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    This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3
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    (Original post by Someboady)
    Hey
    Use part (i) in reverse
    take a=3 and b = 2. take your n as 2014.
    Then you can make that equal to the brackets given in part (i)

    i.e.

    (3-2)(3^2014 + .... + 2 ^ 2014)

    The first bracket is equal to 1. This implies that the number is prime as a prime number is only divisible by one and itself.
    Hope this helps
    Hey thanks for your help, your way makes sense to me but the mark scheme writes it as (3^5-2^5)(3^2010.........2^2010) could you explain how they reached this?? thanks again
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    (Original post by Mystery.)
    Yeah I am just doing it, what did you get?
    a,b,c,a,c,b,d,d,b,b

    Posted from TSR Mobile
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    (Original post by joodaa)
    Hey thanks for your help, your way makes sense to me but the mark scheme writes it as (3^5-2^5)(3^2010.........2^2010) could you explain how they reached this?? thanks again
    Oh balls, sorry my answer was incorrect (I deleted the post)... I'm having a look at it now xD.. I just did the question and I thought I'd gotten it right without checking the solutions. Sorry! Working on it right now!
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    (Original post by Mystery.)
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    This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3
    Rather than giving the triangle a side length, express the area of the triangle as a function of the radius of the circle
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    Oh and how is 9^log3x = 4^log2x????
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    (Original post by AlphaQuark)
    Rather than giving the triangle a side length, express the area of the triangle as a function of the radius of the circle
    ok i'll try
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    (Original post by Mystery.)
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    This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3
    I got area of triangle as x^2/4 where x is side of traingle
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    (Original post by Someboady)
    Oh balls, sorry my answer was incorrect (I deleted the post)... I'm having a look at it now xD.. I just did the question and I thought I'd gotten it right without checking the solutions. Sorry! Working on it right now!
    ooo okay no worriess
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    (Original post by Mystery.)
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    This is n't even a particularly difficult question but I keep getting 90% which is obviously wrong...I made the triangles side equal to 2 then got area of circle as pi/3 and area of triangle as 2/sqrt3
    This is a guess but...

    The radius of the circle seems to be about 1/3 of the height of the triangle.
    Assume the triangle has length l. Then the height of the triangle is \sqrt{l^2 - (\frac{1}{2}l)^2} = \frac{\sqrt{3}}{2}l

    The radius of the circle: r \approx \frac{\sqrt{3}}{{6}}l and hence the area A_c \approx \frac{3\pi}{36}l^2= \frac{\pi}{12}l^2

    The area of the triangle: A_t \approx \frac{\sqrt{3}}{4}l^2

    Dividing through we get \frac{A_c}{A_t} \approx \frac{4\pi}{12\sqrt{3}} = \frac{\pi}{3\sqrt{3}} \approx 60\%
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    (Original post by Nonamebzja)
    I got area of triangle as x^2/4 where x is side of traingle
    how?
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    (Original post by some-student)
    This is a guess but...

    The radius of the circle seems to be about 1/3 of the height of the triangle.
    Assume the triangle has length l. Then the height of the triangle is \sqrt{l^2 - (\frac{1}{2}l)^2} = \frac{\sqrt{3}}{2}l

    The radius of the circle: r \approx \frac{\sqrt{3}}{{6}}l and hence the area A_c \approx \frac{3\pi}{36}l^2= \frac{\pi}{12}l^2

    The area of the triangle: A_t \approx \frac{\sqrt{3}}{4}l^2

    Dividing through we get \frac{A_c}{A_t} \approx \frac{4\pi}{12\sqrt{3}} = \frac{\pi}{3\sqrt{3}} \approx 60\%
    How do you do the actual percentage without a calculator?
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    (Original post by Mystery.)
    how?
    Sorry (root3 × x^2)/4
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    (Original post by Mystery.)
    ok i'll try
    Ok so I did that but I still have pi from the area of circle formula
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    (Original post by Mystery.)
    How do you do the actual percentage without a calculator?
    Pie/3root3 i just did 1/root3 which is root3/3
    (1.7×100)÷3 approx 60%😂
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    (Original post by Mystery.)
    Oh and how is 9^log3x = 4^log2x????
    9^{\log_3x} = 3^{2\log_3x} = 3^{\frac{2\log x}{\log 3}} = (3^{\frac{1}{\log 3}})^{2\log x} = 10^{2\log x} (10 assuming that log is to the base of 10... the reason for this is because 3^{\frac{1}{\log 3}} is like taking the \log 3 root of 3, which will go in 10 times, assuming log is to the base of 10)

    4^{\log_2x} = 2^{2\log_2x} = 2^{\frac{2\log x}{\log 2}} = (2^{\frac{1}{\log 2}})^{2\log x} = 10^{2\log x}

    They are equal
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    (Original post by Mystery.)
    How do you do the actual percentage without a calculator?
    The circle is the incircle of the triangle. There is a formula saying that area = (a+b+c/2) x radius of incircle (I'm pretty sure u could get a solution without these formulas but I'm lazy)

    thus the ratio is pi*r^2/area = pi*r^2/sr where s is the semiperimeter (a+b+c/2)

    r's cancel then we get pi*area/s^2. s=3a/2 area = sqrt3 / 4 * a^2

    put back into formula then after lots of cancelling sqrt3*pi/9 which is roughly 3sqrt3/9 = sqrt3/3 which is about 2/3 so 60%
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    (Original post by Mystery.)
    Oh and how is 9^log3x = 4^log2x????
    So 9 = 3^2
    9^log3x=3^log3x^2 which is same as x^2
    Same with rhs
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    Thanks so much everyone.
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    (Original post by joodaa)
    Could someone explain q2 part IV of the 2015 paper? I dont understand how you can factorise it the way the mark scheme has
    cheersss
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