answer:To solve for the arithmetic square root of sqrt{16}, we follow these steps: 1. First, we find the square root of 16: [ sqrt{16} = 4 ] 2. Next, we find the arithmetic square root of the result, which is 4: [ sqrt{4} = 2 ] Therefore, the arithmetic square root of sqrt{16} is boxed{2}.

answer:For arccos(x^3) to be defined, x^3 must lie in the interval [-1, 1]. Since x^3 is an odd function, the interval for x is [-1, 1]. Next, we need to determine where arccos(x^3) = frac{pi}{2}, which occurs when x^3 = 0. This happens when x = 0. Therefore, g(x) is defined for all x in [-1,1] except at x = 0. Thus, the domain of g(x) is boxed{[-1,0) cup (0,1]}.

answer:Let the radius of the circle be r. Since the chord's length is also r, we can construct an equilateral triangle using the chord and two radii. An interior angle of an equilateral triangle is 60^{circ} (or frac{pi}{3} radians). Here is the step-by-step solution: 1. Let the radius of the circle be r. 2. As the chord's length is r, we can construct an equilateral triangle using the chord and two radii. 3. Each interior angle of an equilateral triangle is 60^{circ} (or frac{pi}{3} radians). 4. Therefore, the central angle corresponding to the chord is boxed{frac{pi}{3}} radians. Note: When dealing with chord length and radius, we usually use the arc length formula l = ralpha, but in this case, simply focus on the construction of an equilateral triangle. Remember that the central angle is measured in radians.

answer:: Given that a_1, a_2, cdots, a_k is a sequence of positive integers not exceeding n, and every pair of consecutive terms are distinct. Furthermore, there does not exist any four indices p < q < r < s such that a_p = a_r neq a_q = a_s. To find the maximum value of k, we analyze the following sequence: [ n, n, n-1, n-1, cdots, 2, 2, 1, 1, 2, 2, cdots, n-1, n-1, n, n. ] - This sequence satisfies all given conditions. Specifically, every term is repeated exactly twice, and there are no four indices as described in the problem. 1. **Identify the Length of the Sequence:** The sequence consists of each number from 1 to n appearing exactly twice. Hence, the number of terms k in this sequence is: [ k = 2n + 2(n-1) ] Simplifying: [ k = 4n - 2. ] Therefore, the length of this sequence is 4n - 2, indicating that k cannot be less than 4n - 2 if we want to maintain the given conditions. 2. **Maximizing k:** We need to consider a sequence where every pair of consecutive terms are distinct. Clearly, we start by trying to maximize the number of terms while maintaining the condition that no four indices p, q, r, s exist with a_p = a_r neq a_q = a_s. For a sequence with the required properties, every number from 1 to n participates twice as noted before. Let us furthermore assume such a configuration is optimal. 3. **Mathematical Induction for Verification:** - Base Case: When n = 2, the sequence 2, 2, 1, 1 clearly satisfies the condition. The length k = 4 cdot 2 - 2 = 6 fits the established formula. - Induction Hypothesis: Assume that for n leq m, the maximum length of the sequence k holds true under the constraint k leq 4n - 2. - Inductive Step: When n = m + 1, consider that a_{k} in {1, 2, cdots, m+1}, Assume a_k = l, 1 leq l leq m+1. If a_1, a_2, cdots, a_{k-1} does not contain l, we can insert l at the beginning: In such a case, adding an item similar to an adjacent one retains the sequence property while increasing k by 1. Since every number can be added to maintain the constraint, if we start from the configuration n = 1, each increase in n (by large induction proof which covers all cases) k=4n-2 is established. 4. **Conclusion:** Thus, the largest number of terms k is given by: [ boxed{4n - 2} ] This formula represents the sequence length maximizing k while satisfying all the given conditions in the problem statement.