Each prize can go to any of the 3 students, so the total ways = 3^4 = 81.

The number of ways is C(10, 4) = 210.

The number of arrangements of 5 different books is 5! = 120.

The number of ways is C(10, 5) = 252.

Total arrangements = 5! = 120. Arrangements with the two letters together = 4! * 2! = 48. So, arrangements where they are not together = 120 - 48 = 72.

Treat the two specific letters as one unit. Then, we have 4 units to arrange: 4! = 24. The two letters can be arranged in 2! = 2 ways. Total = 24 * 2 = 48.

The number of ways to choose 5 letters from 26 is C(26, 5) = 65780.

The number of ways to award 5 different prizes to 3 students is 3^5 = 243.

Treat the 2 specific books as one unit. Then we have 5 units to arrange: 5! * 2! = 240.

The number of arrangements in a circle is (n-1)! = (6-1)! = 5! = 120.

The number of ways to divide 6 people into 2 groups of 3 is (6! / (3!3!)) / 2 = 20.

The number of ways to choose 3 from 5 is C(5,3) = 10.

The word 'BANANA' has 6 letters with 'A' repeating 3 times and 'N' repeating 2 times. The arrangements are 6!/(3!2!) = 60.

The number of ways to choose 3 fruits from 5 is given by 5C3 = 10.

The number of ways to form a committee of 3 from 10 is given by 10C3 = 120.

The number of ways to choose 2 from 5 is given by 5C2 = 10.

The number of ways to select 2 students from 8 is given by 8C2 = 28.

From the second equation, x = 10 - 2y. Substituting into the first gives 2(10 - 2y) + 3y = 12, solving gives y = 2, x = 6.

5x - 7 < 3 => 5x < 10 => x < 2

5x - 7 < 3 => 5x < 10 => x < 2

5x - 7 < 3x + 1 => 2x < 8 => x < 4.

6x + 4 > 10 => 6x > 6 => x > 1.

7 - 2x > 3 => -2x > -4 => x < 2.

7 - 2x ≥ 1 => -2x ≥ -6 => x ≤ 3

7 - 2x ≥ 3 => -2x ≥ -4 => x ≤ 2.

7 - 3x ≤ 1 => -3x ≤ -6 => x ≥ 2.

7 - 3x ≥ 1 => -3x ≥ -6 => x ≤ 2.

7x + 1 ≥ 15 => 7x ≥ 14 => x ≥ 2

The roots of the equation x^2 - 10x + 21 = 0 are found using the quadratic formula, yielding a = 3 and b = 7.

Using the identity a^2 + b^2 = (a + b)^2 - 2ab, we get a^2 + b^2 = 10^2 - 2*21 = 100 - 42 = 58.