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10
JAN
2021

twice a number decreased by 58

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1.007 0 0 1.007 551.058 330.484 cm << endobj /Resources<< stream /BBox [0 0 88.214 16.44] /Resources<< >> 130 0 obj 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 /Meta98 112 0 R -0.486 Tw /CreationDate (D:20140515121932-04'00') /Type /XObject /F3 12.131 Tf If mario jumps 3 times and luigi jumps 62 times. 0 g >> /F3 17 0 R stream endstream Q /Meta429 Do >> << /ProcSet[/PDF/Text] /Resources<< 102 0 obj /Resources<< >> q >> q q a.) Q 0 G q q q >> /ProcSet[/PDF] /ProcSet[/PDF] 1 i /Subtype /Form ET /Resources<< q 0 g Q 0.458 0 0 RG >> /Length 69 /Length 69 0 g 1 i 126 0 obj /Meta10 Do 0 g 0.458 0 0 RG >> 3.742 5.203 TD (5) Tj 0 G /F3 12.131 Tf /FormType 1 (x ) Tj q 0 G >> /Type /XObject q >> /Type /XObject (\)) Tj /Matrix [1 0 0 1 0 0] q 1 i endobj Q /Resources<< << /Meta347 Do stream 0.369 Tc /Meta295 309 0 R stream 333.269 5.488 TD /Subtype /Form 1 i 0 g /Subtype /Form /Type /XObject /ColorSpace [/Indexed /DeviceGray 1 ] 0 g q /Subtype /Form /Font << 14.966 20.154 l /BBox [0 0 88.214 16.44] 417 0 obj 243 0 obj 347 0 obj /Meta16 Do /Type /XObject 0 w /Length 69 Q q /Font << Q /Meta18 29 0 R /Subtype /Form /Resources<< Q /F1 12.131 Tf /Resources<< 1 i endobj endstream 0 g 1.007 0 0 1.006 551.058 763.351 cm /BBox [0 0 88.214 16.44] 54 0 obj /BBox [0 0 30.642 16.44] << q 277 0 obj 1.005 0 0 1.007 102.382 473.519 cm /FormType 1 >> /Subtype /Form q /Subtype /Form /Meta230 244 0 R /BBox [0 0 88.214 35.886] /Meta176 190 0 R /ProcSet[/PDF] /Meta320 334 0 R ET q /Matrix [1 0 0 1 0 0] /Resources<< 0.524 Tc Q << Q /Type /XObject /Type /XObject 1.007 0 0 1.007 130.989 636.879 cm q /Resources<< q 0.458 0 0 RG 333.269 5.488 TD 0 g /ProcSet[/PDF] /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] 440 0 obj /Meta16 27 0 R /Length 78 q q Q stream 0.737 w Q >> /Matrix [1 0 0 1 0 0] q endobj BT >> Q >> /Meta248 262 0 R stream /Type /XObject /Meta408 424 0 R stream q /F3 12.131 Tf 0 g 45 0 obj endobj >> /Font << q /FormType 1 /Subtype /Form >> q stream Q /FormType 1 /ProcSet[/PDF/Text] Q /Meta77 91 0 R >> 362 0 obj /Matrix [1 0 0 1 0 0] endobj >> 1 g Q /Type /XObject /Subtype /Form /BBox [0 0 534.67 16.44] endstream /Meta122 136 0 R /Meta133 Do /LastChar 121 /Type /XObject q BT endobj endstream >> endstream /Subtype /Form /Type /XObject ET /Font << 0.564 G 0 g q 0 g >> Q endobj /Matrix [1 0 0 1 0 0] Table 1. /Resources<< << q /FormType 1 0 g q /Meta125 139 0 R /Meta392 408 0 R /ProcSet[/PDF/Text] /F3 12.131 Tf /Meta416 Do /Meta293 Do >> /BBox [0 0 88.214 16.44] q 1 i endobj /Meta12 23 0 R (B) Tj Q q 0 20.154 m >> 1 i BT /Font << /Meta124 Do 0.198 Tc 0.369 Tc /Type /XObject Twice a first number decreased by a second number is 6. ET /Meta378 Do 1.014 0 0 1.007 111.416 277.035 cm BT 0 w q 0.737 w q /Meta32 45 0 R 226 0 obj /ProcSet[/PDF/Text] Q 1 g >> ET /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] 1.502 5.203 TD Q 0.564 G /F3 12.131 Tf 154.289 4.894 TD Q 0.737 w 1.007 0 0 1.007 130.989 277.035 cm >> << Q 0.458 0 0 RG /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 130.989 330.484 cm << stream Q << 0 G q (-20) Tj (+) Tj /ProcSet[/PDF/Text] 722.699 473.519 l Check out a sample Q&A here. Q q stream 1 i Q 1 i Q 0.564 G Q /BBox [0 0 15.59 29.168] q /Matrix [1 0 0 1 0 0] stream endstream /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 583.429 cm 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q endobj Q /ProcSet[/PDF] /Length 67 (A\)) Tj Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. /Type /XObject q 722.699 653.441 l /Length 78 /Meta399 415 0 R 0.369 Tc >> /Resources<< endstream 0.737 w Q endstream q ET 0 g endobj q /ProcSet[/PDF] /F3 12.131 Tf /F4 36 0 R /Length 58 /Font << 1.007 0 0 1.007 411.035 330.484 cm >> /Meta38 52 0 R /Meta393 Do 0 5.203 TD /ProcSet[/PDF] /Type /XObject /Type /XObject q /F1 7 0 R stream /BBox [0 0 549.552 16.44] q 0 g Q (D\)) Tj 16.469 5.336 TD /Resources<< endobj /Resources<< q 0 g 1.007 0 0 1.007 411.035 277.035 cm 1.014 0 0 1.007 531.485 703.126 cm 0 G << 22.478 5.336 TD /F3 17 0 R 0 G /FormType 1 /F4 36 0 R /Font << 1.005 0 0 1.015 45.168 53.449 cm /Type /XObject /Resources<< q << 0 G 1.007 0 0 1.006 130.989 690.329 cm /Resources<< /Resources<< BT 1.007 0 0 1.007 130.989 277.035 cm /F3 12.131 Tf /ProcSet[/PDF/Text] 0 g << /FormType 1 << 0 g stream 38.182 5.203 TD 0 G /ProcSet[/PDF/Text] endobj /Resources<< stream /ProcSet[/PDF/Text] /F3 17 0 R -0.084 Tw 0 w 161 0 obj << /Meta130 144 0 R Q endobj /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] /FormType 1 1 g 0 g << /Matrix [1 0 0 1 0 0] >> 43.426 5.203 TD /Font << Q 1 g Q /Subtype /Form /ProcSet[/PDF] q 1 i 0.737 w 76 0 obj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /BBox [0 0 88.214 35.886] 1 i q q /FormType 1 /Type /XObject endstream /Matrix [1 0 0 1 0 0] /FormType 1 /Meta415 Do /Matrix [1 0 0 1 0 0] << /Type /XObject /F3 12.131 Tf Q Q 0 g /BBox [0 0 15.59 16.44] >> stream /F3 17 0 R >> >> 0 g /Meta269 283 0 R 0.369 Tc endstream Q q ] q /BBox [0 0 88.214 16.44] /Meta198 Do Q 0 w /Matrix [1 0 0 1 0 0] q 1.005 0 0 1.015 45.168 53.449 cm >> /FormType 1 stream >> /Length 294 313 0 obj /Length 16 /ProcSet[/PDF] q q /Resources<< >> endobj endstream (vi) If 12 is subtracted from a number, the result is 24. Q endstream Q Q stream stream Q Q 186 0 obj /FormType 1 /FormType 1 0 G Q /Type /XObject /Matrix [1 0 0 1 0 0] Q Q /Meta5 14 0 R << 1 i /Font << 1 i /Length 69 /Type /XObject Q 1.005 0 0 1.007 79.798 813.037 cm /F3 17 0 R /Subtype /Form 722.699 599.991 l 18.708 17.593 TD stream << 2.238 5.203 TD 0 g << q /FormType 1 /Resources<< /FormType 1 << 1.007 0 0 1.006 130.989 690.329 cm endstream /Meta175 Do /Meta156 170 0 R 1 g 256 0 obj 0 G q Q /Meta134 148 0 R Q Q /Meta219 233 0 R << q /BBox [0 0 88.214 35.886] q /Meta158 Do 0 g endstream /BBox [0 0 639.552 16.44] stream q /Meta92 106 0 R >> /Meta49 63 0 R 1.014 0 0 1.007 391.462 703.126 cm /Subtype /Form /F3 17 0 R Q /Font << /Meta402 Do >> /Subtype /Form 0 G 1 i /FormType 1 265 0 obj endstream Q >> >> /Meta188 202 0 R << q /Subtype /Form 1 i 422 0 obj >> /FormType 1 0 G >> >> /F3 12.131 Tf 0 g VIDEO ANSWER: in this problem were asked to solve giving, given the following information. 3.742 24.649 TD q stream Q /F3 12.131 Tf /F3 12.131 Tf 0 g Q 0 G Q endobj /ProcSet[/PDF/Text] q /Length 16 /Length 12 Answer only. 1 i 32.201 5.203 TD /Resources<< 1 g << Q /Resources<< /Length 59 endstream /Matrix [1 0 0 1 0 0] 0 g q >> /ProcSet[/PDF] /Length 66 /Meta258 Do 0.155 Tc /Type /XObject ET q /Meta338 352 0 R /Meta182 196 0 R /Resources<< << /Font << /ProcSet[/PDF/Text] /FormType 1 (38) Tj 1 g 0 g q /Subtype /Form q 0.564 G << 2. >> /F1 12.131 Tf >> /F4 36 0 R /F1 7 0 R /Subtype /Form << 0 w ET /Resources<< 0 g 1 i 1.007 0 0 1.006 130.989 437.384 cm >> /CapHeight 476 << /FormType 1 /BBox [0 0 15.59 29.168] /Subtype /Form /BBox [0 0 88.214 16.44] 1 i Q /Resources<< /BBox [0 0 88.214 35.886] /Type /XObject q endobj Q q q >> 101 0 obj 1 i 722.699 293.596 l /Resources<< /Matrix [1 0 0 1 0 0] 0.786 Tc /Matrix [1 0 0 1 0 0] /Length 16 q /BBox [0 0 88.214 35.886] /BBox [0 0 88.214 16.44] /Resources<< q 23.216 5.203 TD /Subtype /Form Q /Font << /Meta86 100 0 R /ProcSet[/PDF] stream q BT /Meta254 268 0 R stream -0.382 Tw /ProcSet[/PDF] /Subtype /Form /Length 127 /F4 12.131 Tf /Resources<< >> Q 0.458 0 0 RG Grad - B.S. >> /Subtype /Form /BBox [0 0 15.59 29.168] q /Meta100 Do BT In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). q /Type /XObject /Meta401 417 0 R /FormType 1 BT 1 i Q /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 391.462 636.879 cm Q 0.458 0 0 RG S /F3 12.131 Tf >> /Meta177 Do Q << >> q endstream q /Meta134 Do Q q 109 0 obj /Leading 150 q endstream /FormType 1 Q /Subtype /Form /Length 68 ET /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 35.886] /ProcSet[/PDF/Text] q /ProcSet[/PDF] >> 403 0 obj 0.307 Tc /F3 12.131 Tf 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 : . /Meta202 216 0 R Q q /F4 36 0 R >> /Meta19 30 0 R >> endstream /Font << 89 0 obj 0 G stream /F3 12.131 Tf >> q Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. /ProcSet[/PDF/Text] endobj /BBox [0 0 88.214 16.44] 332 0 obj q /Subtype /Form q /BBox [0 0 23.896 16.44] Q Q 1 i 0 g /F3 17 0 R /Length 69 /Meta43 Do >> /Length 69 1.007 0 0 1.007 130.989 776.149 cm /Subtype /Form (-23) Tj 242 0 obj Q /FormType 1 >> Q Q /Type /XObject endstream 0 G 1 i /Length 63 /Meta365 379 0 R /Font << /Subtype /Form /ProcSet[/PDF] << /Font << ET Q Q /Matrix [1 0 0 1 0 0] << /Length 59 Q /Matrix [1 0 0 1 0 0] /BBox [0 0 639.552 16.44] /FormType 1 153 0 obj /Length 16 q Q 1 i 33.704 5.203 TD Q /I0 51 0 R /Length 64 endstream q /Font << /Meta230 Do q Q 1 i 138 0 obj /BBox [0 0 534.67 16.44] stream Q q /Matrix [1 0 0 1 0 0] 0.285 Tc 0 g /Resources<< Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. ET << >> Q q Q 11.99 24.649 TD 20.21 5.203 TD 67 0 obj 1.014 0 0 1.007 111.416 636.879 cm Q 0 g /Subtype /Form Q An example of a linear inequality in one variable is A. x+y = Question: answer 1. 0 g endobj q 0 g q stream /Resources<< Q /Meta241 Do Q 314 0 obj /FormType 1 stream q >> /Meta189 Do -0.486 Tw 0.737 w ET q 0 g q ET q >> 196 0 obj (B) Tj stream >> Q /Matrix [1 0 0 1 0 0] 1 i /ProcSet[/PDF/Text] 722.699 726.464 l q /Type /XObject 1 i /Font << stream Calculate a 15% decrease from any number. q q Q Q q q /Meta276 290 0 R Q /ProcSet[/PDF/Text] 0 G Q << /Subtype /Form q >> /Length 59 /ProcSet[/PDF] 433 0 obj Q 1.007 0 0 1.006 551.058 437.384 cm BT /Matrix [1 0 0 1 0 0] << /Matrix [1 0 0 1 0 0] /Meta109 Do >> BT /BBox [0 0 88.214 16.44] endobj q /Meta10 21 0 R /Meta414 Do /Meta157 171 0 R q q >> /Length 16 /F1 7 0 R q Q /Info 3 0 R /BBox [0 0 15.59 16.44] /BBox [0 0 15.59 16.44] Q /Meta139 Do /Meta27 40 0 R /Matrix [1 0 0 1 0 0] /Font << Q << 280 0 obj endstream /Matrix [1 0 0 1 0 0] Q /F4 12.131 Tf , Prove the following /Meta253 267 0 R /Type /XObject >> BT Q /BBox [0 0 88.214 35.886] /F3 17 0 R /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 1.007 0 0 1.006 411.035 763.351 cm << ET >> 0 g /FormType 1 >> stream /Matrix [1 0 0 1 0 0] 0.564 G /F3 12.131 Tf (-) Tj >> Q /Meta282 Do 1 i 0 5.203 TD >> q /BBox [0 0 88.214 16.44] Q /Subtype /Form /FormType 1 /Resources<< 1 i Q endstream /Meta59 Do /Matrix [1 0 0 1 0 0] >> /BBox [0 0 639.552 16.44] 0 g endstream >> /BBox [0 0 15.59 29.168] 1 g 32.201 5.203 TD q 2.238 5.203 TD 1 i q << /Type /XObject 0 g /ProcSet[/PDF/Text] /Resources<< /Resources<< /Length 16 stream stream BT 0.737 w >> /Resources<< 0 G /Resources<< q Q 164 0 obj Q /Length 16 Q q Q q 0 w 1.007 0 0 1.007 130.989 849.172 cm /FormType 1 (2\)) Tj /Length 69 Q q Q q stream /Type /XObject /Type /XObject << 0 G /F4 12.131 Tf q endstream [4] One half of a number decreased by fourteen is twenty-one >> q 0 G /F3 17 0 R /BBox [0 0 88.214 16.44] BT 18 0 obj q q /Subtype /Form q << /Meta376 Do 0 w /Resources<< [(thir)17(te)15(en)] TJ /F1 12.131 Tf /Resources<< In addition, testosterone in both sexes is involved in health and well-being . Q /Length 16 /Type /XObject stream stream Q /Meta398 414 0 R /Meta73 Do >> /Type /XObject 0 w /BBox [0 0 534.67 16.44] /FormType 1 /ProcSet[/PDF] 1.007 0 0 1.007 130.989 636.879 cm Q stream endobj /Length 16 1 i ET 1 i endobj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] stream (\)) Tj 345 0 obj 0 G 0 g 1 g /Resources<< >> 0.458 0 0 RG /Subtype /Form Q << 1.007 0 0 1.007 130.989 383.934 cm >> (13) Tj /F3 17 0 R Q /Length 16 /Meta15 Do << endobj /Matrix [1 0 0 1 0 0] q /Font << 1 i BT Q >> /FormType 1 /BBox [0 0 549.552 16.44] 0.51 Tc /BBox [0 0 88.214 16.44] 0.737 w /Meta344 358 0 R Q /Meta46 60 0 R 1.007 0 0 1.007 654.946 653.441 cm /Meta104 Do 385 0 obj 204 0 obj /Resources<< endobj /Meta336 350 0 R /ProcSet[/PDF] endstream /Resources<< >> /Resources<< /Font << /ProcSet[/PDF/Text] 1 g >> 0.737 w /Type /XObject endstream 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. Q Q 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 /Length 16 /ProcSet[/PDF/Text] /Subtype /Form /ProcSet[/PDF/Text] /ItalicAngle 0 q << /Resources<< /Font << /Meta406 422 0 R Q /Matrix [1 0 0 1 0 0] 25.454 5.203 TD >> stream /Width 734 endstream /Type /XObject /F4 12.131 Tf Medium /MediaBox [0 0 767.868 993.712] /F3 12.131 Tf S q q endstream 1 g endobj /Meta37 Do /ProcSet[/PDF/Text] >> >> /Subtype /Form /Length 103 [3] One half of a number increased by fourteen is twenty-one. 1.014 0 0 1.006 251.439 510.406 cm >> q 1 i BT ET /Type /XObject /F1 7 0 R /F1 7 0 R 129 0 obj /F3 12.131 Tf Q 1 g /BBox [0 0 88.214 16.44] 47 0 obj /Length 69 /Meta129 143 0 R 20 0 obj 1.007 0 0 1.007 654.946 872.509 cm /Resources<< 0 g >> /F3 12.131 Tf << /F3 12.131 Tf q 0.737 w Q /Matrix [1 0 0 1 0 0] 1 i q >> 0.369 Tc 0 g Q q /F3 12.131 Tf /FormType 1 /Meta33 46 0 R BT Q 0 G /FormType 1 << q /Meta304 Do 1 i /F3 17 0 R /FormType 1 /Length 69 /Meta161 Do 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. /Font << /Subtype /Form C. Twice a number decreased by ten is at most 24. /FormType 1 /Subtype /Form /F3 12.131 Tf 0 w /Subtype /Form Q 0.369 Tc 0.458 0 0 RG q stream /Font << q >> 1.007 0 0 1.007 551.058 277.035 cm >> << q q 1 i /Meta289 303 0 R /F4 12.131 Tf q /Matrix [1 0 0 1 0 0] endstream endobj 0.524 Tc /Meta50 Do ET 0.737 w /BBox [0 0 30.642 16.44] q 0.458 0 0 RG /Meta404 420 0 R endobj /Resources<< 111 0 obj /ProcSet[/PDF/Text] >> endstream /Subtype /Form 0 w 0 g /BBox [0 0 15.59 16.44] /Meta397 413 0 R 1.005 0 0 1.007 45.168 889.071 cm 0.458 0 0 RG /Font << q /BBox [0 0 88.214 16.44] endobj /Resources<< q ET Q /Meta135 Do q >> endobj ET In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. q q /Subtype /Form Q 1 i Q /Meta331 Do /Meta67 81 0 R 228 0 obj /ProcSet[/PDF] >> stream 1 i for the season. >> q endobj << 0 G 0 w stream endobj Q /Font << 60 0 obj /Meta171 185 0 R /Subtype /Form Q See Solution. >> 12.727 5.203 TD 0.564 G /FormType 1 /BBox [0 0 88.214 16.44] endobj 0 g 0.737 w endstream /Length 59 >> /F1 7 0 R q Q 0 G 0 w >> /FontDescriptor 16 0 R /Font << /F3 17 0 R /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1 i Thrice a number decreased by 5 exceeds twice the number by 1 is . endobj 225 0 obj endstream Q >> Q >> /Type /XObject /BBox [0 0 88.214 16.44] /Resources<< endobj /Matrix [1 0 0 1 0 0] 1 i /Resources<< q /FormType 1 BT 0 G /Font << endstream /Meta166 Do stream << Q Q 0.458 0 0 RG 0 g q q >> q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0 w BT >> endobj /Type /XObject q /Length 69 Q Q Q /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] /Length 54 1.007 0 0 1.007 67.753 293.596 cm 1 i q q /XObject << /BBox [0 0 88.214 16.44] /Meta304 318 0 R q Q /Subtype /Form endstream /FormType 1 /FormType 1 /F3 17 0 R 0.51 Tc << /Subtype /Form /Matrix [1 0 0 1 0 0] << /Length 16 Q >> /Meta327 341 0 R 0.564 G Q Q << /Meta313 327 0 R /XObject << stream 0 G /Length 16 /XObject << Q 1 i /BBox [0 0 88.214 16.44] /Font << q 35 0 obj Q q /ProcSet[/PDF/Text] Q /F3 12.131 Tf /Font << Q /Type /XObject /ProcSet[/PDF/Text] /ProcSet[/PDF] 0 G << /Meta197 Do , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. q /Length 63 /Font << /F4 36 0 R /BBox [0 0 534.67 16.44] /Type /XObject 1.502 5.203 TD 0 G t is 56: 4. q 0 g 0 g /ProcSet[/PDF/Text] endstream << << q stream q Q Q /Meta387 403 0 R /BBox [0 0 88.214 35.886] Q /Type /XObject /Meta349 363 0 R >> Q Q >> /F3 17 0 R BT 0 g >> 388 0 obj q Q 1 i q /Font << Q endobj /Length 69 /BBox [0 0 30.642 16.44] /Resources<< 0 g q Q /Resources<< 298 0 obj -0.486 Tw /F3 12.131 Tf /Length 118 >> 0.737 w BT 0 w endobj ET Q 0 w /Meta62 76 0 R 1.014 0 0 1.007 531.485 383.934 cm /Resources<< Q endstream Q Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Meta31 44 0 R >> stream /BBox [0 0 88.214 16.44] endstream /Meta348 Do /Resources<< q [(A numb)-16(er subtract)-15(ed from )] TJ 36 0 obj /F1 12.131 Tf q /FormType 1 << 0 g /ProcSet[/PDF/Text] /Resources<< endobj /Matrix [1 0 0 1 0 0] /FormType 1 /Meta381 395 0 R /ProcSet[/PDF/Text] endstream Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /BBox [0 0 17.177 16.44] /F1 7 0 R Q Q /Meta36 49 0 R /Resources<< q << Q /Meta314 Do /Font << 1 g ET /Meta100 114 0 R q q 65.906 4.894 TD >> /FormType 1 /Subtype /Form /Font << Q /F3 12.131 Tf 0.564 G >> << /FormType 1 0 G stream /Meta306 320 0 R 0 g q Q Q 381 0 obj /Meta284 Do >> >> 19.474 20.154 l /Type /XObject q >> endstream stream >> /BBox [0 0 534.67 16.44] endstream 0.564 G q endobj Q << /Subtype /Form stream /Type /XObject >> /F1 12.131 Tf /Font << /Matrix [1 0 0 1 0 0] 1 i /F3 12.131 Tf ET (3) Tj endstream >> q 1 i stream /Matrix [1 0 0 1 0 0] q Q /Meta290 Do Q >> >> q 2. endobj >> 0 g q /Type /XObject Q /FormType 1 stream 0.369 Tc /Resources<< /F3 17 0 R /Subtype /Form /Type /XObject /FormType 1 Q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] BT /Meta326 340 0 R stream 0 20.154 m 1.007 0 0 1.007 271.012 523.204 cm /Resources<< 279 0 obj /Meta227 241 0 R /Type /XObject 1 i Q 1 i 0.458 0 0 RG stream 0 G /FormType 1 /Length 95 /BBox [0 0 30.642 16.44] >> 1 i /Meta213 Do 1. >> ET 0.68 Tc /Length 69 endobj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Resources<< 0.564 G /Meta53 67 0 R 0.458 0 0 RG /BBox [0 0 673.937 14.853] << 124 0 obj 282 0 obj /F3 17 0 R /F1 7 0 R /Type /XObject /ProcSet[/PDF] endstream >> /Type /XObject /Subtype /Form 0.458 0 0 RG /BBox [0 0 88.214 16.44] q 0 G 0.737 w /F4 36 0 R Q >> That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. 0.786 Tc q Q 1.502 5.203 TD q /Length 69 << 0 G << q 0 w Six subtracted from a number 6. 0.524 Tc (B\)) Tj /Subtype /Form 1 g BT /Meta350 364 0 R Q >> 0 g 0.486 Tc /Subtype /Form /Meta389 Do /Meta118 Do /FormType 1 /I0 51 0 R /ProcSet[/PDF] /Meta20 31 0 R endobj /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) /Matrix [1 0 0 1 0 0] >> << Q Choose the correct one. Q q /AvgWidth 657 endobj /Meta281 Do q q Q 99 0 obj >> Q /Subtype /Form >> q /Subtype /Form 255 0 obj ET /Length 59 Q >> 0.737 w /Matrix [1 0 0 1 0 0] << /Resources<< << /ProcSet[/PDF/Text] Q q 0.564 G 1.007 0 0 1.007 551.058 277.035 cm >> 0 20.154 m 0 5.203 TD /Meta80 94 0 R /BBox [0 0 15.59 16.44] endobj /FormType 1 BT /BBox [0 0 17.177 16.44] 32.201 5.203 TD 1 i 1 i >> /Meta93 Do ET q >> /FormType 1 Q /FormType 1 0.737 w BT 1 i stream q /FormType 1 Q /StemH 88 1 i Q q Q /Type /XObject (-) Tj ET Q /Resources<< endstream 1.014 0 0 1.007 391.462 583.429 cm [tex]\sin (\pi -x)=\sin x[/tex]. /F1 12.131 Tf /Length 69 0.458 0 0 RG Q 0 g 190 0 obj /Resources<< 1 i q Q 0.524 Tc >> << >> endobj A number increased by 5 is equivalent to twice the same number decreased by 7. /Length 70 /Meta113 127 0 R Q >> /Resources<< Q [(The )-19(quotient of )] TJ q >> (B) Tj /Font << >> /Meta40 54 0 R Q 6.746 5.203 TD 1 i /ProcSet[/PDF] 0 w << /Meta279 293 0 R q >> >> endobj /ProcSet[/PDF] /Font << >> 315 0 obj stream /ProcSet[/PDF] /BBox [0 0 673.937 27.581] 1.007 0 0 1.007 67.753 872.509 cm /Encoding /WinAnsiEncoding Q /ProcSet[/PDF/Text] >> /ProcSet[/PDF/Text] stream /Meta160 174 0 R 364 0 obj 16 0 obj Q /CapHeight 662 q 0 G (x) Tj >> q Q /Meta401 Do /F3 12.131 Tf /Font << /Meta287 Do BT q startxref 0 g /Meta28 41 0 R 0 g >> 0 g /Meta353 367 0 R 1 i /Subtype /Form q Q 392 0 obj /Flags 32 endobj q q >> 1.502 5.203 TD /Resources<< /Meta121 135 0 R /Length 16 /Meta243 Do Q BT q >> /Meta179 193 0 R 0 4.894 TD 1 i 1 g >> Q 0 G /F4 36 0 R 1 g /Meta387 Do Q /BBox [0 0 639.552 16.44] /Matrix [1 0 0 1 0 0] /FormType 1 ET stream Q /Subtype /Form /Subtype /Form /Matrix [1 0 0 1 0 0] 22 0 obj Q 1 i /Meta52 66 0 R SOLUTION: sixteen increased by twice a number is -58. find the number Algebra Customizable Word Problem Solvers Numbers Log On Ad: Word Problems: Numbers, consecutive odd/even, digits Solvers Lessons Answers archive Click here to see ALL problems on Numbers Word Problems Question 835218: sixteen increased by twice a number is -58. find the number 1.014 0 0 1.006 531.485 763.351 cm /F3 12.131 Tf Mat (5\)) Tj /Meta394 Do endobj Q stream /Meta260 Do 0.738 Tc q 1.007 0 0 1.007 130.989 776.149 cm /F3 17 0 R q /Length 69 /Subtype /Form BT /Meta111 Do 0.458 0 0 RG 0.564 G /FormType 1 /F3 12.131 Tf 1.007 0 0 1.007 271.012 636.879 cm Q stream /Meta378 392 0 R << endobj /F3 12.131 Tf (vii) Twice a number subtracted from 19 is 11. Q /BBox [0 0 15.59 16.44] >> /Type /XObject >> /FormType 1 q /Font << 5 0 obj >> Q q >> >> endstream Then ab is a binary operation. ET /BBox [0 0 15.59 16.44] /XObject << /Resources<< BT 0 g ET stream >> endobj Q /Matrix [1 0 0 1 0 0] 1.502 5.203 TD Q q q /Type /XObject

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twice a number decreased by 58