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A = 2*(pi)*(r)^2+2*(pi)*h=2*(pi)*(r)^2+2*(pi)*(r^2)/k, since r=kh dA/dr = 4(pi)r+4(pi)r/k Btw i havent seen the question, just judging by what he wrote up there, this is what should happen

YOu have to apply throughout year 12, each scholarship usually has different deadlines. So try asking your career advisor and search the internet for the scholarships that you can apply for in your chosen field.

Hey guys, how do you get to the leaf lab?

There's some sort of introductory lecture for maple. http://www.maths.unsw.edu.au/students/current/Week0S1.pdf Also see for wednesday, is says something about math1151 is that meant to say math1141? Thanks.

Nope.

1stly: 2ndly: Tertiary Education > NSW / ACT > University of New South Wales > Faculties of Science and Medicine > Lecture full

You're forgetting about Davis( I think that's his name? )- the guy that likes chloe. I'm pretty sure he's the new villain.

That's not his solution, his saying that your solution is equivalent to mine, as we both assumed i to be a constant.

He could have got a scholarship, who knows.

But isnt "i" a constant?, from what I gather it is always equivalent to \sqrt{-1}

Considering the fact that he posted in the UNSW forum, he can't possibly go there......

\int e^{(1+i)x}=\frac{1}{1+i}.e^{(1+i)x}+c

\frac{2\sqrt{3x+1}}{5}=2\\\therefore\sqrt{3x+1}=5\\3x+1=25\\3x=24\\x=8

Nah, we didn't have mechanics in our trial , except we quite comfortably finished it afterwards.

This line should be, du/dt=-e-t

Re: help with Series/trigonometry question and differentiating an equation with logs? 1) i) s_\infty =\frac{a}{1-r}\\\text{ Now, to find r: } r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=-tan^2\theta/1=-tan^2\theta\\ a=1\\\therefore s_\infty = \frac{1}{1+tan^2\theta}=\frac{1}{sec^2\theta}\text{, where...

\int t.e^{-t}.dt\text {, Since this is a product of two different functions, you need to use integration by parts }\\\text{let: } u=t,u'=1\text{ ,you let u =t as it's simpler to integrate the exponential}\\v=-e^{-t},v'=e^{-t}\\\text{ integration by parts: } uv-\int vu'.dt\\\therefore \int...

Don't worry, there's a repeat tonight at 10:40 :P

The new syllabus hasn't been implemented yet, Also I think that was a draft copy? Not entirely sure.

\text{u'v+vu'}=(x+1)^3+3x(x+1)^2\\=(x+1)^2[(x+1)+3x)\text{,By factoring out the }(x+1)^2\\=(x+1)^2(4x+1)