Cleopatra uploaded a funny cat video on her website, which rapidly gains views over time. $\newline$ The relationship between the elapsed time, $t$ , in days, since Cleopatra uploaded the video, and the total number of views, $V(t)$ , is modeled by the following function: $\newline$ $V_{\text {day }}(t)=580 \cdot(1.17)^{t}$ $\newline$ Complete the following sentence about the weekly rate of change in the number of views. $\newline$ Round your answer to two decimal places. $\newline$ Every week, the number of views grows by a factor of

Full solution

Q. Cleopatra uploaded a funny cat video on her website, which rapidly gains views over time.

\newline

The relationship between the elapsed time,

t

, in days, since Cleopatra uploaded the video, and the total number of views,

V(t)

, is modeled by the following function:

\newline

V_{\text {day }}(t)=580 \cdot(1.17)^{t}

\newline

Complete the following sentence about the weekly rate of change in the number of views.

\newline

Round your answer to two decimal places.

\newline

Every week, the number of views grows by a factor of

Understand function and question: Understand the function and what is being asked. $\newline$ The function $V(t) = 580 \times (1.17)^t$ models the number of views as a function of time in days. We are asked to find the weekly rate of change in the number of views, which means we need to find the factor by which the views grow every $7$ days.
Calculate factor for one week: Calculate the factor for one week. $\newline$ Since one week is equivalent to $7$ days, we need to calculate the value of the function for $t = 7$ . $\newline$ $V(7) = 580 \times (1.17)^7$
Perform calculation for $t = 7$ : Perform the calculation for $t = 7$ . $V(7) = 580 \times (1.17)^7$ To find the growth factor for one week, we only need to calculate $(1.17)^7$ , since $580$ is a constant multiplier and does not affect the rate of change. $(1.17)^7 \approx 2.2233421$
Round result to two decimal places: Round the result to two decimal places. $\newline$ The growth factor for one week, rounded to two decimal places, is approximately $2.22$ .