Recent content by Flappy

Hmm even on another laptop it says i have reached my viewing limit. Can anyone at least "print screen" the pages? I hate to keep bothering hehe

Oh, looks like I already used my view limits for the book. I'll use my sister's laptop to see if I can view them there. Thanks.

Unfortunately google books only shows the first 40 pages of the book. If anyone can scan the lab pages it would be greatly appreciated :)

by Lloyd, 3rd edition? I need a big favor. Can anyone scan Laboratory #16 from the book? It's centripetal acceleration of an object in circular motion. I loan the book to someone and I forgot I need to do the Pre-Lab before Monday. Anyone that can help out I would really appreciate it.

Homework Statement \frac{(kn)!}{(kn+k)!} I was thinking: (kn)! = 1*2*3...(kn) (kn+k)! = 1*2*3...(kn)(kn+k) and I would be left with 1/kn+k But my book has the answer as: \frac{1}{(kn+k)(kn+k-1)...(kn+1)} How can I arrive to that?

Can't use l'hospital's rule since the problem is given as a limit h--> 0. They don't give f(x) "Whoa, wait, what happened to the rest of the denominator, and other term in the numerator?" They cancel out, don't they?

Homework Statement \lim_{h\rightarrow0}= \frac{tan(\frac{\pi}{4}+2h)\ - tan\frac{\pi}{4}}h The Attempt at a Solution Here's what i was able to work out. Use the trig addition formula: \lim_{h\rightarrow0}= \frac{(\frac{tan\frac{\pi}{4}+ tan2h}{1-tan\frac{\pi}{4}tan2h}) -...

Ah, i think i see it. There's a common factor of x^(-5/3)

Hmm I tried to do this: e^{-x^{2}}[ -\frac {8}{3}x^{1/3} - \frac {2}{9}x^{-5/3} + 4x^{7/3} - \frac {2}{3}x^{1/3}] I multiplied the -2x and then took out the e^(-x^2). Would this be right? I'm not sure where you're getting x^(-5/3)

Homework Statement Find the 1st and 2nd derivative: f(x) = x^{1/3}* e^{-x^2} The Attempt at a Solution f'(x) = x^{1/3} * -2xe^{-x^2} + e^{-x^2} * \frac {1}{3}x^{-2/3} I simplified this to: [e^{-x^2}]*[-2x^{4/3} + \frac {1}{3}x^{-2/3}] Also to find the x values is -2x^{4/3} + \frac...

I see, i see. Well, thank you for all the help, appreciated.

Just a question, where did the pi/2 multiplication come from?

I believe you are supposed to get 0. Unless I am not following you. I did the derivative using rules and ended up getting 0. Doing the derivative I got -sin\frac{{\pi}x}{2} * \frac{{\pi}}{2} Then by plugging in 2 you would get 0 * pi/2 which is 0.

Dam, this problem is just nasty. Practicing latex here. this is after you multiply by the conjugate right? \lim_{h\rightarrow0}= \frac{1 - cos(\frac{{\pi}h}{2})^2}{h(1+cos(\frac{{\pi}h}{2})}

Yeah, I see what you did. I was a little confused at first. I forgot to add the 2 under the division on my last 2 lines. Thank you for taking the time by the way.