∆ABC has vertices at A(12, 8), B(4, 8), and C(4, 14). ∆XYZ has vertices at X(6, 6), Y(4, 12), and Z(10, 14).
∆MNO has vertices at M(4, 16), N(4, 8), and O(-2, 8).
∆JKL has vertices at J(14, -2), K(12, 2), and L(20, 4).

are congruent. A
is a single rigid transformation that maps the two congruent triangles.

y= -

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

table 3 represents a direct variation function. : )