Question

Hello, I need some help with this precalculus question for my homework, please HW Q16

Answer 1

we have the function

$A(t)=A_0*e^{-0.0244t}$

Convert into an equivalent function of the form

$y=a(b)^t$

so

$A(t)=A_0e^{-0.0244t}=A_0(e^{-0.0244})^t=A_0(0.9759)^t$

therefore

the equivalent function is

$A_0=A_0(0.9759)^t$

The rate of decay is equal to

b=1-r

b=0.9759

r=1-b

r=1-0.9759

r=0.0241

The answer Part a is -0.0241 (which is negative because is a decay rate)

Part b

For t=10 years

A_0=500 grams

substitute

$\begin{gathered} A(t)=500*e^{-0.0244*(10)} \\ A(t)=392\text{ grams} \end{gathered}$

The answer part b is 392 grams

Part c

For A(t)=200 grams

Find out the value of t

substitute given values

$\begin{gathered} 200=500e^{-0.0244(t)} \\ \frac{200}{500}=e^{-0.0244(t)} \end{gathered}$

Apply ln on both sides

$\begin{gathered} ln\frac{200}{500}=lne^{-0.0244(t)} \\ \\ ln\frac{2}{5}=-0.0244t \end{gathered}$

t=37.6 years

The answer part c is 37.6 years

Part d

For A(t)=A_0/2

substitute

$\begin{gathered} \frac{A_0}{2}=A_0e^{-0.0244(t)} \\ \frac{1}{2}=e^{-0.0244(t)} \end{gathered}$

Apply ln on both sides

$ln\frac{1}{2}=lne^{-0.0244(t)}$

t=28.4 years

The answer part d is 28.4 years