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    (Original post by Heffalump .)
    Does it go up in 10s, or can you get like 98 UMS?
    You can get an odd number of UMS yes. The 10% UMS is just the grade bracket, so you can get anywhere in the bracket. So, for example, if you got 79% UMS - a high B - you'd probably try a remark to get the A at 80% UMS, whereas if you got 74% UMS this is a mid B, and so probably wouldn't remark.
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    (Original post by Madmatician)
    It would be around 98/100 UMS
    (Original post by aoxa)
    You can get an odd number of UMS yes. The 10% UMS is just the grade bracket, so you can get anywhere in the bracket. So, for example, if you got 79% UMS - a high B - you'd probably try a remark to get the A at 80% UMS, whereas if you got 74% UMS this is a mid B, and so probably wouldn't remark.
    Thank you!
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    (Original post by Achroite)
    Anyone???
    Since the roots where 0 and 3 you should have had a curve touching at 0 and coming up through the point (3,0)

    Earlier in the chat there is a electronically drawn graph of this.

    If I understand what you said properly you lost a maximum of 1 or 2 marks
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    (Original post by Heffalump .)
    You'd get 1/3 for that
    Ahh that's a shame, thanks
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    Ugh, that didn't go too well.

    Does anyone remember the very first question. What was the equation of that line we had to find the gradient for (the very first question: 1a)? Not the answer by the way, just the line. I think it was called AB or something.
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    (Original post by Heffalump .)
    Does it go up in 10s, or can you get like 98 UMS?
    Because I think I got around 73/75 which just scrapes 100UMS, would that be 90UMS or like 98?
    AQA have a UMS calculator which calculate UMS for each year depending on mark, obviously 2015 wont be on there!
    http://www.aqa.org.uk/exams-administ...t-marks-to-ums

    You can put in marks to help you understand it
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    (Original post by Madmatician)
    Yeah I meant 5root2 - 1, sorryIt doesn't equal 7
    Sorry I meant root((5root2-1)^2-1) don't know how I wrote it the other way
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    (Original post by excaliburnus)
    Ugh, that didn't go too well.

    Does anyone remember the very first question. What was the equation of that line we had to find the gradient for (the very first question: 1a)? Not the answer by the way, just the line. I think it was called AB or something.
    -3/5 I think
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    (Original post by B82)
    Sorry I meant root((5root2-1)^2-1) don't know how I wrote it the other way
    I think you may of had to simplify your answer more than that, though I cant talk since I left it as root 49
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    (Original post by 2014_GCSE)
    -3/5 I think
    Do you know the equation of that line?
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    (Original post by JackC3001)
    Question 1

    (a) -3/5
    (b)(i)perp. gradient = 5/3
    in the form mx + ny + p = 0
    5x - 3y + 1 = 0

    Question 2

    7+root15
    n=7 (Had to be stated separately)

    Question 3

    a) y-6 = -10(x+1)
    y = -10x - 4

    bi) 108/5 or 21.6
    bii) 162/5 or 32.4

    Question 4

    a) (x+1)^2 + (y-3)^2 = 50
    bi) C is (-1, 3)
    bii) r = √50 = 5√2
    c) k= 8, -2
    d) min. distance = √49 = 7

    Question 5

    a) p = 3/2, q = -¼
    bi) (-3/2, -1/4)
    bii) x=-3/2
    c) y = x^2 - x + 4

    Question 6

    ai) Surface Area = 2πrh
    48π = 2πrh - 2πr^2
    h = 24/r - r/2

    aii) V = πr^2h
    V = πr^2 * (24/r - r/2)
    V = 24πr - (π/2 * r^3)

    bi) 24π - (3πr^2) / 2
    bii) 24π - (3πr^2) / 2 = 0 when r = ±4
    r = 4 as positive value
    d^2V/dr^2 = 3πr
    when r = 4, second derivative = -12π, therefore maximum

    Question 7

    a) draw x^2(x-3) where line touches at 0 and crosses at 3
    bi) Remainder = 36
    bii) R = 0 therefore is a factor
    biii) (x-2)(x^2-5x+10)
    biv) b^2 - 4ac < 0 therefore no real roots for (x^2-5x+10)
    (x-2) only root when x = 2

    Question 8

    a) show that = x^2 + 3(k-2)x -13-k=0 by setting two equations equal to each other
    b) b^2 - 4ac < 0 therefore 9k^2 -32k -16 < 0
    c) -4/9 < k <4

    These are my answers for the C1 paper - pretty sure they are correct - please let me know if I need to change anything

    for the cylinder question, i got the same answer as everyone else but i forgot to include the pi in it as i thought it would complicate the answer.
    How many marks do you think i would loose?
    thanks
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    (Original post by excaliburnus)
    Do you know the equation of that line?
    5x - 3y + 1 = 0
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    If u want to no solutions message me...
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    (Original post by kas69)
    If u want to no solutions message me...
    I can give method
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    (Original post by B82)
    Sorry I meant root((5root2-1)^2-1) don't know how I wrote it the other way
    Was it out of 2? If so, I think you'll get 1 mark for the method
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    (Original post by excaliburnus)
    Do you know the equation of that line?
    It was 5y = -3x +7 since it went through (9, -4) or something
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    (Original post by Sail last)
    I think you may of had to simplify your answer more than that, though I cant talk since I left it as root 49
    alright thanks!
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    (Original post by excaliburnus)
    Ugh, that didn't go too well.

    Does anyone remember the very first question. What was the equation of that line we had to find the gradient for (the very first question: 1a)? Not the answer by the way, just the line. I think it was called AB or something.
    5y + 3x = 7
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    Anyone think they know the raw mark for an A?
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    (Original post by 2014_GCSE)
    5x - 3y + 1 = 0
    I did
    -5x+3y-1=0
    do you think I still got all the marks?:L
 
 
 
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