A roast is removed from the freezer at a refrigerator and left on the counter to defrost. The temperature of the roast was -4 degrees centigrade when it was removed from the freezer and t hours later, was inc. at a rate of T'(t) = 7e^.35t. find a formula for the temp of the roast after hours answer is 16-20e^-.35t i dont get where the 16 coming from
what i have so far \[\int\limits7e^-.35t dt\] \[\int\limits7+ \int\limits e^-.35t\] 7[1/.35e^-.35] my answer -20e^-.35+c
ok uhm let's start by correcting your integration\[\int 7e^{-.35t}dt=7\int e^{-.35t}dt\]
try again for me please =)
oops i didnt see u had a mission :P
haha it's cool
are u there @elskydiablo
yes
can you please answer me then, please redo the problem showing your steps
\[7\int\limits e ^{-.35t } dt\] 7[-1/.35e^-.35] 7/-.35[e^-.35] -20.e^-.35?
ur messing up the integration for e
how?
You are writing it incorrectly but achieving a correct answer
can you please write your first step in LaTex
LaTex?
The equation option at the bottom of the dialogue box. It is actually a math script named latex.
here copy this to start \[7\int\limits e ^{-.35t } dt\]
oops [[7\int\limits e ^{-.35t } dt]]
dan change the first and second to last bracket to a \
ok i see
\[7\int\limits e ^{-.35t} dt\] \[7\int\limits e ^{-.35t}\dt] \[7\int\limits 1/-.35e ^{-.35t}\dt] \[7/-.35 e ^{-.35t}\] is this right so far?
\[7\int\limits e ^{-.35t } dt\] \[=(7/-0.35)e ^{-.35t } + c \] \[=-20e ^{-.35t } + c\]
ya u got it right now, u kept forgetting the t
now apply that inital condition to it
T(0)=-4
yes i see what i did wrong.
yea that's right now you need to find your c, right? and you know that T(0)=-4
So what is your next step?
in words, you don't have to type it
t(0)=-20e^-.35(0)+c=-4
right next what should you do?
-20+c=-4 c=16
yep! congrats you did it!
thank you so much
np
listen to my elsky! u need to become MORE SERIOUS ABOUT THESE THINGS!
i expect you to be doing boundary value laplace PDE EQUATIONS by the time i see u again
good bye
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