Power (P) is given by P = I^2 * R. Thus, P = (4 A)^2 * 2 Ω = 16 W.

The possible blood types from AO x BO are A, B, AB, and O.

The rank of matrix B is the number of non-zero rows in its row echelon form, which is 3.

The determinant of B is calculated as (2*7) - (3*5) = 14 - 15 = -1.

The determinant of C is 0 because it has a row of zeros.

Using the determinant formula for 3x3 matrices, det(C) = 1(3*0 - 1*1) - 0 + 2(-1*1 - 3*2) = 0 - 0 - 12 = -12.

The determinant of an upper triangular matrix is the product of its diagonal elements. Here, det(C) = 1 * 1 * 1 = 1.

C^2 = C * C = [[1*1 + 2*3, 1*2 + 2*5], [3*1 + 5*3, 3*2 + 5*5]] = [[11, 16], [18, 35]].

Using the identity tan A = sin A / cos A, we find sin A = √(1 - cos²A) = √(1 - (5/13)²) = 12/13. Thus, tan A = (12/13) / (5/13) = 12/5.

Using the Pythagorean identity, sin B = √(1 - cos²B) = √(1 - 0.6²) = √(1 - 0.36) = √(0.64) = 0.8.

Cosine of 60° is 1/2, hence angle B = 60°.

Using the Pythagorean identity, sin B = √(1 - cos²B) = √(1 - (5/13)²) = √(1 - 25/169) = √(144/169) = 12/13.

The angle C for which cos C = 0.5 is 60°.

Cotangent is 1 when the angle A is 45°, as cot A = cos A/sin A = 1 when both are equal.

cot A = cos A/sin A = 3/4 implies sin A = 4/5 using the identity sin²A + cos²A = 1.

cot D = 3/4 implies tan D = 4/3. Therefore, sin D = 4/5 (using the identity sin²D + cos²D = 1).

cot θ = 1 implies tan θ = 1, which occurs at θ = 45°.

The trace of a matrix is the sum of its diagonal elements. Trace(D) = 2 + 2 = 4.

The inverse of D is given by (1/det(D)) * adj(D). Here, det(D) = (4*3) - (2*1) = 10, and adj(D) = [[3, -2], [-1, 4]]. Thus, D^(-1) = (1/10) * [[3, -2], [-1, 4]].

The inverse of D is given by (1/det(D)) * adj(D). Here, det(D) = (4*3) - (2*1) = 10. The adjugate is [[3, -2], [-1, 4]]. Thus, D^(-1) = (1/10) * [[3, -2], [-1, 4]].

Matrix E is singular because its determinant is 0 (1*4 - 2*2 = 0).

The inverse of the identity matrix F is itself, so F^(-1) = F.

The determinant of F is calculated as (1*4) - (2*2) = 4 - 4 = 0.

The second row is a multiple of the first row, so there is only one linearly independent row. Therefore, the rank of F is 1.

The trace of F is the sum of the diagonal elements: 1 + 5 = 6.

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.

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.