twice a number decreased by 58twice a number decreased by 58

endstream 1 i Q q Most questions answered within 4 hours. /Meta77 91 0 R 1.007 0 0 1.007 130.989 636.879 cm >> 1.007 0 0 1.007 130.989 523.204 cm >> >> You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. 1 i Q endobj >> Q /BBox [0 0 30.642 16.44] 0 g /BBox [0 0 673.937 14.853] BT /ProcSet[/PDF] >> 128 0 obj /Length 64 >> Q Q q /F4 36 0 R /Resources<< 0.737 w >> /Matrix [1 0 0 1 0 0] 0.737 w 1.007 0 0 1.007 271.012 383.934 cm 0.737 w /FormType 1 /Font << endobj >> q >> << /F3 12.131 Tf endstream [(Negativ)16(e )] TJ /Subtype /Form 0.369 Tc /F3 17 0 R Q Q >> 0 g BT /F3 17 0 R endobj << /Meta273 Do 1.005 0 0 1.007 79.798 813.037 cm Q q q -0.056 Tw /ProcSet[/PDF/Text] /Meta103 117 0 R /BBox [0 0 88.214 16.44] /ProcSet[/PDF] >> 0 g /FormType 1 /ProcSet[/PDF] 1 g q /Resources<< /F4 36 0 R << /F4 12.131 Tf /ProcSet[/PDF] (\(x ) Tj /Type /XObject /Resources<< /ProcSet[/PDF] >> stream q 1 i /ProcSet[/PDF] endobj ET /LastChar 45 << >> /Length 16 q /Length 70 148 0 obj Q << 35 0 obj /Meta356 Do Calculate a 15% decrease from any number. /Type /XObject q Q /ProcSet[/PDF/Text] /Meta411 Do /ProcSet[/PDF] << stream /Subtype /Form /Subtype /Form /ProcSet[/PDF/Text] 23.216 5.203 TD >> >> /FormType 1 /Resources<< endobj /Length 69 /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 551.058 277.035 cm stream 0 g 0 g q 1 i /Resources<< /Matrix [1 0 0 1 0 0] 0 g BT 1 i endobj >> /Length 151 Q /Type /XObject ET Q 98 0 obj /Meta314 328 0 R stream Q /Length 60 >> stream 1 g /Meta156 170 0 R q (40) Tj /Length 69 /Subtype /Form Q /BBox [0 0 15.59 16.44] Q 0.737 w /Resources<< 101.849 5.203 TD much as how 8, Last . **Note: You could choose any variable you want. >> 1 i In humans, testosterone plays a key role in the development of male reproductive tissues such as testes and prostate, as well as promoting secondary sexual characteristics such as increased muscle and bone mass, and the growth of body hair. BT stream q /Resources<< /Resources<< << 1.007 0 0 1.007 271.012 277.035 cm /BBox [0 0 534.67 16.44] Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /Meta80 94 0 R Q /Type /XObject q Q q /ProcSet[/PDF/Text] 1.502 5.203 TD >> /BBox [0 0 88.214 16.44] /ProcSet[/PDF] 0 g /Meta369 Do /Meta226 240 0 R q 26.219 5.336 TD q /Meta378 392 0 R /Type /XObject endstream endobj ET /Matrix [1 0 0 1 0 0] 0.524 Tc Q 1.007 0 0 1.007 411.035 849.172 cm Q 0.68 Tc Q 13 0 obj 0 G >> 0.458 0 0 RG 1 i /F3 17 0 R /Meta260 Do q q stream Q /F4 36 0 R Q 1 i /ProcSet[/PDF/Text] /Type /Font ET /Resources<< Q q /F3 17 0 R endobj /Type /XObject >> /Matrix [1 0 0 1 0 0] /Meta32 Do 1.502 7.841 TD endobj 0.425 Tc /FormType 1 0 g << Q >> /Meta62 Do /Meta261 Do << /F3 12.131 Tf /Font << /ProcSet[/PDF/Text] /Resources<< 0 G endstream q Q 0.458 0 0 RG BT 0.463 Tc /Length 12 Q /Subtype /Form 0 G ET endstream 1.014 0 0 1.007 531.485 330.484 cm /ProcSet[/PDF/Text] Q >> << Q 1 i stream The result is 8 less than 10 times the number. ET BT /FormType 1 endobj q /Subtype /Form 1.007 0 0 1.007 411.035 636.879 cm /BBox [0 0 15.59 16.44] 0.737 w endobj Q 1.005 0 0 1.007 79.798 862.723 cm >> 0 w 1 i 52 0 obj /F3 12.131 Tf (x) Tj 1 i q /Subtype /Form q >> >> 1.007 0 0 1.007 130.989 277.035 cm Q /Resources<< /FormType 1 /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] 177 0 obj Q q endobj q endobj q /Length 16 0 G q 0 w /Subtype /Form 0.369 Tc /Subtype /Form q stream 0 g Q 0 g /BBox [0 0 15.59 16.44] q Q /ProcSet[/PDF/Text] q /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf 0 w /Type /XObject /FormType 1 /Type /XObject /F3 17 0 R >> /F3 12.131 Tf 0 G >> 0 G q BT Q 0 G q >> /Length 69 /BBox [0 0 88.214 16.44] q /FormType 1 /ProcSet[/PDF/Text] /FormType 1 0.458 0 0 RG 0 g q 293 0 obj >> /Subtype /Form 1 i 1.007 0 0 1.007 45.168 829.599 cm /Length 79 /Font << >> /Resources<< /Matrix [1 0 0 1 0 0] /Font << /BBox [0 0 88.214 16.44] BT /Font << 83 0 obj /Meta286 300 0 R /BBox [0 0 639.552 16.44] /Resources<< << 0.458 0 0 RG 165 0 obj 0 g ( \() Tj /BBox [0 0 88.214 16.44] /FormType 1 q q 0.458 0 0 RG q /ProcSet[/PDF/Text] /FontBBox [-90 -216 1195 800] /Type /XObject Q /Meta18 Do /Matrix [1 0 0 1 0 0] 95 0 obj /ProcSet[/PDF] stream /Type /XObject /ProcSet[/PDF/Text] q /Type /XObject << /BBox [0 0 15.59 16.44] endobj q Q /Subtype /Form /Subtype /Form BT /ProcSet[/PDF/Text] Q Q 0 g 1 i endstream endobj 0 w 0 4.78 TD endobj /Font << << q q q 0.564 G q 206 0 obj 672.261 653.441 m 102 0 obj /Length 70 /Resources<< 1 i /Length 12 Q >> 1.007 0 0 1.007 551.058 636.879 cm Q /F3 12.131 Tf ET /F3 12.131 Tf /XObject << /Length 69 >> q BT q /F3 17 0 R 0.737 w /Subtype /Form q Q 0.425 Tc endobj 0 G /Ascent 1050 /FormType 1 /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] (40) Tj 0.68 Tc q q /FormType 1 q /Meta175 189 0 R /FontName /PalatinoLinotype-Bold stream /Meta72 Do /Resources<< q endstream 0 4.894 TD /Type /XObject /Type /XObject Q /FormType 1 /Meta183 Do /Resources<< /Length 68 /BBox [0 0 88.214 16.44] /Length 69 /Length 16 a.) /Meta92 106 0 R 209 0 obj /BBox [0 0 88.214 16.44] endobj q 0 w BT q stream /Meta16 Do q q >> Q >> /Meta216 Do 672.261 400.496 m 1 g /Font << 0 g 1.007 0 0 1.007 551.058 583.429 cm endstream /Subtype /Form /Meta109 Do q endstream Q Q 20.21 5.203 TD /Font << BT >> stream /Font << /BBox [0 0 639.552 16.44] /Meta285 Do /BBox [0 0 88.214 16.44] /Font << 0 g (5\)) Tj endstream /ColorSpace [/Indexed /DeviceGray 1 ] 240 0 obj /Matrix [1 0 0 1 0 0] q 12.727 24.649 TD 0.737 w /FormType 1 stream stream endstream /ProcSet[/PDF/Text] 0.564 G 0 g /Meta381 395 0 R Get a free answer to a quick problem. /Matrix [1 0 0 1 0 0] /F3 17 0 R 68 0 obj q /Type /XObject /Meta19 Do /Meta126 Do /Meta265 Do /FormType 1 q stream ET endobj >> stream q 0 G endstream Q 1 i /Matrix [1 0 0 1 0 0] q 0 g /FormType 1 q 1 i /Length 54 endstream /Meta393 Do /ProcSet[/PDF/Text] 0 g /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] >> Q endobj /Meta417 Do >> q q 1 g /ProcSet[/PDF] q ET ET Q 43 0 obj 230 0 obj Step 1/1. stream /Meta373 387 0 R (1\)) Tj q Q /ProcSet[/PDF] /Subtype /Form << /Resources<< endstream Q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] 1.007 0 0 1.007 45.168 779.913 cm Q q >> /FormType 1 Q /I0 51 0 R q 30 0 obj << Q stream /Meta101 115 0 R Q 1 i 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 endstream /Meta33 46 0 R /Subtype /Form 1 g BT /Font << /Subtype /Form >> >> /Type /XObject Q stream BT 1 i >> 0 g /Length 16 0 g /BBox [0 0 15.59 16.44] >> 549.694 0 0 16.469 0 -0.0283 cm endobj /Type /XObject Q Q Q /Subtype /Form 0 w q q >> /Meta95 Do << 1.007 0 0 1.007 271.012 277.035 cm 0 G /Meta237 251 0 R endstream /F3 12.131 Tf /Meta26 39 0 R q Twice a number decreased by 8 gives 58. 0 g >> 0.297 Tc << ET Q /Length 16 BT 417 0 obj 407 0 obj >> /Subtype /Form /ProcSet[/PDF/Text] 257 0 obj /Meta228 242 0 R Q (\)) Tj Q 221 0 obj /Meta396 412 0 R /Length 54 Q q /Length 16 q endobj /I0 Do 1 g q /Matrix [1 0 0 1 0 0] 19 0 obj /F4 12.131 Tf 416 0 obj Q /Type /XObject /BBox [0 0 88.214 16.44] q 1.007 0 0 1.007 551.058 383.934 cm /Resources<< /Matrix [1 0 0 1 0 0] /Meta35 48 0 R /Meta344 Do 1.007 0 0 1.007 271.012 636.879 cm /Font << endobj 0.737 w /Subtype /Form 1 i /FormType 1 q endstream ET 1.014 0 0 1.006 531.485 510.406 cm /Type /XObject 1 i /FormType 1 >> >> /Meta21 Do 0 G Q Q /XObject << 1 i q /Meta166 Do /ProcSet[/PDF/Text] endstream 1.007 0 0 1.007 411.035 383.934 cm , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. 1.005 0 0 1.007 102.382 293.596 cm /Meta291 Do q >> 0.564 G Q /I0 51 0 R /Length 16 << Q 0 w /Length 69 /Meta365 379 0 R /Font << /ProcSet[/PDF/Text] endobj Q /BBox [0 0 88.214 16.44] 0.564 G /FormType 1 /ProcSet[/PDF] q /Resources<< /Meta91 Do /Resources<< /ItalicAngle 0 /Length 69 /Meta112 Do /Font << 0 g endstream q (x) Tj << /F3 12.131 Tf /F3 17 0 R /FormType 1 endobj Q /BBox [0 0 88.214 16.44] q /F3 12.131 Tf 0 G /Resources<< 1.007 0 0 1.007 271.012 277.035 cm /Type /XObject ET >> Q /Subtype /Form 1.007 0 0 1.007 411.035 383.934 cm q endobj >> 1 g 0.737 w 23.952 4.894 TD 0.738 Tc 0 g /Length 78 /BBox [0 0 15.59 29.168] /Font << q endstream Q Q /Meta115 129 0 R (-20) Tj >> >> >> Q Twice a first number decreased by a second number is 6. >> >> 0 g Find the number.#MathsDoubt Support my work atUPI ID - mathsdoubt@jio /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] 0 g 1.007 0 0 1.006 130.989 437.384 cm endstream BT (-) Tj q >> /Type /XObject /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] stream /Subtype /Form 0.369 Tc Q /Subtype /Form >> 1 i /FormType 1 /Length 69 0 w Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . /Subtype /Form q 0 G endstream /BBox [0 0 30.642 16.44] Q (5) Tj 1.007 0 0 1.007 411.035 583.429 cm 0 g /FormType 1 Q << /ProcSet[/PDF/Text] [(A number )-17(divided by )] TJ endstream >> endstream [(thir)17(te)15(en)] TJ /Type /XObject endobj endstream /Font << >> 0 G /F4 12.131 Tf /Meta127 141 0 R /Resources<< /Meta360 374 0 R << >> /Matrix [1 0 0 1 0 0] /Length 66 1.007 0 0 1.007 411.035 636.879 cm /BBox [0 0 88.214 16.44] << Q /F3 17 0 R q Q 1 i q /Meta403 Do stream /Matrix [1 0 0 1 0 0] >> /F3 12.131 Tf 1 i /Subtype /Form /Matrix [1 0 0 1 0 0] /Type /XObject 0.51 Tc /Resources<< endobj /Matrix [1 0 0 1 0 0] /Font << Q /Resources<< /F1 7 0 R q /ProcSet[/PDF/Text] /Length 70 /Length 60 Q stream (C\)) Tj 0 G >> stream << BT /BBox [0 0 673.937 16.44] q /F3 17 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Length 12 /BBox [0 0 88.214 16.44] /Length 65 << /FormType 1 stream /FormType 1 /ProcSet[/PDF/Text] Q q /ProcSet[/PDF] Q stream endstream Q Q 59 0 obj Q q endstream << 549.694 0 0 16.469 0 -0.0283 cm >> /Subtype /Form /Type /XObject q >> /FormType 1 /FormType 1 /BBox [0 0 534.67 16.44] 0 g /Meta263 Do q 0 G /F3 12.131 Tf 20.21 5.203 TD << /BBox [0 0 15.59 16.44] q Q stream 323 0 obj BT >> /F3 12.131 Tf Q /ProcSet[/PDF] stream Q Q 0.564 G q q stream /Meta227 241 0 R q q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] 1 i Q Q Three times a number equals fifteen 3. stream << endstream stream q Q /Font << BT BT q /F1 7 0 R 163 0 obj 205 0 obj >> /Resources<< >> /FormType 1 /Meta316 330 0 R /BBox [0 0 17.177 16.44] >> endstream /Font << 3.742 5.203 TD 226 0 obj 1 g Q Q 6.746 5.203 TD /Font << /Subtype /Form /Type /Font Q /F4 12.131 Tf q /ProcSet[/PDF] /Meta80 Do >> q /Resources<< q Q (D) Tj Q /Resources<< /FormType 1 ET stream 1 i ET stream /Length 60 Q 0 G /Length 65 0 G << /Resources<< /Length 58 /FormType 1 q /Length 69 ET 0.564 G endstream 0.738 Tc 1.007 0 0 1.007 654.946 400.496 cm (-) Tj >> /Length 69 /Resources<< 1 i << /FormType 1 /F4 36 0 R 0.564 G q /Meta257 271 0 R /Meta180 Do 672.261 799.486 m endobj >> 1.502 5.203 TD 0 G q q /BBox [0 0 88.214 16.44] 0.564 G /F3 12.131 Tf >> /Length 12 q endstream /F1 7 0 R endobj endstream 89 0 obj 16 0 obj q >> /F3 17 0 R Q << q /Resources<< >> /FormType 1 438 0 obj /Font << Q endstream 0.737 w /Meta214 228 0 R 1.014 0 0 1.007 391.462 277.035 cm endobj Q >> /ProcSet[/PDF/Text] 1.014 0 0 1.007 111.416 277.035 cm 0 g /Resources<< q 0.564 G stream /Resources<< /ProcSet[/PDF/Text] stream 1.014 0 0 1.007 111.416 583.429 cm /ProcSet[/PDF/Text] q /ProcSet[/PDF] >> /Meta190 Do /ProcSet[/PDF/Text] /Length 54 /ProcSet[/PDF/Text] /FormType 1 q >> 0.786 Tc /FormType 1 << You can specify conditions of storing and accessing cookies in your browser. /BBox [0 0 30.642 16.44] 0 g Q /Meta426 442 0 R /Resources<< << >> 443 0 obj %PDF-1.4 /Length 69 /Subtype /Form /Resources<< >> ET >> endstream /Meta348 Do 0 G ET >> >> q ET (38) Tj /XObject << /Length 16 endstream Q Q /Type /XObject /Subtype /Form >> ET 0 g /BBox [0 0 88.214 16.44] (-8) Tj /Type /XObject q q >> Q q q 1 i Q /BBox [0 0 534.67 16.44] 1.014 0 0 1.007 391.462 636.879 cm << /Matrix [1 0 0 1 0 0] (x) Tj /Meta16 27 0 R endstream /Type /XObject >> q q 1.014 0 0 1.007 391.462 776.149 cm q /Matrix [1 0 0 1 0 0] >> q Q stream Q endobj /Resources<< >> /Meta7 18 0 R /ProcSet[/PDF/Text] 0 G 1 i Q /Length 68 Q 1.007 0 0 1.007 130.989 330.484 cm /F3 17 0 R >> /FormType 1 1 g /Subtype /Form /Resources<< /Meta119 Do q 20-n c.) n+20 d.) 20+n 3.) /Resources<< >> /Subtype /Form q /FormType 1 328 0 obj >> /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /ProcSet[/PDF] 0.737 w << /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf stream (5) Tj /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 271.012 523.204 cm >> /Meta358 Do q /Resources<< q /Resources<< 0 G 0 5.203 TD 176 0 obj /Meta335 Do /F4 36 0 R >> 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. << /Type /XObject 1.005 0 0 1.007 79.798 796.475 cm << endobj endobj 0 w endobj /Resources<< endstream /BBox [0 0 15.59 29.168] ET 120 0 obj q /Length 69 Q 1 i /Subtype /Form /Meta337 Do (+) Tj 302 0 obj /F3 17 0 R q 0 w 0.17 Tc 0 g 20.21 5.203 TD /Length 96 /Matrix [1 0 0 1 0 0] q stream 1.007 0 0 1.007 130.989 636.879 cm /Matrix [1 0 0 1 0 0] 6.746 5.203 TD 18.708 17.593 TD q (8\)) Tj /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 271.012 583.429 cm 1 i q (-) Tj 1 i /Length 118 Q /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] q stream Twice a number decreased by another number: "twice a number decreased by another number" 2*(x-y), "twice a number, decreased by another number"(2*x)-y, It's possible David. /F3 12.131 Tf endstream q >> Q endstream q /Meta42 Do 0.564 G /Meta159 Do >> 1.014 0 0 1.006 251.439 437.384 cm >> /FormType 1 /BBox [0 0 88.214 16.44] q Solution: /Length 69 /FormType 1 endstream 58 0 obj /ProcSet[/PDF/Text] q endobj Q 276 0 obj 0.271 Tc q /I0 51 0 R Q endobj /Type /XObject Q precision and actual right or wrong answers. /Meta427 443 0 R 1.014 0 0 1.007 111.416 636.879 cm 0 g << 1 i /FirstChar 32 >> 0 G /Resources<< 25.454 5.203 TD /Subtype /Form >> /BBox [0 0 88.214 35.886] /Font << 0.458 0 0 RG /ProcSet[/PDF] >> ET 1.014 0 0 1.007 391.462 636.879 cm >> 1 i q Q >> 0 20.154 m 0 g >> /Meta279 Do /ProcSet[/PDF/Text] q /FormType 1 stream 1 i /Encoding /WinAnsiEncoding /Matrix [1 0 0 1 0 0] 179 0 obj >> Q /FormType 1 /F3 12.131 Tf /Meta242 256 0 R 0 g 1 i >> >> 429 0 obj >> /Resources<< /Length 59 Q << /Length 16 stream /Subtype /Form q 1.007 0 0 1.007 130.989 277.035 cm /FormType 1 /Type /XObject /Resources<< /Subtype /Form >> /Resources<< >> /Resources<< endobj /F3 12.131 Tf Q /F1 12.131 Tf 0 g /Resources<< Q << /Meta186 Do q 0 G /Type /XObject Q /F3 17 0 R 1.502 5.203 TD << 0.271 Tc q endobj 0 g 1.014 0 0 1.007 531.485 776.149 cm stream endobj /F3 17 0 R >> BT q Q endstream << /Subtype /Form Q >> << >> /Meta268 282 0 R stream /Subtype /Form q /Font << 0 g /Font << >> /BBox [0 0 88.214 16.44] /Subtype /Form endobj Q /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 251.439 277.035 cm ET Q >> q 0.425 Tc q >> /FormType 1 (-20) Tj 1 i 1 i q Question. q << Q 0.369 Tc q 0 g A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. /BBox [0 0 534.67 16.44] endstream << ET /Length 59 /ProcSet[/PDF/Text] endobj q /Font << 0.227 Tc 1 i /BBox [0 0 88.214 35.886] 334 0 obj /Matrix [1 0 0 1 0 0] q /Length 65 q << q endobj q (11) Tj BT << /F3 12.131 Tf 0.564 G 1.005 0 0 1.007 102.382 872.509 cm << /Meta347 361 0 R Q /Meta386 402 0 R q >> 0 w BT Q >> /Meta256 Do q /Meta413 429 0 R /Meta382 396 0 R 1.007 0 0 1.006 411.035 510.406 cm Q 0 g /BBox [0 0 15.59 16.44] /Meta165 Do >> >> /FormType 1 q /F3 12.131 Tf BT /Length 70 (D) Tj q 1 i 1 i Q endobj 0 G 1 i endstream 1.014 0 0 1.007 531.485 277.035 cm