Answer:
We have the matrix ![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D)
To find the eigenvalues of A we need find the zeros of the polynomial characteristic 
Then
![p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda](https://tex.z-dn.net/?f=p%28%5Clambda%29%3Ddet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4-%5Clambda%26-4%26-4%5C%5C0%26-8-%5Clambda%26-4%5C%5C0%268%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-8-%5Clambda%26-4%5C%5C8%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29%28%28-8-%5Clambda%29%284-%5Clambda%29%2B32%29%5C%5C%3D-%5Clambda%5E3-8%5Clambda%5E2-16%5Clambda)
Now, we fin the zeros of 
.
Then, the eigenvalues of A are 
of multiplicity 1 and 
of multiplicity 2.
Let's find the eigenspaces of A. For : 
.Then, we use row operations to find the echelon form of the matrix
![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
2.
Therefore,
For 
: 
.Then, we use row operations to find the echelon form of the matrix
![A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%2B4I_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%26-4%26-4%5C%5C0%268%268%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
Then,
Answer:
angle y = 49°
Step-by-step explanation:
If AB = AD and BC = DC, then the two triangles are isosceles triangles. The altitude from the top to the base divides the top angle into two equal angles, with 90° angles at the base. If angle C is divided into two equal angles (41°) and at the base is a 90° angle, that makes angle CBD 49°. On an isosceles triangle, the two angles opposite to the equal sides are identical. So angle y is identical to angle CBD.
Answer:
it could be the number in 2015 is 2,330
Step-by-step explanation:
Just add the number of the students in 2014-2016.
1,030+ 1,300=2,330
Answer:
The sum of 3 and the quotient of y and 2.
Step-by-step explanation:
