/Resources<< /Meta278 292 0 R Q BT Find the number. /Resources<< >> /Resources<< /Meta115 129 0 R q 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 /ProcSet[/PDF/Text] 0 G 0 G /BBox [0 0 88.214 16.44] >> BT Q (D) Tj /Length 59 /Type /XObject >> 0 g 0 g 1.014 0 0 1.007 111.416 703.126 cm /Font << /BBox [0 0 88.214 35.886] /ProcSet[/PDF] endobj stream Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. endstream stream << 151 0 obj q q stream Q endstream /BBox [0 0 88.214 16.44] endstream >> /FormType 1 Q /Length 65 /Font << << stream 1 i Tamang sagot sa tanong: 1.) 0 G /F3 12.131 Tf >> /F4 36 0 R 1.005 0 0 1.007 102.382 293.596 cm /Type /XObject << q /Meta170 Do /FormType 1 26 0 obj 1 i Q Q /ProcSet[/PDF/Text] Q q /Matrix [1 0 0 1 0 0] Q Twice a number decreased by . 0 w 0.458 0 0 RG (x ) Tj 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 370 0 obj Q /Subtype /Form /Length 16 /Matrix [1 0 0 1 0 0] endstream endstream stream q So we have twice of a mystery number decreased by three, and that is all going to be 31. 397 0 obj 1.007 0 0 1.007 271.012 636.879 cm >> /Length 69 1 g 0 g /FormType 1 /Length 16 q stream /Meta34 47 0 R q 0 G q endobj Q /Font << << /Length 69 /Meta319 Do /Subtype /Form /Type /XObject 0.737 w >> >> endobj endstream >> >> BT 51 0 obj /Subtype /Form /F1 12.131 Tf /AvgWidth 657 /Matrix [1 0 0 1 0 0] endobj /Meta409 Do Q 1 g Q << D. Twice a number decreased by ten is less than 24. stream /Type /XObject /F3 12.131 Tf q 1 g 0 G 157 0 obj >> Q 0 G 242 0 obj Q << Q Q 0 g 0 G stream /Subtype /Form endobj /BBox [0 0 15.59 16.44] 1 g /F4 12.131 Tf 1.007 0 0 1.007 130.989 523.204 cm /Type /XObject /Font << q endobj 1 g endstream 188 0 obj /Meta175 Do >> /F1 12.131 Tf >> /Subtype /Form (5) Tj Q /BBox [0 0 534.67 16.44] ET /Meta172 186 0 R 178.979 5.203 TD q Q stream /FormType 1 >> << 1 g (-) Tj endobj endobj /Resources<< /Font << endobj /ProcSet[/PDF/Text] 1 i 1.005 0 0 1.007 102.382 872.509 cm /Resources<< /Font << /FormType 1 Q /Resources<< /Length 139 Q /BBox [0 0 88.214 16.44] q q /Meta182 Do 0.425 Tc stream >> endstream << /ProcSet[/PDF/Text] 1.007 0 0 1.007 271.012 776.149 cm endstream /Subtype /Form 0 w endobj 1.007 0 0 1.007 130.989 277.035 cm /FormType 1 1.007 0 0 1.007 130.989 583.429 cm >> stream << 0 g Q S /F3 12.131 Tf Q /Subtype /Form /Meta420 Do >> /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 277.035 cm 0 G /F3 17 0 R q 0 g /Meta203 Do stream >> /ProcSet[/PDF] /Meta35 48 0 R endstream /Matrix [1 0 0 1 0 0] 0 G >> 0 w q q endobj /Length 64 ET Q 1 i 20.21 5.336 TD /Font << Q 0 w /BBox [0 0 639.552 16.44] q /Type /XObject /Meta13 24 0 R ET q q endstream /ProcSet[/PDF/Text] /Type /XObject /Matrix [1 0 0 1 0 0] 257 0 obj stream >> Q 174.501 5.203 TD /FormType 1 Thrice a number decreased by 5 exceeds twice the number by a unit. Q >> 1. 352 0 obj /Resources<< Q /FormType 1 /Meta275 289 0 R /Meta214 Do 0 G 0 w /BBox [0 0 88.214 16.44] /F3 12.131 Tf /Length 16 0 g 0 G q 1.007 0 0 1.007 130.989 383.934 cm stream /ProcSet[/PDF/Text] (3) Tj 1.005 0 0 1.015 45.168 53.449 cm << Q 0.737 w 1.007 0 0 1.007 551.058 330.484 cm /Length 59 1.005 0 0 1.015 45.168 53.449 cm /Resources<< >> /Meta381 Do /Matrix [1 0 0 1 0 0] endstream /F3 12.131 Tf 0.564 G 0.786 Tc 1.014 0 0 1.007 111.416 330.484 cm S /Meta301 315 0 R Q endobj BT 12 0 obj endstream endobj BT Q /Resources<< /BBox [0 0 88.214 16.44] 0.564 G q BT /Resources<< << << /FormType 1 q Q /Meta126 140 0 R [(A number )-17(divided by )] TJ 0 g 0 w endstream Q Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio 46 0 obj (D\)) Tj /Matrix [1 0 0 1 0 0] stream /ProcSet[/PDF] 50 0 obj Ten divided by a number 5. >> /Meta263 277 0 R << >> /Font << /Length 16 endstream 0.524 Tc 1 i 0 G /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 0 w q q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 330.484 cm /Length 69 1 i q 1 i >> q << >> /BBox [0 0 639.552 16.44] /Font << 12.727 5.203 TD /FormType 1 stream q Q /Resources<< Q 0.37 Tc /BBox [0 0 15.59 29.168] q endstream /F3 17 0 R 195 0 obj /Meta202 Do 371 0 obj ] /Subtype /Form /Subtype /Form q /Meta18 Do q << 1 i /Type /XObject /BBox [0 0 15.59 16.44] Q /Length 70 /ProcSet[/PDF/Text] 0 g q >> 0 G /Subtype /Form /Type /XObject 1 i /Meta393 Do stream q Q Q Q stream Q 1 i /Type /XObject Q >> /F3 12.131 Tf Q /Length 63 Advertisement Answer No one rated this answer yet why not be the first? /Resources<< /Type /XObject q Q /ProcSet[/PDF/Text] BT /ProcSet[/PDF/Text] Q 1 i BT q endobj /FormType 1 /FormType 1 0 G 1 i /FormType 1 0.838 Tc Q >> q /ProcSet[/PDF] /Meta334 348 0 R /F3 17 0 R /F3 12.131 Tf 1 i Q 1.007 0 0 1.007 67.753 726.464 cm >> >> 0 g >> 1 i /FormType 1 endstream /Meta157 Do BT /Meta53 Do >> >> /ProcSet[/PDF/Text] /Meta175 189 0 R 0 g 0 w q 1 i >> /Font << << 1.014 0 0 1.006 251.439 510.406 cm /F3 12.131 Tf Q /ProcSet[/PDF/Text] /FormType 1 /Matrix [1 0 0 1 0 0] Q 1 i 0 G /Matrix [1 0 0 1 0 0] << q Q stream /Type /XObject /FormType 1 /Length 54 /Meta98 Do 0 g endstream Q ET /ProcSet[/PDF] /Type /XObject Q 0 G 443 0 obj /ID [] S 1 i /ProcSet[/PDF/Text] q >> /Type /XObject 0 G /FormType 1 /BBox [0 0 88.214 16.44] q /Matrix [1 0 0 1 0 0] 0 g endobj /FormType 1 q /BBox [0 0 88.214 16.44] 0.786 Tc /Font << /Matrix [1 0 0 1 0 0] << /Font << endstream endobj q /Subtype /Form /ProcSet[/PDF] /ProcSet[/PDF/Text] /F1 7 0 R 0 g /Subtype /Form Five times the sum of a number and four 7. /ProcSet[/PDF/Text] /Meta271 285 0 R Percent Change = (Decrease First Value) x 100% >> /I0 Do endstream 0 4.894 TD /Font << 429 0 obj >> q BT >> q 1.007 0 0 1.007 45.168 779.913 cm 101.849 5.203 TD q q 1 i >> 1 i q /Matrix [1 0 0 1 0 0] Find an answer to your question Twice a number decreased by 8gives 58. ET 0 G stream /Resources<< q >> >> /Length 16 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 -0.067 Tw /F3 17 0 R /FormType 1 >> stream endobj Q >> 0 g Q << BT 722.699 347.046 l 1 g 0 g q q endstream << 1.005 0 0 1.007 79.798 829.599 cm 322 0 obj q q /Meta148 162 0 R 3.742 5.203 TD BT 1 i ET q /Type /XObject >> /Type /XObject q /Matrix [1 0 0 1 0 0] >> endstream Q q stream /Meta233 Do Q 672.261 599.991 m endobj 0.737 w 0.564 G /Meta197 211 0 R 0 g Q 0.524 Tc q /Type /XObject /FormType 1 1 i /Type /XObject stream q >> /Type /XObject /ProcSet[/PDF] >> Q /BBox [0 0 88.214 16.44] endobj /Matrix [1 0 0 1 0 0] /FormType 1 q 1 i /Meta277 291 0 R Q >> Q Q stream Q /BBox [0 0 88.214 16.44] /Meta127 141 0 R q >> stream /FormType 1 /FormType 1 >> 152 0 obj 0.369 Tc Q /BBox [0 0 88.214 16.44] /F3 17 0 R 164 0 obj Q 0 w stream /Type /XObject Q 1 g 4.506 8.18 TD /Subtype /Form Q [(-3)-16(20)] TJ Q endstream /Resources<< /FormType 1 0 g Q << stream Thrice a number decreased by 5 exceeds twice the number by 1. 149 0 obj 1.007 0 0 1.006 130.989 437.384 cm /F3 12.131 Tf /BBox [0 0 15.59 16.44] Q /Length 54 >> endstream 1.007 0 0 1.007 130.989 523.204 cm /F3 12.131 Tf /Type /XObject (+) Tj 1.007 0 0 1.007 271.012 776.149 cm << q 1.007 0 0 1.007 67.753 546.541 cm /Meta280 Do 0 w 1.007 0 0 1.007 411.035 277.035 cm 1 i 0 G /ProcSet[/PDF/Text] BT endobj Twice a number decreased by another number: "twice a number decreased by another number" 2*(x-y), "twice a number, decreased by another number"(2*x)-y, It's possible David. /Meta75 89 0 R /F3 17 0 R 287 0 obj 0.297 Tc 1.014 0 0 1.007 531.485 277.035 cm /Meta37 Do 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. >> /ProcSet[/PDF/Text] 0 5.203 TD >> 4.506 24.649 TD 1 i /Resources<< endobj /Subtype /Form /Resources<< Q /Resources<< Q Q << /Meta99 Do /ProcSet[/PDF/Text] /Resources<< -0.486 Tw /ProcSet[/PDF/Text] 0 G Q 104 0 obj 1.007 0 0 1.007 271.012 849.172 cm Q 0.737 w Q /Meta127 Do Q 1 i endobj 316 0 obj ET endstream (40) Tj q /Descent -216 /Type /XObject 0 g << /Subtype /Form /Type /XObject 0 5.203 TD /Meta217 231 0 R endobj endobj >> 1 g 0.564 G endobj >> /F3 17 0 R /FormType 1 /F3 12.131 Tf stream Q endobj stream /F3 12.131 Tf /Length 12 0 g 1.007 0 0 1.007 411.035 583.429 cm 0 g /Font << /F3 17 0 R 0.51 Tc (x) Tj BT stream >> /Matrix [1 0 0 1 0 0] q 0.458 0 0 RG BT /Subtype /Form Q /Meta209 Do /Matrix [1 0 0 1 0 0] q 0.737 w Q /Font << /Meta324 338 0 R Q 1 i Q /Meta156 170 0 R The sum Of twice a nu4ber What is the number? << /F1 7 0 R /Resources<< Q endstream ET Q endobj 118.317 5.203 TD /Meta61 Do /ProcSet[/PDF] 1 i /ProcSet[/PDF] q << 0 g 0.51 Tc /FormType 1 << endstream 0.001 Tw /Resources<< q >> /FormType 1 /BBox [0 0 88.214 16.44] /Meta298 Do /F3 17 0 R /Meta206 Do /ItalicAngle 0 << /Matrix [1 0 0 1 0 0] q decreased by A number decreased by twelve X - 12 subtracted from Six subtracted from a number X - 6 Multiplication ( x ) times Eight times a number 8x the product of The product of fourteen and a number 14x twice; double Twice a number; double a number 2x multiplied by A number multiplied by negative six 6x Q 1 g /Resources<< /Length 64 Q 0 G (6\)) Tj /Meta301 Do (38) Tj Q Q << /BBox [0 0 15.59 29.168] /Meta257 Do /Matrix [1 0 0 1 0 0] endstream /Type /XObject 0.737 w q /Length 59 1 i (2\)) Tj /Subtype /Form 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) ET 0.737 w << q 0.737 w 1 i /Meta411 Do Q endstream /F3 12.131 Tf 0.838 Tc << 1.007 0 0 1.007 411.035 849.172 cm stream (-11) Tj Q /Meta315 329 0 R Q 0.198 Tc >> /Meta68 Do /FormType 1 1 i ET q >> q /ProcSet[/PDF/Text] /Resources<< << >> 0 G A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. /BaseFont /TestGen-Regular 1.007 0 0 1.006 551.058 836.374 cm /ProcSet[/PDF/Text] /Type /XObject 0 w endobj >> 2.238 5.203 TD -0.463 Tw /Length 59 /Length 65 /F3 12.131 Tf >> /Length 69 q /Type /XObject stream /Matrix [1 0 0 1 0 0] q >> /MissingWidth 252 >> /BBox [0 0 88.214 16.44] /Meta87 Do >> NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. /F3 17 0 R /BBox [0 0 88.214 16.44] Q << /Meta242 Do endobj 0.838 Tc /Matrix [1 0 0 1 0 0] /Length 70 q << Q /ProcSet[/PDF] 124 0 obj q /Meta190 204 0 R /Resources<< /Subtype /Form 0.369 Tc endstream Q 0 g /F3 12.131 Tf q 43.426 5.203 TD Q /Meta236 250 0 R /Subtype /Form BT << q /BBox [0 0 88.214 16.44] << q 1 i ET 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 54 0 obj 1.007 0 0 1.007 67.753 726.464 cm 18.708 17.593 TD /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] Q /F4 12.131 Tf Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. Q 1.502 5.203 TD 1 i 0 g 293 0 obj << 0.738 Tc 0.786 Tc stream 1 i 0.458 0 0 RG q /Resources<< << 0.564 G 0 w >> stream /F1 12.131 Tf /ProcSet[/PDF] >> q /FormType 1 BT Q /ProcSet[/PDF/Text] stream /BBox [0 0 15.59 16.44] q /Meta274 Do /Resources<< 0 G 0 g /FormType 1 /Subtype /Form endobj /FormType 1 q /Meta379 Do Q /BBox [0 0 88.214 16.44] Q /Subtype /Form Q q /F3 17 0 R /Meta20 31 0 R /Meta134 Do /F3 12.131 Tf /Resources<< /Subtype /Form /Subtype /Form Q 1 i /Font << endobj endobj (-11) Tj Q /Type /XObject q 441 0 obj << endstream -0.062 Tw -0.008 Tw ET Q 0.37 Tc /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.502 5.203 TD Q Q /Meta119 133 0 R /F3 17 0 R q for the season. /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Meta380 Do endstream /Type /XObject /Subtype /Form << 1 i 0.786 Tc 1 g /Meta298 312 0 R << /Pages 1 0 R /I0 51 0 R << stream /Length 68 1.007 0 0 1.007 67.753 599.991 cm q /FormType 1 Q 1 i /ProcSet[/PDF/Text] endobj 310 0 obj 0 w Q endstream (viii) A number divided by 8 gives 7. 9.723 5.336 TD 0 G /Subtype /Form /FormType 1 /ProcSet[/PDF/Text] stream 0 g /Subtype /Form >> Q 1 i 1 i 0.786 Tc q 1.014 0 0 1.007 111.416 776.149 cm >> /Matrix [1 0 0 1 0 0] q 1.014 0 0 1.006 111.416 510.406 cm 231 0 obj Q Q >> /Subtype /Form 0.737 w /Subtype /Form 1 i stream 405 0 obj >> Q 1 i 0 20.154 m /BBox [0 0 88.214 16.44] ET Q 0.486 Tc /Type /Pages 1.007 0 0 1.007 271.012 330.484 cm q << /Meta92 Do Q Twice 4 bananas is 8. 1 i -0.058 Tw q q /Length 68 BT Q q /Meta40 54 0 R >> 1 i /Length 16 13.493 5.203 TD endobj 0 g >> /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /Meta81 Do 0.458 0 0 RG /Type /XObject /Type /XObject /BBox [0 0 534.67 16.44] 205.199 4.894 TD 342 0 obj q /Encoding /WinAnsiEncoding Q endstream endstream Q << [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /Subtype /Form /ProcSet[/PDF] q 0.369 Tc /Type /XObject (40) Tj q BT /Meta94 108 0 R /F3 17 0 R /Type /XObject Q q endobj endobj << endobj stream >> Q 215 0 obj /F3 17 0 R Q /Subtype /Form /Resources<< q >> /Meta293 307 0 R q q /Resources<< >> >> /F3 12.131 Tf q q 1 i >> S /FormType 1 q >> 0 G Q q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] q q 0 g Q 264 0 obj /BBox [0 0 534.67 16.44] q /Type /XObject 0 w >> /Meta313 327 0 R 611 556 611 611 389 444 333 611 556 833 500]>> /Meta42 56 0 R q S q /F3 17 0 R /Resources<< /Resources<< ET 1.005 0 0 1.007 79.798 846.161 cm /FormType 1 Q >> /MissingWidth 250 stream /Meta52 Do Five times a number, decreased by 58, is -23 Find the number. >> >> 1 i << /F3 17 0 R /Meta49 63 0 R /F3 17 0 R /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 551.058 583.429 cm /Meta254 Do A number increased by 5 is equivalent to twice the same number decreased by 7. /F3 17 0 R /Resources<< 0.564 G /Resources<< Q 1.007 0 0 1.007 45.168 746.789 cm << >> /Resources<< /Resources<< -0.16 Tw endstream /ProcSet[/PDF/Text] 33 0 obj Q /Length 69 /Meta231 245 0 R /Meta16 Do >> 1 i q 1.005 0 0 1.007 102.382 546.541 cm /BBox [0 0 88.214 16.44] /BBox [0 0 534.67 16.44] Q ET 1.007 0 0 1.007 271.012 523.204 cm (\(x ) Tj endobj >> q 0 G /Length 12 (D\)) Tj stream /Type /XObject 0 G endstream 0 G ET q /FormType 1 1 i Q /Subtype /Form stream 0 G /BaseFont /PalatinoLinotype-Bold -0.486 Tw /Meta323 337 0 R /Type /XObject 0 G Q /Meta133 Do 1.007 0 0 1.007 551.058 523.204 cm >> 0 g 0.463 Tc 21.713 20.154 l 1 i 1 i Q Q /Type /XObject >> /Resources<< Q 0.269 Tc 330 0 obj BT q /ProcSet[/PDF] 0.737 w q q /F1 7 0 R q (x ) Tj /Type /XObject /Length 59 >> /F3 12.131 Tf endobj (x) Tj 1 i /ProcSet[/PDF] >> q q Q 96 0 obj q /Length 74 Q 0 4.894 TD /Subtype /Form 0 g 245 0 obj << [(1)-25(0\))] TJ 1 i endstream 76 0 obj /ColorSpace [/Indexed /DeviceGray 1 ] /Meta83 Do /ProcSet[/PDF/Text] Q /Length 68 << 1.007 0 0 1.007 67.753 473.519 cm Q That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. /Type /XObject /Type /XObject endobj endstream << >> /FormType 1 /ProcSet[/PDF] /Length 96 endobj /Subtype /Form 344 0 obj /Subtype /Form 1 g /Length 19882 /BBox [0 0 534.67 16.44] 0 g /ProcSet[/PDF/Text] q Q >> q 0 20.154 m /ProcSet[/PDF] In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. Q /Meta110 Do Q /F3 12.131 Tf 1.007 0 0 1.006 130.989 437.384 cm 0 G /Matrix [1 0 0 1 0 0] Q /Type /XObject ET >> q >> 1 i >> endstream /BBox [0 0 88.214 35.886] /FormType 1 /F3 12.131 Tf BT /BBox [0 0 88.214 16.44] BT 0 5.203 TD << Q 299 0 obj /Resources<< /Meta388 404 0 R 52.412 5.203 TD q 0.28 Tc /BBox [0 0 15.59 16.44] >> q >> /Subtype /Form 15.731 5.336 TD Q q >> /Matrix [1 0 0 1 0 0] ET Q 0 G BT q /FormType 1 Q Q /F3 12.131 Tf Q /Resources<< >> stream Q 1.005 0 0 1.007 102.382 599.991 cm q Q /Type /XObject 0 G (-4) Tj /Meta363 377 0 R /Resources<< /Matrix [1 0 0 1 0 0] endstream >> endobj q endobj 0.458 0 0 RG 346 0 obj /Meta25 38 0 R q 1 i q Q /ProcSet[/PDF] 0 g Q /BBox [0 0 88.214 35.886] 1 i 0 G Q /Meta292 306 0 R 1 i /FormType 1 /Meta424 440 0 R stream /BBox [0 0 17.177 16.44] /Type /XObject /Meta160 Do << /ProcSet[/PDF/Text] stream q 0.458 0 0 RG /ProcSet[/PDF/Text] ET endstream /Meta354 Do Q -0.486 Tw Q /Meta373 Do 35.206 4.894 TD endobj /Resources<< 0.297 Tc /Subtype /Form Q /BBox [0 0 88.214 16.44] >> 355 0 obj q 1.502 24.339 TD 30.699 5.203 TD Q 17.234 5.203 TD /Font << Q << /BBox [0 0 30.642 16.44] >> /F3 12.131 Tf 0 5.203 TD /Subtype /Form ET /Meta424 Do endstream Q 0 5.203 TD >> c Site 5 is not included in this number. Q endobj 0 g /Matrix [1 0 0 1 0 0] Q 0 G /F4 12.131 Tf /Resources<< Q /Type /XObject q q /F3 12.131 Tf /Meta276 290 0 R /Type /XObject 1 g 0 G /Resources<< >> /Subtype /Form >> /Type /XObject 1.007 0 0 1.007 551.058 277.035 cm /FormType 1 /Length 16 /Length 118 Q 25 0 obj /Meta40 Do >> Have a nice day! ET /BBox [0 0 639.552 16.44] /Resources<< ( \() Tj /Type /XObject endstream /Length 95 /XObject << /BBox [0 0 15.59 16.44] << 0 G /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] >> 297 0 obj >> q /AvgWidth 401 Q /ProcSet[/PDF/Text] Q BT /Font << /Font << Q 0.369 Tc /Meta123 Do 1 i -0.486 Tw 549.694 0 0 16.469 0 -0.0283 cm /BBox [0 0 88.214 16.44] /F1 12.131 Tf /Length 16 /Resources<< 25.454 5.203 TD stream >> q /Matrix [1 0 0 1 0 0] Q /Contents [399 0 R] /F4 36 0 R /BBox [0 0 534.67 16.44] 1.005 0 0 1.007 102.382 653.441 cm 1.007 0 0 1.007 45.168 813.037 cm /Meta228 242 0 R 1.007 0 0 1.007 130.989 383.934 cm >> /Font << 1.007 0 0 1.007 411.035 383.934 cm 1 i /Subtype /Form 0.737 w /Resources<< q q 0 G >> A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. /Meta306 Do stream /Meta196 Do Q /Subtype /Form /Type /XObject /FormType 1 >> q endobj Q 1 i Q 0.486 Tc 426 0 obj >> q /Matrix [1 0 0 1 0 0] >> /FormType 1 /Meta401 417 0 R /Length 106 1.005 0 0 1.007 45.168 889.071 cm /Subtype /Form Q BT 162 0 obj /FormType 1 /ProcSet[/PDF/Text] /Length 59 0.564 G stream /BBox [0 0 15.59 29.168] Q endobj q q /Matrix [1 0 0 1 0 0] 0 g /F4 12.131 Tf stream endstream /Font << >> endobj /Meta300 Do 0.737 w 1 i 1 i /Matrix [1 0 0 1 0 0] /F1 7 0 R >> Twice a number decreased by 8 gives 58. /BBox [0 0 15.59 29.168] >> /Length 67 endstream /Resources<< << endstream /F3 17 0 R q q Q 1 i Q 0.458 0 0 RG Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. ET 0 g >> /Resources<< >> >> << /Matrix [1 0 0 1 0 0] (-4) Tj (x ) Tj /FormType 1 1 i /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] /Type /XObject BT (A\)) Tj /Type /XObject Q endstream << /Resources<< >> 0.738 Tc Q ET >> /Length 245 BT /Matrix [1 0 0 1 0 0] q /BBox [0 0 88.214 16.44] 0 g /Type /XObject /Subtype /Form /Type /XObject BT Q endstream stream 0.458 0 0 RG stream Q q stream 266 0 obj /Length 118 /Type /XObject /Subtype /Form q /ProcSet[/PDF/Text] /Resources<< Q /F3 17 0 R /Matrix [1 0 0 1 0 0] /F1 14.682 Tf BT 1 i << q >> /Meta383 397 0 R Q 0.838 Tc /Resources<< /F3 12.131 Tf 1.014 0 0 1.007 531.485 636.879 cm

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