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The fifth question is simply the use of the conjugate root theorem because a complex number and its conjugate give us double the real portion of the complex number and that the product gives us the squared form of the real portion added to the squared form of the imaginary portion.

With the fourth image, the question you gonna ask yourself is how can we get from \tan{\alpha} to \tan{4\alpha}. Now the first step we need to do is to recognise the double angle formula for the tangent. Noting that the double angle formula for tangent is...

However, if you are planning to organise events with just friends from UNSW then that is the better decision. Unless you want to be the black sheep in your friend group.

Where do you live? This can be a good question to ask because you can set the decision on which uni to choose.

I guarantee this will soothe your day.

For part b you are going to have to use the fact that 1+2+3+4+...+n=\frac{n(n+1)}{2} along your line of working to complete the question.

I do not understand the steps written with a question mark. Everything else is right.

U sure because I find that questionable. Unless they have a knack for it they will quickly get defeated. Unless you want to play Damsel in distress with her.

With the calculus part of the topic, a few sneaky shortcuts will surely make your life easy. For example with partial fractions instead of listening to the teacher unless the teacher knows it themselves, I suggest the very subtle Heaviside cover-up method. Turns partial fractions into a joke...

One of them but also note there is another solution. Notice when I said the absolute values it means that you can have -x+2 > \frac{1}{x} when x is less than 2 so therefore what happens there is multiply by x^{2} giving us -x^{3}+2x^{2}-x > 0 factorising gives us -x(x^{2}-2x+1) > 0 changing the...

Have a look at your graph and also the part where I got x(x^{2}-2x-1) > 0. There you can focus on all the values of x that give a positive value

The absolute value plays an important role.

Okay, are you doing Maths extension II? If you do so or not note that the absolute values represent the length of distance from 0 to a certain equation so in this case, what we have here is that the value has to be either bigger than \frac{1}{x} or smaller than -\frac{1}{x}. With that knowledge...

Funny enough Dr Pender for quite a while waged war with ACARA with how the syllabus is handled so I can see that he takes the education of NSW students very seriously.

Note instead here you can state that 9(................) = 3m - (a_k+a_{k-1}+...+a_0) where 3m - (a_k+a_{k-1}+...+a_0) has to be divisible by 9 and there it is obvious that it is also divisible by 3 and solved. ak+ak-1+...+a0 is divisible by 3. To me this statement becomes redundant when...

Very funny that someone mentioned PDHPE. Never in my lifetime have I heard PDHPE get mentioned in the subject of scaling subjects for students who want to do medicine. So I would actually steer away from that. Does not scale well as intended.

I asked that question too.

So what made you want to go into psychiatry is there something in you about knowing why people are behaving in a certain and that you want to help them out. If so then keep that desire within you when you become a psychaiatrist.

What makes you want to do medicine?

Now prepare to witness the mind of a mathematician. The entire concept revolves around the fact that with a pattern where if we are finding something that is purely imaginary then for example we have \left(\cos{\pi}\right)^{n}=0 which is simply \cos{n\pi}=0 according to De Moivre's theorem...