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:
gravitons nor gravitational fields exists. they are both just abstract concepts that try to explain something that has not yet been conceived. want to know something else? electric fields dont exist either!
heh, saphic_yellow. how did you think of the physics exam. i took it. i didnt think it was bad at all. unfortunately, when i view a test as easy, that is usually not the outcome. :frown: im too impatient to test. i hope you did well, though.
are you trying to do:
\int\frac{dx}{e^x+e^{-x}}
if so, for a=b=c:
\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C
i hope that helps
isnt question three a geometric series?
\sum_{n=0}^{\infty}20(\frac{9}{10})^n
if thats the case, what do you know about geometric series that makes them converge or diverge?
arg, i just went to a bc calculus session at my school. we did a lagrange error problem. the whole concept baffles me for some reason. im sure many wont be on the ap test, because i managed a 5 on the practice one; however, i would love to learn to do it. anyone want to do a sample problem...
7/2 exponent is like my square root to the 7th
-\frac{(\sqrt{81-x^2})^2}{7}=-\frac{(81-x^2)^{7/2)}{7}
also, we had to use trig substitution, so that left out the easy 81-x^2 sub
im hoping i worked this out right; its long:
\int x(81-x^2)^{5/2}dx
the integral contains a^2-x^2, so i set x=asin\theta. that would make x=9sin\theta and dx=9cos\theta d\theta:
\int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta...
correct! they have the same masses, so the cue ball transfers all of its energy to the eightball making it move forward and the cue ball come to a halt.
hehe, yea i left out the 2...so its half the integral when you work it out. this computer at work was frustrating me, so i forgot it. (my excuse :rofl: )