a hot bowl of soups cools according to Newtons law of cooling. its temperature (in degrees Fahrenheit) at time t is given by T(t)=68+144e^-0.4t, where t is given in minutes.
A. What was the initial temperature of soup?
B. What is the temperature of the soup after 15 minutes?
C. How long after serving is the soup 125 degrees Fahrenheit?

Question

a hot bowl of soups cools according to Newtons law of cooling. its temperature (in degrees Fahrenheit) at time t is given by T(t)=68+144e^-0.4t, where t is given in minutes. 
A. What was the initial temperature of soup?
B. What is the temperature of the soup after 15 minutes?
C. How long after serving is the soup 125 degrees Fahrenheit?

cwrw238 · Answer

initial temperature of the soup occurs when t = 0
so plug in t = o into the formula to get the answer to A.

anonymous · Answer

what the function is T(t)=68+144e^-0.04

cwrw238 · Answer

for B plug in t =15

anonymous · Answer

i just dont know what to do with e

cwrw238 · Answer

for A:

T(0) = 68 + 144e^0

whats the value of e^0 ?

cwrw238 · Answer

what is the value of any number to the power 0?

cwrw238 · Answer

3^0 = 
2^0 =  
10 ^0 
x^0 = ? etc
all = 1

cwrw238 · Answer

so 144 e^0 = 144 * 1 
and
t(0) = 16 + 144 = 212

thats the answer to A

anonymous · Answer

ok so for be i get |dw:1389055537264:dw|