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Thread: solve variables in terms of variables

  1. #1
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    solve variables in terms of variables

    Help would be appreciated.

    1.)

    Solve this formula for the variable $\displaystyle \mu$:

    z = $\displaystyle \frac{x-\mu}{\mu}$



    2.)

    Solve this formula for the variable R:

    $\displaystyle \frac{E}{e}$ = $\displaystyle \frac{R+r}{R}$


    3.)

    The intensity of light as measured from a light bulb varies inversely as the square of the distance "d" from the bulb. Suppose that I is 90W/m^2 when the distance is 5 m. What would the light intensity be 10m from the bulb?


    4.) Write a radical expression with an index of 3 and a radicand of (x - 5)^9. Simplify it as much as possible.



    Help with an explanation of the steps would be very much appreciated. I'm completely lost and didn't get a chance to meet with the instructor after class because of work.
    Last edited by dazed&confuzed; Feb 5th 2007 at 06:16 PM.
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  2. #2
    Junior Member AlvinCY's Avatar
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    Question 1: Solve this formula for the variable $\displaystyle \mu$:

    $\displaystyle z = \frac{x-\mu}{\mu}$
    $\displaystyle z = \frac{x}{\mu} - \frac{\mu}{\mu}$
    $\displaystyle z = \frac{x}{\mu} - 1$
    $\displaystyle z + 1 = \frac{x}{\mu}$
    $\displaystyle \mu = \frac{x}{z + 1}$

    Question 2: Solve this formula for the variable R:

    $\displaystyle \frac{E}{e} = \frac{R+r}{R}$
    $\displaystyle \frac{E}{e} = \frac{R}{R} + \frac{r}{R}$
    $\displaystyle \frac{r}{R} = \frac{E}{e} - 1$
    $\displaystyle \frac{r}{R} = \frac{E - e}{e}$
    $\displaystyle r \times e = R \times (E - e)$
    $\displaystyle R = \frac {r\,e}{E - e}$

    Question 3: The intensity of light as measured from a light bulb varies inversely as the square of the distance "d" from the bulb. Suppose that $\displaystyle I = 90 W/m^2 $ when the distance is 5 m. What would the light intensity be 10m from the bulb?


    We know that I varies inversely to the square of the distance, d. In another words:

    $\displaystyle I = k \times \frac{1}{d^2}$ for some constant k.

    We can work k out by substituting $\displaystyle I = 90, d = 5$

    $\displaystyle 90 = k \times \frac{1}{25}$
    $\displaystyle k = 2250$, therefore,
    $\displaystyle I = 2250 \times \frac{1}{d^2}$

    For d = 10, $\displaystyle I = 2250 \times \frac{1}{100} = 22.5 W/m^2$

    Question 4: Write a radical expression with an index of 3 and a radicand of (x - 5)^9. Simplify it as much as possible.

    This question simply refers to: $\displaystyle \sqrt[3]{(x - 5)^9}$

    $\displaystyle \sqrt[3]{(x - 5)^9}$
    $\displaystyle =[(x - 5)^9]^\frac{1}{3}$
    $\displaystyle =(x - 5)^3$ This would be the simplest form without expanding that.
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  3. #3
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    Thanks for the response.

    I have a question regarding number 3. How did you come about I = 2250? Thats the step that I got lost on.
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  4. #4
    Junior Member AlvinCY's Avatar
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    Quote Originally Posted by AlvinCY View Post


    Question 3: The intensity of light as measured from a light bulb varies inversely as the square of the distance "d" from the bulb. Suppose that $\displaystyle I = 90 W/m^2 $ when the distance is 5 m. What would the light intensity be 10m from the bulb?


    We know that I varies inversely to the square of the distance, d. In another words:

    $\displaystyle I = k \times \frac{1}{d^2}$ for some constant k.

    We can work k out by substituting $\displaystyle I = 90, d = 5$

    $\displaystyle 90 = k \times \frac{1}{25}$
    $\displaystyle k = 2250$, therefore,
    $\displaystyle I = 2250 \times \frac{1}{d^2}$

    For d = 10, $\displaystyle I = 2250 \times \frac{1}{100} = 22.5 W/m^2$
    I didn't say I = 2250...

    Ok, let me take you through it,

    $\displaystyle I = k \times \frac{1}{d^2}$ for some constant k.

    This is what we're given in the question, Intensity varies indirectly to the square of the distance, so we know $\displaystyle I$ and $\displaystyle \frac{1}{d^2}$ are related... and so I've introduced a factor of $\displaystyle k$ in there

    Make sense so far?

    Then we're given that $\displaystyle I = 90$ when $\displaystyle d = 5$, so I substituted $\displaystyle I$ and $\displaystyle d$ into the equation above, yielding:

    $\displaystyle 90 = k \times \frac{1}{25}$, so $\displaystyle k = 90 \times 25 = 2250$...

    So now that we know $\displaystyle k$ we can replace
    $\displaystyle I = k \times \frac{1}{d^2}$ with
    $\displaystyle I = 2250 \times \frac{1}{d^2}$ by substituting $\displaystyle k$ into the equation.

    And now it's a matter of finding $\displaystyle I$ when $\displaystyle d=10$ using this new equation!

    Does that make sense? If not, post here again... and explain to me what you don't understand.
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  5. #5
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    Ooooo okay. I just wasn't paying attention to details. I realized what you were doing after I typed that response, I should have looked at it again. Thanks for the response though.
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