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    (Original post by beanigger)
    ok ive made corrections


    you use area of triangle = 1/2ab and bobs ur uncle
    **** me are you serious? I did that at first with the wrong co-ordinates, got the right co-ordinates then assumed that method was still wrong. I thought E^x would be curved but now I realise then it woudn't be a ****ing triangle

    I'm an arse

    I guess that doesn't matter, I'd need another 15 marks probably to get an A* and to get an A overall I need C's in the other papers. A* would be out of reach regardless
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    yeah lol from O to A is a straight line
    so is O to B so that forms a right angle
    hence triangle
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    (Original post by Sirelis)
    fak i got like 40 if I'm lucky
    Same lmao
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    (Original post by beanigger)
    yeah lol from O to A is a straight line
    so is O to B so that forms a right angle
    hence triangle
    Honestly I thought that the first time, then when I saw it was getting nowhere I changed my mind. Looking through this mark scheme I got between 45 and 49, Having gotten 70+ on every past paper I've done in the past week. Can't believe this.
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    Any ideas as to what the grade boundaries will be like ? considering last years was 57 for an A, hopefully this years will be lower.
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    roughly the same, but it will not be lower than 55 for an A
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    (Original post by irMike)
    Honestly I thought that the first time, then when I saw it was getting nowhere I changed my mind. Looking through this mark scheme I got between 45 and 49, Having gotten 70+ on every past paper I've done in the past week. Can't believe this.
    Well I think I got from around 60-64 assuming the worst; but that's assuming the way im distributing marks in questions I got wrong is correct; so that I get the method marks at least. By the way for the question leading up the sinx=2/3 or something; was that a question with a quadratic equation where you had to reject one of the values or am I remembering wrong. AND WHAT WAS THE QUESTION ITSELF????????
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    2e - integrate 5 then do your answer minus the answer from simpsons rule
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    (Original post by Anon998)
    2e - integrate 5 then do your answer minus the answer from simpsons rule
    For that one you just get the answer from the previous question - 5(1.7-0.5); and since that is the value of the integral of the negative x^x you just change signs and it gives you 1.51 or something.
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    (Original post by Anon998)
    2e - integrate 5 then do your answer minus the answer from simpsons rule
    Ahhh that's how you do it, I just did the simpsons rule again haha, for 2 marks will this even gain anything?
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    well aqa maths dont uusally give a lot of follow on marks, its usually just one M1 mark for applying a certain method so make sure you add it up right
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    (Original post by matty9)
    Ahhh that's how you do it, I just did the simpsons rule again haha, for 2 marks will this even gain anything?
    Not sure because I think it said using the previous answer but I don't remember jack ****.
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    59 a*
    53 a
    47 b
    41 c
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    (Original post by beanigger)
    well aqa maths dont uusally give a lot of follow on marks, its usually just one M1 mark for applying a certain method so make sure you add it up right
    Do you remember the question for the one about the area of the triangle :C. I managed to get m and n=3 but I don't know how. I just wanna check what I messed up to see how many marks i lost.
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    (Original post by Parhomus)
    Not sure because I think it said using the previous answer but I don't remember jack ****.
    Yes it said 'hence find an estimate' but wasn't sure what to do so did simpsons ruls again.
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    b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]

    I got gg(x) = x^4
    as g(x) = 1/x^2

    can you double check that one, not sure it's right
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    (Original post by Sayless)
    b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]

    I got gg(x) = x^4
    as g(x) = 1/x^2

    can you double check that one, not sure it's right
    fg(x) means f(g(x)) so when you got the equation; g(x) wouldn't be 1/x^2 it would 1/x because the x in the f(x) just means the input and so g(x) could only be 1/x. I understand why it would make sense for it to be 1/x^2.
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    (Original post by Sayless)
    b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]

    I got gg(x) = x^4
    as g(x) = 1/x^2

    can you double check that one, not sure it's right
    I put the same but it's wrong.

    g(x) was 1/x because I forgot that in functions the entire x is replaced so it wasn't 1/x^2 can't remember the question so kinda hard to explain.
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    (Original post by matty9)
    Yes! I got sinx = 2/3 then used pythag to get cosx = root5/3

    Then subbing these into the orginal equation gave me -1/5 root 5 but I don't know how correct that is
    sub into sin^2(x) +cos^2(x) = 1
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    (Original post by alikhaliq)
    sub into sin^2(x) +cos^2(x) = 1
    Why sub into that?? The eqn was something like f(x) = sec x + tan x so you had to do (1 / cos x) + (sin x / cos x) right?
 
 
 
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