Respuesta :
Step 1
State N(T) and T(t)
[tex]\begin{gathered} N(T)=24T^2-125T+73 \\ T(t)=9t+1.7 \end{gathered}[/tex]T is the temperature of the food
t is time in hours when the food is outside the refrigerator
Step 2
Find N(T(t))
[tex]\begin{gathered} N(T(t))=24(9t+1.7)^2-125(9t+1.7)\text{ + 73} \\ N(T(t))=24(9t+1.7)(9t+1.7)-125(9t+1.7)+73 \end{gathered}[/tex][tex]\begin{gathered} N(T(t))=24(81t^2+15.3t+15.3t+2.89)-1125t-212.5+73 \\ N(T(t))=24(81t^2+30.6t+2.89)-1125t-139.5_{} \\ N(T(t))=1944t^2+734.4t+69.36-1125t-139.5 \end{gathered}[/tex][tex]N(T(t))=1944t^2-390.6t-70.14[/tex]Step 3
Find the time when the bacteria count reaches 24722
[tex]\begin{gathered} N(T(t))=24722 \\ 24722=1944t^2-390.6t-70.14 \\ 1944t^2-390.6t-70.14-24722=0 \\ 1944t^2-390.6t-24792.14=0 \end{gathered}[/tex]Step 4
Find t using the quadratic formula.
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{where} \\ a=1944 \\ b=-390.6 \\ c=-24792.14 \end{gathered}[/tex][tex]\begin{gathered} t=\frac{-(-390.6)\pm\sqrt[]{(-390.6)^2-4\times1944\times-24792.14}}{2\times1944} \\ t=\frac{390.6\pm\sqrt[]{152568.36+192783680.6}}{3888} \end{gathered}[/tex][tex]\begin{gathered} t=\frac{390.6\pm\sqrt[]{192936249}}{3888} \\ t=\frac{390.6\pm13890.14935}{3888} \\ t=\frac{390.6+13890.14935}{3888}=3.67303224\text{ hrs} \\ or \\ t=\frac{390.6-13890.14935}{3888}=-3.472106314\text{ hrs} \end{gathered}[/tex]Since time cannot be negative t = 3.67303224 hrs
