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Math Help - Solving for X

  1. #1
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    Solving for X

    Q:Solving the given demand equation for x in terms of p,

    p= -0.02x+400 (o< or = x < or =20000)

    text book answer: x=f(p) = -50p + 20000

    I got x = -p/0.02 + 20000

    I don't understand how is it 50p.

    Thanks
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  2. #2
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    Re: Solving for X

    Quote Originally Posted by hovermet View Post
    Q:Solving the given demand equation for x in terms of p,

    p= -0.02x+400 (o< or = x < or =20000)

    text book answer: x=f(p) = -50p + 20000

    I got x = -p/0.02 + 20000

    I don't understand how is it 50p.

    Thanks
    p = -0.02x + 400

    note that 0.02 = \frac{2}{100} = \frac{1}{50} ...

    p = -\frac{1}{50} x + 400

    p - 400 = -\frac{1}{50} x

    -50p + 20000 = x
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  3. #3
    Senior Member
    Joined
    Nov 2010
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    Clarksville, ARk
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    Re: Solving for X

    Quote Originally Posted by hovermet View Post
    Q:Solving the given demand equation for x in terms of p,

    p= -0.02x+400 (o< or = x < or =20000)

    text book answer: x=f(p) = -50p + 20000

    I got x = -p/0.02 + 20000

    I don't understand how is it 50p.

    Thanks
    1/(0.02) = 50
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