f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = 0, 2, -2. f(0) = 16, f(2) = 0, f(-2) = 0. Minimum value is 0.

The mode is the number that appears most frequently, which is 4.

Mode is the number that appears most frequently. Here, mode = 2.

Mode is the number that appears most frequently. Here, mode = 5.

Mode = 3 (most frequent value).

The mode is 4 as it appears most frequently (3 times).

Mode is the number that appears most frequently. Here, mode = 2.

Mode is 3 as it appears most frequently (3 times).

The modulus |z| = √((-3)^2 + (4)^2) = √(9 + 16) = √25 = 5.

The total number of subsets is 2^n = 2^2 = 4. Proper subsets exclude the set itself, so there are 4 - 1 = 3 proper subsets.

The empty set has exactly one subset, which is itself (∅).

The number of subsets of a set with n elements is 2^n. Here, n = 3, so 2^3 = 8.

The number of subsets of a set with n elements is 2^n. Here, n = 5, so 2^5 = 32.

The number of subsets of a set with n elements is 2^n. Here, n = 5, so the number of subsets is 2^5 = 32.

The number of subsets of a set with n elements is 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.

Integrating gives y = 2x^2 + C. Using the initial condition, C = 1.

The general solution is y = Ce^(2x). Using the initial condition y(0) = 1, we find C = 1.

The general solution is y = Ce^(2x). Using y(0) = 5 gives C = 5.

The power set of a set is the set of all its subsets. For A = {a, b}, the power set is {∅, {a}, {b}, {a, b}}.

The power set of the empty set contains only the empty set itself: {∅}.

The power set of F = {a} is {∅, {a}}.

The power set of G = {1} is {∅, {1}}.

The power set of a set with n elements has 2^n elements. For {1, 2}, the power set is {∅, {1}, {2}, {1, 2}}.

cot^(-1)(0) = π/2

sec^(-1)(2) = π/3, since sec(π/3) = 2.

There are 26 red cards in a deck of 52. Probability = 26/52 = 1/2.

There are 4 aces in a deck of 52 cards. Therefore, the probability P(ace) = 4/52 = 1/13.

The possible outcomes for a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) which gives us 6 favorable outcomes. Total outcomes when rolling two dice = 6 * 6 = 36. Therefore, P(sum of 7) = 6/36 = 1/6.

Even numbers on a die: 2, 4, 6. Probability = 3/6 = 1/2.