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2 years ago

What graph represents the equation x=2

Mathematics

1 answer:

guajiro [1.7K]2 years ago

3 0

Option 1. The equation tells us that line is going through the x-axis at (2,0) in a straight line.

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I WILL GIVE BRAINLIEST!!!!

posledela

The answers to the questions are:

1. x = 4

2. x = 1.5

3. x = 1.875

4. x = 1

<h3>2 step equation:</h3>

1. 4x + 3 = 19

4x = 19 - 3

4x = 16

divide by 4

x = 4

<h3>2 step equation w/ fractions:</h3>

(4/3)x + 5 = 17

= 1.33x + 5 = 7

Take like terms

1.33x = 7-5

1.33x = 2

divide through by 1.33x to get 2

x = 2/1.33

x = 1.5

<h3>3. Distributive Property:</h3>

4x(6 - 2) - 10 = 20

Multiply and open the bracket

24x - 8x - 10 = 20

Take like terms

16x = 20+10

16x = 30

x = 30/16

= 1.875

<h3>4. Decimals:</h3>

4.3x + 0.7 = 5

4.3x = 5 - 0.7

4.3x = 4.3

x = 1

Read more on distributive properties here:

brainly.com/question/2807928

#SPJ1

7 0

2 years ago

Vincent rolls two dice. What is the probability his results are a multiple of four or the

Afina-wow [57]

Answer:

Sample space: 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 6.1, 6.2, 6.3, 6.4, 6.5, 6.6.

So the mutiples are 1.3, 2.2, 2.6, 3.1, 3.5, 4.4, 5.3, 6.2, and 6.6

Greather than 6 is 1.6, 2.5, 3.4, 3.6, 4.3, 4.5 4.6, 5.2, 5.4, 5.5, 5.6, 6.1, 6.3, 6.4, and 6.5.

So the prop is 24/36 or 2/3=66%

4 0

3 years ago

Find the value of x.

enot [183]

X is 10 take that’s known 80 degrees and the fact that it is a right angle means the whole this is 90 degrees so therefore x is 10 90=80+_

6 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

I need help I'm stuck

ankoles [38]

Then put in some butter that will get you out

8 0

3 years ago