The range of the function y = tan^(-1)(x) is (-π/2, π/2).

The rows of A are linearly dependent, hence the rank is 2.

The rank of the matrix is 2, as the rows are linearly dependent.

The real part of 5 - 7i is 5.

The real part is 2 * cos(π/3) = 2 * 1/2 = 1.

The real part is Re(z) = 4 * cos(π/3) = 4 * 1/2 = 2.

The real part of z = 5 - 2i is 5.

The real part of z = 5 - 4i is 5.

The real part of z = 5 - 6i is 5.

The real part is Re(z) = 5cos(π/3) = 5 * 1/2 = 5/2.

Resultant = (2 + (-1), 3 + 4) = (1, 7).

Resultant = (1 - 1, 2 - 2) = (0, 0)

A · B = 3*0 + 4*0 + 0*5 = 0.

i · j = 0, since they are orthogonal.

The scalar product is 3*4 + 4*3 = 12 + 12 = 24.

Scalar product = 4*1 + (-3)*1 + 2*1 = 4 - 3 + 2 = 3.

Scalar product = 5*(-2) + (-3)*4 = -10 - 12 = -22.

Scalar product = 5*1 + 5*2 + 5*3 = 5 + 10 + 15 = 30.

A · B = 0*1 + 1*0 + 0*1 = 0.

A · B = 1*1 + 1*1 + 1*1 = 1 + 1 + 1 = 3.

A · B = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

A · B = 2*0 + (-1)*4 + 3*(-2) = 0 - 4 - 6 = -10.

A · B = 4*0 + 0*5 + (-3)*2 = 0 - 6 = -6.

K · L = 0*1 + 1*0 + 0*1 = 0 + 0 + 0 = 0.

Scalar projection = (3*1 + 4*0) / √(1^2) = 3

Scalar projection = (A · B) / |B| = (3*1 + 4*0) / 1 = 3.

Scalar triple product = (1, 2, 3) · ((4, 5, 6) × (7, 8, 9)) = 0.

Scalar triple product = A · (B × C) = 1.

Scalar triple product = a · (b × c). Since b and c are linearly dependent, b × c = 0, hence the scalar triple product is 0.

Using the product rule and chain rule, the second derivative is f''(x) = e^x (sin(x) + cos(x)).