Using the limit property, lim (x -> 0) (sin(5x)/x) = 5. The function is continuous at x = 0.

Using L'Hôpital's Rule, lim x→2 (x^2 - 4)/(x - 2) = lim x→2 (2x)/(1) = 4.

The area between the curves is given by ∫(from 0 to 1) (x - x^2) dx = [x^2/2 - x^3/3] from 0 to 1 = (1/2 - 1/3) = 1/6 = 0.5.

The area under the curve is given by ∫(from 1 to 2) 3x^2 dx = [x^3] from 1 to 2 = (8 - 1) = 7.

Using the binomial theorem, the coefficient of x^2 in (2x - 3)^4 is given by 4C2 * (2)^2 * (-3)^2 = 6 * 4 * 9 = 216.

Using the power rule, f'(x) = 4x^3 - 12x^2 + 12x - 24.

Using the derivatives of sine and cosine, g'(x) = cos(x) - sin(x).

Using the distance formula: d = √[(2 - (-1))² + (2 - (-1))²] = √[9 + 9] = √18 = 3√2.

Using the distance formula: d = √[(4 - (-2))² + (5 - (-3))²] = √[(4 + 2)² + (5 + 3)²] = √[36 + 64] = √100 = 10.

Using the distance formula: d = √[(6 - 0)² + (8 - 0)²] = √[36 + 64] = √100 = 10.

The eigenvalues are found by solving the characteristic equation det(G - λI) = 0. This gives λ^2 - 8λ + 7 = 0, which factors to (λ - 1)(λ - 7) = 0, hence λ = 1, 7.

The slope m = (7 - 3) / (4 - 2) = 2. Using point-slope form: y - 3 = 2(x - 2) gives y = 2x + 1.

f'(x) = -2x + 6; setting to 0 gives x = 3; f(3) = -3^2 + 6(3) - 8 = 1.

Solving the equations simultaneously, we find the intersection point is (1, 1).

A · B = (2)(1) + (3)(2) + (1)(3) = 2 + 6 + 3 = 11

First derivative f'(x) = 16x^3 - 6x^2 + 1. Second derivative f''(x) = 48x^2 - 12x.

First derivative f'(x) = 3x^2 - 6x; second derivative f''(x) = 6x - 6.

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

Using the binomial theorem, the coefficient of x^4 in (a + b)^n is given by nCk * a^(n-k) * b^k. Here, n=6, a=x, b=-2, and k=2. Thus, the coefficient is 6C2 * (1)^4 * (-2)^2 = 15 * 4 = 60.

Mean = (1 + 2 + 3 + 4 + 5) / 5 = 3. Variance = [(1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)²] / 5 = 2.

The product of the roots of the cubic equation ax^3 + bx^2 + cx + d = 0 is given by -d/a. Here, d = -6 and a = 1, so the product is -(-6)/1 = 6.

By substituting x = 2 into the equation, we find that 2 is a root since 2^3 - 6(2^2) + 11(2) - 6 = 0.

f'(x) = cos(x) - sin(x). At x = π/4, f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 0.

In the equation x^2 = 4py, we have 4p = 16, thus p = 4. The distance from the vertex to the focus is 4.

The polynomial can be factored as (x-1)^3, indicating that it has one real root with multiplicity 3, hence all roots are real and equal.

The polynomial can be factored as (x - 1)^3, indicating that all roots are real and equal.

The discriminant D = b^2 - 4ac = 4^2 - 4(2)(2) = 16 - 16 = 0.

The discriminant is 0, indicating that the roots are real and equal.

For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0. Solving gives k = -8.

To ensure continuity at x = 2, k(2) + 1 must equal 3. Thus, k = 1.