Let the solutions be a and b.
a = -2; b = -10
a + b = -2 + (-10) = -12
ab = (-2)(-10) = 20
(x - a)(x - b) = 0
(x - (-2))(x - (-10)) = 0
(x + 2)(x + 10) = 0
x^2 + 10x + 2x + 20 = 0
x^2 + 12x + 20 = 0
-h = 12
h = -12
4k = 20
k = 5
It would be a reflection across the x acess
Answer:
(A) It would have to lie on AB because DC would rotate halfway around M. B would lie on D and A would lie on C.
(B) AD would lie on the BC line. D would lie on B and A would lie on C.
(C) The Lines CD and AD would only intersect if one line had to rotate 180 degrees around M.
Answer:
I did this before it is C
Step-by-step explanation:
