Question

question 5 the number of bacteria in a refrigerated food product is given by ( n(t) = 24t^2 - 111t + 54 ), ( 5 < t < 35 ), where ( t ) is the temperature of the food. when the food is removed from the refrigerator, the temperature is given by ( t(t) = 7t + 1.9 ), where ( t ) is the time in hours. find the composite function ( n(t(t)) ): ( n(t(t)) = ) find the time when the bacteria count reaches 5728. time needed = hours question help: video message instructor

Explanation:

Step1: Substitute $T(t)$ into $N(T)$

$N(T(t))=24(7t+1.9)^2 - 111(7t+1.9) + 54$

Step2: Set $N(T(t))=5728$

$24(7t+1.9)^2 - 111(7t+1.9) + 54 = 5728$

Step3: Simplify the equation

Let $x=7t+1.9$, then:
$24x^2 - 111x + 54 - 5728 = 0$
$24x^2 - 111x - 5674 = 0$

Step4: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=24$, $b=-111$, $c=-5674$:

$$\begin{align*} x&=\frac{111\pm\sqrt{(-111)^2-4\times24\times(-5674)}}{2\times24}\\ &=\frac{111\pm\sqrt{12321+544704}}{48}\\ &=\frac{111\pm\sqrt{557025}}{48}\\ &=\frac{111\pm746.34}{48} \end{align*}$$

Take positive root: $x=\frac{111+746.34}{48}\approx29.09$

Step5: Solve for $t$

$7t+1.9=29.09$
$7t=29.09-1.9=27.19$
$t=\frac{27.19}{7}\approx4$
Verify $T(4)=7\times4+1.9=29.9$, which fits $5

Answer:

1. $N(T(t))=24(7t+1.9)^2 - 111(7t+1.9) + 54$
2. 4 hours