twice a number decreased by 58

244 0 obj /Length 70 /Meta238 252 0 R 0 g /Type /XObject BT q 1.502 5.203 TD 0 g Q (-4) Tj /F3 12.131 Tf /Type /XObject /Length 60 >> >> 1.007 0 0 1.007 45.168 779.913 cm /Type /XObject Q Q /Matrix [1 0 0 1 0 0] q q ET /F3 17 0 R /BBox [0 0 15.59 16.44] /Length 57 1 i ET q /Font << Q /Font << 28 0 obj Q Q /Resources<< /F1 7 0 R /Meta222 Do 1 i >> Q /Meta307 321 0 R 0 4.894 TD 406 0 obj /Resources<< BT /BBox [0 0 639.552 16.44] /F3 12.131 Tf /Meta17 28 0 R Q /Meta147 Do /F4 12.131 Tf q q /F3 12.131 Tf >> >> /Length 69 0.524 Tc Q If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. /ProcSet[/PDF/Text] /Meta389 405 0 R Q 1.007 0 0 1.007 130.989 523.204 cm /Meta162 176 0 R Q /F3 12.131 Tf >> endstream Q 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. Q << /Subtype /Form Q Q /ProcSet[/PDF/Text] q Q 1.007 0 0 1.007 45.168 862.723 cm 1.007 0 0 1.007 271.012 277.035 cm 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. 1 i stream /Subtype /Form << 0.737 w endstream /F3 12.131 Tf q 0 g 1.005 0 0 1.007 102.382 310.158 cm 363 0 obj Q >> /Matrix [1 0 0 1 0 0] Q /Length 16 q Q q 0 G /Meta51 Do ET endobj q )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] q /FormType 1 /Meta8 19 0 R stream That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. [(Fiv)25(e ti)18(me)16(s)] TJ /FontDescriptor 16 0 R 0 g /Matrix [1 0 0 1 0 0] Q Q 0.458 0 0 RG /F3 17 0 R /ProcSet[/PDF] ET 435 0 obj 0 g (58) Tj >> Q /Meta210 224 0 R 0.737 w /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 330.484 cm /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 654.946 400.496 cm Q 0.458 0 0 RG >> Q /F3 17 0 R /Meta216 Do Q << /I0 51 0 R /F1 7 0 R 0.045 Tw q 1 i endobj /Type /XObject >> /FormType 1 >> /Subtype /Form /FormType 1 /Length 69 /FormType 1 q /ProcSet[/PDF/Text] q Q the quotient of twenty and a number a.) endstream << 23.216 5.203 TD /FormType 1 /Subtype /Form Q 1.014 0 0 1.006 111.416 437.384 cm 1.007 0 0 1.007 130.989 277.035 cm 389 0 obj 0 5.203 TD Q q So let's go ahead and identify a v /ProcSet[/PDF/Text] /Length 69 Q /F4 12.131 Tf q /Meta183 Do 260 0 obj /F3 17 0 R endobj /Meta171 Do endobj 0 g (x) Tj >> /Length 16 1 i /Length 65 0.564 G /XObject << >> 0.458 0 0 RG >> 1.005 0 0 1.007 79.798 829.599 cm /Meta170 184 0 R endobj >> /Meta243 257 0 R /F3 17 0 R q Q /BBox [0 0 17.177 16.44] 0 w BT 0.838 Tc (D) Tj Q 0 g Q 0 g << 722.699 293.596 l /Meta147 161 0 R 0.458 0 0 RG 0.458 0 0 RG q 20.21 5.203 TD Q /F3 12.131 Tf Q BT endobj /CapHeight 662 stream q Q Twice a number decreased by ten is at least 24. Q Q /Font << ET q ET /Meta90 104 0 R 0 G 1 i << ET endobj /BBox [0 0 30.642 16.44] << [(1)-25(0\))] TJ /ProcSet[/PDF/Text] Q /Subtype /Form 1.007 0 0 1.007 551.058 383.934 cm /Meta205 219 0 R << (x) Tj Q endobj /Subtype /Form The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. /Type /XObject >> endobj endobj Q 111 0 obj /Type /XObject 1 i Q 0 5.203 TD 1.014 0 0 1.007 391.462 583.429 cm q BT Q >> /Resources<< 0 g /Resources<< 169 0 obj stream Q /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g >> 209 0 obj 1 i Q endobj /BBox [0 0 88.214 16.44] << 0 w /Subtype /Form (2) Tj 0.458 0 0 RG << /Matrix [1 0 0 1 0 0] 21.713 20.154 l stream /Length 59 q /Resources<< Q /FormType 1 1.007 0 0 1.007 411.035 849.172 cm /Length 59 1.005 0 0 1.015 45.168 53.449 cm >> Q /Subtype /Form Q >> /Font <> /ProcSet[/PDF/Text] /Subtype /Form /Font << Q 0 w /Matrix [1 0 0 1 0 0] stream 20.21 5.203 TD Q q >> /F3 17 0 R /FormType 1 1.005 0 0 1.007 102.382 653.441 cm /Type /XObject /Font << Q 1.502 24.339 TD endobj ET endstream q q /Type /XObject Q /Subtype /Form /ProcSet[/PDF/Text] Q Q /Subtype /Form /BBox [0 0 639.552 16.44] /Subtype /Form 0.68 Tc /Resources<< >> 0 G endstream stream /Type /XObject Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. Q 0.564 G q /BBox [0 0 673.937 68.796] >> << << BT Q stream Twice Gail's age: 2g 58 decreased by twice Gail's age 58 - 2g President of MathCelebrity. /Length 118 /BBox [0 0 88.214 16.44] 1 i q q stream >> Q (3) Tj q /Meta37 Do Q stream 393 0 obj /ProcSet[/PDF/Text] /ProcSet[/PDF] >> endobj /Type /XObject q /FormType 1 /Subtype /Form /Type /FontDescriptor 0.458 0 0 RG endobj 0 G /FormType 1 1.007 0 0 1.006 130.989 437.384 cm 310 0 obj << ET 0 w /Resources<< Q /BBox [0 0 15.59 29.168] 1 i /Font << 0 G /Matrix [1 0 0 1 0 0] /Font << /Resources<< 1.007 0 0 1.006 130.989 437.384 cm /Subtype /Form Q Q /Resources<< /Matrix [1 0 0 1 0 0] >> Q /ProcSet[/PDF] 1 i 1.005 0 0 1.013 45.168 933.487 cm Q /Type /XObject >> stream >> << /BBox [0 0 88.214 35.886] 0.737 w /Meta269 283 0 R /FormType 1 /Meta206 220 0 R q stream /Length 16 Q >> >> 0 w endobj 426 0 obj 0.524 Tc >> /Meta339 353 0 R /Length 69 >> 0 g 1 i /Resources<< /Length 69 /Type /XObject Q /Subtype /Form q /Subtype /Form q BT /F1 7 0 R /Meta331 Do endstream /Matrix [1 0 0 1 0 0] ET Q 1.007 0 0 1.007 130.989 523.204 cm q Q 0 g q /F3 12.131 Tf /F3 17 0 R /Font << Q /Matrix [1 0 0 1 0 0] >> /Type /XObject /Subtype /Form q endstream /F3 12.131 Tf 0 g << q 23.952 4.894 TD /Length 16 stream >> /Subtype /Form >> /Type /XObject 196 0 obj 16.469 5.203 TD /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] 0 w /Matrix [1 0 0 1 0 0] endobj /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] << (A\)) Tj /Resources<< /FormType 1 q /Font << endobj 0.369 Tc /ProcSet[/PDF/Text] stream /Meta399 415 0 R Q /Subtype /Form >> BT The result is 8 less than 10 times the number. /Meta223 237 0 R 0 w /Subtype /Form /Meta189 203 0 R /Type /XObject (8\)) Tj /Meta29 42 0 R << << >> >> /BBox [0 0 88.214 16.44] endobj /Meta427 Do q /FormType 1 /FormType 1 endobj /Meta238 Do /Length 12 Q 0 g /Meta175 Do 20.21 5.203 TD endobj 1. >> 1.007 0 0 1.007 551.058 330.484 cm /Resources<< /Resources<< 0 5.203 TD q endobj /Font << /F3 12.131 Tf q Q q /FormType 1 >> << q Q /Resources<< Q << /Meta260 274 0 R 1 i 1 i /F3 17 0 R 1.007 0 0 1.007 271.012 636.879 cm 0 5.203 TD 0 w /Length 57 Q /Meta325 Do q BT /Length 59 << 1.007 0 0 1.007 551.058 703.126 cm 275 0 obj 0 w >> /ProcSet[/PDF/Text] 1.007 0 0 1.006 411.035 437.384 cm 341 0 obj /Length 16 >> ET endobj q 1 i endobj 1.005 0 0 1.007 102.382 653.441 cm /Length 69 Q 0 g q ( \() Tj 0 G /FormType 1 stream q /Subtype /Form << /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /Meta322 336 0 R /Type /XObject /Matrix [1 0 0 1 0 0] q endstream << [tex]\sin (\pi -x)=\sin x[/tex]. /Font << 26.957 5.203 TD 25.454 5.203 TD /FormType 1 >> /Length 59 stream q /Subtype /Form 0 g endobj >> /Subtype /Form q 1 i 38.182 5.203 TD q >> /Meta294 Do >> Q q /Length 69 /BBox [0 0 15.59 16.44] >> >> stream /Type /XObject /Meta194 208 0 R /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 1.502 24.649 TD endobj /Meta110 124 0 R 0 g ET 549.694 0 0 16.469 0 -0.0283 cm /BBox [0 0 88.214 16.44] (11) Tj /ProcSet[/PDF] q /BBox [0 0 88.214 16.44] >> q q (B) Tj /Meta425 441 0 R /Meta138 Do Q q Q /Type /XObject >> /Matrix [1 0 0 1 0 0] /Resources<> /FormType 1 q 1 i endstream /Width 734 /Meta321 Do 1.502 7.841 TD 11.99 24.649 TD /Matrix [1 0 0 1 0 0] q 0 g q /BBox [0 0 88.214 16.44] /ProcSet[/PDF/Text] q << 308 0 obj endstream stream We are asked to find the number, so, we could assign the number as "x". >> /Length 60 q 1 i 337 0 obj /Length 16 /Matrix [1 0 0 1 0 0] Q endstream /Font << Q 16.469 5.203 TD /FormType 1 << >> ET /Subtype /Form >> q /Resources<< Q q /FormType 1 stream /Matrix [1 0 0 1 0 0] 0.737 w q q 0 G endstream /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] ET >> /Matrix [1 0 0 1 0 0] 1 i 0 G BT Q >> 0 20.154 m /Meta219 233 0 R /BBox [0 0 15.59 16.44] 0 g /F3 17 0 R /Subtype /Form (B) Tj stream >> << 1 i /ProcSet[/PDF/Text] endstream /Resources<> /Font <> /Meta399 Do 0.564 G q /ProcSet[/PDF/Text] 1 i /Meta115 Do /Meta119 Do << 0 G /FormType 1 /FormType 1 0 G q >> 0 w 325 0 obj /ProcSet[/PDF/Text] 0 G 1 i q /Font << /Meta236 250 0 R /ProcSet[/PDF/Text] /BBox [0 0 15.59 29.168] S q >> /Subtype /Form (B\)) Tj endstream >> /Font << Q -0.092 Tw /F3 17 0 R Q /Resources<< 277 0 obj 148 0 obj BT Q >> /Resources<< << endstream 1 i >> endobj Q /Meta175 189 0 R 0 5.203 TD /Meta343 Do 1 g >> Q Q Q 0.564 G 0 g << stream Q << /Length 16 0 G /Meta354 Do 1.014 0 0 1.007 251.439 523.204 cm /Meta241 255 0 R >> Q stream /F3 17 0 R endstream Q Patients' reasons for declining screening were not collected . Q q 139 0 obj q /Type /FontDescriptor /F3 17 0 R endobj Q q /F3 12.131 Tf /F3 17 0 R 0 g >> 184 0 obj /BBox [0 0 15.59 29.168] q endobj Find the number. /Font << /Subtype /Form /Subtype /Form endstream Two speeding tickets could increase your rate by 58% at your next renewal. 0.564 G q 1 i 1 i >> /Font << 1.007 0 0 1.007 130.989 583.429 cm 0 g /Meta268 282 0 R Q /Subtype /Form Q stream >> 0 g 3.742 5.203 TD 52.412 5.203 TD q /F3 17 0 R Q endstream /Meta32 45 0 R /Subtype /Form 1.007 0 0 1.007 411.035 330.484 cm 1.007 0 0 1.007 551.058 703.126 cm >> >> Q 0 g /Length 118 /Length 69 0 G /Type /XObject Select the correct mathematical statement for the following equation. endobj /FontDescriptor 6 0 R endobj 1 i 9.723 5.336 TD 1 i /Subtype /Form /Meta424 440 0 R 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 stream q << /Type /XObject Q 0 G /BBox [0 0 88.214 16.44] 409 0 obj /F3 12.131 Tf Q /F3 12.131 Tf /FormType 1 endstream /Meta370 384 0 R Formula - How to Calculate Percentage Decrease. /Length 69 /Matrix [1 0 0 1 0 0] 368 0 obj Q 1.005 0 0 1.007 79.798 730.228 cm Q >> /Subtype /Form 0 G /ProcSet[/PDF/Text] endobj q stream q 128 0 obj << /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] (9\)) Tj Q 1 i stream /Resources<< /Meta228 Do /FormType 1 /Meta146 160 0 R /Resources<< q >> 26.219 5.203 TD BT /Meta258 Do Q /Length 69 /Resources<< BT /Meta386 Do 0 g /BBox [0 0 88.214 16.44] 0.458 0 0 RG How many points did Kobe score in the season? /Resources<< 1 i >> q q stream q /BBox [0 0 88.214 16.44] 0.458 0 0 RG endobj 0 G /Resources<< /Meta383 Do 1 i 1 i /Subtype /Form /Meta110 Do /Subtype /Form /F3 17 0 R >> /Meta26 39 0 R /Meta143 Do 0 g << 1 g Q q /Resources<< 1 i /FormType 1 Q Q /Matrix [1 0 0 1 0 0] /Meta292 306 0 R q 157 0 obj 0 G endobj /Type /XObject /ProcSet[/PDF/Text] BT /ProcSet[/PDF] /F1 7 0 R /Meta409 425 0 R >> endstream 1 i q 1 i /Meta225 Do /I0 Do >> 0 g /Font << 0.458 0 0 RG >> /Resources<< Q /Meta348 Do >> Q 287 0 obj /ProcSet[/PDF] /Meta95 109 0 R Q 1.007 0 0 1.007 130.989 277.035 cm Q (38) Tj much as how 8, Last . << 0.425 Tc /BBox [0 0 88.214 16.44] 0 g >> /Length 12 >> /Resources<< /BBox [0 0 15.59 16.44] << /Length 16 stream /Meta408 424 0 R 172 0 obj /FormType 1 /BBox [0 0 30.642 16.44] /Meta204 Do /Matrix [1 0 0 1 0 0] q (C\)) Tj endstream q /FormType 1 1 i /Meta337 351 0 R 1 i Q >> 262 0 obj Q q q 0.564 G endobj >> >> /Matrix [1 0 0 1 0 0] Find the number 1 See answer Advertisement /ProcSet[/PDF] /F3 12.131 Tf /Length 58 Q /Meta178 Do (-20) Tj /Meta240 Do q Q >> stream endstream << 0 G 0 g /FormType 1 stream /Meta153 Do Q /XObject << q ET /Resources<< >> >> endstream q 1.007 0 0 1.007 551.058 636.879 cm Q >> 0 G /FormType 1 >> 0 G stream /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 383.934 cm /Type /XObject /F3 12.131 Tf /Meta209 223 0 R endstream stream Q q /Matrix [1 0 0 1 0 0] 1 i /Type /XObject >> /FormType 1 0 g /BBox [0 0 88.214 16.44] q /Font << q /Font << /Meta219 Do /ProcSet[/PDF/Text] 0 g endstream 1.014 0 0 1.006 391.462 836.374 cm /Length 69 /FormType 1 >> BT 127 0 obj 0 g -37 VI 2. q /Meta16 Do /Subtype /Form , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. q Q /Meta351 365 0 R /Meta169 183 0 R 164 0 obj Q 1 g /Resources<< endobj q /Length 69 /F3 12.131 Tf /Matrix [1 0 0 1 0 0] >> 1 i q /Subtype /Form 20.21 5.203 TD Q Q 0 g << >> Q Q >> /Font << /FormType 1 51 0 obj /Meta17 Do endstream /F1 12.131 Tf Q /FormType 1 /Meta345 Do /Resources<< /Type /XObject 0.564 G /BBox [0 0 534.67 16.44] /Meta134 148 0 R 408 0 obj stream >> 1 g /Resources<< q Q q endobj 1.005 0 0 1.007 45.168 889.071 cm stream /Type /XObject /Type /XObject /Font << stream Q 0.738 Tc << -0.486 Tw q /FormType 1 1.005 0 0 1.007 102.382 400.496 cm endstream /BBox [0 0 88.214 16.44] ET /Resources<< endobj q >> q Q q (x ) Tj Q endstream /Matrix [1 0 0 1 0 0] /FormType 1 Q /Font << /Length 104 endobj Q 0.564 G 379 0 obj q (40) Tj endstream 0.458 0 0 RG << q /Length 59 /BBox [0 0 88.214 35.886] Q endobj 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 Calculate a 15% decrease from any number. /FormType 1 /FormType 1 /F3 12.131 Tf q 1.014 0 0 1.006 111.416 836.374 cm stream endstream Q q >> 0 g /Subtype /Form /FormType 1 /F3 17 0 R Q endobj /Type /XObject /Length 79 q (13) Tj /Matrix [1 0 0 1 0 0] 0 G /Meta196 Do /FormType 1 Q >> 0 G /FormType 1 BT 1.005 0 0 1.006 45.168 879.284 cm /Meta47 Do /Meta83 97 0 R /Subtype /Form 16.469 5.336 TD /BBox [0 0 17.177 16.44] q endstream Q /Meta212 Do q Q /Type /XObject Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. Q /BBox [0 0 15.59 16.44] Q 0 g Q Q /ProcSet[/PDF/Text] endstream /Length 69 /FormType 1 /BBox [0 0 673.937 15.562] Q stream q 0 g 0 g ET Q endstream Q q >> S Q 0 5.203 TD q Q 1.007 0 0 1.007 271.012 330.484 cm q << 0.458 0 0 RG Q q stream /Length 80 /Subtype /Form >> /Meta22 Do endstream 1.007 0 0 1.007 551.058 583.429 cm q /Subtype /Form endstream (3\)) Tj Q /ProcSet[/PDF/Text] q endobj stream 0 G << /Meta113 Do q Q /Meta224 238 0 R stream /Meta271 285 0 R 0.564 G ( \() Tj 0 G << /Matrix [1 0 0 1 0 0] q Q /Meta247 261 0 R Q 1.007 0 0 1.006 551.058 763.351 cm /ProcSet[/PDF/Text] /Meta89 Do stream /Type /XObject 0.737 w (\(x ) Tj endstream << >> /Font << 1.007 0 0 1.007 551.058 383.934 cm /Resources<< q (D) Tj 1.014 0 0 1.007 391.462 330.484 cm 1.014 0 0 1.007 251.439 583.429 cm /Subtype /Form /ProcSet[/PDF/Text] endstream q /F3 12.131 Tf /FormType 1 /Subtype /Form q /Matrix [1 0 0 1 0 0] >> >> 116 0 obj /F3 17 0 R /Resources<< /FormType 1 125 0 obj /Meta319 333 0 R 0.425 Tc /FormType 1 endstream << /FormType 1 1 i q 1 g endstream 11 0 obj /Meta374 Do q BT stream q >> Q q (x) 6 times a number is 5 more than the number. /ProcSet[/PDF] /Length 16 0.738 Tc >> Q /Meta266 Do 0.37 Tc 0.564 G q /Matrix [1 0 0 1 0 0] 1.502 5.203 TD /Type /Font /Meta294 308 0 R 0 g Q Q /Ascent 1050 >> Q /Length 59 Q /F3 17 0 R endobj 0 g Q /Meta410 426 0 R << /Subtype /Form /BBox [0 0 88.214 16.44] >> /FormType 1 endstream (-) Tj /Meta275 289 0 R Q /Resources<< endstream Q q /Subtype /Form >> stream Q endstream endobj << 0.737 w /Matrix [1 0 0 1 0 0] /FormType 1 Q /F4 36 0 R 1.007 0 0 1.007 271.012 523.204 cm /Meta29 Do What word phrase can you use to represent 5x + 2? q ET >> 1.014 0 0 1.006 531.485 690.329 cm /Subtype /Form /Meta347 Do (11) Tj stream stream /Meta215 229 0 R Q >> 1 i endobj q /Font << /Subtype /Form 1.007 0 0 1.007 67.753 473.519 cm /BBox [0 0 30.642 16.44] /F3 12.131 Tf 0.737 w /Flags 32 Q 0.737 w 1.014 0 0 1.007 391.462 450.181 cm /Resources<< endobj q >> endobj >> ET /Matrix [1 0 0 1 0 0] Twice = two times, double. /FormType 1 >> /ProcSet[/PDF/Text] Q 1.005 0 0 1.013 45.168 933.487 cm >> 0.737 w /Type /XObject /Resources<< q There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. Q /Subtype /Form /Meta264 Do (-) Tj /Subtype /Form 0 G /Meta26 Do endobj Q /BBox [0 0 15.59 29.168] /Type /XObject 135 0 obj 6.746 5.203 TD /Resources<< q /Subtype /Form q 12.727 5.203 TD << Q (B) Tj BT /F3 12.131 Tf /BBox [0 0 88.214 16.44] /Type /XObject >> 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 1 i >> /Type /XObject q 318 0 obj >> 249 0 obj 1 i 0 G BT Q /Length 69 >> ET /Meta42 56 0 R 67 0 obj endstream 0 G Q 0 5.203 TD /FormType 1 /Font << 2.238 5.203 TD 0 g 0.564 G 1 i Q Translate 2(x-58) into mathematical phrase. /Type /XObject Q >> 1.014 0 0 1.007 531.485 277.035 cm ET 1 i /FormType 1 /Resources<< Q /Type /XObject /Meta91 105 0 R q /Font << 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. Q stream Q >> << q 0.68 Tc q << /FormType 1 /ProcSet[/PDF] 1 i /FormType 1 ET /Resources<< Medium /Meta62 Do 1 i /Length 59 /Meta137 151 0 R 322 0 obj q 35 0 obj /Subtype /Form q 0 5.203 TD /Font << /Font << (C\)) Tj 0 g /Length 67 /Font << /F3 17 0 R (-20) Tj /Meta39 53 0 R /Subtype /Form q /Length 64 Q Q 1.007 0 0 1.006 411.035 690.329 cm /BBox [0 0 88.214 35.886] /Type /XObject /F3 17 0 R /Type /XObject /Font << (C\)) Tj Q 0 G 265 0 obj /FormType 1 /Resources<< Q endobj endobj q BT endobj 1 g endstream 1.005 0 0 1.007 102.382 473.519 cm >> q /BBox [0 0 88.214 16.44] >> 0 G 1.007 0 0 1.007 130.989 636.879 cm /Subtype /Form the other number. >> /Meta101 115 0 R /Resources<< q >> Q ET 0 g /ProcSet[/PDF] >> 0.564 G 220.931 4.894 TD BT /Length 80 /Resources<< /ProcSet[/PDF/Text] /Meta2 Do 0.737 w Q /Meta31 Do q Q q endobj 0.486 Tc 1.502 5.203 TD 1 g endstream /Meta54 Do Q Q q >> /FormType 1 q /Resources<< 0 g q q 1 i /Type /XObject /Matrix [1 0 0 1 0 0] q /Length 78 /Meta224 Do /ProcSet[/PDF] q 1.007 0 0 1.007 130.989 383.934 cm Q /ProcSet[/PDF] /BBox [0 0 30.642 16.44] 115 0 obj /Font << /Meta167 181 0 R q 0.737 w /Type /XObject /Matrix [1 0 0 1 0 0] /Type /XObject >> 1 i Q q 0 w endobj SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. endstream Q >> /ProcSet[/PDF] /Resources<< /Subtype /Form 1 i /Resources<< stream 1.007 0 0 1.007 411.035 583.429 cm q endstream stream >> Q /Subtype /Form Q 0.786 Tc 0 G Q Q q 1.007 0 0 1.007 411.035 383.934 cm << /Matrix [1 0 0 1 0 0] q endstream 1 i /BBox [0 0 549.552 16.44] 1 i /Length 16 /Subtype /Form /Matrix [1 0 0 1 0 0] /FormType 1 /Resources<< /Subtype /Form q Q 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. 0 g /FormType 1 Get link; Facebook; Twitter; /ProcSet[/PDF] ET 1 i /Subtype /Form 270 0 obj S ET /F3 17 0 R q Q BT /Length 85 /BBox [0 0 549.552 16.44] 0.737 w 0 G Q /Meta275 Do 1.007 0 0 1.007 654.946 473.519 cm >> 0.737 w /Resources<< q 1.007 0 0 1.007 271.012 523.204 cm >> << /Resources<< Q endstream /Meta286 Do endobj /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /F3 17 0 R /Subtype /Form 0 g 0 G 6.746 5.203 TD Q /Meta50 Do /ProcSet[/PDF/Text] 0.737 w /F1 7 0 R 0 g 0 g 0.458 0 0 RG /ProcSet[/PDF/Text] /Length 68 endstream /Subtype /Form 0.564 G 0 w endobj /BBox [0 0 88.214 35.886] 0 g Q >> endstream /Meta153 167 0 R 0 g q BT << endobj << /Subtype /Form Q 1 i /Resources<< >> 1 i 1 i /Type /XObject 1 i 1.502 5.203 TD /Meta391 407 0 R /Meta288 302 0 R 0 g >> 1.005 0 0 1.007 102.382 726.464 cm /FormType 1 /Meta366 380 0 R endobj 77 0 obj 0.369 Tc /FormType 1 /Size 447 243 0 obj 0 g /Subtype /Form endstream 364 0 obj 1.007 0 0 1.007 551.058 277.035 cm /ProcSet[/PDF/Text] 1 g q Ten divided by a number 5. /BBox [0 0 88.214 16.44] 0.269 Tc /Type /XObject q stream 1 i /Meta158 172 0 R /Meta145 159 0 R >> q /Type /XObject /Type /XObject q /BBox [0 0 88.214 16.44] /Font << stream /ProcSet[/PDF] /Meta382 396 0 R /Resources<< /I0 51 0 R q >> >> /Matrix [1 0 0 1 0 0] /Meta211 Do stream << stream 1.502 5.203 TD Q /F3 17 0 R endstream for the season. /FormType 1 0 G >> 0 G ET /Resources<< >> Q /BBox [0 0 88.214 35.886] /Subtype /Form /F1 12.131 Tf /F1 12.131 Tf /FormType 1 (-4) Tj 1 i stream /Length 118 endobj 0 20.154 m >> /Length 69 /XHeight 477 q q << q >> << /F3 12.131 Tf /Font << >> 35.206 4.894 TD stream /Subtype /Form /ProcSet[/PDF] Q << D. b = 4 2. Q endstream q 1.007 0 0 1.007 654.946 799.486 cm /Matrix [1 0 0 1 0 0] /F3 17 0 R Q endstream 383 0 obj /Meta290 304 0 R Q >> (D\)) Tj Q Advertisement Answer No one rated this answer yet why not be the first? 0 G ET /Subtype /Form 377 0 obj ET >> /Matrix [1 0 0 1 0 0] q endobj ET ET >> /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0 w << Q 0 g >> /Length 69 /Meta11 Do Negative thirteen decreased by 3 times a number x. 288 0 obj /BBox [0 0 88.214 16.44] << /BBox [0 0 17.177 16.44] endobj /Font << /Resources<< q /Meta30 43 0 R (vii) Twice a number subtracted from 19 is 11. endobj >> /Meta239 Do /ProcSet[/PDF/Text] 0 5.203 TD 22.478 5.336 TD precision and actual right or wrong answers. endobj << %PDF-1.4 q 20-n c.) n+20 d.) 20+n 3.) /Meta428 Do 250 0 obj /Resources<< /I0 51 0 R 53 0 obj >> 0 g 1 i 0.564 G ET stream >> q 0 g q /Subtype /Form q Q 0 G >> /Meta18 Do /Type /XObject >> /ProcSet[/PDF] q Q /Subtype /Form 1 i q /Subtype /Form 57.656 5.203 TD Q q /Meta407 Do /Type /XObject Q /Font << 1.014 0 0 1.007 531.485 849.172 cm /Matrix [1 0 0 1 0 0] >> /Subtype /Form stream /Length 69 Q ( \() Tj endstream 0.564 G /Length 189 /FormType 1 q Q In addition, testosterone in both sexes is involved in health and well-being . /Length 68 q /ItalicAngle 0 Q 0.737 w Q /Length 69 /Meta160 Do >> 0.297 Tc /F1 12.131 Tf Q Q 1 i /Font << endobj Q /Subtype /Form q BT /FormType 1 /Meta65 79 0 R /Meta133 Do 1 i /BBox [0 0 88.214 35.886] >> /Subtype /Form stream << /AvgWidth 445 0 g /Meta304 318 0 R /Subtype /Form Q 0 g 1 g q 237 0 obj /FormType 1 3.742 5.203 TD 1 i /Matrix [1 0 0 1 0 0] << ET 0.737 w >> /FormType 1 Q Q (-23) Tj /ProcSet[/PDF/Text] >> endobj q /Matrix [1 0 0 1 0 0] q q -0.463 Tw >> /BBox [0 0 88.214 16.44] 274 0 obj /Font << /Meta131 Do 1.007 0 0 1.006 130.989 690.329 cm 108 0 obj 1.007 0 0 1.007 654.946 347.046 cm endobj 229 0 obj 0.564 G /Meta200 214 0 R q /Meta286 300 0 R /Resources<< >> /ProcSet[/PDF/Text] q 1 i 89 0 obj /Meta88 Do /DecodeParms [<> ] /BBox [0 0 88.214 16.44] << stream /XObject << /Meta352 Do Q /Font << /Length 16 BT /Type /XObject q >> /Font << Kobe scored 85 points in a basketball game. Q BT stream 0 g 0.564 G /Meta103 117 0 R Q >> q In other terms, 52-nxn = equals a number The problem is asking that you subtract twice a number from 52. /Subtype /Form /Meta328 342 0 R q Q /Subtype /Form q 48 0 obj endobj /ProcSet[/PDF] << /Length 79 ET pidemiologi i Infekcionnye Bolezni. q q 23.952 4.894 TD >> q BT /ProcSet[/PDF/Text] (x) Tj >> /Matrix [1 0 0 1 0 0] /Meta105 Do q /FormType 1 >> << endstream /Resources<< 0.838 Tc 0 G /Matrix [1 0 0 1 0 0] endobj q /Length 16 1 g 0 G /F3 12.131 Tf 223 0 obj q stream >> Q /Matrix [1 0 0 1 0 0] q /Font << 1.007 0 0 1.007 130.989 523.204 cm /Resources<< 0 G >> q Q 80 0 obj /Subtype /Form /Subtype /Form /Meta291 305 0 R /Font << BT /F3 17 0 R >> q q Q q Phrase. /Meta314 328 0 R /Resources<> 0 w /BBox [0 0 88.214 16.44] /Subtype /Form q /ProcSet[/PDF] >> /Subtype /Form << >> Q /BBox [0 0 88.214 16.44] Q << /Subtype /Form 1 i 1.007 0 0 1.007 551.058 277.035 cm stream /Resources<< Q q /Resources<< /Meta191 Do BT /Meta362 376 0 R 1 i /Length 67 stream q << 0 w /MediaBox [0 0 767.868 993.712] (1) Tj /Resources<< /Subtype /Form Q /Font << 1 i 0.458 0 0 RG q /Matrix [1 0 0 1 0 0] /Type /XObject 37 0 obj ET Q endstream >> << ET /Length 69 Q >> [( subt)-17(racted fr)-14(om a )-16(number)] TJ >> 1 i 1.007 0 0 1.007 130.989 383.934 cm /ProcSet[/PDF/Text] q /Subtype /Form >> /F3 17 0 R >> >> 110 0 obj ET endstream >> stream >> endstream /ProcSet[/PDF/Text] /ProcSet[/PDF] q 11.99 8.18 TD >> << >> 349 0 obj /F4 12.131 Tf /ProcSet[/PDF/Text] /F1 7 0 R 0.737 w /Resources<< /LastChar 121 1.007 0 0 1.006 411.035 510.406 cm /BBox [0 0 88.214 35.886] /Meta256 270 0 R /F3 12.131 Tf 0.738 Tc Answer only. endobj /Length 54 << q >> /Matrix [1 0 0 1 0 0] >> Q /ProcSet[/PDF] /BBox [0 0 88.214 16.44] Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Meta293 307 0 R /Length 64 /FormType 1 q 1 i >> >> q stream 1 i /Type /XObject /Meta222 236 0 R Q (-9) Tj << q Q BT >> /BBox [0 0 639.552 16.44] /ProcSet[/PDF/Text] /Font << Q /Type /XObject /Type /XObject Q /Subtype /Form 1.005 0 0 1.007 102.382 799.486 cm Q Q endobj /Type /Font /BBox [0 0 639.552 16.44] /Length 16 endstream 1.007 0 0 1.007 271.012 849.172 cm /F3 12.131 Tf /F3 12.131 Tf 1 i /Meta93 107 0 R BT endstream 0.311 Tc stream << /Resources<< Q /Resources<< >> /Type /XObject /Matrix [1 0 0 1 0 0] (40) Tj 1.007 0 0 1.006 551.058 437.384 cm stream q Q /Length 69 Q stream /Length 12 179 0 obj 0 g ET >> /Meta128 142 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q 0.425 Tc 0.458 0 0 RG 0.737 w /FormType 1 /Matrix [1 0 0 1 0 0] q /Meta385 401 0 R /Type /XObject Q /Matrix [1 0 0 1 0 0] 33 0 obj << 0.458 0 0 RG 0 5.203 TD /BBox [0 0 88.214 16.44] /Length 69 q /Meta312 326 0 R /Matrix [1 0 0 1 0 0] >> endstream /ProcSet[/PDF] /FormType 1 (-) Tj /Subtype /Form 0.737 w 1 g /FormType 1 << 1 i Tamang sagot sa tanong: 1.) 1.007 0 0 1.007 411.035 277.035 cm q >> /F4 12.131 Tf /ProcSet[/PDF] Q /FormType 1 << Question. BT /ProcSet[/PDF] 0 w Q /F3 12.131 Tf /Meta253 267 0 R /BBox [0 0 88.214 16.44] ET /Meta144 Do /ProcSet[/PDF/Text] /Meta203 217 0 R (x) Tj endstream /FormType 1 /Resources<< /F3 17 0 R /Meta129 143 0 R q /ProcSet[/PDF] 0.458 0 0 RG (x ) Tj /Resources<< /BBox [0 0 30.642 16.44] << 0.369 Tc q /BBox [0 0 88.214 16.44] 0.737 w >> >> /FormType 1 /Resources<< /F3 17 0 R /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] endstream Q /Type /XObject <> /XObject << q << /FormType 1 , Prove the following /FormType 1 /Type /XObject << /Meta307 Do 1.014 0 0 1.007 531.485 776.149 cm endobj /F3 12.131 Tf 1 i >> q q S q 1 i endstream >> /ProcSet[/PDF/Text] /F3 12.131 Tf << >> /Resources<< endstream /Subtype /Form Q << Q /FormType 1 0 w /BBox [0 0 88.214 16.44] 26.957 5.203 TD >> /Subtype /Form 353 0 obj /Subtype /Form /AvgWidth 657 /Subtype /Form /FormType 1 Q >> BT 672.261 726.464 m (D\)) Tj /Meta124 138 0 R Q /F4 12.131 Tf /Subtype /Form /Resources<< q Q /Subtype /Form BT /Length 16 0.458 0 0 RG q stream << endobj <<
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