/Resources<< /BBox [0 0 534.67 16.44] /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] /Type /XObject An example of a linear inequality in one variable is A. x+y = Question: answer 1. 198 0 obj 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 q << Q 0 G endstream Q endstream endstream endobj Q /Matrix [1 0 0 1 0 0] q /Type /XObject /Subtype /Form endstream 1 i /BBox [0 0 673.937 68.796] /Resources<< /Meta336 Do endstream /Resources<< q /Meta383 Do ET /ProcSet[/PDF] /Matrix [1 0 0 1 0 0] >> /Meta418 Do endobj Q S q /F3 12.131 Tf Q stream /Length 59 [(The )-19(quotient of )] TJ 1 g /Matrix [1 0 0 1 0 0] /Subtype /Form (+) Tj /Meta122 Do /Length 16 /Resources<< >> 0 g /F3 12.131 Tf (3) Tj /Meta427 443 0 R Q 237 0 obj /Matrix [1 0 0 1 0 0] >> BT 83 0 obj /Subtype /Form endstream ET endstream /Resources<< /BBox [0 0 88.214 16.44] /Meta88 Do 0.382 Tc /Length 59 Q -0.486 Tw q >> /Resources<< /Type /XObject 1 g 0 G /Font << 1 i q /FormType 1 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 . 71 0 obj /Font << /Meta20 Do Q >> << /Matrix [1 0 0 1 0 0] 1.007 0 0 1.006 411.035 437.384 cm endobj /BBox [0 0 88.214 35.886] /Matrix [1 0 0 1 0 0] ET q /Resources<< ET ET /Subtype /Form 1.007 0 0 1.007 271.012 849.172 cm /Resources<< What word phrase can you use to represent 5x + 2? q /F4 36 0 R 0 w /Resources<< ET 190 0 obj << /Resources<< >> /Length 69 Solution: Let the number be x. 0.297 Tc 343 0 obj >> q /F3 12.131 Tf >> << /Meta138 Do << /FormType 1 >> /F3 17 0 R 0 w q /Meta148 Do /Meta19 Do Q /Resources<< /Type /XObject 0.838 Tc /Type /XObject Q 1.005 0 0 1.007 102.382 599.991 cm >> stream Q stream /BBox [0 0 17.177 16.44] Q >> endobj endobj 3.742 5.203 TD Q /Font << /Meta114 128 0 R /Matrix [1 0 0 1 0 0] endstream /FormType 1 q Twice a number decreased by 58! >> 0 w 0 g pidemiologi i Infekcionnye Bolezni. Q 0.307 Tc /Resources<< Example 1: Use the tables above to translate the following English phrases into algebraic expressions. /Type /XObject << q >> /BBox [0 0 639.552 16.44] /Resources<< 0 g /Meta17 Do 0 G 234 0 obj >> 88 0 obj /Type /XObject 0 833 610 0 0 0 667 778 0 1000 0 0 0 0 0 0 /ProcSet[/PDF/Text] endobj Q /Meta115 129 0 R q 174.501 5.203 TD 0.458 0 0 RG q BT endobj /Resources<< ET >> 0.425 Tc >> q 0 g /F3 12.131 Tf 0.737 w q Q Q n 11 or n 11. SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. A: Given: A number increased by 5 is equivalent to twice the same number decreased by 7. q >> Q /FormType 1 endobj /Meta329 343 0 R >> 0.564 G /Length 54 /Matrix [1 0 0 1 0 0] /F3 17 0 R q endstream 1 i /AvgWidth 401 1.007 0 0 1.007 551.058 277.035 cm ET Q /Matrix [1 0 0 1 0 0] >> q /Meta135 Do /FormType 1 >> >> >> stream stream /Length 69 /ProcSet[/PDF/Text] >> >> endobj stream q Q 0 G Q >> /Matrix [1 0 0 1 0 0] endstream (C\)) Tj /F3 17 0 R q /F1 7 0 R /Type /XObject << if the solution of an equation is x=-2, what could the original equation be? /FormType 1 /F3 12.131 Tf >> /Subtype /Form /FormType 1 << Q << endstream /Resources<< Q stream /Length 80 q 1 i BT >> endstream /Meta182 196 0 R endobj /Contents [446 0 R] /Meta358 372 0 R /Resources<< endobj endstream /Font << /Resources<< /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] q (\)) Tj (58) Tj Q q 0 G >> /Type /Font (+) Tj /Subtype /Form 1 i 0 g endstream Q q BT /BBox [0 0 88.214 35.886] q 367 0 obj Q >> stream /Type /XObject q q >> /Type /XObject 0 5.203 TD /Resources<< - 9737014. >> >> q 0 g /Meta340 354 0 R 54 0 obj q endstream Q >> << >> Q 0 G /ProcSet[/PDF] BT 1 i BT /Matrix [1 0 0 1 0 0] << 152 0 obj 0.425 Tc Q /BBox [0 0 15.59 16.44] /F3 12.131 Tf /BBox [0 0 88.214 16.44] /Type /XObject >> /Meta69 Do 0 5.203 TD /Matrix [1 0 0 1 0 0] 0.198 Tc >> q endstream 0.458 0 0 RG q /F4 12.131 Tf endobj 0 g q 1 i /Meta8 19 0 R Q /Meta198 212 0 R /Type /XObject >> q q Q Q /Length 108 /Subtype /Form Q Mat /Subtype /Form >> /Length 63 /Type /XObject stream ET << q >> q /Length 16 /Matrix [1 0 0 1 0 0] BT 0 g -0.486 Tw 0.564 G q Q /Resources<< /BBox [0 0 15.59 16.44] 0 g /ProcSet[/PDF/Text] 1 i /Matrix [1 0 0 1 0 0] /Length 69 q q /Type /XObject << endobj q /F3 17 0 R [(F)-22(ive)] TJ /F3 12.131 Tf /FormType 1 /BBox [0 0 88.214 16.44] Q q /Type /XObject /Subtype /Form (x) Tj >> BT q /F4 12.131 Tf /Meta98 Do q 1 i All steps. q /Length 59 >> Q >> << (-23) Tj q (-) Tj Q 0.737 w endstream Q /ProcSet[/PDF] 1 i -0.084 Tw /ProcSet[/PDF/Text] << /F3 12.131 Tf /MaxWidth 2000 1 g stream 722.699 872.509 l 57 0 obj Q endstream 1.007 0 0 1.007 411.035 849.172 cm /Meta184 Do /Type /XObject -0.16 Tw /BBox [0 0 534.67 16.44] /Length 54 Q /Font << q 0 g 0 g /FormType 1 Q BT Q 0 g /ProcSet[/PDF/Text] /Meta426 Do >> 14 0 obj /Resources<< << stream /Matrix [1 0 0 1 0 0] 0 w 0.369 Tc 1.014 0 0 1.007 111.416 383.934 cm q q /BBox [0 0 88.214 16.44] Q endobj /F3 17 0 R >> Q 0 g /FormType 1 TJ 0 5.203 TD Q /Subtype /Form << << >> 1.007 0 0 1.007 130.989 383.934 cm 1.005 0 0 1.013 45.168 933.487 cm /Subtype /Form 9 + x. fourteen decreased by a number p. 14 - p. seven less than a number t. t - 7. the product of 9 and a number n. << /Meta198 Do /Matrix [1 0 0 1 0 0] >> 0.458 0 0 RG 1 g /Meta64 Do 1.007 0 0 1.007 551.058 636.879 cm /Type /XObject Let's now proceed and solve for \large {d} d and afterward, check if the value we get indeed makes the equation true. q endstream ( x) Tj >> /ProcSet[/PDF/Text] 2 See answers pharry1800 pharry1800 Answer: 2n-58 Step-by-step explanation: olivbreadh olivbreadh Answer: 2x-116 or 2(x-58) Step-by-step explanation: Transalate it to numbers and operations: => 2(x-58) => 2x-116 You won't have a solid number since its not an equation. /Resources<< BT 263 0 obj >> /BBox [0 0 30.642 16.44] /ProcSet[/PDF/Text] endobj 0 g Q 0.68 Tc /Resources<< endstream >> Q stream /Meta66 80 0 R 2.238 5.203 TD /Type /XObject /BBox [0 0 88.214 35.886] /FormType 1 0 w 13.493 5.203 TD q /Type /XObject Q << q >> /BBox [0 0 30.642 16.44] /F3 12.131 Tf /BBox [0 0 17.177 16.44] Q /FormType 1 /Font << q BT stream /Descent -299 >> q endstream BT /Resources<< (5\)) Tj /Subtype /Form endstream /Length 16 q 1.007 0 0 1.007 551.058 583.429 cm /Type /XObject 1 i /Matrix [1 0 0 1 0 0] 0.369 Tc /Meta84 98 0 R 1.005 0 0 1.007 45.168 916.925 cm /Font << >> /Type /XObject Q [(A number )-17(divided by )] TJ 0.564 G << /FormType 1 >> /Matrix [1 0 0 1 0 0] >> Q endstream /Matrix [1 0 0 1 0 0] >> /Meta295 Do /Resources<< /Matrix [1 0 0 1 0 0] /Meta170 184 0 R q q /BBox [0 0 88.214 35.886] 0 g q endobj 0 G >> >> 0.564 G >> 1 i q 238 0 obj 0 g /Subtype /Form /Resources<< /ProcSet[/PDF] Q >> /F3 17 0 R 1.502 8.18 TD /Type /XObject /Font << /Length 95 412 0 obj Q Q >> /BBox [0 0 15.59 16.44] >> endobj /Resources<< >> 0 5.203 TD BT ET q 1.007 0 0 1.007 271.012 849.172 cm /F3 12.131 Tf 1.005 0 0 1.007 102.382 400.496 cm /FormType 1 >> endstream endobj q << stream >> q /BBox [0 0 15.59 29.168] << >> >> /Matrix [1 0 0 1 0 0] (2) Tj >> /ProcSet[/PDF] stream /F3 17 0 R q q 0 g If LtitnS6S . /Type /XObject /Type /XObject q /Subtype /Form /ProcSet[/PDF/Text] /Length 12 /Meta53 Do [tex]\sin (\pi -x)=\sin x[/tex]. 1.014 0 0 1.007 251.439 583.429 cm 32.201 5.203 TD /BBox [0 0 15.59 16.44] /Meta280 Do q /ProcSet[/PDF/Text] ( \() Tj /Type /XObject 0.564 G 1.007 0 0 1.007 271.012 523.204 cm ET 178.979 5.203 TD /ProcSet[/PDF/Text] q endobj 1.005 0 0 1.007 102.382 293.596 cm /Meta365 379 0 R /Subtype /Form 239 0 obj /F1 12.131 Tf q 1.007 0 0 1.007 67.753 872.509 cm endobj /Meta196 Do 0 g /Type /XObject /Subtype /Form q /Resources<< << /FormType 1 /Meta74 88 0 R 89 0 obj q stream /BBox [0 0 15.59 16.44] << 672.261 872.509 m endobj /Type /XObject q BT 0 G /Length 64 Q >> endobj stream q 1 i 0.737 w /Length 69 q Q >> /Resources<< 1 i q 1.007 0 0 1.006 551.058 836.374 cm /Length 59 q A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. /Font << 52.412 5.203 TD Q endstream q 3.742 5.203 TD /Meta389 405 0 R /BBox [0 0 88.214 16.44] 1 i << /Subtype /Form /ProcSet[/PDF/Text] << Q q /Length 16 /Meta316 Do 1.007 0 0 1.007 411.035 636.879 cm >> /Font << /Matrix [1 0 0 1 0 0] 1.007 0 0 1.007 67.753 347.046 cm endobj Q /FormType 1 Q /Length 16 << ET q endobj /BBox [0 0 88.214 16.44] /Subtype /Form 1.005 0 0 1.007 102.382 670.003 cm /Ascent 976 stream >> >> 1 i /ProcSet[/PDF/Text] stream >> /Length 99 >> /Matrix [1 0 0 1 0 0] Q /FormType 1 /Matrix [1 0 0 1 0 0] 50 0 obj 91 0 obj /Type /XObject /Subtype /Form /Subtype /Form 0 g /Subtype /Form /Meta82 96 0 R Q /F3 17 0 R Q >> /F3 17 0 R /Subtype /Form /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 /Resources<< /Type /XObject /Font << /F4 36 0 R /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] endstream 1. /ProcSet[/PDF] >> q /Length 64 1.014 0 0 1.006 531.485 763.351 cm q /BBox [0 0 30.642 16.44] /F3 17 0 R 1 i 1 i 0 G 1.007 0 0 1.006 551.058 763.351 cm << /Length 65 0 g 0 g /F3 17 0 R /Length 16 1.014 0 0 1.006 251.439 437.384 cm /FormType 1 32.939 5.203 TD q 0.737 w stream q /BBox [0 0 88.214 16.44] 1 i >> /FormType 1 q 0.269 Tc /Type /XObject 30.699 4.894 TD /Matrix [1 0 0 1 0 0] Q /Length 69 349 0 obj /Resources<< stream 0.458 0 0 RG /BBox [0 0 549.552 16.44] 0.564 G /Matrix [1 0 0 1 0 0] Q 1 i /Resources<< /Length 68 1.007 0 0 1.007 45.168 796.475 cm /FormType 1 << 1 i /Meta47 61 0 R /Type /XObject /ProcSet[/PDF/Text] >> /Length 69 0 G >> /Resources<< stream >> Q 1 i Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 /ProcSet[/PDF/Text] /ProcSet[/PDF/Text] >> 2. /F3 17 0 R endstream /Subtype /Form endstream /FormType 1 /Matrix [1 0 0 1 0 0] ET q 115 0 obj /F1 7 0 R /Resources<< 0.369 Tc 0 g (B\)) Tj 1.007 0 0 1.007 130.989 523.204 cm /Meta284 298 0 R 200 0 obj Q /Meta22 33 0 R /FormType 1 BT endobj 1.007 0 0 1.007 67.753 799.486 cm << 0 G 0.524 Tc So we have twice of a mystery number decreased by three, and that is all going to be 31. 213 0 obj /BBox [0 0 88.214 16.44] 0 w /Meta129 Do << /Type /XObject Q 1.005 0 0 1.007 79.798 829.599 cm q /Resources<< /ProcSet[/PDF] /F3 17 0 R q /Type /XObject /Resources<< q /Font << 1.007 0 0 1.007 551.058 636.879 cm /FormType 1 >> endstream 1 i Q 13.464 5.203 TD That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -.