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True that, you should change tan^2 to sin^2/cos^2 then simplify to get \frac{sin^2 \theta}{tan^2 \theta} Then sub the equations from my 2nd line.

xy = tan^2 \theta - sin^2 \theta x + y = 2tan \theta \ \ \ y-x = 2sin \theta xy = (\frac{x+y}{2})^2 - (\frac{y-x}{2})^2

I'm not sure what you mean by "New" HSC Math courses (you mean progressing to senior or a complete revamp in the content?) Anyhow, all I can say is that you general math is basically learning applied "real life" math like finding areas and doing some basic finance stuff you might encounter in...

My parents and relatives would talk about how my mum and dad did really well in math back at school in China. I'm not sure if it's legit or not haha but I was able to gasp at most concepts a bit better than the average student so I guess it's a factor.

That's not even true since you're squaring a number less than 1 0<ln(2)<ln(e) = 1 \\ \\ ln(2) > (ln(2))^2 2ln(2) > (ln(2))^2 \Rightarrow 2 > ln(2) At the moment I still can't figure the lower bound either.

I think you mean \frac{1}{2}(e^2-1) Also I'm not sure how you got that integral cause the integral is suppose to be simply \int_0^2 e^x \ \ dx

In order to find the horizontal asymptote, you need to take the limit as the x-value approaches infinity. You also need use this concept. lim_{x \rightarrow 0} \frac{sinx}{x} =lim_{x \rightarrow \infty} \frac{sin(\frac{1}{x})}{\frac{1}{x}} = 1 lim_{x \rightarrow \infty} \frac {x^2...

Note that is says distinct points so that already implies p \neq q You can easily see that's true if you simply observe the equation mQRmPR=-1. which leads to (p+1)(q+1) = -4 and you'll understand that p and q can't be -1 as you'll get zero. However also observe that both p+1 and q+1 can't...

"Smile life could be worse, you could be married to *random weird name*" My Highschool Math Teacher.

Yeah that's because this not your straight forward "top minus bottom" kind of question. First you should draw out the curves, then find the point of intersection (which is x = 1 by inspection) The area given above is \int_0^1 \ x^3 \ \ dx + \int_1^{\frac{4}{3}} \ \ -3x + 4 \ \ dx = \frac{5}{12}

Yeah the report shows that time. But I actually received my email at 5:39 PM.

Alright zID 501 Issued at Tue Dec 4 16:47:02 2018 Received the email at 5:39 PM Hehe a pretty crap semester for me, but I'm just happy that I survived all of them and thus my undergraduate life is officially over.

The hype is almost ready

You didn't account for the other 2 possible girls.

a) For the first 3 spots you have to select 3 boys from 5 and understand that the order amongst those 3 matters so it's 5P3, then for the remaining 6 people it's just 6! (5P3)*6! b) First you select 2 girls from 4, so that's 4C2, you form that with John to make one group, also multiply that...

You don't use mathematical induction in HSC unless otherwise stated. in iii), you differentiate part i) again, then multiply both sides by x, sub x = -1, then add this with part ii) then divide by 2.

2013 graduate, can confirm I lost way more marks in Q15 than 16.

Is there a link for the actual paper?

Lol this guy He's talking about actually being able to evaluate logs of negative numbers, but only in the complex field log(z) = ln|z| + iArg(z) But in HSC 2U and 3U you would only integrate with real numbers, iArg(z) would always cancel out, thus you're left with just the absolute value...

Perms and combs all the way obvously