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    For the area of the triangle question, couldn't you have integrated the formula of the tangent using 25/8 and 0 as the boundaries? I got the same answer (39.0625) but people are saying that you couldn't do that
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    (Original post by lilrabbits)
    And the last one is 260 instead of 261 right?
    I got 261, but I rounded up
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    What a beautiful paper , I don't think the grade boundary for an A will be any less than 60.


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    (Original post by Chloebbbb)
    Urrgh how many marks do you think I would have lost? I got all the points right and plotted it correctly just stupidly did a curve
    Probably one mark. But it might have affected your gradient and intercept, what did you get?
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    (Original post by Studious_Student)
    For the area of the triangle question, couldn't you have integrated the formula of the tangent using 25/8 and 0 as the boundaries? I got the same answer (39.0625) but people are saying that you couldn't do that
    in some past papers they have shown that method, you should be fine
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    I think I got a like 50%. Worst exam ever
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    (Original post by ScalyJake)
    What intercept and gradient did you guys get for the log graph question? Obviously they'll all be a bit different, but roughly.

    I got 0.65 for the intercept and 0.185 for the gradient
    EXACTLY the same woooø
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    (Original post by Studious_Student)
    For the area of the triangle question, couldn't you have integrated the formula of the tangent using 25/8 and 0 as the boundaries? I got the same answer (39.0625) but people are saying that you couldn't do that
    yes, you can do this , you're right
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    For the trapezium integration question, I stupidly integrated the value before the brackets and carried on from there, how many marks will I lose accounting for method marks
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    (Original post by WuMyster)
    Also. My integrated area was massive. 144cm^2. I redid the integration 100 times but I don't see my mistake.
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    Sup, did exactly the same, you and I both integrated the value before the brackets right? You're not meant to do that, if you carried on and multiplied it by 50 then we'll probably get W for method marks
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    My reaction to that C2 exam:




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    i thought it was ok, hope the A boundary stays at 56 tho
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    (Original post by Studious_Student)
    Also if anyone was wondering, I got 0.6 as my log a and 0.21 as my gradient. In the end I got y = 3.891 x 10^0.21t
    How did you get the last answer when w=4? using this i got something like 25.11 but everyone got 260 ??
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    Did anyone get 4. Something for the sum of the sequence when A is 5, I'm sure it's not 25
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    (Original post by justnaomihere)
    How did you get the last answer when w=4? using this i got something like 25.11 but everyone got 260 ??
    I got 261(rounded up poorly) but all you had to do was just substitute w as 4 to get 260. something, then it has to be an integer so it is 260.
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    For the tangent did it specifically ask for where the line crosses the axis, I never wrote these in the first part but used them in the area question
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    (Original post by Studious_Student)
    I got 261(rounded up poorly) but all you had to do was just substitute w as 4 to get 260. something, then it has to be an integer so it is 260.
    Why can't you round up?
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    (Original post by Mr Moon Man)
    Why can't you round up?
    It asked for reported cases or viruses detected iirc, so you had to go with the lowest integer the decimal could provide. I'm not 100% sure on this though.
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    (Original post by Studious_Student)
    For the area of the triangle question, couldn't you have integrated the formula of the tangent using 25/8 and 0 as the boundaries? I got the same answer (39.0625) but people are saying that you couldn't do that
    I got the same thing using area of a triangle formula 1/2*b*h

    so 1/2*25*25/8

    Your method would take longer and isnt as efficient but is perfectly valid
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    (Original post by ineedA)
    Did anyone get 4. Something for the sum of the sequence when A is 5, I'm sure it's not 25
    It shouldn't be 4. Talk me through your method.
 
 
 
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