y(t) = x(t - 2) + x(t + 4)
y(t) = (t - 4)x(t + 1)
y(t) = (t + 4)x(t - 1)
y(t) = (t + 5)x(t + 5)
Answer is Right!
Answer is Wrong!
The correct answer is A.
A casual system is a system in which the output depends only on the present and past inputs. In other words, the output at any time $t$ cannot depend on the future inputs.
Option A is the only option that satisfies this condition. In option A, the output $y(t)$ is a linear combination of the past inputs $x(t-2)$ and $x(t+4)$. The future inputs do not appear in the expression for $y(t)$.
Option B is not a casual system because the output $y(t)$ depends on the future input $x(t+1)$.
Option C is not a casual system because the output $y(t)$ depends on the future input $x(t-1)$.
Option D is not a casual system because the output $y(t)$ depends on the future input $x(t+5)$.