The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 2.

Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0. Critical points are x = 2.

f'(x) = 6x^2 - 18x + 12. The critical points are x = 1 and x = 2. Test intervals show increase in (0, 3).

Find f'(x) = 6x^2 - 18x + 12, set to 0. Local maxima at x = 2 gives f(2) = 6.

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 7 = -5.

The x-coordinate of the vertex is given by x = -b/(2a) = 12/(2*3) = 2.

The vertex is at x = -b/(2a) = 2. f(2) = 3.

The vertex occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

The vertex x-coordinate is found using -b/(2a) = 12/(2*3) = 2.

f'(x) = 2x + 2. Setting f'(x) = 0 gives x = -1. f(-1) = 1.

f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f(2) = 1 is a local minimum.

Using the determinant formula, det(E) = 1*(1*0 - 4*6) - 2*(0*0 - 4*5) + 3*(0*6 - 1*5) = 1*(-24) - 2*(-20) - 15 = -24 + 40 - 15 = 1.

Determinant of E = 1(1*0 - 4*6) - 2(0*0 - 4*5) + 3(0*6 - 1*5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1.

Determinant of E = (1*4) - (2*2) = 4 - 4 = 0.

The discriminant must be non-negative: 2² - 4*1*k ≥ 0, which simplifies to k ≤ 1.

The discriminant must be less than zero: 6² - 4*1*k < 0, which simplifies to k > 9.

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, thus k ≤ 9. The minimum value of k is -9.

Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas, which is the right side (2 moles of SO3).

Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas, which is the right side in this case.

Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas, which is the right side (2 moles of SO3).

Removing SO3 will decrease its concentration, causing the system to shift to the right to produce more SO3 in order to re-establish equilibrium, according to Le Chatelier's principle.

Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift towards the side with fewer moles of gas, which is the right side in this case.

Increasing the volume decreases the pressure, and according to Le Chatelier's principle, the equilibrium will shift to the side with more moles of gas, which is the left side in this case.

Increasing the concentration of a product (C) will shift the equilibrium to the left to counteract the change, according to Le Chatelier's principle.

Increasing the concentration of B will shift the equilibrium to the left to counteract the change, according to Le Chatelier's principle.

Decreasing the volume increases the pressure, and according to Le Chatelier's principle, the equilibrium will shift to the side with fewer moles of gas, which is the right side in this case.

Kp is expressed in terms of partial pressures, so for the reaction, Kp = (P_NH3)^2 / (P_N2)(P_H2)^3.

Kp is expressed in terms of the partial pressures of the gases involved in the reaction. For this reaction, Kp = (P_NH3)^2 / (P_N2)(P_H2)^3.

Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas. Here, it shifts to the right, producing more C.

The equilibrium constant Kc is given by the expression Kc = [C]^3 / ([A]^2[B]), where the concentrations are raised to the power of their coefficients.