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  1. #1
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    Anyone here know Pythagorean theorem?

    If you know please do this question
    <---------- Is That Right?

  2. #2
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    This is 5th grade bs.. lel, anyways
    The Pythagorean theorem states that the square of the hypotenuse (the side opposite to the right angle) is equal to the sum of the squares of the other two sides
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    Question has nothing to do with the theorem, but mostly sin cosine and tangent

    Quote Originally Posted by HeroicXPharaoh View Post
    This is 5th grade bs.. lel, anyways
    The Pythagorean theorem states that the square of the hypotenuse (the side opposite to the right angle) is equal to the sum of the squares of the other two sides
    damn if only he asked for the definition

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    Quote Originally Posted by Illuminarly View Post
    Did you refresh the thread and see my comment? No way you replied that quickly


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    (a^2)+(b^2)=(c^2)
    (x^2)+(10^2)=((x+6)^2)
    (x^2)+100=((x^2)+12x+36)

    Move your quadratic or degree 2 polynomial to which ever side and set it equal to zero. Solve for your x values.
    Typing it on the computer is rather confusing.

    Edit: You should get +/- 16/3


    - - - Updated - - -

    Quote Originally Posted by Royce View Post
    Question has nothing to do with the theorem, but mostly sin cosine and tangent



    damn if only he asked for the definition
    "Question has nothing to do with the theorem, but mostly sin cosine and tangent"
    You are not given an angle. Your sides have "x" as the variable.
    If you do not have an angle, that will be a y
    For example:
    You will get siny=x/x+6
    You can't solve that unless you introduce another equation for substitution or elimination.
    Last edited by Hugo Boss; 01-22-2015 at 08:29 PM.

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    The only correct answer is 16/3. There is no second solution since:
    (x+6)^2=x^2+10^2
    x^2+12x+36=x^2+100 |-x^2
    12x+36=100 |-36 |:12
    x=16/3=5.3333333333333333333333333333333333333333333333 33333333333333333333333333333333333333333333333333 3333333333333333333

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  10. #8
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    Quote Originally Posted by Hugo Boss View Post
    (a^2)+(b^2)=(c^2)
    (x^2)+(10^2)=((x+6)^2)
    (x^2)+100=((x^2)+12x+36)

    Move your quadratic or degree 2 polynomial to which ever side and set it equal to zero. Solve for your x values.
    Typing it on the computer is rather confusing.

    Edit: You should get +/- 16/3


    - - - Updated - - -



    "Question has nothing to do with the theorem, but mostly sin cosine and tangent"
    You are not given an angle. Your sides have "x" as the variable.
    If you do not have an angle, that will be a y
    For example:
    You will get siny=x/x+6
    You can't solve that unless you introduce another equation for substitution or elimination.
    I seen 90 degress and dun goofed

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    Quote Originally Posted by Illuminarly View Post
    Did you refresh the thread and see my comment? No way you replied that quickly


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