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Math Help - Precal app problem

  1. #1
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    Precal app problem

    There are 2 tanks of water. The first tank holds 200 liters and will be drained by pump at a constant rate completely in 5 hours. The second tank holds 900 liters and will be drained by opening a plug on the bottom, this tank will also drain in 5 hours. The second tanks volume will decrease according to toricellis law which is V(t)=Vo(1-t/T)^2

    When will the volumes of the 2 tanks be equal?
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  2. #2
    Eater of Worlds
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    The amount of water in the first tank at time t will be 200-40t

    The amount of water in the second at time t is 900(1-\frac{t}{5})^{2}

    Set them equal and solve for t. You will get two solutions.
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  3. #3
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    thanks, so far I've got to -45t+t^2/25=676
    does that sound right to you? I dont know where to go from here any pointers?
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  4. #4
    Eater of Worlds
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    Just a quadratic.

    200-40t=900(1-\frac{t}{5})^{2}

    200-4t=36t^{2}-360t+900

    -36t^{2}+320t+700=>36t^{2}-3200t+747=>4(9t^{2}-80t+175)=>9t^{2}-80t+175=0

    Now, solve the quad. Use the quadratic formula. May be the best bet.

    But you could factor.

    Divide by 9 and rewrite:

    t^{2}-\frac{35}{9}t-5t+\frac{175}{9}

    (t^{2}-\frac{35}{9}t)-(5t-\frac{175}{9})

    t(t-\frac{35}{9})-5(t-\frac{35}{9})=0

    \boxed{(t-5)(t-\frac{35}{9})=0}
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  5. #5
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    Quote Originally Posted by galactus View Post
    Just a quadratic.

    200-40t=900(1-\frac{t}{5})^{2}

    200-4t=36t^{2}-360t+900

    -36t^{2}+320t+700=>36t^{2}-3200t+747=>4(9t^{2}-80t+175)=>9t^{2}-80t+175=0

    Now, solve the quad. Use the quadratic formula. May be the best bet.

    But you could factor.

    Divide by 9 and rewrite:

    t^{2}-\frac{35}{9}t-5t+\frac{175}{9}

    (t^{2}-\frac{35}{9}t)-(5t-\frac{175}{9})

    t(t-\frac{35}{9})-5(t-\frac{35}{9})=0

    \boxed{(t-5)(t-\frac{35}{9})=0}
    how did you get from here

    to here?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    how did you get from here

    to here?
    It's just a typo:
    200 - 40t = 900 \left ( 1 - \frac{t}{5} \right )^2

    200 - 40t = 900 \left ( 1 - \frac{2t}{5} + \frac{t^2}{25} \right )

    200 - 40t = 900 - 900 \cdot \frac{2t}{5} + 900 \cdot \frac{t^2}{25}

    200 - 40t = 36t^2 - 360t + 900

    He used the correct equation to proceed to the next line.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    It's just a typo:
    200 - 40t = 900 \left ( 1 - \frac{t}{5} \right )^2

    200 - 40t = 900 \left ( 1 - \frac{2t}{5} + \frac{t^2}{25} \right )

    200 - 40t = 900 - 900 \cdot \frac{2t}{5} + 900 \cdot \frac{t^2}{25}

    200 - 40t = 36t^2 - 360t + 900

    He used the correct equation to proceed to the next line.

    -Dan
    k thanks
    Last edited by bilbobaggins; October 13th 2007 at 09:56 PM.
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