The term containing x^3 is C(6,3) * (2)^3 * (5)^(6-3) = 20 * 8 * 125 = 20000.

The term containing x^3 is C(7,3) * (2)^4 = 35 * 16 = 560.

The term independent of x is C(5,5) * (2x)^0 * (-3)^5 = 1 * 1 * (-243) = -243.

The term independent of x occurs when the powers of x cancel out. The term is C(3,0)(4)^3 + C(3,1)(-3x)(4)^2 + C(3,2)(-3x)^2(4) + C(3,3)(-3x)^3 = 64 - 36 + 0 + 0 = 28.

The term independent of x occurs when the powers of x cancel out. This can be calculated using the binomial expansion.

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

The term independent of x occurs when k = 3, which gives C(6,3) * (x/2)^3 * (-3)^3 = 20 * (1/8) * (-27) = -67.5.

The term independent of x occurs when k = 4. Thus, the term is C(8,4) * (x/2)^4 * (-3)^4 = 70 * (1/16) * 81 = 256.

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.

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

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

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = -1, 1. f'(x) > 0 for x > 1.

f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). f'(x) > 0 for x in (0, 3).

f'(x) = 4x^3 - 12x^2. Setting f'(x) = 0 gives x = 0, 3. Testing intervals shows local minima at (0, 2).

Using the standard limit lim (x -> 0) (tan(kx)/x) = k, we have k = 5. Thus, lim (x -> 0) (tan(5x)/x) = 5.

Factoring gives (x - 1)(x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^2 + x + 1. Evaluating at x = 1 gives 3.

Factoring gives (x - 1)(x^3 + x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Evaluating at x = 1 gives 4.

f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x = 0 or x = 2. f(2) = 5 is a local maximum.

Find f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 12 is a local maximum.