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

Part A

The most correct option is;

b. y = 128 (0.989) x + 72

Part B

Ella can drink the tea after 7:22

Step-by-step explanation:

Here we have the temperature variation with time given in an exponential equation as follows;

Let the temperature at time x (in minutes) = y

Therefore, y = a * mˣ + b

Where:

b = Shift of the curve or the limit value of the decreasing exponential function as x → ∞

When y = 200, x = 0

Therefore. 200 = a * m⁰ + c = a + c

We note that c is the shift of the graph, the value upon which temperature increases = final temperature = 72°F

Hence a = 200 - 72 = 128°F

When y = 197, x = 2 minutes

Therefore, 197 = 128·m² + 72 =

m² = (197 - 72) / 128 = 125/128

m = √ (125/128) = 0.98821

Hence the exponential equation of cooling is presented in the following equation;

y = 128 * (0.98821) ˣ + 72

Therefore, the most correct option is b. y = 128 (0.989) x + 72

Part B

When the temperature is 172°F we have;

172 = 128 * (0.989) ˣ + 72

∴ (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

Hence Ella can drink the tea after 7:22.