The determinant of F is calculated as (2*3) - (0*0) = 6.

Det(F) = 2(0*1 - 2*1) - 1(1*1 - 2*0) + 3(1*1 - 0*0) = 2(0 - 2) - 1(1) + 3(1) = -4 - 1 + 3 = -2.

Using the determinant formula, det(F) = 2(0*1 - 2*1) - 1(1*1 - 2*3) + 3(1*1 - 0*3) = 2(0 - 2) - 1(1 - 6) + 3(1 - 0) = -4 + 5 + 3 = 4.

Using the determinant formula, det(F) = 2*(0*1 - 2*4) - 1*(1*1 - 2*3) + 3*(1*4 - 0*3) = 2*(-8) - 1*(-5) + 3*4 = -16 + 5 + 12 = 1.

The determinant of F is (2*3) - (1*1) = 6 - 1 = 5.

If 1 is included, we can choose from the remaining elements {2, 3, 4}, which has 2^3 = 8 subsets.

The total number of subsets of a set with n elements is 2^n. For F, n = 4, so there are 2^4 = 16 subsets. Proper subsets are total subsets minus the set itself, so 16 - 1 = 15.

The power set of a set with n elements has 2^n elements. For F, n = 3, so the power set has 2^3 = 8 elements.

The subsets containing exactly 2 elements are {a, b}, {a, c}, and {b, c}. So, there are 3 such subsets.

The even numbers between 1 and 10 are 2, 4, 6, and 8. Therefore, F = {2, 4, 6, 8}.

The size of the power set is 2^n where n is the number of elements in the set. Here, n = 3, so the size is 2^3 = 8.

The subsets of F that do not contain 'y' are {∅, {x}}, totaling 2 subsets.

Since F is to the left of G and H is to the right of F, it follows that F must be to the left of H.

The function is discontinuous at x = 1 because it leads to division by zero.

Substituting x = 4 into the function gives f(4) = 2(4) + 3 = 8 + 3 = 11.

f(5) = 2(5) + 3 = 10 + 3 = 13.

f(-1) = 2(-1)^2 + 3(-1) - 5 = 2 - 3 - 5 = -6.

f(2) = 2(2^2) - 3(2) + 1 = 8 - 6 + 1 = 3.

Find f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. Testing intervals, f(x) is increasing on (1, 3).

f(3) = 2^3 = 8.

f(1) = 3(1) + 2 = 5. Since f(x) is a linear function, it is continuous everywhere.

f(2) = 3(2) + 2 = 8. Since f(x) is a polynomial, it is continuous everywhere.

f(-2) = 3(-2) - 4 = -6 - 4 = -10.

f(-2) = 3(-2) - 4 = -6 - 4 = -10.

To find the inverse, set y = 3x - 4, solve for x: x = (y + 4)/3, thus f^(-1)(x) = (x + 4)/3.

First, find f(1) = 3(1) - 5 = -2. Then, f(-2) = 3(-2) - 5 = -6 - 5 = -11.

Substituting x = 1 into the function gives f(1) = 3(1)^2 + 2(1) - 5 = 3 + 2 - 5 = 0.

First, find f'(x) = 6x + 2. Then, f'(2) = 6(2) + 2 = 12 + 2 = 14.

First, find f'(x) = 12x^2 - 4x + 1, then differentiate again to get f''(x) = 24x - 4.

The first derivative f'(x) = 10x + 3, and the second derivative f''(x) = 10.