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# MAT119 M3WA3 ANSWERS

AU

GPA 3.8

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Date Created: 11/15/15

MATH 119 Written Assignment 3 Section 25 2 5y 1y3 gt 0 Change the inequality to equality and solve 5y1y30 5y 10 or y30 1 or 3 y 5 y 1 Put in values below 3 between 3 and g and above 5 The solution is 00 3 or G 00 lt I gt Bgt 4 z26z gt 8 Change the inequality to equality and solve z26z 8 z26z80 z4Xz20 z40 or z20 z 4orz 2 Put in values below 4 between 4 and 2 and above 2 The solution is 004 0r 2 0 lt gt C gtgt 4 2 8 3x2 5x gt 2 Change the inequality to equality and solve 3x2 5x2 3x2 5x 20 3x1x 2 0 3x10 or x 20 1 x or x2 3 1 1 Put in values below between E and 2 and above 2 The solution is 00 or 2 0 lt lt gt lt gt gt 2 12 r2 9rlt0 Change the inequality to equality and solve r2 9r0 rr 90 r0 or r 90 r0 or r9 Put in values below 0 between 0 and 9 and above 9 The solution is 0 9 lt lt gtgt 16 2x4x2 9 s 0 Change the inequality to equality and solve 2x4x2 90 2x4x3x 30 2x40 or x30 or x 30 x 2 or x 3 or x3 Substitute values into the appropriate intervals to get the solution 00 3 or 2 3 20 2m3 7m2 gt 4m Change the inequality to equality and solve 2m37m24m 2m37m2 4m0 m2m27m 40 m2m lm 4 0 m0 or 2m 10 or m40 1 m0 or m or m 4 Substitute values into the appropriate intervals to get the solution 4 0 Of a 0 x2 4 X 36 gt 0 Change the inequality to equality and solve x2 40 or x0 x2x 20 or x0 x20 or x 20 or x0 x 2 or x2 or x0 Substitute values into the appropriate intervals to get the solution 2 0 0r 2 0 Section 31 2 This relation does not define y as a function of x 6 The rule x y2 3 does not define y as a function of x defines a function 8 The rule y 2x3 12 For f x 2x52 the domain is all real numbers since x can take any real value The domain is 00 0 16 For gx 2 you cannot have a x x zero in the denominator so solve the equation x2x 20 to get x1 orx 2 Which are excluded from the domain The domain is all real numbers except 1 and 2 This gives the notation 002 0r 2 1 0r 1 0 2x 3 x lt 4 x2 1 4 s x s 10 can take on x values less than 4 excluding 4 and x2 1 takes on all x values from 4 to 10 including 4 and 10 This gives the domain 22 For fX 2x3 24 For f x 0 the function Will always be 0 for all values of x ie the constant function a f4 0 b f 3 0 c f27 0 d f 49 0 32 fxi x1 a f 4 4 is not defined b f 3 3 2 z 7321 31 c f 27 27 2 is not defined 27 1 2x 4 x S 1 34 fx 3 1ltxlt4 x1 x24 a f4415 b f 3 2 3 4 10 c f27 3 d f 49 2 49 4 138 44 fx24x fxh fx h 24xh 24x h 24x4h 2 4x h 4 h 50 fx 052x2 264x99 a f10 1774 In 2010 the GDP will be about 1774 trillion b f15 2556 In 2015 the GDP will be about 2556 trillion c f 25 49 In 2025 the GDP will be about 49 trillion Section 32 2 For gx 3 x the graph is a straight line With slope 1 and yintercept 3 A L 5 I I L I I I I 8 6 4 2 2 x 5 x S 1 6 For y 2 3x x gt1 you have the graph L y fxx5 fx23x D 14 For f x lxl 2 you have the graph L y fxabsx 2 5 VM 30 For gx 3 x 4 you have the graph y W 5 I I I I I I 8 6 394 2 M 6 8 5 02x 0 s x s 500 48 Tx 10 04x 500 500 lt x s 3000 110 05x 3000 x gt 3000 200 150 I I I I I I I I X 500 1000 1500 2000 2500 3000 3500 4000 4500 52 The year 2006 corresponds to x 16 The two points that you want to use are 0 3206 and 16 8451 Using the slope formula you have 8451 3206 m 16 0 z 3278 Since the yintercept is 0 3206 you have the linear equation y 327896 3206 Which gives the graph Section 33 2 Let x be the number of miles and Cx be the total charge for x miles Then you have the function C x 8x 95 6 The fixed cost is 2000 Which gives the beginning function of C x mx 2000 It costs 5000 to produce 40 items Which gives 5000 40m 2000 3000 40m m 75 This gives the function C x 75x 2000 C100 80100 12000 100 100 10 5000 200 The average cost per item when 100 items are produced is 200 C1000 801000 12000 1000 1000 The average cost per item when 1000 items are produced is 92 51000 92 C 1 0000 8010000 12000 10000 10000 The average cost per item when 10000 items are produced is 8120 510000 8120 16 a Let x1 y1 0 222000 and let x2 y2 6 300000 This gives the slope m 300 000 222 000 6 0 the yintercept is 0 222000 which gives the function y 13000x 222 000 13000 You know that b f 12 1300012 222 000 378 000 In 12 years the house will be worth 378000 18 Cx 68x 450 000 a The fixed costs are C 0 450 000 b The slope of C x 68x 450000 is 68 so the marginal cost is 680 c The cost of producing 50000 disks is C 50000 6850000 450000 790 000 The cost of producing 600000 disks is C 600 000 68600000 450000 4 530 000 d The average cost per book when 50000 are produced is 5 50000 M 50000 6850000 450000 1580 50000 The average cost per book when 50000 C 500 000 5 500000 are produced 1s 500 000 68500 000 450000 500 000 770 34 You know that revenue is R X 80x and cost is C x 50x 2400 a The breakeven point is when revenue and cost are the same ie when R C This gives 80x 50x 2400 30x 2400 x 80 b L y 8000 6000 400 2000 g 5 a g 12 gt c When x 60 R60 8060 4800 C60 5060 2400 5400 42 Exports and imports were the same in the middle of 2003 50 80 60 40 20 X 102030405060708090 b The equilibrium demand happens supply equals demand This gives 3 3 81 9 81 4Q q36 c To find the equilibrium price substitute the equilibrium demand into either equation 3 Both equilibrium demand and equilibrium price are confirmed by the graph in part a Section 34 14 E 16 C 20 D 30 For gxx210x9 al andb10 b 10 x2 261 21 y 52 lO 59 y 25 509 y l6 The vertex is 5 16 40 For f x x52 the vertex is 5 0 and the axis is x 5 The graph is 46 Consider fx 3x2 24x 46 fx 3x2 24x 46 fx 3x2 8x 46 fx 3x2 8x16 16 46 fx 3x2 8x16 4648 fx 3x 42 2 The vertex is 4 2 and the axis is x 4 VM 42 Section 35 2 a f10 1060 101050 500 He sells 500 cases if he spends 10 minutes With each customer f 20 2060 20 2040 800 He sells 800 cases if he spends 20 minutes With each customer f45 4560 45 4515 675 He sells 675 cases if he spends 45 minutes With each customer 800 20800 45675 600 39 10500 1390 2390 30 4390 5390 6390 7390 c The vertex gives the number of minutes per customer xaxis needed to sell the maximum number of cases yaxis C x x60 x Cx60x x2 a l andb60 xb6030 2a 2 l C30 3060 30 302 900 The vertex is 30 900 d He needs spend 30 minutes With a customer to sell the maximum of 900 cases 8 Since supply is equal to demand you have IN UI Q N H Q N 1 O Q N H 1 O Q QN U H H E a o 0 Since 10 is meaningless in this context the equilibrium quantity is q 10 l The equilibrium price is p g 102 20 10 a vN L y 5000quot 4000quot 2000quot 1000quot X I I I I L I I r 20 40 60 80 q2 200 10q 3200 q210q 30000 b q60q 500 q600 or q 500 q 60 orq50 Since 60 is meaningless in this context the equilibrium quantity is q 50 The equilibrium price is p 502 200 2700 The equilibrium point is 50 2700 c As found in part b the equilibrium quantity is 50 and the equilibrium price is 2700 20 a The total number of seats is 100 If x is the number of unsold seats then the number of sold seats is given by the expression 100 x b Since each unsold seat increase the 50 price by 1 the price per seat is given by the expression 50 x c Let R be the revenue Then total revenue is found my multiplying the number of sold seats by the price of each seat Thus the revenue function is R x 100 x50 x d You will need to use the vertex formula Rx 100 x50 x Rx 5000 50x x2 a 1 and b 50 50 x m 25 R25 5000 5025 252 R25 5625 The vertex is 25 5625 which means that to maximum revenue 25 seats need to be unsold e The maximum revenue is 5625 22 a The income per pound is 40 2x cents per pound after x weeks b The number of pounds per tree is 100 5x after x weeks c The total revenue is Rx 40 2x100 5x Rx 4000 200x 200x 10x2 Rx 4000 10x2 cents per tree after x weeks d Using a 10 andb0 x 210 0 so the vertex is 0 4000 Thus the peaches should be harvested now ie in 0 weeks e The maximum revenue is 4000 cents 40 per tree

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