Turn on thread page Beta
    Offline

    18
    ReputationRep:
    (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:
    Offline

    19
    ReputationRep:
    (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
    Offline

    18
    ReputationRep:
    (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?
    Offline

    19
    ReputationRep:
    (Original post by DylanJ42)
    would it get more attention if it had its own thread?
    it would I guess ...
    Offline

    18
    ReputationRep:
    (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
    Offline

    17
    ReputationRep:
    (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
    Offline

    19
    ReputationRep:
    (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 ...
    Offline

    19
    ReputationRep:
    (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:
    Show
    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°?
    Offline

    19
    ReputationRep:
    (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:
    Show
    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
    Offline

    19
    ReputationRep:
    (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...
    Offline

    19
    ReputationRep:
    (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
    Offline

    18
    ReputationRep:
    (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.
    Offline

    19
    ReputationRep:
    (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
    Offline

    18
    ReputationRep:
    (Original post by TeeEm)
    I am sorry
    dont be sorry, i am just relieved that i didnt get wrecked by a gcse question :laugh:
    Offline

    11
    ReputationRep:
    (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.
    Spoiler:
    Show
    radius = 2\sqrt{2}
    Offline

    11
    ReputationRep:
    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.
    Offline

    19
    ReputationRep:
    (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...
    Offline

    22
    ReputationRep:
    (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:
    Show
    Everything except \displaystyle \frac{1}{1 \times \frac{1}{2k+1}} = 2k+1 cancels.
    Offline

    19
    ReputationRep:
    (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:
    Show
    And I should really learn latex soon...
    Offline

    22
    ReputationRep:
    (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:
    Show
    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...?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: June 22, 2016

1,096

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.