Setting y = 0 in the equation gives 4x + 20 = 0, thus x = -5.

Setting y = 0 in the equation gives 5x - 10 = 0, thus x = 2.

Set y = 0 in the equation: 2x + 6 = 0 => x = -3.

Let θ = tan^(-1)(1). Then, cos(θ) = 1/√(1 + tan^2(θ)) = 1/√(1 + 1) = 1/√2.

Using the triangle with opposite = 3 and adjacent = 4, hypotenuse = 5. Thus, cos(tan^(-1)(3/4)) = 4/5.

Using the right triangle definition, cos(tan^(-1)(5/12)) = adjacent/hypotenuse = 12/13.

cos^(-1)(0) = π, since cos(π) = 0.

sin(cos^(-1)(1/2)) = √(1 - (1/2)^2) = √(3/4) = √3/2.

Using the right triangle definition, sin(tan^(-1)(3/4)) = opposite/hypotenuse = 3/5.

Using the identity, sin(tan^(-1)(x)) = x/√(1+x^2).

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

Since 5π/6 is in the range of sin^(-1), sin^(-1)(sin(5π/6)) = 5π/6.

sin^(-1)(sin(π/3)) = π/3, since π/3 is in the range of sin^(-1).

sin^(-1)(sin(π/4)) = π/4

sin^(-1)(√3/2) + cos^(-1)(1/2) = π/2

If sin(x) = 1/√2, then x = π/4, thus tan(sin^(-1)(1/√2)) = tan(π/4) = 1.

Let θ = sin^(-1)(3/5). Then sin(θ) = 3/5 and using the Pythagorean theorem, cos(θ) = 4/5. Therefore, tan(θ) = sin(θ)/cos(θ) = (3/5)/(4/5) = 3/4.

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

tan^(-1)(1) = π/4 and tan^(-1)(√3) = π/3. Therefore, π/4 + π/3 = π/2.

tan^(-1)(√3) = π/3, since tan(π/3) = √3.

The integral evaluates to 1.

The integral evaluates to 6.

f'(x) = e^x + 1/x. At x = 1, f'(1) = e + 1.

The determinant is calculated as \( 2(0*1 - 2*2) - 1(1*1 - 2*3) + 3(1*2 - 0*3) = 2(0 - 4) - 1(1 - 6) + 3(2) = -8 + 5 + 6 = 3 \).

The determinant of the identity matrix is 1.

The determinant is 0 because the first column is repeated.

The determinant is 0 because the rows are linearly dependent.

The determinant evaluates to -12.

The determinant is 0 because the rows are linearly dependent.

The determinant is 0 because the rows are linearly dependent.