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Math Help - Quadratic equation

  1. #1
    Junior Member
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    NWT, Canada
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    Question Quadratic equation

    When bicycles are sold for $300 each, a cycle store can sell 70 in a season. For every $25 increase in the price, the number sold drops by 10.
    a) Represent the sales revenue as a function of the price.
    b)The total sales revenue is $17,500. How many bicycles were sold? What is the price of one bicycle?
    c) What range of prices will give a sales revenue that exceeds $18,000?
    I'm stuck on b
    R= revenue
    n= number of bikes
    p= price
    For a
    Price| Number of Bikes sold
    300 | 70
    325 | 60
    350 | 50
    So find the slope
    \frac{(70-50)}{(300-350)}=-0.4
    Plug that into y=mx+b
    70=-0.4(300)+b
    70=-120+b
    70+120=190
    b=190
    So that means the function to find n(number of bicycles sold) is -0.4p+190
    Since Revenue is r=n*p then part a is r=p(-0.4p+190) which simplifies to r=-0.4p^2+190p
    So to solve part b?

    17500=-0.4p^2+190p
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  2. #2
    Junior Member
    Joined
    Feb 2008
    From
    NWT, Canada
    Posts
    51
    I got it
    17500=-0.4p^2+190p
    \frac{17500}{-0.4}=-43750
    -43750=p^2+(\frac{190}{-0.4}p
    0=p^2-475p+43750
    0=(p-125)(p-350)

    p=125, 350
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