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sonofjohn
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5) As you can see ln(sin(1)) +1/ln.5 does not yield one of the above answers. Can someone tell me where I went wrong?
sonofjohn said:isn't the second derivative of f^2x = 2?
sonofjohn said:could I possibly multiply them and get f(2x^2)?
sonofjohn said:it should be 2f'(x)f'(x) + f"(x)2f(x) = 42. I messed up the chain rule again :(
sonofjohn said:Now 7 is another problem I have yet to face. I see that t is in seconds so I think I should multiply t by 60 to convert it to minutes. Should I then plug 1 in for y and take the derivative?
sonofjohn said:I see where the 60 seconds is coming from in the e^60k but not where the y/2 comes from on the other side of the equation.
jgens said:Write your equation in terms of y and then solve.
Edit: try this. We know y = e^kt. If we let y designate the initial amount of the isotope then y = e^k. Now based on the information we may express the half life as: y/2 = e^60k or y = 2e^60k. Equate the two and solve.