Final answer:
The given expression is not an instance of Ax2.
Step-by-step explanation:
The given expression is (Vx)(Vy)x = y → (Vy)y = y an instance of Ax2. To determine if this is true, we need to analyze the equation and compare it with the definition of Ax2.
In the given expression, (Vx)(Vy)x = y, (Vx) and (Vy) represent the components of a vector along the x and y axes, respectively. The expression can be rewritten as (Ax)(Ay)x = y, where (Ax) represents the x-component of the vector and (Ay) represents the y-component of the vector.
In Ax2, Ax represents the x-component of the vector and 2 represents the exponent. Therefore, (Ax)(Ax) = Ax2. In the given expression, we have (Ax)(Ay)x = y. Since (Ay)x cannot be reduced to Ax2, we can conclude that the given expression is not an instance of Ax2.