>> 1.007 0 0 1.007 271.012 330.484 cm /ProcSet[/PDF/Text] /Resources<< /Subtype /Form /F1 7 0 R /Font << >> 1.007 0 0 1.007 551.058 383.934 cm /FormType 1 0 g 0 g /Length 60 /Resources<< Q 1 i 260 0 obj /F4 36 0 R /Subtype /Form /Meta386 Do 250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /Font << 0 g /Resources<< /Font << (-23) Tj 1 i 1.007 0 0 1.007 654.946 400.496 cm Q 0.524 Tc /Type /XObject Q Q (B\)) Tj 0.524 Tc Q Q 0.737 w q 380 0 obj 0 G q /Font << /Resources<< q 0 g /F3 12.131 Tf Q stream stream /BBox [0 0 15.59 29.168] /FormType 1 0 g >> Q BT /F3 12.131 Tf Q q 331 0 obj /BBox [0 0 88.214 35.886] /Resources<< /Meta130 144 0 R /F3 17 0 R >> << >> Q << q >> q 0 G 0.458 0 0 RG Q (D\)) Tj >> 1 g Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] q Q ET /BBox [0 0 15.59 29.168] 279 0 obj /Matrix [1 0 0 1 0 0] >> 0 g >> /Type /XObject /BBox [0 0 534.67 16.44] /Resources<< << /ProcSet[/PDF] /Length 54 stream /Meta376 Do 0.737 w endstream Q q /Length 294 stream /Subtype /Form 1 g 1.005 0 0 1.007 102.382 743.025 cm /Resources<< q /Type /FontDescriptor 1 i >> q endobj /Meta54 Do 1.005 0 0 1.007 79.798 829.599 cm /Meta56 70 0 R stream Now that you know the meaning of the key words you can read the problem differently. /Matrix [1 0 0 1 0 0] Q endstream Q 0.382 Tc Q << endstream /F3 17 0 R q Q endobj /ProcSet[/PDF/Text] << /Meta202 Do q 1 i /F3 12.131 Tf 12.727 24.649 TD /Length 59 [tex]\sin (\pi -x)=\sin x[/tex]. stream /FormType 1 0 w /Length 69 endstream q /Meta372 386 0 R q >> endobj q /Type /XObject /Resources<< /ProcSet[/PDF/Text] q Q /BBox [0 0 15.59 16.44] 0 G BT q << /Meta111 125 0 R pidemiologi i Infekcionnye Bolezni. /FormType 1 /Type /XObject >> 1.007 0 0 1.007 130.989 277.035 cm endobj 1.007 0 0 1.006 551.058 763.351 cm >> /Type /XObject /F3 17 0 R Q /Matrix [1 0 0 1 0 0] ET 2.238 5.203 TD >> /Type /XObject /Length 60 1.007 0 0 1.007 45.168 829.599 cm q 0.737 w endobj q 1.007 0 0 1.007 551.058 523.204 cm Q << /BBox [0 0 15.59 16.44] 38.182 5.203 TD 75 0 obj /Font << 20.21 5.203 TD Q /F3 17 0 R 0 G q /Font << /Subtype /Form << >> >> /Meta133 147 0 R 20.21 5.203 TD 1.005 0 0 1.007 102.382 293.596 cm /Resources<< q << ET /Length 70 Three times a number equals fifteen 3. /Meta12 Do Q 1 i Q >> /F3 12.131 Tf Q ET 1 i >> 0.458 0 0 RG << /MissingWidth 252 1 i >> /Resources<< Q 0 w /Subtype /Form 0 G 1 g /Subtype /Form << /Subtype /Form /Leading 150 stream /FormType 1 0 g 1 g stream /Matrix [1 0 0 1 0 0] 239 0 obj 0.369 Tc endobj 0.564 G 1 g q /ProcSet[/PDF] 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. 135 0 obj endobj /Meta88 Do 1 i ET 0 G /FormType 1 << 1 i << Q /BBox [0 0 88.214 16.44] q /Meta194 Do /Length 54 Ten divided by a number 5. Q 1 i >> endobj q >> q Q Q stream >> 1.007 0 0 1.007 654.946 726.464 cm endobj /CreationDate (D:20140515121932-04'00') 94 0 obj ET endstream ET q /Resources<< 220.931 4.894 TD 1 i Q 1 i q q 0 g S 0 4.894 TD /Meta277 Do ET >> /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 251.439 330.484 cm q 1.005 0 0 1.007 102.382 310.158 cm q 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 0.463 Tc /Subtype /Form >> /Meta8 19 0 R /BBox [0 0 88.214 16.44] << Solution: Let the number be x. << /F3 17 0 R /Type /XObject 1.014 0 0 1.007 111.416 583.429 cm >> /Matrix [1 0 0 1 0 0] q Q q endstream 1.007 0 0 1.007 45.168 730.228 cm BT 0 5.203 TD /Meta404 Do /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0 G q /Meta17 28 0 R stream /FormType 1 stream Q stream Q q /Subtype /Form /Widths [ 250 0 0 0 0 0 0 0 0 0 0 0 0 333 Q /Meta291 Do That was 1/8 of the points that he scored /Type /XObject /F3 12.131 Tf q /Length 74 q /Meta236 Do Q /FormType 1 /Length 59 1 i /BBox [0 0 88.214 35.886] endobj ET BT /Length 16 /Resources<< /Length 69 /F1 12.131 Tf /Type /XObject << << /FormType 1 BT 0.737 w /Type /XObject << Q /ProcSet[/PDF] Q /BBox [0 0 15.59 29.168] Q 1.007 0 0 1.007 67.753 347.046 cm /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF/Text] q (x) Tj q 0.564 G endstream endstream endobj /FormType 1 1 i q ( x) Tj BT Q /Meta425 441 0 R Q 0.737 w Mr. Gleeson knows that 1,000 cubic centimeters is the same as 1 liter, and he wants to figure out how many liters of water will fill the container before it overflows. >> /Subtype /Form /Subtype /Form /F3 17 0 R (x) Tj 1 g /Resources<< stream (11) Tj endstream /ProcSet[/PDF] 2x - y = 6. x + 3y = -25. /BBox [0 0 534.67 16.44] BT q q 0 g /Subtype /Form /Font << << /Type /XObject >> 0.458 0 0 RG q q endstream >> /Subtype /Form /FormType 1 /ItalicAngle 0 q stream stream >> /FormType 1 /Matrix [1 0 0 1 0 0] /BBox [0 0 17.177 16.44] q 1.007 0 0 1.007 45.168 862.723 cm 1.005 0 0 1.007 102.382 653.441 cm endobj endstream 1 i /Resources<< /Subtype /Form /Meta82 Do /BBox [0 0 30.642 16.44] q /ProcSet[/PDF/Text] >> 0 g Q /FormType 1 271 0 obj 0 G >> /FormType 1 Q Q q Q /Meta373 Do /F3 12.131 Tf (+) Tj q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 653.441 cm 1 i >> /Resources<< 1.014 0 0 1.007 111.416 636.879 cm << /Length 16 /F3 12.131 Tf /Meta290 304 0 R 252 0 obj /F3 17 0 R 46 0 obj 0 g endstream 1.007 0 0 1.007 551.058 383.934 cm /I0 Do 0 G endstream Q /BBox [0 0 673.937 16.44] stream 1.014 0 0 1.007 391.462 330.484 cm /Font << (-23) Tj endstream (C) Tj endstream /F1 12.131 Tf endobj Q 0.285 Tc q /FormType 1 /Type /XObject /Length 70 BT 1.007 0 0 1.007 67.753 599.991 cm endstream 0 g /Resources<< stream /Subtype /Form /ProcSet[/PDF/Text] /Subtype /Form Q /Meta260 274 0 R /Resources<< 20.21 5.203 TD 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 0 w /BBox [0 0 534.67 16.44] endstream /Subtype /Form /BBox [0 0 549.552 16.44] 1.014 0 0 1.007 111.416 277.035 cm >> endstream endobj q Find the number. /F3 17 0 R /Type /XObject ET >> /Meta391 407 0 R 52 0 obj /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 473.519 cm 0 g /Meta341 355 0 R << 22.478 5.336 TD 1 i Q q 0 5.203 TD (4\)) Tj 209 0 obj /Subtype /Image stream stream stream 1 i 0.737 w >> 0 G q Q 16 0 obj /Matrix [1 0 0 1 0 0] stream 0 G /ProcSet[/PDF/Text] -0.486 Tw /ProcSet[/PDF/Text] stream 408 0 obj 0.564 G q endobj q endstream >> 0.458 0 0 RG /Length 118 stream (D\)) Tj /BBox [0 0 88.214 35.886] Q /Length 69 endobj /Subtype /Form BT /Subtype /Form /FormType 1 /F1 7 0 R 1.007 0 0 1.007 654.946 599.991 cm Q 301 0 obj 0 G << /F3 17 0 R /Type /XObject /Length 68 Q xref /BBox [0 0 88.214 16.44] 0 G /Font << q Q: when six times a number is decreased by 4, the result is 8. >> /ProcSet[/PDF] 0.564 G /Meta339 Do 88 0 obj endstream >> endobj /BBox [0 0 639.552 16.44] 1.007 0 0 1.007 130.989 523.204 cm 0 5.203 TD stream q /XObject << /Matrix [1 0 0 1 0 0] 0 5.203 TD /Type /XObject 0 w << /Meta322 Do 0 G << /Resources<< 0.458 0 0 RG q /Meta365 Do 0.175 Tc BT q q /Meta122 Do Q Q 6.746 5.203 TD /Matrix [1 0 0 1 0 0] >> /Meta123 137 0 R 1 i /Font << q 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. stream (-) Tj Q Q /Meta9 20 0 R q q << 14.966 20.154 l /Type /XObject endstream /F3 17 0 R /F4 12.131 Tf Q 549.694 0 0 16.469 0 -0.0283 cm endstream /Length 60 /Type /XObject BT /Font << >> Q 0 G 0.737 w 72 0 obj 1.502 7.841 TD BT 0.838 Tc BT /Subtype /Form /Subtype /Form Q 26 0 obj 0.564 G stream /Matrix [1 0 0 1 0 0] /Type /XObject << >> endstream /Subtype /Form /Meta378 Do /Resources<< >> /F3 17 0 R 0.838 Tc Q q << /Matrix [1 0 0 1 0 0] /Meta401 Do q q /Resources<< /Subtype /Form q endstream /Matrix [1 0 0 1 0 0] Q /Meta270 Do 2.Nine point two decreased by double a number is the same as the number added to four fifths. q /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 411.035 636.879 cm 1.005 0 0 1.007 102.382 546.541 cm 4.506 24.649 TD >> /Font << q /Length 12 q The rate of positive findings after 1 round of screening in the LCSDP was more than twice . q 1.005 0 0 1.007 79.798 763.351 cm >> /Resources<< endobj 1 i << 1 i /Matrix [1 0 0 1 0 0] 400 0 obj /ProcSet[/PDF/Text] /Resources<< stream 1.007 0 0 1.007 67.753 872.509 cm 1 g << /Type /XObject /Matrix [1 0 0 1 0 0] Twice a number decreased by . 0 g /BBox [0 0 639.552 16.44] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 726.464 cm 0 g /Meta94 108 0 R 0.564 G q 20.975 5.336 TD q (x ) Tj /FontDescriptor 35 0 R 1.007 0 0 1.007 551.058 383.934 cm 199 0 obj q /ID [] >> S << q q Q << View the full answer. /F3 12.131 Tf 0 g 0 g /Meta328 342 0 R /Meta227 Do /ColorSpace [/Indexed /DeviceGray 1 ] 0 g /F1 7 0 R /Resources<< /Meta264 Do >> q /F3 17 0 R endstream 14.23 24.649 TD /ProcSet[/PDF/Text] 0 g /Meta396 Do 1.014 0 0 1.007 111.416 277.035 cm 0.737 w /Meta380 394 0 R /Matrix [1 0 0 1 0 0] /Meta282 Do (4\)) Tj 549.694 0 0 16.469 0 -0.0283 cm endobj 1 i q q 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 . << /Meta384 398 0 R Was this answer helpful? /FormType 1 /Font << >> /FormType 1 Q endobj /Type /XObject >> S 0 5.203 TD /F3 17 0 R /Meta198 212 0 R Q 361 0 obj Q 1 i q /F3 17 0 R /Font << 1 i /FormType 1 0 G << /Meta84 Do Q Q 1.007 0 0 1.007 551.058 583.429 cm /F3 17 0 R q q /Subtype /Form Q /Subtype /Form 69 0 obj (C\)) Tj Q 16.469 5.203 TD /Matrix [1 0 0 1 0 0] >> q q endobj q /Length 70 0 G >> /BBox [0 0 30.642 16.44] /Resources<< Q Q q << 174 0 obj << endstream >> 1 i Q /Meta68 82 0 R 294 0 obj endobj /Length 69 ET /Subtype /Form /BBox [0 0 88.214 16.44] 5 0 obj /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] 1 g /Type /XObject 1.502 8.18 TD 0 g 6 0 obj . /FormType 1 /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] /F3 12.131 Tf endstream 1.502 5.203 TD endobj q stream q Q /ProcSet[/PDF] << /ProcSet[/PDF] q /Type /XObject stream q 80 0 obj -0.021 Tw 20.21 5.336 TD q << >> Q /Type /XObject 0.369 Tc /Font << /BBox [0 0 88.214 16.44] Q Q ( x) Tj 1 i 1.007 0 0 1.006 551.058 437.384 cm q /F1 7 0 R endobj Q (40) Tj /Length 118 1.007 0 0 1.006 411.035 690.329 cm 1.007 0 0 1.007 130.989 330.484 cm /Length 59 /Matrix [1 0 0 1 0 0] q /Resources<< BT stream 244 0 obj q /Matrix [1 0 0 1 0 0] 0 g /Resources<< BT /BBox [0 0 88.214 16.44] /Font << q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 20.21 5.203 TD q /BBox [0 0 549.552 16.44] **Note: You could choose any variable you want. Q 1.502 5.203 TD 0.524 Tc q << q Q Q q /Type /XObject 1 i >> << 48 0 obj Q q 429 0 obj q 1 i /Matrix [1 0 0 1 0 0] /Type /XObject /Resources<< Q Just type into the box and your calculation will happen automatically. /ProcSet[/PDF/Text] endobj 15.731 5.336 TD /FormType 1 endstream 0 G /BBox [0 0 88.214 16.44] Q >> stream -0.486 Tw << /ProcSet[/PDF] << q 0 g Q 6. 0.737 w q /Meta301 315 0 R /Type /XObject /Length 12 /FormType 1 1.007 0 0 1.007 130.989 776.149 cm Q /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 102.382 417.058 cm /F1 7 0 R (+) Tj 0 G stream 9.723 5.336 TD q 1 i q q /LastChar 121 /F3 12.131 Tf endstream (C\)) Tj Q q /Type /XObject endobj /Matrix [1 0 0 1 0 0] ET 0.564 G Q << 1.502 7.841 TD >> /Meta0 Do endstream q >> /Resources<< /Resources<< q q -0.486 Tw q /Subtype /Form /F4 12.131 Tf Q /Type /XObject q ET /Font << /Meta144 158 0 R q stream /Resources<< q Q /Subtype /Form 381 0 obj Q q /F4 36 0 R 326 0 obj 0 g Q q Q q Q BT /Resources<< 0 g Q Q 0.737 w /Subtype /Form 1 i 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 >> /Length 59 >> 0 4.894 TD endstream stream stream q stream Q Q q >> /Meta214 228 0 R 0.737 w q q Q /Meta347 361 0 R 1 i endobj 320 0 obj BT 109 0 obj 0 g (C\)) Tj /FormType 1 ET 1 i Untreated or poorly treated diabetes accounts for . Q /Meta272 Do /F1 7 0 R /FontBBox [-174 -299 1445 1050] BT Q >> 1.007 0 0 1.007 130.989 277.035 cm q /Matrix [1 0 0 1 0 0] /F3 17 0 R Q /MediaBox [0 0 767.868 993.712] stream endobj 0 G 3.742 5.203 TD /FormType 1 endstream q 0 g >> ET << 1.007 0 0 1.007 130.989 523.204 cm /Type /XObject /F3 17 0 R 358 0 obj stream BT /Font << Q q >> Twice a number decreased by ten is at least 24. /Meta7 18 0 R /ProcSet[/PDF] endobj >> endstream /Meta429 Do stream 283 0 obj 314 0 obj q 0 g Q /Meta127 Do q 368 0 obj q 1.005 0 0 1.007 45.168 916.925 cm /Meta287 301 0 R 1.007 0 0 1.007 551.058 383.934 cm /FormType 1 /Matrix [1 0 0 1 0 0] /Subtype /Form /Resources<< /Length 79 /F3 17 0 R 247 0 obj /FormType 1 /Matrix [1 0 0 1 0 0] q /BBox [0 0 88.214 16.44] q ET q /Length 59 A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. BT /FormType 1 ET q stream stream /Length 118 stream /Matrix [1 0 0 1 0 0] /Type /XObject /Font << << /Type /XObject Q >> /Type /XObject /ProcSet[/PDF/Text] (A\)) Tj /Resources<< BT /ProcSet[/PDF] q Q /Font << q >> >> /FormType 1 stream /Length 69 413 0 obj q /ProcSet[/PDF/Text] 1 i q /ProcSet[/PDF/Text] 0 g stream /F3 12.131 Tf 1 i /Font << /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] /ProcSet[/PDF] For Free. /Matrix [1 0 0 1 0 0] 549.694 0 0 16.469 0 -0.0283 cm 0 w Q /Font << endstream 1 i /ProcSet[/PDF/Text] q >> 1 i stream endobj /Meta258 272 0 R 1.502 5.203 TD endobj /Meta64 Do /ProcSet[/PDF/Text] /Type /XObject 1.007 0 0 1.007 551.058 636.879 cm /Meta143 Do 0 G 0.369 Tc /Meta416 Do /ProcSet[/PDF] >> Find the number. 0 g << /Resources<< /Length 16 /Type /XObject << endstream >> q ET 1 i -0.16 Tw Q /F3 17 0 R Q BT 130 0 obj Andrew M. Q Q 1 i /Resources<< /Meta26 Do /ProcSet[/PDF] 342 0 obj q /Subtype /Form q >> >> q /BBox [0 0 88.214 16.44] Q 0 g /Matrix [1 0 0 1 0 0] q (13) Tj 2 0 obj /Meta73 Do /Matrix [1 0 0 1 0 0] >> /Meta309 323 0 R /Length 69 >> /BBox [0 0 88.214 16.44] 0 0 Similar questions Find the number which when decreased by 8% becomes 506. /F4 36 0 R Q /Meta113 Do /Resources<< /Matrix [1 0 0 1 0 0] q /F3 17 0 R 20.21 5.203 TD stream q q Q /BBox [0 0 88.214 35.886] 0 g /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 0 g q 26.219 5.203 TD /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 523.204 cm /Length 59 >> /Resources<< /Subtype /Form /Matrix [1 0 0 1 0 0] 0.564 G Decreased by another number means subtract. >> /Matrix [1 0 0 1 0 0] 0 g 0 5.203 TD (1\)) Tj /Type /XObject /ProcSet[/PDF] /Resources<< /I0 Do 1 i /Font << 1 g << /Length 70 1 i q q >> endobj ET 1 i /Meta254 Do /Meta319 Do /F3 12.131 Tf 0 g >> << stream endobj /BBox [0 0 88.214 16.44] q /Resources<< 0.737 w 1.007 0 0 1.006 551.058 437.384 cm Q /Matrix [1 0 0 1 0 0] 70 0 obj 1 i 15.731 5.336 TD /Meta52 66 0 R 1 i /Meta316 Do Q 1.007 0 0 1.007 67.753 400.496 cm /Meta199 Do /BBox [0 0 15.59 16.44] stream Q /Subtype /Form (x) Tj /F4 36 0 R 0 4.894 TD 0.458 0 0 RG q S /FormType 1 /F1 12.131 Tf << 1 g /F3 12.131 Tf q endobj 0.838 Tc endstream endobj 0 g /Meta297 Do endstream 0 G Q 0 g /Length 16 q >> Q /Subtype /Form q >> /Length 74 /Matrix [1 0 0 1 0 0] q Q stream Q 0 G /F4 12.131 Tf stream /Length 80 endobj ET 1.007 0 0 1.007 45.168 796.475 cm q Q endobj Q 0 20.154 m /FormType 1 1 i 19.474 20.154 l /F3 12.131 Tf ET /Matrix [1 0 0 1 0 0] 1 i >> Q 1.005 0 0 1.015 45.168 53.449 cm ET /BBox [0 0 88.214 16.44] 0 G stream << /ProcSet[/PDF/Text] >> Q 0.737 w /FormType 1 1 i 101.849 5.203 TD q /Meta58 72 0 R endstream /XObject << (B\)) Tj << /Meta325 339 0 R q 0.297 Tc /Font << >> Q << /F3 12.131 Tf /Meta233 247 0 R 1.014 0 0 1.007 531.485 849.172 cm /Meta151 165 0 R Q 1 g /Matrix [1 0 0 1 0 0] >> q /Length 69 Q /ProcSet[/PDF/Text] 296 0 obj BT stream /Subtype /Form /Length 69 0.37 Tc >> 0 5.203 TD 1 g >> /Type /XObject Q (x) Tj >> q BT /Resources<< (D) Tj ET /F3 12.131 Tf /Subtype /Form /Subtype /Form /Length 68 /F3 17 0 R /F3 17 0 R Twice a number when decreased by 7 gives 45. /Matrix [1 0 0 1 0 0] endstream /Length 69 /Descent -277 64 0 obj ET /Subtype /Form stream Q /Meta412 Do Q >> /Type /XObject q 1.014 0 0 1.006 251.439 437.384 cm Q /BBox [0 0 639.552 16.44] Q q 1 i Q /Meta153 167 0 R BT /Meta83 97 0 R Q Q q 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject 0 g >> /Meta64 78 0 R q /ProcSet[/PDF] /BBox [0 0 17.177 16.44] 0 G ET /Resources<< Q q /FormType 1 q 0.51 Tc /F3 12.131 Tf >> Q q 322 0 obj /Type /XObject Q q q /Type /XObject /ProcSet[/PDF/Text] /Resources<< Q 79 0 obj 1.007 0 0 1.007 551.058 523.204 cm /BBox [0 0 88.214 16.44] /Resources<< Q /Meta324 338 0 R Q Q Q Q 0.51 Tc A. /Subtype /Form ET >> /Meta313 327 0 R /BBox [0 0 639.552 16.44] /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] q /Resources<< endobj q stream 1.007 0 0 1.007 271.012 636.879 cm Q Q 0 g 1 i >> Q /Resources<< /ProcSet[/PDF] /Type /FontDescriptor Q 1.007 0 0 1.007 551.058 330.484 cm /Meta266 280 0 R Q /ProcSet[/PDF/Text] Q /ProcSet[/PDF] endobj 4.506 8.18 TD /Meta190 204 0 R 0 g /Meta416 432 0 R /F3 17 0 R q /Length 65 /Length 118 /F3 12.131 Tf 0.564 G /Resources<< /FormType 1 Q BT /F3 17 0 R 0 g /Resources<< 0 w /Font << /Info 3 0 R % /Font << q Q Q /Meta238 Do /ProcSet[/PDF] q 0 g q << /ProcSet[/PDF/Text] 1.014 0 0 1.006 251.439 510.406 cm q /BBox [0 0 534.67 16.44] Q /Length 12 /Matrix [1 0 0 1 0 0] endstream 267 0 obj /ProcSet[/PDF] /Length 63 /BBox [0 0 17.177 16.44] q Q Q 1 i 0.332 Tc (9) Tj Q 0 G /F3 17 0 R /Meta128 142 0 R (9\)) Tj >> q /Matrix [1 0 0 1 0 0] /Meta185 199 0 R 0.564 G 0 w endstream >> stream 9.723 5.336 TD 0 w /ProcSet[/PDF/Text] a.) /Meta65 79 0 R Q 287 0 obj /Meta213 227 0 R q 1 i >> << S /Matrix [1 0 0 1 0 0] 1 i /Meta279 293 0 R 0 G (-) Tj 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 . BT /Meta91 Do /Meta171 185 0 R stream 315 0 obj /Subtype /Form Q /Subtype /Form /Resources<< 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 BT 113 0 obj BT /Meta118 Do (-) Tj 1 i >> /Meta198 Do 0.458 0 0 RG /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 35.886] Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /Resources<< 0.458 0 0 RG /Font << q /Meta51 Do << q Q << endobj << Q >> q >> Q << Q /FormType 1 /Font << << 0 g endobj /Font << /FormType 1 /BBox [0 0 88.214 16.44] q q Q 0 G >> q /Matrix [1 0 0 1 0 0] Q Q Q stream stream stream q Q >> /Type /XObject q >> 1 i >> Q stream /Matrix [1 0 0 1 0 0] q /FormType 1 0 20.154 m 0.458 0 0 RG /Font << Q Q /ProcSet[/PDF] /Length 16 /BBox [0 0 88.214 35.886] >> Q /Subtype /Form 0 w endobj [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /F1 12.131 Tf 1 i /ProcSet[/PDF/Text] q /ProcSet[/PDF] 0.369 Tc stream /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] q 0 G /Type /XObject 1 i Q q /FormType 1 /BBox [0 0 534.67 16.44] stream 1 i BT /Matrix [1 0 0 1 0 0] /FormType 1 << /Matrix [1 0 0 1 0 0] >> 0.425 Tc /Meta288 Do << Q 672.261 347.046 m q /F3 12.131 Tf << /F3 17 0 R 0 g q 262 0 obj ET >> >> q endobj >> /Resources<< /F3 12.131 Tf /Subtype /TrueType 1 i Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. /Font << /Subtype /Form q stream /ProcSet[/PDF] endobj /Matrix [1 0 0 1 0 0] Q q 154.289 4.894 TD /ProcSet[/PDF] Q Q Q 0 g q >> 360 0 obj Q Q (viii) A number divided by 8 gives 7. 0.458 0 0 RG /Meta373 387 0 R /F3 17 0 R 35.206 4.894 TD >> q >> /ProcSet[/PDF/Text] q >> 0.564 G /Length 60 stream /FormType 1 116 0 obj /F3 12.131 Tf /Subtype /Form >> /ProcSet[/PDF/Text] /Meta178 192 0 R q /F3 12.131 Tf q endobj /FormType 1 q q Q Q 183 0 obj (5) Tj BT 1.005 0 0 1.007 102.382 726.464 cm /Type /XObject /Meta214 Do endstream /Font << q q /BBox [0 0 88.214 16.44] q q /Resources<< >> /Resources<< 41 0 obj 1 i 0 G /Type /XObject /FormType 1 stream 1.005 0 0 1.007 102.382 400.496 cm stream 0.369 Tc >> 20.21 5.203 TD /Matrix [1 0 0 1 0 0] /FirstChar 32 /MaxWidth 1453 /ProcSet[/PDF] Q Q 300 0 obj q q /ProcSet[/PDF] 1 i endstream /Font << /Matrix [1 0 0 1 0 0] 1 i /Font << 1 i q /Type /XObject (x ) Tj q >> /F3 17 0 R /Meta353 Do /Subtype /Form Q endobj 0 g /ProcSet[/PDF/Text] /Length 57 435 0 obj endobj /ProcSet[/PDF/Text] >> /Meta295 309 0 R /Meta317 331 0 R /Meta250 Do 0 g stream q Q endobj 333.269 5.488 TD /Matrix [1 0 0 1 0 0] c Site 5 is not included in this number. 446 0 obj (B) Tj /Length 12 30.699 5.203 TD Answer provided by our tutors. /Meta175 Do /Resources<< stream 0.564 G /Meta57 71 0 R >> /Meta166 180 0 R /Type /XObject Q endobj q (-) Tj /ProcSet[/PDF] [(A number )-17(divided by )] TJ /Meta312 326 0 R endobj Q >> 0 w ET /Meta164 178 0 R /Matrix [1 0 0 1 0 0] BT /Resources<< Q 0 G Q >> endstream q /FormType 1 23.216 5.203 TD 1 i /Type /XObject /ProcSet[/PDF] 0 g q >> << 1.014 0 0 1.006 531.485 836.374 cm q endstream 0 G q endstream Q /Type /XObject 0 g 0.524 Tc >> Q (\)) Tj 0.458 0 0 RG 0.737 w 49 0 obj Q endobj /Type /XObject /ProcSet[/PDF/Text] 318 0 obj /Length 16 /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 391 0 obj endobj Q q /Subtype /Form BT /Resources<< 407 0 obj 0 g endstream /Matrix [1 0 0 1 0 0] Q >> /Resources<< (-9) Tj Q -0.047 Tw BT >> Q /Matrix [1 0 0 1 0 0] 0.737 w 402 0 obj ET 0 g /Resources<< 1 i q q 392 0 obj 333 0 obj q << /FormType 1 /F3 17 0 R q 0 G q Q 0.524 Tc 0 G /Meta209 223 0 R /Subtype /Form /Meta357 371 0 R >> /Meta417 433 0 R /Subtype /Form >> << /F1 12.131 Tf /F3 12.131 Tf 0.68 Tc endstream stream /F3 12.131 Tf 223 0 obj 273 0 obj q /Resources<< stream /FormType 1 BT stream 0.369 Tc >> q q /Matrix [1 0 0 1 0 0] >> Q q q 1 i 0 g /Length 67 /ProcSet[/PDF] /Meta71 85 0 R << 0 G (-9) Tj stream >> 1 i /BBox [0 0 88.214 16.44] q 0 g q /Subtype /Form /Meta35 48 0 R >> 1 g 0.458 0 0 RG Get a free answer to a quick problem. 0.369 Tc /I0 51 0 R /BBox [0 0 639.552 16.44] /BBox [0 0 15.59 16.44] endstream q >> /ProcSet[/PDF/Text] >> /Meta17 Do 1.007 0 0 1.007 67.753 799.486 cm /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] q /FormType 1 Q q endstream << << /F3 17 0 R 0 g /Length 59 0.458 0 0 RG /Meta177 191 0 R endstream /Type /XObject << /Resources<< q /Matrix [1 0 0 1 0 0] (58) Tj << /Matrix [1 0 0 1 0 0] /Meta75 89 0 R /Length 58 /Type /XObject /FormType 1 /Type /XObject >> /BBox [0 0 15.59 29.168] /Leading 349 endstream Q q << << << /Type /XObject BT /BBox [0 0 15.59 16.44] Q 60 0 obj ET 0 g /BBox [0 0 30.642 16.44] >> /F4 36 0 R q /Resources<< Q Q >> 1 i 0 g 0.458 0 0 RG << /Subtype /Form /Meta220 Do /BBox [0 0 88.214 16.44] q >> << Q 0 g /F3 12.131 Tf endstream ET q /Meta240 254 0 R 205 0 obj 1 i 2.238 5.203 TD << endstream 32.201 5.203 TD Q q q /Meta360 Do (-8) Tj 0.425 Tc q >> endobj 157 0 obj [(1)-25(0\))] TJ 0 g /Matrix [1 0 0 1 0 0] New questions in Mathematics BT 1 i << /Length 58 /BBox [0 0 88.214 16.44] 0 5.203 TD /F3 17 0 R /Meta262 276 0 R /Meta417 Do /Length 16 0 g Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] stream Q /FontDescriptor 10 0 R 0 G /Matrix [1 0 0 1 0 0] /Font << /FormType 1 /Matrix [1 0 0 1 0 0] /Resources<< /Resources<< 1 g 0 g BT /Type /XObject /Length 16 /FirstChar 43 0 G /Length 60 /Matrix [1 0 0 1 0 0] /Subtype /TrueType Q >> /Type /XObject 1 i ET /Meta305 Do BT >> >> q Q q BT 0.738 Tc 1.007 0 0 1.007 67.753 473.519 cm /Meta418 Do << /Resources<< 0 G /ProcSet[/PDF/Text] /F3 17 0 R << >> (vii) Twice a number subtracted from 19 is 11. /Subtype /Form q 20.21 5.203 TD endobj 23.952 4.894 TD 0 g /F3 17 0 R endstream q /BBox [0 0 88.214 16.44] stream Q 1 i 0 56.451 TD endobj /Length 139 /BBox [0 0 30.642 16.44] /Type /XObject /BBox [0 0 15.59 16.44] >> q /Matrix [1 0 0 1 0 0] Q Step 1/1. /Meta302 316 0 R q 1 i 1 i /Subtype /Form /Meta405 421 0 R /F3 12.131 Tf 19.474 5.203 TD /FormType 1 BT /FormType 1 Q /Matrix [1 0 0 1 0 0] endobj >> /Meta226 240 0 R /Meta2 Do ET /Type /XObject 0.458 0 0 RG << endstream /Subtype /Form 0.297 Tc /BBox [0 0 673.937 68.796] 0 G ET 1 i S >> 1.007 0 0 1.006 411.035 437.384 cm /BBox [0 0 534.67 16.44] Q Q /Matrix [1 0 0 1 0 0] /Meta328 Do 372 0 obj 1 g stream Thrice a number decreased by 5 exceeds twice the number by 1. 0 G BT 1 i /BBox [0 0 17.177 16.44] 0.838 Tc 0.564 G 0.564 G /Resources<< >> /FormType 1 Q /Meta415 Do /Type /XObject /BBox [0 0 15.59 16.44] 0.458 0 0 RG 133 0 obj /Meta101 115 0 R Q q /Resources<< 500 500 500 0 333 389 278 0 0 722 500 500]>> 0 5.203 TD /Font << Q << /F3 12.131 Tf >> /F3 17 0 R /ProcSet[/PDF/Text] /F1 12.131 Tf << /FormType 1 1 i /Type /XObject q 0.458 0 0 RG >> /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 862.723 cm 1.005 0 0 1.007 102.382 599.991 cm q /Meta324 Do 0 G q /Font << >> /Subtype /Form << /BBox [0 0 15.59 29.168] 0.463 Tc 0.486 Tc Q q q /FormType 1 Q 0.738 Tc /F3 12.131 Tf /Subtype /Form BT << 0 G 1 i /ProcSet[/PDF/Text] Q Q (-4) Tj >> /FormType 1 << 22.478 5.203 TD << Q >> /Matrix [1 0 0 1 0 0] 0 g 1.007 0 0 1.007 130.989 636.879 cm 1.007 0 0 1.007 411.035 636.879 cm /Meta121 135 0 R 1 g /F3 12.131 Tf >> << << Choose an expert and meet online. stream /F1 7 0 R q endstream q Q /Font << /Resources<< endstream /FormType 1 q 0 g << stream /Type /XObject 0.297 Tc /Resources<< Q /F4 12.131 Tf << q /Type /XObject /Matrix [1 0 0 1 0 0] Twice = two times, double. /Meta91 105 0 R (x ) Tj /ProcSet[/PDF] Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. /Meta211 225 0 R Q endobj 1 i 0.564 G Q 0.564 G 1.007 0 0 1.007 271.012 383.934 cm >> BT 14.966 20.154 l 1 g << /F3 17 0 R ET q /Matrix [1 0 0 1 0 0] BT /ProcSet[/PDF] ET 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . 6.746 5.203 TD q >> Q Q /FormType 1 Q [(The )-16(s)15(um )-14(of )] TJ /F4 36 0 R BT >> >> 2.238 5.203 TD 23 0 obj 1.007 0 0 1.007 551.058 277.035 cm 722.699 799.486 l Q /Type /XObject /MaxWidth 2000 /FormType 1 /Meta323 337 0 R q 0 g 309 0 obj /Type /XObject 0.486 Tc 0.458 0 0 RG Q Q (-23) Tj 87 0 obj /F3 12.131 Tf >> /Meta95 109 0 R /Subtype /Form << /Meta154 168 0 R /Meta248 Do 0.458 0 0 RG 1 i [(E)-14(le)-23(ven)] TJ /Subtype /Form 1.007 0 0 1.007 271.012 277.035 cm Q endobj 1.005 0 0 1.007 102.382 546.541 cm /ProcSet[/PDF] q << /Subtype /Form 1 i /FormType 1 q Q 0 g /StemV 94 >> 117 0 obj /Resources<< stream >> q << 0.564 G >> endstream /FormType 1 722.699 872.509 l /ProcSet[/PDF/Text] /Meta168 182 0 R << q /Descent -299 /Resources<< /Meta127 141 0 R /Meta240 Do endstream ET Q /Matrix [1 0 0 1 0 0] << /Length 54 >> ET 1 i << 1.007 0 0 1.007 411.035 330.484 cm Q 1.005 0 0 1.007 79.798 746.789 cm q q /Type /XObject /F3 17 0 R q /Length 68 /Font << 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 << 0.486 Tc 73 0 obj /Subtype /Form /BBox [0 0 88.214 16.44] Q 549.694 0 0 16.469 0 -0.0283 cm 0.458 0 0 RG /Length 12 3.742 5.203 TD /Meta285 Do /Font << 0.564 G /FormType 1 394 0 obj q /Matrix [1 0 0 1 0 0] /FormType 1 q /Matrix [1 0 0 1 0 0] ET /Font << Q endobj ET (D\)) Tj Q Q /BBox [0 0 17.177 16.44] /Length 63 ET Q endobj /Length 68 q /Matrix [1 0 0 1 0 0] /Subtype /Form 1 i /Flags 32 /Subtype /Form /Matrix [1 0 0 1 0 0] /Type /XObject /Subtype /Form /Subtype /Form 26.957 5.203 TD /FormType 1 /Meta408 424 0 R S stream /FormType 1 /F3 17 0 R , Prove the following endstream [(Negativ)16(e )] TJ /StemH 94 Testosterone is the primary sex hormone and anabolic steroid in males. 0 g 0.564 G 0.458 0 0 RG /Length 70 q 1 i stream (x) Tj Q (B\)) Tj 1.007 0 0 1.006 551.058 437.384 cm q stream << q 2. 0 G /Length 16 stream endstream /Resources<< q 351 0 obj /Length 16 >> /BBox [0 0 88.214 16.44] 0 g /ProcSet[/PDF] BT w/Honors. >> /Type /XObject /BBox [0 0 534.67 16.44] >> >> 0.524 Tc /Resources<< /ProcSet[/PDF/Text] q /ProcSet[/PDF/Text]
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