2x + 2y = 10
x + y = 5
x = 5 - y
Substitute for x in the other equation:-
(5 - y - 3)^2 + (y + 2)^2 = 16
(2 - y)^2 + (y + 2)^2 = 16
4 - 4y + y^2 + y^2 + 4y + 4 = 16
2y^2 + 8 = 16
2y^2 - 8 = 0
y^2 = 4
y = +/- 2
consider 2x + 2y = 10
when y = -2, 2x = 10 +4 giving x = 7
when y = 2, 2x = 10 - 4 giving x = 3
so solution is x = 7, y = -2 and x = 3, y=2.
Yes, I believe so, because the transformation has not in any way altered the image, but simply moved it.
Answer: A-(2,6)! :)
Explanation:
y < x^2 + 6
6 < 2^2 + 6
6 < 4 + 6
6 < 10 IT WORKS
y > x^2 - 4
6 > 2^2 - 4
6 > 4 - 4
6 > 0 IT WORKS
(Could I also get brainliest please?)
The answer i B just finished this.
