Answer:
Step-by-step explanation:
I solved this using initial conditions and calculus, so I hope that's what you are doing in math. It's actually NOT calculus, just a concept that is taught in calculus.
The initial condition formula we need is
![y=Ce^{kt}](https://tex.z-dn.net/?f=y%3DCe%5E%7Bkt%7D)
Filling in our formula with the 2 conditions we are given:
and ![85=Ce^{15k}](https://tex.z-dn.net/?f=85%3DCe%5E%7B15k%7D)
With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k. If
then
which, by exponential rules is the same as
![C=65e^{-10k}](https://tex.z-dn.net/?f=C%3D65e%5E%7B-10k%7D)
Since that value of C is the same as the value of C in the other equation, we sub it in:
![85=65e^{-10k}(e^{15k})](https://tex.z-dn.net/?f=85%3D65e%5E%7B-10k%7D%28e%5E%7B15k%7D%29)
Divide both sides by 65 and use the rules of exponents again to get
which simplifies down to
![\frac{85}{65}=e^{5k}](https://tex.z-dn.net/?f=%5Cfrac%7B85%7D%7B65%7D%3De%5E%7B5k%7D)
Take the natural log of both sides to get
![ln(\frac{85}{65})=5k](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B85%7D%7B65%7D%29%3D5k)
Do the log thing on your calculator to get
.2682639866 = 5k and divide both sides by 5 to find k:
k = .0536527973
Now that we have k, we sub THAT value in to one of the original equations to find C:
![65=Ce^{10(.0536527973)}](https://tex.z-dn.net/?f=65%3DCe%5E%7B10%28.0536527973%29%7D)
which simplifies down to
![65=Ce^{.536527973}](https://tex.z-dn.net/?f=65%3DCe%5E%7B.536527973%7D)
Raise e to that power on your calculator to get
65 = C(1.710059171) and divide to solve for C:
C = 38.01038064
Now sub in k and C to the final problem when t = 23:
which simplifies a bit to
![y=38.01038064e^{1.234014338}](https://tex.z-dn.net/?f=y%3D38.01038064e%5E%7B1.234014338%7D)
Raise e to that power on your calculator to get
y = 38.01038064(3.434991111) and
finally, the temp at 23 minutes is
130.565