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Math Help - Algebra Problems

  1. #1
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    Exclamation Algebra Problems

    Variation Functions can involve non-integer exponents

    1. Assume that the length of time it takes to do a major construction project varies inversely with the 0.7 power of the number of workers on the project. Suppose that 100 workers can construct an office building in 250 days. Write the particular equation expressing time in terms of number of workers, and use it to predict the number of workers needed to construct the building in just 180 days.

    Variation functions can involve more than two variables.

    2. Building a 200 foot long bridge across Scorpion Gulch, where the water is 8 feet deep, costs $700,000. Assume that the cost varies directly with the length of the bridge and directly with the square of the depth of the water. Write the particular equation expressing cost in terms of length and depth. Use the equation to predict the cost of a bridge downstream from the present one, where the water is only 5 feet deep, but the length must be 600 feet.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Nimmy
    Variation Functions can involve non-integer exponents

    1. Assume that the length of time it takes to do a major construction project varies inversely with the 0.7 power of the number of workers on the project. Suppose that 100 workers can construct an office building in 250 days. Write the particular equation expressing time in terms of number of workers, and use it to predict the number of workers needed to construct the building in just 180 days.

    Variation functions can involve more than two variables.

    2. Building a 200 foot long bridge across Scorpion Gulch, where the water is 8 feet deep, costs $700,000. Assume that the cost varies directly with the length of the bridge and directly with the square of the depth of the water. Write the particular equation expressing cost in terms of length and depth. Use the equation to predict the cost of a bridge downstream from the present one, where the water is only 5 feet deep, but the length must be 600 feet.
    Here is one way.

    1.)
    t ---> 1/[x^(0.7)]
    t = k/[x^(0.7] -------------(1)

    When x=100workers, t=250days, so,
    250 = k/[100^(0.7)]
    k = 250[100^(0.7)] = 6279.716079
    Hence,
    t = (6279.716079)/[x^(0.7)] -----answer.

    When t=180days, what is x?
    180 = (6279.716079)/[x^(0.7)]
    x^(0.7) = (6279.716079)/180
    x^(0.7) = 34.88731155
    Raise both sides to their 1/(0.7) power,
    x = (34.88731155)^(1/(0.7))
    x = 159.8861038
    Or,
    x = 160 workers --------answer.

    ---------------------------------------
    2.)
    C --->L, d^2
    C = k(L)(d^2) ------(2)

    When L=200ft and d=8ft, the C is $700,000, so,
    700,000 = k(200)(8^2)
    k = (700,000)/(200*64) = 54.6875
    Hence,
    C = (54.6875)(L)(d^2) ----------------answer.

    Then, when L=600ft and d=5ft, what will be C?
    C = (54.6875)(600)(5^2) = $820,312.50 -------------answer.
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