The maximum number of distinct functions is |B|^|A| = 3^5 = 243.

The number of distinct functions from set A to set B is given by |B|^|A|. Here, |B| = 3 and |A| = 5, so the maximum number of distinct functions is 3^5 = 243.

The number of ways to choose 3 elements from 5 is given by the combination formula C(5, 3) = 10.

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

The subsets containing '1' can be formed by including '1' and choosing from the remaining elements {2, 3}. There are 2^2 = 4 subsets, but we need to exclude the empty subset, so there are 4 - 1 = 3 subsets containing '1'.

The subsets with exactly 2 elements are {1, 2}, {1, 3}, and {2, 3}. So, there are 3 such subsets.

The subsets containing exactly 2 elements are {x, y}, {x, z}, and {y, z}. So, there are 3 such subsets.

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

g(-1) = 3(-1) + 2 = -3 + 2 = -1.

The number of ways to choose 2 elements from 3 is given by the combination formula C(3, 2) = 3.

The total number of subsets is 2^3 = 8. Subtracting the empty set gives 8 - 1 = 7.

The power set of H is {∅, {x}, {y}, {x, y}}. The subsets of H are {∅, {x}, {y}}.

h'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so x = 1 and x = -1 are critical points.

h(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.

h(1) = 1^3 - 3*1 = 1 - 3 = -2.

If 'a' is excluded, we can form subsets from {b, c}, which has 2^2 = 4 subsets.

A relation is a function if every element in the domain (set A) maps to exactly one element in the codomain. Here, 1 maps to 2, and 2 maps to 3, but 3 has no mapping, so R is not a function.

A relation R is reflexive if every element in set A is related to itself. Since R contains (1, 1), (2, 2), and (3, 3), R is reflexive.

A relation R is transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also be in R. Here, (1, 2) and (2, 3) are in R, but (1, 3) is not, so R is not transitive.

R is not reflexive as (1,1), (2,2), (3,3) are not in R. It is symmetric as (2,3) implies (3,2) is not in R. It is transitive as (1,2) and (2,3) implies (1,3) is not in R. Thus, R is not all of the above.

R is reflexive because it contains all pairs (a, a) and symmetric because (1,2) implies (2,1).

R is not a partial order because it is not transitive; (1,2) and (2,2) do not imply (1,2).

R is reflexive and symmetric, but not transitive. Thus, both 1 and 2 are true.

R is neither reflexive, symmetric, nor transitive as it does not satisfy any of the properties.

R is not symmetric as (b,c) does not imply (c,b) is in R. It is reflexive and transitive.

R is neither reflexive, symmetric, nor transitive.

A relation is reflexive if every element in the set is related to itself. Here, R includes (1, 1), (2, 2), and (3, 3), so R is reflexive.

x = cos^(-1)(-1/2) = 2π/3

If x = cos^(-1)(1/2), then x = π/3. Therefore, sin(x) = sin(π/3) = √3/2.

If x = cos^(-1)(1/2), then cos(x) = 1/2, which corresponds to x = π/3. Therefore, sin(x) = sin(π/3) = √3/2.