Recent content by p53ud0 dr34m5

Q=C_{\rho*v}^y i agree with whozum. you should just use logarithms. it doesn't matter about the subscript.

wouldnt the braking create a greater magnitude of friction or create friction if there wasnt any initially present?

i agree with OlderDan. h(x)=f(g(x)) if you have that, then: h'(x)=f'(g(x))*g'(x) so, sine is not the derivative of itself.

antigravitation, the heim theory looks quite interesting. i love to look at new theories, because i have little faith in the current theories. unfortunately, the computer I am on does not have acrobat reader, so i am unable to view the files. the computer at my job = not cool. when i get...

im always looking for a challenge, so i think ill challenge myself with real analysis. the description you provided makes it seem like calculus with proofs. that ought to be fun :smile: . ill talk to my bc teacher and see what she thinks. thanks quasar987 :wink: .

what type of problems are involved in real analysis and abstract algebra? hehe, guess the wrong section to post this...moderators feel free to move it :smile:

i just finished calculus bc ap at my school, and i have no more math classes to take there; i need all 4 years of math to graduate with honors :confused: . I am planning on taking a college course in math next year, but i don't know what math course to take. i was wondering if any of you had...

shouldnt it be: \int sec^2x~tanx~dx=\int\frac{sinx}{cos^3x}dx then, you could do u-sub and set u=cosx and go from there? *Edit* dextercioby, you put sinx instead of cosx. i think tanx = sinx/cosx *Edit*

have you figured out the velocity in terms of the period yet?

1) you have \frac{dy}{dx}=2x+1[/tex] so, you set up the integral: \int dy = \int(2x+1)dx solve the indefinite integral, plug in for x and y, and solve for C. 3) this a u-sub problem, where u=3x^2-7 and [itex]du=6xdx 2) use the same concept as in #1

yes, the first one looks fairly simple. what do you know about parametric equations and their derivatives?

NaF having a smaller radius could account for its higher boiling point. If an atom has a smaller radius, that means the elements are closer together, so more heat is needed to separate them.

well, here's what i do to find the point of intersection of two lines. i just set them equal to each other.

hmmm, let's see: \frac{1}{2}mv^2+\frac{1}{2}I\omega^2=mgh \frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{3}mr^2\frac{v^2}{r^2})=mgh \frac{1}{2}mv^2+\frac{1}{3}mv^2=mgh 3mv^2+2mv^2=6mgh 5v^2=6gh \frac{5v^2}{6g}=h h=2.125~meters yea, i got the same thing as cde42003. :confused:

what is the inertia of a hollow sphere? I=\frac{2}{3}mr^2?