The number of bacteria in a refrigerated food product is given by \( N(T)=29 T^{2}-164 T+80,6

Ask by Curry Gray. in the United States

Oct 28,2024

To find the time when the bacteria count reaches 6326, we start with the composite function \( N(T(t)) \): \[ N(T(t)) = 725 t^2 - 298 t - \frac{3031}{25} \] We need to set this equal to 6326 and solve for \( t \): \[ 725 t^2 - 298 t - \frac{3031}{25} = 6326 \] First, we can rearrange the equation: \[ 725 t^2 - 298 t - \frac{3031}{25} - 6326 = 0 \] Next, we need to express 6326 in terms of a fraction with a denominator of 25: \[ 6326 = \frac{6326 \times 25}{25} = \frac{158150}{25} \] Now, substituting this back into the equation gives: \[ 725 t^2 - 298 t - \frac{3031}{25} - \frac{158150}{25} = 0 \] Combining the constant terms: \[ 725 t^2 - 298 t - \frac{3031 + 158150}{25} = 0 \] Calculating \( 3031 + 158150 \): \[ 3031 + 158150 = 161181 \] Thus, we have: \[ 725 t^2 - 298 t - \frac{161181}{25} = 0 \] To eliminate the fraction, we can multiply the entire equation by 25: \[ 25(725 t^2) - 25(298 t) - 161181 = 0 \] This simplifies to: \[ 18125 t^2 - 7450 t - 161181 = 0 \] Now we can use the quadratic formula to solve for \( t \): \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 18125 \), \( b = -7450 \), and \( c = -161181 \). Calculating the discriminant \( b^2 - 4ac \): \[ b^2 = (-7450)^2 = 55506250 \] \[ 4ac = 4 \times 18125 \times (-161181) = -11600000000 \] Thus, \[ b^2 - 4ac = 55506250 + 11600000000 = 11655506250 \] Now we can find \( t \): \[ t = \frac{7450 \pm \sqrt{11655506250}}{2 \times 18125} \] Calculating \( \sqrt{11655506250} \): \[ \sqrt{11655506250} \approx 10793.5 \] Now substituting back into the formula: \[ t = \frac{7450 \pm 10793.5}{36250} \] Calculating the two possible values for \( t \): 1. \( t = \frac{7450 + 10793.5}{36250} \approx \frac{18243.5}{36250} \approx 0.503 \) 2. \( t = \frac{7450 - 10793.5}{36250} \approx \frac{-3333.5}{36250} \) (this value is negative and not physically meaningful in this context) Thus, the only valid solution is: \[ t \approx 0.503 \text{ hours} \] So, the time when the bacteria count reaches 6326 is approximately: \[ \text{Time Needed} = 0.503 \text{ hours} \]