1.007 0 0 1.007 271.012 776.149 cm >> q 0.737 w /FormType 1 Q Q This site is using cookies under cookie policy . 1 i >> Q endstream endstream 0 G Q 0.564 G 184 0 obj 0 g 288 0 obj 98 0 obj /Font << q Q /Subtype /Form endstream (x) Tj Q /Subtype /Form stream 1 i /F3 12.131 Tf 0.786 Tc 0 g /FormType 1 << q Q 0 g BT << >> 167 0 obj 197 0 obj /FormType 1 Q 1 g 0 G BT q 279 0 obj 0.738 Tc BT q endobj Q q 89 0 obj /Subtype /Form 1.007 0 0 1.007 551.058 703.126 cm /Font << /BBox [0 0 17.177 16.44] /Meta217 231 0 R Q /ProcSet[/PDF] /Meta168 Do 281 0 obj stream Q /Matrix [1 0 0 1 0 0] 1 i >> ET 0 5.203 TD Q q /F4 12.131 Tf /Matrix [1 0 0 1 0 0] Q /Resources<< /Type /XObject Q Kobe scored 85 points in a basketball game. /Resources<< 199 0 obj /Font << q /ProcSet[/PDF] 0.564 G Q /FormType 1 0.51 Tc /Resources<< /Matrix [1 0 0 1 0 0] q >> Q /BBox [0 0 534.67 16.44] (5) Tj /Subtype /Form 1.007 0 0 1.007 551.058 277.035 cm endstream q q 0.564 G q ET q << >> /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /Subtype /Form /Length 68 1 i /Meta224 238 0 R 1 i << BT Q 1.007 0 0 1.006 130.989 437.384 cm endobj endstream stream Q 1 i q endobj << /BBox [0 0 15.59 16.44] q q >> endstream /ProcSet[/PDF/Text] 1 i Q q 1.502 5.203 TD /Subtype /Form /Type /XObject /Subtype /Form >> Two speeding tickets could increase your rate by 58% at your next renewal. 250 0 500 500 500 500 500 500 0 0 500 500 278 0 0 0 /BBox [0 0 30.642 16.44] 0 w /Font << q /ProcSet[/PDF/Text] 0.369 Tc /Meta177 191 0 R /Type /Catalog >> 47.933 5.203 TD /Font << Q >> 1 g /ProcSet[/PDF] 20.21 5.203 TD 0 G 1.008 0 0 1.007 654.946 293.596 cm 0.297 Tc /Font << /ProcSet[/PDF/Text] /Length 103 /Meta305 319 0 R 0.425 Tc endobj >> Q << /Matrix [1 0 0 1 0 0] BT /Matrix [1 0 0 1 0 0] /FormType 1 Q 0 5.203 TD /Subtype /Form << /Type /XObject Q >> /Meta185 Do /Subtype /Form 0 5.203 TD /FormType 1 Q ET stream /BBox [0 0 88.214 16.44] /Resources<< (-11) Tj /F4 12.131 Tf >> /Meta248 262 0 R Q q /BBox [0 0 88.214 16.44] endobj 1 i endstream Thirthy is equal to twice a number decreased by four = solve and check the equation? endobj 0.564 G /Type /XObject (A\)) Tj /FormType 1 stream 0 g 0.838 Tc >> stream 0 4.894 TD 1 g /F1 12.131 Tf /ProcSet[/PDF] /BBox [0 0 88.214 16.44] q (-) Tj 1.007 0 0 1.007 130.989 277.035 cm /Resources<< /Resources<< >> stream 1 i /Meta47 Do /Resources<< 1.007 0 0 1.007 67.753 599.991 cm /F3 12.131 Tf 0.271 Tc 0 g q endstream >> q >> Q q q /Type /XObject >> /Meta422 438 0 R /FormType 1 >> 0 G BT /Meta127 141 0 R Q Q Q Q /Font << /ProcSet[/PDF/Text] 1 i << 0.738 Tc endobj Q /ProcSet[/PDF/Text] << /Subtype /Form 1 g /BBox [0 0 88.214 16.44] 0 G endstream /Subtype /Form /Subtype /Form /BBox [0 0 17.177 16.44] /Matrix [1 0 0 1 0 0] Q ET q /Length 78 BT /ProcSet[/PDF] /Meta52 66 0 R >> /Type /XObject 98.843 5.203 TD endstream /Resources<< endstream endstream 1 i /BBox [0 0 88.214 16.44] /FormType 1 /F3 12.131 Tf 0 g 0 g 0 g /FormType 1 /Font << /BBox [0 0 88.214 16.44] /Subtype /Form /Type /XObject >> /ProcSet[/PDF/Text] Q >> 0.458 0 0 RG 1.007 0 0 1.006 411.035 437.384 cm /ProcSet[/PDF] /Resources<< 1 i /Font << Q /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] Q /ProcSet[/PDF/Text] Q 0 G /Length 78 q /Resources<< /Subtype /Form Q q 1.005 0 0 1.007 102.382 726.464 cm 0.369 Tc 1 i 0.786 Tc /F3 17 0 R 0 g q (C\)) Tj >> >> The result is 8 less than 10 times the number. /Matrix [1 0 0 1 0 0] 0 G /Meta5 Do /F4 36 0 R 1.007 0 0 1.007 130.989 277.035 cm /Subtype /Form 1.014 0 0 1.007 251.439 523.204 cm /F3 12.131 Tf q BT q /Matrix [1 0 0 1 0 0] /Meta407 Do ET /Type /XObject >> /Length 96 /BBox [0 0 15.59 29.168] /Meta237 251 0 R stream ET /BBox [0 0 30.642 16.44] /ProcSet[/PDF] >> 0.564 G 1.007 0 0 1.007 130.989 583.429 cm /Resources<< /Resources<< Q /Subtype /Form /Type /XObject 0 G /Type /XObject /Type /XObject /Resources<< Q /Subtype /Form 0 g /ProcSet[/PDF/Text] (B\)) Tj Q >> Q /FormType 1 /Subtype /Form 1 i endstream << endstream /Matrix [1 0 0 1 0 0] stream Q /BBox [0 0 88.214 35.886] 101.849 5.203 TD Twice a number decreased by . >> /XObject << Q /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] 0 5.203 TD /BBox [0 0 17.177 16.44] /BBox [0 0 17.177 16.44] Q >> ET Q Negative thirteen decreased by 3 times a number x. A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. Q 1.502 5.203 TD /Resources<< /Meta376 Do /Meta244 258 0 R a.) 0 w 4.506 24.649 TD << Q /F3 12.131 Tf >> 1 i << /MissingWidth 252 >> Q 23.952 4.894 TD << /F3 17 0 R >> /BBox [0 0 534.67 16.44] /F3 12.131 Tf (-) Tj /Font << >> /Meta335 349 0 R Just type into the box and your calculation will happen automatically. /FormType 1 stream /ProcSet[/PDF/Text] 0.564 G << q /ProcSet[/PDF/Text] 1 i /F1 12.131 Tf 0 G /F1 12.131 Tf << 1.007 0 0 1.007 130.989 636.879 cm q q >> /Meta130 144 0 R /F3 12.131 Tf << >> q 1.007 0 0 1.007 271.012 849.172 cm >> /Meta98 Do D. Twice a number decreased by ten is less than 24. if the solution of an equation is x=-2, what could the original equation be? BT 0 5.203 TD /Font << endstream >> << /Font << q endstream 1.502 5.203 TD /Matrix [1 0 0 1 0 0] /F4 12.131 Tf ET stream << (B\)) Tj /Meta199 Do /Subtype /Form q 1.007 0 0 1.007 271.012 277.035 cm /Resources<< Q q 0 g /FormType 1 Q /Meta123 Do /Contents [399 0 R] /BBox [0 0 88.214 35.886] /F3 12.131 Tf q /ProcSet[/PDF] /ProcSet[/PDF/Text] 0.737 w >> /Meta281 295 0 R /Length 139 /Length 294 stream 1.007 0 0 1.007 271.012 330.484 cm 1.007 0 0 1.006 411.035 510.406 cm Q endobj /Subtype /Form /BBox [0 0 17.177 16.44] Q Q /Resources<< 22.478 5.336 TD /Meta335 Do >> endstream stream << Q (x) Tj /Meta257 271 0 R 1 i Q endobj /Subtype /Form stream /Ascent 891 165 0 obj /Subtype /Form 0.458 0 0 RG 0 g << Q Q 0.68 Tc /Resources<< /F3 17 0 R Q Q /Meta122 136 0 R /Length 69 endobj 0 w 100 0 obj q Q Q << /F3 17 0 R >> BT 0.486 Tc /Type /FontDescriptor endstream /Type /XObject /ProcSet[/PDF/Text] /Type /XObject /BBox [0 0 673.937 15.562] 1.502 7.841 TD q /Resources<< 0 G q /Meta25 38 0 R >> q 1.005 0 0 1.007 79.798 846.161 cm q /Type /XObject >> 191 0 obj Q q << 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 : . Q << >> /Matrix [1 0 0 1 0 0] /F3 17 0 R /BBox [0 0 88.214 35.886] q 1 g /Matrix [1 0 0 1 0 0] BT >> /Matrix [1 0 0 1 0 0] /Meta190 204 0 R Q >> 355 0 obj 0.369 Tc 1.014 0 0 1.007 111.416 636.879 cm 1.007 0 0 1.007 67.753 872.509 cm 1.007 0 0 1.007 411.035 849.172 cm endstream 0 g /Resources<< stream /ProcSet[/PDF/Text] /Meta421 Do 278 0 obj /Meta250 Do Q q >> q 0.458 0 0 RG /ItalicAngle 0 >> /Matrix [1 0 0 1 0 0] /F3 17 0 R Q ET /F1 12.131 Tf /Length 69 0.68 Tc /F4 12.131 Tf /ProcSet[/PDF/Text] /Resources<< ET << 1 i >> /Length 16 >> /Meta278 Do q /ProcSet[/PDF] (C\)) Tj >> Q 0 g Q 265 0 obj >> 0 g %%EOF. 321 0 obj /Resources<< q Q /Meta277 291 0 R endobj q /Font << /Length 65 /Subtype /Form (13) Tj 1.007 0 0 1.007 551.058 523.204 cm /ProcSet[/PDF] q 178 0 obj Q /BBox [0 0 30.642 16.44] ET 241 0 obj /Font << q 2x - y = 6. x + 3y = -25. Q (+) Tj /FormType 1 Q /Subtype /Form 1 i 0 G /ProcSet[/PDF] /FormType 1 endobj /BBox [0 0 88.214 35.886] << /Meta407 423 0 R The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. endstream /F4 36 0 R endstream 1 g q q /Type /XObject >> /Matrix [1 0 0 1 0 0] ( decreased by ) Tj /F3 12.131 Tf /Resources<< 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. Q /Length 69 stream ET 1 i Was this answer helpful? /Subtype /Form /Meta275 Do Q stream << q /Meta68 82 0 R /Meta317 Do 0.524 Tc Q q /Type /XObject /Resources<< /ProcSet[/PDF] Q ET stream /Meta240 Do /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] q Q q 170 0 obj q /BBox [0 0 30.642 16.44] /Meta67 Do /Meta288 Do q /Matrix [1 0 0 1 0 0] 1 i endobj 0 g /Resources<< Q endobj << q 1 g 1 i << /Length 67 q >> /FormType 1 BT << /BBox [0 0 30.642 16.44] BT Q /Subtype /Form 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. Q BT stream Q 0 w q 0.737 w /Resources<< BT /F3 17 0 R 68 0 obj endstream /Font << /Meta85 99 0 R q q /Meta169 Do 1 i 151 0 obj /Type /XObject /Meta328 342 0 R /F3 17 0 R endstream q Q Find the number. /Font << /Type /XObject Q 0 g /Meta202 Do Q endstream Q 1.014 0 0 1.007 531.485 636.879 cm q 1.007 0 0 1.007 271.012 277.035 cm 109 0 obj /Font << /F1 7 0 R 313 0 obj /ProcSet[/PDF] /Resources<< Q /Font << 300 0 obj q /Resources<< /ProcSet[/PDF] -0.486 Tw 1 g /BBox [0 0 88.214 16.44] /Resources<< << endstream q /FormType 1 /Filter [/CCITTFaxDecode] q << Q 228 0 obj >> 1 i q /Type /XObject /Subtype /Form /Meta117 Do q 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. >> >> /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 796.475 cm endobj << stream /Subtype /Form 1 i ET Q >> /Meta100 Do endobj /Matrix [1 0 0 1 0 0] twice - means that a number (the unknown value) is multiplied by 2 2 With these in mind, let's write our algebraic equation. endobj /ProcSet[/PDF] 311 0 obj >> q Q /Font << q stream /Meta392 Do q Q q Q << 74 0 obj q /Meta157 171 0 R /Meta220 Do /Font << >> Q /Subtype /Form 1 i >> Q /Type /XObject >> /FormType 1 q 0 20.154 m q 417 0 obj /Matrix [1 0 0 1 0 0] /Font << 1 i /F1 12.131 Tf endstream 411 0 obj q stream 0 5.203 TD endstream 0.738 Tc q BT /Meta210 Do /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 411.035 330.484 cm /Meta206 Do ET 1 i 0.737 w 1.007 0 0 1.007 551.058 330.484 cm << q /Subtype /Form /FormType 1 1 i /Type /XObject /FormType 1 /Length 16 (38) Tj Q /Matrix [1 0 0 1 0 0] q 0 5.203 TD 0 G 1.007 0 0 1.007 271.012 849.172 cm (x) Tj >> Q endstream << /F3 12.131 Tf 157 0 obj endobj Q /Matrix [1 0 0 1 0 0] endobj 0 5.203 TD 0.458 0 0 RG /CapHeight 692 q stream /Length 69 q /Resources<< /Meta349 363 0 R /Meta331 345 0 R /Type /XObject /Type /XObject >> q Q Q >> >> /Length 60 endstream >> /F3 12.131 Tf BT stream ET 1 i /F1 14.682 Tf endstream /BBox [0 0 673.937 14.853] >> /F3 12.131 Tf q 0 g 1 i /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] 1 i q q Q Q 1.005 0 0 1.007 102.382 599.991 cm 0 G 1.007 0 0 1.007 411.035 583.429 cm q q /Meta225 239 0 R 1 i /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /Subtype /Form /Meta190 Do /Meta59 Do /Length 16 stream ET /Length 69 Q 1 i Expert Answer. /FormType 1 /BBox [0 0 88.214 16.44] endstream Q /Meta419 Do -0.03 Tw /Resources<< endstream 0 w 0.786 Tc BT >> /Type /XObject /Length 69 /Meta284 Do 49 0 obj /Meta142 Do /Meta15 Do << 0.369 Tc endobj 11.99 8.18 TD endobj (-) Tj In other terms, 52-nx The problem is asking that you subtract twice a number from 52. /Meta179 Do /Length 73 Q >> << /Type /XObject q /Matrix [1 0 0 1 0 0] /FormType 1 q >> Q q /Type /XObject q 1.007 0 0 1.006 130.989 437.384 cm /F3 12.131 Tf >> 0 g << Q /Meta251 265 0 R /ProcSet[/PDF] << /Meta183 197 0 R 1 i /Font << (C\)) Tj 1.007 0 0 1.007 271.012 523.204 cm /Resources<< << endobj >> 0.425 Tc /Matrix [1 0 0 1 0 0] 426 0 obj /Meta226 Do 1 i /Resources<< endstream q Q 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 /Matrix [1 0 0 1 0 0] /Resources<< /ProcSet[/PDF] /Meta75 89 0 R 0 g /FormType 1 /BBox [0 0 30.642 16.44] /Resources<< [(Testnam)19(e: 1.12 T)16(RAN)16(SLATING )17(ALG EXPRES)21(SIONS 2)] TJ /ProcSet[/PDF/Text] >> >> /I0 51 0 R /FormType 1 0 g /FormType 1 >> /Subtype /Form /I0 Do q q /Resources<< /Meta270 Do << q /BBox [0 0 88.214 35.886] >> 0 5.203 TD >> >> Q -0.486 Tw /Type /XObject /Type /XObject (x ) Tj Q /ProcSet[/PDF/Text] ET 0 g q /Meta191 Do Q << ET q 0 w >> q /Meta116 Do 1 i 1.014 0 0 1.006 531.485 836.374 cm >> >> 1 i Q /Length 16 /Subtype /Form 1.014 0 0 1.006 391.462 510.406 cm ET /Meta202 216 0 R /Meta136 Do /F3 12.131 Tf /Type /XObject 1.007 0 0 1.007 654.946 400.496 cm BT /Meta154 Do Q /BBox [0 0 88.214 16.44] /Meta148 Do /ProcSet[/PDF/Text] endobj << >> 0 G /ProcSet[/PDF/Text] /Resources<< Q >> 1 i endobj /Matrix [1 0 0 1 0 0] 0 G /Meta394 Do 0.564 G 0 g << endobj /Meta209 223 0 R /Type /XObject >> 21.713 20.154 l Q 81 0 obj /FormType 1 Q q /Meta368 Do >> 212 0 obj /F4 36 0 R >> /Length 13 /Matrix [1 0 0 1 0 0] q /Meta155 169 0 R 0 w /Resources<< 0 w /Type /XObject /Meta330 Do /ProcSet[/PDF] /FormType 1 /Meta29 Do /F4 36 0 R /Matrix [1 0 0 1 0 0] 396 0 obj /Font << 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. ET 0.737 w << /Subtype /Form /F3 12.131 Tf 0 g /FormType 1 endobj Q >> Formula - How to Calculate Percentage Decrease. 333.269 5.488 TD >> 319 0 obj /Resources<< 1.007 0 0 1.006 551.058 763.351 cm q /FormType 1 /Type /XObject Q 1 i q q Q /Type /XObject Q /Resources<< stream /Resources<< q 0 g 0 G /BBox [0 0 88.214 16.44] 43.426 5.203 TD >> /Meta254 Do A link to the app was sent to your phone. 140 0 obj q q /FormType 1 >> /Subtype /Form Q q /Length 69 /Subtype /Form 0 w endstream q /Meta111 Do /Type /XObject 218 0 obj Q /ProcSet[/PDF] 0.737 w /Type /XObject Q /Matrix [1 0 0 1 0 0] endstream 1 i << /F1 12.131 Tf << /Subtype /Form (9\)) Tj /ProcSet[/PDF/Text] /FormType 1 endobj 1 i /Subtype /Form /BBox [0 0 88.214 35.886] /Subtype /Form 1 i ET 0 G (-) Tj BT /Meta213 227 0 R q /Length 16 >> 1.007 0 0 1.007 654.946 546.541 cm endstream (D\)) Tj 0 w stream /Resources<< 0 g 1.007 0 0 1.007 271.012 776.149 cm /Resources<< /Meta261 Do /FormType 1 /Matrix [1 0 0 1 0 0] 582 546 601 560 395 424 326 603 565 834 516 556]>> 0 w /ProcSet[/PDF/Text] /XObject << /ProcSet[/PDF/Text] /F3 12.131 Tf 142 0 obj q Twice = two times, double. /Meta60 74 0 R 1 i /Subtype /Form 0 g q 0 G /Subtype /Form 0.458 0 0 RG /Matrix [1 0 0 1 0 0] /Length 12 383 0 obj 1.007 0 0 1.006 130.989 437.384 cm /Font << S Q /ProcSet[/PDF/Text] stream endobj 41 0 obj Q endobj << stream stream /Length 16 /Resources<< /F3 12.131 Tf /Resources<< ET 1 g /Length 16 356 0 obj /CapHeight 476 /Type /XObject q 0.738 Tc 21 0 obj /F3 17 0 R 0 g 0.738 Tc << /Subtype /Form q 52 0 obj /StemV 94 endobj endobj Q 0.564 G /BBox [0 0 88.214 16.44] 243 0 obj endstream 0 5.203 TD /FormType 1 0 g q >> /Matrix [1 0 0 1 0 0] Q 0 5.203 TD BT /BBox [0 0 88.214 16.44] q 1 i 0 g /ProcSet[/PDF/Text] 205.199 4.894 TD q /Length 57 /ProcSet[/PDF/Text] /Resources<< q S 1 i q /Matrix [1 0 0 1 0 0] >> %PDF-1.4 /Meta239 Do /Length 70 q /FormType 1 >> Q 55 0 obj Q /F3 12.131 Tf >> Q ET Choose the correct one. /FormType 1 Q /Type /XObject /BBox [0 0 15.59 29.168] q << stream >> /F2 11 0 R /Meta269 283 0 R Check out a sample Q&A here. q Q 672.261 546.541 m /ProcSet[/PDF/Text] 422 0 obj /Matrix [1 0 0 1 0 0] >> /BBox [0 0 15.59 16.44] >> 0.564 G << >> (x ) Tj >> 1.007 0 0 1.007 411.035 583.429 cm /ProcSet[/PDF/Text] 1.007 0 0 1.007 67.753 293.596 cm /FormType 1 1.014 0 0 1.006 531.485 510.406 cm /ProcSet[/PDF/Text] 0 5.203 TD /F1 12.131 Tf Q /Length 95 q /Font << BT /Type /XObject [(The )-16(s)15(um )-14(of )] TJ endstream /ProcSet[/PDF/Text] Diabetes, also known as diabetes mellitus, is a group of common endocrine diseases characterized by sustained high blood sugar levels. /Subtype /Form 1.007 0 0 1.007 271.012 636.879 cm /Type /XObject /Resources<< /Type /XObject /Meta152 166 0 R >> 260 0 obj ET /F3 12.131 Tf /Resources<< /ProcSet[/PDF/Text] 0 G >> Q Q 0 G endstream stream /Resources<< 333.269 5.488 TD endstream /Matrix [1 0 0 1 0 0] endstream 193 0 obj stream 1 i q endstream /Resources<< /F3 17 0 R 1 i >> q ET /Length 69 endstream Q 22.478 5.336 TD 0 G (x) Tj endstream stream /Length 69 /Type /Font q Q Q /FormType 1 << 1.005 0 0 1.007 102.382 653.441 cm >> /F3 17 0 R /Meta367 Do /Meta119 133 0 R Q stream >> /Resources<< /Resources<< endstream 0.737 w 1 i /F3 17 0 R Q /Length 69 << /BBox [0 0 639.552 16.44] Q Q stream /BBox [0 0 88.214 16.44] Q >> 1.014 0 0 1.006 391.462 763.351 cm /Subtype /Form BT endobj /Meta336 350 0 R endstream /Subtype /Form ET /Matrix [1 0 0 1 0 0] /Meta215 Do q /Type /XObject << /Font << 1 i let 'x' and 'y' represent the numbers. endobj endobj q q /Font << >> 0 g Q /Meta216 230 0 R << q /Meta46 Do 1.005 0 0 1.007 102.382 616.553 cm /Subtype /Form (x) Tj 0.524 Tc >> 137 0 obj q /Length 59 /Meta299 Do /Ascent 1050 stream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 308 0 obj /BBox [0 0 88.214 16.44] 16.469 5.336 TD 0 g stream /F3 12.131 Tf ET 1 g q /Type /XObject Q 0.737 w /Font << /Resources<< 0.524 Tc 0 w /Meta213 Do q /ProcSet[/PDF] 0.564 G /Type /Font /Meta314 Do /Type /XObject BT /ProcSet[/PDF/Text] q /FormType 1 /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /BBox [0 0 88.214 35.886] /Subtype /Form 0 w Q q stream /Subtype /Form /Resources<< 1 i 0.564 G 47.933 5.203 TD /F3 17 0 R BT /Resources<< 1 i q 0.458 0 0 RG 0 g q 0 5.203 TD /Resources<< 0.737 w /Subtype /Form /Subtype /Form ET >> /Resources<< /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Type /XObject Q /Resources<< /Meta323 Do 0 G /FormType 1 -0.058 Tw /Resources<< S q >> q (11) Tj much as how 8, Last . (5\)) Tj q stream /F3 12.131 Tf endobj /Font << >> /Length 59 /Type /XObject /Matrix [1 0 0 1 0 0] q /Meta55 Do 1 i 1.005 0 0 1.007 102.382 799.486 cm /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0.564 G q q 1.007 0 0 1.007 411.035 277.035 cm /Subtype /Form q endobj 108 0 obj endstream stream Q 0 g >> 323 0 obj /Size 447 Q >> stream /Font << (A\)) Tj 229 0 obj /Type /XObject Q 0 g /Matrix [1 0 0 1 0 0] stream q (-4) Tj 328 0 obj endobj 0 5.203 TD q [( t)-14(imes a num)-16(ber)] TJ 1.007 0 0 1.007 130.989 583.429 cm Q q ET 0 w Q Q /Matrix [1 0 0 1 0 0] stream /Meta295 Do Q /FormType 1 >> /Length 16 /Resources<< , Prove the following >> /F3 17 0 R endobj 433 0 obj 1 i /ProcSet[/PDF/Text] 1 i /FormType 1 q (\(x ) Tj >> /FormType 1 /Resources<< /Font << 0 g << /Meta305 Do /Length 59 endobj Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] endobj 0 w stream << /F3 17 0 R /Meta328 Do /Resources<< << 0.737 w /BBox [0 0 88.214 16.44] /Type /XObject /BBox [0 0 23.896 16.44] << 23.216 5.203 TD Q /BBox [0 0 88.214 16.44] q /F1 7 0 R endstream 0.737 w /F3 12.131 Tf /Type /XObject 1.007 0 0 1.007 130.989 277.035 cm endobj Q q /Font << Q Q q Q >> /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] 1.007 0 0 1.007 45.168 829.599 cm stream /Type /XObject 0 G 1.007 0 0 1.006 411.035 763.351 cm /Parent 1 0 R /Length 69 /Length 65 q 27.693 5.203 TD stream /FormType 1 q 0 g /XObject << /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.006 551.058 437.384 cm Q /Type /XObject - 9737014. 1 i q /ProcSet[/PDF] five times the sum of a number x and two b.) /Length 68 /Type /XObject /Resources<< /Meta150 Do q Q ET 95 0 obj 297 0 obj q /Meta58 72 0 R /Length 118 /ProcSet[/PDF/Text] Q 0.369 Tc ET /Meta326 340 0 R /F3 17 0 R >> endobj /Matrix [1 0 0 1 0 0] ET 1 i 0 G >> 259 0 obj 0 g >> 1 i /F3 12.131 Tf >> 0 w 1 g 1.007 0 0 1.007 551.058 277.035 cm << /Type /XObject /Font << /FormType 1 440 0 obj endstream /Font << 1 i >> 1.007 0 0 1.007 411.035 636.879 cm /F3 12.131 Tf /Meta361 375 0 R /Length 64 >> /Resources<< 20-n c.) n+20 d.) 20+n 3.) endobj ET 1 i /Resources<< q 85 0 obj 1 i /Type /XObject /ProcSet[/PDF/Text] >> 0 g >> 421 0 obj 0 G (1) Tj 0 g (-) Tj << 1 i stream /F1 12.131 Tf /F4 36 0 R 11.99 24.649 TD 1.005 0 0 1.007 102.382 293.596 cm >> /Resources<< 1.007 0 0 1.007 271.012 636.879 cm /Resources<< /BBox [0 0 88.214 35.886] q /Meta19 Do /Font << /Meta149 Do Q stream >> BT /BBox [0 0 549.552 16.44] >> >> /Subtype /Form q /F4 36 0 R /Matrix [1 0 0 1 0 0] /FormType 1 /FormType 1 30 0 obj 0 g endstream 23.952 4.894 TD 0 5.203 TD /Resources<< Q >> q /Length 16 q << 0.458 0 0 RG /BBox [0 0 88.214 16.44] stream /F3 17 0 R /BBox [0 0 88.214 35.886] 0.297 Tc >> stream 1 g >> /FormType 1 /FormType 1 /Type /XObject 58 0 obj endobj /F3 12.131 Tf Q q Q /F4 12.131 Tf endobj >> /Subtype /Form 0 g /BBox [0 0 88.214 16.44] /ProcSet[/PDF] 0 g /Matrix [1 0 0 1 0 0] 1 i 275 0 obj /Length 69 19 0 obj /Meta147 Do /Length 59 >> Q endobj q 1.007 0 0 1.007 130.989 849.172 cm >> (iv) A number exceeds 5 by 3. Q /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] /Meta135 149 0 R q q /Type /XObject Q endstream /F3 12.131 Tf /Meta344 Do 3.742 5.203 TD /Length 16 /Length 65 0 g q 1 i /F4 36 0 R q 0 w /Resources<< /Length 69 q 416 0 obj /BBox [0 0 639.552 16.44] /FormType 1 Q >> Q Q Q 0 G /Meta377 Do /FormType 1 q 0 G 0.564 G >> /Resources<< 0.838 Tc 549.694 0 0 16.469 0 -0.0283 cm /Length 78 Q >> << q Q endobj Q /Encoding /WinAnsiEncoding >> /Meta216 Do /Type /XObject 1 g /Type /XObject >> >> q 0.51 Tc 1.005 0 0 1.007 79.798 746.789 cm 0 g /Subtype /Form True False q endstream endstream 1.502 5.203 TD BT >> Select the correct mathematical statement for the following equation. /Meta357 371 0 R >> /Subtype /Form 192 0 obj /Meta177 Do 1 i /F3 12.131 Tf q /Meta65 Do 0 w q 1 i Q /Matrix [1 0 0 1 0 0] Q /BBox [0 0 88.214 16.44] 64 0 obj /Subtype /Form 25.454 5.203 TD /F3 12.131 Tf q Q endstream (-) Tj /Subtype /Form /FormType 1 /F3 17 0 R /Matrix [1 0 0 1 0 0] Q Q BT Q /Matrix [1 0 0 1 0 0] 0.737 w << 400 0 obj 217 0 obj /Meta95 Do q /ProcSet[/PDF] >> Q /Meta270 284 0 R /Length 69 221 0 obj q >> /Font << q /Meta248 Do Q stream 1 i 1 i /FormType 1 /Type /XObject endobj /Length 54 /F3 17 0 R q Q /Meta23 34 0 R >> >> q q Q q /Resources<< Q >> Q endobj 0.68 Tc 1.502 5.203 TD /FormType 1 0 G /F3 17 0 R 398 0 obj /Meta71 85 0 R /FormType 1 0 g 0 G << endobj >> /Meta383 Do stream /Length 118 /Subtype /Form endobj /BBox [0 0 15.59 29.168] /Type /XObject q /BBox [0 0 88.214 35.886] q /F3 12.131 Tf /Meta417 433 0 R 0 g /Meta9 Do 1 g Q 0 w Q Q BT q /Resources<< BT /Subtype /Form /Length 12 /BBox [0 0 639.552 16.44] 0.37 Tc /F3 12.131 Tf 1.007 0 0 1.006 411.035 510.406 cm q >> /Meta287 301 0 R >> endobj /ProcSet[/PDF] q endstream 0.458 0 0 RG Q /ProcSet[/PDF] 1.014 0 0 1.007 531.485 277.035 cm The results found were expressed mainly through tables and graphs as the main resources of the statistical language. >> /Matrix [1 0 0 1 0 0] 0 g /AvgWidth 445 /Meta101 Do 1 i 0 g >> ET /Matrix [1 0 0 1 0 0] Q 1.007 0 0 1.007 551.058 636.879 cm /Length 16 stream 1.005 0 0 1.007 79.798 730.228 cm 0.564 G /Matrix [1 0 0 1 0 0] 1 i stream >> Q Q /Subtype /Form 246 0 obj 1 i /Length 69 endobj q 0.564 G Q Q /ProcSet[/PDF/Text] 0.241 Tc 6.746 24.649 TD 1 i 0 5.203 TD /Resources<< stream Q >> Q Q /F3 12.131 Tf Thrice a number decreased by 5 exceeds twice the number by a unit. Q 1.007 0 0 1.007 67.753 872.509 cm q /Meta184 Do
