T = 20 + 80e−0.06t
The temperature T in degrees centigrade of a cooling object is given by the formula
T = 20 + 80e−0.06t
where t is the time in minutes. Find the value of t when T = 40◦C.
(Check your answer). Give the first two digits after the decimal point
know the answer is 23.1049
but
0.06t=ln(80/40)
goes to t = (50/ln(80/40))/3 but gives me 35.27 what is well off
+37084 OK...you are given most everything
T=40(given)= 20 + 80e^(-.06t) I had to assume the '-.0.6t' was an exponent for the 'e' fxn
40= 20 + 80e^(-.06t) Subtract 20 from each side
20 = 80 e^(-.06t) Divide both sides by 80
.25 = e^(-.06t) Take natural log (ln) of both sides
ln(.25) = -.06t Simplify
-1.39629 = -.06t dividey both sides by -.06
t=23.10 minutes
Yah?
