Magnitude = √(6^2 + 8^2) = √(36 + 64) = √100 = 10. Unit vector = (6/10, 8/10) = (0.6, 0.8).

Magnitude |v| = √(4^2 + (-3)^2) = √(16 + 9) = 5. Unit vector = (4/5, -3/5).

Using the binomial theorem, (1 + 2)^4 = C(4,0) * 1^4 * 2^0 + C(4,1) * 1^3 * 2^1 + C(4,2) * 1^2 * 2^2 + C(4,3) * 1^1 * 2^3 + C(4,4) * 1^0 * 2^4 = 1 + 8 + 24 + 32 + 16 = 81.

(1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i.

(1 + i)^4 = (√2 e^(iπ/4))^4 = 4 e^(iπ) = 4(-1) = -4.

Using the binomial theorem, (1 + 1)^10 = 2^10 = 1024.

Using the binomial theorem, (1 + 2)^10 = 3^10 = 59049.

Setting ax + 1 = 2 and x^2 + a = 2 at x = 1 gives a = 0.

Setting ax + 1 = 3 and 2x + a = 3 at x = 1 gives a = 2.

Setting the two pieces equal at x = 2 gives us 2a + 1 = 1. Solving for a gives a = 0.

Setting the two pieces equal at x = 2: 2a + 1 = 2^2 - 3. Solving gives a = 2.

Set the left-hand limit equal to the right-hand limit and their derivatives at x = 2.

Setting the left limit (1 + a) equal to the right limit (3), we find a = 2.

Setting 1 + b = 2 + 3 gives b = 4.

Setting 1 + b = 2 gives b = 1.

Setting the left-hand limit equal to the right-hand limit gives c = 1.

Setting the two pieces equal at x = 2 gives 4 = 4 + c, hence c = 0.

To ensure continuity at x = 1, we set the left limit (1 - 3 + 2 = 0) equal to the right limit (1 + 1 = 2), leading to c = 2.

Setting limit as x approaches c from left equal to 4 and from right gives c = 1.

cos(60°) = 1/2.

cos(tan^(-1)(1)) = 1/√2

cos(tan^(-1)(3)) = 3/√10

Using the identity, cos(tan^(-1)(3/4)) = 4/5

cos^(-1)(-1/2) = 2π/3, since cos(2π/3) = -1/2.

cos^(-1)(0) = π/2, since cos(π/2) = 0.

i^4 = (i^2)^2 = (-1)^2 = 1.

For no real roots, the discriminant must be less than 0: k^2 - 4*1*16 < 0, which gives k < 8.

For both roots to be negative, k must be positive and k^2 > 36, thus k > 6.

f'(x) = 2kx + 2. At x = 0, f'(0) = 2. The function is differentiable for any k, but k = 0 gives a constant function.

The function is a polynomial and is differentiable for all k, hence k can be any real number.