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    (Original post by aoifea100)
    I think the issue with that question was that it cancelled to 2-X whereas the other bracket was 3x - 6, so most people probably didn't realise to take -3 out instead of a positive 3?
    A lowered grade boundary would be amazing
    Instead of 2-x I did -(x-2) which cancelled when you took three out of (3x-6) to get 3(x-2). We ended up with the same answer though
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    (Original post by ozmo19)
    aoifea100


    Did anyone get the same answers as these?

    1)2-√5

    2)i) (2,4)
    ii) (4,3)
    iii) a= -4

    3)x= (7/2)

    4)a) -1/(3x)
    b) (2,3)

    5)ii) (x+4)(2x-3)(x-6)
    b)c= -3/2

    6)a)a= -1, b= -8
    b)ii) x+12y-44=0

    7)ii)w=15cm, h=20cm
    8)a)(1-3√5)/2 ≤ k ≤ (1+3√5)/2
    b) b^2 -4c = 1
    I think most of ours correlates, if I remembered the question I would be able to help.

    When I saw the rationalise the denominator first question, I thought we were in for a good one!



    Yes, low grade boundaries are a must!
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    I'm guessing I got around 60-65 does anyone have a clue what that would roughly be?


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    (Original post by ozmo19)
    aoifea100

    Did anyone get the same answers as these?

    1)2-√5

    2)i) (2,4)
    ii) (4,3)
    iii) a= -4

    3)x= (7/2)

    4)a) -1/(3x)
    b) (2,3)

    5)ii) (x+4)(2x-3)(x-6)
    b)c= -3/2

    6)a)a= -1, b= -8
    b)ii) x+12y-44=0

    7)ii)w=15cm, h=20cm
    8)a)(1-3√5)/2 ≤ k ≤ (1+3√5)/2
    b) b^2 -4c = 1

    can you post a solution to 7i and 6a! Can't remember the qu for 6a and I'm going round in circles still for 7i!!! Thanks for posting answers is put my mind t ease!!!
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    (Original post by Sparklez10)
    can you post a solution to 7i and 6a! Can't remember the qu for 6a and I'm going round in circles still for 7i!!! Thanks for posting answers is put my mind t ease!!!
    Ill talk you through the method
    for 6a
    you find dy/dx and equal it to 0
    Sub in x=2 and get an equation
    Sub in x=-4/3 and get an equation
    Solve simultaneous equations.

    for 7i
    [(40x30)/2] = wh + 0.5 x w x (40-h) + 0.5 x h x (30-w) (i think)
    rearrange for h
    you end with h=40 - (4w/3)
    Then A=wh so sub in h.
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    Okay thank you so much! I have got method marks for both but just couldn't get the final answers!
    do you reckon grade boundaries will be high??
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    (Original post by Sparklez10)
    Okay thank you so much! I have got method marks for both but just couldn't get the final answers!
    do you reckon grade boundaries will be high??
    No, i think they'll be relatively low, or so I'm hoping they are!
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    (Original post by John5678)
    Hows everyone feeling for C1 Maths for CCEA tomorrow?

    Hadn't seen a CCEA thread, so decided to make one!
    First time I hear about this board!
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    (Original post by M14B)
    First time I hear about this board!
    Its a Northern Irish board
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    (Original post by M14B)
    First time I hear about this board!
    Solely Northern Irish board!
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    (Original post by John5678)
    lmg1 ozmo19


    How did yous find it? All was good at the start; q7 the proof was dodgy, I didn't get the final answer; but most of my working was broadly similar, so hopefully a few marks there.
    Q8 ii I think, where b2-4c = n and (n+1) was such a dodgy question, didn't have a clue there.
    Other than that, not bad- but definitely on the hard side.


    Your thoughts on how it went?
    It wasn't that bad, it could've went a lot worse
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    (Original post by ozmo19)
    Ill talk you through the method
    for 6a
    you find dy/dx and equal it to 0
    Sub in x=2 and get an equation
    Sub in x=-4/3 and get an equation
    Solve simultaneous equations.

    for 7i
    [(40x30)/2] = wh + 0.5 x w x (40-h) + 0.5 x h x (30-w) (i think)
    rearrange for h
    you end with h=40 - (4w/3)
    Then A=wh so sub in h.
    Would I lose a lots marks because in Q.6a I multiplied out -4/3 to get -4 and subbed this into equation and for final answers I got -3 and -8. I completely forgot to divide it by 3.
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    (Original post by ozmo19)
    No, i think they'll be relatively low, or so I'm hoping they are!
    What do you think the A will be??
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    (Original post by aoifea100)
    ozmo19 yeah most of these answers look similar, definitely glad you got that -1/3x, I thought I messed that up a lot
    For 8a I did it < in place of ≤ which I know is wrong now, but I got the same answer so I should at least get most of the marks for that part
    Everything but 8b looks the same as what I got
    I done exact same thing for Q.8a even though I had only revised it that morning but I got the right answer. For the question about the product of linear factors how many marks would you lose for forgetting to add (x+4)?
    I have to admit going through the paper I lost a few stupid marks but my working out was mostly fine (except for Q8b).
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    (Original post by Sparklez10)
    What do you think the A will be??
    It would probably be similar to last years paper. Do you know what the grade boundaries were for June 2015.
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    Does anyone have a solution for this?
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    (Original post by lmg1)
    It would probably be similar to last years paper. Do you know what the grade boundaries were for June 2015.
    I'm not sure sorry! I will ask around and see!! How did you find it
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    Could you please put up the method you did for 8 part b?
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    (Original post by ozmo19)
    Name:  image.jpeg
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    Does anyone have a solution for this?
    I did get 1 but I can't remember how I got it exactly. But to start with I did that b^2-4C>0
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    that paper effed me up lol
 
 
 
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