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    (Original post by Galileo Galilei)
    Mathematical logic
    That's very helpful.
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    What exam board are you?
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    (Original post by zed963)
    That's very helpful.
    I am just kidding, I worked it out using the information given in the question
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    (Original post by Galileo Galilei)
    I am just kidding, I worked it out using the information given in the question
    Explain please.
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    (Original post by zed963)
    That's very helpful.
    He literally subbed x=p into the equation for l1.
    (Original post by Super199)
    How do I do part d? Any hints on how to start?
    Thanks
    I did this question, but I can't remember where, so if you could tell me the paper you're doing OP, that would be great.


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    (Original post by zed963)
    Explain please.
    Plug the value of P into the equation for L1 to find Y
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    If C has x-coordinate p, then it has y-coordinate -p/2 + 6.


    Therefore: (A,C) = √((2 - p)^2 + (5 + p/2 - 6)^2) = 5,


    √(4 - 4p + p^2 + (p/2 - 1)^2) = 5,


    √(4 - 4p + p^2 + (p^2)/4 - p + 1) = 5,


    √(16 - 16p + 4p^2 + p^2 - 4p + 4)/2 = 5,


    √(5p^2 - 20p + 20) = 10,


    5p^2 - 20p + 20 = 100,


    p^2 - 4p + 4 = 20,


    p^2 - 4p - 16 = 0.



    Hope this helps
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    (Original post by H0PEL3SS)
    He literally subbed x=p into the equation for l1.

    I did this question, but I can't remember where, so if you could tell me the paper you're doing OP, that would be great.


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    Your working out makes sense.
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    (Original post by BuddingAchiever)
    If C has x-coordinate p, then it has y-coordinate -p/2 + 6.


    Therefore: (A,C) = √((2 - p)^2 + (5 + p/2 - 6)^2) = 5,


    √(4 - 4p + p^2 + (p/2 - 1)^2) = 5,


    √(4 - 4p + p^2 + (p^2)/4 - p + 1) = 5,


    √(16 - 16p + 4p^2 + p^2 - 4p + 4)/2 = 5,


    √(5p^2 - 20p + 20) = 10,


    5p^2 - 20p + 20 = 100,


    p^2 - 4p + 4 = 20,


    p^2 - 4p - 16 = 0.



    Hope this helps
    Can you explain that?
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    (Original post by H0PEL3SS)
    He literally subbed x=p into the equation for l1.

    I did this question, but I can't remember where, so if you could tell me the paper you're doing OP, that would be great.


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    Jan 2009 . I've solved it now so all good. One more question I couldn't do was part c. Any hints on how to do this?
    I can't seem to work it out.
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    My solution.
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    (Original post by Super199)
    Jan 2009 . I've solved it now so all good. One more question I couldn't do was part c. Any hints on how to do this?
    I can't seem to work it out.
    Draw a triangle and think about it
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    (Original post by Galileo Galilei)
    Draw a triangle and think about it
    Using \frac{n}{2}(2a+(n-1)d I can't seem to get it into that form.
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    (Original post by Galileo Galilei)
    But that isnt Q10 C? Which question are you talking about?
    This one haha
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    (Original post by Super199)
    This one haha
    Got ya, do you want the solution or hints mate?
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    (Original post by Galileo Galilei)
    Got ya, do you want the solution or hints mate?
    A solution I guess. I mean I know what to do but I can't seem to get it in that form.
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    (Original post by Super199)
    Using \frac{n}{2}(2a+(n-1)d I can't seem to get it into that form.
    Just find an expression for n using the sum formula. You should double 2750 to get rid of the n/2. From there, expand and simplify.
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    For part C, take the equation of sum of n

    n/2(2a+(n-1)d)
    then substitute numbers

    n/2(-17.5(2)+2.5n-2.5)=2750

    times 2750 by 2

    n(-35+2.5n-2.5)= 5500

    expand
    -35n+2.5n^2-2.5n=5500

    times by 2

    -75n+5n^2=11000
    5n^2-75n-11000=0
    n^2-15n-2200=0

    Therefore

    n^2-15n-55x40
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    (Original post by BuddingAchiever)
    For part C, take the equation of sum of n

    n/2(2a+(n-1)d)
    then substitute numbers

    n/2(-17.5(2)+2.5n-2.5)=2750

    times 2750 by 2

    n(-35+2.5n-2.5)= 5500

    expand
    -35n+2.5n^2-2.5n=5500

    times by 2

    -75n+5n^2=11000
    5n^2-75n-11000=0
    n^2-15n-2200=0

    Therefore

    n^2-15n-55x40

    you can just check that 55x40=2200....if you can't do this question, at least you will be able to do part d
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    (Original post by Super199)
    A solution I guess. I mean I know what to do but I can't seem to get it in that form.
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