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theone
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#41
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#41
(Original post by integral_neo)
didnt know that tan (x/2) will work as well.. how did u realise this?
It's a pretty common method for integrating with a number and a linear (i.e. no powers) trig function in the denominator if it's not obvious how to do it.
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Euler
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#42
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(Original post by theone)
It's a pretty common method for integrating with a number and a linear (i.e. no powers) trig function in the denominator if it's not obvious how to do it.
Are you doing Edexcel maths and F maths? what modules have u taken and what UMS did u get for each and which ones u will taking in June? r u gonna do extension paper?

do this one:

r = 3cox(2x)

find the length between the two parallel (to the initial line) tangents to the graph
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lgs98jonee
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#43
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(Original post by JamesF)
For the first one, write the consecutive numbers
n(n+1)(n+2)(n+3), then if n is odd, n+1, and n+3 are multiples of 2. Either n+1 or n+3 is a multiple of 4. n or n+2 must be a multiple of 3.

n - even. Same arguements.

2) A great trick i read. The highest power of a prime into a factorial is found as follows.
Sum from i=1 to infinity of [20/2^i]
Where [x] is the greatest integer function.

Lol, except it doesnt seem to work.
is i supposed to be sqrt(-1)?
wot year u in james?
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theone
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#44
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(Original post by integral_neo)
Are you doing Edexcel maths and F maths? what modules have u taken and what UMS did u get for each and which ones u will taking in June? r u gonna do extension paper?

do this one:

r = 3cox(2x)

find the length between the two parallel (to the initial line) tangents to the graph
I'm not going to go through this again I remember it from my P4 exam and took me half a side and I don't have time to regurgitate. I'll pm my marks/modules.
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Euler
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#45
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(Original post by theone)
I'm not going to go through this again I remember it from my P4 exam and took me half a side and I don't have time to regurgitate. I'll pm my marks/modules.
lolllll I spent so much time on it and got the right equation after differentiation but i just couldnt solve that equation then that was the only question i messed up on P4
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JamesF
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#46
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No, i is not sqrt(-1). Read what i said about the sum. Using it for this question we get

k = (sum from i=1 to infinity) [20/2^i]
2^5 > 20 so we only need to consider the cases, when i is 1,2,3,4.
[20/2] = 10
[20/4] = 5
[20/8] = 2
[20/16] = 1

Sum these, 10+5+2+1 = 18.

Im in Lower 6th.
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JamesF
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#47
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For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4
[20/25] = 0
So the answer is 4.
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Euler
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#48
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(Original post by JamesF)
For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4
[20/25] = 0
So the answer is 4.
the answers to both questions are indeed right
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lgs98jonee
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#49
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(Original post by JamesF)
For the second part of that question. For 10's to divide 20!, 2 and 5 must divide, so the highest power of 10 to divide 20! depends on the power of 5's to divide 20!, since there will be a higher number of 2's than 5's.

20/5 = 4
[20/25] = 0
So the answer is 4.
that is v. good...have u been taught that then? is so, wot module?
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lgs98jonee
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#50
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james did u do the maths challenge?
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JamesF
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#51
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It was in some university lecture notes on elementary number theory.
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lgs98jonee
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#52
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(Original post by JamesF)
It was in some university lecture notes on elementary number theory.
wow....where did u get them from and did u do the senior maths challenge?
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JamesF
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#53
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#53
No I didnt do SMC.
I cant remember where i found them, but google will throw up loads of stuff if you search.
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