Question

the function $f(t) = 830(0.9935)^{\frac{t}{7}}$ represents the change in a quantity over $t$ days. what does the constant 0.9935 reveal about the rate of change of the quantity?
answer attempt 1 out of 2
the function is $quad$ exponentially at a rate of $square$% every
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decaying
growing

Explanation:

Step1: Identify the type of exponential function

The general form of an exponential function is \( f(t) = a(b)^t \), where if \( 0 < b < 1 \), it is a decay function, and if \( b>1 \), it is a growth function. Here, \( b = 0.9935 \), which is less than 1, so it's a decay function.

Step2: Calculate the decay rate

The decay rate \( r \) is found using the formula \( b = 1 - r \). So, \( r = 1 - 0.9935 = 0.0065 \). To convert this to a percentage, we multiply by 100: \( 0.0065\times100 = 0.65\% \).

Answer:

The function is decaying exponentially at a rate of \( 0.65\% \) every (the time period here is related to the exponent, but from the function \( f(t)=830(0.9935)^{\frac{t}{7}} \), the base's effect is over \( \frac{t}{7} \), but for the rate per the base's cycle, we found the rate as above). So filling in the blanks: decaying, \( 0.65 \).