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    for the cylinder question did anyone else get R as 4?
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    (Original post by Mentalmirz)
    for the cylinder question did anyone else get R as 4?
    Yes. which with d^2V/dr^2 implies max.
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    (Original post by Lizzie33)
    Ah thank you! But wait sorry to be a pain but was the original qn

    X^2 - 3x +2
    Or
    X^2 + 3x +2
    I hope it was x^2 + 3x + 2
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    (Original post by tanyapotter)
    yeah, what i did was the graph starts at the bottom left hand side quadrant, touches the origin, goes back down, has a minimum at the bottom right quadrant, then goes back up crossing the x axis at +3
    Yeah same) woopa
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    (Original post by Mentalmirz)
    for the cylinder question did anyone else get R as 4?
    Yes, you had to say that it was +4 and -4 and therefore say that it was +4 as they wanted the positive value.
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    Trapezium - parallel side sum was (30+6) x 1/2 x height

    The height was 2 -(-1)

    Therefore I got 48 -108/5
    Anyone else?
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    Anyone think that other then that damn cylinder question, it wasn't that bad? Plus this being a bit harder means that Core 2 should be easier.. Should
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    for the cubic graph i showed a positive x^3 graph but went through 0 and touched at 3 how many marks would this be?
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    (Original post by student0042)
    Yes. which with d^2V/dr^2 implies max.
    I put that it was a max awesome thanks!
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    (Original post by rehmh05)
    Trapezium - parallel side sum was (30+6) x 1/2 x height

    The height was 2 -(-1)

    Therefore I got 48 -108/5
    Anyone else?
    h was 3
    0.5(3)(30+6)=54
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    Somehow I ended up with 4pi, and 12pi^2 for the last part of Q6...any idea how I did that?
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    Sniffed a naughty line of lem before the paper, so it was sweet as
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    (Original post by Jupers)
    h was 3
    0.5(3)(30+6)=54
    damn !
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    Are you sure you are right??? Most of your answers match mine apart from 1c,3bi and bii

    (Original post by student0042)
    1bi. 4x-3y+1=0 (I thought it was this form)
    1c. A(9,-4)
    3a. y-6=-10(x+1)
    3bi. 108/5
    3bii. 162/5
    4a. (x+1)^2 + (y-3)^2 = 50
    4bi. C(-1,3)
    4bii. 2√5
    4c. k=-8, 2
    4d. minimum distance =7
    5a. p = 3/2, q = -¼
    5bi. (-3/2, -1/4)
    5bii. 5c. y = x^2 - x + 4
    6bii. =-12π therefore max (r=+4)
    7a. draw x^2(x-2)
    7bi. R = 36
    7bii. R= 0 therefore root
    7biii.(x-2)(x^2-5x+10)
    7biv. x=2
    8a. Show that x^2 + 3(k-2)x -13-k=0
    8b. 9k^2 -32k -16 < 0
    8c. -4/9 < k <4
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    (Original post by rehmh05)
    Trapezium - parallel side sum was (30+6) x 1/2 x height

    The height was 2 -(-1)

    Therefore I got 48 -108/5
    Anyone else?
    I got 54 I found it as a triangle and a rectangle though my final answer came out as 32.4...
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    Can someone tell me if my working is right here:
    (4,k)
    (x+1)^2 + (y-3)^2 = 50
    (4+1)^2 + (k-3)^2 = 50
    25+k^2-6k+9=50
    k^2-6k-16=0
    (k-8)(k+2)
    k=8
    k=-2
    is this how everyone else worked it out? if not, how did you guys do it?
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    (Original post by MD 2 4 40)
    Sniffed a naughty line of lem before the paper, so it was sweet as
    the cylinder question must have made you psychotic
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    The cylinder one:

    48pi = piXr^2 + 2xpiXrxh (pi cancels)
    48 = r^2 +2rh
    48-r^2 = 2rh
    h = (48-r^2)/2r
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    Was the original translation equation:

    X^2 - 3x +2
    Or
    X^2 + 3x +2

    (Before you completed the square originally)


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    (Original post by Heffalump .)
    Can someone tell me if my working is right here:
    (4,k)
    (x+1)^2 + (y-3)^2 = 50
    (4+1)^2 + (k-3)^2 = 50
    25+k^2-6k+9=50
    k^2-6k-16=0
    (k-8)(k+2)
    k=8
    k=-2
    is this how everyone else worked it out? if not, how did you guys do it?
    Yep thats how I did it (but my quadratic was wrong hahah)
 
 
 
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