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3 years ago

Help me please i need this answered fast

Mathematics

1 answer:

Lana71 [14]3 years ago

5 0

Answer:

Step-by-step explanation:

the one on the bottom because

the orientation of the points do not change

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WILL GIVE BRAINLIST SO PLZ HELP

DochEvi [55]

Hello!

D Would be your answer! Because At Midnight, the temperature was -8*f. At noon, 23*F, so

|-8-23| |-3||= 31 There you go! May I have Brainliest?

4 0

3 years ago

How do you do this??????

slava [35]

Answer:

2:1

Step-by-step explanation:

There are 14 spoons (not sure what they are) and 7 eyes so it is 14:7

simplified it is 2:1

Hope this helps :)

4 0

3 years ago

Read 2 more answers

A sum of money is deposited at a bank at a rate of 12½ annum simple interest. In how many years would the deposited money be dou

Dmitrij [34]

Answer:

It would take 10 years for the given sum of money be doubled at the given simple interest rate.

Step-by-step explanation:

A 10% interest would be added to the the principal amount after each year. So the interest would reach 100% i.e. equal to the principal amount in 10 years.

7 0

3 years ago

Which best describes how to add 9.73 and 21.6

Marta_Voda [28]

Answer:

Step-by-step explanation:

9.73

21.60

31.33

8 0

2 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