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ivanzaharov [21]
3 years ago
5

Need help ASAP now plz now z now now now

Mathematics
1 answer:
Viktor [21]3 years ago
6 0
Root root 20^2 - 4^2
X = 2
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Marshall is buying snacks for a party. He spent$4727 on the snacks. If the sales tax rate is 6%, what was Marshall’s total bill?
riadik2000 [5.3K]
Her total bill was $18.09
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3 years ago
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(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca
Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation det(A-\lambdaI)=0 where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6

So the characteristic polynomial is \lambda^2-5\lambda+6=0.

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0

So \lambda=3, \lambda=2

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If \lambda=3 we get the following matrix

\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right].

Since both rows are equal, we have the equation

x-2y=0. Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case \lambda=2, using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that A=PDP^{-1}. We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0
3 years ago
Rewrite in the form . Then find q(x) and r(x).
gizmo_the_mogwai [7]

Answer:

q(x)= 2x+5

r(x)=6

b(x)=x+4

Step-by-step explanation:

I have found what is missing in your question: Rewrite 2x^2+13x+26/x+4 in the form q(x)+r(x)/b(x).

First, we have to divide the polynomial  by x+4 and remember that is x times 2x.

By multiplying this we are going to get 2x^{2} + 8x and now we are doing separation of it and getting 2x^{2}  + 13x + 26 and getting 2x^{2}  + 13x + 26/ x + 4 = 2x^{2} +8x/x+4=5x+26/x+4=2x + 5x+26/x+4

Because 5 is showing x times 5 we are going to multiply (x + 4) with 5 and that is 5x + 20.

5x+26/x+4 = 5x + 20/x+4 = 6/ x+4=5 + 6/x+4

Now when we can get more simplified we are having:

2x^{2} +13+26/x+4=2x + 5= 6/x+4

3 0
3 years ago
A new car worth $20,000 loses 20% of its value every year. Is the value of the car represented by a linear or exponential functi
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B) Exponential. Because it keeps losing 20% of what the previous years amount was so it decreases by a smaller factor each year

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The following steps are used to rewrite the the polynomial expression, 3(x+4y)+5(2x-y)
posledela
The answer is 13x + 7y

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