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ahrayia [7]
3 years ago
14

What is 2x+y=11 5x+3y=29

Mathematics
2 answers:
Dafna11 [192]3 years ago
8 0
2x +. y = 11
5x + 3y = 29

6x + 3y = 33
5x + 3y = 29


x = 4
2*4+y=12

y=3

(4,3)
dolphi86 [110]3 years ago
5 0
1) x = -y/2 + 11/2

2) x = -3y/5 + 29/5
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A magician asks two volunteers to each draw a card from a standard deck of cards. What is the probability that the first card is
mars1129 [50]

Answer:

6.37%

Step-by-step explanation:

A deck of cards have 52 cards.  There are 4 suits of 13 card each, those suits are Hearts, Clubs, Diamond and Spades.

The probability that the first card is a heart is:

P(Hearts) = \frac{13}{52}

Now, the probabily that the second card is a diamond is:

P(Diamond | First card is Heart) = \frac{13}{51}

The Probability that the first card is a heart and the second one a diamond is given by:

P(Hearts)×P(Diamond | First card is Heart)

= \frac{13}{52}\frac{13}{51}

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3 years ago
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What is the relationship between the tables, equations and graphs as it relates to rate of change/slope?
Mazyrski [523]

Answer:

What is the relationship between equations and graphs?

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

Step-by-step explanation:

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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

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Masteriza [31]

Answer:

  • See below

Step-by-step explanation:

<u>Given equation:</u>

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<u>Convert this to standard form by completing the square:</u>

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ryzh [129]

Answer:

Step-by-step explanation:

e

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