Answer:
[tex]\frac{dV(t)}{dt} =[/tex] - 1675.38
Step-by-step explanation:
In 2017, the vakue of the kitchen equipment was $14550
V(0)=$14550
Its value after then was modelled by [tex]V(t)=14550e^{-0.158t[/tex]
We are required to find the rate of change in value on January 1, 2019
[tex]V(t)=14550e^{-0.158t[/tex]
[tex]\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t[/tex]
[tex]\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t[/tex]
[tex]\\Let u= -0.158t,\frac{du}{dt}=-0.158[/tex]
[tex]\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}[/tex]
[tex]\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}[/tex]
In 2019, i.e. 2 years after, t=2
The rate of change of the value
[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}[/tex]
=[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.316}[/tex]= - 1675.38
