V = (pi)r^2h / 3
= (pi)10^2(81) / 3
= (pi)(100)(81) / 3
= (pi)(8100) / 3
= (pi)2700
So C.
What goes in goes out. This means none of the current is lost. So if 0.7amps+0.68amps+0.47amps = 1.85 amps comes in, then 1.85 amps goes out. So far we have 0.8 amps+0.55 amps going out= 1.35 amps. This means we need an additional 1.85-1.35amps going out= 0.5 amps for the third outgoing current. Cheers!
The matrix R represents the reflection matrix for the provided vertices, and graph A represents the pre-image and the image on the same coordinate grid.
<h3>What is the matrix?</h3>
It is defined as the group of numerical data, functions, and complex numbers in a specific way such that the representation array looks like a square, rectangle shape.
We have vertices shown in the picture.
Form a matrix using the vertices:
![= \left[\begin{array}{ccc}-3&5&6\\7&3&-5\\\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%266%5C%5C7%263%26-5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To reflect over the x-axis, multiply by the reflection matrix:
![= \left[\begin{array}{ccc}1&0\\0&-1\\\end{array}\right]\left[\begin{array}{ccc}-3&5&6\\7&3&-5\\\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%266%5C%5C7%263%26-5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\rm R= \left[\begin{array}{ccc}-3&5&6\\-7&-3&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Crm%20R%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%265%266%5C%5C-7%26-3%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The above matrix represent the reflection matrix.
Thus, the matrices R represents the reflection matrix for the provided vertices, and graph A represents the pre-image and the image on the same coordinate grid.
Learn more about the matrix here:
brainly.com/question/9967572
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THE NEW COODINATES
U( -4,-6)
Now( 6,-4)
S( -8,-10)
Now (10,-8)
T(-3,-10)
Now(10,-3)