Question

A loaf of bread is removed from an oven and is left sitting in the kitchen to cool. Its temperature t minutes after being removed from the oven is given by the equation T(t) = 103e-0.0182t + 73. Find the temperature of the kitchen (assuming the temperature remains constant), the initial temperature of the bread when it is removed from the oven, and the temperature of the loaf of bread after 90 minutes. Temperature of the kitchen: a0 Initial temperature of the loaf of bread: a1 Temperature of the loaf of bread after 90 minutes (rounded to the nearest whole degree): a2

Answer 1

initial tempuratire is at t=0

T(0)=103e^0+73=103(1)+73=103+73=176 degrees

after 90 mins

T(90)=103e^-0.0182*90+72

T(90)=103e^-1.638+73

T(90)=19.6589+73

T(90)=92.6589 degrees

Answer 2

A. 73.

B. 176.

C. 93.02.

Explanation:

We have been given an equation
`T(t)=103e^(-0.0182t)+73`, which represents the temperature of loaf, t minutes after being removed from the oven.

A. To find the temperature of kitchen we need to find the temperature of bread as time approaches to infinity.

As time, t, will approach positive infinity our expression
`e^(-0.0182t)` will approach to 0.

`T(\infty)=103e^(-0.0182*\infty)+73`

`T(\infty)=103*0+73`

`T(\infty)=0+73`

`T(\infty)73`

Therefore, the temperature of kitchen is 73.

B. To find the initial temperature of the bread we will substitute t=0 in our given function as when bread was removed from the oven time was 0.

`T(0)=103e^(-0.0182*0)+73`

`T(0)=103e^(0)+73`

`T(0)=103*1+73`

`T(0)=103+73`

`T(0)=176`

Therefore, the initial temperature of loaf of bread was 176.

C. To find the temperature of the loaf of bread after 90 minutes we will substitute t=90 in our given equation.

`T(90)=103e^(-0.0182*90)+73`

`T(90)=103e^(-1.638)+73`

`T(90)=103*0.1943683905943277+73`

`T(90)=20.0199442312157531+73`

`T(90)=93.019944\approx 93.02`

Therefore, the temperature of the loaf of bread after 90 minutes was 93.02.