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IceJOKER [234]
2 years ago
10

HELP ME PLEASE YOU WILL GET A BRAINLIST I NEED HELP ASAP!!A figure is located at (2, 0), (2, −2), (6, −2), and (6, 0) on a coord

inate plane. What kind of 3-D shape would be created if the figure was rotated around the x-axis? Provide an explanation and proof of your answer to receive full credit. Include the dimensions of the 3-D shape in your explanation.
Mathematics
1 answer:
max2010maxim [7]2 years ago
8 0

When the shape is rotated around the x-axis, a 3-D shape would be formed The 3-D shape is a cone with a radius 2 and a height 4

<h3>How to determine the 3-D shape?</h3>

The above coordinates represent a triangle that has one of its sides on the x-axis.

So, when the shape is rotated across the axis, the other two lines would whirl around the side on the x-axis.

This movement would form a cone with the following dimensions

Height = Distance (2, 0) and (6, 0) i.e. 4 units

Radius = Distance (2, 0) and (2, −2) i.e. 2 units

Hence, the 3-D shape is a cone with radius 2 and height 4

Read more about three dimensional shapes on:

brainly.com/question/12280037

#SPJ1

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Answer: 60 mph.

(486+316+638)/24

The distances given by the table. The time driven is 24 (given by the problem). So the average speed is simply the total distance divided by the time driven.

Step-by-step explanation:

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Malcolm bought 6 bowls for $13.20. What is the unit rate?
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Amy pays $1.71 in postage to mail a CD to a friend. She uses 41-cent stamps and 6-cent stamps. How many of each stamp did Amy us
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3 41-cent stamps and 8 6-cent stamps

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8 0
3 years ago
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0
3 years ago
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