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y=x? lol

did it again and got this. I will post up a solution if people want a bit later.

82.7% correct to 1dp? this is wrong, i made a mistake

the first one

IMO Here is the easiest way to understand how these two functions are simplified. tan(90-x)=cot(-x). Think of a right angled triangle. For whichever value x is defined as, the other acute angle will be 90-x. Therefore, the tan of one angle (Opposite/adjacent) is obviously equal to...

Watch out with this, it does change speed when entering the medium perpendicular to the surface.

all you need to do is simple trig. For the first one, the answer is 6/cos(the angle) the second one, 2x6tan(the angle)

99.5-99.8

I solved the question, yet i'm not sure if my proof is the correct way to prove it. To start the proof, would it be ok to say "Let angle BAE be a right angle" and go from there?

http://cpms.anu.edu.au/sites/cpms.jagws01.anu.edu.au/files/ANUMD_2007_group.pdf just had a look at the first 4 questions. Are these questions aimed at year 3 students???

Re: 2012 HSC MX2 Marathon i agree, impressive. I rate that question hard/10

You probably cannot, because who are they going to rank you against if you have not done any assessments in relation to the year group you are in. I did the same thing with MX1 this year, yet I decided straight away after I received my mark that I would take the course again.

woah, some people overcomplicated the solution. Again, if you haven't solved it, dont look. \frac{\pi r^2}{0.5ab}=2\pi \\ \frac{r^2}{ab}=1\\ r^2=ab\\ but~a=2rsin(\theta ),~and~b=2rcos(\theta)\\ \therefore r^2=4r^2sin(\theta )cos(\theta )\\ 1=2sin(2\theta )\\ sin(2\theta )=\frac{1}{2}\\ 2\theta...

<a href="http://www.codecogs.com/eqnedit.php?latex=a~monic~polynomial~with~roots~as~such\\ \\ (x-tan(x))(x-tan(-x))(x-tan(y))(x-tan(-y))(x-tan(z))(x-tan(-z))\\ \\ =(x-tan(x))(x@plus;tan(x))(x-tan(y))(x@plus;tan(y))(x-tan(z))(x@plus;tan(z))\\ \\...

ok, well we know from trig that tan(-x)=-tan(x). So factorise the 6 roots into 3 quadratic roots, all of which will be difference of 2 squares. Then just selectively expand to get the coefficient of x^2 term and that will be your sum. the solution will be. sum of the roots 4 at a time =...

are you implying that there are 6 roots to the polynomial, tan(x), tan(-x), tan(y), tan(-y), tan(z), tan(-z)?

Re: 2012 HSC MX2 Marathon integrate: (2x^2+13x+16)/((2x+1)^2(x-2)) put that statement into wolframalpha, and you will get a perfect solution

I re-did the question, and i did get 0.0790 again. Dunno why i changed it originally

shit, i dont even remember, ill have a look through my solution sheets.

I had that originally!! but then i changed my solution cos I had something wrong with it and then i got my other solution, but maybe I was right the first time...