twice a number decreased by 58

Q stream /Type /XObject Q q 0 G BT 0 g /BBox [0 0 88.214 16.44] /Resources<< q Q q BT 437 0 obj /Filter [/CCITTFaxDecode] stream q /Meta341 355 0 R stream stream 174 0 obj BT >> Q 1.007 0 0 1.007 45.168 730.228 cm (-23) Tj Q Find the number. q 1 g /Subtype /Form /F3 12.131 Tf 0 G ET /ProcSet[/PDF/Text] 1 i Q /FormType 1 /Subtype /Form /BBox [0 0 15.59 29.168] /F3 17 0 R >> /ProcSet[/PDF] /Resources<< Q (9\)) Tj (5) Tj /Matrix [1 0 0 1 0 0] 0 g 1 g /Type /XObject BT /F3 12.131 Tf 1.005 0 0 1.007 79.798 829.599 cm /Matrix [1 0 0 1 0 0] /BBox [0 0 30.642 16.44] /Meta318 332 0 R /Subtype /Form endstream 0 5.203 TD /Type /XObject /BBox [0 0 88.214 16.44] /Meta5 14 0 R Q Q /Matrix [1 0 0 1 0 0] /Meta20 31 0 R /F3 17 0 R 0 G /ProcSet[/PDF] 1.502 5.203 TD 0 g 234 0 obj /Length 16 /Font << 0.458 0 0 RG /Length 16 117 0 obj /F3 12.131 Tf q stream /Font << Q q stream q stream /Length 78 << /FormType 1 (x ) Tj Q 40 0 obj Q >> Q >> /Length 58 /F3 17 0 R /Meta120 134 0 R << q /Font << ET Q Q endobj 0.458 0 0 RG /Length 69 /Resources<< stream 1 i A link to the app was sent to your phone. ET /Matrix [1 0 0 1 0 0] 332 0 obj 0 G q /XHeight 471 q /Matrix [1 0 0 1 0 0] << /F3 12.131 Tf Q /Type /XObject endobj /FormType 1 /Font << /Type /XObject 0 0 500 500 500 500 500 500 500 500 500 500 0 0 0 0 q /BBox [0 0 673.937 68.796] >> endobj stream 1 i /BBox [0 0 88.214 16.44] 0 g Q /Meta397 Do 3.742 8.18 TD /Meta409 Do stream /Matrix [1 0 0 1 0 0] /Resources<< /F3 17 0 R Q /Encoding /WinAnsiEncoding Q /Resources<< /Subtype /Form 0.786 Tc Q 345 0 obj Q >> ET Q 16.469 5.336 TD q << (x ) Tj stream /BBox [0 0 88.214 35.886] /ProcSet[/PDF] q q /F3 17 0 R /CapHeight 662 /F3 12.131 Tf Q /ProcSet[/PDF] >> >> >> endstream /FormType 1 >> /F3 17 0 R ET 0 G /F3 17 0 R 0 g Q 49 0 obj 0.369 Tc /FormType 1 Q /Subtype /Form << stream /F3 17 0 R /ProcSet[/PDF/Text] q /F3 12.131 Tf 1 i 1.014 0 0 1.007 531.485 383.934 cm /Font << /Type /XObject 0.369 Tc /Subtype /Form 1.007 0 0 1.007 67.753 599.991 cm 0.564 G /Meta124 Do endstream q >> Q D. Twice a number decreased by ten is less than 24. 549.694 0 0 16.469 0 -0.0283 cm Q 96 0 obj Q >> /Type /XObject >> /Subtype /Form q 197 0 obj endstream (2) Tj /FormType 1 0.458 0 0 RG a and b or something else.***. >> /ProcSet[/PDF/Text] 1 i /Meta140 154 0 R 0 g 0.737 w q 0.737 w 32.201 5.203 TD endobj stream 0.838 Tc q stream /Subtype /Form /Type /XObject All steps. q q q 0 G /Font << /Font << endobj /Subtype /Form endstream 1 i q endobj Q q /FormType 1 /StemV 94 (B\)) Tj 0 G endobj >> /F3 17 0 R /Matrix [1 0 0 1 0 0] Q Q /Font << BT >> q endobj 17.234 5.203 TD 0 w Q /Type /XObject Q /Meta8 19 0 R >> /F3 12.131 Tf /Length 12 ET [tex]\sin (\pi -x)=\sin x[/tex]. 20.21 5.203 TD 0.425 Tc Q << 25.454 5.203 TD (B\)) Tj 347 0 obj /F3 17 0 R >> endstream /FormType 1 0 g endobj endobj /Matrix [1 0 0 1 0 0] 20.21 5.203 TD /Font << BT 0 g 0 G /Length 69 /Subtype /Form >> /Type /XObject /BBox [0 0 15.59 16.44] >> stream q q 1 i q Q Q /Font << /Meta71 Do /ProcSet[/PDF] /Meta18 Do /Length 245 /Meta51 Do /Length 59 /FormType 1 << 0.737 w >> /Meta236 250 0 R /F1 12.131 Tf /ProcSet[/PDF/Text] << endstream 0 G endobj endstream q Q endstream /F3 12.131 Tf endstream /F1 12.131 Tf Q Q /Font << >> 4.506 8.18 TD endobj /Meta304 318 0 R q /Matrix [1 0 0 1 0 0] /F3 17 0 R /FormType 1 /Meta169 Do q 1.007 0 0 1.007 271.012 523.204 cm 1 i q 0 g -y. q Q 204 0 obj << /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] endstream Q 1 i 0 g ET Q endstream /Resources<< 722.699 653.441 l 1 i /Resources<< 421 0 obj endstream /Subtype /Form There was a 2,769 mmol/L decrease in blood glucose levels after treatment with rectal ozone, which shows metabolic control. 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 0 g 0.175 Tc /BBox [0 0 88.214 35.886] 1 i q >> stream /BBox [0 0 17.177 16.44] q 0 5.203 TD 161 0 obj stream << 1.007 0 0 1.007 271.012 277.035 cm 0.458 0 0 RG endstream /FormType 1 1 i /Meta252 266 0 R BT 1 i /F3 12.131 Tf >> /Matrix [1 0 0 1 0 0] /Resources<< /Meta219 233 0 R /Matrix [1 0 0 1 0 0] BT stream q Q 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. /I0 Do /ProcSet[/PDF] /Length 69 /Subtype /Form q 2. /F3 17 0 R /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Meta75 89 0 R >> >> /Subtype /Form /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] Q /Resources<< /Resources<< q /F3 12.131 Tf 0.737 w /Type /XObject 75 0 obj /I0 Do q endstream q /F1 12.131 Tf /Subtype /Form BT /Type /XObject /F3 12.131 Tf /Subtype /Form 54.679 5.203 TD 0.564 G /ProcSet[/PDF/Text] (x) Tj /Length 54 /AvgWidth 401 BT 0 w >> /ProcSet[/PDF/Text] Q Q /F4 36 0 R /Meta260 Do /FormType 1 /F3 12.131 Tf Q endstream BT /XObject << q 0 g q Q >> /Subtype /Form /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) >> /F1 12.131 Tf 0.737 w q /Encoding /WinAnsiEncoding Q >> >> 722.699 726.464 l A. /Type /XObject first we change the sentence in formula as the following, i think this is clear &easy to understand .i hope it helps you, This site is using cookies under cookie policy . /F3 17 0 R /FormType 1 0 G BT 0 G stream << /Font << /Length 104 >> 44 0 obj >> 386 0 obj 0.458 0 0 RG /Meta297 Do BT q 0 g /Meta66 Do q 6.746 5.203 TD 0.458 0 0 RG /BBox [0 0 534.67 16.44] 0 G /Resources<< stream >> How many points did Kobe score in the season? 0 G 1 i /Meta157 Do /CreationDate (D:20140515121932-04'00') 0.458 0 0 RG 1 i endobj /Subtype /Form endstream q MetS-Z quartiles and their associated risks are presented in Fig. >> Q /FontDescriptor 6 0 R endstream /Subtype /Form /Meta60 Do stream << /BBox [0 0 88.214 16.44] q Q Three times a number equals fifteen 3. 0 G Q 0 G >> /Type /XObject /Type /XObject >> Q ET /Type /XObject /Meta87 101 0 R /Meta346 Do /FormType 1 stream /F1 7 0 R Q /Type /XObject 0 g Q /Subtype /Form (A\)) Tj q stream /F1 12.131 Tf /Meta189 Do Q Table 1. 1.007 0 0 1.007 411.035 849.172 cm endobj 162 0 obj ET /Meta411 427 0 R /F3 12.131 Tf The sum Of twice a nu4ber What is the number? /Resources<< 179 0 obj << stream /F3 17 0 R Q 205.199 4.894 TD /StemV 94 q >> /ProcSet[/PDF] /Type /XObject /F3 17 0 R /Matrix [1 0 0 1 0 0] q << Q Q 1 i endobj Q >> 0.458 0 0 RG q /FormType 1 >> endstream 0 5.203 TD 0.524 Tc Q ET q q Q /F3 12.131 Tf 0.564 G /FormType 1 BT 0 w /Matrix [1 0 0 1 0 0] /Meta309 323 0 R 1.007 0 0 1.007 551.058 636.879 cm 1.007 0 0 1.007 271.012 636.879 cm Q 1.005 0 0 1.007 102.382 310.158 cm Q /Type /XObject 1.014 0 0 1.006 111.416 437.384 cm q (5) Tj /Meta305 Do Six subtracted from a number 6. 1 g (-) Tj /BBox [0 0 15.59 16.44] BT >> /FormType 1 /Meta167 181 0 R /ProcSet[/PDF] /Meta367 381 0 R Q /Resources<< 154.289 4.894 TD /MaxWidth 1397 /F3 17 0 R /Meta349 363 0 R /Subtype /Form endstream /Resources<< /Meta200 214 0 R q q /Meta172 Do /ProcSet[/PDF] Q 426 0 obj /Type /XObject >> >> 1 g >> 1.014 0 0 1.007 531.485 330.484 cm /XObject << /Meta2 9 0 R q Q q /F3 17 0 R /Matrix [1 0 0 1 0 0] 140 0 obj 380 0 obj /Meta427 Do Q >> 0.486 Tc Q /Resources<< 195 0 obj endobj ET 1 i /Font << 0.024 Tw ET q 32.201 5.203 TD << Q 0.564 G q /Type /XObject Q 1 i /Meta214 228 0 R /Font << /I0 51 0 R /Meta191 Do /Matrix [1 0 0 1 0 0] /Meta58 72 0 R Q /Font << BT /BBox [0 0 88.214 35.886] << Q Q stream endobj /Subtype /Form >> q /Type /XObject >> endstream Q Q 12.727 5.203 TD 0 G /Type /XObject 1.007 0 0 1.007 130.989 277.035 cm Q 1 i (x) Tj 1 i /FormType 1 >> q >> /Length 69 /F1 7 0 R /Length 118 /F3 12.131 Tf /Type /XObject q 0 G /Meta64 Do q stream q Q (-11) Tj 0 G q stream /F3 17 0 R /Meta312 326 0 R q /Length 12 q q >> /Meta106 Do BT /Font << >> 0 g /Matrix [1 0 0 1 0 0] >> /ProcSet[/PDF] 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 . stream 0.564 G Q Q >> /Matrix [1 0 0 1 0 0] Q Q 143 0 obj /ProcSet[/PDF/Text] /F3 17 0 R 1 i BT /Subtype /Form /Matrix [1 0 0 1 0 0] 3.742 5.203 TD /Matrix [1 0 0 1 0 0] q /F3 17 0 R 261 0 obj endobj /BBox [0 0 88.214 16.44] /Meta316 330 0 R /ProcSet[/PDF/Text] 1.014 0 0 1.006 251.439 763.351 cm >> /Meta364 378 0 R >> /Font << /FormType 1 Q (-) Tj /Length 16 /FormType 1 1.014 0 0 1.006 391.462 763.351 cm /Type /XObject /ProcSet[/PDF/Text] << Q Q 1 i (x) Tj >> >> 0.458 0 0 RG /Leading 253 Twice a number would be 2x. Q >> << BT /F4 36 0 R 0 G q /Meta233 247 0 R 0 g /FormType 1 endstream 16.469 5.336 TD q 0.369 Tc /FormType 1 Q q /Matrix [1 0 0 1 0 0] Q >> q >> >> q 206 0 obj 1.014 0 0 1.007 251.439 583.429 cm /Length 54 /Meta226 Do ET /ProcSet[/PDF/Text] /FormType 1 /Type /XObject /Meta272 286 0 R /Type /XObject q /Meta89 Do 436 0 obj /BBox [0 0 88.214 16.44] /F3 17 0 R endobj endobj Q /ItalicAngle 0 /FormType 1 Q q /Resources<< BT Q endstream >> /BBox [0 0 88.214 16.44] >> /Type /XObject endstream Q /Resources<< /FormType 1 364 0 obj Q endstream 19.474 20.154 l /Subtype /Form 322 0 obj /Meta419 Do endstream Q /Meta301 Do << stream /ProcSet[/PDF] /Meta32 Do 140781 Q /F1 7 0 R 1.007 0 0 1.007 67.753 293.596 cm ET /Subtype /Form >> q 1.502 8.18 TD 0.458 0 0 RG >> << q q 1.007 0 0 1.007 551.058 583.429 cm /Meta12 Do endstream stream q /Subtype /Form /Resources<< /Type /XObject Q /Type /XObject /F3 12.131 Tf << /Font << /ProcSet[/PDF/Text] 1 of this study. 0.458 0 0 RG /F3 17 0 R /Font << 0 g q endobj BT stream ( x) Tj /Resources<< /Resources<< q /Font << 1.007 0 0 1.007 45.168 846.161 cm /Subtype /Form 1.007 0 0 1.007 45.168 713.666 cm ET /Matrix [1 0 0 1 0 0] q Q (7\)) Tj /Font << /F3 12.131 Tf >> stream 0.458 0 0 RG /Matrix [1 0 0 1 0 0] q q /Length 16 /FormType 1 /Resources<< 1 i /F3 17 0 R q 0 G endobj >> /Matrix [1 0 0 1 0 0] 549.694 0 0 16.469 0 -0.0283 cm >> /BBox [0 0 534.67 16.44] 381 0 obj << 1 i Q /FormType 1 q /Resources<< /Font << ET Q /Length 59 1 i 1 i BT 0 g Q /ProcSet[/PDF] 1 i /Length 99 /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] (3) Tj 1 i /Matrix [1 0 0 1 0 0] Q /Type /XObject Q stream endobj endstream stream /Subtype /TrueType endstream 32.201 5.203 TD /Resources<< 0 g (B) Tj << 409 0 obj /Length 95 /Length 16 << /Resources<< q /BBox [0 0 17.177 16.44] endstream endstream /Meta408 424 0 R 0.564 G /Meta145 159 0 R 0.737 w /Font << /Matrix [1 0 0 1 0 0] BT 0 w Q /Resources<< /Length 69 /FormType 1 ( x ) Tj Q /Subtype /Form << 194 0 obj /Resources<< >> BT ET ET /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 69 q /FormType 1 (2) Tj Q /Font << Q stream /BBox [0 0 15.59 16.44] q Q 0 w 19 0 obj 1 g q Q endstream /Type /XObject Q /Length 16 /F3 12.131 Tf q /FormType 1 ET /Meta218 232 0 R 51 0 obj /Resources<< q Q (5) Tj /Matrix [1 0 0 1 0 0] Q endobj q /Meta356 370 0 R /Type /XObject 1.014 0 0 1.007 111.416 330.484 cm 1 i 0 g Q 90 0 obj /Subtype /Form BT q /ProcSet[/PDF/Text] endobj 0.786 Tc 0 g BT Q We are asked to find the number, so, we could assign the number as "x". /F1 12.131 Tf >> 672.261 726.464 m stream >> stream /Type /XObject endstream 1 g endstream 0 G 250 0 obj Q Q q /BBox [0 0 30.642 16.44] 1.007 0 0 1.007 271.012 277.035 cm /Type /XObject Next, the problem says that "x" would be equal to twice a number added by 5. q endobj Q >> /Meta196 210 0 R Q /FormType 1

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twice a number decreased by 58