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Math Help - rearranging formula

  1. #1
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    rearranging formula

    Hi All;
    I need to make a2 the subject of a1 sqrt 2gh/(a1/a2)-1 = q

    firat divide bt a1 giving sqrt 2gh/(a1/a2)-1 = q/a1

    square both sides giving 2gh/(a1/a2)-1 = qsq / a1sq

    add one to both sides giving 2gh/(a1/a2) = qsq / a1sq + 1

    stuck here.

    Thanks.
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  2. #2
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    Divide by a2?

    Please use parentheses to clarify meaning. In this case, a1 sqrt(2gh/(a1/a2)-1) = q would have been helpful.
    Please don't use implicit multiplication. Use an effective symbol. In this case, a1 * sqrt(2gh/(a1/a2)-1) = q would have been helpful.
    Use "^" to indicate exponentiation. 2gh/(a1/a2)-1 = (q^2) / (a1^2) = (q/a2)^2
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  3. #3
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    Dont understand divide by a2?
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  4. #4
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    I am troubled by this...

    \frac{2gh}{\frac{a1}{a2}} = 2gh\cdot\frac{a2}{a1}

    Okay, you talked me out of it. Instead of that, multiply by \frac{a1}{2gh}
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  5. #5
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    Here is where the formula is Powered by Google Docs

    Please scroll down to question 6 and there answer is at the bottom of the page.
    I think there answer is wrong just let me know if it is.

    Thanks.
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  6. #6
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    Quote Originally Posted by anthonye View Post
    Hi All;
    I need to make a2 the subject of a1 sqrt 2gh/(a1/a2)-1 = q

    firat divide bt a1 giving sqrt 2gh/(a1/a2)-1 = q/a1

    square both sides giving 2gh/(a1/a2)-1 = qsq / a1sq

    add one to both sides giving 2gh/(a1/a2) = qsq / a1sq + 1

    stuck here.

    Thanks.
    The givn problem is
    A1*sqrt{2gh/[(A1/A2)^2 - 1]} = q
    Now simplify.
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  7. #7
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    Dont understand I'm having trouble making a2 the subject.
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  8. #8
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    q^2/A1^2 = 2gh/[(A1/A2 )^2 - 1]

    (A1/A2)^2 - 1 = 2gh*A1^2/q^2

    (A1/A2)^2 = 1 + 2gh*A1^2/q^2

    A1^2/A2^2 = [q^2 + 2gh*A1^2]/q^2

    A2^2 = q^2A1^2/[q^2 + 2gh*A1^2]
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  9. #9
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    Why in the fourth step does q^2 appear in numerator and denominator.
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  10. #10
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    Let A1 = A and A2 = B; then Sarigama's (which is correct) becomes:
    B^2 = q^2 A^2 / (q^2 + 2gh * A^2)
    So complete this way:
    B = qA / SQRT(q^2 + 2gh * A^2)

    You need practice with these:
    try and solve this for n:
    q = a * SQRT[m / (n - 1)]
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  11. #11
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    Quote Originally Posted by anthonye View Post
    Why in the fourth step does q^2 appear in numerator and denominator.
    Example: if x = 1 + a/y, then x = (y + a) / y
    That's quite basic: are you attending math classes?
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  12. #12
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    Hello, anthonye!








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  13. #13
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    Ok so 1/y is equal to its reciprocal y/1 which is just y I take it thats correct.

    Because y is unknown yes.
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  14. #14
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    Ok Wilmer I get n=ma^2 - 1/ mq^2
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  15. #15
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    Quote Originally Posted by anthonye View Post
    Ok Wilmer I get n=ma^2 - 1/ mq^2
    NOOOooooo.......PLUS you forgot brackets: n = (ma^2 - 1) / mq^2 ;
    VERY important; hope you know why...

    We have:
    q = a * SQRT[m / (n - 1)]

    Divide by a:
    q / a = SQRT[m / (n - 1)]

    Square both sides:
    q^2 / a^2 = m / (n - 1) ; still with me?

    Cross multiply:
    q^2(n - 1) = ma^2

    Expand:
    nq^2 - q^2 = ma^2

    Isolate the n term:
    nq^2 = ma^2 + q^2

    Divide by q^2:
    n = (ma^2 + q^2) / q^2 : dig them brackets on left !!

    Amen!
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