answer:First, let's establish the total number of students in each category: 1. Total Music club members = frac{1}{4} times 40 = 10 2. Total Science club members = frac{1}{5} times 40 = 8 3. Total Sports club members = 8 From the Venn diagram, we establish the equations: - For Music: 6 + x + z + w = 10 - For Science: 5 + x + y + w = 8 - For Sports: 2 + z + y + w = 8 - Total students with clubs = 40 - 7 = 33 So: (6 + x + z + w) + (5 + x + y + w) + (2 + z + y + w) - 2(x + y + z) - 3w = 33 or 13 + 2x + 2y + 2z + 3w = 33, solving into 2x + 2y + 2z + 3w = 20 Next, subtract individual clubs' totals: - From Music: x + z + w = 4 - From Science: x + y + w = 3 - From Sports: z + y + w = 6 Substituting z + y + w = 6 into 2x + 2y + 2z + 3w = 20 after solving for x + z + w = 4 and x + y + w = 3: - Solve x = 3 - w, y = 6 - z - w, and z = 4 - x - w and substitute into the inequalities. Substitute these into the cleaned up equation: (3-w) + (6-z-w) + (4-(3-w)-w) + 3w = 6 Simplifying, we find z = 0, x = 2, y = 5, and w = 1. Thus, there is boxed{1} student in all three clubs.

answer:1. **Prime Factorization of 40,000**: The prime factorization of 40,000 is 40,000 = 2^5 cdot 5^3. 2. **Determine Forms of Elements in Set S**: Elements of S are of the form 2^a5^b where 0 leq a leq 5 and 0 leq b leq 3. 3. **Consider Products of Two Distinct Elements**: Consider two distinct elements 2^a5^b and 2^c5^d from S. The product is 2^{a+c}5^{b+d}. Ensure (a, b) neq (c, d). 4. **Relate to Divisors of 40,000^2**: We calculate 40,000^2 = (2^5 cdot 5^3)^2 = 2^{10} cdot 5^6. This has divisors in form 2^p5^q where 0 leq p leq 10 and 0 leq q leq 6. The total number of divisors is (10+1)(6+1) = 77. 5. **Exclude Non-distinct Products**: Exclude products like: - 1 = 2^0 cdot 5^0. - 2^{10}5^6 (cannot be formed as it would require the maximum power of each factor). - 2^{10} and 5^6 as they also cannot be distinctly formed. These exclusions account for 4 divisors. 6. **Calculate Number of Valid Products**: Subtract the 4 non-valid products from the total 77 divisors of 40,000^2, resulting in 77 - 4 = 73. 7. **Verification**: Each other number in form 2^p5^q where not both p and q are maxed (p=10, q=6), can be written as the product of two distinct elements from S. Thus, the number of numbers that are the product of two distinct elements of S is 73. The final answer is boxed{D}

answer:# Problem: 现有 4 种不同颜色的信号灯泡，每种不同的颜色表示不同的信号。假设每种亮色的信号灯泡足够多，若要在如下图所示的某机场信号塔的每个顶点 (P, A, B, C, A_1, B_1, C_1) 上各安装一个灯泡，要求同一线段两端的灯泡不同颜色，则不同的安装方法有多少种。 1. **Identify the total number of positions to place bulbs:** There are 7 points where bulbs need to be placed: ( P, A, B, C, A_1, B_1, C_1 ). 2. **Assign colors to each bulb:** We have 4 different colors available, labeled as ( C_1, C_2, C_3, C_4 ). The bulbs need to be placed such that no two connected points (line segment endpoints) have the same color. 3. **Calculate the number of valid color assignments:** a. **Choose color for point ( P ):** We have 4 choices for point ( P ). b. **Choose color for point ( A ):** For point ( A ), we have 3 remaining choices (since it must differ from ( P )). c. **Choose color for point ( B ):** For point ( B ), we have 3 remaining choices (since it must differ from ( P ), but has no constraint from ( A ) at this moment). d. **Choose color for point ( C ):** For point ( C ), we have 3 remaining choices (since it must differ from ( P ), but has no constraint from ( A ) or ( B )). e. **Choose colors for ( A_1, B_1, C_1 ):** For points ( A_1, B_1, C_1 ), we again consider each point separately keeping in mind the connected points. - ( A_1 ) has 3 choices because it cannot be the same as ( A ). - ( B_1 ) has 3 choices because it cannot be the same as ( B ). - ( C_1 ) has 3 choices because it cannot be the same as ( C ). 4. **Multiplication of choices** considering the constraints at each step: The total number of ways to assign the colors under the given constraints is: [ 4 text{ (choices for } P) times 3 text{ (choices for } A) times 3 text{ (choices for } B) times 3 text{ (choices for } C) times 3 text{ (choices for } A_1) times 3 text{ (choices for } B_1) times 3 text{ (choices for } C_1) ] 5. **Simplify the expression:** [ 4 times 3^6 = 4 times 729 = 2916. ] # Conclusion: The total number of different valid installation methods is: (boxed{2916}).

answer:Since f(x)=lg frac {x^{2}+1}{|x|} (xneq 0), we have f(-x)=lg frac {(-x)^{2}+1}{|-x|}=lg frac {x^{2}+1}{|x|}=f(x), thus the function is an even function, and its graph is symmetric about the y-axis; hence (1) is correct; When x > 0, f(x)=lg frac {x^{2}+1}{x}, it is a decreasing function on (0,1] and an increasing function on [1,+infty); When x < 0, f(x)=lg frac {x^{2}+1}{-x}, it is a decreasing function on (-infty,-1] and an increasing function on [-1,0); Therefore, (2) is incorrect, (3) is correct; When x=±1, the function attains its minimum value lg 2, and it has no maximum value, hence (4) is correct, (5) is incorrect; Thus, the answer is: boxed{①③④} By analyzing the given function's expression, we can determine the function's parity, monotonicity, and extremal values, which leads to the answer. This problem is a comprehensive application of the function graph and properties, where identifying the function's parity, monotonicity, and extremal values is key to solving it.