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Is a(t) = 1200(1 + .03)t a decay function?

Is a(t) = 1200(1 + .03)t a decay function?
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Is a(t) = 1200(1 + .03)t a decay function?

User Mohsin Mehmood
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1 Answer

5 votes

Answer:

Explanation:

No, it is an exponential growth function because 1.03>1. An exponential decay function would have a rate less than 1.

User Shawntell
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