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UNOFFICIAL MARKSCHEME Edexcel Maths Calculator paper 09/06/2016 Watch

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    (Original post by Certified)
    no it wasn't because 56 was corresponding, then its angles on a straight line..
    I got 55
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    24) A) -1/3x+4

    B0 2x/(x+1)(x-1)

    5) 0.09, 0.36

    13) -1,1,-1

    11) 42.28
    12) 144
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    (Original post by rayestar)
    u mean pentagon righttt? :d
    yessss is it 144 omggg yas
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    (Original post by Jf1234)
    What if we put 9/20 instead of 0.45 for 1b?
    I did this it will still give u a mark
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    What did everyone get for the pentagon question?
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    Fast baller..star sigining for the paki internetional team also porn all the way and btw 1+1=69
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    What is the predicted grade boundaries for an A* for this year? And have grade boundaries ever been more than 170/200 for an A*?
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    (Original post by LilyKitty)
    I don't think so because the angle has to be between the two sides (7 &8) for the formula of triangle area to work and x was another angle so you had to do cosine to find another side and then sine rule too before you could get x. Not sure though. xx
    What? I got 80.4 but used cousins twice. Is that all right?
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    (Original post by Rajive)
    I got 40.0?
    Yeah that's what I got too , seems to be a few different answers people got though so not sure if it's right
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    (Original post by 34908seikj)
    Wasn't the last question like 1/2 * 7 * 8 Sin(X) = 18


    56Sin(x) = 36

    Sin^-1(36/56) = X

    x = 40

    ???

    I think I did it wrong then.
    I got the same and I thought it's right but apparently not since everyone else got 80.4 or 80.1


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    GUYS WHAT DO YOU THINK THE GRADE BOUNDARIES WILL BE I REALLY WANT AN A BUT I THINK I GOT AROUND 120-130 OUT OF 200?!?!?! IS it possible?
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    (Original post by ellamoon_22)
    This is correct. I forgot to do the inverse but i shall at least get 3/4marks i think
    This is correct for the angle at B, but the question asked for the angle at A.
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    80.1 is also ok for the last question right?
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    3.8 is right
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    did u guys get 84638 bars ?
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    last question-
    find angle bac

    first find angle abc
    18=0.5x7x8xsinabc
    18=28sinabc
    18/28= sinabc =0.642...
    sin-1(0.642...)=abc=40.0052...

    then find length AC
    ac^2=7^2+8^2-3x7x8cos40.00
    ac^2=27.209...
    ac=5.216...

    then find angle bac
    18=0.5 x5.216...x7xsinbac
    18/18.25...=sinbac=0.985...
    sin-1(0.98...)=bac=80.3755...
    =80.4
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    (Original post by 34908seikj)
    Wasn't the last question like 1/2 * 7 * 8 Sin(X) = 18


    56Sin(x) = 36

    Sin^-1(36/56) = X

    x = 40

    ???

    I think I did it wrong then.

    you had to find the area, then use the cosine rule then the sine rule to get 80.4
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    (Original post by samiuls)
    yep its correct
    W h a t!! !! I got 4! ! Time to cry salty tears.
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    I got 40.0 too...
    Working back from 1/2absinC
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    (Original post by hannah2360)
    What did everyone get for the pentagon question?
    144 degrees
 
 
 
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