Question 1130983
A cup of coffee is heated to 180°F and placed in a room that maintains a temperature of 60°F. The temperature of the coffee after t minutes is given by 
{{{T(t) = 60 + 115e^(-0.042t)}}}.
:
(a) Find the temperature, to the nearest degree, of the coffee 20 minutes after it is placed in the room._________°F 
{{{T(t) = 60 + 115e^(-0.042*20)}}}
{{{T(t) = 60 + 115e^(-.84)}}}
T(t) = 60 + 115(.4317)
T(t) = 60 + 49.6
T(t) = 109.6 degrees after 20 min
:
(b) Determine when, to the nearest tenth of a minute, the temperature of the coffee will reach 145°F.______________min
{{{60 + 115e^(-0.042t)= 145}}}
{{{115e^(-0.042t)= 145 - 60}}} 
{{{115e^(-0.042t)= 85}}}
{{{e^(-0.042t)= 85/115}}}
{{{e^(-0.042t)= .739}}}
using nat logs
-.042t*ln(e) = ln(.739)
ln(e) = 1, therefore
-.042t = -302
t = {{{(-301)/(-.042)}}} 
t = +7.2 minutes it will be 145 degrees