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Yep.

Use u = rt. [x-1]

Lol. It is incomprehensible why you would think that you are going to be studying so intensely such that it would be impossible to sit for your L's. Haha

∫sin3(2x)dx Let u=cos(2x)=>du/dx=-2sin(2x)=>dx=-du/2sin(2x) [using Slidey's substitution] I = -1/2∫sin2(2x)du Using sin2(2x) = 1 - cos2(2x); I = -1/2∫1-cos2(2x)du = -1/2∫1-u2du = -1/2(u - u3/3) = -1/2[cos(2x) - cos3(2x)/3] + c

Haha ...ouch. Good luck man

∫ln(x)dx = xln(x) - x + c

Except he asked to find the integral of sin(x^0) lol. So my answer is correct. :ninja:Granted though, if I knew this was a question directly from an exam, I would of assumed degrees. I assumed that perhaps he stumbled upon this in his working and forgot that x^0 = 1.

Re: How many hours do you have to put in a day to get the best mark in year 11 and 12 It should be more than enough for year 11, so long as you use your time wisely.

y = ex + logex y' = (1+1/x)ex + logex At x = 1; y' = 2e1 + loge1 = 2e, since loge1 = 0.

int. sin(x0) dx = int. sin(1) dx = xsin(1) + C

I think you should probably drop it, if it's not going to count towards your UAI anyway.

I'm pretty sure it's 'finishable', it's all a matter of speed :ninja:.

Re: How do i solve this?? cos2@=1/2 2@=pi/3, 5pi/3, 7pi/3, 11pi/3 [We adjust the domain for 2@: it becomes 0≤2@≤4pi] Then: @=pi/6, 5pi/6, 7pi/6, 11pi/6 Notice I found the other answers by considering where the cosine function is positive in the unit circle, which is the first and fourth...

Well I'm glad that I'm not doing statistics :p

Apparently, yes that's what I've heard as well.

I got the same answer as namburger. Maybe the answers are wrong?

Is anyone else getting tested on prelim stuff for their half yearlies? As in, testing anything from prelim as well as on what we have learnt in year 12? I'm simply wondering if this is standard procedure across most schools. And yes, I am well aware that we can be tested on anything from the...

y=x(x-2)3 y'=(x-2)3+3x(x-2)2=(x-2)2(4x-2) y'=0 will give the stationary points: For y'=0, x=2,1/2. y''=2(x-2)(4x-2)+4(x-2)2=(x-2)(8x-4+4x-8)=(x-2)(12x-12) From this we find x=1, is a point of inflection, and x=2 is a horizontal point of inflection [as it satisfies y'=0 and y''=0], which we can...

Q. If log72 = 0.36 and log73 = 0.56, find the value of log714. log714 = log7(2*7) = log72+log77 = 1+0.36 = 1.36

Re: How long are your periods?? Yeah we normally have 75 minute periods.