log_5(125) = log_5(5^3) = 3.

log_5(25) = 2 and log_5(5) = 1. Therefore, 2 - 1 = 1.

sec(sin^(-1)(1/2)) = 1/cos(π/6) = 2.

sec(sin^(-1)(3/5)) = √(34)/3

Using the triangle with opposite = 1 and adjacent = √3, hypotenuse = 2. Thus, sec(tan^(-1)(1/√3)) = 2.

sin(tan^(-1)(x)) = x/√(1+x^2)

sin^(-1)(1/2) = π/6 and cos^(-1)(1/2) = π/3. Therefore, sin^(-1)(1/2) + cos^(-1)(1/2) = π/6 + π/3 = π/2.

sin^(-1)(1/2) = π/6 and sin^(-1)(√3/2) = π/3. Therefore, π/6 + π/3 = π/2.

sin^(-1)(1/2) = π/6, since sin(π/6) = 1/2.

tan^(-1)(1) = π/4, thus tan^(-1)(1) + tan^(-1)(1) = π/4 + π/4 = π/2.

Using the formula tan^(-1)(a) + tan^(-1)(b) = tan^(-1)((a+b)/(1-ab)), we have tan^(-1)(1) + tan^(-1)(2) = tan^(-1)((1+2)/(1-1*2)) = tan^(-1)(3/-1) = π - tan^(-1)(3) = π/4.

tan^(-1)(1) = π/4, since tan(π/4) = 1.

The 5th term is given by C(6,4) * (x)^4 * (2)^2 = 15 * x^4 * 4 = 240.

The coefficient of x^3 is C(6,3) * (-1)^3 = 20 * (-1) = -20.

The coefficient of x^5 is C(7,5) * (3)^2 = 21 * 9 = 189.

The coefficient of x^5 is C(8, 5) * (3)^3 = 56 * 27 = 1512.

The determinant of matrix A is 0 because the rows are linearly dependent.

The determinant of the matrix is 0 because the rows are linearly dependent.

The determinant is 0 because the first column is repeated.

The determinant is \( 1*4 - 2*3 = 4 - 6 = -2 \).

Calculating the determinant gives -12.

det = (2)(5) - (3)(4) = 10 - 12 = -2.

Subtract 3 from both sides: 2x = 8. Then divide by 2: x = 4.

Solving for x: 3x = 21 => x = 7.

Using the quadratic formula, x = [8 ± sqrt(64 - 48)] / 4 = [8 ± 4] / 4, giving x = 3 or x = 1.

Solving for x: 3x = 12, thus x = 4.

Expanding gives 5x - 10 = 3x + 4. Rearranging gives 2x = 14, thus x = 7.

Rearranging gives 5x - 2x = 12 + 3 => 3x = 15 => x = 5.

Rearranging gives 5x - 2x = 6 + 3 => 3x = 9 => x = 3.

The solutions to z^2 = -1 are z = i and z = -i.