Answer:
Point form: (5,-3)
Equation form: x=5, y=-3
Step-by-step explanation:
Hope this helps <3
Answer:
The answer is "19.71 million".
Step-by-step explanation:
![\to P(t) = \frac{19.71}{1+61.22 e^{(-0.03513t)}}](https://tex.z-dn.net/?f=%5Cto%20P%28t%29%20%3D%20%5Cfrac%7B19.71%7D%7B1%2B61.22%20e%5E%7B%28-0.03513t%29%7D%7D)
where
![\to p= \text{population in million} \\\\\to t= \text{time in years}\\](https://tex.z-dn.net/?f=%5Cto%20p%3D%20%5Ctext%7Bpopulation%20in%20million%7D%20%5C%5C%5C%5C%5Cto%20t%3D%20%5Ctext%7Btime%20in%20years%7D%5C%5C)
when t is increases, then
when decrease so that p(t) continually grows in size to reach the optimum population which you find if t is quite high
![\to P_{max} = \lim_{ \ t \to \infty} \frac{19.71}{1+61.22 e^{(-0.03513t)}}](https://tex.z-dn.net/?f=%5Cto%20P_%7Bmax%7D%20%3D%20%20%5Clim_%7B%20%5C%20t%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B19.71%7D%7B1%2B61.22%20e%5E%7B%28-0.03513t%29%7D%7D)
![when \ t \to \infty \ , 0.035 t \to \infty \ , e^{(0.03513t)} \to e^{\infty} \to \infty\ \ e^{\frac{1}{0.035t} = \ e^{-0.035t}} \to 0](https://tex.z-dn.net/?f=when%20%5C%20%20t%20%5Cto%20%5Cinfty%20%5C%20%2C%20%200.035%20t%20%5Cto%20%20%5Cinfty%20%5C%20%2C%20%20e%5E%7B%280.03513t%29%7D%20%5Cto%20e%5E%7B%5Cinfty%7D%20%5Cto%20%5Cinfty%5C%20%5C%20e%5E%7B%5Cfrac%7B1%7D%7B0.035t%7D%20%3D%20%5C%20e%5E%7B-0.035t%7D%7D%20%5Cto%200)
![= \frac{19.71}{1+0}\\\\= \frac{19.71}{1}\\\\ = 19.71 \ million](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B19.71%7D%7B1%2B0%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B19.71%7D%7B1%7D%5C%5C%5C%5C%20%3D%2019.71%20%5C%20million)
Answer:
b slow
Step-by-step explanation:
Answer:
111°
Step-by-step explanation:
Supplementary means to add the angles up to 180°
(f*g)(x) = (x-4)(2x+1) = 2x² + x - 8x - 4 = 2x² - 7x - 4
(f-g)(x) = x-4 - (2x+1) = (x-4)-2x-1 = -x - 5
(f+g)(3) = 3 - 4 + 2*3 + 1 = -1 + 6 + 1 = 6