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# Year 13 Maths Help Thread

1. (Original post by solC)
Well since we know the points are on the circle the general point on that circle is . Using the exponential form works too as Physicsmaths said

Let

And the range of this is so K is 4
Thanks but why did u divide the rhs fraction by cos2+sin2?
2. (Original post by youreanutter)
Thanks but why did u divide the rhs fraction by cos2+sin2?
when you make the denominator real by multiplying top and bottom by that's what you get.
3. (Original post by Ano123)
I'll volunteer to be a helper. My purpose in this life is to serve...maths to people?
i've retaken maths 3 times now, i can't do it at all like i've been revising all the time - i dont even do half of my other work (science and other lessons) as im trying to pass maths but i cant do it and i need to pass for the apprenticeship i want to do
4. Am I the only one who finds that the 'definite integrals involving modulus function questions' in STEP I papers are some of the easiest questions that they can throw at you?
5. (Original post by savarna)
i've retaken maths 3 times now, i can't do it at all like i've been revising all the time - i dont even do half of my other work (science and other lessons) as im trying to pass maths but i cant do it and i need to pass for the apprenticeship i want to do
Is this GCSE or A level?
6. How were they able to make lamda equal that?

7. (Original post by amelienine)
How were they able to make lamda equal that?

ln both sides, leads you to -40lambda = ln(1/2)
Divide by -40, you get lambda = -ln(1/2)/40
-ln(1/2) = ln(1/2^-1) = ln(2)
8. (Original post by amelienine)
How were they able to make lamda equal that?

Take logs of both sides and remember
ln0.5 = ln(1/2) = ln(2^-1) = -ln2
9. (Original post by k.russell)
ln both sides, leads you to -40lambda = ln(1/2)
Divide by -40, you get lambda = -ln(1/2)/40
-ln(1/2) = ln(1/2^-1) = ln(2)
(Original post by solC)
Take logs of both sides and remember
ln0.5 = ln(1/2) = ln(2^-1) = -ln2
Thank you!! (I've just learnt how to multi-reply woohoo)

10. Took this from the mark scheme, don't know how T = 93 tho? Put it in my calculator and I'm getting a different answer.
11. (Original post by amelienine)
Took this from the mark scheme, don't know how T = 93 tho? Put it in my calculator and I'm getting a different answer.
That's wrong, it does not go to the second line.
12. (Original post by RDKGames)
That's wrong, it does not go to the second line.
then how do you find what T is?
13. (Original post by amelienine)
then how do you find what T is?
If you know your exponentials and natural logarithms, you may notice that and just divide both sides by it for T.

The -1 in front of ln becomes the exponent when moved inside the logarithm, so we take the reciprocal of the fraction as the result.
14. (Original post by RDKGames)
If you know your exponentials and natural logarithms, you may notice that and just divide both sides by it for T.

The -1 in front of ln becomes the exponent when moved inside the logarithm, so we take the reciprocal of the fraction as the result.
Oh thank you! I'm resitting my C3 this year and I've forgotten so much of it

And it turns out it wasn't ln(2/40)! It was (ln(2))/40, which gave out a completely different answer... oi I'm hopeless
15. this question is really ****ing me up, I've tried doing it SO many times using LCM and stuff but can't seem to get the right answer :

part a) A christmas tree has been decorated with a set of lights thatcontain two circuits. On each individual circuit, all the lights come and go off at the same time, but the two circuits are out of sync.

Circuit 1: lights ON for 4 seconds, and OFF for 9 seconds.
Circuit 2: lights ON for 7 seconds and OFF for 2 seconds,

20 seconds ago, both sets of lights came ON at exactly the same time. In how many seconds from now will both sets of lights go OFF at the same time ?

part b) consider the same christmas tree as above but with the following circuits:
Circuit 1: Lights ON for 3 seconds, OFF for 5 seconds
Circuit 2: lights ON for 2 seconds, OFF for 6 seconds
Ten seconds ago, both sets of lights went ON at exactly the same time. In how many seconds from now will both sets of lights go ON at the same time?

answers below, I just don't know how to get to them :
Spoiler:
Show
part a: 23
part b: they will never go OFF at the same time
16. (Original post by medhelp)
this question is really ****ing me up, I've tried doing it SO many times using LCM and stuff but can't seem to get the right answer :

part a) A christmas tree has been decorated with a set of lights thatcontain two circuits. On each individual circuit, all the lights come and go off at the same time, but the two circuits are out of sync.

Circuit 1: lights ON for 4 seconds, and OFF for 9 seconds.
Circuit 2: lights ON for 7 seconds and OFF for 2 seconds,

20 seconds ago, both sets of lights came ON at exactly the same time. In how many seconds from now will both sets of lights go OFF at the same time ?

part b) consider the same christmas tree as above but with the following circuits:
Circuit 1: Lights ON for 3 seconds, OFF for 5 seconds
Circuit 2: lights ON for 2 seconds, OFF for 6 seconds
Ten seconds ago, both sets of lights went ON at exactly the same time. In how many seconds from now will both sets of lights go ON at the same time?

answers below, I just don't know how to get to them :
Spoiler:
Show
part a: 23
part b: they will never go OFF at the same time
Just say
'I don't believe in christmas, question is inaccurate qed'

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17. 0 = x(9-2x^2)^(1/2)
x = root (9/2)
= root 9 / root 2
a = 3 / root 2

I have a feeling this isn't right, though. ' As a multiple of root 2 '

could I do 3root2/2 ( by mutlitplying both parts of fraction by root 2 ) or does this not make a difference?
18. (Original post by JaredzzC)

0 = x(9-2x^2)^(1/2)
x = root (9/2)
= root 9 / root 2
a = 3 / root 2

I have a feeling this isn't right, though. ' As a multiple of root 2 '

could I do 3root2/2 ( by mutlitplying both parts of fraction by root 2 ) or does this not make a difference?
Your method seems right, yes they probably want it as 3√2/2
19. Wut
20. (Original post by jamestg)
Wut
Take a logarithm of base 3 for both sides and rearrange for

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