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3 years ago

1. In 2011, Molly and Jerry both went to

Mathematics

1 answer:

True [87]3 years ago

6 0

Answer:

2017

Step-by-step explanation:

Molly goes every 2 years

Jerry goes every 3 years

LCM(2,3) = 2 x 3 = 6

2011 + 6 = 2017

The answer is 2017

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This question is only for boys<br> what are the signs that a boy likes you

pantera1 [17]

Answer: He leans towards you whenever you are sitting close to one another.

He rarely turns his back on you.

He smiles a lot and looks at you keenly.

He maintains eye contact with you.

He finds an excuse to touch you whenever he has the chance.

Step-by-step explanation:

5 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

The rectangle below has an area of x^2-9x

Romashka-Z-Leto [24]

Answer:

(x+3)m

Step-by-step explanation:

Area of a rectangle = Length × width

Given the area = (x²-9)m²

Width = (x-3)m

Length of the rectangle = Area/Width

Length of the rectangle = x²-9/x-3

= x²-3²/x-3

Note that according to different of two square, a²-b² = (a+b)(a-b)

Therefore x²-3² = (x+3)(x-3)

Length of the rectangle

= (x+3)(x-3)/x-3

= x+3

The length of the rectangle is (x+3)m

5 0

3 years ago

A box is to be constructed with a rectangular base and a height of 5 cm. If the rectangular base must have a perimeter of 28 cm,

Irina18 [472]

Answer:

V = (5)(14 L - L^2) cm^3

Step-by-step explanation:

Let the dimensions of the rectangular base be W by L and the height be H.

The perimeter must be 28 cm, so 28 cm = 2(W) + 2(L). This reduces to

14 cm = W + L, which can be solved for either W or L. Solving for W:

W = 14 cm - L

Then the area of the rectangular base is A = W*L, or A = (14 cm - L)(L), or

A = 14L - L^2.

The volume of the box is then V = A*H.

Because H = 5 cm, the volume is V = (5 cm)A, or

V = (5 cm)(14L - L^2) cm^2

This is a quadratic equation. Putting it into standard form yields:

V = (5)(14 L - L^2) cm^3. This is the desired quadratic formula.

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3 years ago

Identify an equation in point slope form for the line parallel to y=4x-9 that passes through (-5,3)

ELEN [110]

Its y=4x+23, yeet on my feet, apex is a defeat

6 0

4 years ago