q >> ET q /Matrix [1 0 0 1 0 0] 0 g /Resources<< 0 g 1 i /F3 17 0 R >> /Length 69 /Length 81 >> stream /Length 69 << q 1 i ET Q /Meta212 226 0 R 0 w /ProcSet[/PDF] 1 i q /BBox [0 0 88.214 35.886] You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8 gives 58 find the number, 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. Q /BBox [0 0 88.214 16.44] /Meta97 111 0 R (5\)) Tj Q Q /Matrix [1 0 0 1 0 0] >> 0 g /Subtype /Form q Q 0 g /Subtype /Form endobj << >> /Meta266 280 0 R /Meta335 Do /F3 17 0 R BT >> q /FormType 1 0.458 0 0 RG /FormType 1 /Resources<< 368 0 obj Q Q endstream Twice 4 bananas is 8. endobj /Length 2252 0.458 0 0 RG 1 i ET endobj Q /Matrix [1 0 0 1 0 0] BT >> /ProcSet[/PDF] /Meta384 Do /FormType 1 1 g 722.699 347.046 l Twice a number when decreased by 7 gives 45. /Length 64 /Length 69 endobj /ProcSet[/PDF] << q << 1 i 549.694 0 0 16.469 0 -0.0283 cm /Resources<< /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] Thrice a number decreased by 5 exceeds twice the number by 1 is . endobj /Subtype /Form /Meta197 211 0 R endobj 0 G q 34 0 obj /Contents [446 0 R] /F3 12.131 Tf /Resources<< 1 i 0 5.203 TD Q q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] q /FormType 1 /F3 17 0 R q /LastChar 120 q 230 0 obj q >> ET 0 5.203 TD q Q q /Meta177 Do /Length 69 Q >> 0.458 0 0 RG q 383 0 obj /F4 36 0 R 0 g 0.564 G >> q 1.014 0 0 1.006 251.439 690.329 cm ET 1 i /Meta274 288 0 R >> endobj >> /Resources<< Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /FormType 1 73 0 obj /FormType 1 >> /F1 12.131 Tf >> /F3 17 0 R /Font << /Matrix [1 0 0 1 0 0] 0.838 Tc /BBox [0 0 88.214 16.44] Q Q Q 1.005 0 0 1.007 102.382 616.553 cm q q q q /F3 12.131 Tf endobj 1 i endobj /Subtype /Form ET Q q /Resources<< q ET /Font << stream q /FormType 1 /ProcSet[/PDF] << /Length 67 q /Subtype /Form /Meta214 Do 167 0 obj /ProcSet[/PDF] q endobj 1 i 1 i q Q 0 w q /ProcSet[/PDF/Text] 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 /Matrix [1 0 0 1 0 0] endstream q /Type /XObject ET /Meta30 43 0 R 1.008 0 0 1.007 654.946 293.596 cm /Meta65 Do /Meta312 326 0 R Q /BBox [0 0 15.59 16.44] /Meta43 57 0 R 1 g /BBox [0 0 639.552 16.44] /Type /XObject >> endstream Q q >> q Q 1.007 0 0 1.007 551.058 277.035 cm Q q << q >> /Font << 0 w 1.005 0 0 1.007 102.382 726.464 cm /Type /XObject ET stream q 0.458 0 0 RG >> /ProcSet[/PDF/Text] Q << >> q /Meta131 145 0 R q /F1 7 0 R endstream /F1 12.131 Tf /Matrix [1 0 0 1 0 0] Q endobj c Site 5 is not included in this number. ET endstream stream Q /Meta111 Do /Subtype /Form << 1.007 0 0 1.007 654.946 546.541 cm /Resources<< /FormType 1 BT /Subtype /Form Q Q /Meta244 Do endstream q 0 g Just type into the box and your calculation will happen automatically. /F4 12.131 Tf 0 G /Length 69 /Meta45 Do q /Meta320 Do /ProcSet[/PDF/Text] stream >> >> /Type /XObject /Resources<< /Matrix [1 0 0 1 0 0] 1.007 0 0 1.006 411.035 763.351 cm /BBox [0 0 88.214 16.44] q /Meta308 Do 1.007 0 0 1.007 271.012 523.204 cm Q (3) Tj Q >> /Meta142 156 0 R 1.007 0 0 1.007 411.035 277.035 cm Q stream 122 0 obj /Meta128 Do endstream q /Length 68 Q /Matrix [1 0 0 1 0 0] >> BT 1.005 0 0 1.007 102.382 293.596 cm 0.458 0 0 RG q /ProcSet[/PDF] >> /FormType 1 /ProcSet[/PDF] /FormType 1 >> Q endstream /Subtype /Form /Subtype /Form Q >> /Meta236 Do 0 G /Font << >> 1 i Q /Subtype /Form 1 g /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] /FormType 1 q /Matrix [1 0 0 1 0 0] stream endobj /Meta28 Do /Subtype /Form q endstream /ProcSet[/PDF] << /Meta202 Do q q 0.564 G 0 5.203 TD /ProcSet[/PDF/Text] endstream 32.201 5.203 TD /Type /XObject << Q >> Q endobj /F4 36 0 R /Font << >> q >> << /Subtype /Form 0 g stream 20.21 5.203 TD q 0 G /ProcSet[/PDF] (2\)) Tj stream 2.238 5.203 TD 1 i /Resources<< 0.68 Tc Q Q /Type /XObject >> 37 0 obj 1 i q >> 1.014 0 0 1.007 391.462 383.934 cm 0 g Q stream /ProcSet[/PDF] /Meta195 209 0 R q /XHeight 471 1 g stream /Meta94 108 0 R /Subtype /Form Q Q >> /BBox [0 0 88.214 16.44] /F3 12.131 Tf [(Fiv)25(e ti)18(me)16(s)] TJ ET /Meta388 404 0 R /F3 17 0 R << /Type /XObject /FontName /PalatinoLinotype-Roman 0 5.203 TD 0.37 Tc 0.564 G 1.007 0 0 1.007 67.753 347.046 cm /F3 17 0 R /Type /XObject /Font << 15 0 obj 0.486 Tc /Resources<< stream 24.718 8.18 TD 1 i >> /Subtype /Form /Length 118 /FormType 1 >> 0.737 w 377 0 obj The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. /Matrix [1 0 0 1 0 0] With this, we get: "3x-8". /Subtype /Form /Type /XObject /Resources<< q endobj /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] -0.382 Tw endstream 0 G /ProcSet[/PDF/Text] 0 g /Type /XObject /Meta5 14 0 R xref >> BT /ProcSet[/PDF] /Length 68 >> << 1.502 5.203 TD ET /BBox [0 0 15.59 16.44] 1 i /Resources<< /Matrix [1 0 0 1 0 0] >> q >> 1.007 0 0 1.007 271.012 703.126 cm /Subtype /Form 0.271 Tc /BBox [0 0 673.937 14.853] /Font << /F1 12.131 Tf Q q Q endstream Q >> Q 0.51 Tc 0 g endstream 0.564 G Q BT Q 0.738 Tc q /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] -0.03 Tw << 0 G >> /BBox [0 0 639.552 16.44] -0.486 Tw >> >> /BBox [0 0 88.214 16.44] << /Matrix [1 0 0 1 0 0] /FirstChar 32 /ProcSet[/PDF] Q /F3 17 0 R /Subtype /Form 1 i 1.007 0 0 1.007 271.012 849.172 cm 0 g BT 1.007 0 0 1.006 551.058 437.384 cm << /BBox [0 0 88.214 16.44] q >> /Type /XObject /Subtype /Form 232 0 obj Q 0 G /Meta93 Do /FormType 1 /ProcSet[/PDF] >> >> >> q /Meta304 Do 0 g /Length 16 Q /Font << Q /Meta332 Do /F3 12.131 Tf q 1.007 0 0 1.007 271.012 330.484 cm 159 0 obj 1.007 0 0 1.007 67.753 347.046 cm >> 1.005 0 0 1.007 102.382 599.991 cm 0 G q Q /FontBBox [-174 -299 1445 1050] stream endobj /Subtype /Form stream Q 1.502 5.203 TD /Matrix [1 0 0 1 0 0] 9.723 5.336 TD Q >> endobj q q >> << /Subtype /Form Q BT >> /Matrix [1 0 0 1 0 0] q 133 0 obj q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 654.946 726.464 cm >> endobj << /Resources<< 1 i q 0 g >> Q 0 g ET q >> Q (D\)) Tj /BBox [0 0 534.67 16.44] 0 g Q endstream Q 0 g q /Meta217 231 0 R << 0 g q Q /Descent -216 /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.007 271.012 277.035 cm stream 1.007 0 0 1.007 271.012 776.149 cm /Meta401 Do Q In the problem above, x is a variable. 0.458 0 0 RG 0 0 Similar questions Find the number which when decreased by 8% becomes 506. >> /Meta317 331 0 R /Length 16 /Subtype /Form q >> q /Resources<< /Meta264 Do 1st step. /BBox [0 0 88.214 16.44] Q >> 0 g /FormType 1 >> q 128 0 obj BT >> 328 0 obj q /Font << BT S /Meta123 137 0 R 14.23 24.649 TD << /Subtype /Form Q 2x - y = 6. /Subtype /Form q >> >> BT 722.699 473.519 l Q /F1 12.131 Tf endobj q /F1 7 0 R /Font << endstream 0 g >> endstream /Resources<< << q 1 g /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Resources<< /Length 12 Q 1 g q stream 0.564 G stream q [(The )-19(quotient of )] TJ >> /Subtype /Form stream stream q /Meta253 267 0 R ET /Meta195 Do stream 0 g << stream 1 g stream /Font << Q q Q Q 0 g Q /Font << 0.458 0 0 RG /Meta389 405 0 R /Meta168 Do q endstream q 1 i 1.007 0 0 1.007 551.058 583.429 cm (1) Tj Q 172 0 obj /F4 12.131 Tf /Font << 0 G /Type /XObject >> BT 1 i q << >> for the season. /Resources<< Q >> stream q /Type /XObject /FormType 1 (x) Tj >> /Font << endstream 338 0 obj >> 1 i /Meta208 Do 1 i 0 g >> /Length 16 [(thir)17(te)15(en)] TJ /Length 16 /Length 65 BT /FormType 1 only about 58% of candidates will agree to be screened. q Q /Subtype /Form 1 i << /Length 16 Q 0 g Q ET Q ET 1 g Q >> /Meta191 205 0 R 241 0 obj 1 i 0.564 G /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] /Type /XObject BT 0 G /Meta67 81 0 R q /BBox [0 0 673.937 27.581] /Meta169 Do /Subtype /Form /Font << 1 g endstream 1.014 0 0 1.007 111.416 383.934 cm 1.014 0 0 1.006 111.416 437.384 cm Q 255 0 obj /Length 57 q q /ProcSet[/PDF/Text] 0.737 w q /Type /XObject >> 0.458 0 0 RG /Meta252 266 0 R stream /F3 17 0 R /Matrix [1 0 0 1 0 0] q 1.007 0 0 1.007 271.012 849.172 cm /Resources<< endobj endstream Q q 1 i >> >> /Meta247 261 0 R << Q /MissingWidth 250 >> >> Q >> stream >> 0.369 Tc << Q /F3 12.131 Tf Q /Meta164 178 0 R 1 i 0 G 1 i S 0 w /Meta123 Do 1.007 0 0 1.006 411.035 763.351 cm 1.007 0 0 1.007 271.012 776.149 cm /Matrix [1 0 0 1 0 0] /Kids [ Q q << 0.458 0 0 RG >> /Meta234 Do /ProcSet[/PDF] stream 1.007 0 0 1.007 45.168 796.475 cm /Type /XObject 1 i >> Q /F3 17 0 R 1 g Q /Type /XObject Q 106 0 obj ET /BBox [0 0 534.67 16.44] /Resources<< ET 0 G endobj Q /Meta375 389 0 R /Meta424 440 0 R 1 i Q /ProcSet[/PDF] 15.731 5.336 TD /Type /XObject 0 G /ProcSet[/PDF/Text] Q /Font << q BT /Resources<< /Type /XObject << /F1 7 0 R /Length 69 126 0 obj /Font << BT 0 g q /Font << /Font << /FormType 1 stream /Meta183 197 0 R /Meta235 Do (+) Tj 0 g q /Type /XObject 1.007 0 0 1.006 551.058 437.384 cm /Type /XObject Twice a number decreased by 8 gives 58 find the number Advertisement Loved by our community 24 people found it helpful Xiphodon Step-by-step explanation: 2x-8=58 2x=66 x =33 Hope it helps Please mark as brainliest Find Math textbook solutions? 0 G ET << Q stream /Resources<< q /Type /XObject 0.458 0 0 RG 0 g /Matrix [1 0 0 1 0 0] q 1 g Q stream /ProcSet[/PDF/Text] Q /BBox [0 0 15.59 16.44] 0.68 Tc 425 0 obj /ProcSet[/PDF] /Resources<< /BBox [0 0 549.552 16.44] Q /Length 59 /Resources<< Q q 0 G 0.458 0 0 RG >> 32.201 5.203 TD /Matrix [1 0 0 1 0 0] /Length 69 /Meta370 Do endstream endobj /Type /XObject 0.786 Tc /F4 12.131 Tf 1.005 0 0 1.007 102.382 347.046 cm /Matrix [1 0 0 1 0 0] 1 i 1 g /Type /XObject Q endobj /Length 12 /Length 58 Q /XObject << 1.007 0 0 1.007 67.753 599.991 cm /ProcSet[/PDF/Text] 0 G /Resources<< Q /Subtype /Form /Descent -299 0 G q >> 401 0 obj endobj /Length 80 endobj /I0 51 0 R /Subtype /Form /Meta338 352 0 R /FormType 1 Q q /Meta257 Do /Resources<< >> /Length 16 3.742 5.203 TD /Type /XObject 1.005 0 0 1.007 79.798 746.789 cm ET /ProcSet[/PDF/Text] 1.007 0 0 1.007 130.989 383.934 cm Example 1: Use the tables above to translate the following English phrases into algebraic expressions. Q /Meta395 411 0 R BT >> /Type /XObject endobj /Type /XObject /Matrix [1 0 0 1 0 0] 90 0 obj endobj 1.007 0 0 1.007 271.012 383.934 cm /FormType 1 1 i /FormType 1 Q 0.458 0 0 RG >> Q Q /Subtype /Form >> /BBox [0 0 88.214 16.44] q /Resources<< q >> 0 g endstream 1.005 0 0 1.007 102.382 653.441 cm /Length 54 1.007 0 0 1.007 271.012 583.429 cm 0 G endstream Q /FormType 1 0.51 Tc q q stream Q 0 G 1.005 0 0 1.007 79.798 813.037 cm /Subtype /Form q /Matrix [1 0 0 1 0 0] /FormType 1 q (5) Tj q 0.564 G /Meta343 357 0 R Q << endobj 204 0 obj >> endobj endobj /Length 58 BT >> stream >> >> q q ET >> (x ) Tj /Resources<< stream endstream stream /Type /XObject /Subtype /Form (-) Tj Q 0 g >> 185.725 5.203 TD BT /Meta316 330 0 R q /Meta391 407 0 R Q endobj Q 1 i Q Q q >> /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Subtype /Form /Resources<< (A\)) Tj Q << q endobj /F3 17 0 R stream /Subtype /TrueType >> 21 0 obj q 1.014 0 0 1.007 111.416 703.126 cm /MissingWidth 250 << (D) Tj 160 0 obj 1.007 0 0 1.007 271.012 523.204 cm S q Q /Subtype /Form 1.007 0 0 1.007 130.989 583.429 cm /Matrix [1 0 0 1 0 0] /Type /XObject q ET q Q 0.564 G /FormType 1 endobj /Length 57 /Meta173 187 0 R q /Font << /Resources<< Q S 0.51 Tc Q 11 0 obj ( decreased by ) Tj 360 0 obj /FormType 1 /Subtype /Form 0.458 0 0 RG q endstream /Length 58 0.458 0 0 RG Q /Length 151 Q 0 g 1.007 0 0 1.007 271.012 277.035 cm Q stream 0 G /Length 118 endobj endobj endstream 0.737 w /Matrix [1 0 0 1 0 0] A link to the app was sent to your phone. q Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Type /XObject /Type /XObject /Type /XObject 334 0 obj 0.458 0 0 RG 0.737 w 1 i 0.458 0 0 RG 0.297 Tc q 261 0 obj 1.014 0 0 1.007 251.439 849.172 cm /Type /XObject /Subtype /Form q 149 0 obj BT 1 i [(1)-25(0\))] TJ >> q 0 g /FormType 1 >> 222 0 obj >> q /Type /XObject >> q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] 10 0 obj /Meta51 Do 0 g >> 0 g (B) Tj /Subtype /Form 1 i /Font << q /Meta414 Do 0 w 36 0 obj 0.564 G /F3 17 0 R /F3 17 0 R 0 g /FormType 1 316 0 obj Q 32.201 5.203 TD /F3 17 0 R /Resources<< 0.458 0 0 RG << 0 g q endobj << q >> 111 0 obj endobj 1.005 0 0 1.007 45.168 889.071 cm q 0.737 w /Meta235 249 0 R >> /Type /XObject /Length 294 /F3 12.131 Tf /Font << 0 g Q /Length 16 /FirstChar 32 q /Matrix [1 0 0 1 0 0] /Length 69 stream /Meta327 341 0 R Q /F3 12.131 Tf 0.458 0 0 RG 0 g /Matrix [1 0 0 1 0 0] q /Resources<< 1.502 5.203 TD >> ( \() Tj stream 1 i 0 g stream q /Resources<< Q endobj /Font << /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] Q 0.227 Tc /Meta162 Do 1 i /Meta317 Do q q 20 0 obj /FormType 1 /ProcSet[/PDF/Text] /Type /XObject /F3 12.131 Tf 0 g Q >> << 404 0 obj >> endstream stream /Meta328 Do 0 G 1 i /StemH 94 /BBox [0 0 534.67 16.44] Q /BBox [0 0 88.214 16.44] Q >> q 0 G 1 i Q 0.458 0 0 RG >> endobj /Type /XObject /Meta274 Do q 265 0 obj q 0.737 w 112 0 obj /Resources<< endobj /ProcSet[/PDF/Text] ET endobj q /Length 78 [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /Type /XObject q /Length 68 /Type /XObject 0 G 1.007 0 0 1.007 551.058 523.204 cm 0 g endstream /Resources<< /Matrix [1 0 0 1 0 0] << /Length 69 0 G >> stream /Meta52 66 0 R Q /Resources<< /Meta417 433 0 R ] 1.007 0 0 1.007 271.012 703.126 cm /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] 0 G /BBox [0 0 673.937 16.44] q 15.731 5.336 TD /ProcSet[/PDF] Q 1 g /Length 16 Q Q 0.369 Tc /Meta70 84 0 R 203 0 obj /Subtype /Form q /ProcSet[/PDF] /Subtype /Form stream /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] /Length 59 /BBox [0 0 88.214 16.44] /ProcSet[/PDF] q 0 g >> /Meta347 Do /FormType 1 0 5.203 TD Q q endstream q q /ProcSet[/PDF] q /ProcSet[/PDF/Text] /F3 17 0 R BT Q /ProcSet[/PDF] 0 G q q Q q 1 i stream /Type /XObject 1 i /Matrix [1 0 0 1 0 0] stream q 0.458 0 0 RG Q 0 g /F3 17 0 R >> /F3 17 0 R Q 0.458 0 0 RG Q /Meta144 158 0 R stream /FormType 1 q BT 124 0 obj stream /Resources<< 0 g /Meta56 70 0 R 0.564 G /Meta314 328 0 R /F3 12.131 Tf /Type /XObject BT q endstream stream /Font << 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) q q /Font << /F3 12.131 Tf q /Subtype /Form /F3 17 0 R /Matrix [1 0 0 1 0 0] Q /MediaBox [0 0 767.868 993.712] BT Q /F1 7 0 R Q Q endstream stream 0 g endobj /Length 16 358 0 obj stream >> endobj /Meta315 Do 1 i q >> 1.007 0 0 1.007 271.012 849.172 cm /Subtype /Form /Meta199 213 0 R (B\)) Tj /Matrix [1 0 0 1 0 0] 0 g /Subtype /Form Q 1 g << >> 0 g q q q /BBox [0 0 549.552 16.44] Q /Subtype /Form (\)) Tj /BBox [0 0 15.59 16.44] /Subtype /Form << q ET /ProcSet[/PDF/Text] /Resources<< /ProcSet[/PDF] q /FormType 1 >> >> /Meta154 168 0 R q 13.493 5.336 TD /Font << q Q /BBox [0 0 15.59 16.44] >> >> stream 1 i /Subtype /Form BT >> 0 g /Subtype /Form /Meta141 Do /Subtype /Form >> /ProcSet[/PDF] 1 i /Length 118 /Subtype /Form Q /Resources<< /ProcSet[/PDF/Text] Q stream /Meta26 Do Q /Font << Q /ProcSet[/PDF] /F3 12.131 Tf /BBox [0 0 17.177 16.44] /Length 78 0 g 1 i /ProcSet[/PDF/Text] << /BBox [0 0 88.214 16.44] >> q Q q 0 g /F3 12.131 Tf /Font << /Subtype /Form stream endobj /BBox [0 0 88.214 35.886] 1.007 0 0 1.006 551.058 437.384 cm q /FormType 1 /Length 59 endobj q Q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] >> /Matrix [1 0 0 1 0 0] endobj q /Resources<< /Meta323 337 0 R /Length 58 /ProcSet[/PDF/Text] >> /BBox [0 0 88.214 35.886] ET 582 546 601 560 395 424 326 603 565 834 516 556]>> (D\)) Tj q 1 g >> endobj >> /Matrix [1 0 0 1 0 0] Q /Resources<< /Resources<< /BBox [0 0 17.177 16.44] /ProcSet[/PDF/Text] /F3 12.131 Tf endobj << >> /Length 58 1.007 0 0 1.007 45.168 862.723 cm q The value of k is: (b) 3 (d) 0 (a) 4 (c) -4 TL ing:, 1)take a graph and draw two perpendicular lines to obtain four uadrants 2)draw any object using straight line 3) write the coordinates of each point o /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 82 0 obj BT q /Font << endobj /Resources<< 26.219 5.336 TD q Q /F3 12.131 Tf /Length 54 0 g Q /BBox [0 0 88.214 16.44] 0 G >> [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ /F4 36 0 R >> 1 i /F4 12.131 Tf 0 g /Length 118 Q /Font << /BBox [0 0 639.552 16.44] endobj /Font << /Type /XObject q 0 w Q Q /Font << (x) Tj 1 g endobj BT >> 0 g Q /Matrix [1 0 0 1 0 0] << /MediaBox [0 0 767.868 993.712] /FormType 1 endobj 0 g /Length 59 (38) Tj /I0 51 0 R 0.564 G /Meta355 369 0 R q << q 1.007 0 0 1.007 130.989 636.879 cm Q q 0 G 1.007 0 0 1.006 551.058 763.351 cm BT 1 i q 1 i BT 1 i /Length 16 51 0 obj q /BBox [0 0 534.67 16.44] /FormType 1 258 0 obj Q /FormType 1 1 i endobj >> 418 0 obj 0.486 Tc Find the number. /Type /XObject >> /FormType 1 1.007 0 0 1.007 551.058 703.126 cm endstream /Subtype /Form BT << S q 1 i ET /Matrix [1 0 0 1 0 0] 345 0 obj /Length 103 Q q /Meta299 313 0 R 1.005 0 0 1.007 79.798 796.475 cm /Meta91 Do /Matrix [1 0 0 1 0 0] 0 g endstream ET << 296 0 obj 0.486 Tc Q (8\)) Tj /Subtype /Form 0.458 0 0 RG (+) Tj 0 G endobj -0.486 Tw /Meta388 Do 0 G q 3.742 5.203 TD /Type /XObject /Subtype /Form endstream (13) Tj 0 g /Resources<< Q 1 i /ProcSet[/PDF] stream Q stream 0.564 G /ProcSet[/PDF] BT /Font << 0 g /BBox [0 0 88.214 35.886] 1 i ET BT /Meta348 Do 0 5.203 TD q endstream endstream /Type /XObject 0.68 Tc 0.524 Tc 2.238 5.203 TD /ProcSet[/PDF] /Type /XObject >> 1.007 0 0 1.006 130.989 690.329 cm /F4 36 0 R /Meta140 154 0 R Q Q /Subtype /Form 395 0 obj 1.007 0 0 1.006 411.035 437.384 cm q endobj q q /Subtype /Form Q /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] << q >> [(The )-16(s)15(um )-14(of )] TJ >> << /Subtype /Form stream 1 i ET endstream stream BT 18 0 obj saugatpandey635 saugatpandey635 22.09.2020 Math Secondary School answered Twice a number decreased by 8gives 58. /Type /XObject /Meta397 Do 1.007 0 0 1.007 67.753 546.541 cm >> endstream q /FormType 1 /F3 17 0 R 0.369 Tc 672.261 599.991 m /Matrix [1 0 0 1 0 0] 0 5.203 TD q 0 G 32 0 obj /FormType 1 >> << /F4 12.131 Tf stream /Meta0 Do 109 0 obj q /FormType 1 /ProcSet[/PDF/Text] 0 g Q >> /Matrix [1 0 0 1 0 0] >> /Meta32 Do q /Resources<< Q >> q q BT 0 g q 1 g /Meta135 149 0 R /F3 17 0 R /Meta102 Do q /Font << << /Meta152 Do endstream 0.564 G /Meta144 Do q 0.369 Tc /FormType 1 1.005 0 0 1.015 45.168 53.449 cm q 1 i /BBox [0 0 639.552 16.44] /Matrix [1 0 0 1 0 0] q 0.564 G 0 G 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 . /Meta365 379 0 R Q Q endstream q q Q BT endobj >> 0.564 G 0 4.894 TD A number = an unknown number which can be represented by a variable, usually x. >> /ProcSet[/PDF] BT 0 g q /Meta310 Do /Type /XObject 1.502 7.841 TD 0 w 40 0 obj 1.005 0 0 1.007 102.382 599.991 cm /FormType 1 endstream /Meta264 278 0 R >> /Subtype /Form ( \() Tj Q /FormType 1 /Filter [/CCITTFaxDecode] << stream /ProcSet[/PDF/Text] >> 0 G q /F1 7 0 R /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 473.519 cm Q /Meta261 275 0 R q 0 g Q Q 298 0 obj 0.737 w Q 1 i 0.564 G /Subtype /Form 1.014 0 0 1.007 531.485 277.035 cm >> >> >> endobj ET 1 g q /F3 17 0 R /Length 16 endstream /Subtype /Form Q /Resources<< >> endobj endobj /ProcSet[/PDF/Text] endstream /I0 Do << endobj Q /Type /XObject q /Subtype /Form -0.008 Tw >> >> (\)) Tj /BBox [0 0 534.67 16.44] stream q << << /BBox [0 0 88.214 16.44] q endobj /F3 12.131 Tf /StemV 94 q /ProcSet[/PDF] 1.007 0 0 1.007 551.058 523.204 cm >> /FormType 1 /Matrix [1 0 0 1 0 0] 1 i >> ( x) Tj q 0 g 174.501 5.203 TD /Meta137 Do Q /BBox [0 0 88.214 16.44] endobj BT 0 g /ProcSet[/PDF/Text] /Type /XObject /Length 16 endobj stream Q Q 0.564 G 1.007 0 0 1.007 654.946 799.486 cm 2.238 5.203 TD /Matrix [1 0 0 1 0 0] 0.564 G /F3 17 0 R /Meta167 Do q Q /Font << >> /Meta424 Do If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. This gives us: "2x+5". Q /Meta6 Do /Type /XObject 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. Q /Type /XObject endobj 238 0 obj /I0 Do Q endobj /BBox [0 0 30.642 16.44] 0 g /ProcSet[/PDF/Text] 1.014 0 0 1.007 531.485 523.204 cm /Resources<< >> 1.005 0 0 1.007 102.382 743.025 cm q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q Q /Meta159 173 0 R q /Meta87 Do 423 0 obj /BBox [0 0 88.214 16.44] /Length 69 /Resources<< /BBox [0 0 639.552 16.44] stream endobj >> Q /Subtype /Form ET BT >> (2\)) Tj /FormType 1 ET stream q endobj (9\)) Tj 6.746 5.203 TD q stream /BBox [0 0 88.214 16.44] /F3 12.131 Tf /Meta137 151 0 R /Type /XObject ET Q ET /F4 36 0 R (x ) Tj 243 0 obj /FormType 1 /Font << 1.007 0 0 1.007 654.946 473.519 cm 1.005 0 0 1.007 79.798 746.789 cm << Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. >> 0 G q >> >> stream << 1 i stream 1 g stream << /Font << q /Resources<< /Length 59 1.007 0 0 1.007 45.168 779.913 cm q 367 0 obj Q /Type /XObject /F3 12.131 Tf q /Type /XObject 0 g /FormType 1 /ProcSet[/PDF/Text] stream /Meta288 302 0 R Q BT 47.933 5.203 TD Q 0 G endobj Q /BBox [0 0 15.59 16.44] -0.486 Tw /Matrix [1 0 0 1 0 0] 1 i /F3 12.131 Tf /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF] q /Subtype /Form /Resources<< /Font << Q Q endobj q << /Length 54 endobj 445 0 obj /Subtype /Form /Resources<< (-) Tj stream 1 i stream /Resources<< /Meta400 416 0 R 0.737 w /Meta192 Do ET 1 g q >> endobj 1.014 0 0 1.006 531.485 763.351 cm Q /ID [] stream /F3 12.131 Tf /BBox [0 0 88.214 16.44] 420 0 obj endobj S /Subtype /Form /FormType 1 /FormType 1 Q << 0.458 0 0 RG /Matrix [1 0 0 1 0 0] /Font << Q 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. q Q endobj /Length 54 /FormType 1 /Font << /Meta181 Do Q << stream q 43 0 obj /FormType 1 /Meta265 279 0 R /Type /XObject 0 w 1.005 0 0 1.007 79.798 862.723 cm Q 91 0 obj /ProcSet[/PDF] /Font << q 1.014 0 0 1.007 111.416 277.035 cm /BBox [0 0 639.552 16.44] q Q 1 g 0 20.154 m stream endobj BT /Subtype /Form 0.425 Tc 1 g (+) Tj endstream (+) Tj /FormType 1 >> /F3 12.131 Tf /ProcSet[/PDF/Text] /Length 58 /F3 12.131 Tf /Matrix [1 0 0 1 0 0] endstream 0 G q 0 g Q /Subtype /Form twice a number x added to 10 = 2x + 10. a number n decreased by five = n - 5. a number and multiplied by 7 = 7y. /FormType 1 1.007 0 0 1.007 67.753 726.464 cm 248 0 obj /Matrix [1 0 0 1 0 0] /Meta181 195 0 R endobj q /Type /XObject >> /Meta10 Do /Length 69 Q /ProcSet[/PDF/Text] /Subtype /Form /Meta322 Do q Q >> /ProcSet[/PDF/Text] /FormType 1 >> /Type /XObject q >> /Meta224 Do << /Meta121 Do q q >> Q BT (A\)) Tj >> 3.742 5.203 TD q Q /F1 12.131 Tf /Meta375 Do endobj stream 0.564 G Q /Meta193 207 0 R q 1.007 0 0 1.007 654.946 473.519 cm Q /BBox [0 0 88.214 16.44] q 1.007 0 0 1.007 130.989 776.149 cm /Meta65 79 0 R >> 3.742 8.18 TD q q endobj 1 i BT 0 g q endobj >> 1.005 0 0 1.013 45.168 933.487 cm /FormType 1 /F3 12.131 Tf /FormType 1 << q endobj q q /F3 17 0 R /F1 12.131 Tf << 0 G /Subtype /Form q /Meta271 Do 0 g 1.007 0 0 1.007 67.753 293.596 cm stream Q 207 0 obj endobj q BT /FormType 1 /Meta248 262 0 R Q 0 5.203 TD >> 1.007 0 0 1.007 411.035 583.429 cm /F3 12.131 Tf BT 0 w 1 g q >> 1 g /Subtype /Form /Meta367 381 0 R endobj >> Q << 1 i 1. q endobj /MissingWidth 252 435 0 obj /Meta227 241 0 R 209 0 obj /I0 51 0 R stream /Type /Page /Length 69 /Subtype /Form Q endobj 30.699 4.894 TD stream 0 g stream 0.269 Tc /Matrix [1 0 0 1 0 0] endobj 0 g Q /Matrix [1 0 0 1 0 0]
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