Recent content by ngogiathuan

Re: What did you think? I put 2x-1 = u, then get sth like 1 / (u+1) u^1/2 Then let u^1/2 = v again (o.O) so u = v^2 sub in and integrate it becomes arctan v i got pi/6 as well cheers :D

Yeah im not sure why we have to break the slices into 2 parts (like terry did) either. I assume they have different radius, but sometimes it doesnt make sense to me. Can someone explain this? Thanks

Not sure if this correct

Im currently working on this question. Havent figured it out yet. It looks pretty hard cause i think we have to find the area bounded the the parabola and its latus rectum. Anyway, give it a try.

Yeah pretty nice question. It looks tricky at a glance though.

Let AB = 4, AC=3, BC=5 and O be the centre of the circle inscribed inside. area of AOB = (1/2) AB * r (r is radius of the circle) area of AOC = (1/2) BC * r area of BOC = (1/2) AC * r so area of triangle ABC = (1/2) (AB+BC+CA) r = 6r However, area of triangle ABC also equal (1/2) AB AC = 1/2 *3...

Here are some hard questions from affinity, from another topic. I post this again so everyone can have a look. By the way, conics2008 u said that the conics and polynomial question here were "pretty straightforward". Im very much appreciated if you could post your working here.

oh i see thanks

:confused: I dont really understand what u meant by component B Can u explain it a bit more. Thanks

From my understanding, the significance of the ratio q/m simply is that it suggests that the rays must have a charge, and a mass ( q and m), so the ray must be particle (instead of being wave) Correct me if im wrong

you have to consider 2 parts of this volume separately, from y=2 to y=0, and the other one from y= 9/4 to y=2, then you find the volume of each slice, and sum up normally. I cant think of any better way The only hard bit of this volume is when u have to find x in term of y, where y= 3x-x^2. by...

Im interested in actuarial studies due to the fact that it involves quite a lot of high level of mathematics (which is my best best best interest) However i painfully admit that im not very good at communication skill, and my english is decent, will it be suitable for this course, and later on...

Im pretty sure that he spent quite some time typing that one up for u, yet ur reply is not very nice to see. I hate cocky people btw

Yep thats exactly right!!! :D The only thing about this working is by letting u = an angle in a right-angled triangle, you already restricted u from 0 to pi/2, which is not the range satisfied the condition that x go from -infinity to +infinity Like Affinity said, if u restrict u from -pi/2 to...

i understand what you're saying but in this case we dont need to use inverse sec function. For example, i have ln(sec u + tan u) right, i want to get back to x, so just sub in tan u = x/a and sec u = sqrt (x^2 + a^2/a) it becomes ln((x+sqrt(x^2+a^2))/a)