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1. Three charges are at the corners of an isosceles triangle as shown in the figure. The +q1 = 5.05 uC and -q1 = 5.05 uC charges form a dipole. A) Find the magnitude and direction of the net force that the q2 = 8.50 uc charge exerts on the diple. B) For an axis perpendicular to the line...

1. Find the radius and the interval of convergence for the series: Ʃ n=2 --> inf : [(-1)nxn]/ [4nln(n)] 2.To find the radius, we use the alternating series test. **an+1/an 3. From the alternating series test I find that the limit as n --> inf = 4. So our radius is 4...

would it be 999,998 terms ?

1. Show that the series is convergent and then find how many terms we need to add in order to find the sum with an error less than .001 Ʃ (-1)(n-1)/ √(n+3) from n = 1---> infinity 2. I took the derivative. 3. f(x) = (x+3)-1/2 f'(x) = -1/2 (x+3)-3/2 Then I set up the...

Series -- convergent or divergent? 1. Determine whether the following series is convergent or divergent. When a series is convergent, find the sum. If it diverges, find if it is infinity, - inf, or DNE. Ʃ [(1/na) - 1/ (n+1)a] 2. we are finding if a >0 3. I know that it...

I had found 1/2 earlier and it was counted wrong.

No i had watched a video where a guy did that so I wasn't sure if it would work in this case. I worked it out doing what you told me to do and came up with the answer 1/4. Can you tell me if that is correct? ∫0->π/4 = ∫1/2 (4sin2 2θ) dθ -∫1/2(2 sin 2θ) dθ 2∫(1/2)(1-cos4θ)dθ -...

I got the answer 1/4. Does anyone know if that is correct before I use my last attempt?

can I solve for half of the area using only one of the functions and then doubling it?

Yeah I did that. And the inner one is 2 sin θ and the outer one is 2 sin 2θ They cross at 0 and π/4 Correct?

Im still having trouble with this problem. Can anyone help? Please I am so lost. Just need some help on setting it up and getting it started

yes. I always get things wrong just because its not in the form the software wanted. Miss when we could turn homework in on paper haha

∫sin2(θ) - ∫ sin2(2θ) since sin2 = 1/2(1-cos 2θ) we plug it in and bring the 1/2 to the front of each integral giving us.. 1/2∫(1-cos 2θ) - 1/2∫(1-cos 4θ) am I heading in the right direction? how do I find the bounds?

the answer was more simple than I thought. x2+y2=9x Thanks!

Well I don't exactly know how to even start this problem but let's see. ∫2sin(2θ) - ∫2 sin(θ) do the bounds change for the first one to π/4? Any hints on how to even start this problem. I am not looking for an answer. I do not understand my books explanation so I am really lost :confused:

1. Given the curves r = 2sin(θ) and r = 2sin(2θ), 0≤θ≤π/2, find the area of the region outside the first curve and inside the second curve 2.not sure which equations to use 3. I got 1 and 1/2 as the area and they were wrong. I do not really know how to work this problem. A...

1. Find a Cartesian equation to represent the curve r = 9 cosθ 2. I know that rcosθ= x and cos θ= x/r 3. I got (x-9/2)^2 +y^2 = (9/2)^2 but its coming up wrong when I put it into our online homework. Can anyone help me?