15 divided by 4 is 3 with a remainder of 3.

15% of 300 is calculated as (15/100) * 300 = 45.

7 divided by 3 gives a remainder of 1.

The nth term is given by 2 + (n-1) * 3. For n=10, it is 2 + 27 = 29.

The common difference is 3. The 5th term is 2 + 4*3 = 14.

The common difference is 3; the 5th term is 2 + 4*3 = 14.

The common difference is 4; thus, the 5th term is 3 + 4*4 = 19.

The nth term is given by a_n = a_1 + (n-1)d. Here, a_5 = 2 + 4*3 = 14.

Area = π * r² = π * 4² = 16π cm².

Area = base * height = 10 * 6 = 60 cm².

Area of sector = (θ/360) * πr² = (90/360) * π(4)² = (1/4) * 16π = 4π ≈ 12.56 cm².

Area = 1/2 * (base1 + base2) * height = 1/2 * (6 + 10) * 4 = 32 cm².

Total sales of Product C = 100 + 150 + 250 + 200 = 700. Average = 700/4 = 175.

The total score of all students is 300 and there are 4 students. Average score = 300/4 = 75.

Compound Interest = P(1 + r/n)^(nt) - P = 1000(1 + 0.05/1)^(1*2) - 1000 = 1000(1.1025) - 1000 = 102.50.

Amount = 2000 * (1 + 0.10)^3 = 2000 * 1.331 = $2662. Compound Interest = Amount - Principal = 2662 - 2000 = $662.

Compound Interest = P(1 + r/n)^(nt) - P = 2000(1 + 0.06)^2 - 2000 = 2000(1.1236) - 2000 = $247.20.

CI = P(1 + r/100)^n - P = 5000(1 + 0.10)^2 - 5000 = 5000(1.21) - 5000 = $1050.

Sales of Product A in Q3 = 150, Product B = 200. Difference = 200 - 150 = 50.

The discriminant is b^2 - 4ac = 3^2 - 4(2)(1) = 9 - 8 = 1.

Discriminant D = b² - 4ac = (-4)² - 4(2)(2) = 16 - 16 = 0.

The discriminant is b^2 - 4ac = 6^2 - 4*3*2 = 36 - 24 = 12.

Distance = √[(x2 - x1)² + (y2 - y1)²] = √[(7 - 3)² + (1 - 4)²] = √[16 + 9] = √25 = 5 units.

The GCD of 12 and 15 is 3, as it is the largest number that divides both.

The GCD of 36 and 60 is 12.

The GCD of 48 and 180 is 12.

GCD(48, 18) = 6.

Sales of Product B increased from 1000 in Q1 to 1200 in Q2, which is a growth of 20%.

Sales grew from 1000 to 1200, which is a growth rate of 20%.

The HCF of 100 and 250 is 50, as it is the largest number that divides both.