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Steve - ouch!

Answers (15)

Oops. confused y̅ with the moments from the y-axis. y̅ = 5/4 ∫ 1/2 y^2 dx = ∫[0,ln5] 1/2 e-2x dx
nice catch. I originally included the 1/2, but then I noticed that the widths were not all the same, and in my PS I left it out. But I'm sure William picked right up on that... ... ...
Actually, the above solution is wrong. The total is not 20. I need to think on it some more.
If it takes them x hours working together, then 1/2 + 1/3 = 1/x The equation reflects the idea of how much of the job each can do in an hour. Add that up, and it shows how much of the whole job gets done in an hour.
Clearly I was not thinking clearly. As usual, Scott is correct.
oops. that is the minor arc. Change 105 to (360-105)
that would of course be s = rθ = 8.4 * 105(π/180)
I meant (- ∞, 2) If we call the two intervals A and B, then my first post was B-A, not A∩B
Your triangle has y=2 r=√229 so, x=15 Now, all you need to remember is that sinθ = y/r cosθ = x/r tanθ = y/x Now just plug and chug, remembering your double-angle identities.
Ouch - my BAAADDDD!
Sorry - I missed the "three"
Good call. I forgot the 4t But 450/e^8 = 0.15 The answer is still big, though.
1=1^2 assume true for n=k: 1+3+...+(2k-1) = k^2 Now check for n=k+1: 1+3+...+(2k-1)+(2k+1) = k^2+(2k+1) = (k+1)^2 So, P(k) => P(k+1) P(1), so P(all k)
1/0.342 = 2.924 √0.0625 = 0.25
scott got it right. My bad.

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