An asteroid is spotted moving directly toward the center of Starbase Alpha. The frightened residents fire a missile at the asteriod, which breaks it into two chunks, one with 2.4 times the mass of the other. The chunks both pass the starbase at the same time. If the lighter chunk passes 1800 m...
Consider the ideal I of Q[x] generated by the two polynomials f = x^2+1 and g=x^6+x^3+x+1
a) find h in Q[x] such that I=<h>
b) find two polynomials s, t in Q[x] such that h=sf+tg
Can someone, please, show me an example of when you are better of with parabolic cylindrical coordinates than with cartesian coordinates when computing a triple integral over a solid?
Given: A massless particle revolving in a circle with a rotational velocity = (2+sin(a))
To Find: Y-axis acceleration
Method #1 (from rotational acceleration)
Y-axis acceleration = (2+sin(a))(cos(a))^2
Method #2 (from Y-axis velocity)
Y-axis acceleration =...
\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = (cos x - sin x)(cos x + sin x)
does it ring a bell now? You have to do something with the numerator.
To prove that lim f(x,y) does not exist, it suffices to show that the limit along one curve into (a,b) differs from the limit along a second curve. If lim f(x,y) does exist, however, then computing limits along individual curves will prove nothing (although, such computations will likely help to...
You can show that the limit is undefined by showing that the limit depends on how you approach zero.
You can show that f(x,y) \longrightarrow \frac{1}{2} if you approach zero along the curve (\sqrt{y},y).
So you have showed that the limit is both 0 and 1/2, i.e. it must be undefined.
You seem to have misunderstood L'Hospitals rule.
L'Hospitals rule states that \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} if \lim f(x) and \lim g(x) are both zero or ±\infty. L'Hospitals rule isn't necessary. Just use the standard limit for \frac{sin(x)}{x}
You are done!
You just showed that I-T is invertible/bijective by showing that (I-T)(I+T) = (I+T)(I-T) = I. Which means, by definition, (I-T)^{-1} = (I+T)
Dr. Math has answered a lot of questions concerning the sum of consecutive squares here. He explains that there are several ways to derive the formula.