Answer:
2/3
Step-by-step explanation:
Let x be the number of books that Millicent has. So the number of books that Harold has is x/2. And the number of book that Harold is bringing to the new home is (x/2)/3 = x/6.
The home has enough room for another half of Millicent's books, which is x/2. So the total books capacity of the new home is

Since x is also the number of books in Millicent's old library capacity, we can conclude that the fraction of Millicent's old library capacity to the new home's library capacity is 2/3, or 2 thirds
Answer:
Option C is correct.
Step-by-step explanation:
-x+3y=2
4x-2y=22
In matrix form is represented as:
![\left[\begin{array}{cc}-1&3\\4&-2\end{array}\right] \left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{c}2&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%263%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2622%5Cend%7Barray%7D%5Cright%5D)
AX=B


|A| = (-1)(-2)-(3)(4)
|A| = 2-12
|A| = -10
Adj A = ![\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-3%5C%5C-4%26-1%5Cend%7Barray%7D%5Cright%5D)
A^-1 = -1/10![\left[\begin{array}{cc}-2&-3\\-4&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%26-3%5C%5C-4%26-1%5Cend%7Barray%7D%5Cright%5D)
A^-1 = 1/10![\left[\begin{array}{cc}2&3\\4&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%263%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D)
X= A^-1 B
X = 1/10![\left[\begin{array}{cc}2&3\\4&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%263%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}2&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2622%5Cend%7Barray%7D%5Cright%5D)
X=1/10![X=1/10\left[\begin{array}{c}2*2+3*22\\4*2+1*22\end{array}\right]\\X=1/10\left[\begin{array}{c}4+66\\8+22\end{array}\right]\\X=1/10\left[\begin{array}{c}70\\30\end{array}\right]\\X=\left[\begin{array}{c}70/10\\30/10\end{array}\right]\\X=\left[\begin{array}{c}7\\3\end{array}\right]](https://tex.z-dn.net/?f=X%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%2A2%2B3%2A22%5C%5C4%2A2%2B1%2A22%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%2B66%5C%5C8%2B22%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D1%2F10%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D70%5C%5C30%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D70%2F10%5C%5C30%2F10%5Cend%7Barray%7D%5Cright%5D%5C%5CX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D7%5C%5C3%5Cend%7Barray%7D%5Cright%5D)
So, x = 7 and y =3
Hence Option C is correct.
Please include the diagrams in that question :)
Answer
C. third choice
SSS congruence postulate