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    anyone remember what question 2ii was? just trying to remember if i got the right answer.

    And also how to do question 5?
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    (Original post by imfailingbadly)
    Can anyone explain 8)(iv) i kinda assumed you did 0.5n(2a+(n-1)d) substituted in a=8 d=3 and the also changed n to 2N and N took them away - but i got really lost
    yeh you did do this so you had

    N(16 +(2N-1)x3) - 1/2N(16+(N-1)x3)=1256

    N(16 + 6N-3) - 1/2N(16 +3N-3) =1256 then multiply by 2 to make life easier

    2N(13+6N) - N(13+3N)=2512

    12N^2 +26N - 3N^2 +13N =2512

    9N^2 + 13N = 2512

    then i had to use the quadratic formula
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    didnt you just have to sub in y=5, then you got x=56, so p=56 ?
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    (Original post by insertusername)
    yeh you did do this so you had

    N(16 +(2N-1)x3) - 1/2N(16+(N-1)x3)=1256

    N(16 + 6N-3) - 1/2N(16 +3N-3) =1256 then multiply by 2 to make life easier

    2N(13+6N) - N(13+3N)=2512

    12N^2 +26N - 3N^2 +13N =2512

    9N^2 + 13N = 2512

    then i had to use the quadratic formula

    i did all of that but then foolish thought it was all wrong and scribbled it out... thanks for the help!
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    (Original post by imfailingbadly)
    didnt you just have to sub in y=5, then you got x=56, so p=56 ?
    yeah so how do you do that?
    i did

    5 = 3x^2 - 4x + 1
    3x^2 - 4x - 4 = 0
    (3x+2) (x-2)
    3x+2=0
    x = -2/3
    and x = 2??
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    (Original post by imfailingbadly)
    i did all of that but then foolish thought it was all wrong and scribbled it out... thanks for the help!
    haha unlucky and no probs, took me forever to do that bloody question, was driving me mad. It seemed far more complicated than it needed to be
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    (Original post by jkkk32)
    yeah so how do you do that?
    i did

    5 = 3x^2 - 4x + 1
    3x^2 - 4x - 4 = 0
    (3x+2) (x-2)
    3x+2=0
    x = -2/3
    and x = 2??
    sorry i meant you sub in x=5, and the question was find P when (5,P)!
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    alsoo anyone remember the numbers involved and question for question 5? i got the wrong answer, it was the integration one, but if anyone has done it correctly would you be able to explain it? thank you
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    (Original post by insertusername)
    haha unlucky and no probs, took me forever to do that bloody question, was driving me mad. It seemed far more complicated than it needed to be
    oh my god - i know what i did wrong, got the quadratic formula wrong! i thought it was divied by 2, not 2a! did you have to use the quadratic formula anywhere else? im worried now!
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    (Original post by imfailingbadly)
    sorry i meant you sub in x=5, and the question was find P when (5,P)!
    DAMN! so all you were meant to do was put 5 in? ARGH! lol i must have read it as (P,5) or something gargh lol
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    (Original post by jkkk32)
    DAMN! so all you were meant to do was put 5 in? ARGH! lol i must have read it as (P,5) or something gargh lol
    haha! well now, looking back im not sure! cause that seems like an easy 3 marks! wait for Mr M to clafify!
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    (Original post by insertusername)
    I also got p as this, there is no way on earth it can be 56 surely, -2/3 gave me the right answer when i substituted it back in

    as for 3 ii) i did get -35/4, you had to do (-1/4)^3 x 35 x 2^4 i think
    hmmm, i got 35/256 by...

    7C3 x 2^4 x (-0.25^3)^2

    Therefore I thought it was 7C3 x 2^4 x -0.25^6 = 35/256

    It's positive and not negative because a negative number raised to the 6 is positive.
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    (Original post by imfailingbadly)
    sorry i meant you sub in x=5, and the question was find P when (5,P)!
    fairly sure it wasn't and that it was (p,5), at least i hope it was

    and as for quadratic formula technically i used it on 2 but only due to my factorising being so poor, i don't think you actually needed to use it on another question but i may be wrong
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    (Original post by insertusername)
    fairly sure it wasn't and that it was (p,5), at least i hope it was

    and as for quadratic formula technically i used it on 2 but only due to my factorising being so poor, i don't think you actually needed to use it on another question but i may be wrong
    ah your probably right! yet another question i got wrong... lets wait for Mr M to tell us!
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    (Original post by mrdude)
    hmmm, i got 35/256 by...

    7C3 x 2^4 x (-0.25^3)^2

    Therefore I thought it was 7C3 x 2^4 x -0.25^6 = 35/256

    It's positive and not negative because a negative number raised to the 6 is positive.
    thats what i did! but apprentely its wrong... damn!
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    (Original post by mrdude)
    hmmm, i got 35/256 by...

    7C3 x 2^4 x (-0.25^3)^2

    Therefore I thought it was 7C3 x 2^4 x -0.25^6 = 35/256

    It's positive and not negative because a negative number raised to the 6 is positive.
    true but originally it was (2-1/4w^2) so you didn't need to actually square the -1/4 as well as cube it
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    (Original post by insertusername)
    true but originally it was (2-1/4w^2) so you didn't need to actually square the -1/4 as well as cube it
    clever.... !

    anyone got the question paper?
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    i have no idea if your right or im right lol

    to finish my answer i got x=2 as the right one, and then when you put that in you get y =5 which is correct. But who knows, we'll wait for Mr M to clarify. i feel so stupid if i read it wrong because i remember purposely thinking oh normally we would just put the x-value in like you did but this time we were given the y-value ....i think



    anyone remember how to do question 5? i got the wrong answer but cant remember the question
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    (Original post by insertusername)
    true but originally it was (2-1/4w^2) so you didn't need to actually square the -1/4 as well as cube it
    ahhhh of course. silly 2 marks lost there.
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    (Original post by imfailingbadly)
    clever.... !

    anyone got the question paper?
    pm me with email, i'll try to scan it in likie 10min or so
 
 
 
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