course Mth 173
Week 3 Quiz # 1 Version 1If you have y = .5t^2 + 56t – 55 and you want to know the rate of depth change between t = x and t = x + ∆x. you first have to find the depth for each. So when t = x the depth is y = .5(x)^2 +56(x) – 55. When t = x + ∆x the depth is y = .5(x + ∆x)^2 + 56(x + ∆x) - 55. The next thing that you do is you simplify y = .5(x + ∆x)^2 + 56(x + ∆x) – 55 by first doing (x + ∆x)^2 and you get y = .5(x^2 + 2x∆x + ∆x^2) + 56(x + ∆x) – 55. So the depth for t = x + ∆x in the most simplified form is y = .5x^2 + x∆x + .5∆x^2 + 56x + 56∆x – 55. Then you have .5x^2 + x∆x + .5∆x^2 + 56x + 56∆x – 55 – .5x^2 – 56x – 55 = x∆x + .5∆x^2 + 56∆x. Then you take x∆x + .5∆x^2 + 56∆x / ∆x = x + .5∆x + 56. As ∆x goes toward zero you are left the rate of depth change being x + 56.
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Week 3 Quiz # 1 Version 1If you have y = .5t^2 + 56t – 55 and you want to know the rate of depth change between t = x and t = x + ?x. you first have to find the depth for each. So when t = x the depth is y = .5(x)^2 +56(x) – 55. When t = x + ?x the depth is y = .5(x + ?x)^2 + 56(x + ?x) - 55. The next thing that you do is you simplify y = .5(x + ?x)^2 + 56(x + ?x) – 55 by first doing (x + ?x)^2 and you get y = .5(x^2 + 2x?x + ?x^2) + 56(x + ?x) – 55. So the depth for t = x + ?x in the most simplified form is y = .5x^2 + x?x + .5?x^2 + 56x + 56?x – 55. Then you have .5x^2 + x?x + .5?x^2 + 56x + 56?x – 55 – .5x^2 – 56x – 55 = x?x + .5?x^2 + 56?x. Then you take x?x + .5?x^2 + 56?x / ?x = x + .5?x + 56. As ?x goes toward zero you are left the rate of depth change being x + 56.
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If you have an initial investment of $760 which is compounded annually at 10% interest. The growth rate is 10% or .1. The growth factor is 1.1. The function is P(t) = 760(1.1)^t
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