Year 13 Maths Help Thread

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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?

Reply 981

amelienine

How were they able to make lamda equal that?

Reply 982

k.russell

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)

Reply 983

solC

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

Reply 984

amelienine

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)

Reply 985

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.

Reply 986

RDKGames

Study Forum Helper

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.

Reply 987

amelienine

Original post by RDKGames

That's wrong, it does not go to the second line.

then how do you find what T is?

Reply 988

RDKGames

Study Forum Helper

Original post by amelienine

then how do you find what T is?

If you know your exponentials and natural logarithms, you may notice that

e^{-\ln(\frac{2}{40})}=e^{\ln(\frac{40}{2})}=20

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.

(edited 7 years ago)

Reply 989

amelienine

Original post by RDKGames

If you know your exponentials and natural logarithms, you may notice that

e^{-\ln(\frac{2}{40})}=e^{\ln(\frac{40}{2})}=20

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 :frown:

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

(edited 7 years ago)

Reply 990

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

Reply 991

username1162972

Original post by medhelp

Spoiler

Just say
'I don't believe in christmas, question is inaccurate qed'

Posted from TSR Mobile

Reply 992

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?

Reply 993

metrize

Original post by JaredzzC

Your method seems right, yes they probably want it as 3√2/2

Wut

Original post by jamestg

Wut

Take a logarithm of base 3 for both sides and rearrange for

\log_3(2)

Reply 996

jamestg

Original post by RDKGames

Take a logarithm of base 3 for both sides and rearrange for

\log_3(2)

Ahhh, thanks!

Reply 997

jamestg

Also, what does this even mean...

Screen Shot 2016-10-18 at 18.48.49BST.png

Reply 998

solC

Original post by jamestg

Also, what does this even mean...

Screen Shot 2016-10-18 at 18.48.49BST.png

If p(1)=0 then (x-1) is a factor of p(x) from the remainder theorem, so we can rewrite p(x) as q(x)(x-1) for some polynomial q(x).
In this case, dividing p(x) by (x-1) gives a remainder of 2, so we need to add two in order to get a remainder of 0, therefore p(x)=(x-1)q(x)-2.

I'm not 100% sure on this though...

(edited 7 years ago)

Reply 999

RDKGames

Study Forum Helper

Original post by jamestg

Also, what does this even mean...

Well firstly we can eliminate options which prevent p(1)=2 and then investigate the remaining ones, you can quickly see that all of them depend on various conditions satisfied by q(1) apart from one which will always be true, so I'd go for that one.