/Matrix [1 0 0 1 0 0] >> BT /Meta187 Do /BBox [0 0 15.59 16.44] /Subtype /Form q >> /Meta254 268 0 R /Meta379 393 0 R 255 0 obj /Type /XObject 1.007 0 0 1.007 271.012 583.429 cm /Subtype /Form /Length 70 /Subtype /Form 0.737 w /BBox [0 0 534.67 16.44] q BT /FormType 1 >> /Subtype /Form 0.564 G /Length 70 << /Subtype /Form /FormType 1 /Length 119 Q q 0 w Q 1.005 0 0 1.007 79.798 779.913 cm 0 g 1.007 0 0 1.007 551.058 703.126 cm << q /Resources<< /F3 12.131 Tf endstream 0.37 Tc /F3 17 0 R endstream q Q /Type /XObject stream /FormType 1 >> q >> 285 0 obj >> /StemH 77 Q q /Type /XObject Q -0.03 Tw /Subtype /Form 0 4.894 TD BT /Subtype /Form Q >> q /ProcSet[/PDF/Text] /ProcSet[/PDF] 0 g /Subtype /Form /Resources<< 10.487 5.203 TD << /FormType 1 /Matrix [1 0 0 1 0 0] [(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 Q >> endstream /FormType 1 Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 551.058 703.126 cm q q Q /F3 17 0 R >> /Matrix [1 0 0 1 0 0] BT /Resources<< /F3 17 0 R /Type /XObject >> endobj Q >> 0 G 0 G /Type /XObject >> /Meta420 Do q 0 G >> 1.007 0 0 1.007 551.058 277.035 cm Q 0 G Q Q /Type /XObject q /Subtype /Form 0 G Q q Q /F1 7 0 R 0 G /ProcSet[/PDF/Text] /Meta336 350 0 R endstream /Matrix [1 0 0 1 0 0] /Resources<< BT >> (38) Tj 9.723 5.336 TD q /Matrix [1 0 0 1 0 0] /Subtype /Form /Meta403 419 0 R 0.737 w 1.014 0 0 1.007 111.416 277.035 cm /Type /XObject << 0 w /Meta161 Do q >> 260 0 obj 0.51 Tc /FormType 1 (7\)) Tj /FormType 1 1 i 0 G 0.227 Tc 0 5.203 TD BT /Length 16 q q /Font << q 1 i 0.564 G << 178 0 obj 1.007 0 0 1.007 67.753 546.541 cm /FormType 1 /Matrix [1 0 0 1 0 0] endobj /Length 16 << ET endobj 1.502 5.203 TD Q q >> << 252 0 obj q /FormType 1 /ProcSet[/PDF/Text] 0.227 Tc /Meta133 Do /ProcSet[/PDF/Text] q Q 20.21 5.203 TD 1 i q q 0.463 Tc Q /FormType 1 0 g /Type /XObject /FormType 1 0.458 0 0 RG /FormType 1 /FormType 1 >> 6.746 24.649 TD /F1 12.131 Tf endstream >> /Matrix [1 0 0 1 0 0] 1 g /Type /XObject endstream q Q ET 0 g Q << 349 0 obj 0 G endstream 1 i /BBox [0 0 88.214 16.44] 98.843 5.203 TD 1.007 0 0 1.007 411.035 383.934 cm /F3 17 0 R 0.564 G 118 0 obj 1.502 5.203 TD endstream q endobj 32.201 20.154 l endobj Q >> >> 0.524 Tc /FormType 1 /Length 67 << Q /BBox [0 0 88.214 16.44] Q 160 0 obj q endobj /BBox [0 0 88.214 16.44] 1 g /FormType 1 /Subtype /Form /Subtype /Form Q 0.737 w (38) Tj stream Q endobj 326 0 obj >> >> /Subtype /Form q D. Twice a number decreased by ten is less than 24. /Meta143 157 0 R /Resources<< /FormType 1 /FormType 1 Q /Font << endstream q q /Ascent 1050 BT Q 1.007 0 0 1.007 551.058 330.484 cm /Type /XObject << /F1 12.131 Tf q >> stream stream 0 G 225 0 obj /Meta23 34 0 R >> /FormType 1 q /BBox [0 0 88.214 16.44] stream 20.21 5.203 TD stream >> stream /FormType 1 >> q q q endstream q 0 g 0.68 Tc Q 0.564 G (x ) Tj q Q Q /Meta10 Do /F3 17 0 R 1 i /ProcSet[/PDF] 1.007 0 0 1.007 654.946 599.991 cm Q /ProcSet[/PDF] Q endstream Question. endobj << BT Q ( \() Tj /Meta302 Do ET q ET Q q q /Length 69 q ET BT /Matrix [1 0 0 1 0 0] Q >> (3) Tj 1.007 0 0 1.007 130.989 776.149 cm /F3 17 0 R /BBox [0 0 30.642 16.44] S 0 g >> 393 0 obj Q 1.005 0 0 1.007 102.382 473.519 cm /F3 17 0 R q /Matrix [1 0 0 1 0 0] /BBox [0 0 673.937 15.562] ET /Resources<< /BBox [0 0 534.67 16.44] 0 g 0 w endstream 0 g BT Q 0.564 G /Type /XObject (2\)) Tj BT q /Subtype /Form /Resources<< << /Meta87 Do 0 G (x) Tj (\(x ) Tj q >> /Resources<< 1 g /F3 17 0 R /Resources<< endstream >> /I0 Do 0 5.203 TD 0 G /BBox [0 0 88.214 16.44] /Type /XObject >> /Meta274 Do /Resources<< Q 1.007 0 0 1.007 271.012 277.035 cm /Font << 5 0 obj << /Resources<< Q Q /Resources<< << /ProcSet[/PDF] Q 1.014 0 0 1.007 391.462 277.035 cm 1.007 0 0 1.007 271.012 849.172 cm 42 0 obj (2\)) Tj q /BBox [0 0 30.642 16.44] stream Q 0 G q /BBox [0 0 534.67 16.44] /ProcSet[/PDF] /Meta214 228 0 R endstream 1 g q /Font << /ProcSet[/PDF] >> 0.458 0 0 RG /Type /XObject /Length 60 Q By the . /Resources<< Q /F4 12.131 Tf Q /FormType 1 /CapHeight 662 /ProcSet[/PDF/Text] Let the 2nd number be y. stream /Type /XObject 0 g << /Matrix [1 0 0 1 0 0] >> ET endstream >> >> >> ET /Subtype /Form Q (11) Tj 1 i stream endobj << q 1 g 1 g /FormType 1 Five times the sum of a number and four 7. q /Meta253 267 0 R BT q Q 1 g q 20.21 5.203 TD 401 0 obj 1.014 0 0 1.007 531.485 583.429 cm 0.178 Tc 93 0 obj BT Q /ItalicAngle 0 Q 0 g /XObject << /FormType 1 0 w Patients' reasons for declining screening were not collected . 0 w 0.737 w Q /FormType 1 BT endobj 0 g 329 0 obj q 13 0 obj /BBox [0 0 15.59 16.44] q /Type /XObject /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 0.737 w /Meta285 299 0 R Q /BBox [0 0 15.59 16.44] >> 672.261 726.464 m 19.474 20.154 l /FormType 1 /FormType 1 Q Q q ET 0 w endobj /Resources<< /Meta129 Do /Subtype /Form ET (11) Tj /Subtype /Form 1.005 0 0 1.007 102.382 872.509 cm endstream q BT 272 0 obj /Type /XObject ET q /ProcSet[/PDF] 0.425 Tc 1.014 0 0 1.007 111.416 330.484 cm Q 1.007 0 0 1.007 411.035 849.172 cm Thrice a number decreased by 5 exceeds twice the number by a unit. >> stream /Meta425 Do /Length 69 q Q q /Subtype /Form /Length 59 Q Q 66 0 obj 1.007 0 0 1.007 271.012 450.181 cm stream /Meta212 226 0 R BT /F3 17 0 R 0.564 G ET /Resources<< /Resources<< /Length 54 Q 213 0 obj /Subtype /Form 1 i endstream BT /Type /XObject stream /Resources<< /Length 16 Q 0 w /Length 16 stream 418 0 obj 1.007 0 0 1.007 411.035 277.035 cm Q q /Matrix [1 0 0 1 0 0] endobj Diabetes, if left untreated, leads to many health complications. /F3 17 0 R /ProcSet[/PDF] /FormType 1 Q 55 0 obj q >> q Q endstream >> stream q q q endobj (D\)) Tj 1.007 0 0 1.007 411.035 330.484 cm /FormType 1 1 i Q 1 i /Meta220 Do /F3 12.131 Tf /ProcSet[/PDF] 20.21 5.203 TD stream Q << >> 0.271 Tc /Subtype /Form /Resources<< /Meta222 Do Q 1 g Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. stream Q /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Resources<< 0.51 Tc endobj q >> /F3 17 0 R endstream Q /Type /XObject /Length 54 0 g /Matrix [1 0 0 1 0 0] /F4 36 0 R ET << 3.742 5.203 TD /Subtype /Form (x) Tj >> 1 i /BBox [0 0 15.59 29.168] /ProcSet[/PDF/Text] /Meta304 318 0 R >> /Meta119 133 0 R /Meta139 Do Q 282 0 obj endstream What is marios jumps times luigis jumps. 6.746 5.203 TD /ProcSet[/PDF/Text] << q stream /F3 12.131 Tf 1.007 0 0 1.007 271.012 776.149 cm q 0 5.203 TD 1 i Q /Resources<< /Resources<< /BBox [0 0 15.59 29.168] /F3 12.131 Tf Q /Subtype /Form /F3 12.131 Tf Q The sum of 18 and tour times a number is -6 Find the number. /Length 12 0 w HOPE HELPS .3. 0 g Q /Meta334 348 0 R /Subtype /Form /Meta265 Do 1.007 0 0 1.007 130.989 583.429 cm >> 0 g endstream /Matrix [1 0 0 1 0 0] >> 185.725 5.203 TD stream (B) Tj 0 w 20.21 5.203 TD /Matrix [1 0 0 1 0 0] endobj 1 i /Meta401 Do >> 407 0 obj q 0 w /Length 74 1.007 0 0 1.007 130.989 277.035 cm Twice = two times, double. 1 i q 1.007 0 0 1.006 411.035 437.384 cm /Resources<< 1 i 394 0 obj Q /XHeight 477 /Font << Q (A\)) Tj q << Select the correct mathematical statement for the following equation. endobj << /BBox [0 0 30.642 16.44] /F4 36 0 R /Meta45 Do Q 1 i Q /FormType 1 q /LastChar 45 /Type /Page >> /FormType 1 >> endstream q ET endstream 1 i /FormType 1 0.838 Tc q >> stream << /BBox [0 0 88.214 16.44] ET /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 g endobj q stream /Matrix [1 0 0 1 0 0] 0 g /BBox [0 0 88.214 16.44] >> << 209 0 obj q /F3 17 0 R 0 G 0.458 0 0 RG ET 0 g stream BT stream q (A\)) Tj q /FormType 1 Q Q /BBox [0 0 88.214 16.44] /Type /XObject endobj 0.311 Tc endstream Q 0 g /BBox [0 0 15.59 16.44] /Meta427 443 0 R /Subtype /Form >> 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. q 19.474 20.154 l >> 1 i /Meta399 Do 1.014 0 0 1.007 111.416 450.181 cm /Meta118 Do q q 0 G /ProcSet[/PDF/Text] 1 i 427 0 obj /ProcSet[/PDF] 1 i /Type /XObject /Resources<< /Subtype /Form /Subtype /Form /BBox [0 0 88.214 16.44] endstream 218 0 obj Q << 0.297 Tc endobj >> /Length 69 /Meta191 205 0 R /F3 17 0 R stream 0 g 1.014 0 0 1.006 111.416 510.406 cm /ProcSet[/PDF/Text] endstream Q 0.564 G /Resources<< 366 0 obj Q Q Q << q /FormType 1 /BBox [0 0 88.214 16.44] >> ET /ProcSet[/PDF/Text] endobj 0 w 1.014 0 0 1.006 531.485 510.406 cm endobj 0.458 0 0 RG endobj >> /Subtype /Form /Matrix [1 0 0 1 0 0] 1 i 281 0 obj Q 354 0 obj q 23.216 5.203 TD << endstream /Font << Q endstream >> /BBox [0 0 88.214 35.886] Q q /FormType 1 >> endobj /Resources<< q /Type /XObject /FormType 1 1.005 0 0 1.007 102.382 347.046 cm >> endobj /Type /XObject >> >> Q /Type /XObject /Meta304 Do 0.564 G /Subtype /Form /Meta238 Do /Matrix [1 0 0 1 0 0] /Meta7 Do /Resources<< 103 0 obj 0 5.203 TD /Matrix [1 0 0 1 0 0] /F3 12.131 Tf [(The )-19(quotient of )] TJ Q /Meta380 Do q /FormType 1 q /Matrix [1 0 0 1 0 0] /Resources<< 1.007 0 0 1.007 654.946 347.046 cm /ProcSet[/PDF/Text] 1 i 324 0 obj endobj /Meta83 Do 0.564 G /Font << Q /BBox [0 0 88.214 35.886] /Font << /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /FontDescriptor 10 0 R /Matrix [1 0 0 1 0 0] /Resources<< /Meta17 28 0 R 0.737 w BT stream >> Answer only. /Type /XObject /Resources<< 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. << 0 g Q /FormType 1 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] << /Matrix [1 0 0 1 0 0] /Subtype /Form ET q /Meta154 168 0 R Q 0 g >> 0 g q 1.007 0 0 1.007 271.012 277.035 cm 0 g /Resources<< /Meta273 287 0 R /Length 64 q If a number is 50%, then it is a half - the same as 0.5 or 1/2. /F1 7 0 R /Type /XObject /Matrix [1 0 0 1 0 0] >> q ET Q stream 0 g 20.21 5.336 TD q Q 296 0 obj 379 0 obj BT /BBox [0 0 88.214 35.886] 25.454 5.203 TD endstream 43.426 5.203 TD /Type /XObject << TJ endobj /Matrix [1 0 0 1 0 0] 0 w q /Length 59 /Meta24 37 0 R Q ( x) Tj /FormType 1 /BBox [0 0 30.642 16.44] << 1.007 0 0 1.007 271.012 703.126 cm endobj Q q /I0 Do /BBox [0 0 88.214 35.886] Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . -0.486 Tw Q stream Q Q BT /Subtype /Form /Font << << /Meta157 171 0 R q If twice a number is decreased by 13, the result is 9. q /Type /XObject >> q /Type /XObject /Resources<< q 0.564 G q /Meta363 Do 1 g 3 0 obj /ProcSet[/PDF/Text] Q /Resources<< Q q /BBox [0 0 88.214 16.44] BT 54 0 obj q q 0 5.203 TD BT /BBox [0 0 15.59 16.44] 1.014 0 0 1.006 251.439 690.329 cm BT << /Meta421 Do /Length 69 /Length 69 1.007 0 0 1.007 271.012 583.429 cm >> Q /Length 95 endobj q endstream stream Q q Q /Font << /ProcSet[/PDF/Text] 0 g /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q ET 1 i q stream -0.486 Tw /Length 69 stream /Font << 35 0 obj /Meta48 Do Q endobj >> 0.458 0 0 RG /Subtype /Form /Type /XObject 1.007 0 0 1.007 551.058 277.035 cm 0.564 G endobj Q >> /Meta377 391 0 R Q >> 20 0 obj /Font << 65 0 obj (A\)) Tj 1.014 0 0 1.007 251.439 636.879 cm << 1 i >> 0.458 0 0 RG endstream /Meta78 92 0 R 115 0 obj /Subtype /Form ET /BBox [0 0 88.214 35.886] Q (vii) Twice a number subtracted from 19 is 11. Q >> /Type /XObject 0 G >> q /F1 7 0 R q /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] 0 G Q /Resources<< 2.238 5.203 TD q >> /ProcSet[/PDF] 19 0 obj endstream Q /ProcSet[/PDF/Text] q /Resources<< /Font << endobj /BBox [0 0 639.552 16.44] /Meta396 Do 1.007 0 0 1.007 271.012 383.934 cm Q endobj endstream /Type /XObject Q (3) Tj /Matrix [1 0 0 1 0 0] BT /F3 12.131 Tf endstream q /Meta282 Do endobj Q 36 0 obj /Meta139 153 0 R /Meta206 220 0 R 1 i 1 i >> /F3 12.131 Tf 0 g >> /Resources<< /Length 16 Q /Meta186 200 0 R [tex]\sin (\pi -x)=\sin x[/tex]. 19.474 20.154 l /FormType 1 Q Q /Matrix [1 0 0 1 0 0] q endobj >> /Type /XObject /BBox [0 0 639.552 16.44] endstream q /Meta382 396 0 R /ProcSet[/PDF/Text] 0 g Q >> Q Q 0 g q 0 G 1.007 0 0 1.006 411.035 690.329 cm << /Type /XObject endobj endobj 0.564 G /Subtype /Form /Meta32 Do /Type /XObject 0 w Q >> /Type /XObject /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 816.048 cm q endstream >> << 0.564 G >> 0.737 w << q 1.007 0 0 1.007 654.946 799.486 cm /BBox [0 0 15.59 16.44] >> /Meta381 395 0 R /Subtype /Form /FormType 1 /Type /XObject 1 i 0 G Get a free answer to a quick problem. >> /BBox [0 0 88.214 16.44] 0 w q << /FormType 1 0 g /Type /XObject Q Q 375 0 obj /F3 17 0 R /ProcSet[/PDF] 0 g << 14 0 obj /Font << /ProcSet[/PDF/Text] endobj q 1.005 0 0 1.007 102.382 799.486 cm 0 w >> /F3 17 0 R q endstream 0 G >> Q /Type /XObject 0 G /FontName /TimesNewRomanPSMT >> /ProcSet[/PDF] Q /F3 12.131 Tf /Meta112 Do 1 i q 1.502 5.203 TD /BBox [0 0 639.552 16.44] >> /FormType 1 >> /F1 7 0 R Q >> /F3 17 0 R 0.737 w /BBox [0 0 88.214 16.44] endobj >> /BBox [0 0 88.214 35.886] 1 i Q /BaseFont /TimesNewRomanPSMT /Resources<< /Font << /BBox [0 0 88.214 16.44] 0.37 Tc endstream /Meta230 Do /Length 206 /Type /XObject /Subtype /Form (x) Tj q 0 g q /ProcSet[/PDF] stream 413 0 obj /Resources<< q Q endstream BT /Meta199 Do /Subtype /Form /BBox [0 0 88.214 16.44] /Meta168 Do /F3 17 0 R Q q /F3 17 0 R Q q << c Site 5 is not included in this number. /BBox [0 0 88.214 35.886] 0.425 Tc q q Q /ProcSet[/PDF] 0 g /Type /FontDescriptor 0 g 0 g /ItalicAngle 0 >> >> /Resources<< /Matrix [1 0 0 1 0 0] BT Q stream q Q Q Q /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 16 /Subtype /Form endstream (\)) Tj /F1 7 0 R q 415 0 obj /Meta322 Do Q /Meta236 250 0 R ET /Length 69 /BBox [0 0 15.59 16.44] /Length 69 /Length 59 /Type /XObject /BBox [0 0 549.552 16.44] /FormType 1 q /Meta23 Do /Meta397 413 0 R q /Type /XObject stream 144 0 obj stream Q 1.007 0 0 1.006 130.989 690.329 cm Expression. stream q Thrice a number decreased by 5 is 3x - 5. /FormType 1 /Matrix [1 0 0 1 0 0] /FormType 1 /Type /XObject /Font << (iii) 25 exceeds a number by 7. Q q endstream /Type /XObject Q /Matrix [1 0 0 1 0 0] q Q /Type /XObject /Resources<< q 1 i /Resources<< 0.458 0 0 RG >> q BT 47 0 obj ET q /Meta64 Do 0.458 0 0 RG 0 G [(The )-16(s)15(um )-14(of )] TJ /BBox [0 0 15.59 16.44] /Resources<< 1.014 0 0 1.007 251.439 523.204 cm 370 0 obj << endobj 360 0 obj BT /BBox [0 0 88.214 16.44] endstream /ProcSet[/PDF] /F1 7 0 R /Length 2252 /Matrix [1 0 0 1 0 0] Q >> /BBox [0 0 639.552 16.44] To find: The. /Length 69 >> >> /Type /XObject q /Matrix [1 0 0 1 0 0] /Resources<< /BBox [0 0 15.59 16.44] q << 1 g /Matrix [1 0 0 1 0 0] /Resources<< q Q 1 g 0.737 w Q >> q 0 G 1 i q BT /BBox [0 0 15.59 16.44] /F3 12.131 Tf BT /ProcSet[/PDF] /F3 12.131 Tf /Font << Q 0.737 w >> /BBox [0 0 30.642 16.44] endobj >> endstream q >> /Meta351 Do 0 5.203 TD stream /F3 17 0 R q /BBox [0 0 88.214 16.44] /Meta145 Do /Subtype /Form 0.458 0 0 RG Q 0 g stream /Matrix [1 0 0 1 0 0] /Length 16 /Matrix [1 0 0 1 0 0] Q Q q /Meta417 Do 1.007 0 0 1.007 551.058 636.879 cm q Q ET >> /Font << /Resources<< 0.564 G q /Length 64 /BBox [0 0 88.214 16.44] >> /FormType 1 /Resources<< q 206 0 obj /F3 17 0 R /Subtype /Form 1 i 0 g /F1 12.131 Tf q /Matrix [1 0 0 1 0 0] 1 g /Meta293 307 0 R BT >> /F3 12.131 Tf 322 0 obj 0 5.203 TD /Length 108 << /BBox [0 0 88.214 16.44] 1 i 1 i /Meta140 Do >> 161 0 obj /F3 17 0 R /Type /XObject /Matrix [1 0 0 1 0 0] endobj 1 g << Q /Font << >> 32.201 5.203 TD endstream q /Meta169 Do /Font << 0.458 0 0 RG 0.51 Tc 0.737 w /FormType 1 q >> /FormType 1 0 G stream /Length 69 /BBox [0 0 534.67 16.44] Solution: Let the number be x. BT /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endobj >> /F3 12.131 Tf /Subtype /Form /FormType 1 /Type /XObject << Q /Subtype /Form >> /Meta288 Do >> /XObject << /Length 69 /Meta184 198 0 R Q /Length 12 0 G stream 0.737 w /Length 65 0.458 0 0 RG endstream endstream /Meta156 170 0 R Q endstream 1 i Q /Type /XObject endstream q 1.007 0 0 1.007 45.168 813.037 cm /Matrix [1 0 0 1 0 0] 0 G /Matrix [1 0 0 1 0 0] Answer: 52 decreased by twice a number in algebric expression Step-by-step explanation: The problem is asking that you subtract twice a number from 52. Q endobj << /Resources<< 1 i >> 0.738 Tc 0.68 Tc Twice a number decreased by . /Type /XObject 0.369 Tc 0 5.203 TD Q /FormType 1 endobj q q Q /Meta196 Do >> q /ProcSet[/PDF/Text] BT 1 i << Q /F1 7 0 R 0 w << /Length 69 1.005 0 0 1.007 102.382 256.709 cm /Length 16 /Resources<< /Subtype /Form /Meta235 249 0 R Q endstream Q /ProcSet[/PDF/Text] ET endobj q q 0 g BT 0 g /BBox [0 0 17.177 16.44] 1st step. 0.68 Tc Q /Meta118 132 0 R 1.005 0 0 1.007 102.382 400.496 cm >> 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 /Meta86 Do >> /Resources<< 0 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /FormType 1 176 0 obj /Subtype /Form /Subtype /Form /Length 54 q /Meta341 Do q 1.007 0 0 1.007 271.012 583.429 cm Answer (1 of 8): Solution: let the number be x. Q endstream /Subtype /Form 1 i q /Length 16 endstream >> /Meta223 237 0 R /Resources<< Q /BBox [0 0 88.214 16.44] >> endstream << 1.005 0 0 1.007 102.382 726.464 cm The first number increased by three times the second number is -25. x + 3y = -25. by solving the system of equations. endobj >> Q 1 i endstream /Type /XObject /Resources<< 0 g Q /Resources<< Q 0 g 414 0 obj /Meta202 Do 1.007 0 0 1.007 411.035 849.172 cm /F3 17 0 R endobj q q stream q q (x) Tj /Subtype /Form q BT Q << /BBox [0 0 15.59 29.168] /I0 51 0 R 1 g endobj << q q >> BT q /BBox [0 0 88.214 16.44] Q Q /Subtype /Form (5) Tj BT /Resources<< 0.369 Tc /Meta60 74 0 R q stream /Meta194 208 0 R /ProcSet[/PDF/Text] /Meta69 83 0 R
Andra Martin Obituary,
Northern California Junior Tennis Rankings,
Articles T
