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Year 13 Maths Help Thread Watch

1. (Original post by RDKGames)
Take a logarithm of base 3 for both sides and rearrange for
Ahhh, thanks!
2. Also, what does this even mean...

3. (Original post by jamestg)
Also, what does this even mean...

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...
4. (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.
5. (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.
6. (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!
7. (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?
8. (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.
9. (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!
10. 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

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11. How do you do part (c) ? The particular solution is
12. 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
13. (Original post by NotNotBatman)

How do you do part (c) ? The particular solution is
when t > 30, x is approximately sin 3t / 18 since your exponential terms gets negligibly tiny.
14. (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

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init plebs everywhere
15. (Original post by Zacken)
init plebs everywhere
Savage!
16. 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?
17. (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 ..
18. (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,
:/
19. (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.
20. (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

Updated: June 26, 2017
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