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    what's the question where the answer is N=24??
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    (Original post by czj1997)
    what's the question where the answer is N=24??
    it was find the value of N if 2750 = 2N above sigma and n = N below sigma.
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    (Original post by czj1997)
    what's the question where the answer is N=24??
    The one about summations from 2n to n where it equaled 2570 or something


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    (Original post by ella_chloe)
    This would be lovely boundaries


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    These are the grade boundaries all the way back to 2005
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    For the last question, i worked in degrees as i find it easier, then converted to radians at the end and got the same answers as the unofficial mark scheme, would that still be full marks?


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    Does anyone remember how many marks the binomial expansion question was worth?
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    How many marks will I drop on the last one for giving 3.33 rad and 12.75 rad?
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    What was q6 about
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    Hi can someone check through this working for the Sum from N to 2N being 2750 question? I can't understand why I don't get the same answer... :/
    Working:0.5(N+1)[2N+3N] = 2750N+1[5N] = 55005N2 + 5N - 5500 = 0N = \dfrac{-5 \pm \sqrt{5^2-4*5*-5500}}{2*5}
    Gave me an answer of like 33.6??I'm wondering where I went wrong...!Thanks!
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    (Original post by ella_chloe)
    For the last question, i worked in degrees as i find it easier, then converted to radians at the end and got the same answers as the unofficial mark scheme, would that still be full marks?


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    yes full marks
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    (Original post by shizhang)
    Does anyone remember how many marks the binomial expansion question was worth?
    No sorry
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    (Original post by jampot98)
    These are the grade boundaries all the way back to 2005
    this gives me hope thank you


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    (Original post by shizhang)
    Does anyone remember how many marks the binomial expansion question was worth?
    7 in total

    4 for the first
    3 for the second
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    (Original post by ella_chloe)
    For the last question, i worked in degrees as i find it easier, then converted to radians at the end and got the same answers as the unofficial mark scheme, would that still be full marks?


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    Yeah it would be


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    (Original post by gabby07)
    Hi can someone check through this working for the Sum from N to 2N being 2750 question? I can't understand why I don't get the same answer... :/
    Working:0.5(N+1)[2N+3N] = 2750N+1[5N] = 55005N2 + 5N - 5500 = 0N = \dfrac{-5 \pm \sqrt{5^2-4*5*-5500}}{2*5}
    Gave me an answer of like 33.6??I'm wondering where I went wrong...!Thanks!
    I did exactly the same thing, how many marks will we get?
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    Xsanda's solutions:
    1i) common ratio:[1]
    r=-2
    1ii) 11th term: [2]
    ar¹⁰ = 3072
    1iii) sum to 20: [2]
    Sum to 20 = -1048575

    2i) trapezium rule: [4]
    21.4
    2ii) improvement: [1]
    more/narrower strips

    3i) shaded area: [4]
    32.1
    3ii) perimeter: [5]
    22.7

    4i) Binomial expansion: [4]
    64+192ax+240a²x²+…
    4ii) Find a: [5]
    a=2/3

    5) Double antidifferential: [7]
    y=4x^(3/2)-7x-3

    6i) Factorise: [4]
    (x-2)(x+5)(x-3)
    6ii) Integrate: [4]
    256
    6iii) Problem with integral: [2]
    negative section of integration counts negatively

    7i) u₂₀: [2]
    62
    7ii) Σ 10 to 20 show that: [3]
    517
    7iii) Σ N to 2N: [6]
    N=24

    8a) log 2^(n-3) = 18000: [4]
    n=17.1
    8b) simultaneous logs: [5]
    x=32, y=8

    9ia) Next solution of cos: [1]
    6π-α
    9ib) Solution with negative: [2]
    3π-a
    9ii) Draw graph: [2]
    graph of sin, half height
    9iii) Solve graphs: [4]
    3.32, 12.7
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    (Original post by xsanda)
    Xsanda's solutions:
    R=-2
    ar¹⁰ = 3072
    Sum to 20 = -1048575

    trapezium: 21.4
    narrower strips

    triangle area: 32.1
    perimeter: 22.7

    expansion: 64+192ax+240a²x²+…
    a=2/3

    y=4x^(3/2)-7x-3
    (x-2)(x+5)(x-3)
    256
    negative section of integration counts negatively

    u₂₀ = 62
    Σ = 517 (show that)

    N=24
    n=17.1
    x=32, y=8

    6π-α
    3π-a
    graph of sin, half height
    3.32, 12.7
    What question is the n =17.1 from?


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    (Original post by GM16)
    What question is the n =17.1 from?


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    the first logarithm one
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    (Original post by GM16)
    What question is the n =17.1 from?


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    See post above this, questions and marks added in
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    (Original post by xsanda)
    Xsanda's solutions:1i) common ratio:[1] r=-21ii) 11th term: [2] ar¹⁰ = 30721iii) sum to 20: [2] Sum to 20 = -10485752i) trapezium rule: [4] 21.42ii) improvement: [1] more/narrower strips3i) shaded area: [4] 32.13ii) perimeter: [5] 22.74i) Binomial expansion: [4] 64+192ax+240a²x²+…4ii) Find a: [5] a=2/35) Double antidifferential: [7] y=4x^(3/2)-7x-36i) Factorise: [4] (x-2)(x+5)(x-3)6ii) Integrate: [4] 2566iii) Problem with integral: [2] negative section of integration counts negatively7i) u₂₀: [2] 627ii) Σ 10 to 20 show that: [3] 5177iii) Σ N to 2N: [6] N=248a) log 2^(n-3) = 18000: [4] n=17.18b) simultaneous logs: [5] x=32, y=89ia) Next solution of cos: [1] 6π-α9ib) Solution with negative: [2] 3π-a9ii) Draw graph: [2] graph of sin, half height9iii) Solve graphs: [4] 3.32, 12.7
    Wouldn't the integration one be -206.5 or -207.5 (can't remember which) because the next question asked why this was not the area under the curve between -5 and 3?
 
 
 
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