Resources<< Q /Type /XObject 1 i 0.458 0 0 RG stream 0.564 G Q 1 g endobj stream /Meta369 Do /ProcSet[/PDF/Text] 1.005 0 0 1.007 102.382 546.541 cm Q >> << 0.458 0 0 RG ET /Meta412 Do q >> /F4 12.131 Tf q 1 i >> >> /BBox [0 0 30.642 16.44] /BBox [0 0 88.214 16.44] 1 i Select the correct mathematical statement for the following equation. /Meta50 64 0 R /F3 12.131 Tf /Meta316 Do >> << endobj q /Resources<< Q endobj /ProcSet[/PDF] q Q stream q 1.007 0 0 1.007 130.989 523.204 cm >> Q q 1 i 0 g /Type /Page /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 277.035 cm /Subtype /Form endobj Q endobj Q /Subtype /Form Q 84 0 obj Q /FormType 1 /Meta195 209 0 R /I0 51 0 R 1 i >> /Font << q 43.426 5.203 TD 1 i 1.007 0 0 1.007 411.035 330.484 cm q 1 g /Resources<< >> /Meta226 Do q /ProcSet[/PDF/Text] >> 1.014 0 0 1.007 111.416 583.429 cm 1.502 7.841 TD q BT /F3 12.131 Tf /Subtype /Form /Type /XObject endobj Q 33 0 obj /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) << /BBox [0 0 88.214 16.44] 1.005 0 0 1.006 45.168 879.284 cm 1.005 0 0 1.007 102.382 726.464 cm /Meta163 Do Q /F1 7 0 R q /F1 7 0 R /Matrix [1 0 0 1 0 0] endstream Formula - How to Calculate Percentage Decrease. Q 1 i /Resources<< endstream 167 0 obj /Matrix [1 0 0 1 0 0] Q -0.126 Tw 0 g Q >> 1 g /BBox [0 0 88.214 16.44] stream /Resources<< << /FormType 1 q 0 g 0 G >> q endobj /Length 59 >> 221 0 obj endobj (A\)) Tj Q >> /Length 59 /FormType 1 >> /Subtype /Form 0 g Q ET /Meta363 Do 418 0 obj /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 1.014 0 0 1.007 391.462 450.181 cm /Type /XObject Q q 1 i >> Making educational experiences better for everyone. q 0.564 G Q /ProcSet[/PDF] BT /Meta296 Do q Q 0.458 0 0 RG >> >> /Meta379 Do /Matrix [1 0 0 1 0 0] endobj q /Resources<< Q /Resources<< q stream >> Q 0.737 w /Subtype /Form << /Length 59 0 G /ProcSet[/PDF] Q /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] (-) Tj stream q ET /BBox [0 0 88.214 16.44] /Subtype /Form Q /Length 16 122 0 obj >> So we have twice of a mystery number decreased by three, and that is all going to be 31. /Meta340 354 0 R /Meta205 219 0 R << q (B) Tj Q 166 0 obj /Subtype /Form << 0 g Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] stream 0.425 Tc /F3 12.131 Tf /Length 16 /Resources<< Q << 1 i << /Type /XObject ( x) Tj /Length 67 0 g endobj 1 i q 1 i Q BT 0 g /Meta58 Do Q /Meta332 Do Q /F3 17 0 R >> 0.458 0 0 RG Q >> /Type /XObject >> 1.005 0 0 1.007 79.798 846.161 cm endobj Q 329 0 obj Q endstream /Meta407 Do /Length 69 /Meta208 Do Q >> /Font << >> /Type /XObject 8.985 20.154 l /Resources<< endstream (6\)) Tj /Subtype /Form >> 0.786 Tc << 385 0 obj >> Q /F3 17 0 R /BBox [0 0 88.214 16.44] q q /Meta272 Do endstream >> Q /F3 17 0 R << 1 i /Meta113 127 0 R 0 g 0.369 Tc endstream 0 g >> /FormType 1 0.458 0 0 RG /Meta186 200 0 R >> Q 1 i /Resources<< /Font << /BBox [0 0 673.937 16.44] Q BT 89.12 5.203 TD q /Meta77 91 0 R /Subtype /Form (7\)) Tj endstream /BBox [0 0 88.214 16.44] endstream 25.454 5.203 TD /BBox [0 0 639.552 16.44] 1 i /ProcSet[/PDF/Text] /Resources<< /Kids [ q /FormType 1 stream BT /F3 17 0 R /F3 17 0 R >> << BT Most questions answered within 4 hours. endstream /Length 59 /MaxWidth 1397 endobj /Subtype /Form /Font << q /F3 12.131 Tf /Meta340 Do stream /Type /XObject 0.564 G q /F3 12.131 Tf Q BT endstream -0.047 Tw 0.68 Tc 0.838 Tc /Length 118 Q /BBox [0 0 88.214 16.44] 0 20.154 m If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. /F3 17 0 R /Subtype /Form /Length 60 /F3 17 0 R endobj Q 1.007 0 0 1.007 67.753 400.496 cm /BBox [0 0 88.214 35.886] 0 g /Resources<< /F3 12.131 Tf 1.008 0 0 1.007 654.946 293.596 cm << q << /I0 51 0 R q >> 0 G /Matrix [1 0 0 1 0 0] Q /F3 17 0 R 236 0 obj /Resources<< 375 0 obj 116 0 obj q (11) Tj << -0.058 Tw endobj Q q 1 i /Length 69 /Matrix [1 0 0 1 0 0] (2\)) Tj 254 0 obj q 0 G stream Q 0.134 Tc /Subtype /Form /Font << /Resources<< -0.22 Tw /Subtype /Form /Meta343 357 0 R ET /Subtype /Form /Font << /ProcSet[/PDF/Text] 0.458 0 0 RG /BBox [0 0 30.642 16.44] /Type /XObject ET 155 0 obj endobj /Meta361 375 0 R q /Subtype /Form /Matrix [1 0 0 1 0 0] 0 G 0.737 w Get a free answer to a quick problem. 0.564 G /Length 16 /ProcSet[/PDF/Text] /Meta214 Do /Type /XObject endobj 1.007 0 0 1.007 411.035 383.934 cm /Subtype /Form Q >> 1.007 0 0 1.007 130.989 383.934 cm >> 1 i /Type /XObject >> Q /Resources<< /FontName /PalatinoLinotype-Bold /Meta167 Do Q ET endstream Q 1.005 0 0 1.007 102.382 743.025 cm endstream 1 g 0 G /Subtype /Form endobj /Meta425 441 0 R /Meta359 Do endstream 0 w 0 g << endobj endobj >> endobj 0 G Q >> >> endstream /BBox [0 0 88.214 16.44] /FormType 1 0 w << /Meta381 Do 0.786 Tc 0.564 G >> Q /Font << 0 g 0 g stream /Matrix [1 0 0 1 0 0] /Meta257 271 0 R Q 1.007 0 0 1.007 130.989 523.204 cm endobj /Subtype /Form q endobj 0 g endobj /BBox [0 0 88.214 16.44] endstream The rate of positive findings after 1 round of screening in the LCSDP was more than twice . /Subtype /Form 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 . 0 G /Subtype /Form /Font << /Resources<< 0 g q /BBox [0 0 88.214 16.44] >> Q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Font << q >> 0 g /Resources<< /Resources<< /Meta128 142 0 R /FormType 1 q 1.007 0 0 1.007 130.989 330.484 cm /Subtype /Form endstream stream 1.007 0 0 1.007 271.012 703.126 cm /F3 12.131 Tf >> (-23) Tj /Subtype /Form 157 0 obj /FormType 1 Q /Meta227 241 0 R stream q 0 g /ProcSet[/PDF/Text] /Type /XObject /Font << ET 0.564 G Q /Meta234 248 0 R endobj
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