The number N of bacteria in a refrigerated food is given by
N(T) = 10T2 − 20T + 700, 1 ¡Ü T ¡Ü 20where T is the temperature of the food in degrees Celsius. When the food is removed from refrigeration, the temperature of the food is given by
T(t) = 5t + 2, 0 ¡Ü t ¡Ü 6 where t is the time in hours.
(a) Find the composition N(T(t)).
N(T(t)) = 1
, 0 ¡Ü t ¡Ü 6
N(t) = 10t^2 - 20t + 700
N(T(t)) = 10T^2 - 20T + 700
= 10(5t+2)^2 - 20(5t+2) + 700
= 250t^2 + 100t + 700
Well, let's find the composition N(T(t)) by substituting T(t) into the equation for N(T):
N(T(t)) = 10(T(t))^2 - 20(T(t)) + 700
Now, let's substitute T(t) with its definition:
N(T(t)) = 10(5t + 2)^2 - 20(5t + 2) + 700
N(T(t)) = 10(25t^2 + 20t + 4) - 100t - 40 + 700
N(T(t)) = 250t^2 + 200t + 40 - 100t - 40 + 700
Simplifying further:
N(T(t)) = 250t^2 + 100t + 700
So, the composition N(T(t)) is given by the equation:
N(T(t)) = 250t^2 + 100t + 700, where 0 ≤ t ≤ 6
To find the composition N(T(t)), we need to substitute the expression for T(t) into N(T).
Given:
N(T) = 10T^2 - 20T + 700
T(t) = 5t + 2
First, we substitute T(t) into N(T):
N(T(t)) = 10(T(t))^2 - 20(T(t)) + 700
Next, substitute the expression for T(t):
N(T(t)) = 10[(5t + 2)^2] - 20(5t + 2) + 700
Expanding the square and simplifying:
N(T(t)) = 10[25t^2 + 20t + 4] - 100t - 40 + 700
N(T(t)) = 250t^2 + 200t + 40 - 100t - 40 + 700
N(T(t)) = 250t^2 + 100t + 700
Therefore, N(T(t)) = 250t^2 + 100t + 700, where 0 ≤ t ≤ 6.
To find the composition N(T(t)), we substitute T(t) into N(T).
Given: N(T) = 10T^2 - 20T + 700
Now, we substitute T(t) into N(T):
N(T(t)) = 10(T(t))^2 - 20(T(t)) + 700
As T(t) = 5t + 2, we substitute that into the equation:
N(T(t)) = 10(5t + 2)^2 - 20(5t + 2) + 700
Now, we simplify the equation:
N(T(t)) = 10(25t^2 + 20t + 4) - 100t - 40 + 700
N(T(t)) = 250t^2 + 200t + 40 - 100t - 40 + 700
N(T(t)) = 250t^2 + 100t + 700
So, the composition N(T(t)) is given by:
N(T(t)) = 250t^2 + 100t + 700, for 0 ≤ t ≤ 6.