2.36 repeating
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Answer:
![N(T(t)) = 1408t^2 - 385.6t - 105.52](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%201408t%5E2%20-%20385.6t%20-%20105.52)
Time for bacteria count reaching 8019: t = 2.543 hours
Step-by-step explanation:
To find the composite function N(T(t)), we just need to use the value of T(t) for each T in the function N(T). So we have that:
![N(T(t)) = 22 * (8t + 1.7)^2 - 123 * (8t + 1.7) + 40](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%2022%20%2A%20%288t%20%2B%201.7%29%5E2%20%20-%20123%20%2A%20%288t%20%2B%201.7%29%20%2B%2040)
![N(T(t)) = 22 * (64t^2 + 27.2t + 2.89) - 984t - 209.1 + 40](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%2022%20%2A%20%2864t%5E2%20%2B%2027.2t%20%2B%202.89%29%20-%20984t%20-%20209.1%20%2B%2040)
![N(T(t)) = 1408t^2 + 598.4t + 63.58 - 984t - 169.1](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%201408t%5E2%20%2B%20598.4t%20%2B%2063.58%20-%20%20984t%20-%20169.1)
![N(T(t)) = 1408t^2 - 385.6t - 105.52](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%201408t%5E2%20-%20385.6t%20-%20105.52)
Now, to find the time when the bacteria count reaches 8019, we just need to use N(T(t)) = 8019 and then find the value of t:
![8019 = 1408t^2 - 385.6t - 105.52](https://tex.z-dn.net/?f=8019%20%3D%201408t%5E2%20-%20385.6t%20-%20105.52)
![1408t^2 - 385.6t - 8124.52 = 0](https://tex.z-dn.net/?f=1408t%5E2%20-%20385.6t%20-%208124.52%20%3D%200)
Solving this quadratic equation, we have that t = 2.543 hours, so that is the time needed to the bacteria count reaching 8019.
![A = \dfrac{bh}{2}](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cdfrac%7Bbh%7D%7B2%7D)
Cross multiply:
![2A = bh](https://tex.z-dn.net/?f=2A%20%3D%20bh%20)
Divide by h on both sides:
Answer:
D 6 seconds
Step-by-step explanation:
Hope this helps!!! Please amrk as Brainlyest
You check if it is even and that the sum of the digits is divisible by 3. That's the divisibility rule of 6.