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3 years ago

Match the number with its opposite. Match Term Definition 9.2 A) 2.9 −2.9 B) −4.1 −1.4 C) −9.2 4.1 D) 1.4

Mathematics

1 answer:

jenyasd209 [6]3 years ago

8 0

We have these opposite pairs

9.2 and -9.2
2.9 and -2.9
1.4 and -1.4
4.1 and -4.1

So all we're doing is matching each positive number with its negative version. In terms of a visual, the opposite of a number is mirrored over 0 on the number line. So for instance, the opposite of 2 is -2, with each being two units away from 0 on the number line.

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What the result of √[(-2a)^ 2]

Tanya [424]

If you take the square root of a number squared number then they cancel each other out and the number stays the same i.e. √[(4)^2] would equal 4.

In this problem the square root and numbered squared cancel out to leave the problem as -2a.

The solution of this problem is -2a

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3 years ago

Bad White [126]

first polygon

ext. angle=180°-120°

=60°

$ext \: ang = \frac{360}{n}$

n=360°/60°

n=6

second polygon

n=2(6)=12

ext. ang= 360°/n = 360°/12° = 30°

int. ang = 180°-30°= 150°

answer is C

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3 years ago

1. What is the greatest multiple of 7 that is less than 60?

hodyreva [135]

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2 years ago

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How do I check 102÷17=6 without multiplying

insens350 [35]

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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