answer:Since a_1=2 and a_3=6 in the arithmetic sequence {a_n}, the common difference is d=2. Let's denote the number to be added as x. After adding x to a_1, a_4, and a_5, the resulting numbers are 2+x, 8+x, and 10+x, respectively. Since these new numbers form a geometric sequence, we have (8+x)^2 = (2+x)(10+x). Solving this equation for x, we obtain x=-11. Therefore, the number that was added is boxed{-11}.

answer:Let's denote the other number as m. We know that the least common multiple (LCM) of two numbers n and m is given by: LCM(n, m) = (n * m) / GCF(n, m) We are given that LCM(n, m) = 56 and GCF(n, m) = 12. We are also given that n = 48. Substituting the given values into the LCM formula, we get: 56 = (48 * m) / 12 To find m, we can solve for it: 56 * 12 = 48 * m 672 = 48 * m Now, divide both sides by 48 to isolate m: m = 672 / 48 m = 14 So, the other number m is boxed{14} .

answer:First, we select one ball and place it in the box with the corresponding number, which can be done in C_5^1 = 5 ways. For example, we could place ball number 5 in box number 5. The remaining four balls can be arranged in a way that none of their numbers match the boxes they are in, which is a sub-problem known as "derangement." In total, there are 9 derangements for four objects, given by: - (2, 1, 4, 3) - (2, 3, 4, 1) - (2, 4, 1, 3) - (3, 1, 4, 2) - (3, 4, 1, 2) - (3, 4, 2, 1) - (4, 1, 2, 3) - (4, 3, 1, 2) - (4, 3, 2, 1) Using the principle of multiplication for counting, the total number of ways to place the five balls into the five boxes, with the requirement that exactly one ball matches its box number, is the product of the number of ways to select the ball that will match its box and the number of derangements for the remaining balls. Thus, we have 5 times 9 = 45 ways. Therefore, the answer is: boxed{45} This problem investigates the comprehensive application of counting principles. The key to solving it lies in the use of "selecting one ball to place in the box with the corresponding number" to satisfy the condition that "exactly one ball’s number matches the number of its box."

answer:(1) Since F_1 and F_2 are the left and right foci of the ellipse, and A is a point on the ellipse such that overrightarrow{AF_{2}} cdot overrightarrow{F_{1}F_{2}}=0, we can represent A as (c, frac{b^{2}}{a}). The slope of AF_{1} is given as frac{sqrt{3}}{12}. From the properties of an ellipse, we have frac{b^{2}}{2ac}= frac{sqrt{3}}{12}. Simplifying, we get 6b^{2}=sqrt{3}ac. Using the relation b^{2}=a^{2}-c^{2}, we can rewrite the equation as 6c^{2}+sqrt{3}ac-6a^{2}=0. Let e=frac{c}{a}, then 6e^{2}+sqrt{3}e-6=0. Solving for e, we get e=frac{sqrt{3}}{2}. Given length of the major axis is 8, so 2a=8, hence a=4. Therefore, c=2sqrt{3}, and b^{2}=4. Thus, the equation of the ellipse is boxed{frac{x^{2}}{16}+ frac{y^{2}}{4}=1}. (2) We have the following system of equations: begin{cases} x^{2}+4y^{2}-16=0 y=kx+ frac{3}{2} end{cases}. Eliminating y, we get x^{2}+4(kx+ frac{3}{2})^{2}-16=0, which simplifies to (1+4k^{2})x^{2}+12kx-7=0. Let x_1 and x_2 be the roots of this quadratic equation. Then, x_1+x_2=frac{-12k}{1+4k^{2}} and x_1 cdot x_2=- frac{7}{1+4k^{2}}. The midpoint D of the segment EF is (-frac{6k}{1+4k^{2}}, frac{3}{2(1+4k^{2})}). From the problem, we know that the slope of BD is -frac{1}{k}. Therefore, frac{-2-frac{3}{2(1+4k^{2})}}{frac{6k}{1+4k^{2}}}=-frac{1}{k}. Simplifying, we get k^{2}=frac{5}{16}, hence boxed{k=pm frac{sqrt{5}}{4}}.