Factoring gives (x + 4)(x - 2) = 0, hence the roots are 4 and -2.

f'(x) = 6x^2 - 6. f'(1) = 6(1)^2 - 6 = 0.

This is a separable equation. Integrating gives y = 1/(C - x).

This is a linear first-order equation. The integrating factor is e^(3x). The solution is y = Ce^(3x) + 2.

This is a separable equation. The solution is y = tan(x + C).

The term containing x^3 is C(5,3) * x^3 * (-1)^2 = 10 * x^3 * 1 = 10.

The term independent of x occurs when the powers of x cancel out. The term is 81.

The term independent of x occurs when the powers of x cancel out. The term is C(5,2)(-3)^2(1)^3 = 45.

Using the binomial theorem, (3 + 2)^3 = C(3,0) * 3^3 * 2^0 + C(3,1) * 3^2 * 2^1 + C(3,2) * 3^1 * 2^2 + C(3,3) * 3^0 * 2^3 = 27 + 54 + 36 + 8 = 125.

5! = 5 × 4 × 3 × 2 × 1 = 120.

9 × 9 = 81 and 5 × 5 = 25, so 81 - 25 = 56.

The discriminant must be negative: 4² - 4*1*k < 0, which gives k > 4, so the minimum value is -6.

For equal roots, the discriminant must be zero: k² - 4*1*16 = 0, thus k² = 64, k = ±8. The value of k can be -8.

For no real roots, the discriminant must be less than zero: k² - 4*1*16 < 0, which gives k > 8.

Setting x = 0 in the equation gives y = 8. Therefore, the y-intercept is 8.

The integral is (5/5)x^5 + C = x^5 + C.

The integral is (6/3)x^3 - 4x + C = 2x^3 - 4x + C.

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

For a monatomic ideal gas, γ = C_p/C_v = 5/3 = 1.67.

At constant volume, no work is done (W=0), and the heat added equals the change in internal energy (Q=ΔU).

Using t = 1 / (k[A₀]) * (1/[A] - 1/[A₀]), t = 1 / (0.02 * 0.5) * (1/0.25 - 1/0.5) = 25 s.

Using the Arrhenius equation, k = Ae^(-Ea/RT). Calculate k using the given values.

Increasing the temperature generally increases the rate of reaction significantly due to higher kinetic energy.

For a zero-order reaction, t = [A₀] / k. Here, t = 0.5 M / 0.1 M/s = 5 s.

For a zero-order reaction, t = [A₀ - A] / k. Here, t = (0.5 - 0.2) / 0.1 = 3 s.

In an isobaric process, the work done is calculated using the formula W = PΔV, where P is pressure and ΔV is the change in volume.

The Ideal Gas Law (PV = nRT) relates pressure (P), volume (V), and temperature (T) for an ideal gas.

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, which gives k ≤ 9, so the minimum value is -9.

The vertex x-coordinate is given by -b/(2a) = -4/(2*-1) = 2.

Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0. Critical points are x = 2.