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Thread: 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
    Grand Panjandrum
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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. As $\displaystyle x^2 + x -1 $ is positive between $\displaystyle x=1$ and $\displaystyle x=2$, the area is:

    $\displaystyle
    A = \int_1^2 (x^2 + x -1) dx
    $

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

    $\displaystyle
    (1+x)(1-2x)^{10}=(1+x)[1+10(-2x)+\frac{10.9}{2!}(-2x)^2+\frac{10.9.8}{3!}(-2x)^3+..]
    $

    and keeping the four terms with the lowest powers of $\displaystyle x$

    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 $\displaystyle PQ$ and $\displaystyle QM$ then, so we can find the area of the triangle.

    to do that, we need to find points $\displaystyle P$ and $\displaystyle Q$

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

    $\displaystyle x^2 - 2x + 2 = -x + 4$

    $\displaystyle \Rightarrow x^2 - x - 2 = 0$

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

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

    when $\displaystyle x = 2$:

    $\displaystyle y = -2 + 4 = 2$

    when $\displaystyle x = -1$:

    $\displaystyle y = 1 + 4 = 5$

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


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

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

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


    Now let's find the lengths of PQ and QM

    By the distance formula:

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

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


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



    Now to find the area of the shaded region:

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

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

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

    $\displaystyle = 4 \frac {1}{2}$

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