Triangle FGH is graphed on the coordinate plane below. On a coordinate plane, triangle F G H has points (negative 6, negative 8), (negative 2, negative 3), (negative 4, negative 2).

The figure is rotated 180° using the origin as the center of rotation. How do the coordinates of the vertices of the preimage compare to the coordinates of the vertices of the image?
(x, y) right-arrow (y, x)
(x, y) right-arrow (negative y, negative x)
(x, y) right-arrow (negative x, negative y)
(x, y) right-arrow (x, negative y)

(x, y) right-arrow (negative x, negative y) ⇒ (x, y) → (-x, -y)

Step-by-step explanation:

Let us revise the rotation about the origin

Let us solve the question

∵ The vertices of triangle FGH are (-6, -8), (-2, -3), (-4, -2)

∵ The figure is rotated 180° using the origin as the center of rotation

→ By using the 2nd rule above

∴ The vertices of its image are (6, 8), (2, 3), (4, 2)

∴ The rule is (x, y) → (-x, -y)

The coordinates of the vertices of the preimage compare to the coordinates of the vertices of the image do (x, y) right-arrow (negative x, negative y) ⇒ (x, y) → (-x, -y)

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval