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    (Original post by Dapperblook22)
    There is another way to prove the stretch by rearranging the first equation given. See attached working, I went through this with a friend after the exam
    That's over complicated I would say, have a look at what I replied before.
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    what would be 90 ums?????
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    (Original post by wil_is_he)
    That's over complicated I would say, have a look at what I replied before.
    I thought it would be the most intuitive, as this just needs rearranging. Each to their own methods however
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    (Original post by Myachii)
    Allow me to explain myself:
    You could do the sum of the first three numbers using the formula (which is what I presume you used), or you could just figure them out (26, 24, 22) because it was an arithmetic series so subtracting two wasn't that hard :P

    I said it was like using a sledgehammer to crack a nut because if you used the formula to calculate the sum of the first three, you made yourself a hell of a lot more work and more room for error.
    Yeah I didn't clock about the first and second terms so I used the sum formula. Think I got it right though so it's calm
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    (Original post by beanigger)

    3)a) differentiate something to get 3x-1/2 -1 [2]
    b) find y co-ordinate of the maximum point M M(9,6) [3]
    c) equation of normal to curve at P i think P was (5,6) y=-2x+13 [2]
    d) the normal to curve at P is translated by (0,k) find value of k, k = 5.5 [3]
    I beleive part d), the vector was (k,0), and the answer (the one I got atleast) was -5.5
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    (Original post by Rowan396)
    I beleive part d), the vector was (k,0), and the answer (the one I got atleast) was -5.5
    Same!
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    Stretch question:

    \sqrt{x^2+9} to 3\sqrt{x^2+1}

    Notice 3\sqrt{x^2+1} = \sqrt{9} \sqrt{x^2+1} = \sqrt{9(x^2+1)} = \sqrt{9x^2+9} = \sqrt{(3x)^2+9}

    So it's a stretch in the x-direction by scale factor \frac{1}{3}.
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    (Original post by Dapperblook22)
    I thought it would be the most intuitive, as this just needs rearranging. Each to their own methods however
    Ye at first glance it does look like a stretch in y direction as a lot of people did, but when you see that the 9 has changed to a 1 you know something not right! This is what I did:

    So we started with y=sqrt(x^2+9)
    and went to y=3*sqrt(x^2+1)
    Ignore the y for now.
    Square both equations to get x^2+9 -> 9*(x^2+1)
    Then we get x^2+9 -> 9x^2+9
    ignore the 9 and look what happened to x
    We went from x^2->9x^2
    We want to see what happened to x, not x^2
    so next sqrt both, and we end up with:
    x-> 3x, therefore a stretch by a third in x direction!
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    (Original post by wil_is_he)
    Ye at first glance it does look like a stretch in y direction as a lot of people did, but when you see that the 9 has changed to a 1 you know something not right! This is what I did:

    So we started with y=sqrt(x^2+9)
    and went to y=3*sqrt(x^2+1)
    Ignore the y for now.
    Square both equations to get x^2+9 -> 9*(x^2+1)
    Then we get x^2+9 -> 9x^2+9
    ignore the 9 and look what happened to x
    We went from x^2->9x^2
    We want to see what happened to x, not x^2
    so next sqrt both, and we end up with:
    x-> 3x, therefore a stretch by a third in x direction!
    yh dont think your right
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    (Original post by wil_is_he)
    Ye at first glance it does look like a stretch in y direction as a lot of people did, but when you see that the 9 has changed to a 1 you know something not right! This is what I did:

    So we started with y=sqrt(x^2+9)
    and went to y=3*sqrt(x^2+1)
    Ignore the y for now.
    Square both equations to get x^2+9 -> 9*(x^2+1)
    Then we get x^2+9 -> 9x^2+9
    ignore the 9 and look what happened to x
    We went from x^2->9x^2
    We want to see what happened to x, not x^2
    so next sqrt both, and we end up with:
    x-> 3x, therefore a stretch by a third in x direction!
    That's another good way of doing it. It was also that pesky 9 which spurred me to think of rearranging . As long as you put down a correct way of doing it, then there should be no worries if the answer is correct
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    (Original post by Rowan396)
    I beleive part d), the vector was (k,0), and the answer (the one I got atleast) was -5.5
    Because k is a vector it is negative of what is in the equation... so it was --k which is +k... if you look at the graph it gave you, it needed to move right not left so had to be positive (just to check)
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    (Original post by zzxxDash53xxzz)
    yh dont think your right
    Sorry to hear that because you got it wrong then!
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    How did you get -2 for a and 10 for d for the arithmetic series question? I got some long numbers like 10.1756 and when I plugged them into the formula I still got 8 so my final answer for the sum question was something like 138?
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    (Original post by Myachii)
    Allow me to explain myself:
    You could do the sum of the first three numbers using the formula (which is what I presume you used), or you could just figure them out (26, 24, 22) because it was an arithmetic series so subtracting two wasn't that hard :P

    I said it was like using a sledgehammer to crack a nut because if you used the formula to calculate the sum of the first three, you made yourself a hell of a lot more work and more room for error.
    Or just add 28 to the sum of the u2 and u3 which was already given
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    (Original post by zzxxDash53xxzz)
    yh dont think your right
    It is a correct way of doing it. If you can't see it this way, then look at my way attached above that gives the same answer
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    (Original post by jake4198)
    Predicted Grade Boundaries

    A - 61
    B - 56
    C - 51

    Was easier than last year but not the easiest paper there's been.
    Hey

    I think I messed up both C1 and C2 papers and I am very upset about my self. I was wondering, to get an overall B grade in AS maths what do you need to get across all papers inclucding M1? And what would you need to get a C overall? Please reply.
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    (Original post by zzxxDash53xxzz)
    yh dont think your right
    Neither did I, but it checks out.
    RIP my fellow y direction brother <3
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    (Original post by 1120015)
    Hey

    I think I messed up both C1 and C2 papers and I am very upset about my self. I was wondering, to get an overall B grade in AS maths what do you need to get across all papers inclucding M1? And what would you need to get a C overall? Please reply.
    Define "messed up"
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    (Original post by wil_is_he)
    Because k is a vector it is negative of what is in the equation... so it was --k which is +k... if you look at the graph it gave you, it needed to move right not left so had to be positive (just to check)
    In x isnt it the opposite, so moving right it a negative value?

    Either way im sure it was (k,0) not (0,k) like in the OP
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    (Original post by 1120015)
    Hey

    I think I messed up both C1 and C2 papers and I am very upset about my self. I was wondering, to get an overall B grade in AS maths what do you need to get across all papers inclucding M1? And what would you need to get a C overall? Please reply.
    For a B overall you need 70% UMS overall the AS level, meaning if you were to get 100% in M1, you would only need 55% in C1 and C2
 
 
 
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