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Math Help - Need allitle help on these questions.

  1. #1
    Junior Member
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    Need allitle help on these questions.

    I'm having trouble on these questions trying to figure them out but think my work is wrong.

    1.Which of the following is equavlant to (2z^3)/(8z^(n-2))
    a. 1/4z^1-n
    b. z^5-n
    c. 1/4z^(5-n)

    2.Find the range of negative squareroot (36-x^2)
    so far I gotten (x+6)=-6 and (x-3)=6 so would it be y:y -6<y<6?

    3. Find all solutions for sin^2=3cos^2
    so would it be 0, pie/2, 1/3 ?
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  2. #2
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    Quote Originally Posted by goldenroll View Post
    I'm having trouble on these questions trying to figure them out but think my work is wrong.

    1.Which of the following is equavlant to (2z^3)/(8z^(n-2))
    a. 1/4z^1-n
    b. z^5-n
    c. 1/4z^(5-n)
    1. choice is c, i wont explain it because i think you wrote in the answer? i get z^(5-n)/4
    2.Find the range of negative squareroot (36-x^2)
    so far I gotten (x+6)=-6 and (x-3)=6 so would it be y:y -6<y<6?

    3. Find all solutions for sin^2=3cos^2
    so would it be 0, pie/2, 1/3 ?
    Number 3 is pi/3

    sin^2x = 3(1 - sin^2x)
    4sin^2x = 3
    sinx = root of 3 / 2
    x = 60
    x = pi / 3

    and i think your number 2 is right because
    you switch x and y and then solver for 36 - y^2 under the square root to be greater than or equal to 0
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  3. #3
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    Quote Originally Posted by goldenroll View Post
    I'm having trouble on these questions trying to figure them out but think my work is wrong.

    1.Which of the following is equavlant to (2z^3)/(8z^(n-2))
    a. 1/4z^1-n
    b. z^5-n
    c. 1/4z^(5-n)

    2.Find the range of negative squareroot (36-x^2)
    so far I gotten (x+6)=-6 and (x-3)=6 so would it be y:y -6<y<6?

    3. Find all solutions for sin^2=3cos^2
    so would it be 0, pie/2, 1/3 ?
    For Q1:

    \frac{2z^3}{8z^{n-2}}=\frac{1z^{3-(n-2)}}{4}=\frac{z^{5-n}}{4}


    Thus the answer is neither of the choices. If the question is written as (2z)^3, then the answer is B. If it's 2z^3, then my answer doesn't fit into any.

    For Q3:

    sin^2x=3cos^2x

    Divide by cos^2x on the both sides, you get: tan^2x = 3

    Square root both sides, you get:  tan x = sqrt(3) or -sqrt(3)

    Therefore the critical angle is pi/3. If you are looking angles between 0 and 2pi, the possible answers are pi/3, 2*pi/3, 4*pi/3, 5*pi/3.

    Hope this helps.
    Last edited by mr fantastic; April 7th 2009 at 06:27 PM. Reason: Fixed the latex (powers)
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  4. #4
    Super Member Showcase_22's Avatar
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    <br />
\frac{2z^3}{8z^{n-2}}=\frac{1z^{3-(n-2)}}{4}=\frac{z^{5-n}}{4}<br />
    That's c isn't it?

    It depends how you read the brackets .
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  5. #5
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    Yes I think
    c should be read as \frac14\:z^{5-n}
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