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Math Help - [SOLVED] 3 problems

  1. #1
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    [SOLVED] 3 problems

    1. What is shadowed surface of cube if its side is a=2\sqrt{2}
    (Image is on attachment).

    2.On the box, we have 5 red balls, 4 dark balls, and 6 white balls. If we catch one ball, what is probability to catch a white one?

    3. If we have a segment [AB], A(5,-1), B(x,2) and it's middle point is O(0,0), what is value of x?

    Thank you.
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  2. #2
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    1. Where is the attachment, can't see it.

    2. 5 + 4 + 6 = 15 balls in total. 6 are white so its 6/15=2/5, ie the probability is 2/5.

    3. Given a line with two points (A,B) and (C,D), the midpoint , (E,F), is found as E=(A+C)/2 and F=(B+D)/2.

    So, in this example we have
    A=5, B=-1, C=x, D=2, E=F=0
    Using basic algebra we find A=-C (subbing in E=0).
    A=5 thus C=-5
    Wait... If you apply the same reasoning we find that B=-D, which is false so I think you may have given some incorrect numbers. If your point A and point B co-ordinates are correct then it is your midpoint that is incorrect.
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  3. #3
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    Ups, sorry. There is attachment.
    Attached Thumbnails Attached Thumbnails [SOLVED] 3 problems-cube.jpg  
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  4. #4
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    That is a rectangle. Area of a rectangle=base times height

    Base=a
    Height= sqrt(2(a^2))=a*sqrt2 (by pythagoras formula)
    if a=2*sqrt2 then,
    Base=2*sqrt2, and
    Height=2*(sqrt2)*(sqrt2)=2*2=4
    Base*Height=4*2*sqrt2=8sqrt2

    Area=8sqrt2
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  5. #5
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    I have just another problem so I don't see very reasonable to open new thread just for that.

    At this square |DE|=2|EC|. Find cosx.
    Attached Thumbnails Attached Thumbnails [SOLVED] 3 problems-sqr.jpg  
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  6. #6
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    Cosine of angle

    Hello GreenMile
    Quote Originally Posted by GreenMile View Post
    I have just another problem so I don't see very reasonable to open new thread just for that.

    At this square |DE|=2|EC|. Find cosx.
    It's generally better to open a new thread, but here's the answer to this one.

    If the length of the side of the square is a, then |DE| = \tfrac{2}{3}a

    \Rightarrow |AE|^2 = a^2 + (\tfrac{2}{3}a)^2 = a^2 + \tfrac{4}{9}a^2 = \frac{13a^2}{9}, by Pythagoras.

    \Rightarrow |AE| = \frac{\sqrt{13}}{3}a

    \Rightarrow \cos x = \frac{|DE|}{|AE|} = \frac{2}{3}a \div \frac{\sqrt{13}}{3}a = \frac{2}{\sqrt{13}}

    Grandad
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