twice a number decreased by 58
twice a number decreased by 58
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BT 0.458 0 0 RG Q /Meta303 Do /Subtype /Form ET endstream >> >> (4) Tj /FormType 1 1st step. >> 1 i endstream >> 0 G q /FormType 1 0 G /ProcSet[/PDF] q /BBox [0 0 88.214 16.44] BT /BBox [0 0 673.937 15.562] /Resources<< stream Q w/Honors. /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] 1.005 0 0 1.007 79.798 746.789 cm stream Q endstream /Resources<< 0.564 G 47.933 5.203 TD Q /Length 69 /BBox [0 0 534.67 16.44] /ProcSet[/PDF/Text] stream Q S /FormType 1 >> /F3 12.131 Tf /Type /XObject /FormType 1 Q 222 0 obj /F3 12.131 Tf 0 w /Length 16 Q 0 G /ProcSet[/PDF/Text] >> Q endobj ET stream /BBox [0 0 534.67 16.44] 185 0 obj q >> 15.731 5.336 TD /BBox [0 0 88.214 35.886] endobj q >> << /Meta308 Do >> /Font << /ProcSet[/PDF/Text] Q 0.564 G (5\)) Tj /Type /XObject BT /F3 12.131 Tf /ProcSet[/PDF] q stream /Type /XObject >> /Length 58 Q >> BT /Meta180 194 0 R >> Q 0 g /Meta318 Do /Meta335 Do Q /Subtype /Form q 29 0 obj 0 G /BBox [0 0 534.67 16.44] Q /Matrix [1 0 0 1 0 0] /Length 107 /FormType 1 1 i /Length 69 1 i << /F3 12.131 Tf Q 0 g stream Q stream endobj 1.007 0 0 1.006 411.035 510.406 cm /Meta233 247 0 R BT S /Resources<< 186 0 obj Q 1.007 0 0 1.007 130.989 636.879 cm q 1 i endstream Q /Subtype /Form Q Q q /Type /XObject ET /Subtype /Form BT /FormType 1 /F3 17 0 R /Matrix [1 0 0 1 0 0] /Meta195 209 0 R /Meta418 434 0 R Q 1.007 0 0 1.007 130.989 277.035 cm /FormType 1 Solution. Q /F3 12.131 Tf /F3 17 0 R /Matrix [1 0 0 1 0 0] 0 g /F3 17 0 R 0 g 0 g 1.014 0 0 1.007 251.439 383.934 cm 1.007 0 0 1.007 45.168 829.599 cm 22.478 5.336 TD endobj >> /ProcSet[/PDF/Text] /Meta426 Do (3\)) Tj 0 G /Length 54 /Matrix [1 0 0 1 0 0] Q stream q >> ET /Meta365 Do ET endstream ET q /FormType 1 << 1 i >> 1 i /ProcSet[/PDF/Text] /F1 12.131 Tf q >> /Font << /Resources<< /BBox [0 0 88.214 16.44] /Type /XObject ET endobj q endobj 0 g << << Q 1 i 0.175 Tc >> Q stream ET endstream /FormType 1 Q 0.458 0 0 RG stream >> /Subtype /Form /Type /Font 402 0 obj endstream /BBox [0 0 88.214 16.44] 206 0 obj q Q Q endstream BT /Type /XObject stream /ProcSet[/PDF/Text] Q 0.458 0 0 RG /ProcSet[/PDF/Text] /Resources<< 0.458 0 0 RG /FormType 1 q /ProcSet[/PDF] /FormType 1 To find: The. 1.014 0 0 1.007 531.485 450.181 cm stream 1 i 0.564 G 1.014 0 0 1.007 111.416 776.149 cm q (8\)) Tj q /F1 12.131 Tf 0.737 w 6.746 8.18 TD 259 0 obj /ProcSet[/PDF] /Subtype /Form >> (\)) Tj q 0.458 0 0 RG stream q q Q /ProcSet[/PDF/Text] q Q >> /FormType 1 /Meta279 Do q 0.738 Tc /FormType 1 q >> 0 G 1.005 0 0 1.007 79.798 713.666 cm /Matrix [1 0 0 1 0 0] 1 i 197 0 obj 45 0 obj 0 G endobj /FormType 1 /Matrix [1 0 0 1 0 0] -0.106 Tw /Length 54 /Font << /Matrix [1 0 0 1 0 0] 0.737 w >> /Meta103 117 0 R Q 0 g << 0 g The result is 8 less than 10 times the number. 107 0 obj /Font << endobj 400 0 R endobj /Meta231 Do >> endstream /Meta265 Do /Length 16 (-9) Tj Q /Matrix [1 0 0 1 0 0] Q /FormType 1 0 g 0 G /FormType 1 /ProcSet[/PDF] q 226 0 obj 0 g /Meta30 Do /Type /XObject Q 0 G stream /Meta23 Do Q 0.738 Tc /F3 17 0 R Q /Meta227 241 0 R >> 0 G 131 0 obj /FormType 1 << /FormType 1 /Subtype /Form /Matrix [1 0 0 1 0 0] /F3 17 0 R /Matrix [1 0 0 1 0 0] 1 i Q << q [(MULTIPLE CHOICE. /ProcSet[/PDF/Text] endobj /ProcSet[/PDF] /Meta14 Do Q q 1.007 0 0 1.007 411.035 277.035 cm /Type /XObject endstream Q /ProcSet[/PDF/Text] 1.007 0 0 1.007 411.035 330.484 cm /Type /XObject 1.007 0 0 1.007 411.035 636.879 cm endobj Q /Matrix [1 0 0 1 0 0] 144 0 obj ET 0 g /Resources<< /Meta313 Do 0 w >> /F3 17 0 R /I0 Do >> /Resources<< /F3 12.131 Tf (3) Tj q /Resources<< >> /FormType 1 q 0 G stream 0 w 0.838 Tc endobj /Subtype /Form /Meta136 Do 112 0 obj << Q /Type /XObject q 1 i >> Q /Meta141 155 0 R BT 0.458 0 0 RG >> /BBox [0 0 15.59 29.168] Q endobj 1.007 0 0 1.007 551.058 523.204 cm endstream /Length 69 1.005 0 0 1.007 102.382 872.509 cm << 1 g /Font << q Q endobj Q endstream 1 i ET /Length 59 endstream /Font << /FormType 1 ET >> 326 0 obj 120 0 obj BT /Meta287 Do q 54.679 5.203 TD endobj /MissingWidth 250 /ProcSet[/PDF] >> 0 g /BBox [0 0 17.177 16.44] stream q endstream Q /Meta166 180 0 R /BBox [0 0 534.67 16.44] >> /Matrix [1 0 0 1 0 0] the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . endobj /Subtype /Form 0 g Q ET ET 0.737 w In addition, testosterone in both sexes is involved in health and well-being . /Subtype /Form Q /FormType 1 20.21 5.203 TD /Resources<< q >> /Resources<< 1.005 0 0 1.007 102.382 726.464 cm stream << 1.014 0 0 1.006 531.485 437.384 cm stream 0 G /FormType 1 /FormType 1 0.564 G Q >> Q q Q 0.737 w 0 4.894 TD Twice a number is decreased by 9, and this sum is multiplied by 4. /Matrix [1 0 0 1 0 0] << /BBox [0 0 639.552 16.44] /Type /XObject /F3 17 0 R stream /Meta20 31 0 R /Meta390 Do Q /F3 17 0 R 1 i 15.731 5.336 TD 1.014 0 0 1.007 251.439 636.879 cm stream /FormType 1 >> /F3 12.131 Tf 281 0 obj endstream /Matrix [1 0 0 1 0 0] /ProcSet[/PDF] << 0 g 16.469 5.336 TD /Meta218 232 0 R >> q q stream /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /Resources<< q /Subtype /Form (\)) Tj /Resources<< (2) Tj 1.007 0 0 1.007 130.989 277.035 cm Q endobj /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Meta21 Do >> 0.564 G /BBox [0 0 15.59 16.44] 1.502 5.203 TD >> /ProcSet[/PDF/Text] /Type /XObject /ProcSet[/PDF/Text] q q >> /BBox [0 0 88.214 16.44] q q 1 i 1 g 1.007 0 0 1.007 551.058 277.035 cm Q 1 i >> /Matrix [1 0 0 1 0 0] Q 0.524 Tc q endstream Q >> Q 0 g >> /Meta267 281 0 R << 32.201 5.203 TD /FormType 1 >> /Meta110 Do /ProcSet[/PDF/Text] /F3 12.131 Tf ET Q /ProcSet[/PDF/Text] Q /Meta225 Do stream /Length 16 Q /Meta1 8 0 R /ProcSet[/PDF] >> /BBox [0 0 88.214 16.44] endobj /FormType 1 q 2x - 15 = -27. /BBox [0 0 549.552 16.44] /Type /XObject /FormType 1 q q >> /ProcSet[/PDF] /F3 12.131 Tf >> /Subtype /Form 20.21 5.203 TD q /FormType 1 /Subtype /Form stream /FormType 1 0.564 G >> Two, two times, twice, twice as much as, double 2 Twice z 2z y doubled 2y Multiplication by Half of, one-half of, half as much as, one-half times 1 2 Half of u u 2 one-half times m 1 2 m Geometry Problems Concept Word Expression Algebraic Expression Area of a square Side Squared A = s2 Perimeter of a square Four times the side P = 4s 0.838 Tc 114 0 obj /Matrix [1 0 0 1 0 0] q /Subtype /Form /Meta117 Do q /BBox [0 0 88.214 16.44] 0 w endobj 0 g 1 i 0 w /Subtype /Form >> /Meta248 Do << /F3 12.131 Tf >> q 1 i 0 G -0.101 Tw q Q Q >> /Type /XObject 9.723 5.336 TD q 1.014 0 0 1.007 251.439 776.149 cm /Meta304 Do Q /Font << Q Q >> 0.369 Tc 0 g /BBox [0 0 30.642 16.44] q q 30 0 obj /Resources<< /Type /XObject q Q << /F3 12.131 Tf stream 1 i 1.005 0 0 1.007 102.382 599.991 cm /Meta252 266 0 R /F3 17 0 R /Font << 0.564 G 0 G /Font << q q /Meta267 Do Have a nice day! Q BT /Subtype /Form 1 i /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] << >> 1.007 0 0 1.007 551.058 277.035 cm /Resources<< /Meta268 282 0 R /Meta400 Do q /F4 36 0 R /Subtype /Form BT 13.493 5.336 TD /Font << ET /Meta323 337 0 R q /Length 294 1.007 0 0 1.007 551.058 703.126 cm /Length 68 /Meta109 Do Q 0.564 G q stream q 0.425 Tc (2\)) Tj /FormType 1 /Subtype /Form endobj Q ET /Length 78 0 g q /Meta185 Do /FormType 1 /Type /XObject /ProcSet[/PDF] q 23.216 5.203 TD Q /ProcSet[/PDF] /Meta357 Do >> /Matrix [1 0 0 1 0 0] Q endstream >> /Type /XObject 0.564 G endstream 0 G q to represent the numbers. 313 0 obj q Q Thirthy is equal to twice a number decreased by four = solve and check the equation? /FormType 1 q /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 411.035 277.035 cm Q /Resources<< /Meta175 189 0 R Q 426 0 obj /Subtype /Form endobj /Type /XObject /Length 16 BT << endstream /ProcSet[/PDF/Text] BT 1.007 0 0 1.007 551.058 636.879 cm q q /Meta270 Do stream Q /Type /XObject >> >> /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 q 1.007 0 0 1.007 271.012 776.149 cm 1.005 0 0 1.007 102.382 872.509 cm 0.737 w /Subtype /Form q 1 i /AvgWidth 401 Q >> >> /Type /XObject endobj 0.458 0 0 RG /Type /XObject /Resources<< 0 20.154 m 1.502 5.203 TD stream Q /FormType 1 Q 0.51 Tc 106 0 obj BT << /Meta159 Do /Subtype /Form Q /ProcSet[/PDF] >> q /Meta145 159 0 R Q Q >> stream 1 i 1 i endstream Q /ProcSet[/PDF] stream /Type /XObject q Q /Type /XObject 1 g /Length 68 1.014 0 0 1.007 111.416 383.934 cm << 0.838 Tc >> 1.007 0 0 1.007 271.012 383.934 cm 1 i 0.369 Tc /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 0 g Q Q 1.007 0 0 1.007 271.012 776.149 cm >> /Length 69 q /Meta350 364 0 R 1 i endstream /FormType 1 0 g (D\)) Tj /Resources<< (iv) A number exceeds 5 by 3. endstream /Meta393 409 0 R /Font << /F1 7 0 R 1 i /Type /XObject Q Q /Resources<< /Type /XObject q q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 2.238 5.203 TD 0 g /Subtype /Form /Font << /Matrix [1 0 0 1 0 0] >> /Subtype /Form q /Font << BT /Font << /Length 57 Q (5) Tj Q /FormType 1 /Font << << /Subtype /Form /ProcSet[/PDF] /F3 12.131 Tf /BBox [0 0 549.552 16.44] 1.007 0 0 1.007 551.058 277.035 cm /Type /XObject /F3 12.131 Tf Q /Font << Q << >> endstream BT stream 148 0 obj /Meta229 Do /Type /XObject /Meta63 77 0 R Q /BBox [0 0 88.214 35.886] 170 0 obj Q /FormType 1 endstream >> endobj Q 15.731 5.336 TD Q /FormType 1 q 1 i /ProcSet[/PDF/Text] 0 G Expert Answer. /Subtype /Form /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 531.485 703.126 cm endobj [(F)-22(ive)] TJ >> 134 0 obj 1.007 0 0 1.007 654.946 599.991 cm /Matrix [1 0 0 1 0 0] 190 0 obj BT 0 g 1.007 0 0 1.007 411.035 636.879 cm /F3 12.131 Tf Q >> q /Matrix [1 0 0 1 0 0] /Type /XObject << /BBox [0 0 15.59 29.168] /Meta379 393 0 R /Type /XObject /Resources<< 0 G >> /Font << Q >> q 1 g endobj 0 G endobj Q endobj /Meta359 Do stream 0 g ET /Subtype /Form Advertisement Loved by our community 50 people found it helpful Madhvendra13 2x -8=58 2x=66 x=662 x=33 Find Math textbook solutions? (11) Tj /FormType 1 /F3 12.131 Tf /ProcSet[/PDF] /F4 36 0 R stream /Length 68 1.007 0 0 1.007 271.012 450.181 cm << << 0.524 Tc /Length 69 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 /Subtype /Form stream /Resources<< Q stream 212 0 obj /Subtype /Form 0 g >> stream q Q stream 1 i endobj 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. q -0.486 Tw /FormType 1 q /Type /XObject << /Matrix [1 0 0 1 0 0] 202 0 obj Q 0.458 0 0 RG Q >> Solution: ET 0 G q q q 3.742 8.18 TD /Meta151 165 0 R BT Q >> 0 g Q /Matrix [1 0 0 1 0 0] 0 5.203 TD endstream /Length 68 /Resources<< 19.474 20.154 l /ProcSet[/PDF] ( x) Tj 0 G stream stream 1 i /F3 12.131 Tf >> >> BT /Font << 1.005 0 0 1.007 102.382 546.541 cm 0 g /Meta276 290 0 R BT /XObject << q 369 0 obj 0.68 Tc /FormType 1 /Type /XObject 1 g This site is using cookies under cookie policy . /Meta43 57 0 R /Meta372 386 0 R /BBox [0 0 30.642 16.44] 327 0 obj 330 0 obj /Meta340 Do In contrast, nonresident alien undergraduate enrollment was 15 percent lower in 2020 than in 2019 (468,900 vs. 548,600). ET q Q q /Meta316 Do q 0 g >> /Meta427 Do Q >> q 0.458 0 0 RG /FormType 1 ET Q Q Q /ProcSet[/PDF] BT endstream endobj Q /FormType 1 0 g (3\)) Tj Q /Subtype /Form 1 i << /Meta385 401 0 R /FormType 1 /Type /XObject >> /Meta330 344 0 R 0.134 Tc /Subtype /Form Q 1 g q /ProcSet[/PDF] Q 1.014 0 0 1.007 391.462 636.879 cm 1.007 0 0 1.007 130.989 776.149 cm q 1 i << /FormType 1 /Length 16 Q (7\)) Tj >> /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] 11.99 8.18 TD the quotient of twenty and a number a.) 1 i BT 0.737 w /FormType 1 >> Q ET /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] Q /Length 57 /Matrix [1 0 0 1 0 0] >> /BBox [0 0 534.67 16.44] q ET BT 1.007 0 0 1.007 271.012 277.035 cm Q Q Q /Meta243 Do /Type /XObject 1.007 0 0 1.007 130.989 849.172 cm endobj /Type /XObject /ProcSet[/PDF/Text] 1.502 5.203 TD 1 i /Subtype /Form Twice a number decreased by another number: 1.007 0 0 1.007 45.168 713.666 cm << 295.086 4.894 TD /Subtype /Form q q /Meta123 137 0 R << /BBox [0 0 534.67 16.44] endstream >> 0 G We are asked to find the number, so, we could assign the number as "x". /Length 69 Q 1 g /FormType 1 1.007 0 0 1.007 271.012 277.035 cm /Type /XObject 0 G >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] >> /Matrix [1 0 0 1 0 0] >> Q /Resources<< 0 g 1.007 0 0 1.007 411.035 849.172 cm endobj 6.746 5.203 TD >> 0 G ( decreased by ) Tj 0 w /FontName /TestGen-Regular (-11) Tj 1 i /Meta92 106 0 R q q 37 0 obj 0 G /BBox [0 0 88.214 16.44] /F3 17 0 R Q Q /Resources<< Q 1 g 0 G /Ascent 1050 >> /Subtype /Form Q endobj >> ET /F3 17 0 R /FormType 1 /Meta52 Do (x ) Tj /Meta362 376 0 R /Matrix [1 0 0 1 0 0] 1.005 0 0 1.009 45.168 905.633 cm Q Q /Subtype /Form /FormType 1 /FormType 1 /Length 69 1 i >> BT 0.369 Tc /Subtype /Form /F3 12.131 Tf /BBox [0 0 639.552 16.44] q /Type /XObject /F3 12.131 Tf Q 0.738 Tc ET >> Q Q q /Type /XObject 1.007 0 0 1.007 130.989 636.879 cm /F3 12.131 Tf Q endstream /F3 12.131 Tf /Subtype /Form << q /ProcSet[/PDF] Q /Resources<< 0.564 G 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) q /Subtype /Form q 1.005 0 0 1.007 102.382 400.496 cm /Type /XObject 1.007 0 0 1.007 45.168 796.475 cm 0 g q /Matrix [1 0 0 1 0 0] Q /Font << endstream endstream 1 i (4\)) Tj q Q /ProcSet[/PDF/Text] [(E)-14(le)-23(ven)] TJ /Meta238 Do 18.708 17.593 TD 1 i endstream 0.564 G /Matrix [1 0 0 1 0 0] q Q /Meta393 Do /Meta332 346 0 R /BBox [0 0 88.214 16.44] /BBox [0 0 15.59 16.44] q q /F1 14.682 Tf /ProcSet[/PDF] 0 g A link to the app was sent to your phone. q /F3 12.131 Tf Q 1 i stream q >> /Meta47 Do /Meta353 Do /Subtype /Form 1.007 0 0 1.007 551.058 330.484 cm q ET /F3 12.131 Tf 1.007 0 0 1.007 271.012 849.172 cm Q 1 i Q >> 0 g /BBox [0 0 88.214 16.44] (4\)) Tj /Meta336 350 0 R /F3 17 0 R /Subtype /Form q >> /Matrix [1 0 0 1 0 0] (\)) Tj BT /Length 59 /ProcSet[/PDF] Q ( \() Tj 1 g the sum of a number and twelve. BT Q /BBox [0 0 88.214 16.44] endobj Q >> /Resources<< /Subtype /Form Q /Matrix [1 0 0 1 0 0] BT q /Type /XObject endobj /FormType 1 9 0 obj 20.21 5.203 TD 30.699 4.894 TD /FormType 1 Find the number. /BBox [0 0 639.552 16.44] 0 g /Subtype /Form q >> /BBox [0 0 639.552 16.44] 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 q /Meta34 47 0 R /Type /XObject /Matrix [1 0 0 1 0 0] endstream Q >> /ProcSet[/PDF/Text] q << q 1 i q /Length 69 109 0 obj endobj /BBox [0 0 17.177 16.44] 5 0 obj stream /BBox [0 0 549.552 16.44] >> /Length 16 Q 0.458 0 0 RG 38.948 5.203 TD Q 379 0 obj 433 0 obj /ProcSet[/PDF/Text] << q Q /Resources<< /Length 16 1.007 0 0 1.006 551.058 437.384 cm 425 0 obj Q BT Q q 0.369 Tc Q Q /BBox [0 0 15.59 16.44] 1 g 1.007 0 0 1.007 45.168 829.599 cm 1 i << << TJ Q stream /Length 16 /Length 16 /Length 12 q endstream /F3 17 0 R q The difference between six and a number divided by nine 10. q BT q Q /F3 12.131 Tf /Type /XObject Q /Type /XObject /ProcSet[/PDF] 0 0 Similar questions Find the number which when decreased by 8% becomes 506. 141 0 obj Q /Length 78 << >> /FormType 1 q /Resources<< 0 G 329 0 obj /Matrix [1 0 0 1 0 0] /Resources<< >> /FormType 1 q /BBox [0 0 88.214 16.44] /Type /XObject /Meta130 Do (x) Tj Q /ProcSet[/PDF/Text] 1 i >> /Matrix [1 0 0 1 0 0] /Type /XObject endstream >> 0.458 0 0 RG ET /Font << q 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved, Algebra Help Calculators, Lessons, and Worksheets. /FormType 1 /Type /XObject /F1 12.131 Tf Find the number. /Subtype /Form q /Meta180 Do q 0 g 0 g endstream stream (9\)) Tj /Meta370 Do 149 0 obj q ET 1 i >> /Length 69 q >> q /Font << 0.001 Tw q Q 1.005 0 0 1.007 45.168 916.925 cm Q /BBox [0 0 673.937 14.853] /F3 17 0 R stream 1.005 0 0 1.007 102.382 670.003 cm /ProcSet[/PDF] >> /F1 7 0 R /Meta51 Do /F3 12.131 Tf Q /Ascent 891 q 0 g Q 1 i /Font << /Resources<< q /Resources<< /FontName /PalatinoLinotype-Roman /F3 12.131 Tf >> /Meta79 Do /Meta363 Do /Resources<< q /BBox [0 0 15.59 16.44] /F3 12.131 Tf Q q 1 i >> q /F4 36 0 R BT >> 1.007 0 0 1.007 411.035 330.484 cm q /Length 68 >> << >> endstream ET >> 0 w endobj /Subtype /Form 0.737 w BT /FormType 1 0.524 Tc /ProcSet[/PDF/Text] /F4 12.131 Tf /ProcSet[/PDF/Text] /Meta232 246 0 R /Matrix [1 0 0 1 0 0] 47 0 obj /FormType 1 0.297 Tc 0 w /F3 17 0 R 1 g endstream /BBox [0 0 88.214 16.44] /Descent -299 0 g /Resources<< q /Type /XObject /Meta226 Do Q /Meta346 360 0 R Q /BBox [0 0 15.59 16.44] /F1 12.131 Tf /FormType 1 /F3 17 0 R /BBox [0 0 17.177 16.44] q q >> 0 G >> /Meta16 27 0 R /Type /XObject q q stream 0 w q q /Matrix [1 0 0 1 0 0] BT >> /Subtype /Form /Subtype /Form /Meta238 252 0 R /F3 17 0 R 1.005 0 0 1.007 79.798 779.913 cm >> 36 0 obj Q >> /Meta193 Do 1 i Q 0 5.203 TD Q endobj /Length 54 << << q endstream stream endstream /Matrix [1 0 0 1 0 0] 0 g endobj /ProcSet[/PDF/Text] /Meta341 355 0 R /Meta54 Do /Type /XObject Q /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] /F3 12.131 Tf 1 i /Meta297 Do /F3 17 0 R Q /ProcSet[/PDF] q >> stream 235 0 obj /Meta237 Do 0.737 w /F3 17 0 R /F3 12.131 Tf /Resources<< << q << 0.458 0 0 RG 1.007 0 0 1.007 271.012 523.204 cm stream 0 g /Type /XObject endstream /FormType 1 /BaseFont /TestGen-Regular /Meta111 Do /Type /XObject 0 w 3.742 5.203 TD << /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] endstream Q /ProcSet[/PDF] q /BBox [0 0 88.214 16.44] >> ET << endobj /Length 69 1 i endstream endstream 4 0 R 1 i >> (+) Tj /Matrix [1 0 0 1 0 0] 1 g q >> Cho)18(ose the one alternative that best complet)19(es the statement or answers the question)15(.)] q /Matrix [1 0 0 1 0 0] 1 i /Length 78 /BBox [0 0 88.214 16.44] << << /Matrix [1 0 0 1 0 0] >> q 133 0 obj 254 0 obj /FormType 1 >> Q /Font << endobj /Matrix [1 0 0 1 0 0] /Meta258 272 0 R 183 0 obj stream 0 g 209 0 obj /FormType 1 /ProcSet[/PDF/Text] stream Q Q stream /ItalicAngle 0 /Meta221 Do Q /F3 17 0 R 0 g /ProcSet[/PDF/Text] /Length 16 /Resources<< /FormType 1 q 3.742 5.203 TD << /Length 69 /Meta216 Do /Type /XObject five times the sum of a number x and two b.) endstream endstream 443 0 obj << >> /BBox [0 0 88.214 16.44] endstream Then ab is a binary operation. 0 g /Resources<< Q stream /Subtype /Form stream /F3 17 0 R Q >> /Type /XObject endstream Q q >> /FormType 1 Q endobj endobj /F3 12.131 Tf /Font << q << 407 0 obj 243 0 obj >> endstream q << /Type /XObject /Subtype /Form q endobj Q >> << the other number. 0.458 0 0 RG q Was this answer helpful? q Q /Contents [446 0 R] q >> /Font << 361 0 obj endobj /FormType 1 /Length 60 /BBox [0 0 30.642 16.44] 0 g /Matrix [1 0 0 1 0 0] Q /Resources<< /F3 17 0 R 0 G 51 0 obj /FormType 1 BT 1.007 0 0 1.007 551.058 330.484 cm /MaxWidth 1453 Q endstream ET /Type /XObject /Matrix [1 0 0 1 0 0] q 0.458 0 0 RG /F3 17 0 R << 1 i 0 G /Subtype /Form 1.005 0 0 1.007 102.382 653.441 cm /Flags 32 Q /F3 12.131 Tf /Length 69 1 i 1 i BT Q endobj /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Resources<< >> endobj ET /Meta45 Do /Meta176 Do endstream 0 g /Meta239 253 0 R stream << 0 g >> [( a )-15(number, decreased by )] TJ /Matrix [1 0 0 1 0 0] /Meta355 369 0 R /FormType 1 /F3 12.131 Tf 1 i /BBox [0 0 30.642 16.44] q Q endstream endobj stream q /F3 12.131 Tf /Length 65 278 0 obj Q Q Q q >> /Resources<< 0 g q /F4 12.131 Tf Q /Matrix [1 0 0 1 0 0] /Meta172 Do >> 0 w /ProcSet[/PDF] >> Q >> 72 0 obj 0 g /Subtype /Form /Meta329 343 0 R /Length 245 0.458 0 0 RG 0 g q ET /FormType 1 1 i /Meta70 84 0 R 147 0 obj endstream << /Length 54 << /Length 58 /ProcSet[/PDF] Q 0 g /Resources<< /Type /XObject /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] stream /Pages 1 0 R q /ProcSet[/PDF/Text] /Length 12 /Type /XObject q 207 0 obj /Matrix [1 0 0 1 0 0] /Font << >> Percent Change = (Decrease First Value) x 100% 3.742 5.203 TD /Resources<< /F3 17 0 R /BaseFont /PalatinoLinotype-Bold /Meta416 Do Q 0.458 0 0 RG >> Q ET /Subtype /Form 1 g /Font << 0 0 0 500 553 444 611 479 333 556 582 291 0 0 291 883 >> 0 G /Resources<< >> Q q >> /BBox [0 0 549.552 16.44] /BBox [0 0 88.214 16.44] q /Subtype /Form /Meta177 191 0 R 0.425 Tc /Meta415 Do /Meta54 68 0 R /FormType 1 0.737 w Medium /Meta164 Do ET /Resources<< endobj ET endobj /Meta170 184 0 R /Meta115 Do endstream 65.906 4.894 TD Q stream >> Q q stream /FormType 1 q 0 w /Resources<< /Meta355 Do /Font << 0 w q q /BBox [0 0 88.214 16.44] 17.234 5.203 TD q /F1 7 0 R /ProcSet[/PDF/Text] /BBox [0 0 88.214 16.44] endobj 0 20.154 m /FormType 1 Q >> /ProcSet[/PDF/Text] /Meta325 Do 0 w Q << 1.005 0 0 1.013 45.168 933.487 cm endobj /Meta101 115 0 R ET 1 g Q 353 0 obj Q /Resources<< 0.458 0 0 RG BT 0 G /Subtype /Form 273 0 obj q /Matrix [1 0 0 1 0 0] >> 1.007 0 0 1.007 130.989 636.879 cm /FormType 1 endstream /F3 17 0 R >> 20.21 5.336 TD q 1 i 251 0 obj >> q /Resources<< q /BBox [0 0 15.59 16.44] endstream /Font << /Length 80 0 g 549.694 0 0 16.469 0 -0.0283 cm /BBox [0 0 88.214 35.886] 1.007 0 0 1.007 654.946 546.541 cm >> Q >> ET q 1 i 65 0 obj BT 1 i 0 g endobj q q /ProcSet[/PDF] endstream /Length 69 /Length 58 stream endobj 0 w /FormType 1 /F3 17 0 R 1 g /Matrix [1 0 0 1 0 0] q On This Page The Division of Cancer Prevention furthers the mission of the National Cancer Institute by leading, supporting, and promoting rigorous, innovative research and traini q q /F3 17 0 R q /Meta349 Do 1 i 722.699 599.991 l BT >> >> /Resources<< 0 g 0.737 w /Type /XObject 0.458 0 0 RG /Subtype /Form q >> Q 1 i ET /Resources<< Q 1 i /FormType 1 Q >> 1 i /Subtype /Form 0.524 Tc 0 5.203 TD /Meta269 Do /Resources<< Q /Length 16 q 1.007 0 0 1.007 551.058 330.484 cm 0 G /Meta247 Do ET /ProcSet[/PDF] 1 g << 38.182 5.203 TD /Length 57 (C\)) Tj /FormType 1 stream 30.699 5.203 TD 0 g /FormType 1 >> >> >> << Q >> q /Meta60 74 0 R q q 0 g /BBox [0 0 534.67 16.44] /Meta398 414 0 R 2.238 5.203 TD Q /BBox [0 0 88.214 16.44] /F3 12.131 Tf Q << stream /Type /XObject 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 1 i BT Q q /Subtype /Form /Meta291 305 0 R 0 G >> 1 i 1 g /FormType 1 >> /Subtype /Form Q 0.737 w 0 G /Meta300 Do 1 i /Encoding /WinAnsiEncoding endobj /Resources<< BT /Length 16 /Matrix [1 0 0 1 0 0] endstream /F3 17 0 R /Resources<< endobj 0 G 0 G 437 0 obj (x) Tj /ProcSet[/PDF/Text] 0.737 w /Font << 4.506 8.18 TD Thrice of a number = 3x. 1.007 0 0 1.007 67.753 599.991 cm Q << >> << Q 439 0 obj /Matrix [1 0 0 1 0 0] >> Q /FormType 1 0 w 1 i 1.007 0 0 1.007 271.012 849.172 cm << 32.939 5.203 TD stream /Matrix [1 0 0 1 0 0] /FormType 1 /Length 59 endstream endstream /FormType 1 Q 1 i ET Q 0 G 2. (7\)) Tj >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] q /Meta101 Do endstream >> 247 0 obj 286 0 obj q /Meta168 182 0 R 189 0 obj >> q endstream /FormType 1 22.478 4.894 TD q /Length 69 >> /F3 12.131 Tf endstream /Type /XObject endstream ( x) Tj 1.502 5.203 TD q /F3 17 0 R Q /Resources<< stream Q 1 i >> Q q /Meta31 Do >> /Meta213 227 0 R 1.502 7.841 TD /F3 12.131 Tf /FormType 1 << << /Subtype /Form 0 w >> /Meta282 296 0 R >> Q /Matrix [1 0 0 1 0 0] ET endobj 722.699 546.541 l Q Q 0.737 w /Meta12 Do 0 G endobj endobj >> /Meta159 173 0 R 0 g q /Type /XObject 1 g 0.178 Tc /Length 294 >> 150 0 obj /Length 54 BT Q /Resources<< endstream /FormType 1 /BBox [0 0 15.59 16.44] q 1 i ET >> ET /F3 17 0 R Q /FormType 1 q q q >> Q endobj /Matrix [1 0 0 1 0 0] 0 G 0 5.203 TD Q 332 0 obj endstream stream BT endstream << Q q q /Meta49 63 0 R 0 g 169 0 obj /F1 7 0 R Q 0.737 w q /Matrix [1 0 0 1 0 0] 1 g Q endstream q 0.564 G Q 159 0 obj endstream 0 w (vi) If 12 is subtracted from a number, the result is 24. 0.737 w /F1 7 0 R Q /XObject << ET /Matrix [1 0 0 1 0 0] /F3 12.131 Tf q /F2 11 0 R Q q /Meta137 151 0 R /Length 59 Q 0 4.894 TD /FormType 1 111 0 obj Q /BBox [0 0 15.59 29.168] endstream 0 g /Meta147 161 0 R 0 g 1.007 0 0 1.007 271.012 277.035 cm 1.007 0 0 1.007 411.035 277.035 cm 1.005 0 0 1.007 102.382 473.519 cm >> endobj 1 i , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. 2. /FormType 1 /F3 12.131 Tf 0.51 Tc q /Length 16 /FormType 1 /Matrix [1 0 0 1 0 0] 33 0 obj /F3 17 0 R /Meta106 Do /Meta44 58 0 R endstream q << /Type /XObject /Type /XObject endobj /Matrix [1 0 0 1 0 0] Andrew M. 1.502 5.203 TD /Meta241 255 0 R Q 1.007 0 0 1.007 45.168 846.161 cm 132 0 obj 1.502 5.203 TD q << q >> /BBox [0 0 534.67 16.44] 0 56.451 TD /Meta214 Do endobj >> q >> /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] BT /ProcSet[/PDF/Text] /Resources<< (B\)) Tj /BBox [0 0 88.214 16.44] q BT 0 g 0 39.216 TD 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. /Matrix [1 0 0 1 0 0] q /Type /XObject >> >> 1 i /Font << q Q 0 g /F4 36 0 R /Resources<< /FormType 1 endstream /Subtype /Form 1.007 0 0 1.007 271.012 383.934 cm /Type /XObject /ColorSpace [/Indexed /DeviceGray 1 ] endstream q endstream Q /FormType 1 endstream >> /Length 59 q /Meta120 134 0 R Q endstream /Meta379 Do Q: when six times a number is decreased by 4, the result is 8. Q /Meta388 Do Q /Matrix [1 0 0 1 0 0] /Subtype /Form Q Q Q endstream 173 0 obj /Meta286 300 0 R /ProcSet[/PDF] q >> /Type /XObject 1.005 0 0 1.007 102.382 347.046 cm /Meta142 156 0 R stream Q /BBox [0 0 88.214 16.44] (A\)) Tj 1.502 5.203 TD /Matrix [1 0 0 1 0 0] q 1 i /Length 54 stream /Meta363 377 0 R q /FormType 1 0.564 G >> /Meta155 Do q /BBox [0 0 88.214 16.44] 285 0 obj q Q q 0 g 1.007 0 0 1.007 271.012 330.484 cm << /Type /XObject << 0 5.203 TD /BBox [0 0 88.214 35.886] /F3 12.131 Tf 309 0 obj /I0 Do /Type /XObject >> Q >> q Q 0 G >> stream /Subtype /TrueType << /Resources<< 0.458 0 0 RG /Subtype /Form q (C\)) Tj q stream Q stream stream Q q q /Resources<< /Font << >> /Subtype /Form endobj /FormType 1 (13) Tj Q stream stream /Meta22 33 0 R 0 w /Subtype /Form /FormType 1 << Q If a number is 50%, then it is a half - the same as 0.5 or 1/2. /Font << stream 0.737 w 0 g /Matrix [1 0 0 1 0 0] /FormType 1 endobj 672.261 872.509 m /Length 99 endobj endstream /F3 12.131 Tf >> 0 G /Type /XObject >> >> /Meta412 Do /Type /XObject 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 1.007 0 0 1.007 551.058 383.934 cm /F3 17 0 R q /BBox [0 0 88.214 16.44] Q q /Font << Q /Meta338 Do 0 g Find the number 1 See answer Advertisement /ProcSet[/PDF] BT /Resources<< Q q /Length 12 /Meta310 Do q BT q /Ascent 1050 /Resources<< << 0 w 0 g /FormType 1 /Resources<< ET >> Q >> Q q Q 0 w /Matrix [1 0 0 1 0 0] (-20) Tj endobj q 0 G BT /Subtype /Form /Matrix [1 0 0 1 0 0] Q q /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] /Font << , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. 1 g /Meta114 128 0 R /Length 58 /BBox [0 0 15.59 16.44] /Type /XObject /ProcSet[/PDF/Text] << /BBox [0 0 88.214 16.44] Q endobj Q /Matrix [1 0 0 1 0 0] /Subtype /Form /BBox [0 0 88.214 16.44] >> 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 0000001548 00000 n 0000001817 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twice a number decreased by 58