The determinant of a 1x1 matrix is simply the value of the single element. Therefore, the determinant of [[5]] is 5.

The determinant of a 2x2 matrix is calculated as ad - bc.

The determinant of a 2x2 matrix is calculated as ad - bc.

The determinant of E is 0 because the rows are linearly dependent.

The determinant can be calculated using the formula for 3x3 matrices. Here, the first column is the same, leading to a determinant of 0.

Using the distance formula: d = √[(3 - 0)² + (4 - 0)²] = √[9 + 16] = √25 = 5.

Using the distance formula: d = √((8 - 0)² + (6 - 0)²) = √(64 + 36) = √100 = 10.

Using the distance formula: d = √[(3 - 0)² + (4 - 0)²] = √[9 + 16] = √25 = 5.

Using the distance formula: d = √[(3 - 3)² + (1 - 7)²] = √[0 + 36] = √36 = 6.

Using the distance formula: d = √[(1 - 5)² + (1 - 5)²] = √[16 + 16] = √32 = 4√2.

The standard form of a circle's equation is (x-h)² + (y-k)² = r². Here, h=2, k=-3, r=4. Thus, (x-2)² + (y+3)² = 16.

The standard form of a circle's equation is (x-h)² + (y-k)² = r². Here, h=3, k=-2, r=5, so (x-3)² + (y+2)² = 25.

Standard form of a circle: (x-h)² + (y-k)² = r². Here, h=2, k=-3, r=5.

Parallel lines have the same slope. Using point-slope form: y - 2 = 3(x - 1) gives y = 3x - 1.

Since parallel lines have the same slope, the equation is y - 1 = 3(x - 2) which simplifies to y = 3x - 5.

Using the slope-intercept form y = mx + b, with m = -4 and b = 0, the equation is y = -4x.

The equation of a line through the origin with slope m is y = mx. Thus, y = -3x.

The derivative f'(x) = 1/x.

In the equation y^2 = 4px, we have 4p = 20, thus p = 5. The focus is at (5, 0).

Integrating both sides gives y = ∫3x^2 dx = x^3 + C.

The integral of cos(3x) is (1/3)sin(3x) + C, where C is the constant of integration.

The integral of e^(2x) is (1/2)e^(2x) + C, where C is the constant of integration.

The integrating factor is e^(∫3dx) = e^(3x).

The latus rectum of a parabola y^2 = 4px is given by 4p. Here, 4p = 12, so p = 3, and the latus rectum is 4p = 12.

Arc length = (θ/360) × 2πr = (60/360) × 2π(5) = (1/6) × 10π = 5π/6 cm.

In a right triangle, the altitude from the right angle to the hypotenuse is equal to the length of the other side. Thus, the altitude is 6 cm.

The length of the diagonal d of a rectangle can be found using the Pythagorean theorem: d = √(length² + width²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.

Diagonal = √(length² + width²) = √(3² + 4²) = √(9 + 16) = √25 = 5 cm

Diagonal = √(length² + width²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

Area = πr². Given area = 50π, r² = 50, r = √50 = 5√2. Diameter = 2r = 10√2 units.