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Cos a = 0 and Sin a = 0, so x = Cos a + i Sin a = 0 + 0i = 0. Same goes for y and z. And your solution is so long :p

Assume true for n = k, so 9^(k+2) - 4^k = 5M, where M is a positive integer. Prove true for n = k + 1, so prove 9^(k+3) - 4^(k+1) is divisible by 5. 9^(k+3) - 4^(k+1) = 9(9^(k+2)) - 4(4^k). But 4^k = 9^(k+2) - 5M, by assumption. So 9^(k+3) - 4^(k+1) = 9(9^(k+2)) - 4(9^(k+2) - 5M) = 5(5^(k+2))...

\\ Cos a + Cos b + Cos c = Sin a + Sin b + Sin c = 0 \\ $Let $x = Cos a + iSina, y = Cos b + iSinb, z = Cos c + i Sin c\\ \therefore x=y=z=0\\ \therefore x^2 = y^2 + z^2 = 0$, and so $ x^2+y^2+z^2 = 0\\ $But $ x^2 = Cis 2a, y^2 = Cis 2b, z^2 = Cis 2c,$ $ \therefore Cis 2a + Cis 2b + Cis 2c = 0\\...

I thought about the difference of two squares. Dude...that is ownage.

HAHAHAHAHAHA! Ok, that was pure genius...

I skipped 2 school years.

\\ a^2 + b^2 \geq 2ab\\ c^2 + d^2 \geq 2cd\\ a^2 + b^2 + c^2 + d^2 \geq 2(ab + cd)\\ $From a^2 + b^2 \geq 2ab$, $ it follows that $ab + cd \geq 2\sqrt{abcd}\\ a^2 + b^2 + c^2 + d^2 \geq 4\sqrt{abcd}\\ $Let $abc = d^3,$ so $ a^2 + b^2 + c^2 + d^2 \geq 4\sqrt{d^4}\\ a^2 +b^2 + c^2+d^2\geq 4d^2\\...

\\ x^5 + x^4 + 1 = (x^5 + x^4 + x^3 + x^2 + x + 1) - (x^3+x^2+x)\\ = \frac{x^6-1}{x-1} - x(x^2+x+1)\\ =\frac{(x^3-1)(x^3+1)}{x-1} - x\frac{x^3-1}{x-1}\\ = (x^3-x+1)\frac{x^3-1}{x-1}\\ = (x^3-x+1)(x^2+x+1)

Making notes for maths? Are you serious about this?

Haha thankyou.

I have no idea what you're saying here. If they're aggressive yet harm no-one, then so be it.

I think you just restated his point, although quite vulgarly.

It's not multi-valued in the case of absolute values.

Unfortunately, a linear relationship only exists if the solid only consists of straight lines, and no curvature.

Re: solving inequalities with x in the denominator Ahh it's ok then. I just don't remember seeing something like this.

I have friends who achieved 100 aligned without a state rank...so it is more than 10. I doubt it would reach more than 20 though.

Re: solving inequalities with x in the denominator Are you sure you're not just talking about the case where one of the boundaries gives the original denominator a value of 0? 1/(x - 4) >= 0. x - 4 >= 0. x >= 4. But x <> 4 since x - 4 <> 0? Or there's something else?

Re: solving inequalities with x in the denominator Can you please show me an example where it creates more answers? Assuming that the denominator is linear, ofcourse...

I'm not sure what you mean by this? |5x-2|/x > 0, automatically x > 0. And for |x+1| + |2x-1| < 5, I just do ++, +-, -+, -- and then substitute back into the equation to see if it works. Case method does not have to be used.

I guess the rigour point is fair enough. Just wondering though, if 10 people get 84/84 in the HSC exam, how do they choose who is ranked first, and who is second...and so on? Do they not check working out? Just a little query I have.