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    (Original post by PreppyNinja)
    You know the g(x)=+15 or something? How did you do that? It's the only question I left
    The question before you have the equation 2x3 + 9x2 - 2x - 24
    So -24 + 15 = -9
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    (Original post by Daniel__)
    Anyone remember the equations of the two lines in question 2, section A? (intersection question).
    Dont remember the equations but i got 7/10 and 1/10
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    (Original post by Magenta96)
    oh wow I think I've slipped up on that too, can't believe I didn't realise the significance of r > 0 in the question!
    Always read the question!!!


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    I think there will be a A1 marking point for the + on that question.
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    (Original post by jojo19apr)
    For the last question I got k=0 and k=4. I got that because y= - x + k, it had a gradient of -1 so the points on the curve must have done too. Then all I could think to do was differentiate (which as a retaker I can do but I know most of you may not have seen that yet), to find the x coordinates of where the gradient was -1 on the graph. Both seemed to match up graphically, so I hoped it worked ok!
    Oh my God. I'm so relieved to see that someone else did that.

    y=1/(x-2)

    y=(x-2)-1

    (chain rule)

    dy/dx = -(x-2)-2
    -(x-2)-2 = -1

    (x-2)-2 = 1

    1/(x-2)2=1

    (x-2)2=1

    x2-4x+3=0

    (x-3)(x-1)=0

    Gradient = -1 when x= 1 or 3

    To find the y coordinates,

    y=1/x-2
    y=1/1-2=-1

    y=-x+k is a tangent at point (1,-1)

    -1=-1+k
    k=0

    y=1/3-2
    y=1

    y=-x+k is also a tangent at the point (3,1)

    1=-3+k
    k=4

    So k=0 or 4

    Hope this helps.
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    (Original post by janicee)
    What did people get on the circle question? I couldn't find out where the circle intersected the y axis so I used the formula and got 2+root11 and 2-root11. Did I get this completely wrong? Gah I'm scared.
    I did both ways, the formula and by setting y and x to 0 according to which intercept I was trying to find. I got the same as you with both methods, I only did both as I doubted myself with the formula thinking I may have gone wrong!!! So reassuringly, we got the same
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    You know the second last question in question 12? Did anybody else use the quadratic formula to solve the question to find the X coordinate?
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    (Original post by Deceived)
    Oh my God. I'm so relieved to see that someone else did that.

    y=1/(x-2)

    y=(x-2)-1

    (chain rule)

    dy/dx = -(x-2)-2
    -(x-2)-2 = -1

    (x-2)-2 = 1

    1/(x-2)2=1

    (x-2)2=1

    x2-4x+3=0

    (x-3)(x-1)=0

    Gradient = -1 when x= 1 or 3

    To find the y coordinates,

    y=1/x-2
    y=1/1-2=-1

    y=-x+k is a tangent at point (1,-1)

    -1=-1+k
    k=0

    y=1/3-2
    y=1

    y=-x+k is also a tangent at the point (3,1)

    1=-3+k
    k=4

    So k=0 or 4

    Hope this helps.
    Everyone came out with different answers and I was so sure of 4 and 0 because both points had a gradient of -1 and I couldn't see it could be any different! Differentiation for once saved me on this
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    (Original post by jojo19apr)
    They won't know grade boundaries until they have all of the final results, then they adjust the boundaries to have a certain number of candidates walking away each grade. Hence for a hard paper they are lowered and vice versa, this seemed a reasonably fair paper so the boundaries may be slightly higher. But the general guide is 80% = A, 70% = B, 60% = C, 50% = D, 40% = E and 40% > U ........ Hope that gave you a vague idea!
    I get that - but what do you all think it will be?


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    for the rearanging question i got r=sqrt3v/pia+pib
    did i do it wrong? i didnt put the pi in the bracket

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    (Original post by Rubyturner94)
    for the rearanging question i got r=sqrt3v/pia+pib
    did i do it wrong? i didnt put the pi in the bracket

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    no, thats right you just distributed pi on the denominator
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    (Original post by janicee)
    Ogod I failed. Well thanks for replying! I think I did okay on the other questions, fingers crossed!
    Don't worry, there was no answer of (x-3)^2+(y-2)^2=20, that was the equation of the circle we were given! We didn't have to use complete the square on this either, it was substituting x and y in for 0 according to which intercept you were finding Hope you're not quite so worried!
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    for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
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    (Original post by janicee)
    Do you mean 3(x-2)^2 -7? Because that's what I got
    Same, it shouldn't be 3x^2 in front, just 3.
    It also said in the form a(x+b)^2+c, so it can't be right
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    for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
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    (Original post by janicee)
    Oh okay I failed. I'm such an idiot. Omg someone shoot me please :'(
    The last question was 12, the one with k in There definitely wasn't a 13!
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    (Original post by Soho32)
    for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
    the answer was 25;

    (1/5)^-2

    (5/1)^2

    5^2 = 25


    since 1/0.04 = 25 then you should be okay even if youe method is a bit of a weird way round it.
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    (Original post by Soho32)
    for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
    I think you had to give the answer as 25.

    (0.2)-2

    =(1/5)-2

    =1/(1/5)2

    =1/(1/25)

    =25

    You may be completely fine but there's a possibility you could lose an A1 mark.
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    (Original post by Soho32)
    for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
    i got 25 which is the same as 1/0.04
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    (Original post by Soho32)
    for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
    I did exactly this and for some reason didn't see that 1/0.04=25.
    Really hope I don't lose a mark for it
 
 
 
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