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The functionf(t)=5300(0.35)^(365t) represents the change in a quantity over t years. What does the constant 0.35 reveal about the rate of change of the quantity?

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Answer:

The constant 0.35 reveals that the function is decaying exponentially at a rate of 65% every day

Step-by-step explanation:

Given the function:


f(t)=5300(0.35)^(365t)

This can be written as


f(x)=5300(1-0.65)^(365t)

The constant 0.35 reveals that the function is decaying exponentially at a rate of 65% every day. Notice that it is compounded 365 times in a year, so it is safe to say it is compounded daily.

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