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2 years ago

Please answer this question :)

Mathematics

2 answers:

Otrada [13]2 years ago

6 0

5.27. I rounded the number cuz if not, it would have gave me a big #.

Scorpion4ik [409]2 years ago

5 0

Answer:

5.2692307692307692307692307692308

Step-by-step explanation:

hope this helps! :)

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Type the correct answer in each box. Spell all words correctly, and use numerals instead of words for numbers.

Andru [333]

Answer:

Nate made a mistake. He should have got 24 as Least common factor.

Step-by-step explanation:

Given number are 8 and 12.

To find the GCF or LCF we have to list multiples.

For LCF

List the multiples of each number until at least one of the multiples appears on all lists.
Find the smallest number that is on all of the lists.
This number is the LCM.

For GCF

List the multiples of each number until at least one of the multiples appears on all lists.
Find the Biggest number that is on all of the lists.
This number is the GCF

The least common multiple of 8 and 12 is 24.

The Greatest common multiply of 8 and 12 is 4.

Hence, Nate is wrong. Nate should have found 24.

[RevyBreeze]

7 0

2 years ago

HELP URGENT PLEASE!!!!!!!!!!!!!!

leonid [27]

Answer:

The supplement of 89° is the angle that when added to 89° forms a straight angle (180° ).

Step-by-step explanation:

4 0

2 years ago

Can someone please help me with this problem??

Ostrovityanka [42]

Answer:

The slope is - $\frac{1}{3}$ and the y intercept is 2.

Step-by-step explanation:

So, we can see that the line moves down one unit every time it progresses to the right three times. Slope is defined by change in y/change in x. For this problem, the change in y is -1 and the change in x is +3, so the slope is $-\frac{1}{3}$ .

Now for the y intercept. The y intercept is the point where the line first passes through the y axis. In this case, it passes through 2. Your y intercept can be written in two ways; as a number (2) or as an ordered pair $(0,2)$ , although people usually write the number.

I hope this was helpful. Good luck!

7 0

2 years ago

Hey can u plz do this I will give Brainiest I got broken up with yesterday's d I'm sad so I can't rly think rn

Lisa [10]

Answer:

Step-by-step explanation:

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

2 years ago