Answer :
To determine the hourly rate of decrease in the volume of the liquid, we can take the derivative of the volume function with respect to time (\(t\)), which will give us the rate of change of volume with respect to time.
Given that the volume function is represented by \(V(t) = 165 \times (0.97)^t\), we can find its derivative \(V'(t)\) as follows:
\[ V(t) = 165 \times (0.97)^t \]
\[ V'(t) = 165 \times \ln(0.97) \times (0.97)^t \]
This represents the rate of change of volume with respect to time.
So, the hourly rate of decrease in the volume of the liquid is \(V'(t) = 165 \times \ln(0.97) \approx -5.445 \) liters per hour.
Therefore, the volume decreases at a rate of approximately 5.445 liters per hour.
Given that the volume function is represented by \(V(t) = 165 \times (0.97)^t\), we can find its derivative \(V'(t)\) as follows:
\[ V(t) = 165 \times (0.97)^t \]
\[ V'(t) = 165 \times \ln(0.97) \times (0.97)^t \]
This represents the rate of change of volume with respect to time.
So, the hourly rate of decrease in the volume of the liquid is \(V'(t) = 165 \times \ln(0.97) \approx -5.445 \) liters per hour.
Therefore, the volume decreases at a rate of approximately 5.445 liters per hour.
