Solve the problem.
-Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after $x$ years can be described by $\mathrm { A } ( \mathrm { x } ) = 100 \mathrm { e } ^ { - 0.01569 \mathrm { x } }$. How much is remaining after 58 years? Round the answer to the nearest hundredth of a milligram.
A) $0.40 \mathrm { mg }$
B) $248.44 \mathrm { mg }$
C) $91.00 \mathrm { mg }$
D) $40.25 \mathrm { mg }$

Question

Quizplus · Accepted Answer

To find the amount remaining after 58 years, substitute $x = 58$ into the given formula: $A(58) = 100e^{-0.01569 \times 58}$. Calculating this gives approximately $40.25$ milligrams.

Solve the Problem xxx Years Can Be Described By A(x)=100e−0.01569x\mathrm { A } ( \mathrm { x } ) = 100 \mathrm { e } ^ { - 0.01569 \mathrm { x } }A(x)=100e−0.01569x

Solve the Problem $x$ Years Can Be Described By $\mathrm { A } ( \mathrm { x } ) = 100 \mathrm { e } ^ { - 0.01569 \mathrm { x } }$