Respuesta :

(a)

[tex]\text{ We are to write the function y=50}e^{-0.21t}\text{ in the form of y=}pa^t[/tex][tex]\begin{gathered} \text{ y=50}e^{-0.21t} \\ y\text{ = 50(}e^{-0.21})^t \\ \text{but }e^{-0.21}=0.810584 \\ \text{Hence, y=50(0.810584)}^t \end{gathered}[/tex][tex]\begin{gathered} \text{ By comparison with y =pa}^t, \\ a\text{ = 0.8106 ( to 4 decimal places)} \end{gathered}[/tex]

(b)

By comaparing

[tex]\begin{gathered} y=50(0.810584)^t^{} \\ \text{with }y=p\mleft(1+r\mright)^t \\ \end{gathered}[/tex]

The value of ( 1 + r ) is 0.810584 which is less than 1. This indicates that the exponential function is a decay function.

Thus, ( 1 + r ) = 0.810584

1 + r = 0.810584

r = 0.810584 -1

r = -0.189416

The negative indicates the it is a decay

in percentage, r = 0.189416 x 100

r = 18.9416%

Hence, The annual decay rate is 18.94 % per year

(c)

The continous

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