Question

consider the equation $y = 4206(e)^{-0.23t}$, which is of the form $y = ae^{rt}$, to fill in the following blanks. the $r$ value for the equation is. therefore, the type of change represented is? because?. note: you can earn partial credit on this problem. preview my answers submit answers show me another you have attempted this problem 0 times. you have 5 attempts left before new version will be requested. you have unlimited attempts remaining. email instructor page generated at 11/29/2025 at 03:36pm cst webwork © 1996-2023 | theme: math4 | ww_version: 2.18 | pg_version 2.18 the webwork project

Explanation:

Step1: Identify the form of the exponential function

The general form of an exponential function is \(y = Ae^{rt}\). In the given equation \(y = 4206e^{- 0.23t}\), we can directly compare it with the general form.

Step2: Determine the value of \(r\)

By comparing \(y = 4206e^{- 0.23t}\) with \(y = Ae^{rt}\), we can see that \(r=-0.23\).

Step3: Determine the type of change

Since \(r=-0.23<0\), the function represents exponential decay. The reason is that when the exponent \(rt\) has a negative coefficient (\(r < 0\)) in the exponential function \(y = Ae^{rt}\), as \(t\) (time, for example) increases, the value of \(e^{rt}\) decreases. Mathematically, if we consider the limit as \(t
ightarrow+\infty\), \(\lim_{t
ightarrow+\infty}e^{-0.23t}=\lim_{t
ightarrow+\infty}\frac{1}{e^{0.23t}} = 0\) (because \(e^{x}\) is an increasing function and \(e^{0.23t}
ightarrow+\infty\) as \(t
ightarrow+\infty\)).

Answer:

The \(r\) value is \(-0.23\). The type of change represented is exponential decay because \(r=-0.23 < 0\).