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Math Help - 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 \mu:

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



    2.)

    Solve this formula for the variable R:

    \frac{E}{e} = \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; February 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 \mu:

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

    Question 2: Solve this formula for the variable R:

    \frac{E}{e} = \frac{R+r}{R}
    \frac{E}{e} = \frac{R}{R} + \frac{r}{R}
    \frac{r}{R} = \frac{E}{e} - 1
    \frac{r}{R} = \frac{E - e}{e}
    r \times e = R \times (E - e)
    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 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:

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

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

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

    For d = 10, 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: \sqrt[3]{(x - 5)^9}

    \sqrt[3]{(x - 5)^9}
    =[(x - 5)^9]^\frac{1}{3}
    =(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 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:

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

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

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

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

    Ok, let me take you through it,

    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 I and \frac{1}{d^2} are related... and so I've introduced a factor of k in there

    Make sense so far?

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

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

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

    And now it's a matter of finding I when 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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