answer:Since x is the arithmetic mean of a and b, we have 2x=a+b, And since x^2 is the arithmetic mean of a^2 and -b^2, we have 2x^2=a^2-b^2, By solving these two equations simultaneously and eliminating x, we get (a+b)(a-3b)=0 Thus, a=-b or a=3b Hence, the answer is: boxed{C} By using the arithmetic mean, we can obtain a system of equations for x, and by eliminating x and simplifying, we get the answer. This question tests the understanding of the arithmetic mean and involves solving a system of equations, which is a basic question.

answer:1. **Length of BC**: Since ABC is equilateral and AB = 6, we have BC = 6. 2. **Position of D on overline{AC}**: D is one-third of the way from A to C, so AD = frac{1}{3}AC = frac{1}{3} times 6 = 2 and DC = frac{2}{3} times 6 = 4. 3. **Length of DE**: Since ADE is equilateral and shares the same side length as AD, we have DE = 2. 4. **Position of G on overline{AE}**: G is the midpoint of AE, so AG = frac{1}{2}AE = frac{1}{2} times 2 = 1 and GE = 1. 5. **Length of EF**: Since EFG is equilateral and similar in proportion to ADE, EF = 1. 6. **Length of FG**: Since G is the midpoint of AE, the side FG = 1. 7. **Length of GA**: Since AG = 1, then GA = 1. Compute the perimeter: [ AB + BC + CD + DE + EF + FG + GA = 6 + 6 + 4 + 2 + 1 + 1 + 1 = boxed{21} ]

answer:First, let's find the cycle of units digits of powers of 3, starting from n=1: {3, 9, 7, 1, 3, 9, 7, 1, ldots}. This cycle repeats every 4 terms: 3, 9, 7, 1. To determine the units digit of 3^n for any integer n, we find the remainder, R, when n is divided by 4: - R=1 corresponds to a units digit of 3, - R=2 corresponds to a units digit of 9, - R=3 corresponds to a units digit of 7, - R=0 corresponds to a units digit of 1. Now, consider 3^{2^{1000}}. As 2^{1000} equiv 0 pmod{4} (since 2^{1000} is a multiple of 4), the units digit of 3^{2^{1000}} is 1. Thus, the units digit of G_n = 3^{2^{1000}} + 2 is 1 + 2 = boxed{3}.

answer:Without loss of generality, place the vertices of the hexagon at (3,0), (frac{3sqrt{3}}{2}, frac{3}{2}), (-frac{3sqrt{3}}{2}, frac{3}{2}), (-3,0), (-frac{3sqrt{3}}{2}, -frac{3}{2}), and (frac{3sqrt{3}}{2}, -frac{3}{2}). Suppose the endpoints of the segment, positioned between vertices A and B, have coordinates (3cos(theta), 3sin(theta)) and ((3-3sin(theta)), 0). The relationship between the x and y coordinates reflects that the segment has length 3, so (3cos(theta)-(3-3sin(theta)))^2 + (3sin(theta))^2 = 9. Using midpoint formula, the midpoint coordinates become: [ left(frac{3cos(theta) + (3-3sin(theta))}{2}, frac{3sin(theta)}{2}right) ] Since each side contributes similarly, and scaling and positioning symmetries, these points describe a circular region around each vertex. The radius of each circle equals frac{3}{2}, which is half of the distance formula result, sqrt{(3cos(theta) - (3-3sin(theta)))^2 + (3sin(theta))^2} = 3. Each of these circular regions has area frac{pi(3/2)^2}{6} because the enclosure influences only frac{1}{6}th of the complete area around each point: [ frac{pi(2.25)}{6} = frac{2.25pi}{6} = frac{3pi}{8} ] Since there are six vertices, and regions are non-overlapping, the total enclosed area A = 6 times frac{3pi}{8} = frac{9pi}{4}. [ boxed{frac{9pi}{4}} ]