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You want the coefficients of sin(x), xsin(x) and cos(x) to be 0. So if you separate it better than I did, you should be able to work it out.

1) x³/(2-x²) = (x³ - 2x + 2x)/(2-x²) = -x + 2x/(2-x²) which should be easier to integrate 2) f(x) = axsin(x) + bsin(x) + cxcos(x) + dcos(x) f'(x) = asin(x) + axcos(x) + bcos(x) + ccos(x) - cxsin(x) - dsin(x) = sin(x)[a-d-cx] + cos(x)[ax + b + c] Since you want f'(x) = xcos(x) for all x, then...

Lego with unlimited resources!! Some of the videos on youtube showing what people have built are just mindblowing.

How about ridiculously long single player modes like Oblivion =O

I think you're mistaking it with arg[a] + arg[b] = arg[ab] So it's the other way around (products and sum)

Thank You!

w + 1/w = 2 cos 2pi/5, but 2pi/5 is acute. So I think you disregard the negative number? so 2cos 2pi/5 = [-1 + sqrt(5)]/2 only? Double angle as you said, giving you cos²(pi/5) = [3+sqrt(5)]/8 Square root then disregard the negative because pi/5 is acute as well.

@ blackops c = +/- sqrt(2) And I think your solution for the last part might look BS because the question is pretty BS lol!

Thanks!!

Hey everyone, My name is Victor and I am currently doing a combined Commerce/ Science (Advanced Mathematics) degree at UNSW, majoring in Actuarial Studies. With my experience in doing the HSC, I can teach you the proper exam techniques and setting out required to improve your marks in...

Colours are probably what we have to work out in later parts of the question. Like 3 red winners, 2 red 1 blue, or whatever

1. For the first one, you either expand (which might take a while) or take a look at coefficients. So for example, coefficient of x²y² on the LHS is 0, while on the RHS is 1+1-c², then solve from there. 2. Is that > or < 0? 3. Find the roots to 1 and 2 and show that they're not equal? (or...

oh no. arj1's solution

This. Start with red blue (showing chance of drawing red as winning or blue as winning). At the end of each branch, draw a new set of red/blue. You have in total 3 groups of branches representing the 3 prizes. Fractions at start would be 1/100, 1/100, changing to 1/99 or staying at 1/100...

This. I think the problem with your solution is this step: arg(z-2) - arg(z+2) = pi/2 therefore z-2/z+2 = i You got rid of the args by 'tan'ing both sides taking tan(pi/2) = i

Wow, I noobed it hard. The one I chose was one of the few that worked (must be lack of sleep). Let's try again x = 1, y = 2, s = 3 => 1.25 =/=0.8888 x = 2, y = 3, s = 5 => 0.3611 =/= 0.32 etc. I'm thinking it only works if x = y s = 2x LHS = 1/x² + 1/x² = 2/x² RHS = 8/(2x)² = 8/4x² = 2/x²...

I think there's 12 hours here? ;) Good Luck with your goal, and don't let anyone discourage you!

x = 1, y = 1, s = 2 LHS = 2 RHS = 4 LHS =/= RHS !!!

Do you mean 1+2sin(\frac{\pi}{7})+sin^{2}(\frac{\pi}{7})-cos^{2}(\frac{\pi}{7})+2icos(\frac{\pi}{7})(1+sin(\frac{\pi}{7}))...

I can't remember whether Jacaranda had it, but here's one from in2physics: