Question

the number of bacteria in a refrigerated food product is given by $n(t)=30t^{2}-135t + 71$, $1 < t < 24$, where $t$ is the temperature of the food. when the food is removed from the refrigerator, the temperature is given by $t(t)=2t + 1.8$, 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 16110. time needed = hours question help: video message instructor post to forum

Explanation:

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

We know $N(T)=30T^{2}-135T + 71$ and $T(t)=2t + 1.8$. Substitute $T = 2t+1.8$ into $N(T)$:
\[

$$\begin{align*} N(T(t))&=30(2t + 1.8)^{2}-135(2t + 1.8)+71\\ &=30(4t^{2}+7.2t + 3.24)-270t-243 + 71\\ &=120t^{2}+216t+97.2-270t - 243+71\\ &=120t^{2}-54t - 74.8 \end{align*}$$

\]

Step2: Solve for $t$ when $N(T(t)) = 16110$

Set $120t^{2}-54t - 74.8=16110$. Rearrange to get a quadratic - equation in standard form $120t^{2}-54t-16184.8 = 0$.
The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 120$, $b=-54$, and $c=-16184.8$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-54)^{2}-4\times120\times(-16184.8)=2916+7768704=7771620$.
Then $t=\frac{54\pm\sqrt{7771620}}{240}=\frac{54\pm2787.76}{240}$.
We take the positive root since time $t\geq0$. So $t=\frac{54 + 2787.76}{240}=\frac{2841.76}{240}\approx11.84$.

Answer:

$N(T(t)) = 120t^{2}-54t - 74.8$
Time Needed $\approx11.84$ hours