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Yeah my mind's probably not working or I'm missing something important but how do you evaluate this \frac{1}{\beta^2} \int_0^{\infty}e^{-\frac{y^2}{2\beta^2}} dy

fortunately HSC doesn't have many questions like those :P

You sure there's no diagram or anything?

Make z the subject of the equation and take the moduli of both sides. You should get |w-1| = 2 Which is a circle centred at (1,0) with radius 2

Re: Extracurricular Integration Marathon wow I can't believe I didn't take out a v when I did it,

Re: Extracurricular Integration Marathon \int \frac{e^{2x}+e^{3x}}{coshx} dx

Assuming it's x^2-8x-2 = \frac{12}{x^2} -8x-1 You eliminate the -8x from both sides and multiply both sides by x^2 to get x^4-x^2-12 = 0 \\ \\ \Rightarrow (x^2-4)(x^2+3) = 0 Well i think you can solve it from here.

well I can't recall everything at the top of my head, but for Projectile: Calculus stuff (diff and int) Quadratic equations (cos you're dealing with parabolic curves) trig stuff SMH: Calculus Auxilliary angle method trig functions all the basic trig stuff

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread what's the final answer?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread a) it's a geometric series with the ratio r = -3 and first term a = 4, use the sum of GP formula with n terms and simplify b) let Sn = -728 and solve for n c) ehh you could always do guess and check, but if not, then...

Re: UNSW chit chat thread 2016 I actually approve of this :D

Re: UNSW chit chat thread 2016 WELCOME TO UNSW EVERYONE WOOOOO. Make sure you don't get lost, and get used to the massive trek between upper and lower campus :P. Oh yeah, whilst it's nice to set some goals in your uni life, you need to understand that there's a lot of things that's going...

STOP DERAILING THE THREAD WITH YOUR TRIVIAL MATHEMATICAL DISPUTES. Anyway for lectures you can just skip your normal allocated times that you enrolled for and rock up into the other streams seeing as they don't mark attendance. (for most faculties that is, best to check if they do check them)

f''(x) = 12ex^2 \\ \\ \therefore f'(x) = \int 12ex^2 = 4ex^3+C Now since the point (0,3) is a stationary point that means f'(0) = 0 = 4e(0) + C \Rightarrow C = 0 \\ \\ \therefore f'(x) = 4ex^3 f(x) = \int 4ex^3 = ex^4 + C since the point (0,3) exist on f(x), then f(0) = 3 = e(0) + C...

Using the log rule for differentiation d/dx ln(f(x)) = f'(x)/f(x) \frac{d}{dx} ln(x-x^2) = \frac{\frac{d}{dx}(x-x^2)}{x-x^2} = \frac{1-2x}{x-x^2}

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread and the formal proof behind that identity is based on my solution, no I'm saying that using a calculator (the ones approved by the board of studies) to take the inverse tan would just give some numbers rounded to like 5 decimal...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Sorry my question is, how would you know that taking the inverse tan of \sqrt{2} -1 give you an exact value of pi/8 (i.e the principle value) without using a calculator? seeing that this isn't one of the common trig identities...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread How in the world does taking the inverse directly yield that answer? One way of obtaining that solution is to consider the double angle formula with tan pi/4 tan(\frac{\pi}{4}) =...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread How I was taught was ln(x) = log_e(x) \\ \\ log(x) = log_{10}(x)

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread it's essentially a straight line y = x defined between -pi/2 and pi/2 (not inclusive since that's undefined, so there would be open circles there) followed by the exact same line between pi/2 and 3pi/2, and another between 3pi/2...