0 G q q 0 w /Meta47 Do /Type /XObject q BT >> Q /BBox [0 0 15.59 16.44] /Meta168 182 0 R >> Q /Resources<< /Resources<< Q q /ProcSet[/PDF/Text] 101.849 5.203 TD q Q 0.737 w stream 0 g Q stream >> endobj Q /Length 69 Q 43.426 5.203 TD /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] >> ET q q endobj 0 g 1 i 0 G << 0 G 0 g /Subtype /Form The value of k is: (b) 3 (d) 0 (a) 4 (c) -4 TL ing:, 1)take a graph and draw two perpendicular lines to obtain four uadrants 2)draw any object using straight line 3) write the coordinates of each point o q /Meta211 Do /Resources<< Q endstream 155 0 obj Q 137 0 obj /Subtype /Form /Subtype /Form /Length 68 /F1 12.131 Tf >> >> The width Of a rectangle is 15 cm and the perimeter is 12 cm. /F3 12.131 Tf /Font << /Matrix [1 0 0 1 0 0] endstream /Matrix [1 0 0 1 0 0] >> 1.007 0 0 1.007 551.058 636.879 cm 206 0 obj stream 187 0 obj If a number is 50%, then it is a half - the same as 0.5 or 1/2. /Font << Q /Meta393 Do 0 g 0 g 191 0 obj /FormType 1 /F3 17 0 R /Font << /Meta179 Do << 0 G Q endobj 1.007 0 0 1.007 654.946 726.464 cm Q 1.014 0 0 1.007 391.462 330.484 cm 1 i /Font << 130 0 obj << /Meta406 422 0 R /Meta96 110 0 R /Matrix [1 0 0 1 0 0] Q stream Q /Resources<< << >> q 1 i 0 G Q /BBox [0 0 88.214 16.44] 0 G /Meta322 336 0 R /Resources<< BT /Matrix [1 0 0 1 0 0] /F1 12.131 Tf /ProcSet[/PDF/Text] /Type /XObject q /Type /XObject ET >> (D\)) Tj Q stream 0.738 Tc /Meta19 30 0 R /Type /XObject /Length 69 /Meta310 Do 380 0 obj >> q 0.369 Tc q /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 523.204 cm /ItalicAngle 0 /Type /XObject 341 0 obj Q /ProcSet[/PDF/Text] /Font << -0.486 Tw 0 g Q /Font << stream /Meta49 Do << 0 g Q /F4 12.131 Tf q /Type /XObject stream BT /Resources<< Q 1 g q /Resources<< /Meta89 103 0 R /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF] >> /Length 106 236 0 obj >> stream Q 672.261 400.496 m 0.564 G (B) Tj 0 g 0.737 w /F3 17 0 R >> /F3 17 0 R /FormType 1 0 G endstream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] /FormType 1 20.21 5.203 TD /Resources<< 1.007 0 0 1.007 130.989 330.484 cm /Type /XObject 0 g /Length 54 /Type /XObject Q 0 G /BBox [0 0 30.642 16.44] << /Font << >> 0 G Q /FormType 1 endstream /Meta287 301 0 R q /Subtype /Form endstream Q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] 0.458 0 0 RG q /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] endstream ET /FormType 1 0 G ET 1.007 0 0 1.007 130.989 523.204 cm /Length 69 >> /Resources<< Q /F4 36 0 R /Matrix [1 0 0 1 0 0] >> >> >> 298 0 obj q q >> 1 i /FormType 1 BT >> 1.502 5.203 TD /Type /XObject endobj ET -0.486 Tw /Subtype /Form >> endstream Q BT /BBox [0 0 88.214 16.44] endstream /ProcSet[/PDF/Text] 22.478 5.336 TD /BBox [0 0 88.214 16.44] /Subtype /Form 1 i /Font << /Type /XObject /Type /XObject 0 g Q Q /Resources<< (C\)) Tj /BBox [0 0 17.177 16.44] /Type /XObject /BBox [0 0 88.214 16.44] /F3 17 0 R q q 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. /Subtype /Form /LastChar 121 /Subtype /TrueType /Encoding /WinAnsiEncoding /BBox [0 0 15.59 16.44] ET /FormType 1 /Resources<< q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] q /Meta148 Do Q Q /Resources<< 0 g /Resources<< (C\)) Tj 1 i Q /FormType 1 /F4 12.131 Tf /Length 78 endobj 1 i /Type /Catalog /I0 Do >> /Matrix [1 0 0 1 0 0] /Meta313 327 0 R /Resources<< >> /ProcSet[/PDF/Text] /Resources<< /Resources<< /Meta51 65 0 R Q stream 1.502 5.203 TD /Matrix [1 0 0 1 0 0] Q 0.737 w stream Q endstream >> /Resources<< /Flags 32 Q /Length 12 Q endstream /Meta169 183 0 R /FormType 1 /Font << >> /FormType 1 Q /FormType 1 /FormType 1 /Type /XObject Q 1 i 16.469 5.336 TD /Meta283 297 0 R 0 w Two speeding tickets could increase your rate by 58% at your next renewal. >> /Subtype /Form q /BBox [0 0 88.214 16.44] /Length 69 >> q 0 G endobj /Font << BT Q q /Subtype /Form /ProcSet[/PDF/Text] endstream 306 0 obj five times the sum of a number x and two b.) /Resources<< 0.737 w /Length 57 0 5.203 TD /FormType 1 /ProcSet[/PDF/Text] /Length 139 /Matrix [1 0 0 1 0 0] /Resources<< /Meta377 Do 30 0 obj Q stream 2x - y = 6. Q /Meta87 Do q endstream /Meta32 Do 1 g Q BT 1.005 0 0 1.007 102.382 653.441 cm Q >> 172 0 obj q /BBox [0 0 88.214 16.44] q BT /F4 36 0 R /Subtype /Form Twice a number when decreased by 7 gives 45. BT 0 4.894 TD q /Length 63 Q 0.486 Tc /Length 104 endobj /Matrix [1 0 0 1 0 0] /Length 69 >> /ProcSet[/PDF/Text] (D\)) Tj /Length 16 /ProcSet[/PDF/Text] >> Q /Type /XObject q << 0.68 Tc /Matrix [1 0 0 1 0 0] 1 i 1 i q /Matrix [1 0 0 1 0 0] /Meta193 Do >> /I0 Do /I0 Do /Matrix [1 0 0 1 0 0] [tex]\sin (\pi -x)=\sin x[/tex]. 197 0 obj /Type /XObject Q q /Subtype /Form n 11 or n 11. q q Q /ProcSet[/PDF] << q endstream >> endobj 0 G /ProcSet[/PDF] 0.68 Tc stream endstream /Meta195 209 0 R /FormType 1 /Matrix [1 0 0 1 0 0] 20.21 5.203 TD endstream q 0 g 1 g /Meta59 Do Q /BBox [0 0 88.214 16.44] >> /Resources<< q q 154 0 obj Q q << /Resources<< /FormType 1 q Q 0 g 1.007 0 0 1.007 551.058 330.484 cm << stream 32.201 5.203 TD /Matrix [1 0 0 1 0 0] Q /BBox [0 0 30.642 16.44] 0 G /FormType 1 /Subtype /Form 0.297 Tc q /Type /XObject Q Q /Resources<< 1 i /FormType 1 /Matrix [1 0 0 1 0 0] q A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. /ProcSet[/PDF] /F3 12.131 Tf /Length 16 133 0 obj >> q /BBox [0 0 88.214 35.886] q q /Matrix [1 0 0 1 0 0] 118 0 obj 1 i q /Subtype /Form endobj q q /Length 58 q /F4 12.131 Tf 3.742 8.18 TD /Meta241 255 0 R >> Q /Meta50 Do /Meta180 Do 0 g >> q endobj Q /Length 12 endstream /Meta248 Do (D) Tj q Q /StemH 94 endobj q ET 0.737 w >> q q endstream 32 = 2a + 8: The quotient of fifty and five more than a number is ten. /Filter [/CCITTFaxDecode] >> q /F3 17 0 R >> 0 g q 221 0 obj endstream endobj << /Type /Font q q /Matrix [1 0 0 1 0 0] Q /Leading 349 /Length 70 0 5.203 TD /Length 69 /Meta219 Do >> >> endobj 0.737 w BT /Meta121 Do /Meta388 404 0 R 1 i /Type /XObject stream Find the number. >> endobj LAIing for a pizza and, soft drink. /Length 16 /Font << -0.463 Tw endobj q 0 G /ProcSet[/PDF] 26 0 obj ET 0 g ET >> /Subtype /Form /Matrix [1 0 0 1 0 0] q Q endstream /F3 17 0 R 47.933 5.203 TD 0.458 0 0 RG << Q /Matrix [1 0 0 1 0 0] Q 549.694 0 0 16.469 0 -0.0283 cm >> endstream Q Q >> 0 5.203 TD q q /F3 12.131 Tf /Subtype /Form >> q endstream >> /FormType 1 << endstream stream stream Q ET /Meta84 Do /Meta102 Do endobj endstream 349 0 obj Q Q: A number increased by 5 is equivalent to twice the same number decreased by 7. >> /Font << BT Q BT 0 w /Resources<< 0 g /Type /XObject stream 1 i 1 i 1.007 0 0 1.007 67.753 293.596 cm algebraic expressions math_celebrity Administrator Staff Member Translate this phrase into an algebraic expression. /Subtype /Form /Resources<< 0 w 0 w /Length 245 /Meta384 Do Q 0 g 0 g /F3 12.131 Tf q /F3 12.131 Tf << /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Type /XObject 0 G ET 0.564 G >> q >> 0.737 w /F3 12.131 Tf ET 427 0 obj q 20.21 5.203 TD >> /Meta182 196 0 R /Length 69 /Length 58 164 0 obj q 1.007 0 0 1.007 551.058 703.126 cm q endobj << >> 0 G /Font << 0 4.894 TD Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. Q Q /Font << Testosterone is the primary sex hormone and anabolic steroid in males. /F3 17 0 R endobj /BBox [0 0 30.642 16.44] Q /Meta414 430 0 R /ProcSet[/PDF/Text] ET 0 G /Resources<< Q /Type /XObject New questions in Mathematics /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] q /Resources<< 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. q /Meta305 Do stream stream endstream q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /F3 17 0 R 0 G 0 G q /Resources<< endstream /Font << Q >> endobj /Meta100 Do q 1 i /BBox [0 0 88.214 16.44] >> BT 0 w /Subtype /Form /FormType 1 /ProcSet[/PDF/Text] 0 5.203 TD << 0.51 Tc >> 186 0 obj >> /Type /XObject stream ET 145 0 obj Q 1 i /F3 17 0 R Q /ProcSet[/PDF] /Font << >> q 252 0 obj q 1.007 0 0 1.007 551.058 703.126 cm endstream >> /Matrix [1 0 0 1 0 0] /FormType 1 0 g /Matrix [1 0 0 1 0 0] /Resources<< q 14.23 24.649 TD /FormType 1 /Meta53 67 0 R Q /Type /XObject 1.007 0 0 1.007 271.012 849.172 cm Q 722.699 799.486 l Q >> >> >> endstream 1 i (2\)) Tj /Resources<< 0.564 G (7\)) Tj /ProcSet[/PDF] Q endstream 0.458 0 0 RG Now that you know the meaning of the key words you can read the problem differently. /FormType 1 /Length 69 /Type /XObject >> Q ET /ProcSet[/PDF/Text] Q 0 G /F3 12.131 Tf q q /F3 17 0 R /Type /XObject q /Subtype /Form /Meta71 85 0 R 180 0 obj >> /DecodeParms [<> ] /Subtype /Form /F3 17 0 R 1.014 0 0 1.007 531.485 383.934 cm Q stream >> /Subtype /Form >> /Subtype /Form ET /Meta295 Do endobj stream 1 i q /Resources<< Q ET Q q (+) Tj 1.007 0 0 1.007 411.035 636.879 cm /FormType 1 (D\)) Tj endstream /Length 88 Q >> /Font << /BBox [0 0 15.59 29.168] /F3 17 0 R q Q >> /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] /FormType 1 ET /Length 16 Q 0 G (A\)) Tj >> /Meta429 Do ET ET 0.458 0 0 RG 0 g 1.005 0 0 1.007 102.382 653.441 cm ET /Resources<< ET /Meta333 347 0 R Q Q q 1.007 0 0 1.007 551.058 277.035 cm q ET Translate 2(x-58) into mathematical phrase. endobj 1 g /Resources<< Q 415 0 obj /Type /XObject q q Q /Type /XObject /Matrix [1 0 0 1 0 0] 1 g 0 g Q /Length 103 /F4 36 0 R << 1 g /ProcSet[/PDF/Text] /FormType 1 /Meta183 197 0 R 1 i endobj >> 353 0 obj endobj ( x) Tj q q /Length 68 /BBox [0 0 88.214 35.886] 1 i q q << endstream /Type /XObject /F3 12.131 Tf Q (x) Tj << -0.084 Tw stream /BBox [0 0 30.642 16.44] stream /Matrix [1 0 0 1 0 0] /Subtype /Form q 41 0 obj endstream q 0.297 Tc Let x be a number. >> >> /Meta359 Do 0 G 0.564 G endstream /Font << /Meta6 15 0 R /BBox [0 0 549.552 16.44] /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] q Q 0 g 722.699 546.541 l /Meta214 228 0 R /Matrix [1 0 0 1 0 0] /Type /XObject /Type /XObject /BBox [0 0 88.214 16.44] endobj 1.502 5.203 TD 0.564 G q endstream /Length 66 >> /Meta294 Do q /I0 51 0 R stream /Font << endobj 0 G Q >> >> /Resources<< /Resources<< /Meta150 Do 258 0 obj stream q q 0 g stream 238 0 obj >> /BBox [0 0 534.67 16.44] Q /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 391.462 636.879 cm 21 0 obj Q /Matrix [1 0 0 1 0 0] 0.564 G Q q endstream /Meta177 Do /Matrix [1 0 0 1 0 0] endstream Q BT /ProcSet[/PDF/Text] endstream << Q /BBox [0 0 15.59 16.44] q Twice a number would be 2x. /Matrix [1 0 0 1 0 0] 1 g /F3 17 0 R q 0 g 0 G 0 g q /Matrix [1 0 0 1 0 0] /Subtype /Form Find the length. 0 g BT endstream /F1 7 0 R endstream /Type /XObject 30.699 5.203 TD << 128 0 obj /F3 17 0 R q /F1 7 0 R Q /Subtype /Form BT [( the )-24(sum of a n)-14(umber an)-14(d )] TJ /Meta254 Do q /Meta357 371 0 R /ProcSet[/PDF/Text] endstream q >> endobj 1 i endobj 376 0 obj Q /Resources<< 0 g /Font << /Length 59 q endstream << 7 0 obj 1 i 0.564 G >> /Length 67 q 0 5.203 TD /Meta373 Do Q 181 0 obj /Matrix [1 0 0 1 0 0] /Font << 214 0 obj A rectangular garden has a width that is 8 feet less than twice the length. >> << /Font << stream 20.21 5.203 TD 1.007 0 0 1.007 271.012 636.879 cm q q /Type /XObject /Subtype /Form /Meta103 117 0 R Q q /Resources<< /Meta388 Do Q (11) Tj stream endstream /Resources<< /Meta114 Do Q endstream >> /Meta285 299 0 R Q Q /Type /XObject /BBox [0 0 88.214 16.44] Q 0 w stream /FormType 1 /Meta44 Do /Meta0 5 0 R stream /Resources<< 0.369 Tc /Resources<< /BaseFont /TestGen-Regular q /F3 12.131 Tf BT /Meta404 420 0 R /Length 59 /FormType 1 Q /Subtype /Form /Resources<< /Font << /Meta35 Do /XObject << /Meta37 Do /FormType 1 /Resources<< q /Meta41 Do /F3 12.131 Tf endobj /F3 12.131 Tf /ProcSet[/PDF/Text] >> 92 0 obj /Matrix [1 0 0 1 0 0] 0.564 G 32.939 5.203 TD Q q /Type /XObject /ProcSet[/PDF] q /Font << stream >> endobj >> Q /XObject << Q /Meta253 267 0 R /FormType 1 Q 1 i 1.007 0 0 1.007 45.168 829.599 cm >> /Length 58 q 1 i 1.007 0 0 1.006 411.035 763.351 cm /Type /XObject 0 g 6.746 5.203 TD 0 5.203 TD /Type /XObject endstream Q 0 g >> /Meta295 309 0 R 1.007 0 0 1.006 411.035 690.329 cm /BBox [0 0 88.214 16.44] /Meta364 Do Q 0 w 407 0 obj << << Q 248 0 obj /Font << q >> /Type /Page 0 g /ProcSet[/PDF/Text] >> >> Q /Meta250 Do endobj /FormType 1 stream 1 i 722.699 599.991 l BT /Type /XObject Q /ProcSet[/PDF/Text] /Resources<< /F3 17 0 R endobj /ProcSet[/PDF/Text] 1 i /Type /XObject /Length 16 /Type /XObject Q /Meta382 Do 0 g /Font << /FormType 1 q BT /Font << /FormType 1 98 0 obj 140 0 obj Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 >> 0 g /Font << 0.738 Tc 1.005 0 0 1.007 102.382 653.441 cm /F3 12.131 Tf /Meta105 Do /Meta405 421 0 R q /F3 12.131 Tf 0 g 1 i /Meta124 138 0 R /Meta11 Do /Meta60 Do /Type /XObject >> Q /Subtype /Form /Length 69 1 i /Type /XObject Check out a sample Q&A here. 0 g BT /Meta115 129 0 R Q Q /BBox [0 0 534.67 16.44] endobj /Matrix [1 0 0 1 0 0] << /FormType 1 (+) Tj q /Subtype /Form endobj ET q stream /Meta311 325 0 R q q (40) Tj q Q Q Q 0 G 20.21 5.336 TD BT << 52 0 obj /Length 118 /Subtype /Form q 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject 409 0 obj /F3 17 0 R /Type /XObject /Meta68 Do /FormType 1 q >> /FormType 1 /Length 16 (x ) Tj (-11) Tj 0 g Q endstream /Meta204 218 0 R BT endstream endstream << 0 G 1.502 5.203 TD 15.731 5.336 TD 25 0 obj stream /Type /XObject /FormType 1 /Type /XObject 0 w /F3 17 0 R /I0 51 0 R /Meta179 193 0 R 365 0 obj 0 g ET endstream endobj /F3 17 0 R 1.014 0 0 1.006 251.439 510.406 cm endstream (1\)) Tj /Subtype /Form 0 w /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] 0.369 Tc /Type /XObject /Resources<< ET 0 G /Meta148 162 0 R /Subtype /Form >> /Subtype /Form q q /Meta288 302 0 R Q Q /Meta228 Do >> /Matrix [1 0 0 1 0 0] /Subtype /Form stream /F1 7 0 R /BBox [0 0 15.59 29.168] /BBox [0 0 673.937 68.796] Q endstream q /Meta212 Do /ProcSet[/PDF/Text] /Subtype /Form /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] 0.524 Tc 285 0 obj q /F3 17 0 R >> 1.014 0 0 1.007 111.416 523.204 cm 0 g 0.486 Tc q 1 i Q /FormType 1 /Resources<< >> /F3 12.131 Tf /F3 17 0 R BT BT /ProcSet[/PDF/Text] Q stream /Length 59 q endobj /Meta304 Do /Length 70 /Type /XObject /FormType 1 /Type /XObject /BBox [0 0 30.642 16.44] /F3 12.131 Tf 0 g /Subtype /Form 1 i endstream 1 i /Meta238 Do /Meta316 Do /Font << Q Q << 1 i /Subtype /Form /Meta31 Do 1 g /F3 17 0 R /BBox [0 0 88.214 16.44] >> 227 0 obj /Type /XObject Q /ProcSet[/PDF] /Length 60 1 i Q (11) Tj 22.478 5.336 TD /Meta292 306 0 R 17 0 obj << 0 g << 0 G /Font << Q Q Q ( \() Tj 1 i /Meta358 372 0 R 0.458 0 0 RG 0 G /Matrix [1 0 0 1 0 0] (\)]) Tj /Length 67 << >> >> /BBox [0 0 88.214 16.44] /ProcSet[/PDF] q Q Q /BBox [0 0 30.642 16.44] (B\)) Tj q q A number divided by six is eight: (k / 6) = 8. /Length 16 1 i << /ProcSet[/PDF/Text] endstream stream Q BT << q Q Q << /Meta239 253 0 R endobj /FormType 1 endstream /Subtype /Form Q >> /FormType 1 BT << endobj 171 0 obj q q q ET q /Subtype /Form Q /Meta88 Do 0.458 0 0 RG q 0 w << 6.746 5.203 TD endobj /Resources<< Q q (C\)) Tj /F3 17 0 R /BBox [0 0 88.214 16.44] << /ProcSet[/PDF/Text] (2\)) Tj 0 g 1.014 0 0 1.006 531.485 763.351 cm 1 g /F1 12.131 Tf q endobj Q q /BBox [0 0 88.214 16.44] 0 g >> /FormType 1 /Resources<< Q BT 0 5.203 TD 0.458 0 0 RG 401 0 obj /Length 58 stream /F3 17 0 R 178 0 obj /Type /XObject 0 w 0 g /Resources<< q /Subtype /Form (C\)) Tj /Meta13 Do << /Matrix [1 0 0 1 0 0] /Resources<< 0 G stream endstream 1.007 0 0 1.007 45.168 713.666 cm 1 i q 0 G /Font << q /ProcSet[/PDF/Text] endobj Q /FormType 1 q /BBox [0 0 30.642 16.44] q /Meta418 434 0 R q /Type /XObject BT /Type /XObject Q stream >> Q 0.458 0 0 RG q /ProcSet[/PDF/Text] /Length 2252 140781 Q /ProcSet[/PDF/Text] 399 0 obj Q /Resources<< endstream Q 0.738 Tc /Subtype /Form endobj >> endobj 0 g endobj /BBox [0 0 639.552 16.44] 56 0 obj Q /F3 12.131 Tf We are asked to find the number, so, we could assign the number as "x". 1 i /ProcSet[/PDF] /Resources<< >> >> BT /Meta53 Do 43 0 obj << Q /Meta144 Do /Type /XObject q /Subtype /Form 1.014 0 0 1.006 111.416 836.374 cm q 0.458 0 0 RG >> /FormType 1 << >> 1 g endobj << q Q /Resources<< /BBox [0 0 88.214 16.44] (2) Tj >> /Meta136 Do /Type /XObject /Type /XObject 0 g /Meta31 44 0 R /Subtype /Form Q /Meta300 Do 260 0 obj 0.737 w 2x - 15 = -27. /Type /XObject >> q q /FormType 1 Q /Meta20 Do endobj endobj /Meta117 131 0 R 1 g Q Six subtracted from a number 6. ET endobj /F3 12.131 Tf /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 /BBox [0 0 15.59 16.44] 0 g 0.458 0 0 RG /Resources<< ET Q endstream /FormType 1 Q >> Diabetes is due to either the pancreas not producing enough insulin, or the cells of the body not responding properly to the insulin produced. Q q endstream Q /Resources<< 3.742 5.203 TD /FormType 1 0 G << /Subtype /Form There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. 1.007 0 0 1.007 271.012 636.879 cm 0 G >> /Meta324 338 0 R << stream >> Q /ProcSet[/PDF/Text] 1.502 5.203 TD /Length 12 /Font << /BBox [0 0 30.642 16.44] q 26.219 5.203 TD /FormType 1 Q ET /Subtype /Form ET >> /FormType 1 /Meta222 Do /BBox [0 0 673.937 16.44] 0 G endobj /Length 16 1 g q >> /BBox [0 0 673.937 14.853] /Resources<< 433 0 obj << ET (2) Tj /Kids [ 23.216 5.203 TD ET /BBox [0 0 15.59 16.44] endstream 1 i << 1.007 0 0 1.007 271.012 583.429 cm /Length 54 /BBox [0 0 534.67 16.44] 0.486 Tc /F3 17 0 R q >> Two fewer than a number doubled is the same as the number decreased by 38. /F3 12.131 Tf 0000000000 65535 f 0000140665 00000 n 0000140732 00000 n 0000000015 00000 n 0000120613 00000 n 0000000126 00000 n 0000000314 00000 n 0000000577 00000 n 0000001009 00000 n 0000001360 00000 n 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