Year 13 Maths Help Thread

Announcements Posted on
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    Take a logarithm of base 3 for both sides and rearrange for \log_3(2)
    Ahhh, thanks!
    Offline

    3
    ReputationRep:
    Also, what does this even mean...

    Name:  Screen Shot 2016-10-18 at 18.48.49BST.png
Views: 19
Size:  20.5 KB
    Online

    2
    ReputationRep:
    (Original post by jamestg)
    Also, what does this even mean...

    Name:  Screen Shot 2016-10-18 at 18.48.49BST.png
Views: 19
Size:  20.5 KB
    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...
    Online

    3
    ReputationRep:
    (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.
    Online

    3
    ReputationRep:
    (Original post by solC)
    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 subtract 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...
    Can't be true. Plugging x=1 into that would leave you with -2 so p(1)=2 is not satisfied.
    Offline

    3
    ReputationRep:
    (Original post by solC)
    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 subtract 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...
    (Original post by RDKGames)
    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.
    Thanks!
    Online

    2
    ReputationRep:
    (Original post by RDKGames)
    Can't be true. Plugging x=1 into that would leave you with -2 so p(1)=2 is not satisfied.
    Ah yeah, didn't notice that. so you would have to add on 2 rather than subtract?
    Online

    3
    ReputationRep:
    (Original post by solC)
    Ah yeah, didn't notice that. so you would have to add on 2 rather than subtract?
    Well going off your attempt it wouldn't make sense because you said "p(1)=0" which is also false so you cannot use the remainder theorem.
    Online

    2
    ReputationRep:
    (Original post by RDKGames)
    Well going off your attempt it wouldn't make sense because you said "p(1)=0" which is also false so you cannot use the remainder theorem.
    Oh right I see what you mean, thanks

    jamestg Ignore what I said!
    Offline

    3
    ReputationRep:
    Erm duno wtf is goingn on above.
    Consider p(x)-2=r(x)
    x=1 is solution to r(x) so r(x)=(x-1)q(x)
    so p(x)=(x-1)q(x)+2


    Posted from TSR Mobile
    Online

    3
    ReputationRep:
    Name:  2nd order DE.png
Views: 21
Size:  40.2 KB

    How do you do part (c) ? The particular solution is  x = (\frac{1}{2} + \frac{4t}{3})e^{-3t} + \frac{1}{18}sin3t
    Offline

    2
    ReputationRep:
    An employer finds that if he increase weekly wages of each worker by £30 and employs one worker less,he reduces his weekly wage bill from £8160 to £7810, taking the weekly wage of each worker as x ,find out the weekly wage of each worker

    how do I do this?
    I can't find a final answer to this either
    Offline

    3
    ReputationRep:
    (Original post by NotNotBatman)
    Name:  2nd order DE.png
Views: 21
Size:  40.2 KB

    How do you do part (c) ? The particular solution is  x = (\frac{1}{2} + \frac{4t}{3})e^{-3t} + \frac{1}{18}sin3t
    when t > 30, x is approximately sin 3t / 18 since your exponential terms gets negligibly tiny.
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    Erm duno wtf is goingn on above.
    Consider p(x)-2=r(x)
    x=1 is solution to r(x) so r(x)=(x-1)q(x)
    so p(x)=(x-1)q(x)+2


    Posted from TSR Mobile
    init plebs everywhere
    • Thread Starter
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    init plebs everywhere
    Savage!
    Offline

    3
    ReputationRep:
    So basically the question says to find the stationary points of a curve which I've found. It then says justify which is a max and which is a min. Surely that is obvious which one is higher from the y coordinates I just worked out?
    Offline

    2
    ReputationRep:
    (Original post by medhelp)
    An employer finds that if he increase weekly wages of each worker by £30 and employs one worker less,he reduces his weekly wage bill from £8160 to £7810, taking the weekly wage of each worker as x ,find out the weekly wage of each worker

    how do I do this?
    You have two unknowns, the number of employees (n) and their weekly wage (x). You can form two equations for the before and after wage bills. You can then solve for x.

    The first equation would be nx = 8160. The rest is left as an exercise for the reader

    If only the question writer knew the difference between less and fewer ..
    Offline

    2
    ReputationRep:
    (Original post by RogerOxon)
    You have two unknowns, the number of employees (n) and their weekly wage (x). You can form two equations for the before and after wage bills. You can then solve for x.

    The first equation would be nx = 8160. The rest is left as an exercise for the reader

    If only the question writer knew the difference between less and fewer ..
    what do you mean by the bit in bold

    okay, so I got the second equation as (n-1) x (x+30) = 7810 and substituted in x = 8160 /n

    I did a load of working out and somehow I've gotten to 3n^2 - 32n - 816 = 0 which seems like hell to factorise so I'm assuming I've done it wrong,
    :/
    Offline

    2
    ReputationRep:
    (Original post by medhelp)
    what do you mean by the bit in bold
    Sorry - I'm a (partial) grammar pedant. It's fewer for countable things, not less.
    (Original post by medhelp)
    I did a load of working out and somehow I've gotten to 3n^2 - 32n - 816 = 0 which seems like hell to factorise so I'm assuming I've done it wrong,:/
    That quadratic doesn't look correct to me.

    You have:
    nx = 8160 and (n-1)(x+30) = 7810

    Expanding the second gives: (n-1)x + 30(n-1) = 7810

    To make the numbers simpler, I then subtracted the second from the first, giving:
    x - 30(n-1) = 350

    We're after x, so substitute for n (from nx=8160), and you should get an equation that you can solve with the quadratic formula, taking the positive root.
    Offline

    2
    ReputationRep:
    (Original post by RogerOxon)
    Sorry - I'm a (partial) grammar pedant. It's fewer for countable things, not less.

    That quadratic doesn't look correct to me.

    You have:
    nx = 8160 and (n-1)(x+30) = 7810

    Expanding the second gives: (n-1)x + 30(n-1) = 7810

    To make the numbers simpler, I then subtracted the second from the first, giving:
    x - 30(n-1) = 350

    We're after x, so substitute for n (from nx=8160), and you should get an equation that you can solve with the quadratic formula, taking the positive root.
    when I do the bit in bold it gives me
    x^2 - 320 x + 244800 = 0

    that almost looks worse than the last quadratic
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: December 7, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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

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