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# Hard maths questions for higher maths GCSE watch

1. (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
2. (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
If people are trying I will delay it ...one or two days
3. (Original post by TeeEm)
If people are trying I will delay it ...one or two days
4. (Original post by DylanJ42)
it would I guess ...
5. (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
6. (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
7. (Original post by langlitz)
Can't even draw this. Was planning on using some linear algebra but it's not going too well
8. (Original post by TeeEm)
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°?
9. (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
10. (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...
11. (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
12. (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.
13. (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
14. (Original post by TeeEm)
I am sorry
dont be sorry, i am just relieved that i didnt get wrecked by a gcse question
15. (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:
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16. Find

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

where is an integer.
Are you sure that this is GCSE level? I don't even know where to start with this...
18. (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:

coughs cancel coughs

Spoiler:
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Everything except cancels.
19. (Original post by Zacken)
Simplify, the denominator is nothing but:

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...
20. (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 then that's any arbitrary integer that does not imply infinity in anyway.

Everything cancels out, but remember that in the there's something that cancels the as well, so you're only left with...?

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Updated: June 22, 2016
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