that's true. Although the real hard thing on jumping is making it look realistic, i.e. as if there was gravity.

Do people click those o.o I always ignore ads.

Well that's a bit too quick, the base syntax was left the same
But yeah the transition from the last beta to 1.1.x was quite a shock to me too

I might write some tutorials too, one time.

^that. But that's prolly 'cause it's actually not dec 21th but somewhere 2145 or so

And of course, it gets f*cked up in windows. It looks like this: ([]° [][] []°) with [] is the N/A block.

So yeah, I had a math exam today and the last question was this: f(x) = x*2^(2x+3), there is a horizontal line y=q that intersects the y axis in A and f(x) in B and C with AB:BC = 1:2. Calculate q in 2 decimals.
graph:

I didn't finish it during the exam (too bad) but this is what I came at afterwards:
f(-x) = f(-3x)  ; final answer is in -x, for why take a look at the graph
-x*2^(-2x+3)=-3x*2^(-6x+3)
-x*2^(-2x)*8=-3x*2^((-2x)^3)*8
-x*2^(-2x)=-3x*2^((-2x)^3)
;;substitute p for 2^(-2x)
-xp=-3xp^3
3p^2=1
p^2=1/3
p=sqrt(1/3)
2^(-2x) = sqrt(1/3)
log2(sqrt(1/3))=-2x
-x = log2(sqrt(sqrt(1/3)))
now input log2(sqrt(sqrt(1/3))) in f and you get y=~-1.83 so q = -1.83
Did I solve this right? Are there faster or better ways to do it?

My lungs almost burst from laughing. no kidding.