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Leto [7]
2 years ago
8

The radius of a circle is 4 feet. what is the angle measure of an arc pi feet long?

Mathematics
1 answer:
julsineya [31]2 years ago
7 0

Answer:

Step-by-step explanation:

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The sum of a number times 6 and 21 is at least 24. <br> what's the inequality
Alex777 [14]

Answer:

x ≥ 1/2

Step-by-step explanation:

6x + 21 ≥ 24

6x ≥ 3

x ≥ 1/2

check: (1/2 x 6) + 21 = 24

4 0
3 years ago
What is the length of TW?
Ann [662]
<h2>32</h2>

Since the length of the line segment TU is 16, the length of the line segment UW should be the same.

Measurement of the whole line segment (TW) should be:

16 + 16 = 32

The length of TW is

<h3>32</h3>

<em>Hope this helps :)</em>

5 0
1 year ago
What are the solutions of the system solve by graphing
Vladimir [108]
Vertex of x^2-2x-1 is (1,-2) so B is true
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If you're cycling at 24 km/hrs how far will you ride in 10 minutes
olganol [36]
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3 years ago
Use matrices to determine the coordinates of the vertices of the rotated figure. Then graph the pre-image and the image of the s
Arisa [49]

Answer:

The coordinates of the vertices of the rotated figure are :

U' (1 , -6), V' (-8 , -4), W' (-5 , 7) ⇒ the right answer is figure (d)

Step-by-step explanation:

* Lets study the matrices of the Rotation by 180°  

- When we rotate a point around the origin by 180° clockwise

 or anti-clockwise, we change the sign of the x-coordinate and

 the y-coordinate of the point

- Then matrix of the rotation 180° is

 \left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]

* Now lets solve the problem

- We will multiply the matrix of the rotation  by each point to

 find the image of each point

- The dimension of the matrix of the rotation  is 2×2 and the

 dimension of the matrix of each point is 2×1,  then the

 dimension of the matrix of each image is 2×1

∵ Point U is (-1 , 6)

∴ U'=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]\left[\begin{array}{ccc}-1\\6\end{array}\right]=

  \left[\begin{array}{ccc}(-1)(-1)+(0)(6)\\(0)(-1)+(-1)(6)\end{array}\right]=\left[\begin{array}{ccc}1\\-6\end{array}\right]

∴ U' = (1 , -6)

∵ Point V is (8 , 4)

∴ V'=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]\left[\begin{array}{ccc}8\\4\end{array}\right]=

  \left[\begin{array}{ccc}(-1)(8)+(0)(4)\\(0)(8)+(-1)(4)\end{array}\right]=\left[\begin{array}{ccc}-8\\-4\end{array}\right]

∴ V' = (-8 , -4)

∵ Point W is (5 , -7)

∴ W'=\left[\begin{array}{ccc}-1&0\\0&-1\end{array}\right]\left[\begin{array}{ccc}5\\-7\end{array}\right]=

  \left[\begin{array}{ccc}(-1)(5)+(0)(-7)\\(0)(5)+(-1)(-7)\end{array}\right]=\left[\begin{array}{ccc}-5\\7\end{array}\right]

∴ W' = (-5 , 7)

* Now look to the figures to find the right answer

∵ The images of the points are U' (1 , -6), V' (-8 , -4), W' (-5 , 7)

∴ The right answer is figure (d)

7 0
3 years ago
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