5! = 5 × 4 × 3 × 2 × 1 = 120.

5^3 = 5 × 5 × 5 = 125.

9 × 9 = 81 and 3 × 3 = 9, so 81 - 9 = 72.

9 × 9 = 81 and 5 × 5 = 25, so 81 - 25 = 56.

9 × 9 = 81 and 5 × 5 = 25, so 81 - 25 = 56.

9 × 9 = 81, then 81 - 7 = 74.

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.

The discriminant must be negative: 4² - 4*1*k < 0, which gives k > 4, so the minimum value is -6.

For equal roots, the discriminant must be zero: k² - 4*1*16 = 0, thus k² = 64, k = ±8. The value of k can be -8.

For no real roots, the discriminant must be negative: k² - 4*1*9 < 0, thus k² < 36, hence k < -6 or k > 6.