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

Filling in our formula with the 2 conditions we are given:
and 
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

Since that value of C is the same as the value of C in the other equation, we sub it in:

Divide both sides by 65 and use the rules of exponents again to get
which simplifies down to

Take the natural log of both sides to get

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:

which simplifies down to

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

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