The intersection of sets A and B is the set of elements that are common to both sets, which is {4, 5}.

The difference P - Q includes elements in P that are not in Q, which is {1, 2}.

The intersection of sets P and Q includes elements that are both even and prime. The only even prime number is 2.

The symmetric difference of sets R and S includes elements that are in either set but not in both. Thus, it is {1, 2, 3, 6, 7}.

The union of sets X and Y includes all unique elements from both sets. Thus, the union is {a, b, c, d}.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 8 and 4th term = ar^3 = 32. Dividing gives r^2 = 4, so r = 2.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 4d = 14, we can find the 3rd term a + 2d = 10.

Let the first term be a. Then, the 3rd term is ar^2 = 27. Thus, a * 3^2 = 27, giving a = 3.

Let the first term be a and the common difference be d. From the equations a + 4d = 15 and a + 9d = 30, we can find d = 3.

Let the first term be a and the common difference be d. From the equations a + 5d = 30 and a + 8d = 45, we can find d = 5.

Using the formula for the nth term, a + 6d = 50. Substituting d = 5 gives a + 30 = 50, hence a = 20.

Let the angles be 2x, 3x, and 4x. The sum of angles in a triangle is 180 degrees. Therefore, 2x + 3x + 4x = 180. Solving gives x = 20, so the largest angle is 4x = 80 degrees.

Area = base × height. Thus, 120 = 15 × height, giving height = 120/15 = 8 units.

In the formula for the area of a quadrilateral, d1 and d2 represent the lengths of the diagonals.

Area of a sector = (θ/360) × πr². Thus, 30 = (θ/360) × π × 25. Solving gives θ = (30 × 360)/(25π) ≈ 72°.

Area of a square = side². Therefore, side = √64 = 8 cm.

The area of a triangle is given by the formula Area = 1/2 * base * height. Thus, 24 = 1/2 * 8 * height, which gives height = 6 cm.

Area of a triangle = (1/2) × base × height. Thus, 50 = (1/2) × 10 × height. Solving gives height = 10 units.

Total = Average × Number of items = 50 × 10 = 500.

Total = Average × Number of items = 50 × 8 = 400.

Sum of the numbers = Average × Number of items = 30 × 5 = 150.

Let the integers be (x-1), x, and (x+1). Their average is (x-1 + x + x+1) / 3 = x.

In base 'b', '10' represents 'b'. In base 'b+1', it becomes '11'.

The term containing x^2 is C(4,2) * x^2 * 2^2 = 6 * x^2 * 4 = 24x^2.

Circumference = 2πr. Therefore, r = circumference / (2π) = 31.4 / (2 * 3.14) = 5 cm.

Using the formula C = 2πr, we find r = C/(2π) = 31.4/(2*3.14) = 5 cm.

C(n,k) represents the number of ways to choose k items from n, which corresponds to the coefficient of x^k.

The decimal number 15 is represented as 1111 in binary.

The decimal number 255 converts to binary as 11111111.

If the diagonals of a quadrilateral bisect each other, it is a property of parallelograms.