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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un this

Loool

nice diagram

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{2} \frac{dx}{1 @plus; x^2} ~,~x=2tan\theta \\\\ \frac{1}{4}\int_{\frac{\pi}{4}}^{0} \frac{2sec^2\theta }{1 @plus; tan^2\theta}~d\theta ~,~ 1 @plus;tan^2\theta \Rightarrow sec^2\theta \\\\ = \frac{1}{4}\int_{0}^{\frac{\pi}{4}} 2~d\theta...

​<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{4} x\sqrt{16 -x^2} dx ~~,~~ x = 4sin\theta \\\\ \int_{0}^{\frac{\pi}{2}} 16sin\theta cos\theta\sqrt{16-16sin^2\theta}~d\theta \\\\ = 64\int_{0}^{\frac{\pi}{2}} sin\theta cos\theta\sqrt{1 - sin^2\theta} ~d\theta \\\\ =...

Definitely the best advice thus far

<a href="http://www.codecogs.com/eqnedit.php?latex=PS = \sqrt{(2ap)^2 @plus; (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 @plus; a^2p^4 - 2a^2p^2 @plus; a^2} \\\ =\sqrt{2a^2p^2 @plus; a^2p^4 @plus; a^2} \\\ = \sqrt{a^2(2p^2 @plus; p^4 @plus; 1)} \\\ =\sqrt{a^2(p^2@plus;1)^2} \\\ =a(p^2 @plus; 1)"...

Answer is correct, you have done something wrong

why

......

ok

Is it (0, a) or can it be (a, 0 )

<a href="http://www.codecogs.com/eqnedit.php?latex=A ( a(p@plus;q) , apq) ~ S(0,a) \\\\ Gradient~ of~ AS = \frac{apq-a}{a(p@plus;q)} \\\\ \frac{apq-a}{a(p@plus;q)} = 0 \\\\ apq - a = 0 \\\\ a(pq-1)=0 \\\\ pq = 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?A ( a(p+q) , apq) ~...

its ez

Re: 2012 CSSA Paper whose gonna do it? lol same

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(a@plus;b)=\frac{tana@plus;tanb}{1-tanatanb} \\\\ tan(a@plus;b@plus;c)=tan(a@plus;(b@plus;c))=\frac{tana@plus;tan(b@plus;c)}{1-tanatan(b@plus;c)}" target="_blank"><img...

:D

1/2 < x , since x=/=1/2

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x@plus;5)(x-3)\geq 0 \\\\ Ta -da"...

oo