The term independent of x occurs when the powers of x cancel out. The term is C(4,2) * (2)^2 * (-3)^2 = 6 * 4 * 9 = 216.

(1 + 3)^5 = 4^5 = 1024.

25 × 3 = 75, then 100 - 75 = 25.

12 × 3 = 36 and 4 × 2 = 8, so 36 - 8 = 28.

12 × 3 = 36, then 36 - 4 = 32.

12 × 5 = 60, then 60 - 8 = 52.

12 × 8 = 96, then 96 - 10 = 86.

12 × 8 = 96, then 96 - 24 = 72.

12 × 8 = 96 and 4 × 6 = 24, so 96 - 24 = 72.

25% of 200 = (25/100) × 200 = 50.

5! = 5 × 4 × 3 × 2 × 1 = 120.

5! = 120 and 4! = 24, so 120 - 24 = 96.

6^2 = 36 and 4^2 = 16, so 36 - 16 = 20.

According to BODMAS, 2 × 5 = 10, then 8 + 10 = 18.

2 + 3 = 5, then 8 × 5 = 40.

8^2 = 64 and 4^2 = 16, so 64 - 16 = 48.

Weight = mass * gravity = 10 kg * 3.7 m/s² = 37 N

Weight W = m * g = 10 kg * 9.8 m/s² = 98 N.

The slopes are m1 = 2 and m2 = -1/2. The angle θ = tan^(-1) |(m1 - m2)/(1 + m1*m2)| = tan^(-1)(5/4) which is approximately 60 degrees.

The coefficient of x^4 is given by 6C4 * (2)^4 * (-3)^2 = 15 * 16 * 9 = 2160.

Centroid = ((0+6+0)/3, (0+0+8)/3, (0+0+0)/3) = (2, 2.67, 0).

Centroid G = ((0+4+0)/3, (0+0+3)/3, (0+0+0)/3) = (4/3, 1, 0).

Centroid = ((0+0+3)/3, (0+4+0)/3, (0+0+0)/3) = (1, 1.33, 0).

Centroid G = ((1+4+7)/3, (2+5+8)/3, (3+6+9)/3) = (4, 5, 6).

Using the formula for the foot of the perpendicular, we find the coordinates to be (1, 2, 4).

f'(x) = 12x^3 - 24x^2. Setting f'(x) = 0 gives x = 0, 2. Check f(1) = 1.

f'(x) = e^x - 2. Setting f'(x) = 0 gives e^x = 2, so x = ln(2).

Using the distance formula, d = √((5 - 2)² + (7 - 3)²) = √(9 + 16) = √25 = 5.

Using the formula for distance from a point to a line, d = |Ax1 + By1 + C| / sqrt(A^2 + B^2), we find d = |2(3) + 3(4) - 12| / sqrt(2^2 + 3^2) = 3.

f''(x) = -2, which is always negative, indicating concave down everywhere.