We have the following function for this problem:
[tex]\begin{cases}f(t)=0.99,t\leq10 \\ f(t)=0.06(t-10)+0.99,t>10\end{cases}[/tex]Where 't' represents the time in minutes, and our 'f(t)' returns the cost.
For f(t) = 6, we want to know the value for 't'. Clearly, it is not in the first interval, so let's check for t >10.
[tex]6=0.06(t-10)+0.99[/tex]Solving for 't':
[tex]\begin{gathered} 5.01=0.06(t-10)_{} \\ 83.5=t-10 \\ 93.5=t \end{gathered}[/tex]Mary can talk for 93 minutes.
