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Katen [24]
3 years ago
5

The radius of circle C is 6 units and the measure of central angle ACB is I radians. What is the approximate area of the entire

circle? What is the approximate area of the entire sector square units square units square units created by central angle ACB? What is the approximate area of the shaded region only?

Mathematics
2 answers:
Elis [28]3 years ago
4 0

Answer:113, 28, 22

Step-by-step explanation:

VARVARA [1.3K]3 years ago
3 0

Answer:

Step-by-step explanation:

see attachment

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Tony wants to purchase a baseball glove for $14.96, a baseball bat for $19.87, and a baseball for $5.37. He figured out that he
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Reasonable because the three items altogether cost about $40
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There are 70 students in the school band. 40% of them are sixth graders, 20% are seventh graders, and the rest are eighth grader
Inessa05 [86]

Answer:

28

Step-by-step explanation:

40% 6th graders of 70 total = 28 6th graders

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100% - 60% = 40% 8th graders

40% are 8th graders therefore the answer is 28

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The mean is <br> A: Sometimes<br> B: always<br> C: never
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this question is not clear.
6 0
3 years ago
Use Gauss-Jordan elimination to solve the following linear system.
marysya [2.9K]
Just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]
mulstiply 2nd row by -1 and add to 3rd
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]
divde last row by 2
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 2rd row by 6 and add to top one
\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 1st row by -1 and add to 2nd
\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]
divide 2nd row by -3
\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]
mulstiply 2nd row by -1 and add to 1st row
\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]
divide 1st row by -3
\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]

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

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
7 0
3 years ago
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