The GCF of 180 and 240 is 60
Answer:
(2.83 , 1 , 4)
Step-by-step explanation:

Rewrite these equations in matrix form
![\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%26-1%5C%5C4%26-2%26-2%5C%5C3%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
we can write it like this,

so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.
We get the inverse of matrix A,
![A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right] \\](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C)
now multiply the matrix with B
![X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\](https://tex.z-dn.net/?f=X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2.83%5C%5C1%5C%5C4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
Sarah
reason
she has the most consistent scores
Start with the equation of a circle whose center is at (h,k) and whose radius is r:
(x-h)^2 + (y-k)^2 = r^2
Substituting the given coordinates of the center:
(x-5)^2 + (y-[-5])^2 = r^2, or (x-5)^2 + (y+5)^2 = r^2
Substituding the given coordinates of a point (6,-2) on the circle:
(6-5)^2 + (-2+5)^2 = r^2
Simplifying:
1^2 + 3^2 = r^2, or 1 + 9 = r^2, or 10= r^2. Then r = sqrt(10).