Question

Will mark brainliest

Answer 1

Answer:

a. The exponential model is  $T(t)=69+(111)e^{-0.1368357021t}$  

b. It takes the cake 9.32 minutes to cool to the desired​ temperature

Step-by-step explanation:

Let us solve it by using the Newton's Law of cooling

$T(t)=C+(T_{0}-C)e^{kt}$ , where

• T(t) is the temperature at any given time
• C is the surrounding temperature

• $T_{0}$ is the initial temperature of the heated object
• k is a negative constant
• t is the time

∵ A cake recipe says to bake the cake until the center is 180 °F

∴ $T_{0}$ = 180 ⇒ initial temperature

∵ The room temperature is 69 °F

∴ C = 69 ⇒ surrounding temperature

- Use the table to substitute t and T to find the constant k

∵ At t = 5 minutes, T = 125 °F

∴ $125=69+(180-69)e^{5k}$

- Subtract 69 from both sides

∴ $56=(111)e^{5t}$

- Divide both sides by 111

∴ $\frac{56}{111}=e^{5k}$

- Insert ㏑ for both sides

∴ $ln(\frac{56}{111})=5k$

- Divide both sides by 5

∴ - 0.1368357021 = k

∴ $T(t)=69+(111)e^{-0.1368357021t}$  

a. The exponential model is $T(t)=69+(111)e^{-0.1368357021t}$  

∵ The cake cool to 100 °F

∴ The desired​ temperature is 100°

∴ T(t) = 100 ⇒ cake's temperature at t minute

- Use the model above to find t

∵ $T(t)=69+(111)e^{-0.1368357021t}$  

∴  $100=69+(111)e^{-0.1368357021t}$

- Subtract 69 from both sides

∴ $31=(111)e^{-0.1368357021t}$

- Divide both sides by 111

∴ $\frac{31}{111} =e^{-0.1368357021t}$

- Insert ㏑ for both sides

∴ $ln(\frac{31}{111})=-0.1368357021t$

- Divide both sides by - 0.136835702

∴ 9.321711938 = t

∴ t ≅ 9.32 minutes

b. It takes the cake 9.32 minutes to cool to the desired​ temperature