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Math Help - Find the Year

  1. #1
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    Find the Year

    According to data from the U.S. Census Bureau, the population P of Cleveland, Ohio (in thousands) in year x can be approximated by the equation P = .08x^2 - 13.08x + 927 , where x = 0 corresponds to the year 1950. In what year (in the past) was the population about 804, 200?
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  2. #2
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    Quote Originally Posted by emfn96 View Post
    According to data from the U.S. Census Bureau, the population P of Cleveland, Ohio (in thousands) in year x can be approximated by the equation P = .08x^2 - 13.08x + 927 , where x = 0 corresponds to the year 1950. In what year (in the past) was the population about 804, 200?
    Set .08(x)^2 - 13.08(x) + 927 = 804,200

    \Rightarrow .08(x)^2 - 13.08(x) - 803273 = 0

    Then use the quadratic formula to find x, the number of years after 1950 this event occurred.

    Note that:

    a=.08

    b=-13.08

    c=- 803273

    \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}

    Now plug and chug, as they say.
    Last edited by Anonymous1; April 5th 2010 at 08:01 PM.
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  3. #3
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    Quadratic Formula

    Quote Originally Posted by Anonymous1 View Post
    Set .08(x)^2 - 13.08(x) + 927 = 804,200

    \Rightarrow .08(x)^2 - 13.08(x) - 803273 = 0

    Then use the quadratic formula to find x, the number of years after 1950 this event occurred.

    Note that:

    a=.08

    b=-13.08

    c=- 803273

    \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}

    Now plug and chug, as they say.
    I had no clue that the quadratic formula was needed to find the answer.
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  4. #4
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    It's not "required" but you did recognise that this is a quadratic equation, didn't you?
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  5. #5
    baz
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    it is straight forward quadractic equation.
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