/Matrix [1 0 0 1 0 0] /Subtype /Form /Matrix [1 0 0 1 0 0] q /Meta121 Do q ET /Subtype /Form 0.564 G endstream 1 i /F3 17 0 R Q /Length 96 BT BT Q stream 0.564 G Q 334 0 obj q 0.458 0 0 RG (1\)) Tj /BBox [0 0 88.214 16.44] Q Q Q 1 i /BBox [0 0 534.67 16.44] q /BBox [0 0 88.214 16.44] << /FormType 1 /F4 36 0 R /BBox [0 0 15.59 29.168] endobj 0 G /Subtype /Form >> 1.502 5.203 TD /Subtype /Form << endobj Q /Type /XObject /ProcSet[/PDF/Text] /Meta403 Do /FormType 1 /I0 51 0 R q Q 236 0 obj q endstream /Meta208 Do << Q Q /ProcSet[/PDF/Text] Q /Resources<< 0 g endobj 1.005 0 0 1.007 45.168 916.925 cm /Resources<< /FormType 1 /Subtype /Form Q q /Resources<< /F3 17 0 R /F1 7 0 R /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] endstream /Subtype /Form /BBox [0 0 88.214 16.44] /Meta151 165 0 R >> q Q /Subtype /Form 1 g 1.014 0 0 1.007 531.485 776.149 cm /Type /XObject Q Q stream 0 G /Meta340 Do /Font << endstream (5) Tj q endobj << /F3 17 0 R stream Q /Meta14 Do >> 0 5.203 TD /BBox [0 0 30.642 16.44] >> stream 1 i 1 i Q /Matrix [1 0 0 1 0 0] endobj >> 1 i w/Honors. /Font << 1.007 0 0 1.007 654.946 799.486 cm Q 0 5.203 TD (+) Tj /Matrix [1 0 0 1 0 0] /F3 17 0 R Q << /Meta244 258 0 R 0.175 Tc Q /FormType 1 /FormType 1 /Subtype /Form 549.694 0 0 16.469 0 -0.0283 cm 0 g >> q BT >> 1 g /ProcSet[/PDF/Text] Q /BBox [0 0 17.177 16.44] /Meta281 295 0 R /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Meta325 Do /Meta16 27 0 R stream Q q /Meta295 309 0 R q /Type /XObject ET /Meta212 Do /BBox [0 0 17.177 16.44] >> /Type /XObject endobj >> q 0 G q 0 g q 1 g endstream /BBox [0 0 88.214 16.44] (x) Tj endstream /Type /XObject endstream /BBox [0 0 15.59 16.44] /Length 16 q /FormType 1 Q /F3 12.131 Tf /FormType 1 Q 0 G /BBox [0 0 15.59 16.44] 0.737 w /Subtype /Form /Length 69 /BBox [0 0 17.177 16.44] >> 1 i /Meta221 Do q 1 i endstream >> << Q /BBox [0 0 88.214 35.886] BT endobj /Length 59 >> ET 175 0 obj /Type /XObject >> q endobj /F3 12.131 Tf q Q 20/n b.) /FormType 1 /F1 7 0 R (11) Tj q 1.005 0 0 1.015 45.168 53.449 cm (B\)) Tj endobj endobj 363 0 obj Q /Meta278 292 0 R /Meta196 Do /Meta294 Do >> q ET /Meta95 Do q q /Subtype /Form -0.486 Tw /ProcSet[/PDF/Text] /Type /Font 0 g /ProcSet[/PDF/Text] endobj /FormType 1 /Subtype /Form 1.007 0 0 1.007 551.058 703.126 cm 143 0 obj 0 g 0.178 Tc 1 i /FormType 1 Q /Meta213 227 0 R /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Subtype /Form q /Type /XObject /Type /XObject << q Q 0 G /Resources<< /BBox [0 0 88.214 16.44] >> /Type /XObject q q /F3 12.131 Tf /FormType 1 Q q Q q endstream q /Resources<< 20.21 5.203 TD /ProcSet[/PDF/Text] 227 0 obj << /Resources<< /ProcSet[/PDF] 180 0 obj Q 0 G /ProcSet[/PDF] 1 i 1.007 0 0 1.007 130.989 636.879 cm Q endobj /F3 17 0 R << /Meta133 147 0 R q /Resources<< 23.216 5.203 TD 0 w >> /Meta19 30 0 R 1 g /Font << /Subtype /Form saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. /Type /XObject 88 0 obj endobj /Resources<< q q q 1.007 0 0 1.007 551.058 523.204 cm 0 g /Subtype /Form BT q /Subtype /Form /Meta312 Do 0.458 0 0 RG /ProcSet[/PDF/Text] /Meta86 100 0 R /F3 12.131 Tf stream 0 g Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio 0 5.203 TD stream q q q 99 0 obj /BBox [0 0 88.214 35.886] ET Q endobj /Subtype /Form endstream /F4 12.131 Tf /F3 12.131 Tf /F3 12.131 Tf 0 g /Type /XObject >> q ET /Font << endstream /Matrix [1 0 0 1 0 0] >> /Resources<< /Length 16 /Subtype /Form /FormType 1 /Matrix [1 0 0 1 0 0] 0 g endobj 0 g /Subtype /Form 0.564 G Q q 1 i q Find the number. /Resources<< 0 g /Length 67 Q /BBox [0 0 639.552 16.44] Q endstream /ProcSet[/PDF] Q /Meta79 93 0 R Q << << stream 1.007 0 0 1.007 271.012 450.181 cm << /ProcSet[/PDF] (11) Tj stream /Resources<< 0 w 1.007 0 0 1.007 411.035 636.879 cm /ProcSet[/PDF/Text] BT >> 187 0 obj Q (-4) Tj Q Get a free answer to a quick problem. What word phrase can you use to represent 5x + 2? /F3 12.131 Tf Calculate a 15% decrease from any number. /Subtype /Form /F3 17 0 R 1 i Q Q 0.564 G << Q 1.007 0 0 1.006 411.035 510.406 cm /Length 68 endobj BT /Resources<< >> /Meta30 Do ( \() Tj >> /Length 69 /Resources<< >> /ProcSet[/PDF] /Subtype /Form 1 i (-8) Tj /F3 12.131 Tf >> /BBox [0 0 88.214 16.44] /FormType 1 0 g << stream Q /FontBBox [-174 -299 1445 1050] Q /Type /XObject 0.68 Tc /Font << 1 i q /Meta192 206 0 R 0 G << 1.005 0 0 1.007 79.798 846.161 cm /Type /XObject /Subtype /Form /Subtype /Form /Subtype /Form /FormType 1 1 g 320 0 obj /FormType 1 /Type /XObject /Resources<< 0.369 Tc /Meta299 Do /ProcSet[/PDF] /BBox [0 0 639.552 16.44] >> q 0 g 92 0 obj endobj /FontDescriptor 6 0 R q /ProcSet[/PDF/Text] Q /Meta165 179 0 R << q q stream /Subtype /Form /Length 69 >> Q -0.058 Tw /ProcSet[/PDF/Text] 95 0 obj >> /Meta358 372 0 R Q >> /F3 12.131 Tf 0 G ET /Matrix [1 0 0 1 0 0] BT endstream Q 1.007 0 0 1.007 271.012 703.126 cm For the lesson, he grabs a glass container shaped like a rectan /FormType 1 >> >> /F3 17 0 R /Subtype /Form q 0 G stream q /Font << q 1 0 obj >> /F3 12.131 Tf q q >> Q /ProcSet[/PDF/Text] /Meta267 281 0 R Q ET Q 0 G Q 1.007 0 0 1.007 271.012 383.934 cm /ProcSet[/PDF/Text] 241 0 obj (-11) Tj /Resources<< /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] /FormType 1 endobj /Resources<< /Subtype /Form Hence, the number is 6. /Meta321 Do /ProcSet[/PDF] q /Meta58 Do >> 0.838 Tc 1 g Q /Meta344 358 0 R Q /BBox [0 0 88.214 16.44] Phrase. 1.005 0 0 1.007 102.382 256.709 cm /Font << 0 G 0 G q /Meta424 440 0 R endstream 0.564 G Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. /Font << stream Answer only. /ProcSet[/PDF] Q q /Font << >> q stream Q >> /Resources<< 1.007 0 0 1.006 411.035 437.384 cm /Type /XObject /ProcSet[/PDF] 0 4.894 TD /FormType 1 285 0 obj /Meta379 Do /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] >> 1.014 0 0 1.007 391.462 277.035 cm /F3 17 0 R /Subtype /Form stream << A. /Subtype /Form >> /Meta67 Do 0 5.203 TD endstream Q /Length 79 /Font << 0 G q BT ET Q endobj /Meta57 71 0 R >> /BBox [0 0 15.59 16.44] q 0 g 1.014 0 0 1.007 111.416 277.035 cm /Matrix [1 0 0 1 0 0] /Length 69 /FormType 1 /BBox [0 0 639.552 16.44] Q 1 g /Length 59 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. /Subtype /Form /BBox [0 0 88.214 16.44] q BT Q stream /Subtype /Form << 1.007 0 0 1.007 130.989 849.172 cm stream ET 0.458 0 0 RG /Meta399 Do BT q /I0 51 0 R << >> endstream q 415 0 obj /Length 16 0.425 Tc ET /Type /XObject /Matrix [1 0 0 1 0 0] q >> /F3 12.131 Tf /FormType 1 >> q q (-11) Tj stream 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. [(A number )-17(divided by )] TJ endstream /Meta383 Do /Type /XObject There were x cookies at the beginning of a party. << /Type /FontDescriptor 203 0 obj 0 g /Matrix [1 0 0 1 0 0] /Length 69 ET 0.458 0 0 RG Q >> /Length 16 >> << /Subtype /Form << /I0 51 0 R /Meta70 84 0 R Q endstream 2.238 5.203 TD >> /BBox [0 0 88.214 35.886] /Resources<< /Matrix [1 0 0 1 0 0] /Meta225 239 0 R 0.68 Tc /Length 294 1.014 0 0 1.007 111.416 776.149 cm Q Q (+) Tj 1 i /Resources<< /Subtype /Form /MissingWidth 250 /Meta231 245 0 R /Font << stream 246 0 obj >> /Meta245 259 0 R q 0.297 Tc /Resources<< stream /Subtype /Form Q /FormType 1 /Resources<< /Meta152 Do %%EOF. 0 4.894 TD >> q /Resources<< /F3 12.131 Tf BT BT << 0.458 0 0 RG 400 0 obj ET << ET /F1 7 0 R 0 g /Subtype /Form 0.564 G 72 0 obj Q /Resources<< 0.564 G 1 i Q 0.68 Tc /Resources<< Q stream /ProcSet[/PDF] 1 i (D) Tj Q 140781 Q BT /Length 69 /Resources<< q q /Meta279 293 0 R 0 w /ProcSet[/PDF] 100 0 obj BT /Matrix [1 0 0 1 0 0] Q /F3 17 0 R /BBox [0 0 88.214 16.44] /Type /XObject << endobj Q /Meta421 Do S endobj 0 g /BBox [0 0 15.59 16.44] q /Length 16 /Font << /FormType 1 Q endstream 0.486 Tc Q /ProcSet[/PDF] /F3 17 0 R 0.564 G Q endobj /Matrix [1 0 0 1 0 0] endstream >> 0 G BT q t is 56: 4. 0.564 G Q /Meta75 89 0 R -0.008 Tw endobj stream /FormType 1 Q /BBox [0 0 88.214 16.44] /Resources<< /Font << 0 G q Q 0 w << Q Q Find the number. >> q 0 g q Q /F3 12.131 Tf /F3 17 0 R /FormType 1 722.699 546.541 l BT >> (4) Tj /Meta2 9 0 R /Meta24 37 0 R >> ET >> 130 0 obj 1 i /Resources<< 1 i >> q Q endstream 0.564 G /Meta82 96 0 R endobj endobj /XHeight 471 /Length 16 0 g /Meta183 197 0 R /Length 118 /Subtype /Form 0 g 30.699 4.894 TD Q /Meta4 Do q stream /Length 54 Q -0.03 Tw 0 g Q /Type /XObject 0 g q q /Matrix [1 0 0 1 0 0] /Meta318 Do /Resources<< << Q ET stream >> 20.975 5.336 TD 414 0 obj endstream Q /BBox [0 0 534.67 16.44] /Meta12 23 0 R Q >> /Meta356 370 0 R 1.005 0 0 1.007 102.382 872.509 cm endobj /FormType 1 /Meta389 405 0 R 1 i 1 g >> /Subtype /Form 26.957 5.203 TD stream [( times )15(a numb)22(er and )] TJ 0 G /Meta78 92 0 R /Length 12 0 G 371 0 obj 0.458 0 0 RG endstream /Type /XObject /Type /XObject /Type /XObject 0 w /Meta306 320 0 R /Meta39 Do q /Meta78 Do 0 g /BBox [0 0 15.59 16.44] /Meta33 46 0 R >> q /Subtype /Form Q q q 1.007 0 0 1.007 130.989 849.172 cm << /BBox [0 0 15.59 29.168] /Meta338 352 0 R /Length 69 endobj 0 5.203 TD q ET 1.007 0 0 1.007 271.012 776.149 cm /Meta312 326 0 R 332 0 obj /BBox [0 0 88.214 16.44] /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] endstream stream /Length 69 /Meta411 Do /Meta248 Do Q /Meta34 47 0 R 3.742 5.203 TD /Subtype /Form /Meta377 Do /Resources<< Q ET >> /Subtype /Form Q /FormType 1 0.564 G Q endobj /Type /XObject /Meta142 Do /BBox [0 0 30.642 16.44] q 1 i /Length 69 /Type /XObject q 1 i 52.412 5.203 TD >> BT >> /ProcSet[/PDF/Text] /F4 12.131 Tf endobj /Length 16 /BBox [0 0 88.214 35.886] Q 1 g /Font << /Length 66 /BBox [0 0 88.214 16.44] BT q stream /FormType 1 0 g >> >> /FormType 1 /Meta326 340 0 R /Meta348 Do endstream q q /F3 17 0 R (2\)) Tj /F3 12.131 Tf Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. >> stream q q endobj /Meta99 Do q >> (viii) A number divided by 8 gives 7. Q -37 VI 2. 0 g BT 1 i /Meta273 Do 1.007 0 0 1.007 271.012 636.879 cm q /Resources<< /Count 2 /Matrix [1 0 0 1 0 0] /Length 59 220.931 4.894 TD /Length 59 1 i Q >> /Font << 155 0 obj Q /Meta243 257 0 R 370 0 obj /ProcSet[/PDF] >> q (-23) Tj 1 g Q /FormType 1 /Resources<< Q /FormType 1 1 g 0 w >> 302 0 obj pidemiologi i Infekcionnye Bolezni. /Meta168 182 0 R /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 g /ProcSet[/PDF/Text] /Resources<< q q q Q /Length 77 Q Q >> /Type /XObject /BBox [0 0 88.214 16.44] q >> >> q << << /Resources<< /Length 16 q /ProcSet[/PDF/Text] q /Resources<< /Encoding /WinAnsiEncoding >> Q >> endstream << 0.458 0 0 RG S >> q /Matrix [1 0 0 1 0 0] 0 g /ProcSet[/PDF] Q >> >> >> the sum of a number and twelve. /BBox [0 0 30.642 16.44] /Meta173 Do /Type /XObject /Subtype /Form /Type /XObject /Meta84 Do /Subtype /Form << >> /BBox [0 0 88.214 35.886] >> endobj /Subtype /Form /XObject << /Length 16 /FormType 1 endobj 1 i /Matrix [1 0 0 1 0 0] /Type /XObject (7\)) Tj /Length 69 (58) Tj (-9) Tj 1 i /Resources<< /XObject << endstream BT The sum of a number and 2 is 6 less than twice that number. /F3 17 0 R q q 1.502 5.203 TD /BBox [0 0 88.214 35.886] /Subtype /Form 76 0 obj BT 0 G >> 3.742 5.203 TD ET 1 i 214 0 obj << /Subtype /Form endstream /FormType 1 q stream /F3 12.131 Tf 1.007 0 0 1.007 67.753 653.441 cm 0.737 w 0 g q (-9) Tj stream 0.425 Tc Q BT << q >> 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . << /Resources<< /Font << /F3 12.131 Tf 0.51 Tc Q /Resources<< q /Matrix [1 0 0 1 0 0] BT /Meta96 Do /Matrix [1 0 0 1 0 0] q stream 1.007 0 0 1.007 551.058 703.126 cm /F4 12.131 Tf >> /Font << /Length 69 >> 0 g Q /Subtype /Form 288 0 obj q q 1.007 0 0 1.006 411.035 763.351 cm >> 23.216 5.203 TD /BBox [0 0 15.59 29.168] /Meta269 Do q 0.458 0 0 RG Q 0.737 w q 1.502 5.203 TD >> >> 183 0 obj 0.738 Tc endobj q 0 w /BBox [0 0 88.214 16.44] endobj 0 G >> << /Resources<< 1 i /Length 69 Q /Subtype /Form 1 Data in this Fast Fact represent the 50 states and the District of Columbia. /Length 69 stream /Resources<< << << q q 16.469 5.203 TD /Meta301 315 0 R 118 0 obj 17.234 5.203 TD /F3 12.131 Tf /Length 118 /Subtype /Form 229 0 obj q 1.007 0 0 1.007 130.989 776.149 cm /FormType 1 14.23 24.649 TD << /BBox [0 0 88.214 16.44] ET 1 i q /FormType 1 q /Font << /Resources<< /Type /XObject /Subtype /Form 0.458 0 0 RG >> If twice a number is decreased by 13, the result is 9. >> 373 0 obj BT 0 G /Meta155 169 0 R q /Type /XObject 0 g Q /Subtype /Form 1 i q endobj /Meta305 Do 1 i /Meta177 Do Q 91 0 obj 433 0 obj stream /BBox [0 0 15.59 16.44] Q q Q >> /Subtype /Form << Q >> >> /F4 36 0 R >> /FormType 1 endstream 0 5.203 TD /F3 12.131 Tf 9.723 5.336 TD /Subtype /Form /FormType 1 See Solution. q q /F3 17 0 R /Font << Q /Type /XObject /Meta262 Do /Resources<< Q 1.007 0 0 1.007 45.168 862.723 cm q /FormType 1 /Matrix [1 0 0 1 0 0] /Length 69 0 G q endobj ET 0.564 G /Length 57 /F1 14.682 Tf 0 g (x) 6 times a number is 5 more than the number. /ProcSet[/PDF] 1 i stream q Q q /Meta193 207 0 R stream /Meta320 Do 1 i q << 13.464 5.203 TD 1.005 0 0 1.007 102.382 799.486 cm /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 251.439 776.149 cm 1 i stream Q 9.723 5.336 TD Q 0.564 G /Matrix [1 0 0 1 0 0] Q ET 1 i endobj BT 1.014 0 0 1.007 391.462 277.035 cm /BBox [0 0 88.214 16.44] >> q /Resources<< 1. q 0.737 w q /BBox [0 0 15.59 16.44] BT Q >> /Meta219 233 0 R q q /Subtype /Form q 0 g endstream /Resources<< /Meta195 Do 1.007 0 0 1.007 45.168 846.161 cm 0.68 Tc endobj ET 0 w /Type /XObject stream 1.014 0 0 1.007 391.462 583.429 cm endstream endobj 1.005 0 0 1.007 102.382 310.158 cm q 430 0 obj /Length 12 >> q q /Type /XObject /Meta172 186 0 R 1 i /F3 12.131 Tf q 1.007 0 0 1.007 551.058 330.484 cm Q 0.564 G /Length 16 1 i /Resources<< /Type /XObject Q /Subtype /Form Q endstream endstream /Meta370 384 0 R /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /F3 17 0 R 1 i Q /F3 17 0 R endstream q q 7 0 obj >> /Font << /BBox [0 0 534.67 16.44] 0 g q Q 0 G BT q endstream /Subtype /TrueType /Length 69 /F3 12.131 Tf /FormType 1 0.458 0 0 RG Q endstream endstream 1 i 0 w 1.005 0 0 1.015 45.168 53.449 cm Q 345 0 obj q ET /Pages 1 0 R >> >> >> >> << /Subtype /Form 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q /Type /XObject /F3 12.131 Tf q /F1 7 0 R >> /Ascent 891 1.007 0 0 1.007 654.946 473.519 cm /Meta382 Do /FormType 1 /Font << /Resources<< /FormType 1 << /Meta290 304 0 R /ProcSet[/PDF/Text] 0 g Q /Font << endobj If a number is 400%, then it is 4 times, the same as 4. stream endobj /Meta6 15 0 R q << q /Resources<< /F4 36 0 R 0.175 Tc ET 0.737 w 1 i endstream Q (A\)) Tj >> endstream /Length 59 q /Leading 253 /Length 80 /Meta176 190 0 R /FormType 1 << 1.007 0 0 1.007 130.989 583.429 cm /Resources<< /Length 59 /F3 17 0 R /Meta91 Do >> /Meta175 189 0 R /ProcSet[/PDF/Text] 0 g << endstream Q Q /FormType 1 /BBox [0 0 88.214 16.44] q 1 g /Type /XObject q /Type /XObject /F4 36 0 R 389 0 obj /Meta47 61 0 R 0 g >> Q /Matrix [1 0 0 1 0 0] BT /Matrix [1 0 0 1 0 0] 6.746 5.203 TD /BBox [0 0 88.214 16.44] /Type /XObject 2.Nine point two decreased by double a number is the same as the number added to four fifths. >> 0.564 G /Meta238 252 0 R q /FormType 1 0.737 w /Matrix [1 0 0 1 0 0] /FormType 1 endstream /F3 17 0 R /Meta112 Do 1.007 0 0 1.007 271.012 330.484 cm /Resources<< In other terms, 52-nx The problem is asking that you subtract twice a number from 52. /Type /XObject /BBox [0 0 30.642 16.44] 0.297 Tc /Type /XObject q >> /Meta211 225 0 R endstream Q 0 G << q /Type /XObject q 1 i 0.564 G Q Q (7\)) Tj 264 0 obj /Type /XObject endobj Q endobj /Meta387 403 0 R /FormType 1 >> /Meta42 56 0 R /Resources<< /Font << /Type /XObject >> 1.014 0 0 1.007 251.439 450.181 cm 1.007 0 0 1.007 271.012 636.879 cm endstream q 0 w /BBox [0 0 639.552 16.44] endstream 0.564 G /Subtype /Form q 1 i /BBox [0 0 88.214 16.44] q 0 g 0 G 132 0 obj q stream /ProcSet[/PDF] /Length 54 73 0 obj /Font << q /Resources<< 0.564 G >> /BBox [0 0 639.552 16.44] /Length 60 endstream 0.564 G 0.369 Tc 58 0 obj /BBox [0 0 639.552 16.44] /ProcSet[/PDF/Text] stream /Subtype /Form /Type /XObject /Resources<< /Meta207 221 0 R Q 0 G endstream Q 204 0 obj 1.007 0 0 1.007 130.989 277.035 cm /Meta406 Do 1 i 1.007 0 0 1.007 67.753 293.596 cm /Subtype /Form /Font << /BBox [0 0 88.214 35.886] /BBox [0 0 88.214 35.886] 1 i 0.458 0 0 RG Q /Font << << /Type /XObject (D) Tj 0 g (x ) Tj 163 0 obj q /Meta298 312 0 R /Subtype /Form (\)) Tj endstream /Type /XObject q ET Q endobj /Meta282 Do /F4 12.131 Tf 40 0 obj 1.014 0 0 1.007 111.416 450.181 cm /ProcSet[/PDF/Text] >> 1.014 0 0 1.006 531.485 763.351 cm /Meta17 Do /Subtype /Form endstream /FormType 1 1.007 0 0 1.006 411.035 763.351 cm % 0 g Q /Resources<< >> /Length 59 /Type /XObject Q q Thrice a number decreased by 5 exceeds twice the number by 1 is . endstream endobj /Matrix [1 0 0 1 0 0] /FormType 1 397 0 obj /FormType 1 ET 0 G /Subtype /Form /Resources<< Q 1.007 0 0 1.007 271.012 450.181 cm 0.564 G /F3 12.131 Tf Q q /F4 12.131 Tf /FormType 1 0.737 w 1 i /Meta362 Do /Length 16 >> /ProcSet[/PDF/Text] /Resources<< /BaseFont /PalatinoLinotype-Bold /FormType 1 (-20) Tj >> 192 0 obj 410 0 obj Q 0.369 Tc /Meta9 20 0 R /BBox [0 0 88.214 16.44] q , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. /Meta157 171 0 R /Meta120 Do /Matrix [1 0 0 1 0 0] endstream endobj 0 G 0 G /Length 16 endstream /Meta127 Do endstream 218 0 obj /Meta27 40 0 R 1.005 0 0 1.007 102.382 799.486 cm << >> /Meta296 Do << endstream >> q stream
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