Area = ∫ from 0 to 3 of x^2 dx = [1/3 * x^3] from 0 to 3 = 9.

The area is given by the integral ∫ (x^2) dx from 1 to 3. This evaluates to [x^3/3] from 1 to 3 = (27/3 - 1/3) = 26/3.

The area under the curve y = x^4 from 0 to 1 is given by ∫(from 0 to 1) x^4 dx = [x^5/5] from 0 to 1 = 1/5.

The area is given by the integral from 0 to 2 of x^4 dx. This evaluates to [x^5/5] from 0 to 2 = (32/5) = 16.

The argument of z = -1 - i is θ = tan^(-1)(-1/-1) = 3π/4.

First five primes: 2, 3, 5, 7, 11. Mean = (2 + 3 + 5 + 7 + 11) / 5 = 28 / 5 = 5.6.

Mean = (12 + 15 + 18 + 21 + 24) / 5 = 90 / 5 = 18.

The coefficient of x^0 is C(4, 0) * (2x)^0 * 3^4 = 1 * 81 = 81.

The coefficient of x^1 is C(5,1) * 2^4 = 5 * 16 = 80.

The coefficient of x^2 is C(6,2) * (3)^2 * (-4)^4 = 15 * 9 * 256 = 34560.

The coefficient of x^3 is C(6,3)(2)^3(-3)^3 = 20 * 8 * (-27) = -4320.

The coefficient of x^3 is C(6,3) * (1/2)^3 = 20 * 1/8 = 2.5.

The coefficient of x^3 is C(6,3) * (2)^3 = 20 * 8 = 160.

The coefficient of x^3 is C(5,3) * (-1)^2 = 10.

The coefficient of x^3 is C(6,3) * (-1)^3 = 20 * (-1) = -20.

The coefficient of x^3 is C(5,3) * (-3)^2 = 10 * 9 = -90.

Using the binomial theorem, the coefficient of x^4 in (3x - 2)^6 is given by 6C4 * (3)^4 * (-2)^2 = 15 * 81 * 4 = 4860.

The coefficient of x^5 is C(8,5) = 56.

The coefficient of x^5 is C(8,5) * (3)^3 = 56 * 27 = 1512.

The coefficient of x^5 is C(7,5) * (-3)^2 = 21 * 9 = -1890.

For the lines to be parallel, the discriminant D must be equal to 0.

The condition for the lines to be perpendicular is given by the relation ab + h^2 = 0.

The condition for the lines to be parallel is given by h^2 = ab.

The conjugate of z = 5 - 6i is z̅ = 5 + 6i.

Centroid = ((x1+x2+x3)/3, (y1+y2+y3)/3) = (9/3, 6/3) = (3, 2).

Centroid = ((1+3+5)/3, (2+4+6)/3) = (3, 4).

The equation y^2 = -12x can be rewritten as (y - 0)^2 = 4p(x - 0) with p = -3, so the focus is at (-3, 0).

Using the formula for foot of perpendicular, we find the coordinates to be (1, 1).

Using the formula for foot of perpendicular, we find the coordinates to be (3, 2).

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x^2 = 1, so x = 1 or x = -1. f(1) = 0, f(-1) = 4. The point is (1, 0).