Answer:
(-4,-4)
YW :))
Or you could of just looked it up cause thats what I did!!
Answer:
multiply the left side of the constant vector by the inverse matrix
Step-by-step explanation:
The matrix equation ...
AX = B
is solved by left-multiplying by the inverse of A:
A⁻¹AX = A⁻¹B
IX = A⁻¹B . . . . . the result of multiplying A⁻¹A is the identity matrix
X = A⁻¹B . . . . . B needs to be multiplied by the inverse matrix
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}-4&1\\3&2\end{array}\right]^{-1}\left[\begin{array}{c}9&7\end{array}\right]=\dfrac{1}{11}\left[\begin{array}{cc}-2&1\\3&4\end{array}\right]\left[\begin{array}{c}9&7\end{array}\right]=\left[\begin{array}{c}-1&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-4%261%5C%5C3%262%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D9%267%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B11%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%261%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D9%267%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%265%5Cend%7Barray%7D%5Cright%5D)
B. looks like the way to go.
If r(x)= 11x, and c(x)=6x +20, then just put that into the equation.
So instead of p(x) = r(x) - c(x)
I would be p(x) = 11x - 6x + 20
Now solve.
p(x) = 5x + 20 is your final answer, so A.
I think the answer is (4) because if you rotate AEC to the left (counterclockwise) and follow by the scale factor of 2 so, CE = 4x2 = 8 and CA = 3x2 = 6. Hope this help :))