Question

the function $f(t) = 7300(1.095)^{60t}$ represents the change in a quantity over $t$ hours. what does the constant 1.095 reveal about the rate of change of the quantity? answer attempt 1 out of 2 the function is $square$ exponentially at a rate of $square$% every growing decaying

Explanation:

Step1: Identify the type of exponential function

The general form of an exponential growth function is \( f(t) = a(b)^{kt} \), where \( b>1 \) indicates growth, and if \( 0 < b < 1 \), it indicates decay. Here, the base of the exponential part is \( 1.095 \), and since \( 1.095>1 \), the function represents growth.

Step2: Calculate the growth rate

To find the growth rate, we use the formula that for \( f(t)=a(1 + r)^{kt} \), where \( r \) is the growth rate (in decimal). Comparing with \( f(t)=7300(1.095)^{60t} \), we have \( 1 + r=1.095 \). Solving for \( r \), we get \( r = 1.095 - 1=0.095 \). To convert this to a percentage, we multiply by \( 100 \), so \( r=0.095\times100 = 9.5\% \). The \( 60t \) in the exponent means the growth occurs every \( \frac{1}{60} \) of an hour (or 1 minute), but the rate per the period corresponding to the base is 9.5% for each "cycle" of the exponent's base. But in terms of the base \( 1.095 \), the growth rate per the unit of the exponent's argument (here, per \( \frac{1}{60} \) hour) is 9.5%. However, the question is about what 1.095 reveals. Since \( 1.095 = 1+ 0.095 \), the 0.095 is 9.5% growth rate for the period corresponding to the exponent's base (i.e., every \( \frac{1}{60} \) hour, but the key is the base shows growth at 9.5% for that period).

Answer:

The function is growing exponentially at a rate of 9.5% every \(\frac{1}{60}\) hour (or per minute, but the rate from the base 1.095 is 9.5% for the period of the exponent's base unit). So filling the blanks: growing, 9.5.