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A fossilized plant contains 0.0411 milligrams of a particular radioactive substance. After 75 years, it contains 0.0397 milligrams. Given that the amount of the substance at time t is given by the formula N(t)= N0e^-kt, where N0 is the initial amount of the substance, what is the half-life of the substance, to the nearest year?

If you can, please explain how to get the answer. thanks :)

Respuesta :

caylus
Hello,

1) how to find the coefficients: (in mg and years)
[tex]N(0)=0.0411 \\ N(75)=0.0397=0.0411*e^{-k*75}\\ k= \dfrac{ln( \frac{0.0397}{0.0411}) }{-75 } =4.6209245...*e^{-4}\\ N(x)= \dfrac{0.0411}{2}=0.0411*e^{-4.6209245...e^{-4}*x} \\ x= ln(0.5)/-4.620924..*e^{-4} =1500.0184... years [/tex]

≈1500



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