/Matrix [1 0 0 1 0 0] 0 G /F3 17 0 R 0 G /Font << Q endstream 87 0 obj 0.68 Tc >> /F1 14.682 Tf /Length 16 Q endobj /Meta214 228 0 R /Type /XObject /FormType 1 1 i q Q 0 G 722.699 546.541 l /Meta316 Do 1 i /FormType 1 endstream /Matrix [1 0 0 1 0 0] << 0 g >> /Font << Q >> stream << stream /FormType 1 q /BBox [0 0 88.214 16.44] 0 4.894 TD /BBox [0 0 88.214 16.44] /Meta85 Do 0.737 w >> q endobj Q 1 i /Type /XObject 0.564 G /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] Thirthy is equal to twice a number decreased by four = solve and check the equation? >> q ET /Length 58 Q 59 0 obj /Subtype /Form 159 0 obj << q q 274 0 obj /ProcSet[/PDF] stream /F3 12.131 Tf /Resources<< endobj 0.564 G Q >> >> q /Meta70 84 0 R /Length 69 /Subtype /Form 1 i endobj 431 0 obj q /Subtype /Form ET 0 g /Subtype /Form (x) Tj endstream /Meta409 Do /FormType 1 /Font << q 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 /Resources<< /Length 58 40 0 obj q 0 g >> /ProcSet[/PDF] >> /Subtype /Form /Meta351 365 0 R /F3 17 0 R endstream /Type /XObject /BBox [0 0 30.642 16.44] Q /Subtype /Form endobj /Type /Font /Subtype /Form 0.524 Tc q /BBox [0 0 88.214 16.44] /Subtype /Form 0.458 0 0 RG 0 w /Meta228 242 0 R 0 g endobj Q >> >> /Meta96 Do /FormType 1 /Meta314 328 0 R 0.458 0 0 RG 0.737 w endobj Notice that we used the variable \large {d} d in our equation to stand for our unknown value. 0 5.203 TD endobj /Length 54 (x) Tj endstream /Subtype /Form /Type /XObject (x ) Tj q Q /Meta11 Do endstream /Resources<< /Meta225 Do >> /Matrix [1 0 0 1 0 0] endobj q >> /Meta421 Do /BBox [0 0 88.214 35.886] /Font << >> 252 0 obj /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] /Resources<< << /BBox [0 0 88.214 16.44] Q /FormType 1 /Subtype /Form (C) Tj BT >> /Matrix [1 0 0 1 0 0] /Type /XObject ET endobj q Q /Matrix [1 0 0 1 0 0] /Resources<< 3.742 24.649 TD /Type /XObject /Subtype /Form Q /FormType 1 stream << 1.007 0 0 1.007 271.012 636.879 cm /BBox [0 0 88.214 16.44] q -0.463 Tw stream /Meta108 122 0 R /I0 Do /Subtype /Form /F3 17 0 R 0.68 Tc 17.234 5.203 TD q /Meta242 Do >> 0 G endstream /Font << >> BT q ET /BBox [0 0 15.59 29.168] >> /ProcSet[/PDF] 0.458 0 0 RG >> /ProcSet[/PDF/Text] >> /F3 17 0 R /Length 69 /Resources<< q stream /Meta123 Do /Resources<< 0 5.203 TD ET /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] BT 6 more than twice a number: 2x+6: two less than a number: x-2: the sum of 9 and a number: 9+x: two less than three times a number: 3x-2: a number subtracted from 12 . 1.007 0 0 1.007 411.035 583.429 cm /Meta190 Do /Meta152 166 0 R q /Length 54 Q q /F3 12.131 Tf endobj /Resources<< /Font << Q /Font << 1.007 0 0 1.006 411.035 510.406 cm 113 0 obj /Meta25 38 0 R 1.007 0 0 1.007 130.989 383.934 cm Q /Type /XObject << 0 g [(1)-25(0\))] TJ endobj /F3 17 0 R /Resources<< 0.737 w << Q 1 i /FormType 1 /F3 12.131 Tf 0 G stream q /Length 67 /Matrix [1 0 0 1 0 0] /FormType 1 q endstream 0 5.203 TD /Resources<< /F3 17 0 R /Meta312 326 0 R Q 1.007 0 0 1.007 551.058 523.204 cm -0.486 Tw endstream /Font << /ItalicAngle 0 /Font << Q 0 g (D\)) Tj << BT /Matrix [1 0 0 1 0 0] /FormType 1 Q Q /Matrix [1 0 0 1 0 0] /Meta238 252 0 R >> Q >> 6. endobj Q /Type /XObject q 282 0 obj 1 g 1 g Q /Type /XObject 392 0 obj /F3 12.131 Tf stream >> stream Q /Length 118 0 g >> Q /Resources<< << /Length 69 << << stream /Meta171 Do /Meta241 Do (\(x ) Tj /F3 12.131 Tf /Matrix [1 0 0 1 0 0] Twice a number when decreased by 7 gives 45. Q /FormType 1 >> 0.737 w q << Q /Length 69 q /Meta243 257 0 R 0 w endstream << q The ratio of a number to fifteen 4. << Q >> /ProcSet[/PDF] (x ) Tj /Subtype /Form 1 i Q q 1.007 0 0 1.007 130.989 636.879 cm Q /Type /XObject endobj /ProcSet[/PDF/Text] 0 w 1.007 0 0 1.007 130.989 636.879 cm BT /Type /FontDescriptor /Type /XObject 0 w >> endstream 0.307 Tc /Length 65 Q q /Resources<< -0.03 Tw 3.742 5.203 TD >> >> Q endobj Q << /F4 12.131 Tf Q >> /Matrix [1 0 0 1 0 0] Twice a number when decreased by 7 gives 45. 288 0 obj 1.014 0 0 1.007 251.439 636.879 cm /Font << stream /Length 16 /FormType 1 1.007 0 0 1.007 654.946 726.464 cm >> >> BT q -0.008 Tw >> /Resources<< q >> /BBox [0 0 88.214 16.44] >> /FormType 1 0 G /Font << endstream (40) Tj /MaxWidth 1397 /Meta240 Do /ProcSet[/PDF] /F4 36 0 R BT /Meta208 Do >> >> /Matrix [1 0 0 1 0 0] /Meta52 Do 0.458 0 0 RG endobj /Length 69 stream 1 i >> q >> /Meta245 259 0 R /Matrix [1 0 0 1 0 0] q stream 1.007 0 0 1.007 45.168 829.599 cm /Font << 1 i endobj q q /BBox [0 0 88.214 16.44] 0.564 G 0 g Q Q /Type /XObject 13 0 obj /ProcSet[/PDF/Text] stream /Matrix [1 0 0 1 0 0] 97 0 obj 0.369 Tc Q /Length 59 Q /Length 58 160 0 obj << ET 1 i >> Two fewer than a number doubled is the same as the number decreased by 38. q endobj Q /Meta153 167 0 R /Subtype /Form /FormType 1 endobj 0 G endstream /Subtype /Form /FormType 1 /ProcSet[/PDF/Text] endstream >> Q /Subtype /Form 0 g q 1.007 0 0 1.007 271.012 703.126 cm Q /Meta144 158 0 R 153 0 obj q /Meta2 9 0 R >> /FormType 1 /Resources<< Q 6.746 5.203 TD >> /BBox [0 0 88.214 16.44] /Parent 1 0 R << 0 g q How many points did Kobe score in the season? /Meta111 125 0 R 323 0 obj /FormType 1 Q 1.014 0 0 1.007 251.439 383.934 cm /ProcSet[/PDF/Text] 1 i endstream /FormType 1 >> stream /ProcSet[/PDF/Text] Q << endstream q /ProcSet[/PDF] >> q /Font << q /Type /XObject Q q endstream >> >> /Meta392 Do Q Q >> << 0.458 0 0 RG << /Matrix [1 0 0 1 0 0] BT /F3 12.131 Tf 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. 432 0 obj /Meta235 249 0 R 0 w /Meta137 151 0 R 213 0 obj Q endstream Q q 0 G stream BT /Type /XObject /Font << /Meta354 368 0 R 0.425 Tc 0.458 0 0 RG >> stream /Type /XObject BT /FormType 1 163 0 obj 1 g << >> S /Meta289 303 0 R /Meta202 216 0 R /Matrix [1 0 0 1 0 0] Q endstream >> BT /Type /XObject /BBox [0 0 88.214 16.44] Q [( and )16(a nu)26(mbe)18(r)] TJ /Font << /Subtype /Form /Resources<< Q 672.261 599.991 m /BBox [0 0 88.214 35.886] /F4 36 0 R >> q /Type /XObject q twice a number decreased by 58 13 - 3x B. twice a number a divided by three = 2a / 3. five times a number x minus four = 5x - 4. thrice the sum of a number x and six = 3 (x + 6) Add your answer and earn points. >> /Subtype /Form 1 g (-20) Tj q 408 0 obj >> Q q Q /Resources<< /Type /XObject 0 g 0.737 w >> Q /Length 69 We are asked to find the number, so, we could assign the number as "x". Q Q ET Q >> /Meta194 208 0 R /Matrix [1 0 0 1 0 0] 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. 0 g /BBox [0 0 30.642 16.44] stream q /Subtype /Form Q /Length 12 (8\)) Tj endobj 111 0 obj 0 w 1 i /Resources<< /Resources<< /Length 16 q Q ET /FormType 1 /Length 69 /Resources<< Get link; Facebook; Twitter; /FormType 1 Q stream << 66 0 obj q Q q 0 G /FormType 1 q 344 0 obj /Font << /FormType 1 >> q /Matrix [1 0 0 1 0 0] 1 i 404 0 obj Q 0 g >> Find the number. /ProcSet[/PDF/Text] >> 1 i >> 1.007 0 0 1.007 551.058 383.934 cm Q -0.486 Tw q /BBox [0 0 15.59 16.44] ET >> /F3 17 0 R /FormType 1 >> /Subtype /Form >> >> q >> /Type /XObject /Resources<< /Meta422 Do ET Solution: >> /Type /XObject q 0 G Q Q 0.737 w 1 i /Meta371 Do >> endobj 115 0 obj -0.382 Tw BT 1.007 0 0 1.007 130.989 583.429 cm /Meta325 Do 20.21 5.203 TD q Q A. Answer provided by our tutors. stream /Resources<< endobj >> 69 0 obj Q endstream ET Q /F3 17 0 R 1 i Q /Matrix [1 0 0 1 0 0] /Length 78 /Resources<< stream endstream (-11) Tj 0 g Q /ProcSet[/PDF/Text] /Resources<< Q >> /Meta375 Do 1 i q /Matrix [1 0 0 1 0 0] Q (-) Tj q /FormType 1 endstream >> 1 g /F3 17 0 R /Font << 0 g 0 G /Matrix [1 0 0 1 0 0] Q /Font << LAIing for a pizza and, soft drink. 57.656 5.203 TD /Font << 0 w endobj /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 0.564 G Explanation: let the number be n. then we can express division in 2 ways. endstream /Resources<< endstream ET endobj >> BT /ProcSet[/PDF/Text] q /Length 16 q q 1 i Q Mat >> q /Type /XObject 1.007 0 0 1.007 271.012 849.172 cm << endobj Q Q 1.502 5.203 TD /Font << /Matrix [1 0 0 1 0 0] /Subtype /Form 339 0 obj /Matrix [1 0 0 1 0 0] /Type /XObject stream Q This site is using cookies under cookie policy . /Matrix [1 0 0 1 0 0] BT /Subtype /Form Q >> /F3 17 0 R Ten divided by a number 5. 1 i >> q q /F3 17 0 R /XObject << (x) Tj endstream /Flags 32 Q /ID [] q Q q /Resources<< endstream >> /Meta393 Do 428 0 obj 0 g Q 0 G 372 0 obj 1 i 0 5.203 TD 20.21 5.203 TD << 0.737 w /Subtype /Form /Meta264 278 0 R BT 0 G BT Q /FormType 1 NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. Q /Meta106 120 0 R 0 G Q >> /Meta258 Do (- 4) Tj /Length 16 /ProcSet[/PDF/Text] Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] 0.838 Tc q >> >> q /Resources<< stream q q 1 i (9) Tj q stream /F1 7 0 R 0 G Q 0.737 w /Meta275 289 0 R endobj q q /Meta379 393 0 R endstream /BBox [0 0 88.214 35.886] q /Meta305 319 0 R /ProcSet[/PDF/Text] q << stream 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 1 i /Meta232 246 0 R Q /Resources<< 0 5.203 TD Q q << q endstream /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] stream /Type /XObject endobj endobj >> /Type /XObject >> Q 0 g /Length 12 q 1.008 0 0 1.007 654.946 293.596 cm endobj 1 i /Length 69 /Length 16 Q /Matrix [1 0 0 1 0 0] 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. Q 1.014 0 0 1.007 251.439 277.035 cm /F3 12.131 Tf BT 0 g stream endstream /Length 59 201 0 obj /Subtype /Form /Meta363 Do ET endobj 1.005 0 0 1.007 79.798 746.789 cm q 0.369 Tc /Type /XObject >> endobj q >> /BBox [0 0 639.552 16.44] 1 g >> /Meta223 237 0 R 183 0 obj /Type /XObject /F3 12.131 Tf /Meta343 357 0 R /Meta205 Do 229 0 obj 0 4.894 TD endstream /Length 69 -0.021 Tw ET q 47.933 5.203 TD q /Matrix [1 0 0 1 0 0] q Q q /XObject << /Matrix [1 0 0 1 0 0] ET Q /FormType 1 1 i /Type /XObject >> /Type /XObject /F3 12.131 Tf 342 0 obj /ProcSet[/PDF] Q /Type /XObject q endstream 0 G 1.005 0 0 1.007 102.382 799.486 cm /F2 11 0 R 1 i /F3 12.131 Tf q q 1 i /Subtype /Form 0 G >> 20.975 5.336 TD /BBox [0 0 639.552 16.44] (D) Tj stream 1.014 0 0 1.006 251.439 690.329 cm 0.486 Tc /Font << ET 0 G << 1.007 0 0 1.007 654.946 653.441 cm /Font << /Resources<< /Type /XObject /Type /XObject /F4 36 0 R Q endstream 1 i 0 w 0 g 0 g endstream q >> q Q /FormType 1 q /I0 Do q /Meta78 Do BT q Q /Type /XObject /Matrix [1 0 0 1 0 0] endstream >> /Type /XObject q 0 G Q Q /BBox [0 0 88.214 16.44] q Q /Length 12 /Meta298 312 0 R q 0.369 Tc /BBox [0 0 30.642 16.44] q /Length 69 Q >> 0.564 G /ProcSet[/PDF/Text] stream /Meta423 Do stream q 1.007 0 0 1.007 654.946 599.991 cm q /Subtype /Form >> 1 i 1 i Q 0 g >> /F1 7 0 R /F3 17 0 R /Type /XObject /FormType 1 /Font << Q endobj 1 i stream Q Q q /Length 69 /Font << endstream >> Q q << /Meta391 407 0 R /Type /XObject endstream /ProcSet[/PDF/Text] /Meta222 Do Q endobj q q /Resources<< 0 G (\)) Tj /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> Q 0.737 w 0.564 G /Resources<< /BBox [0 0 673.937 68.796] 1 i stream Q 1.007 0 0 1.007 411.035 277.035 cm /Subtype /Form /Subtype /Form Q 1 i Q /Meta312 Do /F3 12.131 Tf endstream /Type /XObject Q 0 g endobj /I0 51 0 R >> /Meta400 416 0 R /BBox [0 0 88.214 16.44] Q >> >> 378 0 obj q BT ET /Subtype /Form /BBox [0 0 88.214 16.44] 0.425 Tc /F3 17 0 R >> /Length 294 /Meta407 Do /Resources<< /BBox [0 0 30.642 16.44] /BBox [0 0 15.59 16.44] q A number = an unknown number which can be represented by a variable, usually x. << /Meta113 Do /FormType 1 /Resources<< /F3 17 0 R /Meta68 Do >> 0 g 1.502 5.203 TD 76.394 5.203 TD endstream if the solution of an equation is x=-2, what could the original equation be? Q Q 0 G endstream Q /Font << /FormType 1 /Font << Q 284 0 obj Q << q 0.458 0 0 RG 1 i 0.564 G >> /Matrix [1 0 0 1 0 0] (13) Tj [(E)-14(le)-23(ven)] TJ 0 g /Subtype /Form 0 g 0 G Q /Meta286 Do 0 w 20.21 5.203 TD (-11) Tj /Resources<< /Subtype /Form 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 0 g >> /ProcSet[/PDF] endstream /Meta344 358 0 R >> endstream 0 g /Matrix [1 0 0 1 0 0] Objective a: Reading and translating word problems 3 There are a couple of special words that you also need to remember. 1.005 0 0 1.007 79.798 829.599 cm /Type /XObject 0 G >> << Q 1 i endobj q /Meta226 Do /ProcSet[/PDF] /BBox [0 0 88.214 16.44] stream 1.007 0 0 1.007 130.989 383.934 cm 0 5.203 TD /Meta429 445 0 R 2 See answers pharry1800 pharry1800 Answer: 2n-58 Step-by-step explanation: olivbreadh olivbreadh Answer: 2x-116 or 2(x-58) Step-by-step explanation: Transalate it to numbers and operations: => 2(x-58) => 2x-116 You won't have a solid number since its not an equation. q 122 0 obj /Matrix [1 0 0 1 0 0] the quotient of twenty and a number a.) >> endstream /FormType 1 q q /Meta27 Do to represent the numbers. /Meta116 130 0 R 291 0 obj /Subtype /Form /Meta197 Do 0 w endobj << 1.502 7.841 TD Q /Length 118 1 i 1.007 0 0 1.007 45.168 763.351 cm 0.564 G BT 0 G /Length 16 q /Length 12 /Meta395 411 0 R q BT Q endstream >> q /Meta227 241 0 R [(thir)17(te)15(en)] TJ >> << TJ /Meta99 Do 0 g stream << ET /Resources<< 0 g q /Matrix [1 0 0 1 0 0] /Font << /Font << (\)) Tj 0 5.203 TD /BBox [0 0 534.67 16.44] ET 0 g >> 0 g 1 i /Font << endobj /Resources<< /FormType 1 0 g 1 i 0 g Q 0 5.203 TD q /BBox [0 0 88.214 16.44] /BBox [0 0 23.896 16.44] q q /Subtype /Form endstream 32.939 5.203 TD 232 0 obj /FormType 1 (-) Tj Q 0 w q q Kobe scored 85 points in a basketball game. ET << BT endobj /F3 17 0 R endstream (-) Tj (x) Tj << /ProcSet[/PDF] q /Resources<< Q /Meta428 444 0 R Q /Type /XObject stream << /FormType 1 << Q endstream 352 0 obj 0 g /Type /XObject 1.014 0 0 1.007 391.462 636.879 cm endobj /Type /XObject >> 19.474 20.154 l 0 G /Length 69 endobj /Subtype /Form Q q 0 w endstream /ProcSet[/PDF] Q Q 30.699 5.203 TD /Length 58 Q /Resources<< Q q Q << 443 0 obj Q /Meta384 Do /F3 17 0 R >> endobj 0 g /ProcSet[/PDF/Text] << ET /F1 12.131 Tf >> >> /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] q 0.737 w Q stream ET endstream 0.564 G >> /Type /XObject /F3 17 0 R >> endobj 0 g /Meta255 269 0 R Q /BBox [0 0 88.214 16.44] ET ET q /Type /XObject ET /Meta59 Do Q Q /Resources<< /FormType 1 /Resources<< 1.007 0 0 1.007 271.012 277.035 cm endstream /ProcSet[/PDF/Text] 101.849 5.203 TD (C\)) Tj 0 g /Font << /Length 69 /ProcSet[/PDF/Text] 0 5.203 TD /Resources<< q /BBox [0 0 88.214 16.44] /Type /XObject /Meta76 Do q endobj Q endobj endobj >> /Length 69 /LastChar 45 q 1.007 0 0 1.007 130.989 776.149 cm 5 0 obj 1.005 0 0 1.007 79.798 829.599 cm >> /BBox [0 0 534.67 16.44] q /ProcSet[/PDF] /Type /XObject endobj Q 0 g 0 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q Q: when six times a number is decreased by 4, the result is 8. 0 g Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 654.946 546.541 cm /FormType 1 /BBox [0 0 30.642 16.44] q /Meta198 Do /Type /XObject 295 0 obj /Length 16 0 g q /Meta243 Do endobj 1 i 0.564 G 0 g ET 233 0 obj q q >> 173 0 obj /FormType 1 /Matrix [1 0 0 1 0 0] q endobj >> /Length 118 /Type /FontDescriptor /FormType 1 /Length 69 [( subt)-17(racted fr)-14(om a )-16(number)] TJ /Resources<< /Meta347 Do /Matrix [1 0 0 1 0 0] Q /FormType 1 /F3 12.131 Tf /FormType 1 >> 1 i /Meta55 69 0 R /Meta124 Do Percent Change = (Decrease First Value) x 100% << << /F3 12.131 Tf Q q /Matrix [1 0 0 1 0 0] endobj q 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 . q q /Type /XObject q q /BBox [0 0 88.214 16.44] << 1.007 0 0 1.007 130.989 330.484 cm 0.737 w stream << ET A number divided by six is eight: (k / 6) = 8. Q stream stream /Type /XObject Q /Meta45 Do /Matrix [1 0 0 1 0 0] 0 g q Q Q BT 0 G /Subtype /Form /Type /XObject ET /Matrix [1 0 0 1 0 0] 401 0 obj /BBox [0 0 534.67 16.44] /BaseFont /PalatinoLinotype-Roman /Parent 1 0 R /Matrix [1 0 0 1 0 0] endstream 85 0 obj endstream /Type /XObject /ProcSet[/PDF] [tex]\sin (\pi -x)=\sin x[/tex]. << Q (C\)) Tj q /FormType 1 /Meta8 Do /Matrix [1 0 0 1 0 0] 0 20.154 m 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? stream 0 G >> /Matrix [1 0 0 1 0 0] BT 0 g 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 . /Meta63 Do /Subtype /TrueType >> stream Q 239 0 obj stream >> >> q /Meta412 428 0 R /Ascent 1050 0 20.154 m /Meta147 161 0 R 20.21 5.203 TD q Q 0 G stream Q 1 g /Matrix [1 0 0 1 0 0] 0.564 G /F3 12.131 Tf /Font << /Font << q S /Length 12 Q 1.007 0 0 1.007 551.058 636.879 cm /Resources<< q ET /Matrix [1 0 0 1 0 0] 0 G /Length 59 BT /Length 16 /Resources<< >> /Type /XObject /Type /XObject [(Negativ)16(e )] TJ /Type /XObject 333.269 5.488 TD 0.524 Tc /FormType 1 36 0 obj << /Meta134 148 0 R q 0.51 Tc /Subtype /Form BT q Q >> stream Q >> /Meta318 Do /Length 59 /ProcSet[/PDF/Text] 1 i 1.007 0 0 1.007 411.035 636.879 cm 250 0 500 500 500 500 500 500 0 0 500 500 278 0 0 0 >> stream /F3 17 0 R 0 g Q /F3 17 0 R ( x) Tj stream ET 4 0 R >> 426 0 obj /Length 59 q << /ProcSet[/PDF/Text] 1.005 0 0 1.007 102.382 726.464 cm /BBox [0 0 673.937 14.853] /Matrix [1 0 0 1 0 0] >> stream q 138 0 obj /Matrix [1 0 0 1 0 0] (-23) Tj 340 0 obj >> Q /F3 12.131 Tf 0.68 Tc /BBox [0 0 88.214 16.44] 1 i endstream q endstream << /Type /XObject stream The quotient of a seven and a number 9. 0.838 Tc Q << /Subtype /Form /Meta277 291 0 R << 65.906 4.894 TD Q /Matrix [1 0 0 1 0 0] BT Q Q Q 0 G ET 0 g q Q /F3 12.131 Tf 14.966 20.154 l /XHeight 476 /ProcSet[/PDF] "49 . /Meta29 42 0 R Q /Matrix [1 0 0 1 0 0] >> /AvgWidth 459 0.737 w /Meta139 153 0 R /Subtype /Form /Subtype /Form >> Q 0 G 1 g q >> stream 1 i 0.458 0 0 RG 0 G q 1.005 0 0 1.007 79.798 730.228 cm /Length 139 0 G 1 i /Resources<< 1 i endstream (B\)) Tj 0 g q endstream BT q You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. (\)) Tj /Resources<< endobj /FormType 1 /Meta221 235 0 R 0 g /Type /XObject /Meta72 86 0 R /Resources<< Q All steps. 0.458 0 0 RG >> /FormType 1 Q /Meta164 Do 1 i 0.458 0 0 RG /FormType 1 /F3 12.131 Tf /FormType 1 /Subtype /Form Q Q /Meta148 162 0 R BT /Meta409 425 0 R /Length 118 /Type /XObject 0.463 Tc /Meta110 124 0 R /Type /XObject /Meta357 Do /F3 17 0 R /BBox [0 0 88.214 16.44] 13.493 5.336 TD q >> endstream 0 g Q << Q /I0 51 0 R Q BT /FormType 1 /FormType 1 >> << /Subtype /Form q /BBox [0 0 88.214 16.44] /Resources<< /BBox [0 0 639.552 16.44] /Meta47 61 0 R /I0 51 0 R 0 g /Subtype /Form Q q /FormType 1 /Resources<< q q >> q /Type /Catalog >> 200 0 obj /F3 12.131 Tf (2) Tj /Font << Q q ET 0 G /ProcSet[/PDF/Text] >> ET /Font << 1.007 0 0 1.006 130.989 437.384 cm /Type /XObject 1 i 2.238 5.203 TD ET 26.219 5.203 TD >> << q /Resources<< /Font << /Resources<< 0 G /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. /Resources<< >> stream 1.014 0 0 1.007 391.462 383.934 cm /Meta256 270 0 R 0.737 w q q Q endobj << >> Q /F3 12.131 Tf BT BT << If you are unsure of the county, call the Administrative Office of the Court at (919) 890-1000.Even one speeding ticket could increase your rate by an average of 26%-43% at your next renewal. /F3 12.131 Tf /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] /Length 69 Was this answer helpful? << /Resources<< >> 0 w q 317 0 obj Q 407 0 obj Q 0 G q /FormType 1 41.186 5.203 TD q endstream endobj /Meta273 287 0 R /Meta260 Do /Meta353 Do BT << /Meta315 329 0 R 1.502 24.649 TD 0 g Q 0 g /Subtype /Form endstream q /Meta326 340 0 R /Meta13 Do /Subtype /Form 1.007 0 0 1.006 551.058 836.374 cm BT ET 0 g 0.564 G Q /F3 17 0 R /Meta21 32 0 R /Font << 0.458 0 0 RG q (+) Tj >> /Subtype /Form 1 i endstream Q /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] q q /Subtype /Form q BT 1.014 0 0 1.007 391.462 776.149 cm q /Subtype /Form /BBox [0 0 15.59 16.44] q BT /BBox [0 0 88.214 35.886] 0 G /Type /XObject 0 w 0.458 0 0 RG /FormType 1 /Meta209 Do q /Length 12 1 i /Resources<< /Length 16 /F1 7 0 R 289 0 obj /Matrix [1 0 0 1 0 0] stream endstream /Length 69 q endstream /Meta330 344 0 R >> 0 g 0.458 0 0 RG BT (B\)) Tj /Matrix [1 0 0 1 0 0] BT /Matrix [1 0 0 1 0 0] BT /Resources<< Q >> (B) Tj Q endobj /ProcSet[/PDF] /FormType 1 /Font << /Matrix [1 0 0 1 0 0] /Font << Q 0.369 Tc q /FormType 1 1.005 0 0 1.007 102.382 599.991 cm >> q 1.007 0 0 1.007 411.035 849.172 cm /Type /XObject /Meta104 Do /Meta138 Do

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