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3 years ago

How to get the radical answer simplified

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

3 0

For 61 you can change it to √25/√49. √25=5 and √49=7 so it would simplify to 5/7.

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find the area and perimeter of a rectangle with a width of 2 1/3inch and a length of 5 2/5inches. show your work please.

aleksandrvk [35]

Area: If you multiply 2 1/3 by 5 2/5, you get an answer of about: 12.6, which means the area is about: 12.6in. squared.
Perimeter: If you add 2 1/3+ 2 1/3+ 5 2/5+ 5 2/5 you get an answer of about: 15.47. So, the perimeter is about 15.47in.

Note: If you aren't allowed to use a calculator on this test, you should probably multiply the numbers yourself so you get show the work and get the answer in a fraction, which is probably what they want. If you are allowed to use calculators, all the information you need is right here.

4 0

3 years ago

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What is the measure of z?<br> an<br> z= [?

Gemiola [76]

what to find

measure of z =1

7 0

3 years ago

(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

Which of the following is equal to the expression below?

rodikova [14]

(8.320)(1)/(3)=

10\root(3)(5)

8 0

3 years ago

Line khas a slope of 2/3. If line m is parallel to line k, then it has a slope of

storchak [24]

Since line m is parallel to line k.

Their slopes are equal.

slope of line k= 2/3 [given]

So; Slope of line m=2/3

Hope it helps...

Regards;

Leukonov/Olegion.

5 0

3 years ago