Question

PRE CALC HELP
A hot bowl of soup cools according to Newtons Law of Cooling. Its temperature (in degrees Fahrenheit) at time t given by < T(t)=68+144e^0.04t. where t is given in minutes.
What is the temperature of the soup after 15 minutes?
thus far I wrote T(15)=68+144e^-0.04(15)

Answer 1

I dont know srry

Answer 2

i am not in calculus

Answer 3

Darn it, thanks anyways

Answer 4

im in precalc.. give just a sec

Answer 5

Thanks:)

Answer 6

ur doing everything correct so far.. ur next step is to multiply and add everything together

Answer 7

basically plug all that u have into ur calculate.. and thats ur answer

Answer 8

c.


125=68+144e^(-0.4t)
(125-68)/144=0.395833333=e^(-.04t)
-0.926762032=.04t
t=23.169
After 23 minutes the soup is 125°F

Answer 9

u dont have to do that, thats finding t, they r trying to find after 15 mins

Answer 10

CAN SOMEONE HELP ON MY QUESTION NOW? I JUST ANSWERED UR QUESTION <33333333333

Answer 11

HELP

Answer 12

Okay thanks :)))

Answer 13

@xoeindi give me a link to your question

Answer 14

ok hold on

Answer 15

Wait I just plugged in what I had to my calculator and I get 330.3 degrees farenheit, but the initial temperature is 212...Something is wrong here.

Answer 16

@16shuston can you help??

Answer 17

was ur calculator in degrees or radians?

Answer 18

It is just in the typo u there...it is -0.04 not the 0.04 as written

\[\large T(t) = 68 + 144e^{-0.04(15)} = ?\]

Answer 19

what kind of calculator do you have? how are you typing in the expression?

Answer 20

You will indeed get that 330 temperature if you do the 0.04 @cutiecomittee123 So make sure you are using the -0.04

Answer 21

oh thanks!! gotcha