answer:To find the maximum size of subset ( S ) of the set ({1, 2, ldots, k}) such that the sum of any ( n ) different elements from ( S ) does not exceed ( m ), we need to follow these steps: 1. **Establish the lower bound for the sum of ( n ) elements:** If ( |S| ge n ) and ( A subset S ) is an ( n )-set, then the sum of the smallest ( n ) elements in ( S ) is at least the sum of the first ( n ) natural numbers: [ sigma(A) ge frac{n(n+1)}{2} ] Therefore, for ( S ) to exist, we must have: [ frac{n(n+1)}{2} le m ] 2. **Determine the upper bound for the sum of ( n ) elements:** If ( |S| = p ge n ), we can write ( S = {x_1, x_2, ldots, x_p} ) with ( x_i ) in increasing order. We have ( x_i ge i ) for all ( i ). Consider the subset ( A := {x_{p-(n-1)}, x_{p-(n-2)}, ldots, x_p} ). The sum of the elements in ( A ) is: [ sigma(A) = x_{p-(n-1)} + x_{p-(n-2)} + cdots + x_p ] Since ( x_i ge i ), we have: [ x_{p-(n-1)} ge p-(n-1), quad x_{p-(n-2)} ge p-(n-2), quad ldots, quad x_p ge p ] Therefore: [ sigma(A) ge (p-(n-1)) + (p-(n-2)) + cdots + p = np - frac{n(n-1)}{2} ] For ( sigma(A) le m ), we need: [ np - frac{n(n-1)}{2} le m ] Solving for ( p ): [ np le m + frac{n(n-1)}{2} ] [ p le frac{m}{n} + frac{n-1}{2} ] 3. **Maximize ( p ):** The maximum size of ( S ) is given by: [ p_{max} = leftlfloor frac{m}{n} + frac{n-1}{2} rightrfloor ] where ( leftlfloor cdot rightrfloor ) denotes the floor function, which gives the greatest integer less than or equal to the given expression. The final answer is ( boxed{leftlfloor frac{m}{n} + frac{n-1}{2} rightrfloor} ).

answer:(1) First, let's simplify each term in the expression: - 0.001;^{- frac {1}{3}} = (0.1)^{3 times (- frac {1}{3})} = 10 - (frac {7}{8})^{0} = 1 - 16;^{frac {3}{4}} = 2^{4 times frac {3}{4}} = 8 - (sqrt {2} cdot 33)^{6} = 2^{frac {1}{2} times 6} cdot 3^{frac {1}{3} times 6} = 72 By substituting these simplified terms back into the original expression, we have: 10 - 1 + 8 + 72 = boxed{89}. (2) First, let's simplify each term in the expression: - log _{3} sqrt {27} = log _{3} 3^{frac {3}{2}} = frac {3}{2} - log 25 = log 100 - log 2 = 2 - log 4 = log 16 - log 2 = 1 - 7^{log _{7}2} = 2 - (-9.8)^{0} = 1 By substituting these simplified terms back into the original expression, we have: frac {3}{2} + 2 + 1 + 2 = boxed{frac{13}{2}}.

answer:Since b=2sqrt{3}, B=120^{circ}, and C=30^{circ}, by the Law of Sines, we have c= frac{bsin C}{sin B}= frac{2sqrt{3} times frac{1}{2}}{frac{sqrt{3}}{2}}=2, thus A=180^{circ}-B-C=30^{circ}, by using the Law of Cosines, we get a^{2}=b^{2}+c^{2}-2bccos A=12+4-2times2sqrt{3}times2timesfrac{sqrt{3}}{2}=4, solving this gives a=2. Therefore, the correct answer is boxed{D}. This problem involves the application of the Law of Sines, the Law of Cosines, and the triangle angle sum theorem in solving triangles, which is considered a basic question.

answer:The circle C: x^2+y^2-2y=0 has its center at (0,1) and a radius of r=1. According to the problem, PQ is a tangent to the circle C: x^2+y^2-2y=0 at point Q, and the minimum length of PQ is 2. Therefore, the minimum distance from the center of the circle to the line, PC, is sqrt{5}. By using the formula for the distance from a point to a line, we get frac{|1+4|}{sqrt{k^2+1}} leq sqrt{5}. Therefore, k leq -2 or k geq 2. Hence, the correct choice is: boxed{text{C}} By utilizing the fact that PQ is a tangent to the circle C: x^2+y^2-2y=0 at point Q, and the minimum length of PQ is 2, we can determine the minimum distance from the center of the circle to the line, and from the formula for the distance from a point to a line, we can find the range of values for k. This problem tests the application of equations of lines and circles, including the tangent to a circle and the formula for the distance from a point to a line, and is considered a medium-level question.