Answer:

<h2>Hope it helps you!!</h2>
P=Ae^kt
225 = 210 * e^(k*(1990-1980)
225/210=e^10k ln(225/210)=10k
k=ln(225/210)/10=0.0069
P = 210*e^(0.0069t)
for 2000 ===> t = 2000-1980=20
P = 210*e^(0.0069*20)
P=241.0749=241
300,022,000.
Don't quote me on that, I'm really not all that good at math.
Answer:
The answer is "9 and 7".
Step-by-step explanation:
Given:
![A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D)
Using formula:

![\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\](https://tex.z-dn.net/?f=%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%5Clambda%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%20%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7-%5Clambda%20%269%5C%5C0%269-%5Clambda%5Cend%7Barray%7D%5Cright%5D%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%287-%5Clambda%29%289-%5Clambda%29%7C%3D0%5C%5C%5C%5C%5Cto%20%287-%5Clambda%29%289-%5Clambda%29%3D0%5C%5C%5C%5C%5Cto%207-%5Clambda%3D0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%209-%5Clambda%3D0%5C%5C%5C%5C)

The answer would be C. hope that helps out :)