Let the original price be x. After a 20% discount, the selling price is 80% of x. Thus, 0.8x = 30. Solving for x gives x = 30/0.8 = 37.5. Therefore, the original price is $36.

Let the original price be x. After a 25% discount, the selling price is x - 0.25x = 0.75x. Setting this equal to $30 gives 0.75x = 30, so x = 30/0.75 = $40.

Let the cost price be x. The selling price is given by x + 0.2x = 1.2x. Setting this equal to $30 gives us 1.2x = 30, so x = 30/1.2 = 25. Thus, the cost price of the shirt is $25.

Let the cost price be x. The selling price is 30, which is 120% of the cost price. Therefore, 1.2x = 30. Solving for x gives x = 30/1.2 = 25. Thus, the cost price is $25.

Let the original price be x. Then, x - 0.20x = 300. Therefore, 0.80x = 300, x = 300 / 0.80 = $400.

Let marked price = x. Then, 0.8x = 800 => x = 800/0.8 = 1000.

Let the original price be x. Then, x - 0.20x = 800 => 0.80x = 800 => x = 1000.

Let the marked price be x. After a 20% discount, selling price = x - 0.2x = 0.8x. So, 0.8x = 800, hence x = 1000.

Let CP = x. Then, 800 = x + 0.20x = 1.20x. Therefore, x = 800 / 1.20 = 666.67.

The time period (T) is the reciprocal of frequency (f). T = 1/f = 1/5 = 0.2 s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 0.5 kg. Thus, ω = √(200/0.5) = √400 = 20 rad/s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 2 kg. Thus, ω = √(200/2) = √100 = 10 rad/s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 50 N/m and m = 2 kg. Thus, ω = √(50/2) = √25 = 5 rad/s.

ω = V_max/A = 2/0.1 = 20 rad/s.

Maximum potential energy (PE) = (1/2)kA^2 = (1/2)(200)(0.1^2) = 1 J.

T = 2π√(m/k) = 2π√(2/200) = 1 s.

The total energy E is proportional to the square of the amplitude. If the amplitude is halved, the new energy will be E/4.

Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.

Total energy in SHM is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 4.

The maximum speed v_max of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.

The maximum speed (v_max) of a simple harmonic oscillator is given by v_max = Aω, where ω is the angular frequency.

The maximum displacement in simple harmonic motion is equal to the amplitude, which is 5 cm.

Time period (T) = Total time / Number of oscillations = 40 s / 20 = 2 s.

The time period T of a simple pendulum is given by T = 2π√(L/g). For L = 1 m and g = 9.8 m/s², T = 2π√(1/9.8) ≈ 2.0 s.

The period T of a simple pendulum is given by T = 2π√(L/g). Rearranging gives L = (T^2 * g) / (4π^2). Using g = 9.8 m/s² and T = 2 s, we find L = (2^2 * 9.8) / (4π^2) ≈ 1 m.

The insect can walk on water due to surface tension, which creates a 'skin' on the surface of the water.

The insect can walk on water because of surface tension, which creates a 'skin' on the surface that can support the weight of the insect.

The insect can walk on water due to surface tension, which creates a 'skin' on the surface that can support the weight of the insect.

40% of 150 calories = 0.40 * 150 = 60 calories from fats.

20% of 150 calories = 0.20 * 150 = 30 calories.