/Subtype /Form 0 g 1.007 0 0 1.007 271.012 277.035 cm (D\)) Tj /Resources<< /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 0 w 1.014 0 0 1.006 531.485 836.374 cm /BBox [0 0 88.214 16.44] 0.297 Tc /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 400.496 cm /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 271.012 849.172 cm 0 g stream 430 0 obj 1 i /FormType 1 << Q Q 0.17 Tc 672.261 599.991 m 12.727 5.203 TD q /Matrix [1 0 0 1 0 0] 1 g /Subtype /Form Q /Type /Font Q Q 278 0 obj -0.03 Tw endstream /Type /XObject /Subtype /Form >> 0.269 Tc endobj 0.737 w q ( x) Tj /Type /XObject /Matrix [1 0 0 1 0 0] endobj 1 i /Font << Mat Q q endstream /Resources<< Q endobj Q 387 0 obj Q << >> /ProcSet[/PDF/Text] q >> >> /ProcSet[/PDF/Text] 1 i /Type /XObject << 0.458 0 0 RG endobj What word phrase can you use to represent 5x + 2? BT 0 g 1 i /Length 69 1.014 0 0 1.007 111.416 636.879 cm 0.737 w /Subtype /Form /F3 12.131 Tf >> q Thrice a number decreased by 5 exceeds twice the number by 1 is . Q >> 47.933 5.203 TD /FormType 1 Q /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /BBox [0 0 88.214 16.44] >> /Resources<< /FormType 1 142 0 obj 0 w q 101.849 5.203 TD q endstream /Subtype /Form << /Meta377 Do /Meta199 213 0 R >> << >> >> 0 g /Type /XObject 143 0 obj q /Meta57 71 0 R /Meta293 307 0 R >> >> /F3 12.131 Tf ET << 0 G endstream >> 138 0 obj 1.007 0 0 1.007 551.058 636.879 cm /Type /Font Q stream 1 i /Subtype /Form /FormType 1 /Subtype /Form 370 0 obj 0.564 G << >> stream stream 40.45 4.894 TD /Meta146 160 0 R /Type /XObject /Matrix [1 0 0 1 0 0] >> << You can specify conditions of storing and accessing cookies in your browser. 0 G /Type /XObject 0.458 0 0 RG 1 i Q Q /Font << /Meta67 Do /FormType 1 /FormType 1 /Meta224 238 0 R /BBox [0 0 88.214 16.44] 0 G (A\)) Tj 1.007 0 0 1.006 551.058 437.384 cm 0 g /Subtype /Form << q 1 g 1 i 0 g /Subtype /Form q Q endstream If a number is 50%, then it is a half - the same as 0.5 or 1/2. 1 i /Font << endstream endstream 1 i /Subtype /Form 333 0 obj 1.007 0 0 1.007 271.012 636.879 cm /Resources<< /Font << >> /Meta362 Do q /Font << /Subtype /TrueType 0 g /Resources<< /Type /XObject /F3 17 0 R 1 i stream /Meta267 281 0 R Twice a number decreased by 58! /Resources<< 0.458 0 0 RG q /Meta390 406 0 R q /F3 12.131 Tf ET endstream << /Font << Q 180 0 obj /Resources<< endobj /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Type /XObject /Meta51 Do ET Tamang sagot sa tanong: 1.) Q Q stream q /BBox [0 0 88.214 16.44] /F3 12.131 Tf endstream /Resources<< q q Q /BBox [0 0 30.642 16.44] q >> ET /Font << endobj Q endstream /Resources<< 1 i endobj /Resources<< /Length 80 /F3 12.131 Tf 1 g In the problem above, x is a variable. ET 178 0 obj S /Font << Q 0.369 Tc >> /BBox [0 0 15.59 16.44] endstream q q << endstream /F1 12.131 Tf endobj ET Q Q endstream /Length 68 /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] 329 0 obj endstream ET Q >> 0 G endstream /Matrix [1 0 0 1 0 0] 1.005 0 0 1.015 45.168 53.449 cm 0.297 Tc /Meta394 410 0 R >> Q /Type /XObject /Meta386 402 0 R -0.486 Tw 0.564 G Q ET endstream 443 0 obj q q Q q Q /Meta221 Do 1.007 0 0 1.007 411.035 849.172 cm /FormType 1 423 0 obj /Length 16 How many points did Kobe score in the season? 1.007 0 0 1.007 411.035 330.484 cm 0 g Q Q >> /Type /XObject /Matrix [1 0 0 1 0 0] BT Q /Resources<< 1.005 0 0 1.007 79.798 763.351 cm /Resources<< /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] Q endstream q stream q endobj /BBox [0 0 17.177 16.44] 299 0 obj /Subtype /Form 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. Q Q endstream 276 0 obj /FormType 1 /BBox [0 0 15.59 29.168] Q << Q /Resources<< 0.737 w stream /Resources<< 30.699 5.203 TD 1 g endstream /Resources<< 0 w q /BBox [0 0 534.67 16.44] endobj << Q 0 g >> /F3 17 0 R /Subtype /Form /BBox [0 0 15.59 16.44] /F3 17 0 R q endobj /Meta328 Do /Type /XObject Q Q << q /Resources<< /Length 12 /Type /XObject >> ET 0 g Q /Resources<< q Answer (1 of 8): Solution: let the number be x. << 20.21 5.203 TD /Meta369 Do >> endobj << 194 0 obj endstream endobj 0.458 0 0 RG 347 0 obj 0 g 1 g /F3 17 0 R ET /FormType 1 /Matrix [1 0 0 1 0 0] Q /Meta135 149 0 R 1 i endstream 0 G q /BBox [0 0 88.214 16.44] q /Type /XObject >> 0.486 Tc /ProcSet[/PDF/Text] q BT 0 w /Length 74 /Resources<< /Length 245 q 6.746 24.649 TD >> /Length 69 /Contents [399 0 R] /Type /XObject >> endstream 1.007 0 0 1.007 411.035 330.484 cm 0 g 1 i ET /Meta226 240 0 R Q endobj /BBox [0 0 88.214 16.44] stream /Type /XObject Q << /FormType 1 Q /Meta212 Do /Matrix [1 0 0 1 0 0] >> endobj Q /Length 16 q endobj 1.007 0 0 1.007 67.753 293.596 cm << 0.737 w endstream /Font << /ProcSet[/PDF/Text] 0 g Q q /Type /XObject 1 i /Type /XObject /ProcSet[/PDF/Text] q Q q Q 1.007 0 0 1.007 654.946 347.046 cm >> Q /Subtype /Form /BBox [0 0 88.214 16.44] q -0.021 Tw /Resources<< endobj /BBox [0 0 88.214 16.44] /Meta326 340 0 R Q 1 i 342 0 obj Q /Matrix [1 0 0 1 0 0] /Type /Font /Resources<< /Meta211 Do /Meta208 Do endstream 1 i Q 154 0 obj q /Type /XObject >> /FormType 1 0 4.894 TD stream >> q 426 0 obj 0.369 Tc << /Subtype /Form 0 G 15 0 obj Q endstream /Matrix [1 0 0 1 0 0] q << BT 1.502 5.203 TD 1 i /Resources<< endobj stream >> /F4 12.131 Tf << Q q (3\)) Tj >> /Meta19 Do /Length 12 /Resources<< /Length 58 >> >> /F3 12.131 Tf 18 0 obj [(-3)-16(20)] TJ /Resources<< << >> << /Type /XObject /Meta41 55 0 R /Length 12 endstream Q BT /Resources<< q /StemV 94 26.219 5.336 TD 1 i 1 g /FormType 1 A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. /Font << /Meta423 Do /Matrix [1 0 0 1 0 0] q q endstream << 0.564 G /Resources<< 58 0 obj >> /F3 12.131 Tf Q 294 0 obj ET /Type /XObject ET q /F3 12.131 Tf 6.746 5.203 TD endobj /FormType 1 Q q Q Q /F3 17 0 R 1 i /FormType 1 (+) Tj ET Q q Q Q /F3 17 0 R Q /FormType 1 Q 1 i /Type /XObject /ProcSet[/PDF] >> >> q << /Meta187 Do Q /Matrix [1 0 0 1 0 0] Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. /BBox [0 0 88.214 16.44] 352 0 obj /Resources<< 104 0 obj Translate 2(x-58) into mathematical phrase. /Font << 0.564 G /Subtype /Form /FormType 1 Q 66 0 obj /Subtype /Form /FormType 1 /Font << >> Q That was 1/8 of the points that he scored /Font << 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 0 g endobj >> >> /FormType 1 /Subtype /Form /Resources<< /Meta96 110 0 R /Font << endstream Q 1.014 0 0 1.007 251.439 703.126 cm /F1 7 0 R << >> 1 i /F3 17 0 R Q Q q >> /Font << 1.007 0 0 1.007 130.989 330.484 cm /Meta367 381 0 R /FontDescriptor 16 0 R /Length 69 endobj Q Q q endobj Q 0.458 0 0 RG /Meta275 Do Q /Subtype /Form q /F3 12.131 Tf q /Meta180 194 0 R 0.737 w /BBox [0 0 549.552 16.44] BT /F3 12.131 Tf q q /Length 16 Q /Font << Q 0 g ET /ProcSet[/PDF/Text] endstream /FormType 1 ET /Font << /Length 16 /Font << /Length 58 283 0 obj q /ProcSet[/PDF/Text] >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Meta161 Do 1.007 0 0 1.006 551.058 836.374 cm Q /ProcSet[/PDF] 259 0 obj /F3 12.131 Tf /Type /XObject 1.007 0 0 1.007 551.058 277.035 cm 0 g /Resources<< /Resources<< Q /Length 69 /FormType 1 0 g Q >> >> q /Length 63 >> /FormType 1 /F3 17 0 R ET q >> /Meta189 203 0 R /Subtype /Form (D\)) Tj q q /Meta313 327 0 R /Size 447 /Meta216 230 0 R /Length 59 2.238 5.203 TD q Q q /F3 12.131 Tf /Type /FontDescriptor /ItalicAngle 0 >> /Type /XObject Q 25.454 5.203 TD endobj /Subtype /Form /BBox [0 0 88.214 16.44] 0 g 0 g Q /BBox [0 0 30.642 16.44] q Expert Answer. >> /Matrix [1 0 0 1 0 0] >> /Subtype /Form Q q stream 2x - y = 6. /F3 17 0 R /Type /XObject /Type /XObject /ProcSet[/PDF] >> >> /Resources<< /Type /XObject endstream (-11) Tj Q /ProcSet[/PDF/Text] /F3 17 0 R 1 g /F3 17 0 R Q /F3 12.131 Tf >> Q 0 G >> Q /Meta11 22 0 R 0 g ET 0 g /Length 16 ET >> /Resources<< 0 w endobj Twice = two times, double. /Font << >> 0 G (- 4) Tj 1.007 0 0 1.007 411.035 330.484 cm >> >> 1 i Diabetes, if left untreated, leads to many health complications. 71 0 obj /Resources<< (9\)) Tj /Resources<< -0.486 Tw /Subtype /Form Q /Length 54 /Meta8 Do 0 w /Meta360 Do /Font << 1.007 0 0 1.007 411.035 383.934 cm /Resources<< 0.425 Tc ET >> 1.007 0 0 1.007 654.946 400.496 cm /BBox [0 0 639.552 16.44] /FormType 1 0.564 G 0 w 0 g /Subtype /Form /Subtype /Form /Meta366 Do q 1.014 0 0 1.007 251.439 383.934 cm Q /Length 67 >> Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. Q endstream Q /BBox [0 0 549.552 16.44] Q /Matrix [1 0 0 1 0 0] 19.474 20.154 l Q /Type /XObject >> q /F3 17 0 R Just type into the box and your calculation will happen automatically. /Font << /BBox [0 0 639.552 16.44] /Font << [(The )-16(s)15(um )-14(of )] TJ 444 0 obj /Length 16 1 i 0 G 1.007 0 0 1.007 551.058 277.035 cm Q stream Q /Font << endstream /Resources<< Q ET /Subtype /Form /Type /XObject /Length 68 Q /Font << 323 0 obj , Prove the following 0 G q /Subtype /Form /Resources<< 0.737 w /FormType 1 /Encoding /WinAnsiEncoding Q Q /Meta50 Do >> /Meta103 Do /Font << >> /Meta429 445 0 R /Type /XObject 3x - 5 = 2x + 1. x = 6. >> /BBox [0 0 15.59 29.168] /Length 118 /Font << /Type /XObject /Resources<< Q /Type /XObject /Length 69 << (B\)) Tj /BBox [0 0 88.214 16.44] /Resources<< 265 0 obj 0 g q 193 0 obj /Font << Q 1.007 0 0 1.007 411.035 636.879 cm endobj endstream 0 5.203 TD Q /Resources<< >> /Resources<< /Type /XObject /Font << Q /Font << /Subtype /Form endstream /ProcSet[/PDF] /Subtype /Form 1 i stream 234 0 obj /Font << q 1 i << >> /BBox [0 0 88.214 35.886] Q 0.369 Tc /FormType 1 0 G (B\)) Tj 0 g 0.486 Tc /BBox [0 0 17.177 16.44] Q /F3 17 0 R /Matrix [1 0 0 1 0 0] q >> (viii) A number divided by 8 gives 7. q Q /FormType 1 414 0 obj ET /FirstChar 32 1.005 0 0 1.007 102.382 563.103 cm 0 g /FormType 1 Q stream q /Resources<< endstream /F3 17 0 R /Matrix [1 0 0 1 0 0] endobj q << /F1 12.131 Tf 0.564 G /Meta29 42 0 R /BBox [0 0 88.214 16.44] 0.001 Tw 21.713 20.154 l q /Type /XObject stream 1 g 0.51 Tc /BBox [0 0 88.214 16.44] >> q /ProcSet[/PDF] << 1.007 0 0 1.007 411.035 330.484 cm 1.007 0 0 1.007 551.058 277.035 cm 311 0 obj /Type /XObject endobj /F3 12.131 Tf /ProcSet[/PDF/Text] /ProcSet[/PDF] /Meta376 Do /Meta157 Do 20.21 5.203 TD 1.005 0 0 1.013 45.168 933.487 cm q /Meta193 207 0 R ET 0.458 0 0 RG /Type /XObject >> BT 0.564 G q /FormType 1 /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] Q endobj /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 271.012 277.035 cm (-11) Tj ET << /Meta71 Do stream /Font << /Meta228 242 0 R >> 43.426 5.203 TD Q stream /Length 65 q q /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] stream 0 g 0 G >> -0.16 Tw /Meta349 Do /Resources<< >> stream /Meta46 60 0 R /Meta15 Do endstream stream 1 g Q /FormType 1 /Font << BT BT Q 0 g /Meta156 Do q /Length 118 0 g >> /Subtype /Form >> endobj q /Length 69 /Length 69 /Resources<< /Subtype /Form endstream BT q /Meta169 Do q Q q 363 0 obj q q /XObject << 0 g 1.007 0 0 1.007 551.058 277.035 cm >> >> /Type /XObject Q BT << /I0 Do >> q /Subtype /Form Q q q << /Meta278 Do BT 0.458 0 0 RG /Resources<< Q /F3 17 0 R q Q 335 0 obj /FormType 1 /ProcSet[/PDF] 0 g /Resources<< Q >> q 0 g Q >> 0.838 Tc /Meta184 Do 0 g /Meta68 Do /Meta172 Do Find the number. endobj 1.014 0 0 1.006 391.462 836.374 cm /Matrix [1 0 0 1 0 0] >> 1.007 0 0 1.007 551.058 383.934 cm /ProcSet[/PDF/Text] (5\)) Tj 0 w endobj q 1. /F4 36 0 R endobj /Meta296 Do stream q 0 g /Length 118 >> 77 0 obj 1.005 0 0 1.007 79.798 779.913 cm endstream /BBox [0 0 17.177 16.44] /Subtype /Form /ProcSet[/PDF/Text] /Type /XObject << endstream /Subtype /Form /ProcSet[/PDF] /BBox [0 0 30.642 16.44] 0 g q /F1 12.131 Tf /Meta245 259 0 R 0 G q endstream 0 g /ProcSet[/PDF/Text] endstream /FormType 1 >> /F3 17 0 R /ProcSet[/PDF/Text] q 1.014 0 0 1.006 391.462 510.406 cm 101 0 obj 0.564 G /Subtype /Form /FormType 1 /Matrix [1 0 0 1 0 0] /F3 12.131 Tf q /Matrix [1 0 0 1 0 0] Q >> 1 i /Font << /Length 12 endobj /ProcSet[/PDF] /Subtype /Form /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] ( x) Tj /Matrix [1 0 0 1 0 0] endobj 672.261 653.441 m 1 i >> /Matrix [1 0 0 1 0 0] /Meta137 151 0 R Q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] (5) Tj Q /Font << BT 0.458 0 0 RG BT q /Font << q /FormType 1 /Parent 1 0 R /F3 17 0 R /FontDescriptor 6 0 R Q /Type /XObject /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Length 69 0.369 Tc 1 i 131 0 obj endobj /F3 12.131 Tf q /ProcSet[/PDF] ET 1.007 0 0 1.007 271.012 277.035 cm q Q q /ProcSet[/PDF/Text] 0 5.203 TD 1 i /Type /XObject endobj A link to the app was sent to your phone. q endobj q /Font << stream /FormType 1 /Subtype /Form Q (5) Tj /Length 54 Answer provided by our tutors. /Resources<< endstream 0 g /Subtype /Form /Resources<< /Subtype /Form Q >> << /Matrix [1 0 0 1 0 0] q q You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. /Type /XObject 1.007 0 0 1.007 411.035 636.879 cm q /Meta104 118 0 R /FormType 1 1 i In other terms, 52-nx The problem is asking that you subtract twice a number from 52. endobj ET 0 G /DecodeParms [<> ] /F4 36 0 R /Resources<< /Meta283 Do Q 1.007 0 0 1.007 551.058 277.035 cm (3\)) Tj /BBox [0 0 15.59 16.44] q /Matrix [1 0 0 1 0 0] 1 g /Resources<< /Resources<< BT >> 1.007 0 0 1.007 271.012 383.934 cm /Meta144 158 0 R q Q Q 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 /Meta84 Do 1 i >> q 1.007 0 0 1.007 130.989 277.035 cm /F1 12.131 Tf Q stream << endobj >> /Length 16 /Meta59 Do 303 0 obj /Type /XObject /Font << endstream 0 5.203 TD /Length 69 0 g /BBox [0 0 88.214 16.44] q q /BBox [0 0 88.214 16.44] >> 1.007 0 0 1.007 45.168 862.723 cm << /F3 12.131 Tf /Subtype /Form Q /Font << Q << Q /ProcSet[/PDF] Q /Matrix [1 0 0 1 0 0] /Length 69 endstream /F3 12.131 Tf 1 i 1.502 5.203 TD >> >> stream /Type /XObject Q endobj >> Q Q Get link; Facebook; Twitter; q /F3 17 0 R Q /F3 12.131 Tf 0.737 w /BBox [0 0 88.214 16.44] q /Font << 0 G q >> Q /FormType 1 1 i Q 0 5.203 TD >> q (4) Tj Q q q Q /Resources<< 0 g endstream stream endstream << Q 0 g 1.007 0 0 1.007 271.012 776.149 cm /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] << /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q q /Resources<< 0 G q q /Resources<< /BBox [0 0 88.214 35.886] (1\)) Tj Q 0 G >> /Subtype /Form 1 i q Q /Type /XObject /ProcSet[/PDF] /Type /XObject >> endstream 48 0 obj ET >> /Meta215 Do An example of a linear inequality in one variable is A. x+y = Question: answer 1. 0 G 20.21 5.203 TD 0.425 Tc /Subtype /Form The sum Of twice a nu4ber What is the number? q q 0 G 23.952 4.894 TD >> q 1.007 0 0 1.007 551.058 330.484 cm 425 0 obj ET 144 0 obj q >> /F1 7 0 R << /Resources<< >> /Matrix [1 0 0 1 0 0] /Type /XObject /BBox [0 0 15.59 16.44] 0 g q >> endstream >> /Font << q /Meta414 Do 4 0 obj ET ET /Subtype /Form ET /F3 12.131 Tf C. Twice a number decreased by ten is at most 24. /Subtype /Form 19.474 20.154 l 0 G 1.007 0 0 1.007 411.035 383.934 cm endstream endstream /FormType 1 ET /Font << q >> 1.005 0 0 1.007 102.382 743.025 cm 1 g >> (C) Tj Q /Meta80 Do >> /F3 17 0 R /FormType 1 << /Length 16 stream endobj >> q 1 i /BBox [0 0 15.59 29.168] /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] 25.454 5.203 TD /Meta358 372 0 R View the full answer. 0.564 G endstream /Meta379 393 0 R 549.694 0 0 16.469 0 -0.0283 cm /BaseFont /PalatinoLinotype-Roman /Type /XObject endobj ET q (-8) Tj /F3 12.131 Tf /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] endstream 0 G endstream q Q /FormType 1 0.737 w /Font << 1 i endstream /Meta411 Do /Meta263 277 0 R 0.458 0 0 RG 1 i Q [(A number )-17(divided by )] TJ 1 g /Length 54 /F3 12.131 Tf ET /Type /XObject >> Q /F1 12.131 Tf endstream stream q /ProcSet[/PDF/Text] << >> 0 g [(The )-19(quotient of )] TJ /Meta240 Do 1 i q 2.238 5.203 TD endobj 0 g 0 G 9.723 5.336 TD Q << Q 0.458 0 0 RG 0 G 0 G /Type /XObject 0.564 G q 0 g q /Meta45 59 0 R 0 g 1.005 0 0 1.007 102.382 400.496 cm /Length 59 0.738 Tc >> /Resources<< << /Type /XObject 0.838 Tc q /Meta49 Do /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Resources<< 1.007 0 0 1.007 551.058 383.934 cm q /Font << -0.067 Tw 0 G /BBox [0 0 88.214 35.886] >> stream /Meta153 Do 0 G /BBox [0 0 17.177 16.44] endobj >> << /Length 59 /Type /XObject Q 0 g >> stream 0.737 w 1.008 0 0 1.007 654.946 293.596 cm 1.007 0 0 1.007 130.989 776.149 cm >> /Length 16 /FormType 1 /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) /Matrix [1 0 0 1 0 0] ET >> Q stream /Matrix [1 0 0 1 0 0] Q 53 0 obj /FormType 1 1 i /Meta429 Do 722.699 546.541 l /Meta38 52 0 R /Meta335 349 0 R >> /BBox [0 0 88.214 16.44] 1 i q /BBox [0 0 88.214 16.44] q /Count 2 /ProcSet[/PDF/Text] Q /F3 12.131 Tf 1 i /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF/Text] 1.005 0 0 1.015 45.168 53.449 cm 327 0 obj /FormType 1 q /Resources<< Q 0 g Q /Matrix [1 0 0 1 0 0] /Length 69 /BBox [0 0 15.59 16.44] << Q >> /Matrix [1 0 0 1 0 0] 1 g ET q /Subtype /Form q /BBox [0 0 15.59 29.168] /Meta253 267 0 R 0 g /Subtype /Form Find the number. /F3 17 0 R stream Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. endstream endobj /Length 69 /Matrix [1 0 0 1 0 0] 122 0 obj Q /Meta121 Do Q 549.694 0 0 16.469 0 -0.0283 cm endstream >> /Resources<< q Patients' reasons for declining screening were not collected . 1.014 0 0 1.007 391.462 636.879 cm 1.005 0 0 1.007 102.382 347.046 cm BT /Subtype /Form << /Font << 1 i /BBox [0 0 30.642 16.44] endstream 0.737 w 1.502 8.18 TD /BBox [0 0 673.937 16.44] 0 g 1 i 0 39.216 TD /ProcSet[/PDF/Text] q Q << 1 i 0 G /Matrix [1 0 0 1 0 0] q >> 1.007 0 0 1.007 551.058 277.035 cm 1 i /Meta361 375 0 R q endstream 1 i q /Matrix [1 0 0 1 0 0] << /BBox [0 0 88.214 16.44] /Meta129 Do /F3 17 0 R >> 1 i BT 1.007 0 0 1.006 551.058 836.374 cm 1 i Q /Type /XObject >> Q /Subtype /Form Q /Meta264 278 0 R endobj 672.261 872.509 m 410 0 obj endobj 0.51 Tc /F3 12.131 Tf q /XObject << stream 0 g >> >> 0 G 1 i Q 1 i /F3 17 0 R Q q /F3 12.131 Tf Q 0 g 0 g 1.005 0 0 1.007 79.798 813.037 cm 0.458 0 0 RG /ProcSet[/PDF] 0 g Find the number. >> endobj << 0 g >> 0 G /Meta304 Do 1.007 0 0 1.007 411.035 383.934 cm 1.014 0 0 1.007 251.439 277.035 cm /Subtype /Form Q Q /Length 12 /Length 54 /Matrix [1 0 0 1 0 0] >> >> /Meta3 Do << endstream Q /Font << 40 0 obj Q 1.007 0 0 1.007 551.058 383.934 cm 1 i 125.064 4.894 TD /Resources<< /F1 7 0 R endobj endobj /F4 12.131 Tf 1.007 0 0 1.007 130.989 523.204 cm stream Q endstream 1.502 5.203 TD /FormType 1 q /Resources<< endstream /F3 12.131 Tf 0.838 Tc /XHeight 476 The quotient of a seven and a number 9. q 0.737 w Q SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. endobj 1 i /Type /XObject /Meta280 294 0 R /Font << /FormType 1 1.007 0 0 1.007 551.058 330.484 cm 0 G /F1 7 0 R 0 g 1.007 0 0 1.007 551.058 636.879 cm Q 1.007 0 0 1.007 130.989 849.172 cm q /Meta75 Do >> /Subtype /Form /Subtype /Form << >> endstream /Length 63 >> 1 i Q 1.014 0 0 1.007 531.485 330.484 cm /BBox [0 0 30.642 16.44] /F3 12.131 Tf /BBox [0 0 88.214 16.44] /Length 118 /Type /XObject /Length 59 ET /Length 12 /Resources<< /Font << stream 54.679 5.203 TD /Type /XObject 1.007 0 0 1.007 551.058 636.879 cm 1.014 0 0 1.007 251.439 450.181 cm BT /FormType 1 endobj >> endstream /Length 245 Q q 1 i Q In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. >> Q Q /Resources<< /Meta180 Do /ProcSet[/PDF/Text] endobj q /Length 54 /ProcSet[/PDF/Text] 0 w /Font << /ProcSet[/PDF/Text] q /Length 59 0 g >> /Meta342 356 0 R q /F3 17 0 R (-8) Tj /Resources<< >> 0 5.203 TD 0.303 Tc ET /ProcSet[/PDF/Text] << 0 g 1 i 217 0 obj q /ProcSet[/PDF] /Subtype /Form Q endobj /F3 12.131 Tf Q 0 g BT q Want to see the full answer? endobj stream This site is using cookies under cookie policy . stream /Type /XObject /Length 59 /Subtype /Form 36 0 obj Q /ProcSet[/PDF/Text] /Resources<< q 1 g q /Subtype /Form Q 351 0 obj q /Resources<< Q /Meta252 Do (13) Tj ET /F1 7 0 R >> endobj stream Q << /Meta250 264 0 R >> Q /Meta185 199 0 R q << 0.51 Tc 1.014 0 0 1.006 531.485 763.351 cm stream /Meta76 90 0 R /ProcSet[/PDF/Text] 271 0 obj /F1 12.131 Tf /FormType 1 0 G 0 g 1.007 0 0 1.007 551.058 523.204 cm >> q Q /FormType 1 0 G /FormType 1 Q /Font << Q q A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. endobj endstream endobj 0 G stream q stream /ProcSet[/PDF/Text] q /F3 12.131 Tf /BBox [0 0 88.214 16.44] q Q q Q /Meta172 186 0 R endobj 1.005 0 0 1.015 45.168 53.449 cm /F1 14.682 Tf /ProcSet[/PDF] /Meta84 98 0 R stream 235 0 obj >> /Subtype /Form Q endstream /FormType 1 Q 1.007 0 0 1.007 67.753 347.046 cm q endstream /Subtype /Form >> 1.007 0 0 1.006 551.058 437.384 cm 220.931 4.894 TD /Type /XObject 1 i /FormType 1 /Meta197 211 0 R /Type /XObject endobj /Subtype /Form /F3 17 0 R 0.737 w /Subtype /Form 0 G >> q stream /Resources<< 20.21 5.203 TD Q q >> 290 0 obj Q /Resources<< /FormType 1 q Q /FormType 1 >> q >> /Resources<< /Matrix [1 0 0 1 0 0] 1 i q 1.014 0 0 1.007 391.462 776.149 cm /Length 69 q /F3 12.131 Tf 223 0 obj ET /Font << /Length 294 Q endstream /ProcSet[/PDF/Text] >> >> /Meta219 233 0 R 1 i << /Meta397 413 0 R 0 g /F1 12.131 Tf BT /Meta202 216 0 R /Type /XObject 0.369 Tc /F3 17 0 R /Type /XObject /FormType 1 0 g 1 i stream Q q >> /ProcSet[/PDF/Text] BT /Subtype /Form Q q endobj Q /BBox [0 0 88.214 35.886] /FormType 1 /F3 17 0 R /F3 12.131 Tf >> /Meta315 Do /Matrix [1 0 0 1 0 0] /Meta329 343 0 R /Meta326 Do q /F3 17 0 R /CapHeight 692 New questions in Mathematics /Meta106 120 0 R >> q << >> /Matrix [1 0 0 1 0 0] 549.694 0 0 16.469 0 -0.0283 cm /Meta66 Do 300 0 obj ET Q Q /Matrix [1 0 0 1 0 0] Q /ProcSet[/PDF/Text] /Subtype /Form q 1.007 0 0 1.007 67.753 653.441 cm /Resources<< /Type /XObject stream 277 0 obj /Meta259 Do /Length 16 q /Meta318 Do >> (A\)) Tj -0.058 Tw 1.007 0 0 1.006 411.035 690.329 cm /F3 17 0 R 0 g saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. 1.007 0 0 1.007 551.058 330.484 cm 0 g (2) Tj << endstream /ProcSet[/PDF/Text] Q gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. Q 0.737 w 1.007 0 0 1.007 130.989 849.172 cm BT /Meta392 Do stream endstream /FormType 1 q 0 G /Font << >> 0 w >> << 0 w Q q >> /FormType 1 endobj /XHeight 447 0.51 Tc q 0 G /Matrix [1 0 0 1 0 0] << 0 5.203 TD >> Q /Length 69 /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Length 16 q Q (38) Tj /Meta217 231 0 R /F3 17 0 R /Meta425 Do Q /Subtype /Form 1.014 0 0 1.007 111.416 450.181 cm ET << /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] /Length 244 /Subtype /Form /Subtype /Form 0.524 Tc << >> >> endobj 0 G /FormType 1 >> stream 1 i 0 g /Matrix [1 0 0 1 0 0] The difference between six and a number divided by nine 10. Q /Resources<< /Meta375 389 0 R Q /Subtype /Form 1 g Q /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta383 Do 0 w >> Q ET q BT 32.201 20.154 l stream stream Q /Type /XObject endstream q (-20) Tj /FormType 1 endstream 0 G Q stream << 0.311 Tc Q /Resources<< /ProcSet[/PDF] BT /Meta30 43 0 R 0.524 Tc /Subtype /Form stream stream 1.007 0 0 1.006 130.989 690.329 cm 0.486 Tc /Leading 150 /FormType 1 20.975 5.336 TD << 1 i /F3 12.131 Tf >> Q /Type /XObject 33 0 obj /Matrix [1 0 0 1 0 0] endstream /Subtype /Form >> /Matrix [1 0 0 1 0 0] stream /ProcSet[/PDF/Text] /BBox [0 0 639.552 16.44] -0.22 Tw BT /BBox [0 0 549.552 16.44] /Subtype /Form 82 0 obj << 0 G Q (x) Tj 0.458 0 0 RG /FormType 1 1 i endobj /Subtype /Form /F1 7 0 R /Resources<< endstream q q /Meta6 15 0 R 401 0 obj /Matrix [1 0 0 1 0 0] ET stream endstream /Resources<< 162 0 obj /Resources<< >> /ProcSet[/PDF/Text] 0 g Q /Type /XObject 0 G << /Type /XObject q Q 1 i /Type /XObject 1 i endobj Q /ProcSet[/PDF/Text] /Length 59 Q q Q /Length 59 >> /Meta89 103 0 R q /F3 17 0 R << 0.271 Tc q 228 0 obj Q /Subtype /Form Q 1 g /Type /XObject /BBox [0 0 30.642 16.44] >> /Matrix [1 0 0 1 0 0] /Meta83 Do /FormType 1 /F3 17 0 R /F1 7 0 R endstream /Matrix [1 0 0 1 0 0] q /Meta145 Do /Length 69 1.014 0 0 1.007 531.485 583.429 cm BT /F3 12.131 Tf q /FormType 1 /Length 59 /FormType 1 endobj /Subtype /Form 1.007 0 0 1.007 271.012 330.484 cm q q /BBox [0 0 15.59 29.168] [( a )-15(number, decreased by )] TJ BT /Subtype /Form q q >> >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Meta206 Do /Subtype /Form q Q 1 i endobj endstream /Length 57 0.737 w 1 i /BBox [0 0 88.214 16.44] q 1.005 0 0 1.007 45.168 889.071 cm /Descent -277 /FormType 1 q /ProcSet[/PDF/Text] Q stream /ProcSet[/PDF/Text] /F4 12.131 Tf 1.014 0 0 1.006 531.485 690.329 cm /Type /XObject /Subtype /Form BT Q /Resources<< Q /Meta29 Do -0.382 Tw >> 0 G q /F3 17 0 R << q /BBox [0 0 88.214 35.886] /Resources<< /Subtype /Form /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /ProcSet[/PDF/Text] /Type /XObject << stream 6.746 5.203 TD /BBox [0 0 15.59 16.44] q /FormType 1 << 0 w >> S /Matrix [1 0 0 1 0 0] -0.084 Tw Q /Length 16 /Font << q stream 0.458 0 0 RG 0.738 Tc 32.201 5.203 TD /F1 12.131 Tf >> (+) Tj 1.007 0 0 1.007 271.012 383.934 cm >> /Resources<< << >> /ProcSet[/PDF/Text] >> 0 g << BT BT /Meta202 Do BT 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. /BBox [0 0 534.67 16.44] q /Type /XObject 1.007 0 0 1.007 271.012 776.149 cm stream endstream Q /Font << /Resources<< q endobj << /Subtype /Form /Meta270 284 0 R endobj Q 1.007 0 0 1.006 411.035 763.351 cm /FormType 1 /F3 12.131 Tf 0 G /Type /XObject endobj [tex]\sin (\pi -x)=\sin x[/tex]. /XObject << 0.564 G 188 0 obj /BBox [0 0 88.214 16.44] q q endobj 0 g q /Type /XObject Q /Length 68 /ProcSet[/PDF/Text] Q 215 0 obj 1 i 0.458 0 0 RG q Q /Meta364 Do Q /Length 59 1 i stream /Resources<< q << /Type /XObject BT /Font << 0.68 Tc Q stream Q /Meta10 21 0 R /Resources<< /BBox [0 0 15.59 29.168] /Font << >> /Resources<< << Q /Font << ET 1 i 549.694 0 0 16.469 0 -0.0283 cm endobj /BBox [0 0 88.214 16.44] q /Type /XObject >> Q Q /Length 79 Q /Meta242 Do
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