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    (Original post by timtjtim)
    Sure, I've attached an image (tangent.png) to make it clearer.

    The area of the integral (dark grey and light grey) was 81/4 or 20.25.

    But we don't want the light grey section (because we want it bounded by the tangent). So, using the coordinate of Q(-5/4,0) we determine that the base of the triangle is 3/4. The height it 24, and (24 x 0.75) / 2 = 9 (second image)

    For my workings, I wrote that 9 = 36/4, and then 81/4 - 36/4 = 45/4 or 11.25
    I got 243/12 for first part?
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    (Original post by ZT006409)
    no it was 9
    The length AC was. The length CT was 9.

    Name:  circle2.png
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    (Original post by beanigger)
    thats wrong because we didnt originally complete the square
    you are jumping to the second half of completing the square which is not what we were asked to do

    completing the square
    1) make sure coefficient of x^2 = 1
    2) take out a factor of x from x^2 and bx
    3) half the coefficient of bx
    4) raise the new b and x to the power of 2
    5) and then minus the square of the new b

    you jumped to number 5 without doing all the before, how is that right
    thats like doing this

    y=x^3 +5x
    dy/dx = 3x^2
    you see where its wrong? The 5x is differentiated twice this is the same as what u done, jumped the gun
    doont ever tell me im wrong , you beanhead
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    (Original post by medicine71012)
    How did u find the coordinates of B
    in the question they said AB was diameter of circle
    so that means C was the midpoint of line AB
    we have co ordinates of C
    to work out midpoint (x1+x2)/2 and (y1+y2)/2

    make the co ordinates of C equal to the respective formula for x and y and u get the value of B
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    (Original post by WhiteTigerKURD)
    I got 11.25
    SAME
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    Integration question seems to be 11.25 then as that is also what I got.
    And KittyKat97, yes it is 45 lol.
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    (Original post by money-for-all)
    doont ever tell me im wrong , you beanhead
    lol
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    (Original post by timtjtim)
    take a look at my image i've attached for you bruv. As you can see, you go to the side 7 and you're uppin and downin by 4 - well 4^2 + 7^2 = 16+49 = 65.

    I've attached a second image, of desmos, showing that the only circle which intersects a has radius of 65 - waddya think? If you're still not getting the hang of these circles, feel free to drop me a private message bro.
    ooooooo , check you fancy pants , desmos and ****, you still clapped tho
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    (Original post by iamastudent123)
    I got 243/12 for first part?
    It is the same 243/12 = 81/4 (divided by 3)
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    (Original post by beanigger)
    in the question they said AB was diameter of circle
    so that means C was the midpoint of line AB
    we have co ordinates of C
    to work out midpoint (x1+x2)/2 and (y1+y2)/2

    make the co ordinates of C equal to the respective formula for x and y and u get the value of B
    Ahhhhh is that the only way and how many marks?
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    (Original post by medicine71012)
    How did u find the coordinates of B
    The line AB is the diameter of the circle.

    We know that A to C is sqrt(65). We also know that this is the radius.

    Hence, C is the midpoint between A and B.
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    (Original post by medicine71012)
    Ahhhhh is that the only way and how many marks?
    that is the way i done it i am not sure of any other ways sorry xd and it was worth 2 marks
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    that's what i got as well
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    I blanked on the question of the circles where the coordinates of B were asked.
    Last minute i realised all i had to was midpoint
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    I thought it was actually horrendous. Missed out 5,6 and part of 7. If I get 100% of what I did I'll get a low B/C. Defo resorting next year
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    (Original post by christineedaley)
    How did you guys find the value of k? (in questions 6 i think) I just guessed and put k=4 :/
    well there were two questions involving k, the one with the circles and k was the radius squared, and the other where coordinates where you simply had to substitute the coordinates in ur equation.

    the one with the circles was root 65
    and the other was -30
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    (Original post by money-for-all)
    oh my days tim tim , you are clapped fam.

    Why you bringing pythagoras into this ? Thats just the same as saying you use logarithms to integrate...
    You had to use pythagoras to work it out. Radius=root 65

    Therefore,
    K=65
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    (Original post by musslih)
    OH MY GOSH that was the hardest paper...... it was kindof harder than Past Year Papers.... Can we discuss some stuff here?

    1. Question 5b, what was your coordinate of B?

    2. For the integration question ( was it Q7? or 8?)
    a)what was your x- coordinate for Q?
    b)what was your equation for the tangent at P?
    c) what was your area under the graph? (referring to the same qn)
    1. the coordinates were 12, -7 i think or something like that
    2a) x coordinate was -5/4
    b)y= 32x+40
    c) that was 135/12 = 45/4 = 11.25
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    Sorry but Tim's right
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    (Original post by Poppy1098)
    I also got root 34
    you had to use pythagaros which got you route 64
 
 
 
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