Setting p = 1 for continuity at x = 0 gives f(0) = 1.

Setting the left limit (1 - 1 = 0) equal to the right limit (2(1) + 1 = 3), we find p = 2.

Setting 1 - 3 + p = 2 + 1 gives p = 4.

Setting -3 + p = 3 gives p = 0.

Setting the determinant to zero and solving gives \( k = 10 \).

The solutions are x = π/4 and x = 3π/4.

The solutions are x = 0, π/2, π, 3π/2.

The solutions are x = nπ/2, where n is any integer.

Factoring gives sin(x)(sin(x) - 1) = 0, so x = 0, π/2, π, 3π/2.

Setting y = 0 in the equation gives 4x + 8 = 0, thus x = -2.

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.

During a phase change, the temperature of a substance remains constant while the substance absorbs or releases heat.

In an isochoric process, the volume of the gas remains constant.

In an isochoric process, the volume of the system remains constant.

In an isothermal process for an ideal gas, the internal energy remains constant because the temperature does not change.

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.