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3 years ago

Please help, in a hurry!!

Mathematics

1 answer:

Tanya [424]3 years ago

6 0

Answer:

First one is infinitely many solutions. Second one is One solution. The third one is No solution.

Step-by-step explanation:

First line both equations have the same line. second one are two completely different equations while the third one are parallel lines which never intersect.

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Triangle ABC is transformed to triangle A′B′C′, as shown below:

Montano1993 [528]

Answer:

C. m<CAB = m<C'A'B'

Step-by-step explanation:

Triangle ABC looks like it was only reflected over the line y = 0. If the triangle was only reflected then the triangles should be congruent to each other.

3 0

3 years ago

If four times the sum of a number and 3 is divided by 2, the quotient is 0: find the number

AveGali [126]

The other number is -3.

Step-by-step explanation:

Step 1; First we develop formulae for the given information. A number is added with 3 and the resulting sum is multiplied by 4. This number is then divided by 2 and the quotient is 0. Assume the unknown number is x, then the given information is as follows

The quotient of $\frac{4(x+3)}{2}$ is 0. So $\frac{4(x+3)}{2}$ = 0.

Step 2; Now we solve the above equation. The denominator is taken to the RHS and the RHS remains zero.

$\frac{4(x+3)}{2}$ = 0, 4(x+3)= 0, 4x+ 12 = 0, 4x = -12, x = $\frac{-12}{4}$ = -3.

So the other number is -3.

3 0

4 years ago

Find the product. (y3)^2 · y^7

Sedbober [7]

Hey there! :D

When powers are outside the parenthesis, you multiply them. When they are not, you add them.

(y^3)^2 * y^7

3*2= 6

y^6*y^7

Add the powers.

y^13= the simplest form

I hope this helps!
~kaikers

6 0

3 years ago

Read 2 more answers

(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

Factor as the product of two binomials.<br><br> x^2 - 4 <br><br> Thank you!!

EastWind [94]

I think the answer is (x-2 ) x ( x+2)

3 0

3 years ago

Read 2 more answers