Answer:
't' lies in the interval (0.2476, 4.038)
Explanation:
Given:
The function for temperature (T) as:
[tex]T=\frac{6t}{t^2+1}+98.6[/tex]
Now, according to the condition given in the question
[tex](\frac{6t}{t^2+1}+98.6)>100[/tex]
or
[tex]\frac{6t}{t^2+1}>1.4[/tex]
or
[tex]{6t}>({t^2+1})1.4[/tex]
or
6t>(1.4t² + 1.4)
or
(1.4t² + 1.4 - 6t) < 0
on solving the quadratic
we get the intervals as:
t = 0.2476 and t = 4.038
Therefore, the 't' lies in the interval (0.2476, 4.038)
14t^2-80t+28<0
7t^2-40t+14<0
Solve by quadratic:
t=[ 40+/-sqr(1600-4*7*14)]/14
=.375 or 5.340
These are the boundaries.
So the temperature is over 100 in the interval (.375, 5.340) hrs.
