tan(45°) = 1, hence x = 45°.

Using the identity tan(x) = sin(x)/cos(x), we can find sin(x) = 3/5 after applying the Pythagorean theorem.

Using the identity tan(x) = sin(x)/cos(x), we can find sin(x) = 3/5 after calculating the hypotenuse using the Pythagorean theorem.

tan(θ) = 1 at θ = 45°.

tan(45°) = 1, so θ = 45°.

Using the identity sin²(θ) + cos²(θ) = 1, we find sin(θ) = 3/5.

tan^(-1)(x) = π/4 implies that x = tan(π/4) = 1.

D can only start after both A and C are completed, making the inference correct.

E can only start after both A and C are completed, making the statement true.

The sequence F, G, H respects the order that F is before G and G is before H.

Since M must be completed before N, and N before O, M must indeed be completed before O.

Task X can only start after both tasks Y and Z are completed, making this statement true.

The sequence B, A, C is invalid as it does not respect the order of A before B.

The sequence A, B, C follows the required order.

The sequence L, M, N is correct as it respects the order of L before both M and N.

If Team A wins 60% of its matches, then in 9 matches, it wins 0.6 * 9 = 5.4, which rounds to 6 matches.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 8 and 4th term = ar^3 = 32. Dividing gives r^2 = 4, so r = 2.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 4d = 14, we can find the 3rd term a + 2d = 10.

Let the first term be a. Then, the 3rd term is ar^2 = 27. Thus, a * 3^2 = 27, giving a = 3.

Let the first term be a and the common difference be d. From the equations a + 4d = 15 and a + 9d = 30, we can find d = 3.

Let the first term be a and the common difference be d. From the equations a + 5d = 30 and a + 8d = 45, we can find d = 5.

Using the formula for the nth term, a + 6d = 50. Substituting d = 5 gives a + 30 = 50, hence a = 20.

According to the Arrhenius equation, an increase in activation energy results in a decrease in the rate constant k.

Using the formula A(t) = A_0 e^(-ζω_nt), we find ζ = 0.2.

Maximum velocity V_max = Aω. If A is doubled, V_max also doubles.

The total energy E in SHM is given by E = (1/2)kA². If A is doubled, E becomes (1/2)k(2A)² = 4(1/2)kA², which is quadrupled.

Total energy (E) in SHM is proportional to the square of the amplitude (A). If A is doubled, E becomes 4A^2, hence it quadruples.

Maximum velocity V_max = ωA. If A is halved, V_max is also halved.

The total energy in simple harmonic motion is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 2^2 = 4.

The total energy of a simple harmonic oscillator is proportional to the square of the amplitude. If the amplitude is doubled, the energy increases by a factor of 2^2 = 4.