A restaurant purchased kitchen equipment on January 1, 2017. On January 1, 2019, the value of the equipment was $14 comma 550

Question

vladimir2022

1 month ago

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A restaurant purchased kitchen equipment on January 1, 2017. On January 1, 2019, the value of the equipment was $14 comma 550

. The value after that date was modeled as follows. V(t)equals 14 comma 550 e Superscript negative 0.158 t a) What is the rate of change in the value of the equipment on January 1, 2019

Mathematics

1 answer:

babunello [11.8K] · Answer 1 · 2026-03-02T09:23:41+03:00

Answer:

$\frac{dV(t)}{dt} =$ - 1675.38

Step-by-step explanation:

In 2017, the kitchen equipment's value was recorded at $14,550.

V(0)=$14550

Its following value is represented by $V(t)=14550e^{-0.158t$ .

We need to ascertain the rate of value change as of January 1, 2019.

$V(t)=14550e^{-0.158t$

$\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t$

$\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t$

$\\Let u= -0.158t,\frac{du}{dt}=-0.158$

$\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}$

$\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}$

As of 2019, which is 2 years later, we set t=2.

The rate of value change is

$\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}$

= $\frac{dV(t)}{dt} =-2298.9e^{-0.316}$ = -1675.38

A restaurant purchased kitchen equipment on January​ 1, 2017. On January​ 1, 2019, the value of the equipment was ​$14 comma 550

A restaurant purchased kitchen equipment on January 1, 2017. On January 1, 2019, the value of the equipment was $14 comma 550