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    (Original post by fpmaniac)
    Can anyone help me on how to solve sigma r=0 to r=1. Someone explained it before but i couldnt fully get it. Maybe give me some examples with n=1,2,3 etc. Thanks
    If you look at the screenshot I have put examples in.

    All you do is add on what r=0 is to the the summation for your soultion.

    Hope this helps!
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    (Original post by iMacJack)
    Should be (12,12) not (27,12)
    Thought it looked too easy

    For some reason I used p=3 when I got the x coordinate, whoops
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    (Original post by Music With Rocks)
    Just to check when rearranging matrices does the term always go to the front so if
    AB=C

    A=B-1C
    and
    B=A-1C

    Or is this wrong? it looks wrong
    It depends on what side your value is on, if we have

    XA = B
    and we have to isolate the 'X'
    then:
    XAA^-1 = BA^-1
    XI = BA^-1
    X = BA^-1

    Basically you apply it to the same side of the term on either side
    (Sorry for the bad explanation!)
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    (Original post by TheMoon)
    I don't understand, isn't it like two right angled triangles which you can use 1/2 base times height with?

    I tried drawing it, obviously I've drawn it very inaccurately (terrible paint skills) since it doesn't look like a normal but is it something like this? Might be wrong. Sorry if I'm completely wrong with how it looks.
    This gets 225 anyway.
    If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
    I drew a diagram and it was pretty much irregular as far as I could tell
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    Would you get credit for using this in the exam? TeeEm, Zacken ?

    (Original post by WhatIsSleep)
    For that PQS question, I found an area formula which helped me work it out..

    (I have attached it as an image but the source is http://www.mathopenref.com/coordtrianglearea.html)

    So A is (3, 0)
    B is (12, 12)
    and C is (27, -18)

    Ax=3, Ay=0
    Bx=12, By=12
    Cx=27, Cy=-18

    Subbing those into the formula, I got 225 units^2

    Honestly though, I tried drawing diagrams and everything...couldn't find a better way of doing it tbh
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    (Original post by Brailey)
    If you look at the screenshot I have put examples in.

    All you do is add on what r=0 is to the the summation for your soultion.

    Hope this helps!
    Where did the 2n come from in the end equation. If u add r=0 shouldnt it be just +2
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    (Original post by WhatIsSleep)
    If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
    I drew a diagram and it was pretty much irregular as far as I could tell
    yeah sorry, I used p=3 for some reason when getting the x coordinate. not sure then

    There has to be an easier way though.
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    (Original post by TrueDAN)
    That is a beastly question haha!! Just done the first part now! How far have you got?
    tbh I think that's fp3. They introduce Loci in FP2 so FP1 takers wouldn't know what a loci of points is
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    (Original post by iMacJack)
    It depends on what side your value is on, if we have

    XA = B
    and we have to isolate the 'X'
    then:
    XAA^-1 = BA^-1
    XI = BA^-1
    X = BA^-1

    Basically you apply it to the same side of the term on either side
    (Sorry for the bad explanation!)
    Oh I think I get what you mean so

    AB=C
    A=CB-1

    AB=C
    B=A-1C


    Is this correct now?
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    (Original post by WhatIsSleep)
    If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
    I drew a diagram and it was pretty much irregular as far as I could tell
    Just draw a diagram, find the gradients of SP and SQ and u will see that they are perpendicular, then its just half the area of the rectangle using length of a line formula to work out the lengths
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    (Original post by TheMoon)
    yeah sorry, I used p=3 for some reason when getting the x coordinate. not sure then

    There has to be an easier way though.
    Maybe it's something that's not on the UK FP1 spec :/
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    (Original post by Music With Rocks)
    Oh I think I get what you mean so

    AB=C
    A=CB-1

    AB=C
    B=A-1C

    Is this correct now?
    Correct
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    (Original post by iMacJack)
    ii A is just finding the Det M, because the det is the scale factor, so ii (a) is 4, isn't it? Aren't you talking about part ii (b)?
    Could you explain a bit more in depth please?
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    (Original post by iMacJack)
    Correct
    Thank you very much you explained it well
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    (Original post by TheRandomGenius)
    6b anyone? (i and ii). I've gone around in circles and ended up in an algebraic mess.

    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
    did you figure it out?
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    Nope.
    (Original post by AmarPatel98)
    did you figure it out?
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    (Original post by connorbarr)
    Just draw a diagram, find the gradients of SP and SQ and u will see that they are perpendicular, then its just half the area of the rectangle using length of a line formula to work out the lengths
    ahh seems you're right :P
    I didn't think to find the gradients of the lines
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    (Original post by TheRandomGenius)
    Nope.
    I've done 6bi, i'm just doing ii. I'll post in a minute.
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    (Original post by TheRandomGenius)
    Nope.

    So you get the tangent at P from part a.

    Similarly you get the tangent at Q.

    The gradient of the tangent at P is 1/p, so similarly, the gradient at Q is 1/q. The two tangents are perpendicular, so you know their gradients must multiply to give -1. Form an eqn from this.

    You are told the intersection point, so form another eqn from this.

    Solve simultaneously to obtain p and q
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    (Original post by TheRandomGenius)
    Nope.
    I think you would make the gradient of p equal to the -1/gradient of q then you would substitute back into the equation to find p and q and substitute the R coordinates
 
 
 
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