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Orlov [11]
3 years ago
5

What is the circumference of a circle

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0
C=2πr

Hoped i helped BYEEEEEEEEEEEEEEEEEEEEEEEEE
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katovenus [111]

Step-by-step explanation:

its 60

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3 years ago
Arrange the entries of matrix A in increasing order of their cofactors values
givi [52]

To find the cofactor of

A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]

We cross out the Row and columns of the respective entries and find the determinant of the remaining 2\times 2 matrix with the alternating signs.


Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|


Ac_{11}=4\times 1- -1\times 2


Ac_{11}=4+ 2

Ac_{11}=6




Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|


Ac_{12}=-(-7\times 1- -1\times -8)


Ac_{12}=-(-7- 8)

Ac_{12}=15




Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|


Ac_{21}=-(5\times 1- 3\times 2)


Ac_{21}=-(5-6)


Ac_{21}=1







A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|


Ac_{23}=-(7\times 2 -8\times 5)


Ac_{23}=-(14-40)


Ac_{23}=26




A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|


Ac_{31}=5\times -1 -4\times 3


Ac_{31}=-5-12


Ac_{31}=-17


A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|


Ac_{33}=7\times 4- -7\times 5


Ac_{33}=28+35


Ac_{33}=63


Therefore in increasing order, we have;

Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63



7 0
3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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kipiarov [429]
3/4 is greater than 2/3 is the right one. Because 1/2 is more than 1/3. -1/4 is not less than -2/3, and -1 is not greater than 3/4s. I hope it helps
6 0
3 years ago
there are four Defenders on a soccer team that this represents 20% of the players on the team which equation can be used to find
gladu [14]
4=(total players on the team) x 0.2
(total players on the team) = 4/0.2
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20% is the same as 0.2.
2 is 20% of 10.
to find 10, divide 2 by 0.2
2/0.2=10
4 is 20% of something
to find 'something', divide 4 by 0.2
4/0.2=20
7 0
3 years ago
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