/BBox [0 0 15.59 29.168] /F3 12.131 Tf Q 1 i 22.478 5.336 TD Q Q ET 1 g endobj 159) n decreased by 28 is equal to 48 160) the difference of m and 27 is 34 161) a number decreased by 9 is 23 162) 13 less than w is equal to 35 163) the difference of a number and 22 is equal to 34 164) a number decreased by 27 is equal to 29 165) the difference of r and 20 is 37-6-You may use this math worksheet as long as you help someone . /Meta134 Do 181 0 obj >> Q /Resources<< /Resources<< /I0 Do /Meta387 Do /ProcSet[/PDF/Text] BT 1.007 0 0 1.006 551.058 763.351 cm /Resources<< q /Meta342 356 0 R << Q << 0 g (D) Tj 0 g /FormType 1 407 0 obj Q /Subtype /Form /Length 58 Q q /FormType 1 1 i /ProcSet[/PDF] /BBox [0 0 17.177 16.44] /Meta238 252 0 R /BBox [0 0 15.59 16.44] stream endstream 0 g /Matrix [1 0 0 1 0 0] /Resources<< ET 1 i /Matrix [1 0 0 1 0 0] 323 0 obj 0 w /FormType 1 0.297 Tc Q endstream q endstream Q /Meta427 Do /Resources<< /Meta375 Do /Type /XObject /Resources<< Q /Length 69 /F3 17 0 R endobj endstream /Meta424 Do 1.007 0 0 1.007 411.035 383.934 cm 1 g q /ProcSet[/PDF] 0.458 0 0 RG Q endstream Q << /Resources<< q /FormType 1 /ProcSet[/PDF/Text] Q /Subtype /Form endstream /Font << BT /F3 17 0 R /Font << >> << /Subtype /Form 1 g endobj 1.007 0 0 1.007 67.753 347.046 cm Q stream /Meta371 Do endstream /F3 12.131 Tf q 0.838 Tc /Font << /BBox [0 0 88.214 16.44] q /Subtype /Form 0 G q /ProcSet[/PDF/Text] >> /Type /XObject Q /Matrix [1 0 0 1 0 0] >> 0 G 166 0 obj 0 G /Resources<< q /Meta195 Do BT 672.261 653.441 m 1 i /ProcSet[/PDF] /F3 12.131 Tf /Resources<< q 0 g BT stream 1.014 0 0 1.007 251.439 849.172 cm q /Subtype /Form endstream q Q /ProcSet[/PDF/Text] endstream >> >> /Matrix [1 0 0 1 0 0] /F2 12.131 Tf /Meta77 Do Q 1 i /Font << endobj /ProcSet[/PDF/Text] 1 g << Q ET >> 0 g q Q q /Matrix [1 0 0 1 0 0] >> /F3 12.131 Tf /FormType 1 >> /BBox [0 0 15.59 29.168] /ProcSet[/PDF/Text] /Length 16 /Meta174 188 0 R 0 g 0 g endobj /Matrix [1 0 0 1 0 0] 0 g stream >> /F3 17 0 R q /Meta25 Do 1.502 5.203 TD Q 0 5.203 TD >> /ProcSet[/PDF/Text] /Meta338 352 0 R /FormType 1 /Meta326 Do 290 0 obj >> 1 g Q q /Matrix [1 0 0 1 0 0] /Meta236 250 0 R /Subtype /Form endstream endobj endstream /ProcSet[/PDF/Text] q >> Q 1.007 0 0 1.007 45.168 796.475 cm >> q >> 302 0 obj q /Meta96 Do /Matrix [1 0 0 1 0 0] >> << 216 0 obj 0 g q /Resources<< /Matrix [1 0 0 1 0 0] BT 1 i /Resources<< 151 0 obj /Type /XObject 0 G /ProcSet[/PDF] BT Q 0 g 0 g 1 i (x) Tj 1 i gular prism that is 60 centimeters long, 20 centimeters wide, and 45 centimeters tall. 0 G Q 1 g endobj 0 g >> /F3 12.131 Tf stream 0.564 G 1.005 0 0 1.007 102.382 653.441 cm /Length 65 q 0 G q ET 0.37 Tc q /Font << stream S 279 0 obj Two speeding tickets could increase your rate by 58% at your next renewal. ] Q 1.007 0 0 1.007 551.058 583.429 cm q /F3 17 0 R /Type /XObject /F3 17 0 R ET /ProcSet[/PDF/Text] Q /Length 58 >> Q q BT >> q 1 i /Type /XObject >> >> (x) Tj /Meta100 Do q q stream /Meta361 375 0 R Q /ProcSet[/PDF/Text] q 0 w q ET /Subtype /Form /BBox [0 0 549.552 16.44] >> /Meta213 227 0 R q >> /ProcSet[/PDF/Text] /Resources<< /F3 17 0 R q /Subtype /Form /Type /XObject endstream /Contents [399 0 R] /Meta60 Do 1 i /Matrix [1 0 0 1 0 0] >> 1 i 0 g /BBox [0 0 88.214 16.44] 0.737 w /Meta198 212 0 R >> >> Q BT BT /Resources<< q q 76.394 5.203 TD 0.369 Tc 1.007 0 0 1.006 551.058 437.384 cm /BBox [0 0 15.59 16.44] /Descent -216 [( and )16(a nu)26(mbe)18(r)] TJ >> /F3 12.131 Tf q ET /Length 77 q stream stream 1.007 0 0 1.007 411.035 383.934 cm 1 i 0.564 G q >> /Resources<< Q endobj /F3 17 0 R (A\)) Tj 2. q << Q 0.458 0 0 RG 0 g << Grad - B.S. 0 G /Type /XObject /Font << /FormType 1 << 0 g 0 g >> /ProcSet[/PDF] /FormType 1 Medium Q q endstream >> /ProcSet[/PDF] /Meta60 74 0 R 97 0 obj /F4 36 0 R >> >> /F3 12.131 Tf >> >> /Matrix [1 0 0 1 0 0] q Q BT stream q 0.458 0 0 RG 1 i /Meta128 Do /Subtype /Form /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /FormType 1 q 1 i 0.564 G /Matrix [1 0 0 1 0 0] 1.014 0 0 1.007 391.462 583.429 cm Q BT /FormType 1 /F3 17 0 R q 1 0 obj 1 g Q 1 i /Resources<< /Meta309 323 0 R /BBox [0 0 15.59 16.44] stream q /Type /XObject /Subtype /Form q (38) Tj >> q 20.21 5.203 TD /FormType 1 /Subtype /Image 63 0 obj 1.007 0 0 1.007 411.035 583.429 cm q 75 0 obj /Length 16 >> /Subtype /Form Q /Meta247 261 0 R Q q /Resources<< 1 i q endstream 0.564 G q >> q /ProcSet[/PDF] /Type /XObject 0 g /Font << stream /Resources<< /Meta184 198 0 R Q /BBox [0 0 88.214 16.44] q Q If you are unsure of the county, call the Administrative Office of the Court at (919) 890-1000.Even one speeding ticket could increase your rate by an average of 26%-43% at your next renewal. q 0.458 0 0 RG >> 0 w endstream q endstream /Type /XObject endobj >> /Length 69 /BBox [0 0 673.937 68.796] 0.68 Tc /Length 68 Q ET endobj /F3 17 0 R /BBox [0 0 88.214 35.886] /Subtype /Form 672.261 347.046 m endstream 0 5.203 TD /Resources<< ET Q 0.458 0 0 RG 0 g /BBox [0 0 88.214 16.44] /Subtype /Form endobj << /Resources<< endstream endobj 1 i /F1 12.131 Tf /Type /Font /BBox [0 0 15.59 29.168] /Meta76 Do /ProcSet[/PDF] endobj 1.007 0 0 1.007 130.989 277.035 cm q /Matrix [1 0 0 1 0 0] q 0 g /F3 12.131 Tf endstream 1 i /Ascent 1050 /Type /XObject 405 0 obj /ProcSet[/PDF/Text] stream /FormType 1 /Subtype /Form 1.502 5.203 TD /ProcSet[/PDF/Text] /Length 16 Q ET 1.005 0 0 1.007 102.382 546.541 cm q /Matrix [1 0 0 1 0 0] >> endobj q 49 0 obj 1 i >> /FormType 1 /ProcSet[/PDF] q 0.458 0 0 RG S 1 i /BBox [0 0 30.642 16.44] stream 1 i /Resources<< q << 0 56.451 TD /Meta192 Do /Subtype /Form /Meta186 200 0 R /Length 69 << /Matrix [1 0 0 1 0 0] /Meta158 172 0 R 2.238 5.203 TD Q q /Length 118 /FormType 1 /Meta51 Do 208 0 obj Q 0 G 0 g Q /BBox [0 0 88.214 16.44] /Meta399 415 0 R Q << (A\)) Tj /Length 118 Q /I0 51 0 R q >> >> /BBox [0 0 30.642 16.44] q [( subt)-17(racted fr)-14(om a )-16(number)] TJ /Font << BT 0 g /Length 60 >> 1.007 0 0 1.007 271.012 636.879 cm q 0 g 0 5.203 TD 0 G 0.458 0 0 RG 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. /Meta328 Do /Subtype /Form /Matrix [1 0 0 1 0 0] q /Meta204 Do 0 g 0 G q 55 0 obj 1 i >> endstream /BBox [0 0 88.214 16.44] >> 1.007 0 0 1.007 551.058 330.484 cm (C\)) Tj 396 0 obj 0.369 Tc /FormType 1 0 g q q 0.737 w q /Meta314 328 0 R 1 i Q >> Q q /Meta339 Do /BBox [0 0 17.177 16.44] 0.786 Tc 0 G /ProcSet[/PDF] /Subtype /Form << Q /Length 16 You can specify conditions of storing and accessing cookies in your browser. >> endobj ET << /StemH 94 >> 0 g q endstream /Resources<< /FormType 1 /Font << /Meta142 Do endstream /FormType 1 Q >> q 1 i 0 g 1 i /Matrix [1 0 0 1 0 0] /FormType 1 /Matrix [1 0 0 1 0 0] Q /Resources<< Q: A number increased by 5 is equivalent to twice the same number decreased by 7. /BBox [0 0 15.59 16.44] /ProcSet[/PDF/Text] >> /ProcSet[/PDF] /Meta314 Do endobj q << 1.005 0 0 1.007 102.382 473.519 cm endobj /Length 68 139 0 obj 0 G << Q Q >> /Subtype /Form /Meta10 21 0 R ET q q /F1 12.131 Tf q Q << << 0 g 1.007 0 0 1.007 551.058 703.126 cm stream /FormType 1 /Meta144 Do >> endstream >> MetS-Z quartiles and their associated risks are presented in Fig. endstream /Subtype /Form 0.311 Tc /Type /XObject /ProcSet[/PDF/Text] 0.458 0 0 RG /Resources<< /Length 118 Q 1.007 0 0 1.006 130.989 437.384 cm >> >> /Subtype /Form /Type /XObject 0.564 G Twice a number when decreased by 7 gives 45. 446 0 obj 0 G /Resources<< ET q /Subtype /Form /Meta295 309 0 R q 0 w << >> 0 G 0 G /FormType 1 endstream >> /ProcSet[/PDF/Text] /Resources<< >> endobj >> >> (38) Tj /Resources<< Q 0 w /Meta8 Do /Resources<< Q BT /BBox [0 0 88.214 16.44] S /FormType 1 /F3 17 0 R Q /F4 12.131 Tf 346 0 obj stream /Matrix [1 0 0 1 0 0] /Type /XObject /Type /XObject 0.564 G q >> /BBox [0 0 88.214 16.44] /Resources<< Q /Resources<< ET (2) Tj /Font << 0.737 w << /Resources<< Q /F3 12.131 Tf q /Resources<< 0.51 Tc Q ET /Type /XObject /Meta248 Do stream q 0 g 1 i << /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 130.989 849.172 cm q /BBox [0 0 88.214 16.44] /Meta268 Do /Subtype /Form /Subtype /Form >> q stream /Length 69 /Type /XObject BT endobj q /F3 17 0 R << q 0 G 0 g 333.269 5.488 TD Q 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 /BBox [0 0 30.642 16.44] 0 g /Matrix [1 0 0 1 0 0] /Meta66 80 0 R (3\)) Tj ET Q 0 G 0.737 w endstream See Solution. Q /Meta238 Do Q 1.007 0 0 1.007 271.012 450.181 cm /Meta324 Do >> q endstream >> q /Resources<< 169 0 obj /ProcSet[/PDF] Q /F1 7 0 R Q q /FormType 1 1 g stream q << Q nine increased by a number x. 0 g /Matrix [1 0 0 1 0 0] q >> ET << Q 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. /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] stream /ProcSet[/PDF] q /Meta175 Do >> /BBox [0 0 88.214 16.44] /Resources<< Q /Subtype /Form >> 0 g stream q stream Q stream >> ( decreased by ) Tj /FormType 1 /ID [] q /FormType 1 >> Q Q Q /Resources<< /Meta249 Do /Meta16 27 0 R /Meta338 Do 1 i 1 i /Type /XObject (-11) Tj q q 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. Find the number. 202 0 obj 1 i << q BT /Subtype /Form Q endobj 0.737 w Q /Resources<< Q 0 5.203 TD /F1 12.131 Tf >> /Matrix [1 0 0 1 0 0] /Resources<< 350 0 obj )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] q Q q 0 g /Meta289 Do /Type /XObject endobj ET /Matrix [1 0 0 1 0 0] Q /FormType 1 q Q /Meta187 201 0 R Most questions answered within 4 hours. /BBox [0 0 15.59 29.168] /Meta355 Do 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 If we let "a number" be represented by the variable x, we can translate the given statement into an inequality as: 2x - 4 <= 26. BT /FormType 1 BT /Subtype /Form /Meta26 39 0 R q /Subtype /Form 1.007 0 0 1.007 67.753 599.991 cm ET q Q << stream Q 0 g endstream /Meta289 303 0 R ET 1.007 0 0 1.007 271.012 383.934 cm stream 1.007 0 0 1.007 551.058 636.879 cm Q /Type /XObject /Matrix [1 0 0 1 0 0] /Resources<< Formula - How to Calculate Percentage Decrease. << /BBox [0 0 17.177 16.44] 0 G endobj A. x+6=8 B. x-6=8 C. x+8=6 D. x-8=6. q >> q /XObject << /Font << /Type /XObject /FormType 1 /F3 17 0 R /Matrix [1 0 0 1 0 0] Q -0.463 Tw (4\)) Tj /F3 17 0 R stream /Matrix [1 0 0 1 0 0] << /FormType 1 BT q << endobj endobj stream /ProcSet[/PDF] >> /FormType 1 (D\)) Tj q Q /Matrix [1 0 0 1 0 0] q endstream endstream New questions in Mathematics >> /Meta408 424 0 R /ProcSet[/PDF] /Subtype /Form >> endobj ET stream 1.014 0 0 1.007 391.462 636.879 cm >> /Matrix [1 0 0 1 0 0] << 0.425 Tc >> << /Resources<< endstream /F4 36 0 R /Matrix [1 0 0 1 0 0] Q 120 0 obj Q /FormType 1 q 1 g /Font << Q endobj 0 5.203 TD /ProcSet[/PDF/Text] /Resources<< /Resources<< stream /FormType 1 endstream Q 1.005 0 0 1.007 79.798 746.789 cm 0 G q /ProcSet[/PDF/Text] /Meta148 Do >> /Resources<< /Resources<< BT 1.005 0 0 1.007 79.798 763.351 cm Q /Matrix [1 0 0 1 0 0] /Resources<< >> << /Font << >> /Meta403 Do 1.005 0 0 1.007 79.798 730.228 cm q Q Q (+) Tj /Length 16 /Meta373 Do Q /Meta135 149 0 R (viii) A number divided by 8 gives 7. endstream 0 G Afterward, we are given the second case, here we see that the number would be three times the number decreased by 8. /Font << q endstream [(A number )-17(divided by )] TJ /Subtype /Form << q /FormType 1 0.458 0 0 RG >> Q q 2.238 5.203 TD q 1.005 0 0 1.007 102.382 417.058 cm 0.564 G /F4 36 0 R 1.007 0 0 1.006 411.035 763.351 cm 158 0 obj q q /Length 68 q Q 0.369 Tc /F3 17 0 R >> /F3 17 0 R /Meta362 376 0 R endstream q 1.007 0 0 1.007 130.989 776.149 cm /Subtype /Form /Subtype /Form ET 418 0 obj /Meta111 Do /ProcSet[/PDF/Text] /F3 17 0 R 0 w 1.007 0 0 1.007 130.989 277.035 cm Q endstream /Meta37 Do /ProcSet[/PDF] 1 i Q /ProcSet[/PDF] >> >> q endobj /BBox [0 0 30.642 16.44] >> 439 0 obj /Meta10 Do /Subtype /Form stream q Q /Matrix [1 0 0 1 0 0] /StemV 88 Q q /Type /XObject << q ET << /Matrix [1 0 0 1 0 0] /Resources<< Q /Meta401 417 0 R 0.463 Tc >> t is 56: 4. /ProcSet[/PDF] five times the sum of a number x and two b.) /Type /XObject Q q 0 g /ProcSet[/PDF/Text] The results found were expressed mainly through tables and graphs as the main resources of the statistical language. /F3 12.131 Tf >> /Length 16 /Font << >> >> Q 1.007 0 0 1.007 67.753 473.519 cm >> 1 i 722 722 556 0 667 556 611 0 0 0 722 0 0 0 0 0 /FormType 1 /F3 17 0 R 0.838 Tc /Font << ET /BBox [0 0 17.177 16.44] Q 1.014 0 0 1.007 531.485 636.879 cm 1.007 0 0 1.007 551.058 330.484 cm Q Q /Meta399 Do 1.005 0 0 1.007 102.382 400.496 cm BT /BBox [0 0 88.214 16.44] 1 g /Length 116 q 1.007 0 0 1.007 130.989 383.934 cm endobj 0.564 G q /Subtype /Form S /BBox [0 0 15.59 16.44] 19.474 20.154 l >> 245 0 obj 0.68 Tc /F3 17 0 R /BBox [0 0 15.59 16.44] /FormType 1 0 g Q 185 0 obj /BBox [0 0 549.552 16.44] stream 1 g /Resources<< >> /ProcSet[/PDF] /BBox [0 0 15.59 16.44] Q ET /FormType 1 stream /F3 12.131 Tf q /Meta160 Do 421 0 obj >> Q >> /Subtype /Form /Type /XObject endobj /I0 Do 0 G /Meta271 Do q q /Resources<< Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. 1 i /Meta41 Do Q Q 1 i Q q 1.005 0 0 1.007 79.798 846.161 cm << /Matrix [1 0 0 1 0 0] q /Meta396 412 0 R >> endstream q /Subtype /Form Q /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] Q /Resources<< /Length 80 stream << >> endobj /F3 17 0 R 1 i BT >> 0.737 w 250 0 obj q /Meta425 441 0 R /Subtype /Form /Matrix [1 0 0 1 0 0] Q q 0.463 Tc q /Meta133 147 0 R ET endobj 393 0 obj Q /Meta2 Do q 0 G /FormType 1 /FormType 1 q /Matrix [1 0 0 1 0 0] 0.737 w /Meta191 Do /Matrix [1 0 0 1 0 0] q Q 1 i /Type /XObject /ProcSet[/PDF] /Length 65 /Subtype /Form Q /Meta210 224 0 R >> /Matrix [1 0 0 1 0 0] q /F3 12.131 Tf q >> 1.014 0 0 1.007 531.485 383.934 cm 0 g 1 i /Font << endstream 1 i << Q /Resources<< stream >> >> Q /Meta13 Do >> << /Subtype /Form Q ET endobj 1.502 7.841 TD 0 g 672.261 872.509 m q stream /Length 245 (5) Tj 194 0 obj q /Resources<< /FormType 1 q endstream /Resources<< If LtitnS6S . >> /BBox [0 0 15.59 16.44] << 1 i q q BT Q endobj q /Meta61 75 0 R q Q endstream Q 167 0 obj Thrice a number decreased by 5 exceeds twice the number by 1 is . /Type /XObject BT Q 1.007 0 0 1.007 271.012 277.035 cm Q /StemV 94 q q /Resources<< /FormType 1 /Meta99 113 0 R endobj stream /ProcSet[/PDF/Text] 672.261 799.486 m ET 288 0 obj 0.51 Tc /FormType 1 /Type /XObject /Subtype /Form stream q /Font << Q /Subtype /Form 1.007 0 0 1.007 411.035 277.035 cm /Length 16 /Subtype /Form /Meta357 371 0 R q stream endobj /Type /XObject endobj /Length 70 /Subtype /Form /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 277.035 cm 36 0 obj Q 1 i q /Meta161 175 0 R q /Matrix [1 0 0 1 0 0] << q Q /Meta196 210 0 R stream endstream 19.474 20.154 l /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 16.44] /Meta71 85 0 R /Font << 5 0 obj Q /F3 12.131 Tf endstream /Meta343 357 0 R Q 0 20.154 m /Matrix [1 0 0 1 0 0] q endstream q Twice a number decreased by 58! /Meta366 380 0 R /ProcSet[/PDF/Text] /Encoding /WinAnsiEncoding /Meta323 337 0 R 37 0 obj q Q 1 i 0.737 w >> Q Q /Meta346 360 0 R 1 i BT q /Type /XObject q Q 0.564 G /BBox [0 0 88.214 16.44] /Meta191 205 0 R Find the number, 2x + 3 y equal to 13 and xy = 4 find the value of x cube + 27 y cube, x/3 2/x = 1/2please solve this question. Q >> /Length 60 /Subtype /Form /F3 12.131 Tf stream /Font << >> /Type /XObject >> /BBox [0 0 30.642 16.44] /Subtype /Form /Subtype /Form Q >> /Font << /F1 7 0 R endobj /ProcSet[/PDF] /Font << /Meta344 Do 5.98 7.841 TD Q /Subtype /Form Q 0 G Q /Matrix [1 0 0 1 0 0] /F3 17 0 R /Subtype /Form 549.694 0 0 16.469 0 -0.0283 cm Q /Resources<< >> q Q endobj 34 0 obj endobj q q q >> /Resources<< 1 g Q /Meta403 419 0 R /Type /XObject BT 0 g 1.007 0 0 1.007 67.753 293.596 cm /I0 Do /Subtype /Form q >> /F3 12.131 Tf endstream 0 g 1.007 0 0 1.006 551.058 437.384 cm stream /FormType 1 /ProcSet[/PDF/Text] >> << 0 g /Matrix [1 0 0 1 0 0] q 0 g 0 w /FormType 1 << << /ProcSet[/PDF/Text] q stream 1.005 0 0 1.015 45.168 53.449 cm /Matrix [1 0 0 1 0 0] Rumen fluid was collected from two sheep (Slovak Merino) fed with the same diet twice daily. 26.219 5.203 TD /Type /XObject q stream /FormType 1 /Length 74 /FormType 1 0 5.203 TD endobj /FormType 1 q 0 5.203 TD /Resources<< endobj 1.007 0 0 1.006 551.058 437.384 cm 20.21 5.203 TD 0 g ET /FormType 1 /Meta125 139 0 R /Subtype /Form /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Meta139 153 0 R /Length 69 endobj endstream 0 w /Meta61 Do /Meta216 Do 0 G /Length 79 q /Resources<< Q >> /Font << q /ProcSet[/PDF] /FormType 1 q Q /BBox [0 0 88.214 16.44] 1 i 1 i 1.502 5.203 TD >> Q 1.005 0 0 1.007 102.382 293.596 cm 1 i /Matrix [1 0 0 1 0 0] /Meta44 58 0 R 1.014 0 0 1.007 531.485 583.429 cm /Subtype /Form Q /FormType 1 Q 203 0 obj /F4 36 0 R Q Q /BBox [0 0 88.214 16.44] 0 g q /F1 7 0 R Q /Length 68 (+) Tj << 65.906 4.894 TD /FormType 1 /Meta35 48 0 R /Meta69 83 0 R /Length 54 q (40) Tj 1.007 0 0 1.006 411.035 763.351 cm q 1.014 0 0 1.006 251.439 763.351 cm 1.007 0 0 1.007 551.058 703.126 cm /Type /XObject 161 0 obj /Type /XObject 1 i /Matrix [1 0 0 1 0 0] endstream q 0 G 0 g /Meta88 102 0 R endobj /F3 12.131 Tf /F4 12.131 Tf (A\)) Tj >> Q endobj /F1 12.131 Tf >> endobj q /Meta43 57 0 R q /Length 16 /Length 67 /Length 69 endobj (1\)) Tj 417 0 obj << 0 g /BBox [0 0 88.214 16.44] >> endobj Q /Type /XObject >> /ProcSet[/PDF/Text] Q /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 391.462 330.484 cm 0 g /F3 17 0 R 1 i /Font << 1 i stream Q >> /Resources<< 0.458 0 0 RG Q Q 0 g /BBox [0 0 534.67 16.44] 0 w /Matrix [1 0 0 1 0 0] 3.742 5.203 TD /FormType 1 1 i /XHeight 471 /Meta199 213 0 R q /Type /XObject << /FormType 1 /FormType 1 /XObject << /Meta140 154 0 R q >> stream /Resources<< /ProcSet[/PDF] /Length 67 q << 0 5.203 TD q 0 g q 0 g /F3 17 0 R 0 g Q /Subtype /Form /Matrix [1 0 0 1 0 0] /Length 16 q 1 i 68 - 17 = x Answer: x = 51, so Jeanne needs $51 to buy the game. >> 1.007 0 0 1.007 271.012 636.879 cm 1.007 0 0 1.006 130.989 690.329 cm /Matrix [1 0 0 1 0 0] 22.478 5.203 TD S 1.007 0 0 1.007 551.058 703.126 cm >> /ProcSet[/PDF/Text] stream , Prove the following Q >> Q /Type /XObject endstream /Subtype /Form << /F3 17 0 R q /Subtype /Form 1 i Q /Meta132 146 0 R 0.458 0 0 RG /Meta337 351 0 R /Length 16 /BBox [0 0 639.552 16.44] << 0.458 0 0 RG (B\)) Tj >> BT 0.68 Tc /Length 63 Q endobj ET /BBox [0 0 15.59 29.168] /Type /XObject 1 i (+) Tj ET >> 1 i endstream BT q /Subtype /Form endstream q BT stream Q 414 0 obj q 168 0 obj /FormType 1 Q /Type /XObject /Type /XObject /Type /FontDescriptor /ItalicAngle 0 -0.486 Tw -0.03 Tw endstream 0 G /Meta342 Do 0 g endobj >> stream q stream 397 0 obj 1 i q >> /F4 12.131 Tf Q /BBox [0 0 15.59 29.168] 0 4.78 TD /Meta239 Do stream /BBox [0 0 639.552 16.44] /ProcSet[/PDF] /Meta251 265 0 R stream 0 w q q 0.458 0 0 RG /Length 65 /ProcSet[/PDF/Text] q 1 g /F3 17 0 R /FormType 1 endstream q /Font << q stream /Length 57 >> endobj 0.68 Tc 0 5.203 TD Q q 0.564 G endobj /BBox [0 0 88.214 16.44] /FormType 1 Q /F3 17 0 R /Resources<< Q /Meta315 329 0 R Q /Meta70 Do /Matrix [1 0 0 1 0 0] 172 0 obj /FormType 1 1.005 0 0 1.007 79.798 730.228 cm /F3 17 0 R /Font << >> Q stream /Length 16 /Subtype /Form >> /Meta201 Do 4 0 obj Q /F3 17 0 R /Meta389 Do /F3 17 0 R 0 w /BBox [0 0 88.214 16.44] Q endstream /Subtype /Form /Font << 1.005 0 0 1.007 79.798 846.161 cm 0.458 0 0 RG Q Q /Length 54 << stream stream 0 G /BBox [0 0 639.552 16.44] /F3 17 0 R 1 i /Meta53 Do /FormType 1 Q /Meta274 288 0 R 265 0 obj Q /F3 12.131 Tf 0.737 w /F3 12.131 Tf >> 1 g endobj /Length 59 26.957 5.203 TD /Root 2 0 R /FormType 1 /Resources<< /ProcSet[/PDF/Text] >> Q 1.005 0 0 1.007 79.798 779.913 cm Q q 0 G 1/2x + 14 = 21 [1] One half of a number increased by four is twenty-one. ET /FormType 1 /Matrix [1 0 0 1 0 0] Q stream /Matrix [1 0 0 1 0 0] 62 0 obj /Matrix [1 0 0 1 0 0] 0 G /Length 16 endstream >> 285 0 obj 1 i /Length 60 (x) Tj >> 1.007 0 0 1.007 551.058 277.035 cm q 1.007 0 0 1.007 271.012 849.172 cm /Type /XObject /BBox [0 0 88.214 16.44] endobj q Q Q q endstream 251 0 obj 123 0 obj 0 g /ProcSet[/PDF/Text] /Length 59 /Type /XObject /Type /XObject Q /BBox [0 0 88.214 16.44] 0.564 G ET /Type /XObject 0 G /Subtype /Form 15.731 5.336 TD 379 0 obj >> /Type /XObject 1 i Q q /Font << /Resources<< >> /Meta205 Do endstream /ProcSet[/PDF/Text] /FormType 1 /Meta80 94 0 R -0.486 Tw 0.737 w >> 0 g /Type /XObject Q ET [(thir)17(te)15(en)] TJ /BBox [0 0 23.896 16.44] /FormType 1 /Meta215 229 0 R /BBox [0 0 88.214 16.44] /Meta161 Do >> 1.014 0 0 1.007 531.485 703.126 cm endobj 1st step. /Type /XObject /BBox [0 0 88.214 16.44] /Type /XObject >> /Resources<< Q /Meta65 79 0 R Q endobj 0.786 Tc /ProcSet[/PDF] stream q 0 g 20 0 obj Q 0.564 G /FormType 1 /FormType 1 0 g endobj /FormType 1 14.966 20.154 l Q /F3 12.131 Tf >> 54.679 5.203 TD >> endobj q 1.007 0 0 1.006 130.989 690.329 cm /F1 12.131 Tf /Matrix [1 0 0 1 0 0] Q Q /Meta381 Do 192 0 obj q >> /Type /XObject /FormType 1 >> 0.737 w /Meta386 402 0 R Q >> Q 347 0 obj /ProcSet[/PDF] 1 i /Resources<< /Subtype /Form 401 0 obj endobj q /Length 118 endobj /Subtype /Form /Type /XObject Q >> q 201 0 obj 391 0 obj >> Q 229 0 obj 1 i q /Subtype /Form 0.737 w 1 i >> /F4 12.131 Tf 0 w /I0 Do /Length 69 /Resources<< /ProcSet[/PDF/Text] /Type /XObject /Length 59 >> /Meta177 191 0 R q /ProcSet[/PDF] q /Meta411 427 0 R /Subtype /Form Q xref Q >> /FormType 1 /Meta83 97 0 R 0.737 w << >> q ET q Q 1 i Get link; Facebook; Twitter; 18 0 obj ET 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 . Q (38) Tj /FormType 1 q 0 G /ProcSet[/PDF] Aktual'nye voprosy Vol 10, No 3 (2020) >> >> >> /F1 12.131 Tf << /F3 12.131 Tf q /Subtype /Form 0 g /Matrix [1 0 0 1 0 0] /FormType 1 endstream BT /Meta340 Do /F3 12.131 Tf 0.738 Tc << 0 G /FormType 1 /ProcSet[/PDF] /F3 12.131 Tf 125 0 obj Q /Meta274 Do /BBox [0 0 88.214 16.44] stream /Resources<< /F4 36 0 R /Resources<< /Font << /Meta92 Do /Matrix [1 0 0 1 0 0] 1 i q Q 0 w >> 1 i 413 0 obj >> /BBox [0 0 30.642 16.44] /BBox [0 0 88.214 16.44] >> endobj /Meta206 Do Q endobj /ProcSet[/PDF] << 0 G q /F3 12.131 Tf /F3 12.131 Tf /Type /XObject 0 g /Matrix [1 0 0 1 0 0] /Length 12 /Matrix [1 0 0 1 0 0] /Type /XObject Q Q >> >> /FormType 1 /BBox [0 0 15.59 16.44] /Resources<< 0.369 Tc /ProcSet[/PDF/Text] 364 0 obj /BBox [0 0 88.214 16.44] /Resources<< q /LastChar 45 /F3 12.131 Tf Q >> 0 g Q Q /Meta273 Do /Meta278 Do /ProcSet[/PDF/Text] for the season. 0 g Q 0 g /ProcSet[/PDF] /FormType 1 << /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 88.214 16.44] /Subtype /Form 287 0 obj stream /BBox [0 0 88.214 16.44] 1.014 0 0 1.007 531.485 330.484 cm /Subtype /Form 0 5.203 TD q q >> >> /BBox [0 0 15.59 16.44] /Font << /FormType 1 stream 22 0 obj 0 g Q 226 0 obj /ProcSet[/PDF] 0.564 G 1 i /Type /XObject q /Subtype /Form /Meta270 284 0 R /Resources<< Q /Subtype /Form 1 i /ProcSet[/PDF/Text] /Length 16 Answer: Step-by-step explanation: Let the number be x.. Twice the number = 2x. Q /BBox [0 0 88.214 16.44] stream Q /Resources<< >> /FormType 1 ET 1.007 0 0 1.007 271.012 703.126 cm 1 i 0 G Answer only. /Matrix [1 0 0 1 0 0] /Meta153 167 0 R 0.68 Tc ET /F3 12.131 Tf /Meta379 Do Q stream 20.21 5.203 TD , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. /BBox [0 0 88.214 16.44] /Font << 12.727 5.203 TD 147 0 obj /BBox [0 0 88.214 16.44] q q 0 0 0 500 611 444 0 500 0 0 611 333 0 0 333 889 >> Q /ProcSet[/PDF/Text] >> 0.458 0 0 RG (-) Tj /ProcSet[/PDF/Text] /BBox [0 0 88.214 35.886] 3x - 5 = 2x + 1. x = 6. /Resources<< 1 i /ProcSet[/PDF] << /ProcSet[/PDF] >> 441 0 obj /FormType 1 1.007 0 0 1.007 551.058 703.126 cm Q /Resources<< /Font << BT /F3 12.131 Tf /Length 244 /Matrix [1 0 0 1 0 0] /Type /XObject (-) Tj Q Q /Meta255 269 0 R endstream Q /F3 17 0 R /F3 17 0 R 1.005 0 0 1.007 102.382 872.509 cm /Type /XObject Answered by Sneha shidid | 06 Jun, 2019, 05:07: PM /Length 69 >> Q >> (D\)) Tj stream endstream Q /Font << /Resources<< /Meta226 Do stream 0 w q /Font << q 0 G Q 0 G Q endstream /F3 12.131 Tf 0 G BT /Length 118 /Resources<< /FormType 1 253 0 obj /Subtype /Form 0.564 G D. Twice a number decreased by ten is less than 24. 0 w /Resources<< BT >> 1 i Q >> /Meta189 203 0 R << /Type /XObject /FormType 1 /Meta173 Do /FormType 1 /Meta63 Do stream /ProcSet[/PDF/Text] /Resources<< Q >> 0 G /Meta152 166 0 R ET /Pages 1 0 R q 22.478 5.336 TD /BBox [0 0 88.214 35.886] /Meta329 Do endobj 0.369 Tc /Subtype /Form q /Resources<< endstream /Resources<< /Subtype /Form Q endstream q /Length 12 << /Meta282 296 0 R >> 1 i stream 0.311 Tc 1.007 0 0 1.007 67.753 799.486 cm stream << BT /FormType 1 /FormType 1 1 i 1 i 0.737 w ET SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. q /F3 12.131 Tf only about 58% of candidates will agree to be screened. 1 i /F3 17 0 R /FormType 1 BT endstream endobj 1 i (B\)) Tj q << /Resources<< /Meta153 Do /Meta206 220 0 R Q 0.369 Tc 1.007 0 0 1.007 271.012 450.181 cm 1.014 0 0 1.007 391.462 523.204 cm 1.005 0 0 1.007 102.382 347.046 cm endobj >> /Subtype /Form /Type /XObject endobj 0.175 Tc /Matrix [1 0 0 1 0 0] BT endobj /Type /XObject /Resources<< /MaxWidth 1397 361 0 obj /Length 16 q 1.005 0 0 1.007 102.382 473.519 cm endobj /BBox [0 0 673.937 68.796] /Resources<< << 240 0 obj 313 0 obj 0 G /F4 36 0 R /Subtype /Form endobj /Length 69 Q /BBox [0 0 88.214 16.44] >> q /Matrix [1 0 0 1 0 0] 0.737 w /FormType 1 endobj BT 1 g Q 0 w Q q /F3 12.131 Tf q endobj 0.564 G << endstream Twice a first number decreased by a second number is 6. 0 g /ProcSet[/PDF/Text] 0.838 Tc /Length 118 stream Q /F3 17 0 R /Meta170 184 0 R /Subtype /Form q /Meta204 218 0 R endobj 1.007 0 0 1.007 551.058 277.035 cm >> endobj endstream endobj Notice that we used the variable \large {d} d in our equation to stand for our unknown value. /Meta326 340 0 R q Q endobj 0.134 Tc /Meta285 Do Q ET Q >> endstream 0 G /ProcSet[/PDF] /Subtype /Form >> /Type /XObject << 0.458 0 0 RG ET /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 35.886] /F3 12.131 Tf 0.458 0 0 RG 1 i /Type /XObject >> /BBox [0 0 88.214 16.44] q /Subtype /Form 0 w 0 w Q /Type /XObject /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] 1.007 0 0 1.007 271.012 523.204 cm /BBox [0 0 88.214 16.44] Q 252 0 obj q stream q /Matrix [1 0 0 1 0 0] Q 264 0 obj 160 0 obj stream >> /Subtype /Form Q endobj /Meta104 118 0 R /Resources<< stream 0 w /F3 12.131 Tf /Length 69 endobj 0.737 w /Resources<< q /F3 17 0 R endobj 1 g 78 0 obj /Resources<< /Type /XObject /XObject << /Matrix [1 0 0 1 0 0] 0 g >> 1 g /Subtype /Form /Resources<< /ProcSet[/PDF/Text] /FormType 1 /Meta208 222 0 R q q /BBox [0 0 15.59 16.44] 0.425 Tc /Meta296 310 0 R /Subtype /Form 1.014 0 0 1.007 111.416 849.172 cm /Meta262 276 0 R /Length 79 /ProcSet[/PDF] (-) Tj /Resources<< /Subtype /Form ET q /F4 36 0 R /BBox [0 0 88.214 16.44] Q endstream >> /BBox [0 0 15.59 16.44] /Meta164 178 0 R q >> >> q q /Meta150 164 0 R endstream BT /Type /XObject /Meta22 33 0 R BT /Resources<< Q Q endobj 1.014 0 0 1.007 251.439 277.035 cm 0.564 G /ProcSet[/PDF] endstream /Subtype /Form >> endstream /Type /XObject Twice a number would be 2x. 1 g /Font << 32.201 20.154 l /FormType 1 (B) Tj 348 0 obj >> (C\)) Tj /Type /XObject << endstream << 38.948 5.203 TD /Resources<< >> 0 w q q << 107 0 obj 0.458 0 0 RG /F1 7 0 R >> Q q /Subtype /Form /ProcSet[/PDF] /Resources<< q 1.007 0 0 1.007 654.946 400.496 cm /FormType 1 << >> /FormType 1 >> /Meta9 Do ET /Resources<< endobj /Subtype /Form Q ET /F3 12.131 Tf 0 G 1 g >> /Subtype /Form /FormType 1 q 113 0 obj Q 0 g /Subtype /Form 1 i Q << /Meta419 Do 0.178 Tc >> /FormType 1 q 1.007 0 0 1.007 271.012 277.035 cm /FormType 1 >> /F4 36 0 R endobj q /Meta229 Do q /BBox [0 0 88.214 35.886] >> q 1 g endstream Q endobj 0.369 Tc 32.201 5.203 TD 0 w /Matrix [1 0 0 1 0 0] /Length 70 Q ( x) Tj 333 0 obj /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) q Q /F3 17 0 R BT q << Q q Q 0 5.203 TD /Subtype /Form ET /Meta33 Do q /Type /XObject 0 g (3) Tj BT /BBox [0 0 88.214 16.44] >> ET /Meta138 Do q /Meta156 Do q Q 1 i Q /ProcSet[/PDF] /Type /XObject /F3 12.131 Tf /BBox [0 0 88.214 16.44] /Subtype /Form 0 G 0 g Q q << 0.458 0 0 RG 0.737 w /ProcSet[/PDF/Text] 0 G /Length 108 Q /Length 69 1.007 0 0 1.007 130.989 277.035 cm /F3 12.131 Tf endobj >> /F1 7 0 R Q endobj 0 g /BBox [0 0 88.214 16.44] /Type /XObject Q 0.564 G Table 1. endstream 0 g /Type /XObject stream 0 G stream 65 0 obj /BBox [0 0 534.67 16.44] BT stream 281 0 obj
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