The mode is 10, which appears 3 times, more than any other number.

The general solution is y = Ce^(2x). Using the initial condition y(0) = 1 gives C = 1, so y = e^(2x).

Solving the equations simultaneously, we find the intersection point is (1, 1).

Solving the equations simultaneously, we find the intersection point is (3, 4).

A · B = (2)(1) + (3)(2) + (1)(3) = 2 + 6 + 3 = 11

A · B = (6)(2) + (8)(3) = 12 + 24 = 36.

A · B = (7)(3) + (-2)(4) + (1)(-5) = 21 - 8 - 5 = 8.

A · B = (7)(3) + (1)(4) + (2)(5) = 21 + 4 + 10 = 35.

First derivative f'(x) = 16x^3 - 6x^2 + 1. Second derivative f''(x) = 48x^2 - 12x.

First derivative f'(x) = 3x^2 - 6x; second derivative f''(x) = 6x - 6.

First derivative f'(x) = 3x^2 - 6x; Second derivative f''(x) = 6x - 6.

Magnitude |A| = √(6^2 + (-8)^2) = √(36 + 64) = 10. Unit vector = (6/10)i + (-8/10)j = (3/5)i - (4/5)j.

Magnitude |D| = √((-3)^2 + 4^2) = √(9 + 16) = √25 = 5. Unit vector = D/|D| = (-3/5)i + (4/5)j = -0.6i + 0.8j.

(1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i - 1 = 2i.

Using the binomial theorem, (1 + 2)^6 = 3^6 = 729.

Using the binomial theorem, (a + b)^4 = C(4, 0)a^4b^0 + C(4, 1)a^3b^1 + C(4, 2)a^2b^2 + C(4, 3)a^1b^3 + C(4, 4)a^0b^4. Substituting a = 2 and b = 3 gives 16 + 4*6 + 6*9 + 4*27 + 81 = 81.

The discriminant must be less than zero: k^2 - 4*1*16 < 0 leads to k < -8.

The coefficient of x^2 in (x + k)^4 is C(4, 2) * k^2 = 6. Thus, 6k^2 = 6, giving k^2 = 1, so k = 1 or -1.

The term containing x^4 is C(6,4) * (2)^2 * x^4 = 15 * 4 * x^4 = 60x^4. Setting 60 = 240 gives k = 4.

C(7, 4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7*6*5)/(3*2*1) = 35.

Using the binomial theorem, the coefficient of x^4 in (a + b)^n is given by nCk * a^(n-k) * b^k. Here, n=6, a=x, b=-2, and k=2. Thus, the coefficient is 6C2 * (1)^4 * (-2)^2 = 15 * 4 = 60.

∫(0 to 1) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 1 = (1/3 + 1) - (0) = 4/3.

∫(0 to 2) (x^2 + 1) dx = [x^3/3 + x] from 0 to 2 = (8/3 + 2) - (0) = 4.

∫(0 to π) sin(x) dx = [-cos(x)] from 0 to π = [1 - (-1)] = 2.

∫(1 to 3) (x^2 - 2x + 1) dx = [(x^3/3 - x^2 + x)] from 1 to 3 = (9/3 - 9 + 3) - (1/3 - 1 + 1) = 0.

∫(1 to 4) (x^3) dx = [x^4/4] from 1 to 4 = (256/4 - 1/4) = 63.

The integral ∫(2x + 1)dx = x^2 + x. Evaluating from 0 to 2 gives (2^2 + 2) - (0 + 0) = 4.

To find the minimum, set f'(x) = e^x + 2x = 0. The minimum occurs at x = 0.

Set f'(x) = e^x - 2x = 0. The solution is approximately x = 1.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting 6x - 12 = 0 gives x = 2.