$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

8 0

1 month ago

Help me asap please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

zmey [24]

Answer:

D. should be the answer

3 0

2 years ago

Read 2 more answers

X+y+z=10<br> 6x+y+2z=37<br> y-z=10

garik1379 [7]

Ignore the scribbles
Search up Elimination and substitution

7 0

2 years ago

Find a formula for the described function. Express the surface area of a cube as a function of its volume v.

Digiron [165]

Let the side of the cube=a

So the volume will be=V=a³

Then surface area of the cube will be=6a²= $\frac{6a^3}{a}$ = $\frac{6V}{a}$

The above expression showing surface area as a function of Volume. (Answer)

4 0

2 years ago