Question

if you pour a cup of coffe that is 200 degrees F and set it on a desk in a room thaat is 68 degrees F and 10 minutes later it is 145 degrees F what temperature will it be 15 minutes after you originally poured it?

Answer 1

Are you just giving your home work to us or?

Idc But any way Can't w8 to see then solve for problems be for i sleep

Answer 2

yeah.. i mean not all my homework but the problems that i couldnt do on my own

Answer 3

Btw you should fan me :P

Answer 4

I do that all the time so i am giving back atm but your not helping me do that are you?

Answer 5

exponential decay function
\[P _{t} = 200e^{-kt} + 68\]
P10 = 145
t = 10
solve for k

Answer 6

Thank you Kassia :D

Answer 7

the negative sign isn't showing for some reason

Answer 8

on the exponent?

Answer 9

yeah or is that just me

Answer 10

i see it

Answer 11

very good cow
i wouldn't have seen to write an exponential equation

Answer 12

how did you know

Answer 13

what was your thought process

Answer 14

Hey myiniayan you should see the problem i did a couple down

Answer 15

i've seen these before
also if you think about how the temp is changing with respect to time, its gradually approaches 68 the longer its out which looks like the exponential graph

Answer 16

so when i solve what does e equal?

Answer 17

oh you will have to take the natural log of both sides

Answer 18

right e^{-t} approaches 0 as t appraoches infinity

Answer 19

so we would have that e{-t}+65 approaches 65 as t approaches infinity

Answer 20

so you seen these in algebra?

Answer 21

yep

Answer 22

im a bit lost e=-65?

Answer 23

upper algebra/pre-calc

Answer 24

yeah this is all precalc

Answer 25

I might have to look into this

Answer 26

kassia, remember you are solving for k
i'll get it started
\[145 = 200e^{-10k} + 68\]
\[77 = 200e^{-10k}\]
\[77/200 = e^{-10k}\]
From here take log of both sides
get k by itself

Answer 27

so we do
\[P_t=(initial)e^{-kt}+(constant) \]
always?

Answer 28

Yes for growth problems make it positive exponent

Answer 29

ok cool thanks cow

Answer 30

I made a mistake, i didn't consider t=0 for initial value
If t=0, P = 200
I have to change it to
\[P_{t} = 132e^{-kt}+68\]
Its like the graph was shifted up 68
sorry kassia, i'll go ahead and solve it
\[145 = 132e^{-10k}+68\]
\[\rightarrow k = 0.0539\]
\[P_{15} = 132e^{-.0539*15}+68\]
\[P_{15} = 126.81\]