/Meta425 441 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] ET 0.564 G /Type /XObject q 0.458 0 0 RG endobj Q 0 g /Subtype /Form /Matrix [1 0 0 1 0 0] 0.838 Tc Q 1 g /Meta332 346 0 R /BBox [0 0 88.214 16.44] (C\)) Tj /Type /XObject endstream /ProcSet[/PDF] 220.931 4.894 TD Q 0 g /FormType 1 /Length 16 /Meta257 271 0 R /BBox [0 0 88.214 16.44] /Meta380 Do stream /BBox [0 0 88.214 16.44] 0.524 Tc /F3 17 0 R 0 g endstream /Matrix [1 0 0 1 0 0] BT endobj Q /Meta318 Do /F3 12.131 Tf 1.005 0 0 1.007 102.382 799.486 cm /BBox [0 0 673.937 16.44] /BBox [0 0 88.214 16.44] /FormType 1 /Meta125 139 0 R /Font << q /F4 12.131 Tf endobj /FormType 1 -0.058 Tw >> 0.737 w q >> 0.425 Tc (\(x ) Tj 1.007 0 0 1.007 654.946 599.991 cm BT /Meta89 103 0 R 0.458 0 0 RG /Matrix [1 0 0 1 0 0] endobj Q Q q endstream /F1 12.131 Tf stream 0 g << /Matrix [1 0 0 1 0 0] /F3 12.131 Tf /Subtype /Form /Meta330 Do 1 i q >> stream Q /Type /XObject (A\)) Tj 1 i /FormType 1 endstream Q /Meta100 Do 0.737 w ET Q >> q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0 G Q >> Q /Meta285 Do Aktual'nye voprosy Vol 10, No 3 (2020) q >> /I0 Do /Subtype /Form /FormType 1 Q /FormType 1 Q /Type /XObject 131 0 obj 0.564 G 0 g /FormType 1 q stream /Meta173 Do [(-1)-16(52)] TJ ET q Q /BBox [0 0 15.59 16.44] 0.737 w 0 w >> /Length 54 >> 0.271 Tc Q endobj /Type /XObject q 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 0.737 w (x ) Tj A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. 0.51 Tc /FormType 1 /Type /XObject 0 w /BBox [0 0 549.552 16.44] 0.737 w 0 20.154 m ET /Matrix [1 0 0 1 0 0] 0 G /Subtype /Form 1 g q 265 0 obj 1 g /Length 294 /Meta11 Do /Length 60 1 i /ProcSet[/PDF/Text] (-8) Tj Q The solution of the equation ax + b = 0 is Solution: (c) The equation is ax + b = 0 ax - b Solution is Question 2. 0 w q /FormType 1 /Length 69 /Type /XObject /F3 12.131 Tf /XObject << /Length 118 0.369 Tc 1 i /Subtype /Form 1.014 0 0 1.006 111.416 690.329 cm /F3 17 0 R 322 0 obj Q /ProcSet[/PDF] /F4 12.131 Tf 0 w /Type /XObject q /F4 12.131 Tf 1.005 0 0 1.007 102.382 293.596 cm endstream << << stream /Length 16 >> /BBox [0 0 30.642 16.44] 1.007 0 0 1.007 271.012 523.204 cm /F3 12.131 Tf /Matrix [1 0 0 1 0 0] q /ProcSet[/PDF/Text] 232 0 obj BT 0.564 G /FormType 1 >> q >> endobj 582 546 601 560 395 424 326 603 565 834 516 556]>> Q endobj q /F3 12.131 Tf /Type /XObject q 0 g >> << Q q >> q /Meta44 58 0 R /Length 65 1.014 0 0 1.007 531.485 849.172 cm 0.458 0 0 RG a and b or something else.***. /Font << stream /Type /XObject 1.005 0 0 1.007 102.382 473.519 cm Q /Meta290 Do >> /F3 17 0 R /Meta240 Do 22.478 5.336 TD << << /Meta296 Do BT Twice a number decreased by . q 1.007 0 0 1.007 271.012 776.149 cm 0 G S /Type /XObject endobj q q 285 0 obj /FormType 1 0 G /F3 12.131 Tf /ProcSet[/PDF/Text] /Font << q /Subtype /Form [(1)-25(0\))] TJ >> /Matrix [1 0 0 1 0 0] << 26.957 5.203 TD Q endstream >> 0 G 2 Data in this Fast Fact may not sum to 15.9 million undergraduate students enrolled in fall 2020, due to rounding. q /Meta382 Do /Subtype /Form >> Q Q endstream /Resources<< /ProcSet[/PDF/Text] BT q >> 1.005 0 0 1.007 102.382 799.486 cm /Matrix [1 0 0 1 0 0] q q BT /Resources<< << stream 0 G 1.007 0 0 1.007 130.989 330.484 cm /Meta202 Do Q 0 G /ProcSet[/PDF] 5.98 7.841 TD /F3 17 0 R q >> Q Q /Meta32 45 0 R << 312 0 obj endstream << 0 G q /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] 0 G /ProcSet[/PDF] /Subtype /Form 1.007 0 0 1.007 45.168 779.913 cm Q 1 i endobj /F3 17 0 R 119 0 obj >> << 1.007 0 0 1.007 271.012 523.204 cm Q Q q /Font << /F4 36 0 R endobj BT /Type /XObject q 1.502 8.18 TD /Resources<< Q /Matrix [1 0 0 1 0 0] stream /F3 12.131 Tf >> endstream Q BT >> stream Q Q /Type /XObject /BBox [0 0 534.67 16.44] /FormType 1 /Meta16 27 0 R /F3 17 0 R /Meta61 75 0 R /ProcSet[/PDF] Q /FormType 1 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 Q >> /BBox [0 0 88.214 16.44] 1 i /Subtype /Form 32.201 5.203 TD 1 i BT endobj /Type /XObject 391 0 obj >> /Type /XObject >> /Type /XObject q /Font << Q /Matrix [1 0 0 1 0 0] stream Q q 1 i endobj /Matrix [1 0 0 1 0 0] 1 i q /Meta210 224 0 R 0.369 Tc 262 0 obj /Font << /Type /XObject Q 0 g >> 1 g /BBox [0 0 88.214 35.886] /F3 17 0 R 500 500 500 0 333 389 278 0 0 722 500 500]>> /BBox [0 0 15.59 16.44] endobj /BBox [0 0 88.214 16.44] << /F3 12.131 Tf /F3 17 0 R /F1 7 0 R >> /FormType 1 stream /Font << q 0.737 w /BBox [0 0 88.214 16.44] /Resources<< /BBox [0 0 88.214 35.886] 438 0 obj /MediaBox [0 0 767.868 993.712] /Resources<< Q q /F3 12.131 Tf 214 0 obj q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] q /Meta46 60 0 R Q 1.007 0 0 1.007 67.753 347.046 cm 57.656 5.203 TD q Q 0.737 w Q Q 1.007 0 0 1.007 551.058 523.204 cm /Subtype /Form endobj 4 less than some number : x - 4 : a number decreased by 10 : y - 10 : 8 minus some number : 8 - t : the difference between a number and 12 : . endobj Q 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/Subtype /Form /Meta277 Do 1.005 0 0 1.007 79.798 779.913 cm /F3 17 0 R /BBox [0 0 88.214 16.44] /Resources<< /ProcSet[/PDF] /Meta124 138 0 R >> Q Q /Length 67 /Subtype /Form stream /Type /XObject << >> q /Meta98 Do /Meta310 324 0 R 1 i /Length 69 0 w Q q /Meta376 Do q 0 g /Length 16 endstream Q 0.68 Tc 0 g /Type /XObject /Font << >> Q /BBox [0 0 15.59 16.44] /Length 16 /Resources<< 20.21 5.203 TD >> 1.007 0 0 1.006 411.035 437.384 cm stream Q q /Subtype /Form 30.699 5.203 TD /Type /XObject /Meta264 278 0 R 0.737 w 1 i 0.458 0 0 RG endstream /Resources<< 1.007 0 0 1.006 551.058 437.384 cm stream q q >> >> q q stream ET >> << Q 247 0 obj stream /Meta377 391 0 R stream Q q 1.007 0 0 1.007 411.035 277.035 cm /Resources<< /Meta259 Do 20.21 5.203 TD /Matrix [1 0 0 1 0 0] Q ET stream 3.742 24.649 TD 0 g /Type /XObject 1 i q q /ProcSet[/PDF] /FormType 1 << /Type /XObject endobj 1.005 0 0 1.007 102.382 816.048 cm /Length 69 /Meta295 309 0 R q /Resources<< stream /Resources<< >> Q /F3 17 0 R Q 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