The power set is the set of all subsets of A. P(A) = { {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }. The number of elements n(P(A)) = 2^n = 2^3 = 8.

n(P(A)) = 2^n

Let n(C) = 24, n(F) = 16, n(C union F) = 35 (each likes at least one). By the formula n(C union F) = n(C) + n(F) - n(C intersection F), we get 35 = 24 + 16 - n(C intersection F), so n(C intersection F) = 40 - 35 = 5. Hence 5 students like both games.

n(A union B) = n(A) + n(B) - n(A intersection B)

n(T) = 150, n(C) = 225, n(T intersection C) = 100. Students drinking at least one = n(T union C) = n(T) + n(C) - n(T intersection C) = 150 + 225 - 100 = 275. Students drinking neither = total - n(T union C) = 600 - 275 = 325.

neither = n(U) - n(A union B)

Union (A union B) is the set of all elements that are in A or in B or in both. Intersection (A intersection B) is the set of elements common to both A and B. Difference (A - B) is the set of elements that are in A but not in B. Example: if A = {1,2,3} and B = {2,3,4}, then A union B = {1,2,3,4}, A intersection B = {2,3}, and A - B = {1}.

A union B = {2,3,4,5,6,7,8}, so (A union B)' = U - (A union B) = {1,9,10}. Now A' = {1,6,7,8,9,10} and B' = {1,2,3,9,10}, so A' intersection B' = {1,9,10}. Since (A union B)' = {1,9,10} = A' intersection B', De Morgan's law is verified.

(A union B)' = A' intersection B'

{x : -3 <= x < 5} includes -3 but not 5, so in interval notation it is [-3, 5). The interval (2, 9] excludes 2 and includes 9, so in set-builder form it is {x : x is a real number, 2 < x <= 9}.

The Cartesian product A x B is the set of all ordered pairs (a, b) with a in A and b in B. A x B = {(1,3), (1,4), (2,3), (2,4)}. The number of elements n(A x B) = n(A) x n(B) = 2 x 2 = 4.

n(A x B) = n(A) x n(B)

Two ordered pairs are equal when their corresponding components are equal. So x + 1 = 3 gives x = 2, and y - 2 = 1 gives y = 3. Hence x = 2 and y = 3.

(a, b) = (c, d) => a = c and b = d

The domain of a relation is the set of all first components (inputs), and the range is the set of all second components (outputs). For R = {(1,2), (2,4), (3,6)}, the domain = {1, 2, 3} and the range = {2, 4, 6}.

For f(x) to be real, the expression under the square root must be non-negative: 9 - x^2 >= 0, i.e. x^2 <= 9, so -3 <= x <= 3. Hence the domain is [-3, 3]. As x varies over this interval, 9 - x^2 ranges from 0 (at x = +/-3) to 9 (at x = 0), so sqrt(9 - x^2) ranges from 0 to 3. Hence the range is [0, 3].

sqrt() defined only for non-negative values

A function f from A to B is one-one (injective) if different elements of A have different images in B, i.e. f(x1) = f(x2) implies x1 = x2. A function f from A to B is onto (surjective) if every element of B is the image of at least one element of A, i.e. the range of f equals the codomain B.

f(2) = 2^2 + 1 = 4 + 1 = 5. f(-1) = (-1)^2 + 1 = 1 + 1 = 2. For f(x) = 10: x^2 + 1 = 10, so x^2 = 9, giving x = +3 or x = -3.

f(x) = x^2 + 1

Using pi radians = 180 degrees: 5pi/6 radians = (5/6) x 180 = 150 degrees. And 240 degrees = 240 x (pi/180) = 4pi/3 radians.

radians = degrees x (pi/180)

360 revolutions per minute = 360/60 = 6 revolutions per second. One revolution corresponds to an angle of 2pi radians. Therefore the angle turned in one second = 6 x 2pi = 12pi radians.

angle = number of revolutions x 2pi

Write 75 = 45 + 30. Using sin(A + B) = sin A cos B + cos A sin B: sin 75 = sin 45 cos 30 + cos 45 sin 30 = (1/sqrt2)(sqrt3/2) + (1/sqrt2)(1/2) = (sqrt3 + 1)/(2 sqrt2) = (sqrt6 + sqrt2)/4.

sin(A + B) = sin A cos B + cos A sin B

Expand each factor: cos(A + B) = cos A cos B - sin A sin B and cos(A - B) = cos A cos B + sin A sin B. Their product is of the form (p - q)(p + q) = p^2 - q^2 with p = cos A cos B and q = sin A sin B. So it equals cos^2 A cos^2 B - sin^2 A sin^2 B. Writing cos^2 B = 1 - sin^2 B and sin^2 A = 1 - cos^2 A: = cos^2 A (1 - sin^2 B) - (1 - cos^2 A) sin^2 B = cos^2 A - cos^2 A sin^2 B - sin^2 B + cos^2 A sin^2 B = cos^2 A - sin^2 B. Hence proved.

cos(A+B)cos(A-B) = cos^2 A - sin^2 B

Using sin^2 x + cos^2 x = 1: sin^2 x = 1 - (9/25) = 16/25, so sin x = +/- 4/5. In the second quadrant sine is positive, so sin x = 4/5. Then tan x = sin x / cos x = (4/5) / (-3/5) = -4/3.

sin^2 x + cos^2 x = 1

Start from the fundamental identity sin^2 x + cos^2 x = 1. Divide both sides by cos^2 x (cos x not equal to 0): sin^2 x/cos^2 x + cos^2 x/cos^2 x = 1/cos^2 x, which gives tan^2 x + 1 = sec^2 x. Hence 1 + tan^2 x = sec^2 x.

1 + tan^2 x = sec^2 x

The principal value for which sin x = 1/2 is x = pi/6. The general solution of sin x = sin a is x = n pi + (-1)^n a, where n is any integer. Here a = pi/6, so the general solution is x = n pi + (-1)^n (pi/6), n belongs to the set of integers.

sin x = sin a => x = n pi + (-1)^n a

The principal value for which tan x = sqrt(3) is x = pi/3. The general solution of tan x = tan a is x = n pi + a, where n is any integer. Here a = pi/3, so the general solution is x = n pi + pi/3, n belongs to the set of integers.

tan x = tan a => x = n pi + a