The derivative f'(x) = 1/x.

The derivative of f(x) = tan(x) is f'(x) = sec^2(x).

In the equation y^2 = 4px, we have 4p = 20, thus p = 5. The focus is at (5, 0).

For the parabola y^2 = 4px, here 4p = 20, so p = 5. The focus is at (5, 0).

Integrating both sides gives y = ∫3x^2 dx = x^3 + C.

The indefinite integral of e^x is e^x + C.

The integral of cos(3x) is (1/3)sin(3x) + C, where C is the constant of integration.

The integral of e^(2x) is (1/2)e^(2x) + C, where C is the constant of integration.

The integrating factor is e^(∫3dx) = e^(3x).

The latus rectum of a parabola y^2 = 4px is given by 4p. Here, 4p = 12, so p = 3, and the latus rectum is 4p = 12.

Arc length = (θ/360) × 2πr = (60/360) × 2π(5) = (1/6) × 10π = 5π/6 cm.

In a right triangle, the altitude from the right angle to the hypotenuse is equal to the length of the other side. Thus, the altitude is 6 cm.

The length of the diagonal d of a rectangle can be found using the Pythagorean theorem: d = √(length² + width²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.

Diagonal = √(length² + width²) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm

Diagonal = √(length² + width²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

Area = πr². Given area = 50π, r² = 50, r = √50 = 5√2. Diameter = 2r = 10√2 units.

Using the median formula: m_a = 1/2 * sqrt(2b^2 + 2c^2 - a^2), where a = BC, b = AC, c = AB. m_a = 1/2 * sqrt(2*6^2 + 2*10^2 - 8^2) = 7 cm.

Using the identity 1 - cos(x) = 2sin^2(x/2), we have lim (x -> 0) (1 - cos(x))/(x^2) = lim (x -> 0) (2sin^2(x/2))/(x^2) = 1/2.

Using the Taylor series expansion for cos(x), we find that lim (x -> 0) (cos(x) - 1)/x^2 = -1/2.

Using the derivative of e^x at x = 0, we find that lim (x -> 0) (e^x - 1)/x = 1.

Using L'Hôpital's Rule, we differentiate the numerator and denominator to find lim (x -> 0) (1/(1 + x)) = 1.

Factoring gives (x - 1)(x + 1)/(x - 1), which simplifies to x + 1. Thus, lim (x -> 1) (x + 1) = 2.

Magnitude |C| = √(5^2 + (-12)^2) = √(25 + 144) = √169 = 13.

The maximum occurs at x = -b/(2a) = -10/(-4) = 2. f(2) = -2(2^2) + 10(2) - 12 = 6.

The vertex is at x = -12/(2*(-3)) = 2. The maximum value is f(2) = -3(2^2) + 12(2) - 5 = 7.

The vertex occurs at x = -b/(2a) = 2. f(2) = -2(2^2) + 8(2) - 5 = 9.

Mean = (4 + 8 + 6 + 5 + 3) / 5 = 26 / 5 = 5.2, rounded down to 5.

Supplementary angles sum up to 180 degrees. Therefore, the angle measures 180 - 110 = 70 degrees.

The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Therefore, the exterior angle = 50 + 60 = 110 degrees.

Complementary angles sum up to 90 degrees. If one angle is 30 degrees, the other angle is 90 - 30 = 60 degrees.