Question

A population of bacteria is changing exponentially according to the function below how can you describe the bacteria population?

B(t)=520(0.4)^1.8t + 11

A.the population is negative
B.the population is growing
C.the population is approaching a constant value
D.the population is decaying

Answer 1

it sis growing, that is obvious by the + sign (i.e. + 11) it is also growing exponentially, i.e (o.4)^1.8t [if we assume t is positive]

Answer 2

`\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{elapsed time}\\ \end{cases} \\\\\\ \stackrel{A}{B(t)}=\stackrel{P}{520}(\stackrel{1-r}{0.4})^(1.8t)+11`

the tell-tale factor on this exponential equations is the "growth factor", namely the parenthesized value.

if the value is greater than 1, is growth, if the value is less than 1, is decay.

notice in this case is 0.4, so the "growth factor" is less than 1.