The subsets containing exactly 2 elements are {x, y}, {x, z}, and {y, z}. So, there are 3 such subsets.

The number of subsets of a set with n elements is 2^n. Here, n=2, so the number of subsets is 2^2 = 4.

g(-1) = 3(-1) + 2 = -3 + 2 = -1.

The number of ways to choose 2 elements from 3 is given by the combination formula C(3, 2) = 3.

The total number of subsets is 2^3 = 8. Subtracting the empty set gives 8 - 1 = 7.

The power set of H is {∅, {x}, {y}, {x, y}}. The subsets of H are {∅, {x}, {y}}.

h'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so x = 1 and x = -1 are critical points.

h(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.

h(1) = 1^3 - 3*1 = 1 - 3 = -2.

I · J = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

I · J = 1*3 + 2*2 + 3*1 = 3 + 4 + 3 = 10.

I · J = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

I · J = 2a + 2b + 2c = 12 gives the equation 2a + 2b + 2c = 12.

If 'a' is excluded, we can form subsets from {b, c}, which has 2^2 = 4 subsets.

Using the grating equation d sin(θ) = mλ, where d = 1/1000 mm = 1 x 10^-6 m, for m=1, we find θ ≈ 30 degrees.

log(2x) = 3 => 2x = 10^3 => x = 1000.

Since 1000 = 10^3, log_10(1000) = 3, thus x = 3.

log_10(20) = log_10(2 * 10) = log_10(2) + log_10(10) = a + 1.

log_10(x) = 2 implies x = 10^2 = 100.

log_2((x + 1)/x) = 1 implies (x + 1)/x = 2 => x + 1 = 2x => x = 1.

log_2(x + 1) = 3 implies x + 1 = 2^3 = 8 => x = 7.

log_2(x) + 2 = 5 => log_2(x) = 3 => x = 2^3 = 8.

log_2(x(x - 3)) = 3 => x(x - 3) = 2^3 = 8 => x^2 - 3x - 8 = 0. Solving gives x = 6.

log_2(x(x-1)) = 3 => x(x-1) = 2^3 = 8 => x^2 - x - 8 = 0. Solving gives x = 5.

log_2(x) = 5 implies x = 2^5 = 32.

log_3(9) = 2 and log_3(27) = 3, thus x = 2 + 3 = 5.

log_3(9) = log_3(3^2) = 2.

log_3(x + 1) = 2 implies x + 1 = 3^2 = 9 => x = 8.

log_3(4x) = 2 => 4x = 3^2 = 9 => x = 9/4 = 2.25.

log_3(x) = 2 implies x = 3^2 = 9.