Question

THE TEMPERATURE T OF FOOD PUT IN THE FREEZER IS T=700/(t^2+4t+10)
where t is the time n hours. FIND THE CHANGE OF T IN RESPECT TO t at each of the following times
a)t=1
b)t=3
c)t=5

Answer 1

take the derivative and plug in the numbers

Answer 2

can you please tell me how??

Answer 3

derivative is
\[T'=-\frac{1400(t+2)}{(t^2+4t+10)^2}\]

Answer 4

the derivative of
\[\frac{1}{f(t)}\] is
\[-\frac{f'(t)}{f^2(t)}\] so it is quick in this case

Answer 5

i used
\[f(t)=t^2+4t+10\]
\[f'(t)=2t+4=2(t+2)\] and got the answer that way. not you just have to substitute in numbers for t

Answer 6

can u do it for me plase i amlost???