It's just the same method the guy in the link above uses.
Basically the light one moves up h0 metres, with a certain acceleration found from F=ma, till the heavy one hits the ground. Then the string becomes loose and the light one becomes a projectile. It's acceleration is now -g. Then...
Multiply the LHS by (1+sinA)/(1+sinA). The top is now (1+sinA)^2 and the bottom is the difference of two squares 1-(sinA)^2 = (cosA)^2
LHS = (1+sinA)^2/(cosA)^2 = [(1+sinA)/cosA]^2 = [1/cosA + tanA]^2
Ask someone at oweek. I took prof eng instead and if it's anything like that course you won't need one.
For next semester you'll definitely want Engineering Mechanics, so if you see anyone selling that pick it up.
I imagine they'll all have lecture slides online. Kopystop (sorry I got the name wrong before) is printing shop off campus. It's a little way down a street off Broadway. Look it up on Google maps.
Definitely don't get the Engineering Computing textbook. There's extensive, easier-to-access and free notes online for MATLAB (not to mention the program's built-in help). For Diff Calc there's notes from kopyshop I believe, so you don't need the textbook. I didn't get the linear algebra one...
Just sub in numbers to see when p(x) goes from negative to positive. It's between 4 and 5, which you might have guessed from part c. That means the modulus sqrt(5/a) is between sqrt(5/5) and sqrt(5/4) or between 1 and sqrt(5)/2.
Another option if you got an ATAR above 97 (which you did) is to go to Sydney, get a distinction average in first year and transfer in second year. You'll be in the same cohort as those who started in law the year before, I believe. Not sure about UNSW's transferring policy.
My Analysis isn't great but I think you've misused the equality sign there.
cat = C
dog = D
4 legs = L
:. C ∈ L
and D ∈ L
That doesn't meant C=D
What I've written in post 10 is, from memory, the transitive property of equality, and is true.
EDIT: Write that next to your proof if you're...