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5 days ago

8

At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A

.M. the temperature of the tea is 197°F. At 7:15 A.M. the temperature of the tea is 180°F. Alicia will drink the tea when its temperature is 172°F.
Part A

Which of the following shows an exponential cooling equation that models the temperature of the tea?

A. y = 72(0.989)x + 128

B. y = 128(0.989)x + 72

C. y = 200(0.989)x + 72

D. y = 128(0.989)x + 200

Part B

When can Ella drink the tea?

Ella can drink the tea after
7:22
7:18

1 answer:

zzz [9K]5 days ago

6 0

Response:

Part A

The best choice is;

b. y = 128(0.989)x + 72

Part B

Ella can consume the tea post 7:22

Detailed clarification:

Here, we observe the temperature variation with time represented in an exponential equation as follows;

Let the temperature at time x (measured in minutes) be y

Thus, y = a × mˣ + b

Where:

b = Curve shift or the limiting value of the decreasing exponential function as x → ∞

When y = 200, x = 0

This leads to 200 = a × m⁰ + c = a + c

Here, c represents the graph's shift, which is the temperature increase = final temperature = 72°F

Thus, a = 200 - 72 = 128°F

When y = 197, at x = 2 minutes

Thus, 197 = 128·m² + 72 =

m² = (197 - 72)/128 = 125/128

m = √(125/128) = 0.98821

Therefore, the exponential cooling equation is given by;

y = 128 × (0.98821)ˣ + 72

Hence the best choice is b. y = 128(0.989)x + 72

Part B

When the tea's temperature reaches 172°F, we have;

172 = 128 × (0.989)ˣ + 72

Thus, (0.989)ˣ = (172 - 72)/128 = 100/128 = 25/32

log(0.989)ˣ = log(25/32)

x·log(0.989) = log(25/32)

x = log(25/32)/log(0.989) = 22.32 minutes

Thus, Ella can consume the tea post 7:22.

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