twice a number decreased by 58twice a number decreased by 58

stream endstream /Meta334 Do (D) Tj /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] >> stream /Length 13 39 0 obj Q This site is using cookies under cookie policy . In addition, testosterone in both sexes is involved in health and well-being . >> stream /Font << 1.007 0 0 1.007 411.035 330.484 cm BT 0 w q /BBox [0 0 88.214 16.44] /FormType 1 << Q /BBox [0 0 534.67 16.44] q 1.014 0 0 1.007 251.439 383.934 cm q 20.975 5.336 TD << Q /Meta296 310 0 R 1 i /FormType 1 endstream endobj endstream 41.186 5.203 TD 1 g stream 19.474 20.154 l >> 368 0 obj >> (vi) If 12 is subtracted from a number, the result is 24. /F3 12.131 Tf /Type /XObject 1 i /ProcSet[/PDF/Text] q 1 i /FormType 1 >> Q Q /F3 17 0 R 1.014 0 0 1.007 111.416 849.172 cm /ProcSet[/PDF/Text] /ProcSet[/PDF] 0 w 1 i 1.502 5.203 TD (B\)) Tj /Subtype /Form /Subtype /Form q /ProcSet[/PDF/Text] stream /ProcSet[/PDF/Text] endobj /ProcSet[/PDF] >> >> 1 i stream 1 i 1 i q /Subtype /Form 45 0 obj /ProcSet[/PDF/Text] endstream At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . /Meta109 123 0 R /Subtype /Form /Meta321 335 0 R /F3 17 0 R ( x) Tj /Font << /FormType 1 D. Twice a number decreased by ten is less than 24. >> /Matrix [1 0 0 1 0 0] >> 0 w Q ET /Subtype /Form /Meta179 193 0 R Q /Meta379 393 0 R stream /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] stream >> 126 0 obj /BBox [0 0 88.214 35.886] q /FormType 1 /Subtype /Form /Meta74 88 0 R /F3 12.131 Tf /Meta7 Do /F1 12.131 Tf /Subtype /Form /I0 Do q q 0.564 G 0 G 1.007 0 0 1.007 551.058 383.934 cm 0.564 G 1 i q /FormType 1 /Subtype /Form /Font << /Resources<< /Meta362 376 0 R /Subtype /Form ET 1 i >> /Type /XObject >> /Matrix [1 0 0 1 0 0] /Resources<< q /Font << q /Matrix [1 0 0 1 0 0] /Type /XObject 1 i stream /Type /XObject /Meta34 Do /Type /XObject q /Resources<< /Type /XObject q /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] stream >> /FormType 1 /Font << 15.731 5.336 TD Q /Resources<< /ProcSet[/PDF/Text] endstream 1 i /Length 54 /Font << /Font << /ProcSet[/PDF] /Meta341 Do The difference between six and a number divided by nine 10. q /Meta115 Do q >> n 11 or n 11. /Meta387 Do /Font << >> Q 0.737 w << q Q /F3 17 0 R /FormType 1 endobj /Length 69 /Meta399 Do stream 1 i /Font << >> /Meta69 83 0 R >> 1 i /F3 12.131 Tf /Subtype /Form 1 i Q q /Meta35 Do /F1 7 0 R 1.005 0 0 1.007 102.382 546.541 cm 0.296 Tc /ProcSet[/PDF/Text] 0 5.203 TD (B\)) Tj /FormType 1 /Length 54 /Matrix [1 0 0 1 0 0] /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 0 4.894 TD 1.005 0 0 1.013 45.168 933.487 cm /Subtype /Form endstream /FormType 1 /Meta165 Do q /Type /XObject 6.746 5.203 TD << /BBox [0 0 534.67 16.44] ET /F3 12.131 Tf /BBox [0 0 88.214 16.44] 0 g q endobj /ProcSet[/PDF] q /FormType 1 q /Meta384 Do 20.975 5.336 TD /F3 17 0 R 286 0 obj /F4 12.131 Tf << endstream 0.68 Tc endobj q /Font << << /BBox [0 0 88.214 16.44] q endstream 0 G (-20) Tj 0.458 0 0 RG (5) Tj ET >> /ProcSet[/PDF/Text] Q /Subtype /Form /BBox [0 0 88.214 35.886] /Subtype /Form Q 152 0 obj q (3\)) Tj endobj 2. >> 12.727 5.203 TD /FormType 1 Q /BBox [0 0 534.67 16.44] 0 g BT /FormType 1 only about 58% of candidates will agree to be screened. << 0 447 42 0 obj /Meta61 Do 0 G Q >> /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 29.168] 0 g BT endobj stream q q endstream q /Type /XObject Q /BBox [0 0 88.214 16.44] << /Type /XObject q q Q /BBox [0 0 88.214 16.44] Q 9.723 5.336 TD 290 0 obj stream endstream xref /FormType 1 Q Q /Meta378 Do 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. /Subtype /Form endobj 1 i 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 Q 0 G /Type /XObject Q q endobj /Length 65 ET q 1.005 0 0 1.007 102.382 293.596 cm q BT 0 w /Meta182 196 0 R /Meta424 Do /Length 85 endobj /F3 12.131 Tf 1.007 0 0 1.007 45.168 846.161 cm 201 0 obj Q 0 4.894 TD /Resources<< Q >> /FormType 1 Q >> q Q /Length 69 q /Meta100 Do /Meta201 Do /F3 17 0 R q 0.786 Tc q /Matrix [1 0 0 1 0 0] /StemV 77 0.458 0 0 RG (iii) 25 exceeds a number by 7. q Q 1 i /Meta37 50 0 R Q BT /FormType 1 /ProcSet[/PDF/Text] >> 0.369 Tc q /BBox [0 0 30.642 16.44] q /Font << /BBox [0 0 88.214 16.44] BT That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. >> Q >> /Length 60 Q /Type /XObject 0.564 G >> q 0.297 Tc q endobj 0.458 0 0 RG /Matrix [1 0 0 1 0 0] ET endstream Q >> Q /Resources<< endobj /Type /XObject /Meta57 71 0 R /F3 12.131 Tf 1.502 5.203 TD 0.382 Tc q >> Q q endstream /Font << /Meta336 Do /BBox [0 0 673.937 14.853] Q /Font << >> (+) Tj q /Length 78 q stream stream (1\)) Tj 0.297 Tc Q /Meta82 96 0 R 1.014 0 0 1.007 391.462 330.484 cm endobj >> /FormType 1 20.21 5.203 TD 0 G 1.007 0 0 1.006 411.035 690.329 cm 0.458 0 0 RG BT /F4 12.131 Tf endstream stream BT /F3 12.131 Tf 0 G Q 0.134 Tc 0 g ET 0.737 w >> endstream /BBox [0 0 534.67 16.44] 0 g >> 366 0 obj /Type /XObject >> endstream 0 w 0 G /Type /XObject q 1.007 0 0 1.007 130.989 636.879 cm /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] Q q Q Q 549.694 0 0 16.469 0 -0.0283 cm << >> /Font << /Subtype /Form /FormType 1 ET /BBox [0 0 673.937 68.796] q /F3 17 0 R /Type /XObject Q Q >> /Subtype /Form Q /BBox [0 0 88.214 16.44] 1 i /Meta182 Do >> 1.007 0 0 1.007 271.012 636.879 cm ET /F3 17 0 R BT /Font << q endstream Q /Resources<< endstream << >> Q q 9 0 obj /Subtype /Form Q 89.12 5.203 TD the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . /Type /XObject /Meta185 Do (D\)) Tj /ProcSet[/PDF] stream /Length 16 << q 1 g Get link; Facebook; Twitter; /Meta26 39 0 R /Meta111 Do /FormType 1 /Meta385 401 0 R 1.007 0 0 1.007 551.058 383.934 cm endstream /Subtype /Form 17 0 obj /Meta29 42 0 R 0 g /ProcSet[/PDF/Text] /Font << q /Type /XObject 0.458 0 0 RG >> << << << q /Meta362 Do /Subtype /Form /BBox [0 0 88.214 16.44] /Meta236 250 0 R /Font << q Q endobj >> q >> /BBox [0 0 15.59 29.168] /Subtype /Form /Meta16 Do 57.656 5.203 TD Q BT /Meta366 Do 1 i 1 i /ProcSet[/PDF/Text] q Q /F3 12.131 Tf 1 g /Length 65 << /ProcSet[/PDF] /Meta260 274 0 R q 69 0 obj Q >> 1.005 0 0 1.007 102.382 473.519 cm Twice a number decreased by . q 75 0 obj endstream 190 0 obj endstream Q endstream Q /Meta355 Do /FormType 1 q endstream 1.005 0 0 1.007 79.798 813.037 cm 1 i /Length 68 /Length 59 /BBox [0 0 88.214 16.44] Q q /BBox [0 0 17.177 16.44] 0 5.203 TD /F3 17 0 R q endstream Q /Length 59 ET 0 w 1 i BT /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 endobj /Meta187 201 0 R BT /Type /XObject >> 1 i ET q endstream /Matrix [1 0 0 1 0 0] >> q 1 i 0.564 G /Type /XObject Q /Type /XObject /FormType 1 /Meta244 Do endstream q ET Q >> 0 5.203 TD /Length 78 BT /Subtype /Form 0 g 0.175 Tc 0.297 Tc >> >> 1 i q q /Meta5 Do [(F)-22(ive)] TJ /Meta236 Do 0.737 w Q q endstream /F3 12.131 Tf /F1 12.131 Tf 255 0 obj /F3 12.131 Tf Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] Q >> 0 g /Subtype /Form BT 1.007 0 0 1.006 130.989 437.384 cm /F3 17 0 R /FormType 1 /Matrix [1 0 0 1 0 0] 0.425 Tc q << 1 i /Type /XObject >> 0 g << /Meta404 420 0 R >> stream /Meta153 Do /F3 12.131 Tf /FormType 1 /Type /XObject 0.737 w /BBox [0 0 88.214 16.44] endstream 175 0 obj Q 0 G /Length 69 q /Font << There were x cookies at the beginning of a party. Q BT 0 G 1.007 0 0 1.007 130.989 523.204 cm 421 0 obj 0.458 0 0 RG >> 0.564 G q Q >> q q (+) Tj Q q Q >> 317 0 obj q Q 0 g << endobj q /Matrix [1 0 0 1 0 0] >> 302 0 obj /Meta117 131 0 R 0 g q /F3 12.131 Tf /FormType 1 0.737 w (-23) Tj /Subtype /Form q /ProcSet[/PDF] >> q /Meta374 388 0 R 0 G /Matrix [1 0 0 1 0 0] 1 g 0 G /BBox [0 0 15.59 16.44] /Meta403 419 0 R /Meta112 Do /Length 69 q /FormType 1 BT /Resources<< Q 1 i /Type /XObject /F3 17 0 R /BBox [0 0 88.214 16.44] 32.201 5.203 TD /Meta135 149 0 R 1.007 0 0 1.007 271.012 523.204 cm If a number is 50%, then it is a half - the same as 0.5 or 1/2. /Matrix [1 0 0 1 0 0] /F3 17 0 R 167 0 obj /ProcSet[/PDF] /Meta324 338 0 R /Length 79 (x ) Tj Q >> /Meta165 179 0 R 0 g /F3 17 0 R Q 0 G /Length 63 0 w q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] endstream If n is "the number," which equation could be used to solve for the number? 0.045 Tw 0 g Q /ProcSet[/PDF] Q ET endobj 20.975 5.336 TD >> 1 i /Type /XObject /Meta205 219 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> endstream /ProcSet[/PDF] 1 i 0.369 Tc /AvgWidth 657 /F4 12.131 Tf /ProcSet[/PDF/Text] -0.092 Tw /BBox [0 0 534.67 16.44] /F1 7 0 R /Resources<< Q q /Matrix [1 0 0 1 0 0] /Meta408 Do 334 0 obj (A\)) Tj /Subtype /Form stream /Type /XObject 0 5.203 TD endobj << q /Type /XObject Q /Meta213 Do stream 1 i BT 0 G Q 1 i stream /FormType 1 endstream Q 0 g /Meta419 435 0 R /Type /XObject q /Meta360 Do /Resources<< /Meta210 Do /Matrix [1 0 0 1 0 0] stream /BBox [0 0 15.59 29.168] /BBox [0 0 88.214 16.44] 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. >> >> /Resources<< BT 0.458 0 0 RG /F3 12.131 Tf 32.201 5.203 TD 1 i /Subtype /Form endobj /F1 12.131 Tf Q 0 G q 0 g >> /FirstChar 32 /Type /XObject stream ET /F3 17 0 R Q w/Honors. stream q 549.694 0 0 16.469 0 -0.0283 cm /FormType 1 endstream /Matrix [1 0 0 1 0 0] q 1 g /Meta342 Do 1.005 0 0 1.015 45.168 53.449 cm 0.737 w Q 0 g 0 g ET endstream 0 5.203 TD /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 0 5.203 TD Q endstream >> 333.269 5.488 TD endstream /Font << /Meta231 245 0 R q >> 1.007 0 0 1.007 271.012 523.204 cm /Meta163 177 0 R q >> q /BBox [0 0 30.642 16.44] /ProcSet[/PDF/Text] Q << /ProcSet[/PDF] /ProcSet[/PDF/Text] /Meta390 Do /FontDescriptor 6 0 R Q >> endobj >> /Meta100 114 0 R >> 0 w Q endobj >> stream >> /BBox [0 0 88.214 16.44] /Length 63 /ProcSet[/PDF/Text] /Resources<< Q 0 5.336 TD >> /Meta396 412 0 R Q /Meta166 Do endobj stream /Meta368 382 0 R Q /ProcSet[/PDF] [(E)-14(le)-23(ven)] TJ /F3 12.131 Tf endobj /ProcSet[/PDF/Text] endobj 0 g /Meta171 Do 1.005 0 0 1.007 102.382 473.519 cm 1 i /ProcSet[/PDF/Text] 0 g q Q 0 G Q Q 354 0 obj 1 i Q /ProcSet[/PDF] /ProcSet[/PDF/Text] /Resources<< /Resources<< /Meta108 Do 0 g q 1.007 0 0 1.007 654.946 546.541 cm 0 w 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. endobj /F3 17 0 R >> /F4 36 0 R 0 g /Matrix [1 0 0 1 0 0] 0.738 Tc q Q /Length 118 /BBox [0 0 17.177 16.44] q (x ) Tj /Font << /Resources<< 9.723 5.336 TD /Meta133 147 0 R stream Q endstream (2) Tj ET Q 1.007 0 0 1.007 551.058 330.484 cm BT >> 0.68 Tc 198 0 obj /Resources<< q /Matrix [1 0 0 1 0 0] q endobj /Meta291 305 0 R 1.007 0 0 1.007 271.012 849.172 cm Q /ProcSet[/PDF] >> 0 G /Meta80 Do >> >> /F3 12.131 Tf /Resources<< 0.458 0 0 RG /F3 17 0 R endobj 372 0 obj /Meta70 84 0 R 0.564 G /Meta377 Do << /Meta152 166 0 R /Meta386 Do 0 g endstream /Type /XObject Q /Resources<< << (2) Tj (\(x ) Tj /F3 12.131 Tf stream << /BBox [0 0 15.59 29.168] 155 0 obj 0 G (v) 5 subtracted from thrice a number is 16. 1.007 0 0 1.007 45.168 813.037 cm q /Meta338 Do Q /Resources<< stream /BaseFont /TestGen-Regular /F1 7 0 R 1 i /ProcSet[/PDF/Text] /Meta167 181 0 R >> /Length 66 q /Meta240 254 0 R /Length 68 Q /Type /XObject >> /FormType 1 274 0 obj 206 0 obj << stream endobj 404 0 obj Q /Matrix [1 0 0 1 0 0] Q 0 0 Similar questions Find the number which when decreased by 8% becomes 506. >> /Length 64 /Length 118 /Font << /Meta337 351 0 R endobj /Type /Font << Q 0 g Q /Meta293 307 0 R endstream /F3 17 0 R << /Subtype /TrueType 12.727 5.203 TD << endstream 1.007 0 0 1.007 45.168 713.666 cm 125.064 4.894 TD q /F3 17 0 R Q /Subtype /Form /ProcSet[/PDF] 14.966 20.154 l << 397 0 obj stream endobj q ET /ProcSet[/PDF] /FormType 1 /Meta330 344 0 R /F3 12.131 Tf /F3 17 0 R /Resources<< /BBox [0 0 30.642 16.44] /BBox [0 0 30.642 16.44] /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] /Resources<< /F3 12.131 Tf ET 0 G /F3 12.131 Tf >> >> 1 i /Matrix [1 0 0 1 0 0] >> ET /F3 17 0 R >> 1 g /BBox [0 0 88.214 35.886] /Length 69 >> /F1 12.131 Tf Q >> >> endstream [( subt)-17(racted fr)-14(om a )-16(number)] TJ endobj << 0 G q 1 i 260 0 obj stream 96 0 obj Q 223 0 obj >> >> Twice a first number decreased by a second number is 6. q /Matrix [1 0 0 1 0 0] /Meta11 Do /Length 70 BT Q >> /Matrix [1 0 0 1 0 0] (iv) A number exceeds 5 by 3. 0 w 0 g /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) /Type /XObject /Subtype /Form /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> (40) Tj Q /Resources<< /Meta309 323 0 R 65.906 4.894 TD /Meta228 242 0 R 0 G /BBox [0 0 534.67 16.44] /Length 69 0 g /Meta46 Do ET /FormType 1 0 G /Resources<< endobj q 1 i q q 1.014 0 0 1.007 111.416 636.879 cm /FormType 1 >> >> Q endobj q /Type /XObject endstream /Resources<< /Resources<< q endobj /ProcSet[/PDF] >> >> 215 0 obj q q stream /BBox [0 0 88.214 16.44] 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 /Matrix [1 0 0 1 0 0] 1 i /Subtype /Form /Meta323 Do /Meta412 Do /Resources<< /Matrix [1 0 0 1 0 0] >> << /FormType 1 /Resources<< stream endobj 326 0 obj 83 0 obj 0 g Q /F3 12.131 Tf Q /BBox [0 0 15.59 16.44] /Length 16 Q /Resources<< endobj /Length 79 208 0 obj << 1 i >> endstream /BBox [0 0 88.214 16.44] stream /Meta151 Do 0 w stream /FormType 1 /Matrix [1 0 0 1 0 0] Q /FormType 1 Q /Subtype /Form /BBox [0 0 88.214 16.44] Q /Font << /Subtype /Form >> /Font << Q q endstream 0 g << /Meta44 58 0 R /F4 36 0 R /FormType 1 << /ProcSet[/PDF] /Length 118 /F3 12.131 Tf /BBox [0 0 639.552 16.44] endstream endobj /BBox [0 0 15.59 16.44] >> 0 w /ProcSet[/PDF/Text] /Subtype /Form BT 0.564 G 0 g /Resources<< /Meta337 Do q q >> /FormType 1 Q 0.737 w 0.786 Tc endstream 0.51 Tc Q stream >> /Subtype /Form 1.502 8.18 TD /Meta409 Do /Resources<< endstream 1 i q q /Meta356 370 0 R endobj /Matrix [1 0 0 1 0 0] /Resources<< q (D\)) Tj /F4 12.131 Tf /Meta394 410 0 R stream Q >> 1.007 0 0 1.007 271.012 330.484 cm q /Length 69 Q 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. /ProcSet[/PDF] /ProcSet[/PDF] q /Font << 1 i 0 g Q q /BBox [0 0 88.214 16.44] q /Length 59 /Meta124 138 0 R (+) Tj /Meta274 Do 20.21 5.203 TD /Length 16 Q /BBox [0 0 534.67 16.44] >> S 0.564 G q q >> /Resources<< /Subtype /Form Then ab is a binary operation. >> /Resources<< 0.51 Tc q 0 5.203 TD 0.024 Tw /FormType 1 stream q stream >> 109 0 obj Q >> 0 G Q 0 G >> << q /Type /XObject >> /Type /XObject BT /Subtype /Form 55 0 obj q Q /Length 59 q BT stream /Parent 1 0 R /Length 59 << /Meta103 117 0 R << /F1 14.682 Tf /Resources<< q /BBox [0 0 88.214 16.44] /Length 118 endstream >> /Subtype /Form /Subtype /Form /Length 99 1 i ET (D\)) Tj Q 395 0 obj endobj /ProcSet[/PDF/Text] q /Meta138 152 0 R /Resources<< >> BT /Font << q 43 0 obj ET endobj /Length 12 q >> q stream >> 0 g >> /F1 7 0 R endobj Q 1.502 5.203 TD /FormType 1 Question 1. /Type /XObject 153 0 obj Q 0 G q /Resources<< /BBox [0 0 15.59 16.44] 0.564 G q q Q: A number increased by 5 is equivalent to twice the same number decreased by 7. /Type /XObject Q 211 0 obj ET You can specify conditions of storing and accessing cookies in your browser, Twice a number, decreased by 58 is less than 112, Mr. Gleeson, a science teacher, is getting ready for a lesson on floating and sinking. 1.005 0 0 1.007 102.382 872.509 cm /Meta174 Do Q /Length 63 1.005 0 0 1.007 79.798 829.599 cm 292 0 obj /ProcSet[/PDF/Text] q >> q Q Q /Subtype /Form q << 0.458 0 0 RG /FormType 1 (B\)) Tj 220.931 4.894 TD stream << << /Font << /ProcSet[/PDF/Text] Q q q /Meta254 Do /Resources<< /Matrix [1 0 0 1 0 0] 1 i endobj << 0 w /Meta397 413 0 R q /Resources<< >> 1.502 8.18 TD endstream >> /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] >> >> /Resources<< BT >> >> Q /Resources<< /Type /XObject 0 g -0.486 Tw (11) Tj q BT /F3 17 0 R /BBox [0 0 30.642 16.44] /Length 63 /Resources<< 0 5.203 TD 0 G 0 g Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 0 g (8\)) Tj << stream << stream /Length 68 0 g Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 131 0 obj >> /Subtype /Form /Resources<< Q q (x) Tj ET /Meta408 424 0 R q /F3 17 0 R 0 g 1 i 0 g /FormType 1 ET -0.008 Tw /Meta159 Do Q 1 i >> /Length 70 /ProcSet[/PDF/Text] Q BT stream stream >> /Subtype /Form 385 0 obj >> q Q endobj /ProcSet[/PDF] stream /FormType 1 722.699 347.046 l Q (C) Tj 1 g endstream /F3 17 0 R /Type /XObject endobj q ET ET stream /Length 12 1 i endstream /F1 7 0 R /FormType 1 q /FormType 1 >> (7\)) Tj Q >> /Meta412 428 0 R q 1 g /Meta56 Do /Meta47 Do /Matrix [1 0 0 1 0 0] q q Q stream Two, two times, twice, twice as much as, double 2 Twice z 2z y doubled 2y Multiplication by Half of, one-half of, half as much as, one-half times 1 2 Half of u u 2 one-half times m 1 2 m Geometry Problems Concept Word Expression Algebraic Expression Area of a square Side Squared A = s2 Perimeter of a square Four times the side P = 4s 0 G endstream /ProcSet[/PDF] q endstream q /Resources<< /F3 12.131 Tf /ProcSet[/PDF/Text] Q /FormType 1 endobj /Resources<< q /Resources<< 357 0 obj /Meta184 Do >> /FontBBox [-568 -307 2000 1007] stream Q Q Q >> q Q /Type /XObject >> endobj /Subtype /Form stream >> /Resources<< /Length 16 /Subtype /Form 1.007 0 0 1.007 271.012 330.484 cm /Meta384 398 0 R /Type /XObject /Meta156 Do /Type /XObject /Matrix [1 0 0 1 0 0] /Meta106 120 0 R stream 0.311 Tc >> /Subtype /Form 1 i endobj 0 G endstream 0 g q endobj 265 0 obj << /Meta69 Do 3.742 8.18 TD /Subtype /Form /Length 16 q q /BBox [0 0 15.59 16.44] 0.458 0 0 RG Q endobj q Use the variable g to represent Gails age. (5\)) Tj Q ( \() Tj 1 i 103 0 obj BT /Type /XObject 0 w q /BBox [0 0 549.552 16.44] /F1 12.131 Tf 1.014 0 0 1.007 251.439 277.035 cm /Font << Q q q endobj /ProcSet[/PDF/Text] << (B\)) Tj ET Q ET You can also contact the clerk of court in the county you received the ticket. ET 1 i ET /Length 70 /BBox [0 0 88.214 16.44] 0 G stream << Q Q Q >> 0.458 0 0 RG /Meta66 Do 212 0 obj /BBox [0 0 88.214 16.44] >> q 1 i /Type /XObject >> ET q /Meta62 Do Q << /Meta156 170 0 R 0 20.154 m 23.952 4.894 TD Q endstream [(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 (+) Tj 0.369 Tc /ProcSet[/PDF/Text] Q 390 0 obj /F3 12.131 Tf Q /FormType 1 0 w /Meta33 Do /FormType 1 /Length 60 0 G Q q /Type /XObject /FormType 1 /BBox [0 0 88.214 16.44] endobj >> /Meta143 157 0 R << Q 419 0 obj endstream /FormType 1 stream /FormType 1 1 g /Meta256 270 0 R q q /BBox [0 0 88.214 35.886] /FormType 1 q q << ET /ProcSet[/PDF/Text] q Q /F3 12.131 Tf 0.564 G /Resources<< endstream 0 g S /Meta78 Do endstream We are asked to find the number, so, we could assign the number as "x". (4\)) Tj /Type /XObject 0 w 0.564 G >> Q BT 1 g 229 0 obj /Meta52 66 0 R 1 i >> /F3 12.131 Tf /Subtype /Form endobj >> /Meta168 182 0 R /Meta304 318 0 R 52 0 obj 1.007 0 0 1.007 654.946 473.519 cm Q Q /FormType 1 /Resources<< endobj 0 G 1 g >> Q /Matrix [1 0 0 1 0 0] /Meta123 Do Q q >> BT 0 G 0 G >> /Meta400 Do stream /BBox [0 0 88.214 16.44] 100 0 obj /F3 12.131 Tf 177 0 obj q /Meta338 352 0 R endobj 549.694 0 0 16.469 0 -0.0283 cm Q (A\)) Tj q Q q /Resources<< /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] BT q q Q << /Matrix [1 0 0 1 0 0] /Meta418 Do q /Meta65 79 0 R /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Q /FormType 1 /Font << 407 0 obj stream 262 0 obj Q stream q /Type /XObject /BBox [0 0 15.59 16.44] 0 g /ProcSet[/PDF/Text] 0 G /ProcSet[/PDF/Text] /Meta164 Do Q /Resources<< let 'x' and 'y' represent the numbers. /Meta174 188 0 R /Resources<< 1.005 0 0 1.007 102.382 799.486 cm You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. /ProcSet[/PDF] /Type /XObject 0.564 G 1 i /Matrix [1 0 0 1 0 0] >> /Type /XObject /Subtype /Form Q << 1.007 0 0 1.006 130.989 690.329 cm /Matrix [1 0 0 1 0 0] q Q /Meta295 Do q /ProcSet[/PDF] /Type /XObject 0.786 Tc 0.737 w ET **Note: You could choose any variable you want. /Matrix [1 0 0 1 0 0] 1 i /FormType 1 0.369 Tc /Type /XObject /FormType 1 q /Meta15 Do 140781 /F1 12.131 Tf 275 0 obj /Length 16 Q endobj /Meta425 Do endstream q /BBox [0 0 88.214 16.44] q /Font << 264 0 obj stream 160 0 obj Q 0 G Q /Encoding /WinAnsiEncoding /FormType 1 0.564 G 25 0 obj >> ET >> /F3 17 0 R stream 30.699 5.203 TD Q BT >> >> /Meta367 381 0 R /Resources<< endstream q << /Resources<< 0 G /BBox [0 0 88.214 16.44] 1 i 1 i /Resources<< >> 0.564 G 331 0 obj 0 G 0 G endobj 0 G 1.007 0 0 1.007 654.946 799.486 cm BT 0 w 1 i /Matrix [1 0 0 1 0 0] We determined the effect of plant oils (rapeseed, sunflower, linseed) and organic acids (aspartic and malic) on the fermentation of diet consisting of hay, barley and sugar beet molasses. 0.486 Tc /Length 54 /Meta422 438 0 R /Type /XObject 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 /Length 69 Q ET /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 2 times x minus 58 C. twice the difference of a number and 5 B. twice a number decreased by 58 D. 2 times the sum of a number and 58 Answer: B. Step-by-step explanation: twice - (2) number - (x) 58-(58) Edukasyon. (A\)) Tj /F3 17 0 R 112 0 obj -0.486 Tw Q endobj BT /Matrix [1 0 0 1 0 0] endstream Q /F3 17 0 R >> /Meta93 Do Q BT q << Q /F4 12.131 Tf (38) Tj Q /Subtype /Form BT Q Q /Length 69 stream /Subtype /Form /Matrix [1 0 0 1 0 0] endobj stream q 0 g q >> 0 5.203 TD 254 0 obj /Subtype /Form /Meta205 Do >> 349 0 obj /FormType 1 Q 0.564 G q /Length 57 0 G /Meta381 395 0 R Q /FormType 1 /Subtype /Form /FormType 1 /ProcSet[/PDF/Text] (40) Tj /ProcSet[/PDF/Text] Q >> 1.007 0 0 1.007 551.058 636.879 cm endobj BT endobj /Subtype /Form Q q 549.694 0 0 16.469 0 -0.0283 cm >> /Meta59 73 0 R q q q /Matrix [1 0 0 1 0 0] /FormType 1 Q /FormType 1 q /Subtype /Form q 23.216 5.203 TD Q /FormType 1 Q 0 g /BBox [0 0 673.937 16.44] q Q << /Font << 1 g /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /Subtype /Form q endstream /BBox [0 0 88.214 16.44] >> /BBox [0 0 88.214 16.44] << q /Subtype /Form >> stream 1 i 1 i << q 1 i endobj /Matrix [1 0 0 1 0 0] endstream Q /Resources<< Q (13) Tj /F3 17 0 R /Meta339 Do endobj ET /Meta418 434 0 R /Matrix [1 0 0 1 0 0] >> q q 1.014 0 0 1.007 531.485 636.879 cm Q q endobj >> /ProcSet[/PDF/Text] nine increased by a number x. q ET 1.007 0 0 1.007 271.012 776.149 cm /BBox [0 0 534.67 16.44] /Meta295 309 0 R q q >> /Font << If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. ET >> 0 g /F3 17 0 R q endobj 76.394 5.203 TD 0 5.203 TD Q 1.005 0 0 1.007 79.798 763.351 cm >> >> Q Q /Resources<< stream 0 g /Meta340 Do The rate of positive findings after 1 round of screening in the LCSDP was more than twice . Q BT << << Q /Meta133 Do /F3 17 0 R /F3 17 0 R q >> /I0 51 0 R Diabetes, if left untreated, leads to many health complications. << /Resources<< 1.007 0 0 1.007 411.035 277.035 cm /F3 17 0 R 1 i 216 0 obj /FormType 1 Q endstream /Meta94 108 0 R /Subtype /Form q Q stream /Matrix [1 0 0 1 0 0] /F1 7 0 R 1 i 0 G 0 g stream >> Q /Type /XObject /Matrix [1 0 0 1 0 0] ET 422 0 obj /ProcSet[/PDF] 1 i >> /Meta108 122 0 R >> Q /ProcSet[/PDF/Text] /F3 17 0 R 135 0 obj q q q 18.708 17.593 TD Q 0 w q /Matrix [1 0 0 1 0 0] >> endobj /Meta414 Do /Matrix [1 0 0 1 0 0] >> stream /Matrix [1 0 0 1 0 0] /Meta227 241 0 R /ProcSet[/PDF/Text] /Length 16 0.458 0 0 RG 145 0 obj /Matrix [1 0 0 1 0 0] %%EOF. >> >> >> Q Q stream >> stream >> /Resources<< Q /Length 12 280 0 obj Q /BBox [0 0 23.896 16.44] Q /Subtype /Form /Meta216 Do 1.007 0 0 1.007 130.989 383.934 cm 415 0 obj Q /F3 17 0 R q Q Q /F3 17 0 R ET 1 i /Resources<< /Font << Q << endobj /BBox [0 0 88.214 16.44] /Meta275 289 0 R 101.849 5.203 TD Q Q /Matrix [1 0 0 1 0 0] Q /F3 12.131 Tf Q Q /Resources<< q /Meta389 Do Q stream Q /Size 447 >> /Font << /FirstChar 32 Q >> 1 g /FormType 1 >> /Meta184 198 0 R /Meta200 Do 11 0 obj /FormType 1 418 0 obj /F4 12.131 Tf Q 1.007 0 0 1.007 411.035 277.035 cm Q 2.238 5.203 TD /Matrix [1 0 0 1 0 0] /Meta331 345 0 R 446 0 obj /Resources<< (-8) Tj 1 i >> 0 w endobj >> 1 g 7 0 obj /Length 16 /Meta38 52 0 R >> Q /F3 17 0 R q q q /Subtype /Form 427 0 obj Q /FormType 1 << /Matrix [1 0 0 1 0 0] Q >> /Matrix [1 0 0 1 0 0] (C\)) Tj endstream Q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0.564 G >> q << /Meta346 360 0 R >> 1.005 0 0 1.007 102.382 490.08 cm /ProcSet[/PDF/Text] Q /F3 12.131 Tf /FormType 1 1.014 0 0 1.007 391.462 849.172 cm /F3 17 0 R q stream /FormType 1 /Subtype /Form >> stream 0 G 20.21 5.203 TD /ProcSet[/PDF/Text] /FormType 1 -0.092 Tw /Subtype /Form /Meta379 Do 0 g 0 g q /Resources<< 0 G 1.007 0 0 1.007 67.753 347.046 cm 246 0 obj q /Resources<< Q /BBox [0 0 15.59 16.44] /Meta149 Do This gives us: "2x+5". endobj 1 g /Length 19882 /BBox [0 0 673.937 15.562] stream q /Length 59 endstream /Type /XObject 0.241 Tc 0 g 0 w q endstream ET /BBox [0 0 88.214 16.44] 8.985 20.154 l /Type /XObject Q 1.007 0 0 1.006 411.035 763.351 cm /FormType 1 >> q /Length 69

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