Recent content by BIRUNI

Who Do you think is gonna get the highest mark? I say there are 3 possibilities: Hoque,Yip, Vafa but from BOS perspective, I would say Vafa is fucking smart...

Refer to Vafa's solution for reasoning.

Vafa, Can we please change our brains?

I am sure his solution is correct but the way he said it may make misunderstanding. He should say in s(n): we know that pluvia would like to stay on topic and the statement is "n people would like to stay on topic" S(1): one person would like to stay on topic which is correct. s(k): k...

That Proof was interesting man Acmilan, How would you do it?

answer:This is not binomial probability; it is a comination of harder 2 unit probabilities and selections in 3 unit. you have 4 team and each team contains 10 player and so you have 40 players. you are going to select 5 players so you can do this in 40c5 ways. i)you have 4 number 6 and 4...

Never Call me that. perhaps your dad and mum call each other bastard.

To know About great kouros look at this: http://en.wikipedia.org/wiki/Achaemenid_dynasty to know about persia's goverments and kings since 3200 BCE look at this: http://en.wikipedia.org/wiki/History_of_Iran I read those stuffs and honestly it was enjoyable for me.

To know About great kouros look at this: http://en.wikipedia.org/wiki/Achaemenid_dynasty to know about persia's goverments and kings since 3200 BCE look at this: http://en.wikipedia.org/wiki/History_of_Iran I read those stuffs and honestly it was enjoyable for me.

I am not persian but i know that great draus is the son or grandson of great korous. They all were Persia's kings

there are so many ways such as t formula, subsidary angles, squaring boths sides and etc but I advise auxilary method you can memorize this asinx+bcosx=rsin(x+alpha) asinx-bcosx=rsin(x-alpha) acosx-bsinx=rcos(x+alpha) acosx+bsinx=rcos(x-alpha) r^2=a^2+b^2 but r>0 alpha=tan^(-1)(b/a)

first one: (a-b)^2=a^2-2ab+b^2>=0 a^2+b^2>=2ab similarly b^2+c^2>=2bc a^2+c^2>=2ac by adding these: a^2+b^2+c^2>=ab+ac+bc (a+b+c)^2-2ab-2ac-2bc>=ab+ac+bc (a+b+c)^2>=3(ab+ac+bc) can you do the rest now?

are these enough? but as soulsearcher mentioned, the simplest and quickes way is graphing.

method3: 2sin^x-1=-cos2x for x between o and pi/4 cosx is between 1 and 0.7071.. for x between o and pi/4 cos2x is betwwen 1 and 0.3826.. so -cos2x is between -1 and -0.3826.. hence from these two cosx>-cos2x or you could alternatively say that cosx+cos2x is obviously biiger than zero so the...

method 2: x is between o and pi/4 then cosx is between 0.7071... and 1 for x between o and pi/4 sin x is between o and 0.7971... so 2sin^2x-1 is between o and -1 from these two results it is obvious that cosx>2sin^x-1