answer:To solve this problem, we first identify the upper vertex of the hyperbola frac{{y}^{2}}{4}-frac{{x}^{2}}{8}=1. The standard form of a hyperbola is frac{{y}^{2}}{a^2}-frac{{x}^{2}}{b^2}=1, where (0, a) and (0, -a) are the vertices on the y-axis for a vertical hyperbola. Comparing, we find a^2 = 4, so a = 2. Thus, the upper vertex is at (0,2). Next, we find the equation of the asymptotes. For a hyperbola of the form frac{{y}^{2}}{a^2}-frac{{x}^{2}}{b^2}=1, the asymptotes are given by frac{y}{a}=pmfrac{x}{b}. Here, b^2 = 8, so b = sqrt{8} = 2sqrt{2}. Therefore, the equations of the asymptotes are y = pmfrac{a}{b}x = pmfrac{2}{2sqrt{2}}x = pmsqrt{2}x or x pm sqrt{2}y = 0. To find the distance from the upper vertex (0,2) to one of its asymptotes, say x + sqrt{2}y = 0, we use the formula for the distance from a point (x_0, y_0) to a line Ax + By + C = 0, which is frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}. Substituting A = 1, B = sqrt{2}, C = 0, x_0 = 0, and y_0 = 2, we get: [ text{Distance} = frac{|1cdot0 + sqrt{2}cdot2 + 0|}{sqrt{1^2 + (sqrt{2})^2}} = frac{2sqrt{2}}{sqrt{1 + 2}} = frac{2sqrt{2}}{sqrt{3}} = frac{2sqrt{6}}{3} ] Therefore, the distance from the upper vertex of the hyperbola to one of its asymptotes is boxed{frac{2sqrt{6}}{3}}, which corresponds to choice boxed{A}.

answer:**1. Substituting y = (x + 2)^2 into x + 6 = (y - 5)^2:** [ x + 6 = ((x + 2)^2 - 5)^2 ] Expanding and simplifying: [ x + 6 = (x^2 + 4x - 1)^2 ] [ x + 6 = x^4 + 8x^3 + 14x^2 - 8x + 1 ] [ x^4 + 8x^3 + 14x^2 - 9x - 5 = 0 ] By Vieta's formulas, x_1 + x_2 + x_3 + x_4 = -8. **2. Now substituting x = (y - 5)^2 - 6 into y = (x + 2)^2:** [ y = ((y - 5)^2 - 4)^2 ] Expanding and simplifying: [ y = (y^2 - 10y + 21)^2 ] [ y = y^4 - 20y^3 + 140y^2 - 420y + 441 ] Setting y on one side: [ y^4 - 20y^3 + 140y^2 - 421y + 441 = 0 ] By Vieta's formulas, y_1 + y_2 + y_3 + y_4 = 20. Hence, the required sum is: [ x_1 + x_2 + x_3 + x_4 + y_1 + y_2 + y_3 + y_4 = (-8) + 20 = boxed{12} ]

answer:Solution: (1) Let the common difference of the arithmetic sequence be d. Since S_1, frac{1}{2}S_3, frac{1}{3}S_5 form an arithmetic sequence, we get S_1 + frac{1}{3}S_5 = S_3, which means a_1 + frac{1}{3} cdot 5a_3 = 3a_2, thus 1 + frac{5}{3}(1+2d) = 3(1+d), solving this gives d=1, therefore a_n = 1 + (n-1) times 1 = n boxed{n} (2) Since {b_1,b_2,b_3} subseteq {a_1,a_2,a_3,a_4,a_5}, that is {b_1,b_2,b_3} subseteq {1,2,3,4,5}, because the sequence {b_n} is an increasing geometric sequence, therefore b_1=1, b_2=2, b_3=4, therefore b_n = b_1left( frac{b_2}{b_1} right)^{n-1} = 2^{n-1} therefore T_n = a_1b_1 + a_2b_2 + a_3b_3 + ldots + a_{n-1}b_{n-1} + a_nb_n Then 2T_n = a_1 cdot 2b_1 + a_2 cdot 2b_2 + a_3 cdot 2b_3 + ldots + a_{n-1} cdot 2b_{n-1} + a_n cdot 2b_n, which means 2T_n = a_1b_2 + a_2b_3 + a_3b_4 + ldots + a_{n-1}b_n + a_nb_{n+1} Subtracting the two equations gives -T_n = a_1b_1 + (a_2-a_1)b_2 + (a_3-a_2)b_3 + (a_4-a_3)b_4 + ldots + (a_n-a_{n-1})b_n - a_nb_{n+1}, which means -T_n = 1 + 2 + 2^2 + ldots + 2^{n-1} - n cdot 2^n = frac{1-2^n}{1-2} - n cdot 2^n = 2^n - 1 - n cdot 2^n = (1-n)2^n - 1, therefore T_n = (n-1) cdot 2^n + 1 boxed{(n-1) cdot 2^n + 1}

answer:First, we need to determine what three consecutive integers multiply to give 336. Let the middle integer be ( n ), then the three consecutive numbers are ( n-1 ), ( n ), and ( n+1 ). Their product can be expressed as: [ (n-1)n(n+1) = 336 ] To solve for ( n ), we start by taking an educated guess or factorizing 336 to find suitable candidates. The prime factorization of 336 is: [ 336 = 2^4 times 3 times 7 ] Trying different values around the cube root of 336 (which is approximately 7), checking ( n = 6 ) seems promising: [ (6-1) times 6 times (6+1) = 5 times 6 times 7 = 210 ] This is incorrect, so let’s try ( n = 7 ): [ (7-1) times 7 times (7+1) = 6 times 7 times 8 = 336 ] This is correct. Since we have found that ( n = 7 ), the consecutive integers are 6, 7, and 8. The sum of these integers is: [ 6 + 7 + 8 = 21 ] Thus, the sum of the three consecutive integers whose product is 336 is (boxed{21}).