The first term a = 2 and the common difference d = 3. The 15th term = a + (15-1)d = 2 + 42 = 44.

The common difference d can be found using the formula for the nth term. The last term is given by a + (n-1)d. Here, 48 = 12 + (8-1)d, solving gives d = 4.

Using the formula for the last term: a + (n-1)d = last term, we have 8 + 5d = 32. Solving gives d = 4.

Let the number of items be n. Then, 15n - x = 12(n - 1). Solving gives x = 18.

Let the number of elements be n. Then, 20n - x = 18(n - 1). Solving gives x = 22.

The number 40 appears most frequently (4 times), so the mode is 40.

Selling Price = Marked Price - Discount = 200 - (15% of 200) = 200 - 30 = $170.

The selling price after a 20% discount on $200 is $200 - ($200 * 0.20) = $200 - $40 = $160.

Discount = 10% of 500 = 50. Selling Price = Marked Price - Discount = 500 - 50 = $450.

Let the marked price be x. Then, 300 = x - 25% of x => 300 = x - 0.25x => 300 = 0.75x => x = 400.

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 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.