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Yea I had that in mind. Crap, what a waste of time and energy. I learnt 3 extra cells.

So there's no chance on comparing a specific cell and with another specified cell?

I kinda know your point but most wickets in international(and the higher grades) cricket are catches. Bowling a batsmen out is pretty hard without any swing or deviation(after bounce). So you kind of have more luck to bowl it almost wide and hope the batter mis-times his shot and get him caught lol

Hope you guys could follow that. If not, it's the part where you learn all about different commerical cells/batteries. Do you really need to know all seven of them? It's a huge section in conquering chem but it hasn't been tested frequently in the hsc. The only question I saw in the past 3-4...

Hahaha, I used that book for year 11. Very ironic name as it doesn't help you understand any topics. Does help you memorise though

It'll be hard to get guess your way for 160. However from what I've heard about UMAT, it tests your logical thinking and all that crap. To me it's either you have it or you don't. The best thing is to prepare I guess...

1/2 integral 2sin2xcos2x and then it would be integrate sin4x

I dunno. I remember seeing one that looked like that in cambridge yr11 3unit. I remember asking you about a circle question though...

OMG! When will any resources be back?!

Let the volume be T T/80 = rate of both pipes Let x be time in minutes T/(x-120) + T/x = T/80 since it's both pipes working together 2x-120 = x(x-120)/80. Solve for x. x will be the larger solution as one of them will give a negative time value

Wow aligning makes your marks look better. Plus uac scales the aligned marks. Big wow to subs that align and scale well

I don't think it's a hard question, you just gotta interpret it correctly. P and Q is really at points a,0 and 0,b. Semi diameters are like the "radius" of the ellipse. OP is a, OQ is b, hence OP^2 + OQ^2 = a^2 + b^2 Oh the points I chose can be interchangable..so P could be 0,b...etc

This probably looks better z1= rcis% z2=rcis% x cis(pi/3) z1^2 = r^2(cis%)^2 z2^2 = r^2(cis%)^2 x cis(pi/3)^2 z1^2 +z2^2 = r^2(cis%)^2 x [1+cis(2pi/3)] =r^2(cis%)^2 x [1/2 + iroot3/2] =r^2(cis%)^2 x cis(pi/3) z1 x z2 = r^2(cis%)^2 x cis(pi/3)

Note the < sign and not <= sign next to the zer0 part. Sicne y=0 is the horizontal asymptote of 1/x^2+1

Well, I wonder how well he/she is going to respond well to that

You can sketch it easily after you've learnt graphs in 4u. Oh and I made a mistake for x^2=>-2, it's no solutions. I got so confused from your > and < sign from your working

x^2>=-2 x>= root2 x<=root2 x^2>=0 All real x It really is just all real x... The graph whould look like a cheap speed bump. It's easy to graph if you do ext2. Just use wolfram

cool thx

Yea...you're almost there just solve the 2 inequalities Oh for the range it's 0 < y <= pi/2 as 1/x^2+1 cannot be 0

For conics, do you have to remember the actual equations? Or should you derive them yourself in the exam? As in for the tangents and normals. It's like prelim parametrics, only more boring