Hello! I am new here. Can someone check to see if my answer is correct?
The number of bacteria in a refrigerated food product is given by N(T)=23T^2-56T+1, 3<T<33, where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T(t)=5t+1.5, where t is the time in hours.
a. Find the composite function N(T(t))
My answer:
N(T(t))= N (5t+1.5)
=23(5t+1.5)^2-56(5t+1.5)+1
(5t+1.5)(5t+1.5)
=25t^2+7.5t+7.5t+2.25
=25t+15t+2.25
23(25t^2+15t+2.25)-56(5t+1.5)+1
=575t^2+345t+51.75-280t-84+1
=575t^2+65t-31.25
b. Find the bacteria count after 4 hours.
My answer:
575t^2+65t-3+.25
N(T(4))
=575(4^2)+65(4)-31.25
=575(16)+260-31.25
=9200+260-31.25
=9460-31.25
=9428.75 bacteria
The number of bacteria in a refrigerated food product is given by N(T)=23T^2-56T+1, 3<T<33, where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T(t)=5t+1.5, where t is the time in hours.
a. Find the composite function N(T(t))
My answer:
N(T(t))= N (5t+1.5)
=23(5t+1.5)^2-56(5t+1.5)+1
(5t+1.5)(5t+1.5)
=25t^2+7.5t+7.5t+2.25
=25t+15t+2.25
23(25t^2+15t+2.25)-56(5t+1.5)+1
=575t^2+345t+51.75-280t-84+1
=575t^2+65t-31.25
b. Find the bacteria count after 4 hours.
My answer:
575t^2+65t-3+.25
N(T(4))
=575(4^2)+65(4)-31.25
=575(16)+260-31.25
=9200+260-31.25
=9460-31.25
=9428.75 bacteria