56.58 divided by 6 would be 9.43 per shirt
Answer: The answer is true :)
Without so much as a quick peek at the picture, it's not possible to even begin
to answer the question. That's the reason, you see, that the picture is printed
right there, along with the question, in the same place from which you copied
this cut-off portion of the question. It's not enough. We need the picture too.
The function that represents the vertical stretch by a factor of 3 of f(x) is given as follows:
g(x) = 3x - 21.
<h3>How are vertical stretches and compression applied to the graph of a function f(x)?</h3>
Vertical stretches or compression to the graph of a function f(x) are applied by the multiplication of a constant a to the function, as follows:
g(x) = a x f(x).
Then the value of the constant determines if it is an stretch or a compression.
- If the absolute value of the constant a is <u>greater than 1</u>, it is a stretch.
- If the absolute value of the constant a is <u>less than 1</u>, it is a compression.
For this problem, we want a vertical stretch with a factor of 3, then the constant is given as follows:
a = 3.
Then the definition of function g(x) is given by:
g(x) = 3f(x)
g(x) = 3(x - 7)
g(x) = 3x - 21.
More can be learned about transformations of functions at brainly.com/question/28810470
#SPJ1
Answer:
Step-by-step explanation:
Consider the matrix 
. We will calculate the correspondent eigenvalues and eigen vector of the matrix. REcall that, to calculate the eigenvalues of a square matrix A, we must solve the following equation 
where I is the identity matrix. In this case we have the following
which gives us the following polynomial (known as the characteristic polynomial of the matrix A).
.
Hence, the only eigenvalue of the given matrix is 
.
For us to calculate the eigenvalue, we want to find a base for the Kernel of matrix 
replacing the value of lambda with the value of the eigen value. REcall that finding the base for the Kernel is solving the associated homogeneus system of the matrix
![\left[\begin{matrix} -2 & -2 \\ 2 & 2 \end{matrix}\right]= \left[\begin{matrix} 0 \\ 0 \end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Bmatrix%7D%20-2%20%26%20-2%20%5C%5C%202%20%26%202%20%5Cend%7Bmatrix%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Bmatrix%7D%200%20%5C%5C%200%20%5Cend%7Bmatrix%7D%5Cright%5D)
which lead us to the equation 
or equivalently, 
. Hence, the solution of the system is of the form (x,y) = (x,-x) = x(1,-1). So the correspondent eigenvector is the vector (1,-1).
