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 im on does not have acrobat reader, so i am unable to view the files. the computer at my job = not cool. :yuck: 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 :rofl: . 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: . im planning on taking a college course in math next year, but i dont know what math course to take. i was wondering if any of you had any...
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*
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
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 seperate them.
hmmm, lets 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: