Calc1SupAns

Calc1SupAns - Answers 1. A = 2r + , r 0 2. (a) A(1) 30,...

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Answers 1. A = 2 π r + π , r 0 2. (a) A(1) 30, A(4) 40, (b) A(x) =40 for x 3 and x 4 . A(x) = 60 has no solution with 1 x 6. (c) A (2) 26, [ A (2)] 2 26 2 = 676 A(2 2 )= A(4) 40 , [A(2)] 2 is greater. 3. If A is a function, then A(2) is its value at 2. If A is a constant, then A(2) is A multiplied by 2. 4. p(1) = 18, p(2) = –6, p(3) = 2 5. Possible answers: (a) She is stopped from about t = 40 to t = 55, which is about 15 seconds. (b) At about t = 110 (c) From about t = 70 to t = 115 (d) From about t = 7 to t = 27, She goes about 1/6 mile. 6. (a) Least inrease in number of PAC's = 111 (from 1986 to 1988), Greatest increase = 898 (from about 1978 to 1980) (b) Least increase in contributions = $10,045,000 (from 1974 to 1976), Greatest increase = $31,403,000 (from 1980 to 1982) (c) $20,604 per PAC in 1974 and $34,653 per PAC in 1988(rounded to the nearest dollar) 7. (a) 10Q(10) = 210 (b) Q(6 + 3 + 1) = 21 (c) Q(100/20) = 7 (d) Q(10)/Q(5) = 3 (e) Q(Q(5) + 3) =21 8. (a) Change = 4300 – 3500 = 800 cigarettes per person, Relative change = 800/3500 = 0.23 (b) Change = 3300 – 4300 = -1000 cigarettes per person, Relative change = -1000/4300 = -0.23 (c) C(1953) = 3700 cigarettes per person, Error = |3700 - 3500| = 200 cigarettes per person, Relative error = 200/3500 = 0.057 (d) The first graph does not show relative change well because the C-axis is not shown for 0 C 3300. 9. P(100) = 3P(50) = 3(2P(1)) = 6P(1), Relative change = P(100) – P(1) P(1) = 5 10. A(1960) = 100(1610)/(1610 + 18581) = 7.97, A(1965) = 14.04, A(1970) = 30.24, A(1975) = 38.44, A(1980) = 45.84, A(t) is the percentage in year t of the km traveled by Japanese in private cars, trains, and busses that were traveled in private cars. 11. Possible answer: Set F(0)=1. Then F(1) = 4F(0) = 4, F(2) = 4F(1) = 16, F(3) = 4F(2) = 64, F(4) = 4F(3) = 256, and F(5) = 4F(4) = 1024. 12. (a) 4 – π 4 X100 = 21.5% (b) π – 2 π X 100 = 36.3% (c) Circumscribed area = 2(inscribed area) 13. y = 36/x 2 - 25, y = 20 – 3x 2 , y = – 24/x, y = x 3 – 5. 14. x -1 dominates x -2 for large positive and large negative. x dominates x -1 for small positive and large negative x. 15. (a) x 3 is the dominant term in the numerator and 7x 2 is the dominant term in the denominator when x is a large positive or negative number. (b) –5 is the dominant term in the numerator and 3x is the dominant term in the denominator when x is close to zero. 16. (a) figure 12 (b) figure 11, b is positive (c) figure 13, a is positive 17. Figure 15, b is negative since y →−∞ as x 0
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18. Figure 17, Both a and b are positive since y = a when x = 0 and y + as x + 19. Figure 14, a is positive since y a as x ± , b is negative because y →−∞ as x 0 + 20. 200 miles + (t hours)(60 miles hours ) = 200 + 60t miles 21. 1 0.05 gallons/mile = 20 gallon 22. V = 20 – (s miles0 ( 0.05 gallons mile )=20 – 0.05s gallons The tank is empty when 20 – 0.05s = 0, which is at 400 miles, so the domain of V as a function of s is the interval 0 s 400. 23. s = 60t miles. Since s = 400 at t = 6 2/3, the domain of s a function of t is 0 t 6 2/3. 24. (a) V = 20 – 3t for 0 t 6 2/3 (b) -3 gallons per hour (c) The volume of gas in your tank decreases 3 gallons every hour. 25. -6 26. f(x) = 10 + 50(x – 3) 27. -1/3 28. The average rate of change of Z(x) with respect to x from x = 0 to x = 5 is 200 – 100 5 – 0 = 20.
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Calc1SupAns - Answers 1. A = 2r + , r 0 2. (a) A(1) 30,...

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