<h2>
The required solution is x = 6 and y = 11 </h2>
Step-by-step explanation:
Given system of equations are
x+5y = 11 and x-y =5
![A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%265%5C%5C1%26-1%5Cend%7Barray%7D%5Cright%5D)
![X=\left[\begin{array}{c}x\\y\end{array}\right]](https://tex.z-dn.net/?f=X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D)
and ![B= \left[\begin{array}{c}11\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
∴AX=B
![adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]](https://tex.z-dn.net/?f=adj%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%7B-1%7D%26%7B-5%7D%5C%5C%7B-1%7D%261%5Cend%7Barray%7D%5Cright%5D)
∴
So,![A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%5Cfrac%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%7B-1%7D%26%7B-5%7D%5C%5C%7B-1%7D%261%5Cend%7Barray%7D%5Cright%5D%7D%7B-6%7D)
![A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20%5Cc%7D%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B5%7D%7B6%7D%7D%5C%20%5C%5C%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B-1%7D%7B6%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%7D)
⇒![\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20%5Cc%7D%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B5%7D%7B6%7D%7D%5C%20%5C%5C%20%20%7B%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7D%26%7B%5Cfrac%7B-1%7D%7B6%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%7D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D11%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
⇒![\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c} {6}\\ {11} \end{array}\right]}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D%20%20%7B6%7D%5C%5C%20%20%7B11%7D%20%5Cend%7Barray%7D%5Cright%5D%7D)
∴ x= 6 and y = 11
The required solution is x = 6 and y = 11
Answer:
201.1 meters squared
Step-by-step explanation:
A= pi*r squared
Diameter= 16, to find the radius we must divide by 2
16/2=8
Radius= 8
A= pi* (8)^2
A= 201.06
The area of the circle is 201.1 meters squared.
Answer:
A
Step-by-step explanation:
Idek.
<h2>QUESTION</h2>
<h2>FIRST YOU HAVE TO DISTRIBUTE:</h2>
<h2>SOLVING FOR OUR ANSWER</h2>
<h2 /><h2>NEW EQUATION/THE ANSWER YOU WERE LOOKING FOR </h2>
<h2>GOOD LUCK ON YOUR ASSIGNMENT AND ENJOY YOUR DAY!</h2>
~
<h2>SIDE NOTE: THE DISTRIBUTIVE FORMULA IS</h2>
:
Answer:
80 is 50% of 160.
Step-by-step explanation:
In order to find this, we take the whole number and divide by the percentage that it represents. This gives us the number we are looking for.
80/50%
80/.50
160
