Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

kiwiberry · « **Reply #1365 on:** July 31, 2017, 09:33:55 pm »

Quote from: Opengangs on July 31, 2017, 09:18:36 pm

Since we want to find when Alan eventually picks a black jellybean, we need to consider every case.
...

Omg thanks so much!!! I get it

Opengangs · « **Reply #1366 on:** July 31, 2017, 09:37:26 pm »

Quote from: kiwiberry on July 31, 2017, 09:33:55 pm

Omg thanks so much!!! I get it

Glad I could help!

raymatar · « **Reply #1367 on:** August 02, 2017, 07:46:25 pm »

Hey can someone please help me with these 2 questions. Thanks in advance.

RuiAce · « **Reply #1368 on:** August 02, 2017, 09:48:08 pm »

Quote from: raymatar on August 02, 2017, 07:46:25 pm

Hey can someone please help me with these 2 questions. Thanks in advance.

$\text{The length of }PQ\text{ is}\\ \begin{align*}PQ&=\sqrt{(q-p)^2+\left(\frac{1}{q}-\frac{1}{p}\right)^2}\\ &= \sqrt{(q-p)^2 + \left(\frac{p-q}{pq}\right)^2}\\ &= |q-p|\sqrt{1+\frac{1}{(pq)^2}}\\ &= |p-q|\times \frac{\sqrt{(pq)^2+1}}{|pq|}\end{align*}$
$\text{The perpendicular distance from }O\text{ to }PQ\text{ is}\\ \begin{align*}d_{O\perp PQ}&= \frac{|0+0-(p+q)|}{\sqrt{1+(pq)^2}}\\ &= \frac{|p+q|}{\sqrt{1+(pq)^2}}\end{align*}$
$\text{So using }A = \frac{1}{2}bh\\ A = \frac{1}{2} \times \frac{|q-p|}{|pq|}\times |p+q|\\ \text{which gives the desired result}.$
Are you having trouble with 7b or 7c?

beau77bro · « **Reply #1369 on:** August 03, 2017, 09:33:51 am »

help please, i really don't understand the process to get here in the solution - maybe you have an alternative? or could u explain how u got there? should i add the solution?

RuiAce · « **Reply #1370 on:** August 03, 2017, 09:47:38 am »

Quote from: beau77bro on August 03, 2017, 09:33:51 am

help please, i really don't understand the process to get here in the solution - maybe you have an alternative? or could u explain how u got there? should i add the solution?

Why not?

beau77bro · « **Reply #1371 on:** August 03, 2017, 03:33:54 pm »

idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way

RuiAce · « **Reply #1372 on:** August 03, 2017, 03:36:39 pm »

Quote from: beau77bro on August 03, 2017, 03:33:54 pm

idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way

That actually didn't cross my mind during the time I had to think about it. I'll probably explain it later tonight though (unless someone gets there before me) because of a few more classes I have today that I want to focus on.

beau77bro · « **Reply #1373 on:** August 03, 2017, 06:57:21 pm »

why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z

Ali_Abbas · « **Reply #1374 on:** August 03, 2017, 07:09:13 pm »

Quote from: beau77bro on August 03, 2017, 06:57:21 pm

why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z

Unless I'm mistaken, I agree the answer should be B.

RuiAce · « **Reply #1375 on:** August 03, 2017, 07:13:29 pm »

Quote from: beau77bro on August 03, 2017, 06:57:21 pm

why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z

Well what did they say it was?

(I haven't forgotten about your binomial question; but I just got home and I'm having dinner first.)

beau77bro · « **Reply #1376 on:** August 03, 2017, 07:56:44 pm »

Quote from: RuiAce on August 03, 2017, 07:13:29 pm

Well what did they say it was?

(I haven't forgotten about your binomial question; but I just got home and I'm having dinner first.)

sorry they said it was C

RuiAce · « **Reply #1377 on:** August 03, 2017, 08:17:14 pm »

Quote from: beau77bro on August 03, 2017, 07:56:44 pm

sorry they said it was C

Wait... but isn't that correct? It goes from iz to z so it should be z - iz = (1-i)z

I feel like they made typos somewhere

RuiAce · « **Reply #1378 on:** August 03, 2017, 08:28:55 pm »

Quote from: beau77bro on August 03, 2017, 03:33:54 pm

idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way

$\text{It's definitely a weird symmetry argument}\\ \text{but it's definitely faster than what I had planned to do.}$
$\text{I trust that you understood the factorisation }(1+x+x^2)^{2n}=x^{2n}(1+x^{-1}+x^{-2})^{2n}$
$\text{It helps to remind ourselves of the basics}\\ \text{By def. }a_r\text{ is the coefficient of }x^r\text{ in }(1+x+x^2)^{2n}$
Also, I believe they intended to write a_2n-r in the place of a_r but made a mistake in transcript
$\text{Now, further by def. }a_{2n-r}\text{ is the coefficient of }x^{2n-r}\text{ in }(1+x+x^2)^{2n}$
$\text{This is equivalent to finding the coefficient of }x^{2n-r}\text{ in }\boxed{x^{2n}(1+x^{-1}+x^{-2})^{2n}}$
$\text{In that boxed equation}\\ \text{If we choose to divide out }x^{2n}\text{ in that term, we have to }\\ \text{adjust by dividing }x^{2n}\textbf{ as well in what we're taking the coefficient of}$
$\text{That's to say, BECAUSE }a_{2n-r}\text{ is the coefficient of }\boxed{x^{2n-r}}\text{ in }\boxed{x^{2n}(1+x^{-1}+x^{-2})^{2n}}\\ \text{It is ALSO the coefficient of }\boxed{x^{-r}}\text{ in }\boxed{(1+x^{-1}+x^{-2})^{2n}}$
______________________________________
$\text{Now we join the other side of the puzzle}\\ \text{The main result is achieved through inferring a }\textbf{comparison.}$
$\text{Because }a_r\text{ is the coefficient of }\boxed{x^r}\text{ in }\boxed{(1+x+x^2)^n}\\ \text{Letting }y=x^{-1} \iff x = y^{-1}\\ a_r\text{ is }\textbf{also}\text{ the coefficient of }\boxed{y^{-r}}\text{ in }\boxed{(1+y^{-1}+y^{-2})^{2n}}$
$\text{Which is of course, equivalent to the coefficient of }x^{-r}\text{ in the same thing above!}\\ \text{ i.e. }\boxed{(1+x^{-1}+x^{-2})^{2n}}$
$\text{Equate and you're done}$

beau77bro · « **Reply #1379 on:** August 03, 2017, 09:22:28 pm »

Quote from: RuiAce on August 03, 2017, 08:28:55 pm

$\text{It's definitely a weird symmetry argument}\\ \text{but it's definitely faster than what I had planned to do.}$
$\text{I trust that you understood the factorisation }(1+x+x^2)^{2n}=x^{2n}(1+x^{-1}+x^{-2})^{2n}$
$\text{It helps to remind ourselves of the basics}\\ \text{By def. }a_r\text{ is the coefficient of }x^r\text{ in }(1+x+x^2)^{2n}$
Also, I believe they intended to write a_2n-r in the place of a_r but made a mistake in transcript
$\text{Now, further by def. }a_{2n-r}\text{ is the coefficient of }x^{2n-r}\text{ in }(1+x+x^2)^{2n}$
$\text{This is equivalent to finding the coefficient of }x^{2n-r}\text{ in }\boxed{x^{2n}(1+x^{-1}+x^{-2})^{2n}}$
$\text{In that boxed equation}\\ \text{If we choose to divide out }x^{2n}\text{ in that term, we have to }\\ \text{adjust by dividing }x^{2n}\textbf{ as well in what we're taking the coefficient of}$
$\text{That's to say, BECAUSE }a_{2n-r}\text{ is the coefficient of }\boxed{x^{2n-r}}\text{ in }\boxed{x^{2n}(1+x^{-1}+x^{-2})^{2n}}\\ \text{It is ALSO the coefficient of }\boxed{x^{-r}}\text{ in }\boxed{(1+x^{-1}+x^{-2})^{2n}}$
______________________________________
$\text{Now we join the other side of the puzzle}\\ \text{The main result is achieved through inferring a }\textbf{comparison.}$
$\text{Because }a_r\text{ is the coefficient of }\boxed{x^r}\text{ in }\boxed{(1+x+x^2)^n}\\ \text{Letting }y=x^{-1} \iff x = y^{-1}\\ a_r\text{ is }\textbf{also}\text{ the coefficient of }\boxed{y^{-r}}\text{ in }\boxed{(1+y^{-1}+y^{-2})^{2n}}$
$\text{Which is of course, equivalent to the coefficient of }x^{-r}\text{ in the same thing above!}\\ \text{ i.e. }\boxed{(1+x^{-1}+x^{-2})^{2n}}$
$\text{Equate and you're done}$

ok so it's kind of a weird comparison/equating of co-efficients for different expansions, i can see what you did but I'm struggling with the reasoning. is it just re-denoting the question so that you can show a new relationship between a with the power? - seems liek they just gave u a question where u just use the answer to find something - i can't see a straightforward process other than changing the powers.

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