Question

The number of bacteria in a refrigerated food product is given by N(T)

4
29T2 - 115T + 52,

When the food is removed from the refrigerator, the temperature is given by T(t) = 6t + 1.5, where t is

the time in hours.

Find the composite function N(T(t)):

N(Tt))

Find the time when the bacteria count reaches 500.

Time Needed =

hours

Answer 1

Explanation:

Given

N(t) = 29t²-115t+52

T(t) = 6t+1.5

To get the composite function N(T(t)), we will follow the steps

N(T(t)) = N(6t+1.5)

To get N(6t+1.5), we will replace t in N(t) with 6t+1.5 as shown:

N(6t+1.5) = 29(6t+1.5)²-115(6t+1.5)+52

N(6t+1.5) = 29(36t²+18t+2.25)-690t-172.5+52

N(6t+1.5) = 1044t²+522t+65.25-690t-172.5+52

N(6t+1.5) = 1044t²+522t-690t+65.25-172.5+52

N(6t+1.5) = 1044t²-168t-55.35

Hence;

N(T(t)) = 1044t²-168t-55.35

To determine the time when the bacteria count reaches 500, we will equate the expression to 500 and find t

500 = 1044t²-168t-55.35

1044t²-168t-55.35-500= 0

1044t²-168t-555.35 = 0

Factorize

t =140±√1630115/1740

t = 0.81

Hence the time reached when the bacteria was 500 hours is 0.81hours