Question

the number of bacteria in a refrigerated food product is given by $n(t) = 25t^2 - 155t + 78$, $6 < t < 36$, where $t$ is the temperature of the food. when the food is removed from the refrigerator, the temperature is given by $t(t) = 2t + 1.7$, where $t$ is the time in hours. find the composite function $n(t(t))$: $n(t(t)) = \square$ find the time when the bacteria count reaches 18338. time needed = $\square$ hours question help: video submit question

Explanation:

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

Replace $T$ in $N(T)$ with $2t+1.7$:
$N(T(t))=25(2t+1.7)^2 - 155(2t+1.7) + 78$

Step2: Expand $(2t+1.7)^2$

Use $(a+b)^2=a^2+2ab+b^2$:
$(2t+1.7)^2=4t^2 + 6.8t + 2.89$

Step3: Distribute coefficients

Calculate each term:
$25(4t^2 + 6.8t + 2.89)=100t^2 + 170t + 72.25$
$-155(2t+1.7)=-310t - 263.5$

Step4: Combine all terms

Sum the expanded terms and constant:
$N(T(t))=100t^2 + 170t + 72.25 - 310t - 263.5 + 78$
$N(T(t))=100t^2 - 140t - 113.25$

Step5: Set $N(T(t))=18338$

Form the quadratic equation:
$100t^2 - 140t - 113.25 = 18338$
$100t^2 - 140t - 18451.25 = 0$

Step6: Solve quadratic equation

Use quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=100$, $b=-140$, $c=-18451.25$:
First calculate discriminant:
$\Delta=(-140)^2 - 4(100)(-18451.25)=19600 + 7380500=7399100$
$\sqrt{\Delta}=\sqrt{7399100}=2720.13$ (approx)
$t=\frac{140\pm2720.13}{200}$
Take positive root (time can't be negative):
$t=\frac{140+2720.13}{200}=\frac{2860.13}{200}=14.30065\approx14.3$
Verify domain: Check $T(t)=2(14.3)+1.7=30.3$, which is between 6 and 36, so valid.

Answer:

$N(T(t))=100t^2 - 140t - 113.25$
Time Needed = 14.3 hours