What are the vertices of ΔA'B'C produced by T⟨−3, 6⟩ (ΔABC) = ΔA'B'C? The graph with X-coordinate marks -4, -2, 0, 2, 4 and Y-coordinate mark -4, -2, -1, 0, 2, 4. -The triangle ABC with coordinates A at (3, 0), B at (3, -2), C at (0, -3).

A. A′(0, 6), B′(0, 4), C′(−3, 3)
B. A′(6, 6), B′(6, 4), C′(3, 3)
C. A′(0, −6), B′(0, −8), C′(−3, 9)
D. A′(6, −6), B′(6, −8), C′(3, 9)

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

whaere is the picture

step-by-step explanation: