Recent content by Abide

I know that haha, as it turns out, in my calculator for some reason you cannot use a and b as algebraic variables, it works perfectly fine using any other variable XD

Can anyone helpme with this? I am not having any luck at all, is this even possible?

Homework Statement So, I know how to solve multiple equations using the cSolve method on the ti 89, but for some reason when I try to solve the following... 80a + 240b = 0 and (80+J79.975)a-80b = 50 by using the following syntax... cSolve(80a + 240b = 0 and (80+J79.975)a-80b = 50...

Ok, so I understand this a lot better now after reading the following link...feeling pretty stupid at the moment... http://www.mathsisfun.com/polar-cartesian-coordinates.html I now understand that the measurement is in Degrees after taking Tangeant inverse of y/x. My question now is, how do...

I don't really know what you mean by explaining the graph...It looks like an elipse, and I believe that polar coordinate measure real and imaginary parts of a number - That may be completely wrong. I am also fully aware that 5/0 is undefined I was just using the definitions presented in my book...

Well, for (5,0) the Polar Coordinates (r,theta) would be r = \sqrt{(5^2)+0^2} = 5 then the theta value would be Tan^{-1}(0) or (-5,0) the Polar Coordinates (r,theta) would be r = \sqrt{(-5^2)+0^2} = 5 then the theta value would be Tan^{-1}(0) for (0,5) the Polar Coordinates (r,theta) would be...

It's an arc in the first quadrant, with a y maximum of 5 and an x maximum of 5. I don't really know what you're asking, I apologize. This is holding me up on every problem I try, I have no clue how you get these values for theta...

This is what I got when I plotted it using wolfram, I assumed giving the max on the Y axis that it would be 5theta, I apologize if this is something simple, I am having a really difficult time understanding it http://www.wolframalpha.com/input/?i=y+%3D+%2825-%28x%5E2%29%29%5E%281%2F2%29

Ok, so it would be 0 to 5\Theta?

Homework Statement Change the Cartesian integral to an equivalent polar integral and evaluate ∫∫dydx The bounds of the first integral (The outermost) are -5 to 5, and the bounds of the second (inner) are 0 to \sqrt{ 25-x^{2}}Homework Equations ∫∫dydx == ∫∫r(dr)(d\Theta) x^{2}+y^{2}=r^{2}...

Ok, but my question was more about whether I am performing the transform correctly , thank you for your response

Homework Statement I'm trying to Solve for an impulse response h(t) Given the excitation signal x(t) and the output signal y(t) x(t) = 4rect(t/2) y(t) = 10[(1-e-(t+1))u(t+1) - (1-e-(t-1))u(t-1)] h(t) = ? y(t) = h(t)*x(t) --> '*' meaning convolution! I am unsure how to take the Fourier...

Homework Statement Find the Impulse response of this system 3y[n]+4y[n-1]+y[n-2] = x[n] + x[n-1] Homework Equations Eigenvalues = -1/3 and -1 hc[n] = k1[-1/3]n+k2[-1]n 3h[n] +4h[n-1] +h[n-2] = δ[n] + δ[n+1] The Attempt at a Solution I know that normally we would plug in hc[n] for two...

Right but that isn't in the terms requested right? We can't have it in terms of output flow..or does y(t) somehow relate to valve resistance?

So then after some algebraic manipulation I have dh/dt= x(t)/A - h/RA This still uses output flow as a parameter, which isn't what the question wants. I'm sorry that I'm so hung up on this problem