Question

Mike heats some soup to 216 F. In order to eat the soup, he decides to let the soup cook in his kitchen. The following function represents the temperature of the soup, located in the kitchen with the air temperature of 73 F, after x minutes, where k is the constant rate at which the soup is cooking. T(x)= 73 +143 *e^(kx) If the temperature of the soup is 180 F after 8 minutes, then what is the approximate constant rate of cooling?

Answer 1

Answer:

-0.04

Step-by-step explanation:

The initial temperature of the soup is 216° F. After 8 minutes, the temperature of the soup is 180° F.

Set T(x) equal to 180, and set x equal to 8. Then, solve for r.

Therefore, the approximate constant rate of cooling is -0.04.

Answer 2

Answer:

Mark me brainliest please

Step-by-step explanation: