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No. The definition of refraction is a change in direction due to a change in its velocity.

%5Cint%20%5Cfrac%7Bx%5E2+3%7D%7B2x%7Ddx=%5Cint%20%5Cfrac%7Bx%7D%7B2%7D+%5Cfrac%7B3%7D%7B2x%7Ddx%5C%5C%20=%5Cfrac%7Bx%5E2%7D%7B4%7D+%5Cfrac%7B3lnx%7D%7B2%7D+C EDIT: Too slow :(

Would you do it first by assuming BAE isn't 90 degrees and showing that x cannot equal z and then showing that they do equal when BAE is 90? I haven't tried the problem yet but I'll have a go.

Will do. There are some past questions available on the website if you want to have a look, but I'll make sure to post any really interesting questions we get. Luckily for us, our school is paying for the whole trip :)

Hey guys, Is anyone else from BOS going to the Canberra Maths Day at ANU towards the end of the month : http://cpms.anu.edu.au/special/maths-day. For others that have gone in the past, what did you think of it? Our school did a similar thing in Year 10 for maths and I thought it was really...

How long do you reckon it would take to multiply out the whole of the RHS and simplify?

Depends on how good you are. I don't want to sound like a douche, but I'm confident I'll get 75% quite easily. If you understand the concepts well and are good at maths, you should be able to get most questions, except maybe the last one, which would give you over 75%.

If you are good at maths and enjoy it, you will be fine (remember you can always drop it if it becomes too hard). At a rank 90 school though, I think you are moer than capable of it. In regards to time, for me personally, 4U doesn't even take that long- maybe half an hour a day, maybe longer...

I know you didn't get what you wanted or expected but just try and move on (I know its hard- I hate it when I under perform in tests). Just try and rape the next test and seize your rightful rank.

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{3\sqrt{2}-4}{3\sqrt{2}@plus;4}=\frac{(3\sqrt{2}-4)(3\sqrt{2}-4)}{(3\sqrt{2}@plus;4)(3\sqrt{2}-4)} \\=\frac{18-24\sqrt{2}@plus;16}{18-16}\\ =\frac{34-24\sqrt{2}}{2}\\ =17-12\sqrt{2}" target="_blank"><img...

As much as you need to understand everything. In year 11 (I did your subjects bar accounting), I did maybe 5 hours a week including homework and I came first at a selective school because it was productive study and I enjoyed my subjects so a lot of the course was intuitive to me. Others studied...

I got 13.50, which is probably the same answer- you might have rounded at some point or something.

Is this Sydney Tech?

Re: HSC 2012 Marathon :) That's not a bad method. I did this: <a href="http://www.codecogs.com/eqnedit.php?latex=2^x=e^{ln2^{lnx}}\\ =e^{ln2lnx}\\ =(e^{lnx})^{ln2}\\ =x^{ln2}\\ \therefore \int_{0}^{1}2^xdx=\int_{0}^{1} x^{ln2}=\left [ \frac{x^{ln2@plus;1}}{ln2@plus;1} \right ]^{1}_{0}\\...

Re: HSC 2012 Marathon :) How did you do it- there's a relatively simple way (note the word relatively...)

Re: HSC 2012 Marathon :) Yep- My answer was different (1/1+ln2) but they are the same thing. Well done :)

Re: HSC 2012 Marathon :) I might be wrong but I think you can- this is a question from Camrbdige and wolfram gives a definite answer. Try this: We know that e^ln(x)=x. Now try expressing 2^x in a similar form and then integrate.

Re: HSC 2012 Marathon :) Oh crap- I know why I said you guys had a answer different to what I thought the answer should me, I mistyped the question- MASSIVE SORRY. The real question is: <a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^{lnx}dx" target="_blank"><img...

Re: HSC 2012 Marathon :) Is the answer 8-5root3? It was a nice question :)

Re: HSC 2012 Marathon :) This isn't the answer- maybe try again or someone else can have a go.