6 - 3x ≤ 0 => -3x ≤ -6 => x ≥ 2.

7 - 3x > 1 => -3x > -6 => x < 2.

8x + 1 ≤ 5 => 8x ≤ 4 => x ≤ 0.5.

x + 2 > 3 => x > 1.

Solving gives sin(x) = -√3/2, so x = 7π/6 and 11π/6.

The solutions are x = π/6 and x = 5π/6.

The solutions are x = π/6 and x = 5π/6.

The sum of the roots is given by -b/a = 12/3 = 4.

Magnitude = √(3^2 + 4^2) = 5. Unit vector = (3/5, 4/5) = (0.6, 0.8).

Magnitude = √(3^2 + 4^2) = 5. Unit vector = (3/5, 4/5, 0) = (0.6, 0.8, 0).

Unit vector = (4, 3) / √(4^2 + 3^2) = (4, 3) / 5 = (4/5, 3/5).

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.