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xenn [34]
3 years ago
15

Please I need help really fast

Mathematics
1 answer:
Arlecino [84]3 years ago
6 0

Answer: 30 degrees

Step-by-step explanation:

The total is 180 degrees.

there is a right angle on the left. A right angle is 90 degrees.

180 - 90 = 90

There is still 90 degrees left, meaning that the right side is a total of 90 degrees.

2x + x = 3x

x = 90/3

x = 30

Therefore,

x = 30 degrees!

Hope it helped!

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Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.
solong [7]

Answer:

A)\,\,det(A)=1

B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

Step-by-step explanation:

det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|

Expanding with first row

det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1

To find inverse we first find cofactor matrix

C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\

C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\

Cofactor matrix is

C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

4 0
3 years ago
A quadrilateral can be inscribed in a circle, if and only if, the opposite angles in that quadrilateral are supplementary.
dimaraw [331]

The opposite angles are equal to are supplementary to each other or equal to each other.

<h3>What is a Quadrilateral Inscribed in a Circle?</h3>

In geometry, a quadrilateral inscribed in a circle, also known as a cyclic quadrilateral or chordal quadrilateral, is a quadrilateral with four vertices on the circumference of a circle. In a quadrilateral inscribed circle, the four sides of the quadrilateral are the chords of the circle.

The opposite angles in a cyclic quadrilateral are supplementary. i.e., the sum of the opposite angles is equal to 180˚.

If a, b, c, and d are the inscribed quadrilateral’s internal angles, then

a + b = 180˚ and c + d = 180˚

by theorem the central angle = 2 x inscribed angle.

∠COD = 2∠CBD

∠COD = 2b

∠COD = 2 ∠CAD

∠COD = 2a

∠COD + reflex ∠COD = 360°

2a + 2b = 360°

2(a + b) =360°

By dividing both sides by 2, we get

a + b = 180°.

Learn more about this concept here:

brainly.com/question/16611641

#SPJ1

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