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    (Original post by TeeEm)
    then it is officially becomes available to any A level student !!
    Definitely it is hard to model ...
    do you have a rough time period for when the solution will go up?

    As you probably guessed from me asking about the bearing of NE earlier I cant even get the first 3 facts modeled :laugh:
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    (Original post by DylanJ42)
    do you have a rough time period for when the solution will go up?

    As you probably guessed from me asking about the bearing of NE earlier I cant even get the first 3 facts modeled :laugh:
    If people are trying I will delay it ...one or two days
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    (Original post by TeeEm)
    If people are trying I will delay it ...one or two days
    would it get more attention if it had its own thread?
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    (Original post by DylanJ42)
    would it get more attention if it had its own thread?
    it would I guess ...
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    (Original post by TeeEm)
    it would I guess ...
    maybe that's an idea, a lot of the really good alevel students probbaly wont see it in a hard gcse questions forum
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    (Original post by TeeEm)
    it would I guess ...
    Can't even draw this. Was planning on using some linear algebra but it's not going too well
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    (Original post by langlitz)
    Can't even draw this. Was planning on using some linear algebra but it's not going too well
    hopefully your face is not like your avatar ...
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    (Original post by TeeEm)
    hopefully your face is not like your avatar ...
    Could you steer us in the right direction, perhaps? I've done a drawing of this, though I'm not sure if that is even how it's supposed to look...
    Spoiler:
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    Are A, C and B all on the same line if the bearing of C from A is 075° and the bearing of B from C is also 075°? And if this is true, how is it possible for the bearing of B from A to be 045°?
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    (Original post by TheOtherSide.)
    Could you steer us in the right direction, perhaps? I've done a drawing of this, though I'm not sure if that is even how it's supposed to look...
    Spoiler:
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    Are A, C and B all on the same line if the bearing of C from A is 075° and the bearing of B from C is also 075°? And if this is true, how is it possible for the bearing of B from A to be 045°?
    terrible typo on my behalf (I am notorious) due to copy and paste
    I will post a new thread
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    (Original post by TeeEm)
    terrible typo on my behalf (I am notorious) due to copy and paste
    I will post a new thread
    I'll probably still have no clue about what's happening...
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    (Original post by TheOtherSide.)
    I'll probably still have no clue about what's happening...
    I changed it here but I will post a new thread
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    (Original post by TeeEm)
    terrible typo on my behalf (I am notorious) due to copy and paste
    I will post a new thread
    thank goodness, i thought i was being really silly. i couldn't understand how to sketch it at all.
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    (Original post by DylanJ42)
    thank goodness, i thought i was being really silly. i couldn't understand how to sketch it at all.
    I am sorry
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    (Original post by TeeEm)
    I am sorry
    dont be sorry, i am just relieved that i didnt get wrecked by a gcse question :laugh:
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    (Original post by atsruser)
    Here's one I've just made up, TeeEm style:

    A sphere of radius 1 is placed on top of a sphere of radius 2 so that the line joining their centres is vertical. Find the radius of the base of the smallest right circular cone in which they will fit, with both spheres in contact with the cone.
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    radius = 2\sqrt{2}
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    Find

    \displaystyle \frac{1}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{2}{5})(1-\frac{2}{7})(1-\frac{2}{9})\cdots(1-\frac{2}{2k+1})}

    where k is an integer.
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    (Original post by atsruser)
    Find

    \displaystyle \frac{1}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{2}{5})(1-\frac{2}{7})(1-\frac{2}{9})\cdots(1-\frac{2}{2k+1})}

    where k is an integer.
    Are you sure that this is GCSE level? I don't even know where to start with this...
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    (Original post by TheOtherSide.)
    Are you sure that this is GCSE level? I don't even know where to start with this...
    Simplify, the denominator is nothing but:

    \displaystyle 

\begin{equation*}\frac{1}{2} \times \frac{2}{3} \times \frac{3}{5} \times \frac{5}{7} \times \cdots \times \frac{2k-3}{2k-1}\times \frac{2k-1}{2k+1}\end{equation*}

    coughs cancel coughs

    Edit, solution if you want to check your answer:
    Spoiler:
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    Everything except \displaystyle \frac{1}{1 \times \frac{1}{2k+1}} = 2k+1 cancels.
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    (Original post by Zacken)
    Simplify, the denominator is nothing but:

    \displaystyle 

\begin{equation*}\frac{1}{2} \times \frac{2}{3} \times \frac{3}{5} \times \frac{5}{7} \times \cdots \times \frac{2k-3}{2k-1}\times \frac{2k-1}{2k+1}\end{equation*}

    coughs cancel coughs
    Alright, so those fractions in the first part cancel each other out, so that you get 1/infinity, and that first part is negligible?

    And then the (2k - 1) terms cancel each other out so that you get (2k - 3)/(2k + 1)?

    I really don't know...
    Spoiler:
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    And I should really learn latex soon...
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    (Original post by TheOtherSide.)
    Alright, so those fractions in the first part cancel each other out, so that you get 1/infinity, and that first part is negligible?

    And then the (2k - 1) terms cancel each other out so that you get (2k - 3)/(2k + 1)?

    I really don't know...
    Spoiler:
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    And I should really learn latex soon...
    There's no infinity involved here. If I'm using k then that's any arbitrary integer that does not imply infinity in anyway.

    Everything cancels out, but remember that in the \cdots there's something that cancels the 2k-3 as well, so you're only left with...?
 
 
 
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