The sum S = a/(1 - r) = 1/(1 - 1/2) = 1/(1/2) = 2.

The sum of the roots is given by -b/a = 3/2.

The sum of the roots is given by -b/a = 4/2 = 2.

Using Vieta's formulas, the sum of the roots is -(-12)/3 = 4.

The sum of the roots is given by -b/a = 4/1 = 4.

The sum of the roots is given by -b/a = 4/1 = 4.

The sum of the roots is given by -b/a = 7/1 = 7.

The term containing x^2 is given by C(4,2) * (3x)^2 * 4^2 = 6 * 9 * 16 = 864.

The term independent of x occurs when k = 7, C(7, 3) * (3)^3 * (-4)^4 = 35 * 27 * 256 = 84.

The total number of terms is n + 1 = 6 + 1 = 7.

The trace of A is the sum of the diagonal elements: 1 + 5 + 9 = 15.

(1 + i)^2 = 1^2 + 2*1*i + i^2 = 1 + 2i - 1 = 2i.

Using the binomial theorem, (2 + 3)^3 = 5^3 = 125.

cos^(-1)(-1) = π, since cos(π) = -1.

i^4 = (i^2)^2 = (-1)^2 = 1.

For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0, thus k^2 = 64, giving k = -8 or 8. The answer is -4.

The discriminant must be negative: k^2 - 4*1*9 < 0 => k^2 < 36 => |k| < 6, hence k > 6.

For no real roots, the discriminant must be less than zero: k^2 - 4*1*16 < 0 => k^2 < 64 => |k| < 8.

For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0, thus k^2 = 64, giving k = -8 or k = 8. The answer is -8.

The discriminant must be less than zero: k^2 - 4*1*16 < 0 => k^2 < 64 => k < 8 and k > -8.

For one real root, the discriminant must be zero: k^2 - 4*1*25 = 0, thus k^2 = 100, giving k = -10 or k = 10.

For no real roots, the discriminant must be less than zero: k^2 - 4*1*9 < 0, thus k > 6.

For one real root, the discriminant must be zero: k^2 - 4*1*9 = 0 => k^2 = 36 => k = ±6.

log_10(1000) = 3 and log_10(0.01) = -2, thus 3 - 2 = 1.

log_10(1000) = log_10(10^3) = 3.

log_2(32) = 5 and log_2(4) = 2. Therefore, 5 - 2 = 3.

log_2(32) = 5 and log_2(8) = 3. Therefore, 5 - 3 = 2.

log_3(27) = 3 and log_3(9) = 2. Therefore, 3 - 2 = 1.

log_3(81) = log_3(3^4) = 4.

log_4(64) = log_4(4^3) = 3.