Final answer:
The equation for the plane passing through the origin and parallel to the given plane x-5y+4z=4 is x-5y+4z=0.
Step-by-step explanation:
The equation of the plane passing through the origin and parallel to the plane x-5y+4z=4 can be found by using the normal vector of the given plane. Any plane parallel to another plane will have the same normal vector. The normal vector for the given plane is (1, -5, 4), as these are the coefficients of x, y, and z in the plane's equation. Since the plane must pass through the origin, its equation has the general form Ax + By + Cz = D, where D=0 because the plane passes through the origin (0,0,0). Therefore, the equation for the plane that we're seeking is simply 1x - 5y + 4z = 0 or x - 5y + 4z = 0.