answer:For an arithmetic sequence, the difference between consecutive terms must be equal. Therefore, we can set up the following equations based on the sequence given: [ (y + 2) - left(-frac{1}{3}right) = 4y - (y+2) ] Simplify and solve these equations: [ y + 2 + frac{1}{3} = 4y - y - 2 ] [ y + frac{7}{3} = 3y - 2 ] [ frac{7}{3} + 2 = 3y - y ] [ frac{13}{3} = 2y ] [ y = frac{13}{6} ] Thus, the value of y that satisfies the given arithmetic sequence is boxed{frac{13}{6}}.

answer:First, we need to find the inverse function g^{-1}(x). Given g(x) = 5x - 3, solve for x: [ y = 5x - 3 ] [ y + 3 = 5x ] [ x = frac{y + 3}{5} ] Thus, g^{-1}(x) = frac{x + 3}{5}. Now, apply g^{-1} twice to the given value 14: [ g^{-1}(14) = frac{14 + 3}{5} = frac{17}{5} ] [ g^{-1}left(frac{17}{5}right) = frac{frac{17}{5} + 3}{5} = frac{frac{17}{5} + frac{15}{5}}{5} = frac{32}{5 times 5} = frac{32}{25} ] Thus, g^{-1}(g^{-1}(14)) = boxed{frac{32}{25}}.

answer:Using the given dimensions, we set up the area equation: [ (3m+8)(m-3) = 76. ] Expanding this, we get: [ 3m^2 - 9m + 8m - 24 = 76, ] [ 3m^2 - m - 24 = 76, ] [ 3m^2 - m - 100 = 0. ] Factoring the quadratic, we find: [ (3m+25)(m-4) = 0. ] This gives two potential solutions for m: m=-frac{25}{3} and m=4. Since m must be a positive integer, the only valid solution is m = boxed{4}.

answer:1. Since f(x) = log_a(1+x) and g(x) = log_a(1-x), where a>0 and a neq 1, we have f(x) - g(x) = log_a(1+x) - log_a(1-x), where a>0 and a neq 1. To ensure the function f(x) - g(x) is meaningful, we need begin{cases} 1+x > 0 1-x > 0 end{cases} Solving this, we get -1 < x < 1, which means the domain of the function f(x) - g(x) is (-1, 1). 2. Since the domain of f(x) - g(x) is (-1, 1), which is symmetric about the origin, let F(x) = f(x) - g(x). Then F(-x) = f(-x) - g(-x) = log_a(1-x) - log_a(1+x) = -[log_a(1+x) - log_a(1-x)] = -F(x). Therefore, f(x) - g(x) is an odd function. 3. From f(x) - g(x) > 0, we get f(x) > g(x), which means log_a(1+x) > log_a(1-x). If a > 1, then begin{cases} -1 < x < 1 1+x > 1-x end{cases} which simplifies to begin{cases} -1 < x < 1 x > 0 end{cases} Solving this, we get 0 < x < 1. If 0 < a < 1, then begin{cases} -1 < x < 1 1+x < 1-x end{cases} which simplifies to begin{cases} -1 < x < 1 x < 0 end{cases} Solving this, we get -1 < x < 0. In summary, if a > 1, the solution set for the inequality is (0, 1), and if 0 < a < 1, the solution set for the inequality is (-1, 0). Therefore, the final answers are: 1. The domain of f(x) - g(x) is boxed{(-1, 1)}. 2. The function f(x) - g(x) is an boxed{text{odd function}}. 3. The range of x for which f(x) - g(x) > 0 is boxed{(0, 1)} if a > 1, and boxed{(-1, 0)} if 0 < a < 1.