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Math Help - An architect drawing !!!

  1. #1
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    Smile An architect drawing !!!

    i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem
    Attached Thumbnails Attached Thumbnails An architect drawing !!!-untitled-1.jpg   An architect drawing !!!-untitled-2.jpg   An architect drawing !!!-untitled-2-3.jpg  
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  2. #2
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    Quote Originally Posted by carlasader View Post
    i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem



    The question as asked makes no sense. There must be at least one
    constraint or additional conditions missing.

    RonL
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  3. #3
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    Quote Originally Posted by carlasader View Post
    i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem


    Has this not already been posted?

    RonL
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  4. #4
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    4. As x^2 + x -1 is positive between x=1 and x=2, the area is:

    <br />
A = \int_1^2 (x^2 + x -1) dx<br />

    5. The first four terms in the expansion of (1+x)(1-2x)^10 may be found by expanding the second term using the binomial theorem and multiplying out:

    <br />
(1+x)(1-2x)^{10}=(1+x)[1+10(-2x)+\frac{10.9}{2!}(-2x)^2+\frac{10.9.8}{3!}(-2x)^3+..]<br />

    and keeping the four terms with the lowest powers of x

    RonL
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  5. #5
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    Wink thank u soo much for the answer

    thanks alot for the answer, and by the way in number 2 there is nothing missing ,its just like that ...i myself am wondering ,anyways i would like to thank u for answering them for me and would like please to answer number 3 thanks the last one left in the home work . thank you , answer only if u can
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by carlasader View Post
    thanks alot for the answer, and by the way in number 2 there is nothing missing ,its just like that ...i myself am wondering ,anyways i would like to thank u for answering them for me and would like please to answer number 3 thanks the last one left in the home work . thank you , answer only if u can
    let's find the distance PQ and QM then, so we can find the area of the triangle.

    to do that, we need to find points P and Q

    these points will be where the graphs intersect, that is, where

    x^2 - 2x + 2 = -x + 4

    \Rightarrow x^2 - x - 2 = 0

    \Rightarrow (x - 2)(x + 1) = 0

    \Rightarrow x = 2 \mbox { or } x = -1

    when x = 2:

    y = -2 + 4 = 2

    when x = -1:

    y = 1 + 4 = 5

    so we have P(-1,5) and Q(2,2)


    Now, slope of line PQ is \frac {y_2 - y_1}{x_2 - x_1} = \frac {3}{-3} = -1

    slope of line QM is \frac {y_2 - y_1}{x_2 - x_1} = 1

    this means, line PQ is perpendicular to QM, and so \triangle PQM is a right-triangle.


    Now let's find the lengths of PQ and QM

    By the distance formula:

    PQ = \sqrt {(2 - (-1))^2 + (2 - 5)^2} = \sqrt {18}

    QM = \sqrt {(1 - 2)^2 + (1 - 2)^2} = \sqrt {2}


    \Rightarrow A_{ \triangle PQM} = \frac {1}{2}bh = \frac {1}{2} \left(  \sqrt {2} \right) \left( \sqrt {18} \right) = 3



    Now to find the area of the shaded region:

    A_{R} = \int_{-1}^{2} \left( -x + 4 - x^2 + 2x - 2 \right) dx

    = \int_{-1}^{2} \left( -x^2 + x + 2 \right) dx

    = \left[ - \frac {1}{3} x^3 + \frac {1}{2} x^2 + 2x \right]_{-1}^{2}

    = 4 \frac {1}{2}

    Since 4 \frac {1}{2} = 1 \frac {1}{2} \cdot 3, we have the desired result
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