f'(1) from left = 2, from right = 2; hence f is differentiable at x = 1.

Mean = 30; Mean Deviation = (|10-30| + |20-30| + |30-30| + |40-30| + |50-30|) / 5 = 10.

The mode is the number that appears most frequently, which is 23.

There is no mode as all values appear only once.

Mean = 6; Mean deviation = (|2-6| + |4-6| + |6-6| + |8-6| + |10-6|)/5 = (4 + 2 + 0 + 2 + 4)/5 = 12/5 = 2.4.

Mean = (4 + 8 + 6 + 5 + 3) / 5 = 26 / 5 = 5.0.

Q1 = 2, Q3 = 4; Interquartile Range = Q3 - Q1 = 4 - 2 = 2.

Mean = (1 + 2 + 3 + 4 + 5) / 5 = 3. Variance = [(1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)²] / 5 = 2.

Mean = 7.5; MAD = (|5-7.5| + |7-7.5| + |8-7.5| + |9-7.5| + |10-7.5|) / 5 = 1.

Mean = 7.5; Variance = [(5-7.5)^2 + (7-7.5)^2 + (8-7.5)^2 + (9-7.5)^2 + (10-7.5)^2] / 5 = 2; Standard Deviation = sqrt(2) = 1.41

The semi-major axis is 4 and the semi-minor axis is 3 after rewriting the equation in standard form.

Substituting x = 1 into the equation gives 1^2 + 2*1 + k = 0 => 1 + 2 + k = 0 => k = -3.

The discriminant must be non-negative: k^2 - 4*1*9 >= 0 => k^2 >= 36 => k <= -6 or k >= 6.

The product of the roots of the cubic equation ax^3 + bx^2 + cx + d = 0 is given by -d/a. Here, d = -6 and a = 1, so the product is -(-6)/1 = 6.

By substituting x = 2 into the equation, we find that 2 is a root since 2^3 - 6(2^2) + 11(2) - 6 = 0.

The discriminant must be non-negative: 4² - 4*1*k ≥ 0, thus k ≤ 4.

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, which gives k ≤ 9, so the minimum value is -9.

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

There are 2 moles of gas on the reactant side (2A).

According to Le Chatelier's principle, increasing the concentration of B will shift the equilibrium to the left, resulting in a decrease in the concentration of A.

Decreasing the volume increases the pressure, and 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 the equilibrium will shift towards the side with fewer moles of gas, which is the right side in this case.

Doubling the volume decreases the pressure, which shifts the equilibrium towards the side with more moles of gas, in this case, the right side.

Doubling the volume decreases the pressure, which shifts the equilibrium towards the side with more moles of gas (B and C).

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 temperature for an endothermic reaction will shift the equilibrium to the left, favoring the formation of HCl.

Increasing the concentration of B will shift the equilibrium to the left to produce more A, according to Le Chatelier's principle.

Decreasing the temperature for an exothermic reaction will shift the equilibrium to the right, favoring the formation of products.

Adding H2 will shift the equilibrium to the right to produce more NH3.

Increasing the pressure will shift the equilibrium to the right, favoring the formation of NH3, which has fewer moles of gas.