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3 years ago

What are these < > =???????????

Mathematics

2 answers:

madam [21]3 years ago

7 0

These Are Less Than(<) and Greater Than(>) Symbols!!

Hope This Helps!!!

Elenna [48]3 years ago

6 0

Answer:

< means less than

> means greater than

= means equal to

≥ means greater than or equal to

≤ means less than or equal to

Step-by-step explanation:

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aliina [53]

Answer:

In response to tissue injury, the body initiates a chemical signaling cascade that stimulates responses aimed at healing affected tissues. These signals activate leukocyte chemotaxis from the general circulation to sites of damage. These activated leukocytes produce cytokines that induce inflammatory responses [7].

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2 years ago

Give another name for Plane L. SELECT ALL THAT APPLY. K P S T L e M N O U R Q c d

tigry1 [53]

Answer:

Plane MNO

Plane PST

Step-by-step explanation:

The plane L can be given another name using uppercase letters only. The plane will require three letters name so that it represents a complete angle. The names can be assigned using the point to point method or randomly. There are two choices available to name the current plane. Plane MNO and Plane PST are the names that can be given to plane L.

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4 years ago

What integer is the opposite of 9? a. 9 b. 9 c. 0 d. 1?

goldenfox [79]

The opposite of a number is that same number, but with the opposite sign.
so the opposite of 9 is -9

7 0

3 years ago

15+25/5x7= 15+25divide5x7=

jenyasd209 [6]

Its either 1.14 or 0.87 Hope this helps!

7 0

3 years ago

Read 2 more answers

A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei

Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

$A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]$

a) To find the characteristic polynomial we calculate $\text{det}(A-\lambda I)=0$ where I is the identity matrix of appropiate size. in this case the characteristic polynomial is

$\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0$

Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

$(4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0$

which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are $\lambda=4,3,-1$

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression $A-\lambda I$ and solve the system $(A-\lambda I)x =0$ by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider $\lambda = 4$ . We get the matrix

$\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]$

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

$(x,y,z) = (5z,5z,z) = z(5,5,1)$

So, the base for this eigenspace is the vector (5,5,1).

If $\lambda = 3$ then the base is (0,4,3) and if $\lambda = -1$ then the base is (0,0,1)

3 0

3 years ago