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Thats one of the many reasons why im choosing Japanese. It still counts as homework :D

Its just says it on the side of your posts (under your name). Its just this work is of a fairly high standard for year 9 students so i was just curious.

Are you in year 9?

This whole problem works on the fact that \frac{x+y}{2} \geq \sqrt{xy} This is called the AM-GM inequality because \frac{x+y}{2} is the arithmatic mean and \sqrt{xy} geometric mean. Note that is is equivilent to x+y \geq 2\sqrt{xy} x+y \geq 2\sqrt{xy} x + 2 \geq 2\sqrt{2x} y+2...

This whole problem works on the fact that \frac{x+y}{2} \geq \sqrt{xy} This is called the AM-GM inequality because \frac{x+y}{2} is the arithmatic mean and \sqrt{xy} geometric mean. Note that is is equivilent to x+y \geq 2\sqrt{xy} The solution is as follows, x+y \geq 2\sqrt{xy}...

Whoops, for 1 (b) I solved \log_{5} \left(x+2\right) - \log_{5}\left(x-2\right) = 1 instead of \log_{5} \left(x+2\right) + \log_{5}\left(x-2\right) = 1 The solution is as follows, \log_{5} \left(\left(x+2\right)\left(x-2\right)\right) = 1 \left(x+2\right)\left(x-2\right) = 5 x^2 -...

Question 1 (a) x^2 - 5x - 24 P: -24x^2 S: -5x We must look for numbers that multiply together to give -24 and sum to give -5. So we know we will have a negative number and a positive number because when multiplied it is the only way you can get a negative number. The factors of 24 are...

Yes, the Cambridge 2 unit text still exists.

Close, but it is actually 14 because 1 x 9 is the same as 3 x 3.

Re: free mx1 trial HSC's, come and get them Thanks so so much!

perhaps, but the point im trying to make is that it is not necessarily true that who ever studies the most comes first.

I did very little study compared to my year 12 friend for the HSC Mathematics half yearly, and i came 1st and he came 10th.

lol...

so is mine, but thats not going to stop me from do these things.

when x = e, y = \frac{1}{e} \therefore \left( e, \frac{1}{e} \right) :D

As no one has been able to solve Trebla problem, then (if you dont mind Trebla), for the sake of furthering the progress of this thread, i will post a new question. given f(x) = \sin\left(\omega x + \phi \right) where \omega > 0 and \left|\phi \right| < \frac{\pi}{2} if...

In the Subject Selection Handbook: "You must be studying the Advanced English Course to do the extension course."

whoops, i forgot to finnish the question! For stationary points, dy/dx is zero \frac{1 - ln x}{x^2} = 0 1 - ln x = 0 ln x = 1 log_{e} x = 1 e^1 = x \therefore x = e Testing both sides of the derivative with e On the left side of e: \frac{dy}{dx} > 0 On the right side of e...

English Extension 1 is basically an english course that introduces some more topics related to english, such as literature or philosophy. It is not an alternative to English advanced, just an extension. It is worth 1 unit. Perhaps you should just give Maths extension 1 a go, and if you dont...

Um, you have to do the advanced English course to do Extension English...