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    (Original post by Rit101)
    This means that the gradient of AB was -3/5
    Hence the parallel line should have the same gradient correct?
    Yeah i got something like that,i dont think its -3/2
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    (Original post by UbaidS)
    how did you get 65 for 5a
    to work out the radius of circle u needed to find the "difference" in x and y directions and use Pythagoras to find the actual length of it
    if you do M1 you would be very familiar with this.
    think of it like this, how do you get from coordinates (10,10) to (5,6)
    well, you go -5 in the x-direction and -4 in the y-direction
    now think of x and y as being horizontal and vertical respectively.
    what is the hypotenuse of that triangle u have? ( hypotenuse = radius of the circle)
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    does anyone remember how many marks question 6 was worth?
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    (Original post by Rit101)
    This means that the gradient of AB was -3/5
    Hence the parallel line should have the same gradient correct?
    Yeah i think i got that too..
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    (Original post by UbaidS)
    Yeah i got something like that,i dont think its -3/2
    its -5/3
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    fixed the gradient mb lol, if anyone could remember any more answers or help me correct this it would help a lot thanks
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    rip wasil's 100 ums dream FeelsBadMan, can we get some 1G's in the chat? Kappa
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    (Original post by UbaidS)
    how did you get 65 for 5a
    I personally used the distance equation between the two points which came out to 4squared (16) and 7 squared (49) I then added these together to recieve 65 however remember 65 is the radius squared so the radius is sqrt of 65 👍
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    plz the gradient is -5/3 how to do people not see this
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    (Original post by ryanredfern)
    rip wasil's 100 ums dream FeelsBadMan, can we get some 1G's in the chat? Kappa
    Kappa123
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    (Original post by Sniperdon227)
    If you can justify why it was 34 ill give you a cookie!
    You can only complete the sq when they give you x^2 +y^2 +6x+8y+3=0 to find the radius, in this case they gave two coordinates!

    by the way that x^2 + y^2 .... was made up
    (x-5)^2-25+(y+3)^2-9=k
    (x-5)^2+(y+3)^2=34
    Thats how i did it
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    (Original post by timtjtim)
    Okay, I've added a little bit.

    1) a)Asked to work out gradient of a parallel line, m= - 3/2
    b) Asked to find co-ordinates of B, B( - 3,4)
    c) Asked to find K, K= - 30

    2) a) simplify (3√5)^2 = 45
    b) Put ((3√5)^2 + √5) / (7 + 3 √5) into m + n√5= 75 - 325

    3) a) y=(x-7/2)^2 - 41/4
    b) Translation of (1/2 , 41/4)

    4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48. p(-3) = -3^3 - 5 * 3-^2 - 8 * -3 + 48 = 0. As p(-3) = 0, (x+3) is a factor of p(x).

    b) Three linear factors goes to (x+3)(x-4)(x-4)
    c) find the remainder when x^3 - 5x^2 -8x + 48 was divided by (x+2) R=20
    d) factorise p(x) using your answer to part c) as R = (x-2)(x^2-3x+14) + 20

    5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
    b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
    c) asked to find the equation of the tangent at A 7x-4y+18=0
    d) asked to find the length of CT = 9

    6) a)

    7) a) I think there were 4 parts to this question I can't remember part a and b
    b)
    c) Definite integral (from 1 to -2) = 81/4 (20.25)
    d) Area of shaded region (integral - triangle area) = 45/4 (11.25)

    8) a) can't remember anything except the answers were k>6 and k<-3/2 (NOT 6 > k > -3/2 )

    Answers that need a question to be assigned to
    - Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
    - k=4 and k=20
    - x = -1±√5
    - d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
    Why was it not 6>k>-3/2
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    Any estimates for 100 UMS?
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    (Original post by Rit101)
    Why was it not 6>k>-3/2
    the actual final inequality before it was solved said that it was greater than 0
    if u draw a diagram the graph is greater than 0 when k>6 and k<-3/2
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    For the one where you had to prove it was a minimum value, I believe you first need to find that it is a stationary point by subbing in to dy/dx, then subbing in to d^2y/dx^2 to show it is a minimum.

    I base this on the fact the question was 4 marks, and it wouldn't be 4 marks for only subbing in to d^2y/dx^2.
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    (Original post by qwertyuiop1998)
    Any estimates for 100 UMS?
    I'm guessing 73.
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    (Original post by UbaidS)
    shouldnt it be 34?!
    No, you don't use the points given, you use the distance equation
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    (Original post by UbaidS)
    (x-5)^2-25+(y+3)^2-9=k
    (x-5)^2+(y+3)^2=34
    Thats how i did it
    That method only works when you are given the equation in the way that was already shown. You HAD to use the fact that it said the point was on the circle to find the radius. I almost made that mistake for k=34 but if you plug in the coordinates for the point on the circle then you see it doesn't make sense.
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    I believe the d^2y over dx^2 answer was for question 8a
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    Anyone know how much question 8 was worth in total??

    I got almost everything else correct from what I'm reading here but I was a dumbass and completely missed out question 8...

    Really afraid I won't get 80ums this year unless it's like 55 or something for an A which is unlikely tbh...
 
 
 
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