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    JAN 2009 EDEXCEL core 2
    i know its a simple question but i just dont seem to understand why you would need to find the gradient of QR to figure out a?

    The points P(−3, 2), Q(9, 10) and R(a, 4) lie on the circle C, as shown in Figure 2.
    Given that PR is a diameter of C,
    (a) show that a = 13,


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    (Original post by Tplox)
    Attachment 401919

    Part a I posted above, my phone is crap lol soz
    Here part b, it's a tray so you don't need the surface area of the top
    Use Pythagoras to find slanted edge
    Part a just form an equation to find volume


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    I don't understand the triangle how do you know the other side is x aswell


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    do you think they'll ask us a complicated binomial expansion question or just easyish ones like they have done in previous years?
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    (Original post by purplefoot)
    do you think they'll ask us a complicated binomial expansion question or just easyish ones like they have done in previous years?
    They would probably try make it harder, most likely they would add x^3=300 etc


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    (Original post by skittles100)
    JAN 2009 EDEXCEL core 2
    i know its a simple question but i just dont seem to understand why you would need to find the gradient of QR to figure out a?

    The points P(−3, 2), Q(9, 10) and R(a, 4) lie on the circle C, as shown in Figure 2.
    Given that PR is a diameter of C,
    (a) show that a = 13,


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    Best thing is to do a quick sketch then you would understand


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    hey everyone! so when doing applied differentiation questions i kind of stuggle to find the Volume or area. I know this may be seem easy, but can someone give me the formula(e) of volume that always works? is it = cross section area times the height?
    so confused, please help
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    (Original post by David_Waddell)
    I am also repeating this exam, I got an E in it last year
    I'm retaking this module too. I got a C last year. Just aiming for an A in c2 and c1.


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    (Original post by David_Waddell)
    I am also repeating this exam, I got an E in it last year
    You know how youre retaking your exam, what predicted grade goes towards uni?

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    (Original post by physicsmaths)
    Yes you can factorise it!!!!!!!!


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    Go on then
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    (Original post by enaayrah)
    2(x^3 - 8) ?
    Well yeah, obviously, but i meant you cant factorise it like you would factorise a three term quadratic
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    (Original post by IIEquinoxII)
    Go on then
    2(x-2)(x^2+2x+4)


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    (Original post by BaranSR)
    Best thing is to do a quick sketch then you would understand


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    I just dont know why you would use the gradient of RQ

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    Has anyone else tried the June 2014 (R) paper, i thought it was extremely difficult compared to other papers
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    (Original post by skittles100)
    I just dont know why you would use the gradient of RQ

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    Because of Circle geometry the line QR is the normal to QP at Q as Q is a right angle so the gradient of QR is the negative reciprocal of QP so if you work out the gradient of QP ((10-2)/(9+3) =2/3) the gradient of QR will therefore be -3/2 so if do ((10-4)/(9-a)=-3/2) you can solve to find a


    To be fair though the only reason i knew this was because you said gradient, the question doesn't make it that obvious that Q's a right angle
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    (Original post by skittles100)
    I just dont know why you would use the gradient of RQ

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    Hello mate, I'll be your tutor for the day.

    Firstly, The line PQ is.. a line. The line QR is a chord of the circle C.

    Now we know that when a line attaches to the chord, like PQ attaching to QR it forms a right angle at the point Q. (90 Degrees at Q). And because it is 90 Degrees we know that the lines are perpendicular to each other.

    Therefore if we work out the gradient of PQ. We know that the gradient of the line QR is going to be Perpendicular to the one of PQ. and we know that

    m1 x m2 = -1 - (gradient 1) x (gradient 2) = -1

    The gradient of line PQ is (y2-y1) // (x2-x1) therefore (10-2) // (9-(-3)) = 2/3

    So the gradient of the line QR is going to be perpendicular to this so:
    Gradient of QR = -1 // (2/3) (// = divided by)
    This equals -3 // 2. Gradient of QR = -3 // 2

    Now you know that to work out the gradient of QR without knowing the gradient of the line perpendicular to it (PQ) you can do (y2-y1) // (x2-x1) So do that with the co-ordinates for QR.. (4-10) // (a-9) = -3 // 2 --> (We already know the gradient of QR)

    Solve this to find a = 13 (4-10) // (a-9) = -3 // 2

    Simple.
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    (Original post by Bezaleel25)
    Because of Circle geometry the line QR is the normal to QP at Q as Q is a right angle so the gradient of QR is the negative reciprocal of QP so if you work out the gradient of QP ((10-2)/(9+3) =2/3) the gradient of QR will therefore be -3/2 so if do ((10-4)/(9-a)=-3/2) you can solve to find a


    To be fair though the only reason i knew this was because you said gradient, the question doesn't make it that obvious that Q's a right angle
    Yes it does make it obvious that Q's a right angle, as QR is a chord of the circle. It provides you with a diagram.

    When a line and a Chord meet (it is a right angle)
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    (Original post by MATTTT)
    Hello mate, I'll be your tutor for the day.

    Firstly, The line PQ is.. a line. The line QR is a chord of the circle C.

    Now we know that when a line attaches to the chord, like PQ attaching to QR it forms a right angle at the point Q. (90 Degrees at Q). And because it is 90 Degrees we know that the lines are perpendicular to each other.

    Therefore if we work out the gradient of PQ. We know that the gradient of the line QR is going to be Perpendicular to the one of PQ. and we know that

    m1 x m2 = -1 - (gradient 1) x (gradient 2) = -1

    The gradient of line PQ is (y2-y1) // (x2-x1) therefore (10-2) // (9-(-3)) = 2/3

    So the gradient of the line QR is going to be perpendicular to this so:
    Gradient of QR = -1 // (2/3) (// = divided by)
    This equals -3 // 2. Gradient of QR = -3 // 2

    Now you know that to work out the gradient of QR without knowing the gradient of the line perpendicular to it (PQ) you can do (y2-y1) // (x2-x1) So do that with the co-ordinates for QR.. (4-10) // (a-9) = -3 // 2 --> (We already know the gradient of QR)

    Solve this to find a = 13 (4-10) // (a-9) = -3 // 2

    Simple.
    Oh wow thanks!! I fully understand it now

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    (Original post by skittles100)
    Oh wow thanks!! I fully understand it now

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    Do we need to know any other circle rules for c2?

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    When solving sine and cosine, I was told to use the radian mode on the calculator, but when I was solving
    sin⁻¹(\sqrt 3 \div 2 ) I ended up getting a wrong answer, so I switched my calculator to degrees mode and got 60{\circ}
    so I'm confused now, when are we suppose to use radian mode?
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    (Original post by FluffyCherry)
    When solving sine and cosine, I was told to use the radian mode on the calculator, but when I was solving
    sin⁻¹(\sqrt 3 \div 2 ) I ended up getting a wrong answer, so I switched my calculator to degrees mode and got 60{\circ}
    so I'm confused now, when are we suppose to use radian mode?
    Look at the range they give you
    If it's in pii, then it's radian mode


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