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1 year ago

Enter the values for the following composite transformation.

Mathematics

2 answers:

julia-pushkina [17]1 year ago

8 0

do u have the answer

Allisa [31]1 year ago

6 0

Why are there hearts and idk sorry :/

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Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22

GuDViN [60]

Answer:

$b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$

Step-by-step explanation:

Consider the given matrix

$A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]$

Let matrix B is

$B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

It is given that

$A=B$

$\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

On comparing corresponding elements of both matrices, we get

$b_{11}=9,b_{12}=-2,b_{13}=3$

$b_{21}=2,b_{22}=17,b_{23}=0$

$b_{31}=3,b_{32}=22,b_{33}=8$

Therefore, the required values are $b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$ .

3 0

1 year ago

Read 2 more answers

which equation has an a-value of –2, a b-value of 1, and a c-value of 3? a. 0 = –2x2 x 3 b. 0 = 2x2 x 3 c. 0 = –2x2 3 d. 0 = 2x2

ehidna [41]

If you would like to find the matching equation, you can do this using the following steps:

ax^2 + bx + c = 0
a = -2
b = 1
c = 3

-2x^2 + x + 3 = 0

The correct result would be a. 0 = <span>-2x^2 + x + 3.</span>

8 0

1 year ago

Read 2 more answers

The orbit of the planet Venus is nearly circular. An astronomer develops a model for the orbit in which the sun has coordinates

klio [65]

Since the Venus orbits round the sun, the sun is the center of the circular path of the revolution of the planet, Venus.

Thus, the distance of the planet, Venus fron the sun is given by the distance between the points (0, 0) and (41, 53).

Recall that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Thus, the distance between the points (0, 0) and (41, 53) is given by:

$d= \sqrt{(41-0)^2+(53-0)^2} \\ \\ = \sqrt{41^2+53^2} = \sqrt{1,681+2,809} \\ \\ = \sqrt{4,490} =67 \ units$

Given that each unit of the plane represents 1 million miles, therefore, the distance from the sun to the Venus is 67 million miles.

8 0

1 year ago

The net of a triangular prism is shown below. What is the surface area of the prism? A. 128 cm^2 B. 152 cm^2 C. 176 cm^2 D. 304

shtirl [24]

Answer:

B. 152 cm²

Step-by-step explanation:

To find the surface area using a net, do this:

Take apart the figure. We see that there are three rectangles and two triangles. Find the area of each ( $A=l*w$ ) and then add the values together: