Determining the momentum of a 80kg football player running at 3.4 meters/second =272kgm/s

Answer the question below

Lina20 [59]

Answer:

kckgkfjjfufufuf8f8ff88r8r8r8r8r8rrffufuufuff

3 0

2 years ago

Sum of three Consecutive integers of 384

Burka [1]

The three consecutive integers are 127, 128, 129

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

Please help !!!!!!!!!

Serggg [28]

Answer:

This is some ton of work LOL

Step-by-step explanation:

A. 4/7×5/5=20/35 We multiply by 5 for A because 35÷5=7 and you need to multiply top and bottom. You can NOT forget that. We do not need to simplify something this time? 3/5X7/7=21/35. Why are we muliplying something new? figure out. no need to simplify. oBVIOUSLY 3/5 IF BIGGER

B. As usual etc...

8 0

2 years ago

Which word shares a word root with remember? revisit reenter medical memory

Vladimir79 [104]

Memory bc when you remember something that means its in your memories?

8 0

2 years ago

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