answer:To prove that the complement of mathbb{Q} times mathbb{Q} in the plane mathbb{R}^2 is connected, we will use the fact that mathbb{Q} times mathbb{Q} is countable and that the complement of a countable set in mathbb{R}^2 is always connected. Since mathbb{Q} times mathbb{Q} is countable, we can write it as a sequence of points: mathbb{Q} times mathbb{Q} = {(q_1, q_2), (q_3, q_4), (q_5, q_6), ldots} Now, let S = mathbb{R}^2 setminus (mathbb{Q} times mathbb{Q}). We will show that S is connected by showing that any two points in S can be connected by a path in S. Let (x_1, y_1) and (x_2, y_2) be two points in S. Since mathbb{Q} times mathbb{Q} is countable, there are uncountably many disjoint paths between (x_1, y_1) and (x_2, y_2) in mathbb{R}^2. Therefore, we can find a path between (x_1, y_1) and (x_2, y_2) that does not intersect mathbb{Q} times mathbb{Q}. This path is contained in S, so S is connected. Therefore, the complement of mathbb{Q} times mathbb{Q} in the plane mathbb{R}^2 is connected.

answer:When dividends are declared, Retained Earnings is decreased (debited) and Dividends Payable is increased (credited). When dividends are paid, Dividends Payable is decreased (debited) and Cash is decreased (credited). Therefore, the declaration and payment of dividends affect the Retained Earnings and Cash accounts in the accounting equation.

answer:We have that p(x+y) leq p(x) + p(y) by the sublinearity of p. Subtracting p(x) from both sides, we obtain the desired inequality: p(x+y)-p(x)leq p(y)

answer:The binomial coefficient binom{n}{k} is defined as the number of ways to choose k elements from a set of n elements, without regard to order. In this case, we have n = 27147/2 and k = 27145/2. We can use the formula for the binomial coefficient to compute its value: binom{n}{k} = frac{n!}{k!(n-k)!} Substituting the values of n and k, we get: binom{27147/2}{27145/2} = frac{(27147/2)!}{(27145/2)!(27147/2 - 27145/2)!} Simplifying the expression, we get: binom{27147/2}{27145/2} = frac{(27147/2)!}{(27145/2)!(1)!} Since 1! = 1, we can further simplify the expression to: binom{27147/2}{27145/2} = frac{(27147/2)!}{(27145/2)!} Finally, we can use the fact that (27147/2)! = (27147/2) times (27147/2 - 1) times (27147/2 - 2) times cdots times 1 and (27145/2)! = (27145/2) times (27145/2 - 1) times (27145/2 - 2) times cdots times 1 to cancel out the common factors in the numerator and denominator, leaving us with: binom{27147/2}{27145/2} = frac{27147/2}{27145/2} Simplifying further, we get: binom{27147/2}{27145/2} = frac{27147}{2} Therefore, the value of the binomial coefficient binom{27147/2}{27145/2} is frac{27147}{2}. The answer is frac{27147}{2}