<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
Here is my work for the problem. Hope this helps!
Answer:
0.6
Step-by-step explanation:
Answer:
4.5
Step-by-step explanation:
Answer:
Step-by-step explanation:
7) The formula for determining the area of a parallelogram is expressed as
Area = base × height.
Length of base = Area/height
Therefore,
Length of base = 7/2 = 3.5 feet
8) The formula for determining the area of a trapezoid is expressed as
Area = 1/2(a + b)h
Where
a and b are the length of the bases
h is the height. Therefore
21 = 1/2(2 + 4)h
21 = 3h
h = 21/3 = 7 inches
9) Area = base × height.
Height = Area/Length of base
Height = 28/14 = 2 inches
10) a and b are 10 inches each.
Area = 1/2(a + b)h
Therefore,
35 = 1/2(10 + 10)h
35 = 10h
h = 35/10
h = 3 inches
