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I have a double exam on the 18th (saturday too FML) and the last exam on the very last day in the exam period, the 20th. Just casually chilling at home doing some maths while the others are out there having a good time.

Mate I haven't even started my exams yet.

To get the area of the triangle, you have to use the perpendicular distance from the line AB to P, that's the height. The base is of course the distance of AB. Then find the max area using the max min theorem. Which turns out to be 3/4 of part i)

Universities wouldn't really care about a horrible mark like that especially if it isn't counted. They only care about ATAR and the marks that are relevant for the degree you would study e.g Math degrees may take your math marks into account. Lol you would have separate certificates for each...

Some one please send link to the paper thanks.

10000(1.2)^{24}

Clearly you're so good at math that you can just wing it through like a boss, why bother studying? But in all seriousness, do you know what your weakest topics are? normally I would spend a bit more time on those compared to my stronger topics (but you should still put some time in those as...

I'll only give out the steps for Q10 $Deriving the equations in terms of t would yield$ \ \ \ x= Vtcos \theta \ \ \ y= Vtsin \theta - \frac{gt^2}{2} \\ \\ $But since the question has no mention of the variable time, we should eliminate it. So simultaneously we should sub the first equation...

Also are you currently doing 3U math? if not then repeating 2U in year 12 should give you a much better much since you're more exposed to a higher level of mathematics.

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So it works like this, I'm currently studying Adv Math/ Economics at UNSW, I decided to combine my math degree with eco since it would give me better job prospects (well that's what I think) and it seemed interesting to me. However as I'm as going through halfway of eco, I'm starting to hate...

I guess we have to step it up a notch with your knowledge of general solution. So the equation could be re-written as the cos functions. \cos(8\theta) = \cos(5\theta+\frac{\pi}{2}) \Rightarrow \\ \\ 8\theta = 2k\pi \pm (5\theta +\frac{\pi}{2}) \Rightarrow \\ \\ \theta =...

Alpha can't be zero, but it doesn't say that theta can't be zero.

Your answer is right, along with theta = 0

Welp I can't seem to use the "hence" method. But for otherwise I just did like sin(5 \theta) = - cos(8 \theta) \\ \\ sin(5 \theta) = -sin(\frac{\pi}{2} - 8 \theta) = sin(8 \theta - \frac{\pi}{2}) \\ \\ \Rightarrow 5 \theta = 8 \theta - \frac{\pi}{2} \\ \\ \Rightarrow \theta = \frac{\pi}{6}

Sorry I don't understand that. I can't give you any info on your specific degree, but I can say that it's not a bad thing if you transfer from UNSW to UTS, I know someone who did that since he unfortunately failed many courses and got an fail average as a result. He's doing alright at UTS I heard.

Really sorry to hear all this :'(

I think it should be L T1 T3, otherwise it doesn't make sense.

Going to be a bit pedantic here, but you shouldn't be using your limits in terms of degree but in radians, it could create a lot of awkward scaling factors especially if you're combining trig functions with some linear function. e.g \int_0^{45^{\circ}} \ \ sinx - x \ dx wouldn't give you the...

Aye Q2 is just asking for the change in the area of the base, not the entire surface area.