/Resources<< /Meta215 Do endstream Q 0 G endstream BT >> q Q /ProcSet[/PDF] /F3 12.131 Tf 0 w 1 i 0.68 Tc q 17.234 5.203 TD /Length 12 stream 0 g q >> /F3 12.131 Tf /Resources<< q /Resources<< 0.737 w /Matrix [1 0 0 1 0 0] stream 195 0 obj stream q Q /Meta77 91 0 R /Matrix [1 0 0 1 0 0] /Type /XObject stream Q q /FormType 1 0 G /ProcSet[/PDF/Text] Q Q Q /Type /XObject q q Q 10 0 obj /Matrix [1 0 0 1 0 0] /Meta263 Do /Meta17 28 0 R /BBox [0 0 534.67 16.44] q 1.005 0 0 1.007 102.382 743.025 cm (7\)) Tj /Type /XObject endstream >> q /Resources<< >> /ProcSet[/PDF/Text] /F3 17 0 R /ProcSet[/PDF/Text] Q 0.51 Tc q /ProcSet[/PDF/Text] Q 0 G /Resources<< /ProcSet[/PDF] 269 0 obj 1 i endobj Q stream 0 G BT BT 0.737 w q 112 0 obj /Resources<< /ProcSet[/PDF/Text] 0 G endobj stream stream 1 g q ET endstream /Resources<< 25.454 5.203 TD >> >> /Resources<< 274 0 obj Q 7) The quotient of 40 and the product of a number and -8 7) A) 40 x - 8 B) -320 x C) 40-8x D)-8x 40 8) Twice a number, decreased by 58 8) A) 2 (x - 58 ) B) 2 x - 58 C) 2 x + 58 D) 2 (x + 58 ) 9) A number subtracted from -20 9) A) -20 x B) -20 + x C) x - (-20 ) D) -20 - x 10) Five times the sum of a number and -23 10) /ProcSet[/PDF/Text] /Length 65 endstream endobj /F4 36 0 R << >> Q /BBox [0 0 88.214 16.44] BT 1.502 8.18 TD 348 0 obj q /Meta341 355 0 R /Matrix [1 0 0 1 0 0] /Meta330 Do Q >> 1.014 0 0 1.007 391.462 583.429 cm q >> /Resources<< >> 356 0 obj 26.957 5.203 TD >> -0.486 Tw /F4 12.131 Tf /ProcSet[/PDF] /Length 16 0.369 Tc /FormType 1 Q /Subtype /Form >> Q ET /Resources<< q /Meta351 Do /F3 17 0 R /F4 36 0 R Q >> q 1.007 0 0 1.007 551.058 330.484 cm 1.007 0 0 1.006 411.035 763.351 cm q BT /BBox [0 0 88.214 16.44] stream Q Q endstream q /FormType 1 /Type /XObject 0 w /Length 69 << Q 0.458 0 0 RG /Type /XObject BT 0.458 0 0 RG /Font << endstream /F3 12.131 Tf q /F3 17 0 R q endstream BT /Type /XObject /Resources<< << 1 i << /Meta272 286 0 R /Length 16 /Meta368 382 0 R 1.007 0 0 1.007 551.058 277.035 cm /BBox [0 0 17.177 16.44] /ProcSet[/PDF] q /Subtype /Form 1 i /Resources<< /Meta366 Do /FormType 1 BT << /ProcSet[/PDF] /F3 12.131 Tf /Matrix [1 0 0 1 0 0] 0 w q 1.007 0 0 1.007 45.168 779.913 cm 1 i /Meta296 Do 1 g /ProcSet[/PDF] 0 g Q stream q /Meta405 Do 0 g /Length 59 stream /BBox [0 0 15.59 29.168] stream /Resources<< endstream Q BT Q /FormType 1 Q q q >> Q Q 1 i /Meta286 300 0 R 431 0 obj /Type /XObject /Subtype /Form 0 g << 1.007 0 0 1.007 551.058 636.879 cm (9\)) Tj 0 g -y. 0 5.203 TD q /Pages 1 0 R endobj /Subtype /Form Q /BBox [0 0 88.214 16.44] /Subtype /Form 0 G 0 g 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 0 G Transcribed Image Text: A number increased by 5 is equivalent to twice the same number decreased by 7. >> /XObject << Q >> 1 i q q q /BBox [0 0 88.214 16.44] q 1 i endstream 0.564 G 1 i >> 166 0 obj >> 0.486 Tc q Q >> (C\)) Tj xref 1.007 0 0 1.007 551.058 383.934 cm >> /Subtype /Form 0 G endobj Q /BBox [0 0 88.214 16.44] 0.564 G endobj Q Q stream 289 0 obj endstream ET /Resources<< /Font << Q endstream /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 0 q >> >> ET Q 0 20.154 m >> q /FontBBox [-568 -307 2000 1007] /ProcSet[/PDF/Text] /Type /XObject /Meta166 180 0 R /Length 60 endstream /Font << /Subtype /Form /Meta159 Do Q q Q /Meta324 Do /Meta13 Do /FormType 1 0 g >> /Font << 1 i 89.12 5.203 TD Q endstream /F1 7 0 R endstream -0.056 Tw 0 g >> Q << q /Matrix [1 0 0 1 0 0] stream /Length 68 /Length 59 1 i /Matrix [1 0 0 1 0 0] q q >> /F1 7 0 R 0.369 Tc Q 1.007 0 0 1.007 551.058 277.035 cm /Meta197 Do /Matrix [1 0 0 1 0 0] << endstream /Meta205 219 0 R BT /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 35.886] /Meta83 97 0 R 6 0 obj 0 G 1.005 0 0 1.007 79.798 763.351 cm endobj >> /Length 67 0 G Q >> Q (-8) Tj /ProcSet[/PDF/Text] q Q , Prove the following endstream /Meta2 Do /Matrix [1 0 0 1 0 0] /FormType 1 << Q 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. /Meta363 377 0 R /Meta165 179 0 R q /Subtype /Form >> /BBox [0 0 30.642 16.44] /Matrix [1 0 0 1 0 0] << endobj /Font << /ProcSet[/PDF/Text] q endstream /Subtype /Form ET 0.737 w >> 1 i 0.564 G 1 i q /BBox [0 0 534.67 16.44] /Meta142 156 0 R 0 g BT 0 g /Matrix [1 0 0 1 0 0] 0.369 Tc >> 0 g 0 G endobj 19.474 5.203 TD 0.564 G q Q 1 g /ProcSet[/PDF/Text] >> /BBox [0 0 88.214 16.44] 391 0 obj 0 G q /Type /XObject >> << Q >> endobj Q >> Q /Length 69 q /Length 69 /Subtype /Form 320 0 obj /Subtype /Form q (D) Tj 0 w 16.469 5.336 TD << endstream /Resources<< BT 0 g 291 0 obj /FormType 1 Q q /Subtype /Form [tex]\sin (\pi -x)=\sin x[/tex]. 0.369 Tc /Length 118 >> endobj stream /Meta320 Do << /ProcSet[/PDF] q Q 0.737 w >> /Meta273 287 0 R endstream >> /Meta243 Do w/Honors. /Resources<< ET endobj q endstream 0 G /Matrix [1 0 0 1 0 0] 0 g q q Easy Solution Verified by Toppr Let the number be x. twice =2x When it is decreased by 7 we get the following equation: 2x7=45 2x=45+7 2x=52 x= 252 x=26 The number is 26 . Q 0.458 0 0 RG q << 28 0 obj BT TJ /FormType 1 /Subtype /Form /FormType 1 /Length 16 Q ET /FormType 1 endobj /ProcSet[/PDF/Text] 0.458 0 0 RG 37 0 obj >> q Q >> /F3 17 0 R 0.458 0 0 RG q /Meta287 Do >> /FormType 1 << /F3 17 0 R endstream 1 i Q /BBox [0 0 88.214 35.886] /Type /XObject /CreationDate (D:20140515121932-04'00') q >> 1.007 0 0 1.007 271.012 703.126 cm >> Q >> q 60 0 obj /Meta400 416 0 R /Length 16 q /F4 36 0 R /Matrix [1 0 0 1 0 0] /Meta294 308 0 R >> << /FormType 1 /Meta316 330 0 R q 9.723 5.336 TD /Length 80 0 w /ProcSet[/PDF/Text] [(Wr)-14(ite th)-23(e phra)-15(se as a v)-17(ari)-14(able e)-21(xpress)-18(ion. stream q Q stream stream q /Matrix [1 0 0 1 0 0] Q Q Q /Type /XObject q /F3 12.131 Tf /Meta82 96 0 R q /Length 54 181 0 obj Calculate a 15% decrease from any number. Now that you know the meaning of the key words you can read the problem differently. /Subtype /Form /Font << endstream /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 35.886] /Meta385 401 0 R 22.478 5.336 TD endstream 0 G /Matrix [1 0 0 1 0 0] >> endobj /BBox [0 0 88.214 16.44] /FormType 1 /F4 12.131 Tf /BBox [0 0 88.214 16.44] /Meta49 63 0 R Q /Resources<< q endobj 339 0 obj q 0 g 1 i << /Meta210 Do Q 0.786 Tc 1.007 0 0 1.007 45.168 730.228 cm /FormType 1 0.564 G Q 412 0 obj >> q /Matrix [1 0 0 1 0 0] >> Twice a number decreased by 8 gives 58 find the number Advertisement Loved by our community 24 people found it helpful Xiphodon Step-by-step explanation: 2x-8=58 2x=66 x =33 Hope it helps Please mark as brainliest Find Math textbook solutions? 38.948 5.203 TD /Meta41 55 0 R /Font << 0 G (8\)) Tj >> /FormType 1 >> /Resources<< Q q /ProcSet[/PDF/Text] /Subtype /Form 288 0 obj /Length 59 /Meta99 113 0 R q << q /Meta109 Do /F3 17 0 R Q 0 w endstream /F3 17 0 R << /Matrix [1 0 0 1 0 0] /Meta208 Do /Resources<< /Type /XObject 0 g >> Q 1.502 5.203 TD q /ProcSet[/PDF/Text] /Font << Q /ProcSet[/PDF] /Resources<< Q q << /Subtype /Form /Meta315 329 0 R q stream 110 0 obj /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 654.946 400.496 cm >> Q 0 w SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. ET 0 g /Resources<< /Resources<< endobj /Font << /Subtype /Form 1 g 1 g q /FormType 1 /Font << /Resources<< Q q Q >> q /ProcSet[/PDF/Text] 0.564 G 1 i BT q 1 g Q endobj /F3 12.131 Tf /ProcSet[/PDF] /Font << /Resources<< /Subtype /Form /F3 17 0 R q 1.014 0 0 1.007 251.439 383.934 cm 0 G /Font << >> 1 g endstream /Subtype /Form /Font << 1.007 0 0 1.007 411.035 277.035 cm endstream stream /Length 16 58 decreased by twice Gails age. 0.564 G >> 0.175 Tc >> /F4 12.131 Tf /BBox [0 0 88.214 16.44] >> 1.014 0 0 1.007 251.439 277.035 cm /Length 16 319 0 obj endstream Q 1.007 0 0 1.007 551.058 523.204 cm /F3 17 0 R /Matrix [1 0 0 1 0 0] /Meta268 282 0 R 184 0 obj >> /Resources<< 1.014 0 0 1.007 531.485 277.035 cm 0 w ET /Resources<< ET 267 0 obj /Meta56 Do q gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. q Q q /Length 59 q Q q Q /F3 17 0 R /F1 7 0 R 1.007 0 0 1.007 411.035 583.429 cm stream /FormType 1 /Type /XObject endstream >> Q q q endstream 419 0 obj /F2 11 0 R 1 i 1 i >> 1 g q (C) Tj q Q q 1 i Q /ProcSet[/PDF/Text] /ProcSet[/PDF] /Meta395 Do 0.458 0 0 RG /Meta260 Do /BBox [0 0 30.642 16.44] q /Subtype /Form /Meta300 314 0 R 2.238 5.203 TD >> 0 g Q Q 0 g 0.458 0 0 RG /BBox [0 0 88.214 16.44] ET q /ProcSet[/PDF] >> /Subtype /Form 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 /Matrix [1 0 0 1 0 0] /BBox [0 0 534.67 16.44] Q /Resources<< /Resources<< 1 i /Type /XObject /Meta236 Do endstream /Resources<< /Meta209 Do 268 0 obj endstream /F3 12.131 Tf Q 167 0 obj 0.369 Tc q /Meta151 Do /Meta218 Do /Type /XObject /Resources<< BT /FormType 1 /Subtype /Form /Encoding /WinAnsiEncoding Q q /Resources<< 722.699 872.509 l q q /ProcSet[/PDF/Text] stream /Meta189 Do /F3 12.131 Tf Q ET /Type /XObject 0 G /Length 69 /Matrix [1 0 0 1 0 0] /Font << 0 G /MissingWidth 250 (-20) Tj /Subtype /Form /BBox [0 0 534.67 16.44] endobj /BBox [0 0 88.214 35.886] 0 G /BBox [0 0 30.642 16.44] q 23 0 obj Q 0.564 G endobj /F3 17 0 R /Resources<< /Resources<< >> Twice a first number decreased by a second number is 6. /FormType 1 Q Q /Matrix [1 0 0 1 0 0] /Subtype /Form 0.737 w 440 0 obj /Type /XObject /BBox [0 0 88.214 16.44] /Length 79 Q /Resources<< q stream 1 i endstream 1.007 0 0 1.007 411.035 636.879 cm 1 i 0.838 Tc /Meta118 132 0 R BT /ProcSet[/PDF/Text] Q 0.458 0 0 RG 0 G 1 i /Meta45 Do /Meta409 425 0 R q << /Length 16 0 G /FormType 1 Q 0 g << /Type /XObject /Matrix [1 0 0 1 0 0] Q /Resources<< /F3 12.131 Tf << q 0.458 0 0 RG 146 0 obj 0 w /ProcSet[/PDF/Text] /Length 59 >> /Resources<< /Resources<< endstream q /Matrix [1 0 0 1 0 0] q >> 1 i /Resources<< Q endobj Q endobj 0 G ET 0 G /Type /XObject << /Matrix [1 0 0 1 0 0] /FormType 1 /Type /XObject 0 g >> q /Font << 20.21 5.203 TD 0.369 Tc Q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 67.753 347.046 cm /Type /XObject 1 i >> endobj endstream 1 i q /Subtype /Form /Subtype /Form >> /Type /XObject S >> Q /Type /XObject /Ascent 976 q 0 g /Meta93 Do q /Resources<< q stream 1.007 0 0 1.007 130.989 776.149 cm stream /Subtype /Form /Matrix [1 0 0 1 0 0] Q /Matrix [1 0 0 1 0 0] q 0 w BT << ET Q q >> endstream /ProcSet[/PDF/Text] 436 0 obj /Meta297 Do /F3 12.131 Tf 1 i 1.007 0 0 1.006 130.989 437.384 cm Q endobj /Meta91 105 0 R q /ProcSet[/PDF/Text] 243 0 obj >> 1 g 0 g /Length 70 Q ET 31 0 obj 0.564 G 23.952 4.894 TD /ProcSet[/PDF] endstream Q q stream stream q >> 0 w >> /Resources<< endstream Q stream 97 0 obj q q /F1 12.131 Tf q endstream stream /Matrix [1 0 0 1 0 0] (5) Tj BT >> /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] << 1 i >> /Length 16 q BT ET 1.014 0 0 1.007 251.439 703.126 cm 1.007 0 0 1.007 654.946 599.991 cm Q 1.007 0 0 1.007 130.989 776.149 cm /F3 17 0 R 1 g Q stream /Type /XObject /BBox [0 0 88.214 16.44] >> q /FormType 1 23.952 4.894 TD /Subtype /Form >> 141 0 obj endstream q 357 0 obj /Type /XObject /Meta346 360 0 R /ProcSet[/PDF/Text] ET Q 0 G /Length 65 >> q q endobj /ProcSet[/PDF] >> endstream q /Meta360 Do q >> 1 i endobj endobj (iv) A number exceeds 5 by 3. Q 144 0 obj /Font << 0.564 G Q q /F3 17 0 R 0 G q /Meta35 48 0 R /Matrix [1 0 0 1 0 0] /F3 17 0 R /Type /XObject /ProcSet[/PDF/Text] endstream 0 G Q ET (-11) Tj q endobj /Length 16 /Length 68 Q 0.458 0 0 RG /ProcSet[/PDF] /Length 69 ET ET /Meta399 Do /ProcSet[/PDF/Text] /Type /XObject 138 0 obj (-9) Tj 672.261 347.046 m q 0 G /Type /XObject /Matrix [1 0 0 1 0 0] 152 0 obj /Length 59 /ProcSet[/PDF/Text] >> 0 g 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. /ProcSet[/PDF/Text] Q /Meta380 394 0 R /Meta25 Do /FormType 1 /Meta6 15 0 R 1.005 0 0 1.007 79.798 746.789 cm >> 1 i /Matrix [1 0 0 1 0 0] stream << /Length 59 << 1.014 0 0 1.006 251.439 437.384 cm Q /BBox [0 0 15.59 16.44] S /Matrix [1 0 0 1 0 0] /F3 17 0 R /FormType 1 /FormType 1 /Meta392 408 0 R << 0 0 0 444 500 444 0 444 0 500 0 278 0 0 278 778 q Q endobj 6.746 5.203 TD q /Font << q Q /F3 12.131 Tf /Length 69 /Length 63 1.007 0 0 1.006 411.035 510.406 cm /Meta81 95 0 R Q Q q endobj /Meta62 Do Q >> Q: A number increased by 5 is equivalent to twice the same number decreased by 7. /Resources<< >> >> /Meta39 Do >> >> /Meta139 Do /Font << BT 328 0 obj 378 0 obj q >> /Matrix [1 0 0 1 0 0] << /Type /XObject [4] One half of a number decreased by fourteen is twenty-one 1.007 0 0 1.007 271.012 523.204 cm ET endstream /F1 12.131 Tf /BBox [0 0 549.552 16.44] q /Meta295 Do << 1 i /Meta252 Do q q /Meta131 145 0 R stream /ProcSet[/PDF] Q /Font << /Length 69 /Type /XObject /Meta288 Do q A number increased by 5 is equivalent to twice the same number decreased by 7. /Type /XObject /ProcSet[/PDF] 0 w stream /Resources<< /Meta207 221 0 R 95 0 obj /Meta375 389 0 R /ProcSet[/PDF/Text] endobj /Matrix [1 0 0 1 0 0] BT endstream >> /Filter [/CCITTFaxDecode] /Meta131 Do Q Q 3.742 5.203 TD stream /ProcSet[/PDF/Text] /Meta404 Do /ProcSet[/PDF/Text] 0.737 w /Subtype /Form ] At inclusion, the mean MetS-Z of female participants was placed within the 3 rd quartile of MetS-Z scores, while the . stream Q 1 i >> /FormType 1 /Meta298 Do . 1 i 0 g endobj -0.486 Tw /Subtype /Form (iii) 25 exceeds a number by 7. /FormType 1 /F3 17 0 R 1.014 0 0 1.007 391.462 330.484 cm Q New questions in Mathematics q q /FormType 1 >> [(-3)-16(20)] TJ /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] ET /F1 7 0 R 0 G /Matrix [1 0 0 1 0 0] /Meta304 318 0 R Q /Meta425 441 0 R q 1.007 0 0 1.007 411.035 636.879 cm Q 1.007 0 0 1.007 411.035 277.035 cm 1 i >> q << q >> Q /BBox [0 0 88.214 16.44] << 0 447 BT Find an answer to your question Twice a number decreased by 8gives 58. ET 1.502 5.203 TD << endstream >> endstream Q /Resources<< 1.014 0 0 1.007 531.485 583.429 cm (38) Tj ET q >> 38 0 obj 0.564 G /Length 244 0.68 Tc Q The difference between six and a number divided by nine 10. << endobj /Matrix [1 0 0 1 0 0] /Meta417 433 0 R 0 g BT Q 0 G << /Resources<< Let the 2nd number be y. /Meta78 Do endstream stream 0.738 Tc endobj /Type /XObject Q q /FormType 1 0 g /Type /XObject << >> Q /Resources<< >> /F3 12.131 Tf Q endobj /FormType 1 /Meta325 339 0 R /ProcSet[/PDF/Text] << Answer link. 0 g /Resources<< /Matrix [1 0 0 1 0 0] /F3 12.131 Tf endobj 446 0 obj stream q Q >> Q /Matrix [1 0 0 1 0 0] Twice a number is decreased by 9, and this sum is multiplied by 4. Q stream 0 g 0 g q /Resources<< Answer only. q Q >> /Length 12 >> endobj q Q >> /Length 16 >> BT five times the sum of a number x and two b.) /Subtype /Form ET /Resources<< endstream >> 0 G /Matrix [1 0 0 1 0 0] 0.564 G /F3 17 0 R endobj /F3 12.131 Tf 1.007 0 0 1.006 551.058 437.384 cm /Matrix [1 0 0 1 0 0] >> /Meta74 Do /Length 16 /Type /XObject 301 0 obj ET endobj /Subtype /Form BT /ProcSet[/PDF/Text] /ProcSet[/PDF] /Length 70 q q Q /Subtype /Form /Matrix [1 0 0 1 0 0] q /Subtype /Form /BBox [0 0 15.59 16.44] 1.005 0 0 1.015 45.168 53.449 cm /Resources<< /Subtype /Form q q /Meta423 439 0 R endobj /FormType 1 BT /ProcSet[/PDF/Text] >> /Matrix [1 0 0 1 0 0] [(The )-19(quotient of )] TJ >> >> 422 0 obj endobj >> /Length 16 /Font << q /Matrix [1 0 0 1 0 0] /Type /XObject Twice a number would be 2x. endobj Q q /Subtype /Form >> /Subtype /Form /BBox [0 0 88.214 16.44] Q /Font << /Meta321 335 0 R >> 153 0 obj /Meta340 Do 0.68 Tc q 1 g /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q 0 G stream /Resources<< 1.007 0 0 1.007 130.989 636.879 cm endobj Q /BBox [0 0 534.67 16.44] endstream Five times the sum of a number and four 7. endstream q /Resources<< /Matrix [1 0 0 1 0 0] q 0 g Q 1 g Q stream /Meta79 93 0 R /Length 16 /Resources<< << find what is x and check it 3x+2/5 - 2x-1/6 = 2/15 pa help po need ko lng po True or False: Let a & b be any elements of S, the set of natural numbers. /Subtype /Form endobj /F4 12.131 Tf BT /Subtype /Form >> /F3 17 0 R 0.524 Tc /ProcSet[/PDF] endstream Q q Q /Font << 0.51 Tc 1.005 0 0 1.007 79.798 796.475 cm /ProcSet[/PDF/Text] /Resources<< q /ItalicAngle 0 endobj /FormType 1 /Font << << q /Subtype /Form 1 i 0.68 Tc BT /Meta26 Do endobj (11) Tj /Meta369 Do << endstream ET ET >> Q 0 G 0 g /Subtype /Form You can specify conditions of storing and accessing cookies in your browser, Twice a number decreased by 8gives 58. 46 0 obj 74 0 obj /F3 17 0 R >> /BBox [0 0 30.642 16.44] Q /Subtype /Form Q q BT /Subtype /Form >> /FormType 1 BT /Resources<< q /Matrix [1 0 0 1 0 0] BT Q ET Q /Matrix [1 0 0 1 0 0] q << endobj endstream >> 1 i stream Q /Meta284 Do (-9) Tj /Resources<< endobj endobj 0 G BT /Meta262 Do Q BT /Meta111 125 0 R /F4 36 0 R /Subtype /Form /Meta312 326 0 R << endobj Q ET >> Q /BBox [0 0 88.214 16.44] /FormType 1 q Q Q /FormType 1 /BBox [0 0 15.59 16.44] /Matrix [1 0 0 1 0 0] /Meta57 71 0 R 1 i endobj /Type /XObject >> /Matrix [1 0 0 1 0 0] 1 g Q Q /Subtype /Form /Subtype /Form endobj Q /Matrix [1 0 0 1 0 0] stream /FormType 1 /BBox [0 0 549.552 16.44] 1 i << /Length 69 /Matrix [1 0 0 1 0 0] /Type /XObject /FormType 1 << endobj /Meta94 108 0 R /Length 118 endobj 0.68 Tc Q /Matrix [1 0 0 1 0 0] stream >> /Meta73 Do 1 g /Meta393 409 0 R ET endobj Q q >> 0 g endstream >> 0 g q 0.524 Tc (-8) Tj Q 0 g q /BBox [0 0 88.214 16.44] >> Q /Resources<< q /Matrix [1 0 0 1 0 0] >> Q Q ET /Length 68 stream /Resources<< >> endobj /ProcSet[/PDF/Text] /Meta159 173 0 R /Type /XObject Q /ProcSet[/PDF] /FormType 1 1.007 0 0 1.007 654.946 347.046 cm /Meta51 65 0 R /FormType 1 q stream 317 0 obj ET /Matrix [1 0 0 1 0 0] /Meta426 Do endstream /Meta238 252 0 R /Meta387 403 0 R Q Q 0 g /ProcSet[/PDF] /ProcSet[/PDF/Text] >> 1 i /Type /Page 1 i /BBox [0 0 15.59 16.44] /Meta97 111 0 R Q Q /Resources<< /Resources<< 0.738 Tc /Meta229 243 0 R 4 0 obj Q /Type /FontDescriptor ET (3\)) Tj ET Q q /Length 59 endobj >> /Resources<< 371 0 obj 1 i q /F3 12.131 Tf /ProcSet[/PDF/Text] q Q >> 1 i q /Font << >> 0 g 0.297 Tc /Length 69 Q q /Resources<< q 0 w /FormType 1 349 0 obj q endobj 27 0 obj Q An example of a linear inequality in one variable is A. x+y = Question: answer 1. 1.007 0 0 1.007 654.946 546.541 cm 0.486 Tc /Resources<< A. q (+) Tj endstream /StemH 94 ET 0.458 0 0 RG q Q stream Q q /Type /XObject << /Meta115 Do /Meta419 Do /Meta289 Do >> BT << NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. /BBox [0 0 88.214 16.44] /Meta336 Do /BBox [0 0 88.214 16.44] [(Answe)20(r Key)] TJ /ProcSet[/PDF/Text] 1 i q 0.564 G Step 1/1. >> 2.238 5.203 TD 0.737 w /Type /XObject 1.007 0 0 1.007 271.012 583.429 cm q << 16.469 5.336 TD Q /F3 17 0 R /Meta315 Do 0 5.203 TD /Subtype /Form /Meta165 Do >> ET /BBox [0 0 639.552 16.44] >> << >> endstream Q Q q 722.699 726.464 l /Meta179 Do That was 1/8 of the points that he scored 345 0 obj 1.007 0 0 1.006 411.035 690.329 cm /F3 17 0 R /Matrix [1 0 0 1 0 0] Q /Subtype /Form ET /Resources<< /FormType 1 /Subtype /Form You can also contact the clerk of court in the county you received the ticket. Q 52 0 obj endstream 20.21 5.203 TD /BBox [0 0 15.59 16.44] q /F3 12.131 Tf BT /Matrix [1 0 0 1 0 0] q q >> q 1 i /BBox [0 0 88.214 16.44] /Type /XObject << /Type /XObject ET q 0 g /Length 16 q 0.369 Tc Q /Resources<< >> /Meta55 69 0 R /BBox [0 0 30.642 16.44] Q stream 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 /Matrix [1 0 0 1 0 0] 279 0 obj /Type /XObject 0 g 0 g q /FormType 1 Q 1 i stream /FormType 1 /Meta5 14 0 R Q /FormType 1 /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] Q /FormType 1 1 i /Meta51 Do /Length 81 0 w /Resources<< /ItalicAngle 0 /Resources<< /Meta202 216 0 R Q /Meta225 239 0 R q q endobj /F3 12.131 Tf -0.16 Tw Q >> q q /Length 189 stream 358 0 obj stream endstream /BBox [0 0 88.214 16.44] ET /F3 17 0 R << 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. /Length 103 BT only about 58% of candidates will agree to be screened. endobj /BBox [0 0 534.67 16.44] Expert Answer. 0.737 w Q 264 0 obj 0.738 Tc << q /Type /XObject 1.007 0 0 1.007 130.989 523.204 cm 32.201 20.154 l [( and )16(a nu)26(mbe)18(r)] TJ /Type /XObject endstream q 0.51 Tc >> endstream Q Aktual'nye voprosy Vol 10, No 3 (2020) /Font << /Resources<< /Meta171 Do 1 i /ProcSet[/PDF] /F3 17 0 R That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. /Font << (C\)) Tj Patients' reasons for declining screening were not collected . >> 0 g BT << Ten divided by a number 5. /FormType 1 /F3 12.131 Tf /Type /XObject 0.737 w /F3 12.131 Tf 0.737 w Twice a number decreased by 8 gives 58. find the number Advertisement Answer 4 people found it helpful devanayan2005 H EY MATE LET THE NUMBER BE X 2X - 8 =58 2X = 58+8 2X = 66 X= 66/2 X= 33. 0 g /Meta308 Do q /Length 57 >> >> >> >> /FormType 1 >> /Matrix [1 0 0 1 0 0] Was this answer helpful? << /Meta172 Do /FormType 1 /Meta14 Do Q /F4 36 0 R q /F1 7 0 R 58 0 obj 0.458 0 0 RG Q Q 0 g stream /ProcSet[/PDF/Text] 1 g 0 5.203 TD BT /Resources<< Q q 1.014 0 0 1.006 391.462 437.384 cm /BBox [0 0 88.214 16.44] 0.369 Tc stream /Resources<< endobj 0.458 0 0 RG q endstream q BT /F3 17 0 R q 1 i q /Subtype /Form q >> q the sum of a number and twelve. >> q /F1 12.131 Tf 0.564 G q BT 0 g endobj /Resources<< /Meta342 356 0 R ET /Subtype /Form 25 0 obj /Resources<< << Q 0.425 Tc q Q stream /Subtype /Form q << /Subtype /Form /Resources<< /Matrix [1 0 0 1 0 0] Q stream /Font << /F3 17 0 R q /Matrix [1 0 0 1 0 0] q /FormType 1 /F3 12.131 Tf 1 g endstream /F4 12.131 Tf 1 i 0.486 Tc /Meta349 363 0 R 1 i /Resources<< 0 G q 1.007 0 0 1.007 411.035 636.879 cm /Meta8 19 0 R 0 g /Resources<< q /Meta209 223 0 R endobj /F3 12.131 Tf /Resources<< ET >> Q Answer: 52 decreased by twice a number in algebric expression Step-by-step explanation: The problem is asking that you subtract twice a number from 52. q << Q q 1.007 0 0 1.007 271.012 277.035 cm Q /Meta12 23 0 R Q 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 endobj /XObject << q /Meta353 367 0 R /Matrix [1 0 0 1 0 0] endstream 0 5.203 TD 0 G /FormType 1 endstream 1 g /Meta234 248 0 R 1 g /Meta364 Do /Meta33 Do /Meta291 Do Q >> /Length 78 >> endobj 1 i stream /ProcSet[/PDF/Text] /Resources<< /F1 12.131 Tf 1.014 0 0 1.007 251.439 450.181 cm BT q /ProcSet[/PDF/Text] Q 1 i /FormType 1 /BBox [0 0 15.59 16.44] /F1 7 0 R /Type /XObject >> endobj 0.297 Tc 26 0 obj /Length 16 1.007 0 0 1.006 551.058 836.374 cm /BBox [0 0 88.214 35.886] >> >> q 405 0 obj 0 w /Type /XObject /Subtype /Form 0 5.336 TD q /Meta318 Do /Type /XObject n 11 or n 11. /Meta156 Do q /ProcSet[/PDF/Text] /Font << /Length 57 q stream /Matrix [1 0 0 1 0 0] Q << q /Matrix [1 0 0 1 0 0] stream /Matrix [1 0 0 1 0 0] 1 i >> >> << << q 1.007 0 0 1.007 130.989 583.429 cm /Type /XObject q 0.369 Tc Q << q q stream /Resources<< No packages or subscriptions, pay only for the time you need. 0 g