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Math Help - 2 problems

  1. #1
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    2 problems

    Alright well basically I've got 2 problems, haven't done algebra in a year, totally forgot how to do this stuff.

    √x/y =4 , which of the following represents y in terms of x?

    a. y = x/2

    b. y = 2/x

    c. y = x/16

    d. y = 16/x



    problem number 2

    The forumla P = 2pi (√L/32) can be used to approximate the period of a pendulum,

    Where L is the pendulum's length in feet and P is the pendulum's period in seconds. If a pendulum's period is 1.6 seconds, what is the length of the pendulum to 3 decimal places?



    I'd appreciate explanation with the answer please, thank you.
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  2. #2
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    Re: 2 problems

    I'll get you started on the answer to the first problem:

    IF:

    √(x/y) = 4

    THEN:

    x/y = 4*4 = 16, so:

    x = 16y

    can you continue?
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: 2 problems

    Please use parentheses. What you wrote is this:

     \frac {\sqrt{x}}y = 4

    but I think what you meant is this:

     \sqrt{\frac x y } = 4

    It makes a difference. Same with the second problem - the correct formula is  P = 2 \pi \sqrt{ \frac L {32}}, not  P = 2 \pi \frac {\sqrt{ L}} {32}

    In both cases you can start by squaring both sides of the equation and then rearranging. For problem 1 to get 'y' as the subject after squaring you will want to multiply both sides of the equation by y, then divide both sides by 4^2. For problem 2 to get L as the subject after squaring both sides you can divide both sides by 2pi and then multiply both sides by 32. Try it and post back with the results you get.
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  4. #4
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    Re: 2 problems

    So is y 16/x?
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: 2 problems

    Let's try your answer and see if it works. If y = 16/x then starting with original equation:

     \sqrt{\frac x y } = 4, substitue 16/x for y:

    \sqrt {\frac {x}{16/x}} = 4. Now group the x terms:

    \sqrt { \frac {x^2}{16}} = 4 . And take the square root of both sides:

     \frac x 4 = 4

    So no, it doesn't seem to work. Starting with what Deveno left you with: x=16y, what happens if you divide both sides by 16?
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  6. #6
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    Re: 2 problems

    For the first problem y=16/x

    and for the second problem here is what your going do:
    P = 2pi(√L/32)
    √(L/32) = P/2pi
    √L = P√32/2pi
    L = (32P^2)/(4pi^2)

    after this you do the calculations and you get L = 2.075
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