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    crap i just realised i got very different answers for 6. i got a=15 and b=32.
    when I substituted these values back in i got the correct answers :confused:
    did anybody else get that?
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    (Original post by JChoudhry)
    pm me with email, i'll try to scan it in likie 10min or so
    can you type out question 5 pleaseeee? or anyone else who knows how to do it.
    thank you!
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    I also got x as -2/3. Im sure the point was (p,5) because after factorising i found x to be -2/3 or 2. pOint A was (2, something) and so p as (-2/3) would fit whereas x = 5 wouldnt?
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    (Original post by HotCoco.)
    Thanks Mr M. Your very fast!
    Collected paper at 15:25 and answers were posted at 15:56 so I was superspeedy (I used a computer for the binomial expansion and the trapezium rule though so I cheated slightly).
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    (Original post by Haruhi)
    Mr M, for question 8 iv do I get any method marks for substituting in the arithmetic sum formulae instead of both sigma signs and also sumstituting in values of a and d? even though I got the answer completely wrong?

    EDIT: Drive safely!... Im always too slow on things -___-;
    You should get at least a couple of marks for that.
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    (Original post by jkkk32)
    DAMN! so all you were meant to do was put 5 in? ARGH! lol i must have read it as (P,5) or something gargh lol

    the co-ordinate was (P,5), not (5,P), you put y as 5 and solve with quadratic equation, so p = 2. defo!
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    Could someone explain 5 please??
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    (Original post by owen4567)
    I also got x as -2/3. Im sure the point was (p,5) because after factorising i found x to be -2/3 or 2. pOint A was (2, something) and so p as (-2/3) would fit whereas x = 5 wouldnt?
    your right. i just looked at the paper.
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    (Original post by insertusername)
    I also got p as this, there is no way on earth it can be 56 surely, -2/3 gave me the right answer when i substituted it back in
    How embarrassing for me. -2/3 is correct. I don't know the difference between y and x.
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    (Original post by Mr M)
    How embarrassing for me. -2/3 is correct. I don't know the difference between y and x.
    was it -2/3 then not 2? 2 seems to fit, giving a yvalue of 5? thank you! i will have got some marks even if it was -2/3 as i had that written down as my other value. thank you
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    (Original post by JChoudhry)
    Could someone explain 5 please??
    i combined the 2 curves then integrated, then substituted 1 and 3 and hoped to hell it was right but it gave me 5 1/3 so lets hope so
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    (Original post by imfailingbadly)
    sorry i meant you sub in x=5, and the question was find P when (5,P)!
    It was (p,5) not (5,p) so -2/3 is correct.
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    what was question 5 again? what were the equations of the curves? and x values where they intersected each other?
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    (Original post by Mr M)
    It was (p,5) not (5,p) so -2/3 is correct.
    yeah i know so why is 2 wrong?
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    5 = 3x^2 - 4x + 1
    3x^2 - 4x - 4 = 0
    (3x+2) (x-2)
    3x+2=0
    x = -2/3
    and x = 2??
    thats what i posted earlier, 2would work though??
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    (Original post by mrdude)
    It's positive and not negative because a negative number raised to the 6 is positive.
    No the power of 6 comes from (-w^2)^3 so the answer is negative.
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    (Original post by rubberdam123)
    what was question 5 again? what were the equations of the curves? and x values where they intersected each other?
    y=x^2+1 and y=11-(9/x^2)
    intersect at 3 and 1
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    (Original post by JChoudhry)
    y=x^2+1 and y=11-(9/x^2)
    intersect at 3 and 1
    Thanks.
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    -2/3 wouldnt work? but 2 would! can someone confirm this or am i calculate illiterate
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    (Original post by jkkk32)
    can you type out question 5 pleaseeee? or anyone else who knows how to do it.
    thank you!
    \int^3_1 ((11 - 9x^{-2}) - (x^2 + 1)) \ dx = \frac{16}{3}
 
 
 
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