Question

http://prntscr.com/4vjhpt

Answer 1

plug in given values
\[T(10) = 72 + 133 e^{-10k} = 195\]
solve for "k" value
then solve following equation for "t", plugging in the k value
\[72 + 133 e^{-kt} = 180\]

Answer 2

how did 195 become 180?

Answer 3

you are told temp is 180 after 10 min ---> equation 1

you are asked how long before temp is 180 ---> equation 2

Answer 4

so it will be\[72+133(2.718)^k(10)=180\]

Answer 5

??
no T(10) = 195 ..... T(x) = 180

Answer 6

what will be the next step?

Answer 7

to solve for unknown "t" value you need to know what constant "k" is

step 1 is solve equation 1 for k

Answer 8

isolate the e^(-10k) term, then take natural log of both sides

Answer 9

or a simpler way of doing it is just solve for "e^-k" since we have to plug that into equation 2 anyway

Answer 10

\[(e^{-k})^{10} = \frac{123}{133}\]
\[e^{-k} = (\frac{123}{133})^{1/10}\]

Answer 11

the fist one looks easier for me.

Answer 12

so i plug in 2.718 for e?

Answer 13

i dont follow .... no just leave it as "e" we're not computing anything yet

Answer 14

o ok

Answer 15

so do you understand how i got value for "e^-k" ?

Answer 16

kinda

Answer 17

now looking at equation 2:

\[72 + 133 (e^{-k})^t = 180\]
plug in value for e^-k
solve for t
\[(\frac{123}{133})^{t/10} = \frac{108}{133}\]

Answer 18

take log of both sides
\[\frac{t}{10} \ln \frac{123}{133} = \ln \frac{108}{133}\]
\[t = \frac{10 \ln(108/133)}{\ln (123/133)}\]

Answer 19

ok

Answer 20

so is that all?

Answer 21

yep, t is solved for.... which is amount of time it takes coffee to cool to 180 degrees

Answer 22

so how many minutes it took?

Answer 23

really? i just went through the solution
do you understand the expression for value of "t" ?

Answer 24

yes

Answer 25

im not gonna do calculator work for you

Answer 26

why not?

Answer 27

im here to help you understand the process....you do the computing, im sure you know how to enter a log into a calculator

Answer 28

yup and thanks