Question 1163600
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A(t)\ =\ A_o(0.88)^t]


For half of the amount remaining, *[tex \Large \frac{A(t)}{A_o}\ =\ 0.5]


So solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A\(t_{0.5}\)\ =\ (0.88)^t\ =\ 0.5]


For *[tex \Large t]


It would take just as long to go from twice as much to the current amount as it takes to go from the current amount to half as much.  Half-life of a radioactive material is a constant for that particular element.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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