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Math Help - Can Anyone help me with these questions?

  1. #16
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    For number 8 a) I got v = 70 but I'll wait to see what others get.

    Just incase you think I just repeated what you said:
    Last edited by r_maths; May 9th 2007 at 01:05 PM.
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  2. #17
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    Re:

    rmath you want to minimize C I haven't delt with British lbs. before!
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  3. #18
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    I got 70 as well by doing, cross multiplying to get 2v^2 = 9800 then I got 70

    Is this what you guys did as well.

    but then thats too little work for 5 marks. Don't you agree?
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  4. #19
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    Re:

    Quote Originally Posted by novadragon849 View Post
    but then thats too little work for 5 marks. Don't you agree?
    What do you mean by this? Test score? The use of the Red pen? I think I know what you are inferring to...
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  5. #20
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    ok then cheers, so how would you go about doing b and c and the steps for it?
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  6. #21
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    Quote Originally Posted by qbkr21 View Post
    rmath you want to minimize C I haven't delt with British lbs. before!
    Don't you just substitute 70 into original equation.

    C = 40

    but then thats too little work for 5 marks. Don't you agree?
    You get:
    1 Mark for stating "For minimum f"(x) = 0"

    so how would you go about doing b and c
    I would of been stuck at this two days ago.
    Look here: http://www.mathhelpforum.com/math-he...entiation.html

    C, look at top of this post. I hope its correct.
    1 Mark for sub.
    1 Mark for right answer.
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  7. #22
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    RE: 8b

    Re:
    Attached Thumbnails Attached Thumbnails Can Anyone help me with these questions?-minimize.gif  
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  8. #23
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    anyone for number 9
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  9. #24
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by novadragon849 View Post
    anyone for number 9
    9)
    a)
    By the Law of Cosines:

    (PR)^2 = (PQ)^2 + (QR)^2 - 2(PQ)(QR)cos(PQR)
    => 108 = 36 + 36 - 72cos(PQR)
    => cos(PQR) = [108 - 2(36)]/(-72) = -1/2
    => PQR = arccos(-1/2)
    => PQR = 2pi/3 radians

    b)
    Area of a sector is given by:
    A = (1/2)r^2(theta)
    where r is the radius and theta is the angle subtended by the arc in radians

    => A = (1/2)(6)^2(2pi/3) m^2
    => A = 12 pi m^2

    c)
    Using A = (1/2)absinC, we have:

    A = (1/2)(PQ)(QR)sin(PQR)
    => A = (1/2)(6)(6)(2pi/3)
    => A = 9sqrt(3) m^2

    ....or you could use the half base times height formula (do you want to see how?)

    d)
    Area of segment = Area of sector - Area of triangle = 22.1

    e)
    Perimeter of PQRS = 6 + 6 + length of arc RSP

    length of arc s is given by:

    s = r(theta) , where r is the radius and theta is the angle in radians
    => s = 6(2pi/3) = 4pi

    => Perimeter = 12 + 4pi = 24.6 m
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