Mathematics Question Thread

[RuiAce]

Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21733200_1324623537666455_1924185311_n.png?oh=755e3ab03cb1b287d4ad471e3535d5c2&oe=59C00DE2
Apparently my friend got it from a 2U Trial paper. If this is not a 2U/3U question, then don't worry about it.
Thanks in advance.

I personally don't believe your friend at all. The \( w^{n+1}=-1 \) thing and proving that it's real hints that this is actually a 4U question. If anything, given the typeface it looks as though it'd be a question from brilliant.org

I'm happy to not worry about it or take more time if it gets shoved into the 4U thread (up to you if you want to do this), but unless your friend cannot clearly identify what paper this is from I will certainly not do it here. It's potentially possible that 2U knowledge is sufficient enough to actually do the question, but it goes beyond the level of understanding required.

My first guess: Geometric series on the top.

Edit: Just had a go at it. I managed to get it to work but I could only finish it off with complex numbers.

(Image removed from quote.)could I pls have help with both parts of the question? Thxx

So the information we have is:
Radius of cone = 2sqrt(3/pi)
Height of cone = 6
Radius of liquid = sqrt(3/pi)
Height of cone = 3


And part 2 is just simple subtraction: 24 - 6 = 18.

[hansolo9]

Hi
How do you do this?

[RuiAce]

Hi
How do you do this?

Note that csc is how \( \LaTeX \) denotes 'cosec', and thus is what is displayed
\[ \text{From continued use of Pythagorean identities} \]
\begin{align*}\sqrt{\frac{1-\sin^2\theta}{\csc^2\theta -\cot^2\theta -\cos^2\theta}} &= \sqrt{\frac{\cos^2\theta}{(\csc^2\theta -\cot^2\theta) -\cos^2\theta}}\\ &= \sqrt{\frac{\cos^2\theta}{1-\cos^2\theta}}\\ &= \sqrt{\frac{\cos^2\theta}{\sin^2\theta}}\\ &= \frac{\cos\theta}{\sin\theta}\\ &= \cot \theta\end{align*}

[davidss]

Hi can someone help with this pls thanks <3

[Mathew587]

Hi can someone help with this pls thanks <3

(Image removed from quote.)

whcich q

[JuliaPascale123]

Hi Guys,
Could I please have working out and answer for this question as I have found others like it and need to practice the skill.



Thanks <3,

Julia

[persistent_insomniac]

Hey guys I don't even know if this question belongs here so sorry if it doesn't but how do you solve:
If logax=4 and logay=5, use log laws to find the exact values of:
logax^2y
logaaxy
logaxy
 
I literally don't know who to ask since theres nothing on my textbook or even on google. Thanks so much in advance!

[RuiAce]

Hey guys I don't even know if this question belongs here so sorry if it doesn't but how do you solve:
If logax=4 and logay=5, use log laws to find the exact values of:
logax^2y
logaaxy
logaxy
 
I literally don't know who to ask since theres nothing on my textbook or even on google. Thanks so much in advance!

If you want to avoid hitting the subscript button, can you please use more brackets to make it clear what we're talking log-base-a of?

[Shadowxo]

Hey guys I don't even know if this question belongs here so sorry if it doesn't but how do you solve:
If logax=4 and logay=5, use log laws to find the exact values of:
logax^2y
logaaxy
logaxy
 
I literally don't know who to ask since theres nothing on my textbook or even on google. Thanks so much in advance!

As Rui said it's a bit difficult to determine what you want to find but the basic log laws should make this question easy

Try using these, in particular the first one, for your problems

[jamonwindeyer]

Hi Guys,
Could I please have working out and answer for this question as I have found others like it and need to practice the skill.
Thanks <3,
Julia

Hi Julia - I think you'll have trouble getting an answer to this here, it doesn't look like anything from a HSC course to me! Where did you find it?

[davidss]

whcich q

the first bit pls xd

[Shadowxo]

Hi can someone help with this pls thanks <3

(Image removed from quote.)

Let CAD be alpha
So ACD = alpha (isosceles)
2alpha + theta = 180°
CBA = alpha (isosceles)
Hence ACB = theta (angles add to 180°)
Hence they're similar as they have the same three angles (alpha alpha and theta)

There are many ways to approach this problem, I just thought angles was the easiest way. Their solution may be different, and you may need to be more specific than I've been

[RuiAce]

Let CAD be alpha
So ACD = alpha (isosceles)
2alpha + theta = 180°
CBA = alpha (isosceles)
Hence ACB = theta (angles add to 180°)
Hence they're similar as they have the same three angles (alpha alpha and theta)

There are many ways to approach this problem, I just thought angles was the easiest way. Their solution may be different, and you may need to be more specific than I've been

It suffices to only prove that two angles are equal for the equiangular test. The third can be assumed to equal by the angle sum

[caitlinlddouglas]

hi i didn't understand the proof for question 16 c) from the 2013 mathematics paper.
 for part ii) it says just to use parts from 1) and 2) but i'm unsure how they've manipulated it! thanks
http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
solutions: http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf

[RuiAce]

hi i didn't understand the proof for question 16 c) from the 2013 mathematics paper.
 for part ii) it says just to use parts from 1) and 2) but i'm unsure how they've manipulated it! thanks
http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
solutions: http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf

________________________________

Observation: AB appears in the denominator twice. Also, AX + BX = AB, which is extremely convenient. This second observation was deduced by looking at the diagram, because clearly, AB is literally just AX and BX joined together.

And hence the intuition to add the equations

Note that clearly XY was just factorised.