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
Answer:
Greater than.
Step-by-step explanation:
Alright, let's factor this to get the answer.
3k^2-10k+7
To find the factors, we want to think "What will add up to -10, and multiply to (+)7?"
Because the leading coefficient is 3, we know that we can take one factor of 7 and multiply it by 3.
Thus, this factors to
(3k-7)(k-1)
(if you FOIL it it should come out to be the original equation)
From this, set both of those [(3k-7) and (k-1)] equal to zero and solve
3k-7=0
3k=7
/3
k=7/3
or
k=1
Answer:
6:15 6 to 15 it's the same o
