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If you prove the n=3 case for AM-GM, as has been suggested, it fall into place pretty quick. This is the method I've learnt for proving it, if anyone is interested: First do the simple part and prove (x + y)/2 ≥ √(xy) ... (1) Let R = (x+y+z)/3 R = (R + x + y + z)/4 &ge...

Haha, I got one as well. It's pretty amusing. The University of Sydney goes to such lengths to butter up students and get them into their courses. It's no wonder that their courses are so popular.

My writing booklet is bigger than yours.

I'm sure this is posted somewhere else as well but remember that BOS hosts a couple of the old standards packages: http://www.boredofstudies.org/mirror/Standards/2002/ENG/ADV/ It'd be really cool to see what they put out for english in 2004, I can't seem to track it down on the internet.

At the end of the tempest Prospero breaks fourth wall and adresses the audience. I said something like "in light of this, now would be an appropriate time to break the impersonal wall of the literary essay... etc... *stuff about what I learnt thanks to journeys*".

I'd been planning to do a trippy Alice in Wonderland kind of thing - and they gave us a caterpillar :D.

Merry HSC eve.

You really notice it here because this website is good at isolating the overachiever demographic.

You'll die of shame if you get under... 80 You won't tell anyone what you got if you get less than.... 60 You will be satisfied with over... 97.75 You'll be bloody estatic with over... 99 Note: if I got the (*) mark I would probably tell people about it. It's almost an acheivement in...

For the purpose of my own vein reassurance and my curiousity as to how accurate these estimate are here are my school ranks for a UAI estimate: Math ext.1....2/84 Math ext.2....1/49 Physics ........1/66 Chemistry ... 3/71 English adv...10/160 English ext1..17/49 I don't know...

With your subjects you'd get over ninety with bottom level band 5's (or band 3's for the math subjects) and you can do that easily with your kind of motivation. Don't worry 'bout it, you'll kick ass.

Slicing deals in areas of similar cross section. If you do a basic "rotate y = x² around the y axis" then you're summing infinitely thin circular discs. The cross sections you take are parallel to the axis along which you integrate. The key thing is to get the area of the cross...

Sticking together the working throughout the thread (plus some additions you get): You have the curves y = 2√(ax) and y = 2x/3. The distance between them is (2√(ax) - 2x/3) so the radius of each semi-circle is r = (√(ax) - x/3). (because the distance between them is the...

A decent amount.

It looks good given that three of us have converged on it.

Haha, our answers agree now that I've added the coefficients up correctly :p.

And it looks like I can't add numbers together :D. My previously posted answer of 2547a²π/20 should probably be 27a³π/20 , that is, if I even added it up correctly this time.

It works out I think: ax + x²/9 - (2√a/3)x^3/2 ... integrate [(ax²)/2 + x³/27 - (4√a/15)x^5/2] from 0 to 9a. [81a³/2 + 27a³ - 972a³/15] The limits include an 'a' term so everything evens...

Whether or not that method is a rotation is more of a conceptual issue.

They're the curves y = 2√(ax) and y = 2x/3 right? So the radius of each semi-circle is r = (√(ax) - x/3). The volume should then be π/2 ∑ r² δx from 0 to 9a as δx-->0 = 2547a³π/20 Anything wrong with that reasoning?