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.

Set P = {2, 4, 6, 8} and set Q = {2, 3, 5, 7}. The union is {2, 3, 4, 5, 6, 7, 8}.

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 difference R - S includes elements in R that are not in S, which is {1, 2}.

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

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_10 = 10/2 * (2*4 + 9*3) = 5 * (8 + 27) = 5 * 35 = 175.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 12 and 4th term = ar^3 = 48. Dividing these gives r^2 = 4, so r = 2.

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 + 3d = 14, we can find a = 6.

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. We have a + 2d = 12 and a + 6d = 24. Subtracting these gives 4d = 12, so d = 3.

Let the first term be a and the common difference be d. From the equations a + 2d = 15 and a + 6d = 27, solving gives d = 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 + 4d = 20 and a + 9d = 35, we can solve for a to find it is 10.

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.

Using the formula A = P(1 + r)^t, we have 1210 = 1000(1 + r)^2. Solving gives r = 0.1 or 10%.

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 = πr². 154 = (22/7)r². Solving gives r² = 154 * 7 / 22 = 49, so r = 7 cm.

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

If the height of a parallelogram is halved, the area is also halved, as area = base × height.

This formula applies to a kite, where the diagonals intersect at an angle.

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

Area = Length * Width. Therefore, Width = Area / Length = 48 / 8 = 6 meters.

Area of a sector = (θ/360) × πr². Thus, 25π = (θ/360) × π(10)². Solving gives θ = 90 degrees.

Area of a sector = (θ/360) × πr². Thus, 25π = (θ/360) × π(5)². Solving gives θ = 90°.

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.