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Just a tip of what's helped me is don't remember all of them. For example, I just remember that sin^2x+cos^2x=1, then if I want to know the other two, I can just divide through by either sin^2x or cos^2x.

I have weetbix without milk or anything, just the biscuits. Coffee with no milk and no sugar.

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{If it is an A.P., then logc-logb=logb-loga}\\ logc-logb=log(\frac{c}{b})\\ logb-loga=log\frac{b}{a}\\ a,b,c\textup{ are in a geometric progression so }\frac{c}{b}=\frac{b}{a} \\\therefore logc-logb=logb-loga \textup{ and it is an...

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Just take the derivative and prove its less than or greater than 0, depending on the question.}\\ \textup{For the first one...}\\ y=2x-\frac{1}{x}\\ \frac{dy}{dx}=2@plus;\frac{1}{x^2}\\ \frac{dy}{dx}\geq 2 \\\textup{ Therefore its...

Going off that, an increasing CAD is associated with an increase in foreign debt to fund it. Thus, as foreign debt increases, investors might feel that it is no longer safe to invest in that particular nation. To curb this, governments employ contractionary fiscal/monetary policy to keep the CAD...

I haven't tried the Sharp, but I've been using the Casio 82 and I think its a very good calculator...

No, you can assume its right angled because since its a tower, by definition it is vertically up. Therefore it is at a perpendicular with the ground, meaning that its a right angle.

It would have been much easier to use sinx=Opposite/hypotenuse giving h=80tan25 straight away

For this, when I say O I mean the base of the tower and T is the top of the tower. (a)- Use trig in triangle OAT (b)- Use the information you worked out in part a in order to do trig in triangle OBT (c)- Use Pythagoras in triangle OAB Good luck

Since you're already decent, I won't bother about the simple stuff. The thing that defines people in short answer is people who get 5/5 vs. 4/5. To help achieve this, you have to give the marker no reason to deduct marks... 1. Make sure you answer the question and make it obvious that you did...

For 1... Take logs of both sides. Seperate the LHS into sums of logs. Turn it into a quadratic by 'pulling' the powers down. Solve the quadratic. Good luck

Re: 2012 HSC MX2 Marathon That's epic :). I just did the Fibonnaci sequence with it. Thanks.

Re: 2012 HSC MX2 Marathon About that. The question said to use induction but you can do it however you want...

Re: 2012 HSC MX2 Marathon <a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16...

Re: 2012 HSC MX2 Marathon Not sure if serious. Regardless it doesn't matter for you. I got 42/55 in my 2U test because I did it without a calculator to see how I'd go (trig became my new enemy...)

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Let the roots be a and 2a. Using sum and product of roots...}\\ (1)- 3a=2k@plus;4 \\ (2)- 2a^2=k^2@plus;3k@plus;2 \\ \textup{Subbing the value of a from (1) }\\ 2(\frac{2k@plus;4}{3})^2=k^2@plus;3k@plus;2\\...

I wrote a practice essay on a similar topic (Discuss the influence of international organisations and contemporary trading blocs and agreements in promoting globalisation). If you want, I can post/send you my essay (it got 20/20). Otherwise, the key points you want to hit are: - Talk about each...

Re: 2012 HSC MX2 Marathon New question: The lengths of the sides of a triangle are the first three terms of an arithmetic progression with common difference d. If the first term is 1, find the set of possible values of d.

You are right that velocity changes as a function of time but acceleration is always constant. Regardless of where the ball is, the gravitational force downwards will lead to a gravitational acceleration downwards of 9.8ms^-1, thus acceleration is constant.

As theta is greater than 90, it also has to be less than 180 (angle sum of triangle)- second quadrant. It can't be 0 as cosx is zero for x=90,270 (both out of range), it can't be 0.5 because cosx is negative in the second quadrant and it can't be 1.5 because cosx is always less than 1. Maybe the...