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x= 12t y = 16t -5t^2 and you know that when it hits the surface, 5y = x (incline = arctan(1/5)) (16t -5t^2)*5 = 12t 25t^2 - 68t = 0 t(25t-68) = 0 t = 0 gives the initial point.. so at pont of impact, 25t - 68 = 0 t = 68/25 = 2.72 x = 32.64 y = 6.528 so the distance up the ramp is 33.28

1. You can probably improve on your understanding of the material 2. People tend to waste time at the beginning.. if you manage to work at the speed you were working at near the end of the exam throughout the whole exam then you will finish alot more.

try deriving it by using the substitution (x - 1 ) = \frac{3}{2}(e^x - e^{-x})

Most people would recommend more practice.. but I thikn it's these things that you will need to develop, how to do it depends on how you learn: 1. Technical - algebraic manipulations, differentiation, integration etc, you'll need to be reasonably quick and not make mistakes with these. Practice...

That's probably the best way without resorting to heavy machinery. alternatively you can find the equation of 2 perpendicular bisectors of the line segments joining 2 points and solving them to get the center.

Try web proxying

Pretty sure you just need to write down the line with the substitution CORRECTLY.

If you understood what was going on, then 2 weeks revisiosn is enough

This is a special case of the Beta function, for positive integers m,n you can prove using integration by parts that B(m+1,n+1) = (int with limits 1-0) x^m(1-x)^n dx = m!n!/(m+n+1)!

have you tried any of these intrgation question you posted ;P?

P'(x) = x^3 - x^2 - 4x + 4 = x^2(x - 1) - 4(x - 1) = (x^2 - 4)(x - 1) = (x+2)(x-2)(x-1) so P'(x) = 0 for x =2,1,-2 P(1) = 23/12 + c P(2) = 4/3 + c P(-2) = c - 28/3 so P(x) > 0 for all stationary points (hence all x since P increases without bounds for large and small...

1. Don't care about complex u's -> looking at real x's hence u's [- A +- sqrt(A^2 - 4(B-2))] / 2 now since A is positive, the negative sign with the square root results in a bigger absolute value if there's a real root for x, then |u| must be >= 2 want |[- A - sqrt(A^2 - 4(B-2))] / 2| >= 2...

something along those lines (sqrt(|x|) - 1/sqrt(|x|) )^2 >= 0 then expand tidy up and consider the case when x < 0

if you solve for u you get u = [- A +- sqrt(A^2 - 4(B-2))] / 2 besides the usual constraint that the discriminant is positive, you have an additional condition that would preclude some roots : |u| > 2 (think why) so the curve is A^2 - 4(B-2) = 0 you can get he straight line from the other...

the limit is obviously 0 when it exists Should put this in extracurricular, out of scope for 4 u

Differentiate to get P'(x) solve P'(x) = 0 then plug back into P(x), etc.

The only problem is that 1/w = conjugate(w) only when |w| = 1. So that doesn't work Minor detail: since (w =/= 0)

well this takes care of the hard part, the rest is just tidying things

"do you think I might be capable?" - No idea, don't have any info to make a guess. But you should do it

try something like floor(v/10,0)*10 for that.