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why wouldn't you be able to?

graphmatica

it's not gonna really get anywhere near pi/2 max: x=0.6522, y=0.5611

the lower limit is easy to find (just use x=-1). the upper limit is obv wrong. it can't rly be found, afaik.

when doing the working you get to: -1/2 \int_{0}^{\pi }cos(nx)dx\\ = -1/2n [sin(nx)]|_{0}^{\pi }\\ but\ if\ n\ =0,\ you're\ dividing\ by\ zero,\ so\ n\neq 0\\ if\ n=0,\ cos(nx)=cos(0)=1\\ -1/2 \int_{0}^{\pi }cos(nx)dx\\ = -1/2 \int_{0}^{\pi }1dx\\ = -1/2[x]|_{0}^{\pi }\\ = -\pi/2\\ \neq 0

n=/=0 for the condition to hold, as you can see from doing the working of the first part were there any more parts to the question? eg. find U(0), hence find U(1)?

the only good prog/metal bands i know of are death/prog, eg. atheist, gorguts, pestilence

having x in terms of t doesn't really make sense does it lol? i would think x is some constant (i.e. can't vary with t) http://mathworld.wolfram.com/AiryFunctionZeros.html if it was solvable exactly it would say there wouldn't it?

iii) since we have the 3 factors of the 3rd degree polynomial, any other polynomial must have the same factors. so, it must be a constant multiple of x(x-5)(x+5), eg. 3x(x-5)(x+5). but as P(x) is monic, the constant out the front con only be one, .'. no other polynomials iv) d is just the...

sin3\theta = sin2\theta cos\theta +cos2\theta sin\theta \\ = 2sin\theta cos^2\theta + 2cos^2\theta sin\theta -sin\theta \\ = sin\theta (4cos^2\theta -1)\\ sin9\theta=sin3\theta(4cos^2 3\theta-1)\\ = sin\theta(4cos^2\theta-1)(4cos^2 3\theta-1)\\ cos3\theta = cos2\theta cos\theta -sin2\theta...

v ~ rt(D) as the depth decreases, the velocity also decreases as energy is constant, and 1/2 mv2 goes down (assuming constant mass?), the mgh part must go up, so height increases C

i get less than \sum_{3}^{2001}\frac{1}{r-2}-\frac{1}{r}-\sum_{3}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = \sum_{1}^{1999}\frac{1}{r}-\frac{1}{r+2}-\sum_{3}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = 1/1 - 1/3 + 1/2 - 1/4 - \sum_{2000}^{2001}\frac{1}{r}-\frac{1}{r+2}\\ = 11/12-...

f(x)=ax^2+bx+c\\ f[f(x)] = a[f(x)]^2 + b[f(x)] + c \equiv [f(x)]^2\\ $equating coefficients gives a=1, b,c=0$\\ \therefore f(x)=x^2

if you draw the parabola, and the circle, then if they're tangent, the line from the origin to @ must be perpendicular to the tangent (circle geo property). the gradient of that line is a@(@-1)/@ = a(@-1) by differentiating we find the gradient of the tangent to be a(2@-1) so a^2(@-1)(2@-1) =...

\int_{1}^{4} (1/x^2+2x-3)dx\ ?

atheist and cynic definitely

let x = 2sin@

why not negative infinity, or -pi/2 for the inverse? (ie. isn't inverse tan not defined for pi/2? its range is -pi/2 < x < pi/2) edit: though i guess you could say that since the function is always positive, it could only be pi/2

ha. maybe say that the integral gives the same value if the upper limit was (pi/2 - h) as h -> 0, h > 0 or something? i dono edit: or break the initial integral into two integrals, one of which contains the 0 limit, and the other has the pi/2 limit. so then you can divide by cos^2x for the...

2nCn = (2n)!/(n!)^2 to choose the n people for each table round tables, so for each table if you hold one person constant there are then n-1 people to be arranged, which can be done in (n-1)! at each table overall: (2n)!/(n*(n-1)!)2 * (n-1)!2 = (2n)!/n2 edit: another way, is just to arrange...