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Trebla may have been thinking the 'converse' was trivial ie. If P is sum of two real poly's then P>=0.

Try to do the harder q's in my MATH1151 Linear Alg hw showing riguorous working and then you may be able to say 4u>1st yr.

I'm in the same situation, I've got so much work that i've had to just scrap some stuff like hw i hadn't completed since week 2 and so on. I'm primarily focusing on assignments and cram when tests come around. So cbf doing work for uni now lol. Anyways, may see you around uni sometime.

Hey, would've replied to ur rep with a rep but i can't (error message pops up). Uni's ok but it's too much work especially for lazy ppl like me. Anywyas, how's civil eng going?

Anyways, 1st yr uni maths >>> 4u maths.

I would argue that Math1131/1141/1151 are conceptually much harder overall than MX2. Plus, the marking is very harsh as you're marked to standards of pure mathematicians not HSC students (in my experience).

I don't understand ur 1st sentence lol. In regards to q, it's pretty useless, i would just get over it and do other q's.

Hey, i haven't been to lectures in ages but i think i might remember you. In regards to SCIF, the subject is okay but there's a lot of irrelevant stuff to be learnt in SCIF eg global warming, AI etc. However, it might be worth it in the long-term as you'll get to major in QR and the titles...

No, but thanks anyway for the suggestion lol

cos(x)=1-x^2/2!+x^4/4!+...=Sigma {k=0,n} (-1)^k.(x^(2k))/(2k)! So what they are asking is find: Sigma {k=2,n) (-1)^k.(x^(2k))/(2k)!>0.001, ie find the error in using cos(x)=1-x^2/2 opposed to the sum stated above. But, common sense dictates the answer is x=0 for f(x)=cos(x)-1+x^2/2=0. So...

Re: The Official Football Thread Yeah, I can't stand Serie A. Has to be the worst league to watch. Anyone hear follow La Liga? I'm still hoping RMad can still win the league

A{segment}=1/2.r^2.@-(1/2.r^2.sin@) =(1/2)(r^2)(@-sin@) dA/d@=(1/2)(r^2)(1-cos@) dA/dt=dA/d@.d@/dt =(1/2)(r^2)(1-cos@).(0.1) =(1/20)(r^2)(1-cos@) Let dA/dt=0: cos@=1 => @=0 Idk, if the above is correct as i haven't done rates of change in ages.

Yeah, self-teaching is really easy as the topics are pretty simple and usually easy to grasp (in regards to math1151). Hence, i haven't attended any lectures since week 2. However, my lecturers are pretty smart (Tran and Du) even though they are hard to understand sometimes lol.

Let u=pi/2-x: du=-dx I=S [0,pi/2] dx/(1+(tan(x))^k) = -S [pi/2,0] du/(1+(cot(u))^k) = S [0,pi/2] (tan(u))^k du/((tan(u))^k+1) = S [0,pi/2] 1-(1/((tan(u))^k+1) du = S [0,pi/2] du -I Thus, 2I=pi/2 => I=pi/4 Nice Q. Edit: Beaten

Hi guys, in regards to Actuarial Studies, is it better to be very adept in Maths & Stats or have a deep understand of the accting and eco aspects of economy/markets/whatever is relevant. Also, is it possible to do a Masters in Actuarial Studies (not saying i will in any certain terms) in just a...

x=Asin(nt+K)+B x'=Ancos(nt+K) x"=-An^2.sin(nt+K) Max speed occurs at x"=0 ie. t=-K/n. x'=An.cos(0)=An. Thus, max speed is An. Time for 1 osscillation is: T=2pi/n v=s/t => v=4A/(2pi/n) => v=2An/pi At max speed: An:2An/pi 1:2/pi => 2 : pi ratio

A=2, T=12.5hrs => n=2pi/(12.5). f=-Acos(@)+B =-2cos( (4pi/25)t)+8, is your equation (i think) but i can't really remember useless maths from HSC. 9=-2cos((4pi/25)t)+8 -1/2=cos((4pi/25)t) 2pi/3=(4pi/25)t 50/12=t Thus, Time is 6+50/12=10+1/6=10:10am

tan(pi/8)=sqrt((2-sqrt(2))/(2+sqrt(2))) =(2-sqrt(2))/sqrt(4-2) [Multiplying by:sqrt(2+sqrt(2))] =sqrt(2)(sqrt(2)-1)/sqrt(2) =sqrt(2)-1

But you state: "find the exact value..."

If there's an exact value for tan(35), i'd like to hear it lol.