Please Help. Will Mark Brainliest Answer. 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. Dropdown possible answers: When t = 0, the temperature of the juice is -0.06, 0, 2.718, 32, 40, 72 degrees. As time increases, -32(2.718)^-0.06t gets close and closer to -0.06, 0, 2.718, 32, 40, 72. So, f(t) gets close and closer to -0.06, 0, 2.718, 32, 40, 72.

When t = 0, the temperature of the juice is 40°.

                    As time increases, gets closer and closer to 0.

                    So, f(t) gets close to 72°

    Function representing the temperature of of the juice at any time 't' is,

1). If t = 0,

          degrees

2). If ,

[As denominator of the fraction becomes larger and larger with the increase in the value of t, value of fraction gets smaller and smaller]

3). if ,

  Therefore, when t = 0, the temperature of the juice is 40°.

                    As time increases, gets closer and closer to 0.

                    So, f(t) gets close to 72°.

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When t= 0

f(t)= 40 degrees

The value of −32(2.718)^−0.06t approach 0 as t increases

If −32(2.718)^−0.06t approach 0 as t increases then f(t)=72−32(2.718)−0.06t approach 72

Step-by-step explanation:

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.

f(t)=72−32(2.718)^−0.06t

When t= 0

f(t)=72−32(2.718)^−0.06(0)

f(t)=72−32(2.718)^(0)

f(t)=72−32(1)

f(t)=72−32

f(t)= 40 degrees

As t increases −32(2.718)^−0.06t

Let t= 1

=−32(2.718)^−0.06(1)

= −32(2.718)^−0.06

= -30.14

Let t = 2

=−32(2.718)^−0.06(2)

=−32(2.718)^−0.12

=−32(0.8869)

= -28.38

The value of −32(2.718)^−0.06t approach 0 as t increases

If −32(2.718)^−0.06t approach 0 as t increases then f(t)=72−32(2.718)−0.06t approach 72

i believe the correct answer is b.