The coefficient of x^3 in (3x + 2)^5 is given by 5C3 * (3)^3 * (2)^2 = 10 * 27 * 4 = 1080.

The product of the roots of the equation ax^2 + bx + c = 0 is given by c/a. Here, c = 6 and a = 2, so the product is 6/2 = 3.

Using Vieta's formulas, the product of the roots is c/a = 10/1 = 10.

The real part of z is -2.

The real part of z = 4 + 5i is 4.

To find the sum of the coefficients, substitute x = 1: (2*1 - 3)^4 = (-1)^4 = 1. The sum of coefficients is 81.

The sum of the coefficients in the expansion of (x + 1)^4 is (1 + 1)^4 = 2^4 = 16.

The sum of the coefficients in the expansion of (x + 1)^n is given by (1 + 1)^n = 2^n. Here, n=8, so the sum is 2^8 = 256.

To find the sum, we add the real parts and the imaginary parts: (2 + 3) + (2 - 4)i = 5 - 2i.

The sum of the roots is given by -b/a = 5/1 = 5.

The sum of the squares of the roots is given by (sum of roots)^2 - 2(product of roots). Here, sum = 5, product = 6. So, 5^2 - 2*6 = 25 - 12 = 13.

Using the binomial theorem, (1 + 2)^5 = 3^5 = 243.

To add the complex numbers, we combine the real parts and the imaginary parts: (2 + 4) + (3 - 2)i = 6 + 1i = 6 + i.

Convert 1/2 to 2/4, then (3/4) + (2/4) = 5/4.

Multiplying three negative ones gives −1.

(−1)^3 = −1.

(−2)² = 4.

According to BODMAS, 4 × 2 = 8, so 3 + 8 = 11.

3√(27) = 3 × 3 = 9.

The discriminant must be less than zero. Thus, k^2 - 4*1*16 < 0 leads to k^2 < 64, giving k < 8 and k > -8.

For equal roots, the discriminant must be zero. Thus, k^2 - 4*1*4 = 0, which gives k^2 = 16, so k = ±4.

For real and distinct roots, the discriminant must be positive: k^2 - 4(1)(16) > 0. Thus, k^2 > 64, leading to k < -8 or k > 8.

For both roots to be negative, k must be negative and |k| > 8. Thus, k = -8.

For both roots to be negative, k must be positive and k^2 > 4(1)(9). Thus, k > 6.

For both roots to be positive, k must be negative and k^2 > 36. Thus, k < -6.

The sum of the roots is -2 + -2 = -4, so k = 4.

For the roots to be equal, the discriminant must be zero. Thus, (-5)^2 - 4(1)(k) = 0. Solving gives k = 6.25.

The sum of the roots is 2 + 4 = 6, so k = 6.

The sum of the roots is 3 + 3 = 6, so k = 6.

log_10(0.01) = log_10(10^-2) = -2.