Answer:
![3801.32 {cm}^{3}](https://tex.z-dn.net/?f=3801.32%20%7Bcm%7D%5E%7B3%7D%20)
Step-by-step explanation:
![V=π {r}^{2} h \\ =π· {11}^{2} ·10 \\ ≈3801.3271](https://tex.z-dn.net/?f=V%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%3D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%CF%80%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%20%7Br%7D%5E%7B2%7D%20%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3Eh%3C%2Fp%3E%3Cp%3E%20%5C%5C%20%3C%2Fp%3E%3Cp%3E%3D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%CF%80%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%C2%B7%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%20%7B11%7D%5E%7B2%7D%20%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%C2%B7%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E10%3C%2Fp%3E%3Cp%3E%20%5C%5C%20%3C%2Fp%3E%3Cp%3E%E2%89%88%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E3801.3271)
Answer: ![\bold{\dfrac{16}{33} = 48\%}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cdfrac%7B16%7D%7B33%7D%20%3D%2048%5C%25%7D)
<u>Step-by-step explanation:</u>
Red or Green
![=\dfrac{red}{total}+\dfrac{green}{total}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7Bred%7D%7Btotal%7D%2B%5Cdfrac%7Bgreen%7D%7Btotal%7D)
![=\dfrac{9}{33}+\dfrac{7}{33}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B9%7D%7B33%7D%2B%5Cdfrac%7B7%7D%7B33%7D)
![=\dfrac{16}{33}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B16%7D%7B33%7D)
≈ 48%
It would be FH
A diameter is basically the length of the line through the centre that touches 2 points on a circles edges
You find the eigenvalues of a matrix A by following these steps:
- Compute the matrix
, where I is the identity matrix (1s on the diagonal, 0s elsewhere) - Compute the determinant of A'
- Set the determinant of A' equal to zero and solve for lambda.
So, in this case, we have
![A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D%20%5Cimplies%20A%27%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1-%5Clambda%26-2%5C%5C-2%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D)
The determinant of this matrix is
![\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bcc%7D1-%5Clambda%26-2%5C%5C-2%26-%5Clambda%5Cend%7Barray%7D%5Cright%7C%20%3D%20-%5Clambda%281-%5Clambda%29-%28-2%29%28-2%29%20%3D%20%5Clambda%5E2-%5Clambda-4)
Finally, we have
![\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}](https://tex.z-dn.net/?f=%5Clambda%5E2-%5Clambda-4%3D0%20%5Ciff%20%5Clambda%20%3D%20%5Cdfrac%7B1%5Cpm%5Csqrt%7B17%7D%7D%7B2%7D)
So, the two eigenvalues are
![\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}](https://tex.z-dn.net/?f=%5Clambda_1%20%3D%20%5Cdfrac%7B1%2B%5Csqrt%7B17%7D%7D%7B2%7D%2C%5Cquad%20%5Clambda_2%20%3D%20%5Cdfrac%7B1-%5Csqrt%7B17%7D%7D%7B2%7D)
Find the scale:
30 miles / 5miles = 6
Multiply km by the scale:
8 km x 6 = 48 km
The answer is 48 km