Recent content by bfpri

Thats log(n^logc/c^logn)= log(n^logc)-log(c^logn)=log(c)log(n)-log(n)log(c)=0. Ok i got it Thanks

Homework Statement Show that n^(logc)/c^(logn) =1 as n->inf where c is a constant greater than 1Homework Equations The Attempt at a Solution Tried L'hospitals. But the logs mess it up. Even if you assume that logc>1 then the top does eventually become a constant (second derivative). However...

ah, its f(2k+2). Thanks I got it now. Just factor out the f(2k) then factor out the f(2k+2) and you end up with f(2k+2)f(2k+2) Thanks!

So i need to prove f0f1 + f1f2...f2n-1f2n=f2n2 where fn is the n-th fibonacci number.. Attempting to solve this i let n=k . Then k=1 and f22=1 which is 1=1 so that is true. Then i show if the statement is true for k then it is true for k+1, that is f0f1 +...

So i have the equation m*(d2x/dt^2)+c*(dx/dt)+kx=0, where d2x/dt^2 is the second derivative. So I'm given that m=10 kg, and k=28 N/m. At time t=0 the mass is displaced to x=.18m and then released from rest. I need to derive an expression for the displacement x and the velocity v of the...

http://img708.imageshack.us/img708/8897/symimage.gif so I did x=atanΘ. which is x=3tanΘ and dx is 3sec^2\Theta. Then it is \sqrt[]{9tan^2\Theta+9}*3sec^2\Theta which evaluates after factoring to \sqrt[]{9sec^2\Theta}*3sec^2\Theta which is then 3sec\Theta*3sec^2\Theta If i take the 9...

bump.

Is 3.7 one of the answers? Just making sure i did it right.

An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half the water out of the aquarium. (The density of water, p, is 1000 kg/m3, and acceleration due to gravity, g, is 9.8 m/s2.) So in order to truly understand how to do these kind of problems I decided...

The terminal side of an angle theta in standard position passes through (-2,-3). What is sec theta? How would you solve this problem without a calculator? Thanks.