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Thanks for taking the time to post it up heh, I get the concept now.

Well from what I understood, Big is located on a line 10km away from the beach, however the nearest it can be to X is if its perpendicular to it.

This method is ingenious... Can you explain why this works though?

http://community.boredofstudies.org/showthread.php?t=280826&p=5990429&viewfull=1#post5990429 Answer to your question, just in case it gets hidden by all the junk in that thread.

Oh hang on I didnt see the other part saying not necessarily together. My bad.

\frac{7!}{2!} Simply treat the word KOALA as a whole, then find the number of different arrangements of this 'seven' element word. The reason you dont have a multiplier of 5! is that we dont want KOALA to have a different order. So we treat it so it cant change order, hence our answer.

I am accelerating 3u at the moment, I will hope to do 4u next year hehe. Thats a journey and a pain for another day.

Harro Spiral's sister.

Heh, thanks for the help guys. I do try and label questions though, but 5.5 questions will receive the the first option, 1 question will receive the second one, and 0.5 questions will recieve the third one, and that's only if its not a lazy made exam. Also, next trial paper I do, I will try and...

So do you spend all 5 minutes on it? I usually spend about 1 min on Q7 during reading time. So is it worth not reading anything else, and just focusing the whole reading time on trying to mentally solve Q7?

Ive done many trial papers and the like. I can wizz through the first 5 questions with relative ease. Its not hard whatsoever. Naturally Q6 takes a bit longer for me, no problemo either. Q7, through practising I can usually get up to here with about 50 min-1 hour remaining. I will split the...

\cos (3x+2x) + \cos(3x-2x) - \cos3x=0 \\ \cos 3x \cos 2x - \sin 3x \sin 2x +\cos 3x \cos 2x + \sin 3x \sin 2x -\cos 3x =0 \\ \\ 2\cos 3x \cos 2x -\cos 3x =0 \\ \\ \cos 3x (2\cos 2x -1)=0 I generally like to prove the identity first though, but its a handy method to do, unless you want to...

When Im doing school work Ill do Grid Books, simply because its much easier to read. I use normal lined books when doing trials. And when Im doing questions on BoS, or anything else math related on the internet, I have a writing pad which I just chuck out the papers for when Im done with them.

James Ruse 3U is easy, try more CSSA papers if you have not already.

You have to be careful though... You said you will drop 2 subjects and do 4U, so Im assuming you will have 10 units when you do 4U. This can be a big problem, if you find out you suck really badly at 4U, then its all for nothing really. I suggest that you DO do 4U, however you keep one of the...

LHS=\frac{2\cos\frac{x}{2}-1-\cos x}{2\cos \frac{x}{2}+1+\cos x} \\ \\ \cos x=2\cos^2 \frac{x}{2}-1 \\ \\ LHS=\frac{2\cos0.5x-1-2\cos^20.5x+1}{2\cos0.5x+1+2\cos^20.5x-1}=\frac{2\cos 0.5x(1-\cos 0.5x)}{2\cos 0.5x(1+\cos0.5x)}=\frac{1-\cos \frac{x}{2}}{1+\cos \frac{x}{2}} \\ =RHS \\ \\ \square

You can use the tags {tex} *Insert Latex here* {/tex}. Except replace {} with [] There is a good beginners guide here: http://community.boredofstudies.org/showthread.php?t=234261 Or, if you want to first try out the latex by pressing buttons and stuff...

1) A+B+C=180 \\ \therefore B+C=180-A \\ \\ LHS=2\sin(B+C)\sinA=2\sin(180-A)sinA=2\sin^2A \ \ \ ($Since$ \ \ \sin(180-x)=\sinx) \\ \\ 2\sin^2A=2-2\cos^2A=1-(2\cos^A-1)=1-\cos2A=RHS 2) \tan x \sin x=\cos x (3\tan x - \sec^2 x) \\ \\ \tan x \frac{\sin x}{\cos x}=3\tan x -1-\tan^2 x \\ \\ \tan^2...

Since we are only regarding amplitude and period, then we can fix the Centre of oscillation to be x=0 \therefore \ddot{x}=-n^2x \frac{1}{2}v^2=\int -n^2 x \ dx \\ \frac{1}{2}v^2=\frac{-n^2 x^2}{2}+C \ \ x=0 \rightarrow v=2 \ \ $Max speed of 2 at C.O.O$ \\ \\ \therefore C=2...

Is this the correct explanation then? \int_0^a 0 \ dx = [0x]_0^a=0 \times a - 0 \times 0=0 Also, Ive read from a number of sources that say that it is that reason that we dont write Cs when doing a definite integral. But if its wrong its wrong I guess...