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    (Original post by PhysicsGal)
    The book multiplied the right hand side by (x+3)/(x+3).

    Now 5/5 = 1, and x/x = 1, anything divisible by itself = 1, thus (x+3)/(x+3) = 1.

    So if we multiply the LHS by this 'fraction', we get y multiplied by (x+3)/(x+3) = y multiplied by 1 = y
    oh right , got it!

    OK

    x = 3root t , y= t(4-t)

    0<eqaul t <eqaul 4

    I am finding it difficult to get my limits what do I do? The x value from the graph is x=0 and x=6 , but we need the limits in terms of t?
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    (Original post by otrivine)
    oh right , got it!

    OK

    x = 3root t , y= t(4-t)

    0<eqaul t <eqaul 4

    I am finding it difficult to get my limits what do I do? The x value from the graph is x=0 and x=6 , but we need the limits in terms of t?
    You do. Use the parametric definition of x.
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    (Original post by Indeterminate)
    You do. Use the parametric definition of x.
    I did but does not work.
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    (Original post by otrivine)
    I did but does not work.
    Of course it does.

    6=3\sqrt{t}

    0=3\sqrt{t}

    You can solve those, right?
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    (Original post by Indeterminate)
    Of course it does.

    6=3\sqrt{t}

    0=3\sqrt{t}

    You can solve those, right?
    Yes gives 2 and 0
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    (Original post by otrivine)
    Yes gives 2 and 0
    TWO and zero?
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    (Original post by Mr M)
    TWO and zero?
    You

    Sorry 2 power 1/2=4
 
 
 
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