Recent content by spideyunlimit

x+2y=3 x^2-4y^2=-33 -33 = (x+2y)(x-2y) got it now?

Righto! I'm sure the poster will get it, just try again carefully.

LOL dude, use a^2 - b^2 = (a+b)(a-b) and u have the a+b eqn. with you.. Get the a-b eqn from the above equation, and solve a+b and a-b for a and b.

Work = dot product of force vector and displacement vector.

It's period will become 8 (obvious). And determine odd or even function: if f(-x) = f(x) then even function if f(-x) = -f(x) then odd function.

DO it again, you'll get 0^4 + n^4

I think you just need to find the hypotenuse using trigonometry.

Super application of mechanics!

Sorry, my bad, I meant to say even function.

See, g(x) is an odd function.. g(x) = g(-x) [because of the mod and square] I think you got it now!

i said sec x instead of cosx because else you get 0 too, but anyways, right! You won't get the afore said series :| Hmmm, Let me think.

quadrant wise... in first quadrant cos is positive so sec is positive too and signum will give 1. in 2nd and 3rd quadrant cos is negative so sec is also negative and signum will give -1 in both second and third quadrants. then in 4th quadrant, positive so +1.

signum (sec(x))

@poster - the method u used is only valid for two linear expressions' product, but for your one you'll have to use Bn instead of just B.

1/[(n^3)+n] = 1 / (n)(n^2 + 1) = 1 + n^2 - n^2 / (n)(n^2 + 1) = (1/n) - [n/(n^2 +1)]