Newton's Law of Cooling
Tf=Ts+(Ti-Ts)e^(-kt) where Tf is temp at time t, Ts is temp of surroundings, Ti is temp of object/fluid. So we need to find k first.
200=68+(210-68)e^(-10k)
132=142e^(-10k)
132/142=e^(-10k)
ln(132/142)=-10k
k=-ln(132/142)/10
k≈0.0073 so
T(t)=68+142e^(-0.0073t) so how long until it reaches 180°?
180=68+142e^(-0.0073t)
112=142e^(-0.0073t)
112/142=e^(-0.0073t)
ln(112/142)=-0.0073t
t= -ln(112/142)/(0.0073)
t≈32.51 minutes
the answer that you are looking for is d
Solution
g(x) = 2
- 11x
Plugging in x = -1 in g(x), we get
g(-1) = 2
- 11(-1)
Now we have to simplify it.
g(-1) = 2*1 + 11
g(-1) = 2 + 11
g(-1) = 13
So the answer is D) 13.
Thank you. :)
Answer:
Step-by-step explanation:
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