The discriminant must be greater than zero: 4^2 - 4*1*k > 0 => 16 - 4k > 0 => k < 4.

For real and equal roots, the discriminant must be zero: 16 - 4k = 0, thus k = 4.

Using Vieta's formulas, the sum of the roots is -b/a = -5/1 = -5.

Using the sum and product of roots: -2 + -4 = -6 and -2*-4 = k => k = 8.

The sum of the roots is -(-2 - 3) = 5, so m = 5. The product of the roots is (-2)(-3) = 6, so n = 6. Thus, m + n = 5 + 6 = 11.

Using Vieta's formulas, p = -(-2 - 3) = 5.

Using Vieta's formulas, p = -(-2 - 3) = 5 and q = (-2)(-3) = 6. Therefore, p + q = 5 + 6 = 11.

The sum of the roots is 0, hence p = -sum = 0.

For equal roots, the discriminant must be zero: p^2 - 4q = 0, hence p^2 = 4q.

For the roots to be equal, the discriminant must be zero. Thus, b^2 - 4ac = 0 => 25 - 4k = 0 => k = 25.

Using Vieta's formulas, the sum of the roots is 7 and the product is p. Thus, 3 * 4 = p, so p = 12.

Let the roots be 3k and 4k. Then, 3k + 4k = 7 => 7k = 7 => k = 1. The product of the roots is 3k * 4k = 12k^2 = p => p = 12.

For equal roots, the discriminant must be zero: k^2 - 4*1*8 = 0, solving gives k = 4.

Using the product of the roots, c = 3 * (-2) = -6.

The condition for equal roots is given by the discriminant b^2 - 4ac = 0.

The sum of the roots is 3 + 4 = 7, hence m = -7.

For equal roots, the discriminant must be zero: p^2 - 4q = 0, hence p^2 = 4q.

Using Vieta's formulas, sum of roots = 1 + 2 = 3 and product of roots = 1*2 = 2. Thus, p = 2.

Using the formula S_n = a(1 - r^n) / (1 - r), we have 81 = 3(1 - r^n) / (1 - r). Solving gives r = 3.

When r = 1, S_n = a(1 - 1^n)/(1 - 1) is indeterminate, but the sum of n terms is na.

The 4th term a_4 = S_4 - S_3 = (3(4^2) + 2(4)) - (3(3^2) + 2(3)) = (48 + 8) - (27 + 6) = 56 - 33 = 23.

The 5th term can be found using a_n = S_n - S_(n-1). Calculate S_5 and S_4, then find a_5 = S_5 - S_4 = (5(5^2) + 3(5)) - (5(4^2) + 3(4)) = 38.

The sum of the roots is given by -b/a = 3. Here, -(-3)/1 = 3, so p can be any value.

The complement of A, denoted A', includes all elements in U that are not in A. Thus, A' = {3, 4, 5}.

The union A ∪ B includes all elements from both sets, which are {1, 2, 3}.

From the first equation, x = 10 - 2y. Substituting into the second gives 2(10 - 2y) - y = 3, solving gives y = 4, x = 2.

Since x = cos^(-1)(1/2) = π/3, then sin^(-1)(1/2) = π/6.

Since cos^(-1)(1/2) = π/3, we have sin^(-1)(√(1 - (1/2)^2)) = sin^(-1)(√(3/4)) = π/3.

cos^(-1)(1/2) = π/3.

Using the identity sin(x) = sqrt(1 - cos^2(x)), we have sin(x) = sqrt(1 - (1/2)^2) = √3/2.