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.

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

Let the angles be 2x, 3x, and 4x. Then, 2x + 3x + 4x = 180 degrees. Therefore, 9x = 180, x = 20. The largest angle is 4x = 80 degrees.

Area = 1/2 * base * height. Therefore, 60 = 1/2 * 10 * height, which gives height = 12 units.

Substituting x = 2: 2(2) + 3y = 12 => 4 + 3y = 12 => 3y = 8 => y = 8/3 ≈ 2.67, closest is 3.

Substituting x = 0: 3(0) + 4y = 12 => 4y = 12 => y = 3.

All values are the same, so variance = 0.

Area = (d1 * d2) / 2 = (8 * 6) / 2 = 48 cm²

Circumference = πd = π(12) = 12π cm.

The distance from the focus to the directrix is 4, so the equation is x^2 = 8y.

Area of square = side². If side is doubled, new area = (2 * side)² = 4 * side², so area quadruples.

Perimeter = 2 * (length + width) = 2 * (4 + 6) = 20 cm

A triangle with sides 3 cm, 4 cm, and 5 cm satisfies the Pythagorean theorem (3² + 4² = 5²), thus it is a right triangle.

Total sum = Mean * Number of elements = 50 * 10 = 500.