The determinant of an upper triangular matrix is the product of its diagonal elements. Here, det(C) = 1 * 1 * 1 = 1.

Using the Pythagorean identity, sin B = √(1 - cos²B) = √(1 - 0.6²) = √(1 - 0.36) = √(0.64) = 0.8.

Cotangent is 1 when the angle A is 45°, as cot A = cos A/sin A = 1 when both are equal.

cot D = 3/4 implies tan D = 4/3. Therefore, sin D = 4/5 (using the identity sin²D + cos²D = 1).

The inverse of D is given by (1/det(D)) * adj(D). Here, det(D) = (4*3) - (2*1) = 10, and adj(D) = [[3, -2], [-1, 4]]. Thus, D^(-1) = (1/10) * [[3, -2], [-1, 4]].

The determinant of F is calculated as (1*4) - (2*2) = 4 - 4 = 0.

f(1) = 3(1) + 2 = 5. Since f(x) is a linear function, it is continuous everywhere.

The second derivative f''(x) = d^2/dx^2(e^x) = e^x.

f'(x) = e^x and f''(x) = e^x. Therefore, f''(0) = e^0 = 1.

lim x→-1 f(x) = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0.

f(1) = 1^3 - 3(1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

First, find f'(x) = 3x^2 - 6x. Then, f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.

Find f''(x) = 6x - 12. Set f''(x) = 0 gives x = 2. The inflection point is (2, f(2)) = (2, 0).

f'(x) = 4x^3 - 12x^2; thus, f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

Using the chain rule, h'(x) = 2e^(2x).

The rank of I is 1 because all rows are linearly dependent.

At 45 degrees East, it is 45/15 = 3 hours ahead. Therefore, it is 12:00 PM + 3 hours = 3:00 PM.

Each degree of longitude represents 4 minutes of time. 75°W = 75 * 4 = 300 minutes = 5 hours. Therefore, it is 12:00 noon - 5 hours = 7:00 AM.

New Delhi is 77°E, which is 77 * 4 = 308 minutes = 5 hours and 8 minutes ahead of GMT. Therefore, 3:00 PM - 5:08 = 9:30 AM.

New York is 5 hours behind London, so 3:00 PM + 5 hours = 8:00 PM in London.

log_2(x) + 2 = 6 implies log_2(x) = 4, so x = 2^4 = 16.

log_3(x) = 4 implies x = 3^4 = 81.

log_4(256) = log_4(4^4) = 4.

log_a(16) = 4 implies a^4 = 16. Since 16 = 2^4, we have a^4 = 2^4, thus a = 2.

log_b(25) = 2 implies b^2 = 25. Therefore, b = 5.

log_x(64) = 6 implies x^6 = 64. Since 64 = 2^6, we have x = 2.

The sum of the roots is 3 + (-1) = 2, hence p = -2.

tan A = sin A / cos A = (1/√2) / (1/√2) = 1.

Using the Pythagorean identity, cos C = √(1 - sin²C) = √(1 - 0.8²) = √(1 - 0.64) = √(0.36) = 0.6.

A · B = |A||B|cos(60°) = 5 * 10 * 0.5 = 25