The slopes are m1 = 2 and m2 = -1/2. The angle θ = tan⁻¹(|(m1 - m2) / (1 + m1*m2)|) = tan⁻¹(5/3), which is approximately 90 degrees.

The left limit as x approaches 1 is 1, the right limit is 2, and f(1) = 2. Since the left and right limits do not match, f(x) is not continuous at x = 1.

At x = 1, f(1) = 2(1) - 1 = 1 and lim x→1- f(x) = 1, lim x→1+ f(x) = 1. Thus, f(x) is continuous at x = 1.

Using the distance formula: d = √[(2 - 2)² + (-1 - 3)²] = √[0 + 16] = √16 = 4.

f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.

f'(x) = 2x; thus, f'(3) = 2(3) = 6.

Setting y = 0 in the equation gives 5x = 10, thus x = 2. The x-intercept is 2.

Using the product rule, f'(x) = 4e^x + 4x^2e^x.

Using the chain rule, f'(x) = 2x/(x^2 + 1).

Using the product rule, f'(x) = x^2 * e^x + 2x * e^x.

Number of attendees from urban areas = 800 * (1 - 0.25) = 800 * 0.75 = 600.

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.