The number of bacteria in a refrigerated food product is given by \( N(T)=21 T^{2}-127 T+86 \), \( 6

Ask by Harrington Gray. in the United States

Feb 03,2025

To find the composite function \( N(T(t)) \), we substitute \( T(t) = 5t + 1.9 \) into the equation \( N(T) = 21T^2 - 127T + 86 \). So, we have: \[ N(T(t)) = N(5t + 1.9) = 21(5t + 1.9)^2 - 127(5t + 1.9) + 86 \] Now, expanding \( (5t + 1.9)^2 \): \[ (5t + 1.9)^2 = 25t^2 + 19t + 3.61 \] Now, substituting that back into \( N \): \[ N(T(t)) = 21(25t^2 + 19t + 3.61) - 127(5t + 1.9) + 86 \] \[ = 525t^2 + 399t + 75.81 - 635t - 241.3 + 86 \] \[ = 525t^2 - 236t - 79.49 \] Thus, the composite function is: \[ N(T(t)) = 525t^2 - 236t - 79.49 \] Next, to find the time when the bacteria count reaches 8570, we set \( N(T(t)) \) equal to 8570: \[ 525t^2 - 236t - 79.49 = 8570 \] \[ 525t^2 - 236t - 8650.49 = 0 \] Now we can use the quadratic formula to solve for \( t \): \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{236 \pm \sqrt{(-236)^2 - 4(525)(-8650.49)}}{2(525)} \] Calculating the discriminant: \[ (-236)^2 = 56196 \] \[ 4(525)(8650.49) = 18294865.8 \] \[ \sqrt{b^2 - 4ac} \approx \sqrt{56196 + 18294865.8} = \sqrt{18351061.8} \approx 4286.2 \] Plug that back into the quadratic formula: \[ t = \frac{236 \pm 4286.2}{1050} \] Taking the positive root: \[ t \approx \frac{4522.2}{1050} \approx 4.3 \text{ hours} \] Thus, the time when the bacteria count reaches 8570 is approximately: \[ t \approx 4.3 \]