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question:13. There are several red and white balls in a bag. If each time 2 red balls and 3 white balls are taken out, when there are no white balls left in the bag, there are still 18 red balls left; if each time 5 red balls and 3 white balls are taken out, when there are no red balls left in the bag, there are still 18 white balls left. How many red balls are there in the bag? qquad
answer:【Analysis】Consider 3 white balls as 1 group, if 2 red balls are allocated to each group, then there are 18 more red balls, if 5 red balls are allocated to each group, then there are 18 div 3 times 5=30 fewer red balls, Therefore, the number of white balls is (1) 8+30 div 5-2=16 groups, and the number of red balls is 16 times 2+18=50.
question:In (triangle ABC), (B=30^{circ}), (AC=2sqrt{5}), (D) is a point on side (AB), (CD=2), if (angle ACD) is acute, and the area of (triangle ACD) is (4), then (BC=)_______.
answer:AnalysisThis problem tests the knowledge of solving triangles, using the area formula to find (sin angle ACD), and then applying the sine and cosine theorems to solve the problem.SolutionAccording to the problem, we have (S_{triangle ACD} = frac{1}{2}CD cdot AC cdot sin angle ACD = 2sqrt{5} cdot sin angle ACD = 4), thus (sin angle ACD = frac{2}{sqrt{5}}).Since (angle ACD) is acute, we have (cos angle ACD = sqrt{1-sin^2 angle ACD} = frac{1}{sqrt{5}}).In (triangle ACD), (AD = sqrt{CD^2 + AC^2 - 2CD cdot AC cdot cos angle ACD} = 4),(frac{AD}{sin angle ACD} = frac{CD}{sin A}), (sin A = frac{CD cdot sin angle ACD}{AD} = frac{1}{sqrt{5}}).In (triangle ABC), (frac{AC}{sin B} = frac{BC}{sin A}), (BC = frac{AC cdot sin A}{sin B} = 4).Therefore, the answer is boxed{4}.
question:## Problem 3Three vertices of a regular 2 n+1 sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.
answer:## SolutionAnswer: (n+1) /(4 n-2).Label the first vertex picked as 1 and the others as 2,3, ldots, 2 n+1 (in order). There are 2 n(2 n-1) / 2 ways of choosing the next two vertices. If the second vertex is 2 (or 2 n+1 ), then there is just one way of picking the third vertex so that the center lies in the triangle (vertex n+2 ). If the second vertex is 3 (or 2 n ), then there are two (n+2, n+3) and so on. So the total number of favourable triangles is 2(1+2+ldots+n)=n(n+1). Thus the required probability is (n+1) /(4 n-2).
question:2. From z^{2}+z+1=0 we get z^{3}=1A=I_{3}+B, where B=left(begin{array}{ccc}0 & 0 & 0 z^{2} & 0 & 0 z & z^{2} & 0end{array}right)I_{3} B=B I_{3} Rightarrow A^{n}=left(I_{3}+Bright)^{n}=I_{3}+C_{n}^{1} B+C_{n}^{2} B^{2}+cdots+B^{n}B^{2}=left(begin{array}{lll}0 & 0 & 0 0 & 0 & 0 z & 0 & 0end{array}right) and B^{k}=O_{3}, for any k geq 3.A^{n}=I_{3}+n B+frac{n(n-1)}{2} B^{2}=left(begin{array}{ccc}1 & 0 & 0 n z^{2} & 1 & 0 frac{n(n+1)}{2} z & n z^{2} & 1end{array}right).
answer:2. (7p) Let z in mathbb{C} with the property z^{2}+z=-1 and the matrix A=left(begin{array}{ccc}1 & 0 & 0 z^{2} & 1 & 0 z & z^{2} & 1end{array}right) in M_{3}(mathbb{C}). Calculate A^{n}, where n is a non-zero natural number.
question:Let the region enclosed by the curve y= sqrt {1-(x-1)^{2}} and the x-axis be denoted as D. If the curve y=ax(x-2) (a < 0) divides the area of D into two equal parts, then the value of a is __( )__.A: - dfrac {3}{8}B: - dfrac {3pi}{16}C: - dfrac {3pi}{8}D: - dfrac {pi}{16}
answer:From y= sqrt {1-(x-1)^{2}}, we get (x-1)^{2}+y^{2}=1 (ygeqslant 0), thus, region D represents the upper half-circle with center (1,0) and radius 1, and the curve y=ax(x-2) (a < 0) divides the area of D into two equal parts, therefore int _{ 0 }^{ 2 }[ax(x-2)]dx= dfrac {pi}{4}, therefore( dfrac {1}{3}ax^{3}-ax^{2}) |_{ 0 }^{ 2 }= dfrac {pi}{4}, therefore a=- dfrac {3pi}{16}, thus, the correct choice is boxed{B}. By identifying region D as the upper half-circle with center (1,0) and radius 1, and using the curve y=ax(x-2) (a < 0) to divide the area of D into two equal parts, we can derive int _{ 0 }^{ 2 }[ax(x-2)]dx= dfrac {pi}{4}, which leads to the conclusion. This problem mainly examines the application of the positional relationship between lines and circles, and solving it hinges on using definite integrals to calculate the area.
question:One. (20 points) Let x, y be non-zero real numbers, and satisfy frac{x sin frac{pi}{5} + y cos frac{pi}{5}}{x cos frac{pi}{5} - y sin frac{pi}{5}} = tan frac{9 pi}{20}.(1) Find the value of frac{y}{x};(2) In triangle ABC, if tan C = frac{y}{x}, find the maximum value of sin 2A + 2 cos B.
answer:(1) Let frac{y}{x}=tan theta. Then from the given information,begin{array}{c}frac{tan frac{pi}{5}+tan theta}{1-tan theta cdot tan frac{pi}{5}}=tan frac{9 pi}{20} Rightarrow tan left(theta+frac{pi}{5}right)=tan frac{9 pi}{20} Rightarrow theta+frac{pi}{5}=k pi+frac{9 pi}{20} Rightarrow theta=k pi+frac{pi}{4}(k in mathbf{Z}) .end{array}text { Hence } frac{y}{x}=tan theta=tan left(k pi+frac{pi}{4}right)=tan frac{pi}{4}=1 text {. }(2) From (1), we get tan C=1.Since 0<angle C<pi, therefore, angle C=frac{pi}{4}.Then angle A+angle B=frac{3 pi}{4} Rightarrow 2 angle A=frac{3 pi}{2}-2 angle B.Thus, sin 2 A+2 cos B=sin left(frac{3 pi}{2}-2 Bright)+2 cos Bbegin{array}{l}=-cos 2 B+2 cos B=-2 cos ^{2} B+2 cos B+1 =-2left(cos B-frac{1}{2}right)^{2}+frac{3}{2} .end{array}Therefore, when cos B=frac{1}{2}, i.e., angle B=frac{pi}{3}, sin 2 A+ 2 cos B achieves its maximum value of frac{3}{2}.