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

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UpStudy AI · Accepted Answer

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:

$$
\text{Time Needed} = 0.503 \text{ hours}
$$
The time when the bacteria count reaches 6326 is approximately 0.503 hours.

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