Answer:
a) 1296 bacteria per hour
b) 0 bacteria per hour
c) -1296 bacteria per hour
Step-by-step explanation:
We are given the following information in the question:
The size of the population at time t is given by:
![b(t) = 9^6 + 6^4t-6^3t^2](https://tex.z-dn.net/?f=b%28t%29%20%3D%209%5E6%20%2B%206%5E4t-6%5E3t%5E2)
We differentiate the given function.
Thus, the growth rate is given by:
![\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdb%28t%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%289%5E6%20%2B%206%5E4t-6%5E3t%5E2%29%5C%5C%5C%5C%3D%206%5E4-2%286%5E3%29t)
a) Growth rates at t = 0 hours
![\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdb%28t%29%7D%7Bdt%7D%20%5Cbigg%7C_%7Bt%3D0%7D%3D%206%5E4-2%286%5E3%29%280%29%20%3D%201296%5Ctext%7B%20bacteria%20per%20hour%7D)
b) Growth rates at t = 3 hours
![\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdb%28t%29%7D%7Bdt%7D%20%5Cbigg%7C_%7Bt%3D3%7D%3D%206%5E4-2%286%5E3%29%283%29%20%3D%200%5Ctext%7B%20bacteria%20per%20hour%7D)
c) Growth rates at t = 6 hours
![\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdb%28t%29%7D%7Bdt%7D%20%5Cbigg%7C_%7Bt%3D6%7D%3D%206%5E4-2%286%5E3%29%286%29%20%3D%20-1296%5Ctext%7B%20bacteria%20per%20hour%7D)
The answer is C: The size of the fish Joe can catch is a function of the strength of his fishing line.
Don’t know if this is right but u cud try this.
$3000*0.12*0.5/100
Answer:
-10,000
Step-by-step explanation:
To solve for x, first simplify 45x^3 to get 91,125, then subtract 91,125 from both sides of the equation. 81,125 - 91,125 is -10,000, so x = -10,000.
15 goes into 975 65 times