The product of the roots is given by c/a = 10/1 = 10.

The product of the roots is given by c/a = 9/3 = 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 is 3.

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

The real part of z 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 two complex numbers z1 + z2 = (2 + 3i) + (4 - 2i) = (2 + 4) + (3 - 2)i = 6 + i.

Using Vieta's formulas, the sum of the roots is -(-8)/2 = 4.

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.

(1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i - 1 = 2i.

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.

(−1)² = 1 and (−2)² = 4, so 1 + 4 = 5.

(−1)³ = −1.

(−2)² = 4.

(−5)² = 25.

−7 + (−3) = −10.

(√(49) + √(16)) = 7 + 4 = 11, so 11 / 5 = 2.2, which rounds to 2.

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

3.5 + 2.8 = 6.3.

3√(27) = 3 × 3 = 9.