Answer:
Step-by-step explanation:
<u>Eigenvalues of a Matrix</u>
Given a matrix A, the eigenvalues of A, called 
are scalars who comply with the relation:
Where I is the identity matrix
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix is given as
![A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D)
Set up the equation to solve
![det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Expanding the determinant
![det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3-%5Clambda%265%5C%5C8%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Operating Rearranging
Factoring
Solving, we have the eigenvalues
Step-by-step explanation:
You can imagine this figure as a rectangle and cube
If you want volume of this irregular figure than you have to do it like this:
V(figure)= V(rectangle)+ V(cube)
V(figure)= a*b*c+ a³
V(figure)= 4*3*(I don't see dimension on the left)+ 3³
V(figure)=12*(I don't see dimension on the left)+ 27
And only you have to to do is to set this dimension which I can't see.
Answer:
Yes, because the x's do not have more than one y.
Step-by-step explanation:
x y
0 0
5 10
6 12
10 20
(That's meant to be a table). Looking at the x and y table values, you can see that none of the x's have two y's. So, this relation is a function.
Answer:
domain: x>3/5
Step-by-step explanation:
First we need to derive our function g(x) to get a new function g'(x)
To do this we will have to apply chain rule because we have an inner and outer functions.
Our G(x) = square root(3-5x)
Chain rule formula states that: d/dx(g(f(x)) = g'(f(x))f'(x)
where d/dx(g(f(x)) = g'(x)
g(x) is the outer function which is x^1/2
f(x) is our inner function which is 3-5x
therefore f'(x)= 1/2x^(-1/2) and f'(x) = -5
g'(f(x)) = -1/2(3-5x)^(-1/2)
Applying chain rule then g'(x) = 1/2 (3-5x)^(-/1/2)*(-5)
But the domain is the values of x where the function g'(x) is not defined
In this case it will be 3-5x > 0, because 3-5x is a denominator and anything divide by zero is infinity/undefined
which gives us x >3/5
