Question

A container of juice is taken from the refrigerator and poured into a pitcher. The temperature of the juice will warm to room temperature over time. The temperature of the juice can be modeled by the following function: f(t)=72−32(2.718)−0.06t, where t is measured in minutes after the juice is taken out of the refrigerator. Use the drop-down menus to complete the explanation of how the function models the juice warming over time.

Answer 1

Answer:

(1) The temperature of the juice at t = 0 is 40.

(2) 0

(3) 72

Step-by-step explanation:

The temperature of the juice can be modeled by the following function:

$f(t)=72-32\cdot(2.718)^{-0.06t}$

Here t is measured in minutes after the juice is taken out of the refrigerator.

(1)

Compute the temperature of the juice at t = 0 as follows:

$f(t)=72-32\cdot(2.718)^{-0.06t}$

$f(0)=72-32\cdot(2.718)^{-0.06\times 0}$

       $=72-32\times (2.718)^{0}\\\\=72-32\times 1\\\\=72-32\\\\=40$

Thus, the temperature of the juice at t = 0 is 40.

(2)

Compute the value of $-32\cdot(2.718)^{-0.06t}$ for increasing values of t as follows:

For t = 10;

$-32\cdot(2.718)^{-0.06t}=-32\cdot(2.718)^{-0.06\times 10}=-17.56$

For t = 20;

$-32\cdot(2.718)^{-0.06t}=-32\cdot(2.718)^{-0.06\times 20}=-9.64$

For t = 30;

$-32\cdot(2.718)^{-0.06t}=-32\cdot(2.718)^{-0.06\times 30}=-5.29$

For t = 40;

$-32\cdot(2.718)^{-0.06t}=-32\cdot(2.718)^{-0.06\times 40}=-2.90$

For t = 50;

$-32\cdot(2.718)^{-0.06t}=-32\cdot(2.718)^{-0.06\times 50}=-1.59$

For t = 60;

$-32\cdot(2.718)^{-0.06t}=-32\cdot(2.718)^{-0.06\times 60}=-0.87$

Thus, as time increases the value of  $-32\cdot(2.718)^{-0.06t}$ gets closer and closer to 0.

(3)

As the time increases the value of  $-32\cdot(2.718)^{-0.06t}$ gets closer and closer to 0.

This leads to f (t) getting closer and closer to 72.