A circle has a circumference of 9π m. What is the area of this circle?

Mathematics

1 answer:

Likurg_2 [28]3 years ago

3 0

Answer:

20.25 m²

Step-by-step explanation:

hope this helps

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Answer:

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A circle has a radius of 3. An arc in this circle has a central angle of 60°.

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In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal. The same is true for a lower

SCORPION-xisa [38]

Answer:

Verified!

Step-by-step explanation:

Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.

Let's apply it for 2x2 matix:

$A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0$

So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)$ .

So, eigenvalues are $\lambda_1=a,\:\lambda_2=c$

Let's apply it for 3x3 matrix:

$A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0$

So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)(\lambda-f)$ .