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Thread: Subject of the formula

  1. #1
    Senior Member Mukilab's Avatar
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    Subject of the formula

    Make r the subject

    $\displaystyle P=\pi r+2r+2a$

    I could only get

    $\displaystyle \frac{P-2a}{\pi}=r+\frac{2r}{\pi}$
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    Make r the subject

    $\displaystyle P=\pi r+2r+2a$

    I could only get

    $\displaystyle \frac{P-2a}{\pi}=r+\frac{2r}{\pi}$
    Take 2a as normal.

    $\displaystyle \pi r + 2r = P-2a$

    You can then factor out an r on the LHS

    $\displaystyle r(\pi+2) = P-2a$

    Can you find $\displaystyle r$ now?
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  3. #3
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Take 2a as normal.

    $\displaystyle \pi r + 2r = P-2a$

    You can then factor out an r on the LHS

    $\displaystyle r(\pi+2) = P-2a$

    Can you find $\displaystyle r$ now?

    $\displaystyle r=\frac{P-2a}{\pi +2}?$
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    $\displaystyle r=\frac{P-2a}{\pi +2}?$
    Indeed. $\displaystyle \pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question
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  5. #5
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Indeed. $\displaystyle \pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question
    But thank you for the supply of extra information ^^
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