<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
To start, you want to determine the inequality. You can do this by setting the number of buying n fish to be less than or equal to her current amount of money, being as it can't go over or she wouldn't be able to afford the fish. This inequality would be written so her amount of money (20) would be greater than or equal to the cost of n fish (2.5n), which would then look as such: 20≥2.5n. To solve this inequality, just solve for n by dividing both sides by 2.5, giving you 8≥n. This would mean that Becky can afford to buy up to 8 fish.
I think it's 1/4 that's my answer
Answer:
Translation
Step-by-step explanation:
Vertical and horizontal translation
Answer:
(?)=70
Step-by-step explanation:
