Recent content by Karma

Construct a 2x2 matrix that is not the zero vector yet satisfies A^2=0 been studying this question for a while...any help?

Well i have two ratios Ta= L(V+c) +L(V-c) and Tb= L / \sqrt{}c^{}2+v^{}2 It asks to just find the ratio of Ta/Tb ... Meaning just a simple division of the Ta over Tb... correct?

One object will be sent to the left and the other object will be sent out to the right. momentum would be conserved in this closed system..

what do you mean?

Consider three bodies with mass m0 m1 and m2 at rest on a frictionless surface. When we insert a compressed spring between bodies 0 and 1 and release the spring we notice that the magnitudes of their accelerations are related by: a1=4a0. When we do the same experiment with bodies 0 and 2 we...

thanks a lot dick..it also states find all values of t with 0<t<20 such taht x(t)=x(0)...so basically all the values of t in this situtation ...are those values that hit 0 on the x-axis ...correct?

And the Period has the equation 2pi/B...what would the "B" represent in my equation dick? would it be pi/3? ... and how would i calculate what the new zeroes of the graph would be?

What does the 1 in the beginning of the equation represent dick?

For the Graph... x(t)=1+2cos(pi/3(t+1)) First i just draw a graph of cos and than... It is just a simple stretch of 1 in the y-direction and a translation of 1 to the left ... correct?

Halls.. My calculator gives me the answer of pi/4 though not 3pi/4 and where does the sqrt of 22 come from?

No the problem does not give the intervals of [-90,90]...just the question... so i can come to the conclusion that 3pi/4 is also correct (as is pi/4)

For the problem sin-(1/sqrt2) ...(sin-.. being arcsin) the answer is pi/4 but is that the only answer because pi/4 lies between [-90,90]? or would it also be right to say PI-4PI=3pi/4? (although i think this is wrong)

Homework Statement evaluate arcsin(cos(7pi/5)) Homework Equations x-pi/2=7pi/5 The Attempt at a Solution sorry don't know where to start.

sec(pi(2x+1))= 2 is what i mean.

The question is solve 2 + cos2x=3cosx for 0<- X <-2pi greater/equal to zero and less than or equal to 2pi. My solution 2+cos2x=3cosx 2+cosx^2-sin^2=3cosx 2+cosx^2-(1-cosx^2)=3cosx 2+2cosx^2-1-3cosx=0 2cos^2-3cosx+1=0 Factoring 2x^2-3x+1=0 (2x-1)(x-1)=0 x=1 or x=1/2...