The question is incomplete. The complete question is as follows:
Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.
![\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C-6%26-9%5Cend%7Barray%7D%5Cright%5D)
· X·![\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%26-3%5C%5C7%26-6%5Cend%7Barray%7D%5Cright%5D)
=<em>I</em>.
First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means
<em>I</em> = ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
So, · X· =
Isolating the X, we have
X·= -
Resolving:
X·= ![\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2-1%268-0%5C%5C-6-0%26-9-1%5Cend%7Barray%7D%5Cright%5D)
X·=![\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%268%5C%5C-6%26-10%5Cend%7Barray%7D%5Cright%5D)
Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,
X=⁻¹·
Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.
So,
·![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
=
9a - 3b = 1
7a - 6b = 0
9c - 3d = 0
7c - 6d = 1
Resolving these equations, we have a=
; b=
; c=
and d=
. Substituting:
X= ![\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B2%7D%7B11%7D%20%26%5Cfrac%7B-1%7D%7B11%7D%20%5C%5C%5Cfrac%7B7%7D%7B33%7D%26%5Cfrac%7B-3%7D%7B11%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
·
Multiplying the matrices, we have
X=![\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B8%7D%7B11%7D%20%26%5Cfrac%7B26%7D%7B11%7D%20%5C%5C%5Cfrac%7B39%7D%7B11%7D%26%5Cfrac%7B198%7D%7B11%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
Debido a restricciones de extensión y la características del ejercicio, recomendamos leer la explicación de esta pregunta para mayores detalles sobre la adición de números <em>enteros</em>.
<h3>¿Cuáles son los resultados de cada suma?</h3>
En este ejercicio tenemos un grupo de sumas con números <em>enteros</em> <em>positivos</em> y <em>negativos</em>, en las cuales se prueba la capacidad del estudiante para realizar varias operaciones en serie (adición, sustracción) y comprender las diferencias entre números <em>positivos</em>, <em>negativos</em> y <em>neutros</em>. Ahora procedemos a determinar el resultado de cada una de las expresiones:
20 + 50 + 30 + 7 = 107
30 + 5 + 2 = 37
- 200 - 50 - 70 - 8 = - 328
- 500 + 100 - 20 + 50 = - 370
10 - 5 = 5
20 + 50 - 25 - 10 = 35
- 100 + 20 = - 80
- 30 + 5 + 4 - 20 + 8 = - 33
- 258 + 8 = - 250
- 10 + 20 + 520 - 100 + 8 = 438
- 20 - 5 - 42 + 3 = - 64
1000 - 200 + 50 + 30 - 45 + 75 - 87 + 90 + 50 - 100 + 50 - 10 = 903
- 400 + 500 - 200 - 50 + 48 + 8 - 47 - 50 = - 191
300 + 20 - 50 + 30 - 84 + 35 - 7 + 20 - 40 + 10 - 45 + 65 + 8 - 55 = 207
800 + 50 - 69 + 8 - 35 + 85 - 54 + 40 + 85 + 74 - 32 - 8 + 65 - 27 = 982
Para aprender más sobre sumas: brainly.com/question/1456841
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Answer:
None, there are no degrees/exponents
Step-by-step explanation:
Hope this is helpful :)
Answer:
Step-by-step explanation:
Properties of a circumcenter;
1). Circumcenter of a triangle is a point which is equidistant from all vertices.
2). Point where perpendicular bisectors of the sides of a triangle meet is called circumcenter of the triangle.
From the picture attached,
9). AG = GB = GC = 21
10). BC = 2(DC)
= 2×16
= 32
11). By applying Pythagoras Theorem in ΔGFB,
GB² = GF² + FB²
(21)² = GF² + (19)²
441 = GF² + 361
GF² = 441 - 361
GF = 
GF = 8.9
12). By applying Pythagoras theorem in ΔGDB,
GB² = DG² + BD²
(21)² = (DG)² + (16)² [BD = DC = 16]
DG² = 441 - 256
DG = √185
DG = 13.6
