answer:AnalysisThis problem tests your understanding of basic properties of hyperbolas. By utilizing the equation of the hyperbola, find the coordinates of the focus and the equation of the asymptote. Then, calculate the distance using the formula for the distance from a point to a line. This is a basic-level problem.Step-by-Step Solution1. Identify the parameters of the hyperbola from its equation: dfrac{{x}^{2}}{16} - dfrac{{y}^{2}}{9} = 1. Here, a = 4, b = 3, and since c^2 = a^2 + b^2, we have c = 5.2. Find the equation of one of the asymptotes. The asymptotes of a hyperbola are given by y = pm dfrac{b}{a}x. In this case, one of the asymptotes is 3x + 4y = 0.3. Find the coordinates of one of the foci. The coordinates of the foci of a horizontal hyperbola are given by (c, 0) and (-c, 0). Here, we'll use the focus at (5, 0).4. Calculate the distance from the focus to the asymptote using the formula for the distance from a point (x_1, y_1) to a line Ax + By + C = 0: d = dfrac{|Ax_1 + By_1 + C|}{sqrt{A^2 + B^2}}. In this case, d = dfrac{|3(5) + 4(0) + 0|}{sqrt{3^2 + 4^2}} = boxed{3}.

answer:Given:- The width of the rectangle is a cm.- The length is 1 cm more than twice its width.Step 1: Express the length in terms of a.- Length = 2 times text{width} + 1 = 2a + 1 cm.Step 2: Calculate the perimeter of the rectangle.- The formula for the perimeter of a rectangle is 2 times (text{length} + text{width}).- Substituting the given values, we get 2 times (a + (2a + 1)).Step 3: Simplify the expression.- 2 times (a + 2a + 1) = 2 times (3a + 1) = 6a + 2.Therefore, the perimeter of the rectangle in cm is boxed{6a+2}.

answer:For any pair of natural numbers (x,y) , we denote by Q_{xy} the unit square with vertices (x-1,y-1) , (x,y-1) , (x,y) , (x-1,y) , and by R_{xy} - the rectangle with vertices (0,0) , (x,0) , (x,y) , (0,y) . If A is any set of pairs of natural numbers, then the symbol Z(A) will denote the following subset of the plane:[Z(A) = bigcup_{(x,y) in A} Q_{xy}]We assume the following:[Z(emptyset) = emptyset]According to the problem statement, the set A belongs to the family W if and only if the set Z = Z(A) satisfies the conditions:1. Z is the union of a finite number of squares Q_{xy} ( x,yin mathbb{N} ),2. Z subset R_{nn} ,3. Q_{ab} subset Z Longrightarrow R_{ab} subset Z .In particular, the empty set (consisting of 0 unit squares) satisfies these conditions.Assume for a moment that the set Z is not empty. Implication (3) then expresses exactly that Z is the union of a certain (finite, positive) number of rectangles of the form R_{ab} - that is, rectangles with two sides lying on the positive coordinate axes. A finite union of such rectangles (contained in R_{nn} ) is a polygon bounded by a closed broken line ON_0M_1N_1ldots M_kN_kO , where: kin mathbb{N} ; all vertices have integer coordinates; O is the origin of the coordinate system; N_0 is a point on the segment OK , where K = (0, n) ; N_k is a point on the segment OL , where L = (n, 0) ; vectors overrightarrow{N_{i-1}M_i} are parallel to the Ox axis and directed in the direction of this axis; vectors overrightarrow{M_iN_i} are parallel to the Oy axis and directed opposite to the direction of this axis ( i = 1, ldots , k ) (see figure 3).om38_1r_img_3.jpgConversely, any polygon Z bounded by such a broken line satisfies conditions (1), (2), (3), and thus corresponds to some non-empty set Ain W . It remains to count these polygons.Let ON_0M_1N_1ldots M_kN_kO be a polygon of the described form. Each of the vectors overrightarrow{N_{i-1}M_i} is the sum of a certain number of unit vectors [1,0] , and each of the vectors overrightarrow{M_iN_i} is the sum of a certain number of unit vectors [0, -1] . Consider the broken line KN_0M_1N_1ldots M_kN_kL (it may be K = N_0 or N_k = L ), oriented from K to L . It consists of 2n unit vectors: each of the vectors [1, 0] and [0, -1] must appear n times (going from K to L we must make n steps to the right and n steps down). The polygon Z is uniquely determined by a path of this type running from K to L .Since we assumed (temporarily) that the set Z is not empty, we must exclude the path from K to L composed sequentially of the vector [0, -1] repeated n times, and then the vector [1,0] repeated n times (i.e., the broken line KOL ). It is natural to now allow this path as well, assuming that it corresponds precisely to the empty set Z (which must also be included in the considerations).So how many admissible paths are there from K to L ? We have 2n unit vectors of two types. If in the sequence (1, 2, ldots, 2n) we fix n positions where we place the vector [1, 0] , then on the remaining n positions the vector [0, -1] must appear. There are as many such paths as there are n -element subsets of a 2n -element set, i.e., binom{2n}{n} . There are as many polygons Z satisfying conditions (1), (2), (3), and thus as many sets A in the family W .

answer:AnalysisThis question examines piecewise functions and the solution of inequalities, which is a basic problem with moderate difficulty.SolutionSolve f(x)leqslant 2,When xleqslant 1, 2^{1-x}leqslant 2, which implies 1-xleqslant 1,therefore xgeqslant 0, hence 0leqslant xleqslant 1;When x > 1, 1-log_2xleqslant 2, which implies log_2xgeqslant -1,therefore xgeqslant dfrac{1}{2}, hence xgeqslant 1;Combining the above, we get xgeqslant 0.Therefore, the correct choice is boxed{text{D}}.

answer:SOlUTIONGiven that A B C D E F G H is a regular octagon, and letting its centre be O, we find that angle A O D=3 times 360^{circ} div 8=135^{circ}.Similarly, given that A P Q is an equilateral triangle, we also have angle A O P=360^{circ} div 3=120^{circ}.Therefore the angle P O D=135^{circ}-120^{circ}=15^{circ}. Now, since 15 is factor of 360 , it is true that P D is a side of a regular polygon inscribed in the circle. The number of sides, n, of this polygon is 360 div 15=24.

answer:7. 7 .Each A_{i} contains 3 binary subsets, and S has a total of mathrm{C}_{n}^{2} binary subsets. Therefore,3 m=mathrm{C}_{n}^{2} text {. }For an element k in S, there are n-1 binary subsets in S that contain k.By condition (1), each of these is contained in exactly one A_{i}, and each A_{i} containing k includes two binary subsets that contain k, so there are frac{n-1}{2} A_{i}'s containing k.Considering A_{i} as points, and connecting each pair of points with an edge, a graph is formed with mathrm{C}_{m}^{2} edges.By condition (2), the edge between A_{i} and A_{j} can be labeled with the unique common element k of A_{i} and A_{j}, and the edge labeled k appears exactly mathrm{C}_{frac{n-1}{2}}^{2} times, thus,mathrm{C}_{m}^{2}=n mathrm{C}_{frac{n-1}{2}}^{2} .Solving equations (1) and (2) yields (m, n)=(1,3),(7,7).text { Since } m>1 text {, we have } m=7 text {. }