A · B = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32

The number of ways is 6C2 * 4C3 = 15 * 4 = 60.

The number of ways to choose 4 students from 10 is given by 10C4 = 210.

Using the identity a^2 + b^2 = (a + b)^2 - 2ab, we have 70 = 12^2 - 2ab. Thus, 70 = 144 - 2ab, leading to ab = 37.

(a + b)² = (2 + 3)² = 5² = 25.

a^2 = 4 and b^2 = 9, so 4 + 9 = 13.

A · B = (2)(4) + (3)(0) = 8 + 0 = 8

a² + b² = 4² + 5² = 16 + 25 = 41.

A · B = (4)(0) + (3)(5) = 0 + 15 = 15

A · B = (5)(-3) + (-2)(4) = -15 - 8 = -23

A · B = (7)(2) + (1)(3) = 14 + 3 = 17

Area = πr². If radius is doubled (2r), new area = π(2r)² = 4πr², which is quadruple the original area.

The order of the product of two matrices is determined by the outer dimensions. Here, A (2x2) and B (2x3) can be multiplied, resulting in a matrix of order 2x3.

A 3x3 matrix has 3 rows and 3 columns, resulting in a total of 3 * 3 = 9 elements.

'c' represents the y-intercept of the line, which is the point where the line crosses the y-axis.

Setting x = 0 in the equation gives y = 8. Thus, the y-intercept is -8.

The y-intercept is the constant term in the equation, which is 6.

In a diagonal matrix, all non-diagonal elements are zero.

The standard form of a parabola that opens to the left is y^2 = -4px.

The perimeter P of a square is given by P = 4 * side. If P = 32 cm, then side = 32/4 = 8 cm.

When a transversal intersects two parallel lines, the same-side interior angles are supplementary, meaning they add up to 180 degrees.

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

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

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

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

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

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

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

The second derivative f''(x) = d^2/dx^2(e^x) = e^x.