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    (Original post by yelash)
    I thought I got that one, but apparently its like 13.7 and I don't remember writing that.. Almost forgot to add the arc length on but spotted it with 10 mins to go :P



    I changed stretch s.f. 1/3 in x-direction after writing what you did. I don't think s.f. 3 in y-direction is right because when you expanded both it was y=X+3 changes to y=3X+3 or something
    f(x) was √x^2 + 9 , or something similar.

    and the transformation was 3√x^2 + 9 which is 3f(x) so answer is stretch s.f. 3 vertical i think


    From BBC Bitesize
    y = af(x) is a stretch, parallel to the y axis, scale factor a
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    (Original post by Alfie Oliver)
    Did anyone get k (in the translation question) as 5.5?
    I did!!!
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    For the binomial I used the c4 method to fully expand whatever the quadratic was - (2+x)^7 ?

    Problem is I completely forgot to times each answer by 2^7 although I did take it out the brackets and write it.

    I then used the indices pairings from both expansions and multiplied any thats powers would add to 10.

    How many marks would I lose?
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    On 8bii if you got up to this bit 5+3cosx=0

    how many marks would you lose?
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    (Original post by akpbrooks)
    I did!!!
    ****, I got 11
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    hi whats the deal with resitting c1 - c4 seeing as this is the last type of this exam paper?
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    (Original post by Excuse Me!)
    f(x) was √x^2 + 9 , or something similar.

    and the transformation was 3√x^2 + 9 which is 3f(x) so answer is stretch s.f. 3 vertical i think


    From BBC Bitesize
    y = af(x) is a stretch, parallel to the y axis, scale factor a
    The new transformation was actually 3√(x^2+1), so I got the answer as stretch SF 1/9 in x direction...
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    (Original post by Excuse Me!)
    On 8bii if you got up to this bit 5+3cosx=0

    how many marks would you lose?
    I'd say 2-3 because you had to say the minimum answer is 2 as cos anything can range between between -1 and 1 and then inverse cos -1 gave you 0 and 2 pie
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    Updated - Unofficial Mark Scheme.
    Here's what I got

    1. b) a=-2
    2. b) x=-0.861
    c) reflection in y-axis
    3. a) 3x^-1/2 - 1
    b) y=6
    c) y + 2x = 13
    d) k = 5.5
    4. b) d = -2, a = 28
    c) Sum to 21 - sum to 3. 168-78 = 90
    5. b) area = 19.9
    c) perimeter = 13.8
    6. a) 65.6
    b) i) translation (0,5)
    ii) stretch in X-direction 1/3 sf
    7. a) p= -10 q= 40 r= -80
    b) -1648
    8. b) Tan(X) = 1 and tan(X) =-5/4 (to the nearest degree) X = 45, 225, 129, 309
    c) lowest = 2, when theta = pi
    9. a) c^(m/2 -6n)
    b) X = 5/2

    I don't exactly remember the answer to 7 b) if anyone could help.
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    Thats exactly the same as what I got so hopefully we both got it
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    (Original post by M.N1997)
    For the 3(x^3 + 1)^1/2 translation, I took the 3 into the square root and times it all by factor nine, this gave me stretch in X direction sf 1/9??
    Your first bit was right. You do take the 3 inside the square to make it a 9. Then you get (9x^2+ 9)^1/2. Which is basically ((3x)^2+9)^1/2. So it's a stretch along x-axis, scale factor (1/3).
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    Anyone know if there's an unofficial mark scheme for statistics 1 Aqa - 25th may 2016
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    (Original post by tomdavis.)
    Updated - Unofficial Mark Scheme.
    Here's what I got

    1. b) a=-2
    2. b) x=-0.861
    c) reflection in y-axis
    3. a) 3x^-1/2 - 1
    b) y=6
    c) y + 2x = 13
    d) k = 5.5
    4. b) d = -2, a = 28
    c) Sum to 21 - sum to 3. 168-78 = 90
    5. b) area = 19.9
    c) perimeter = 13.8
    6. a) 65.6
    b) i) translation (0,5)
    ii) stretch in X-direction 1/3 sf
    7. a) p= -10 q= 40 r= -80
    b) -1648
    8. b) Tan(X) = 1 and tan(X) =-5/4 (to the nearest degree) X = 45, 225, 129, 309
    c) lowest = 2, when theta = pi
    9. a) c^(m/2 -6n)
    b) X = 5/2

    I don't exactly remember the answer to 7 b) if anyone could help.
    9 a) was 3^(m/2 -6n)
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    For 2c, instead of putting reflection in the y axis I put stretch scale factor -1? from y=0.2^x to y=5^x
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    (Original post by bheaton)
    Anyone know if there's an unofficial mark scheme for statistics 1 Aqa - 25th may 2016
    Isn't S1 on June 01?
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    (Original post by tomdavis.)
    Updated - Unofficial Mark Scheme.
    Here's what I got

    1. b) a=-2
    2. b) x=-0.861
    c) reflection in y-axis
    3. a) 3x^-1/2 - 1
    b) y=6
    c) y + 2x = 13
    d) k = 5.5
    4. b) d = -2, a = 28
    c) Sum to 21 - sum to 3. 168-78 = 90
    5. b) area = 19.9
    c) perimeter = 13.8
    6. a) 65.6
    b) i) translation (0,5)
    ii) stretch in X-direction 1/3 sf
    7. a) p= -10 q= 40 r= -80
    b) -1648
    8. b) Tan(X) = 1 and tan(X) =-5/4 (to the nearest degree) X = 45, 225, 129, 309
    c) lowest = 2, when theta = pi
    9. a) c^(m/2 -6n)
    b) X = 5/2

    I don't exactly remember the answer to 7 b) if anyone could help.
    Was K not 6.5??? for question 3)d)
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    (Original post by Excuse Me!)
    f(x) was √x^2 + 9 , or something similar.

    and the transformation was 3√x^2 + 9 which is 3f(x) so answer is stretch s.f. 3 vertical i think


    From BBC Bitesize
    y = af(x) is a stretch, parallel to the y axis, scale factor a
    Yep, I got the same as you did.
    But they said the transformation was 3√x^2 + 1 not 3√x^2 + 9 which totally confused me.

    I remember when I looked at the question it was 3√x^2 + 9, right??

    I just really didn't know whether I read the question wrong or not.
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    http://www.thestudentroom.co.uk/show....php?t=4117603

    Just gonna put this here in case anyone didn't see.
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    I realised I did the expansion correct and then managed to add up the coefficients incorrectly wtf..........
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    (Original post by bheaton)
    Anyone know if there's an unofficial mark scheme for statistics 1 Aqa - 25th may 2016
    Isn't S1 on 8th June for AQA?
 
 
 
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