answer:Given: - There are 6 students, each having at least 3 friends within the group. - Friendship is mutual. - A new book can be given to a student such that it will eventually be read by all other students. We need to prove that the teacher can always give the new book to one of the students such that it will be passed to all other students. Step-by-Step Solution: 1. **Existence of a Chain:** - Since each student has at least 3 friends, we can construct a chain of friends starting from any student. - Let us pick any student and label them as student (1). Let's denote one of their friends as student (2). - Since student (2) has at least 3 friends and one of these friends (student (2)) is not student (1), we can choose another friend of student (2) and label them as student (3). 2. **Constructing the Chain:** - Continue this process, selecting new friends such that no friend is repeated within the current portion of the chain. - This can be done as long as there are additional friends who have not been chosen previously. 3. **Chain Length Considerations:** - Continue this process until we either include all students or we find some students who are not in the chain. 4. **Ensuring the Chain Covers All Students:** - Assume that our chain only contains some subset of the 6 students. Suppose it contains only 4 students initially. - Consider a student (N) who is not in the chain. (N) must have at least 3 friends. - Verify if (N) is friends with any of the students at either end of the chain. - Since (N) has 3 friends, if (N) is not friends with the first or last student in the current chain, then (N) must be friends with someone in the middle of the chain. 5. **Adjusting the Chain:** - If (N) is friends with more students in the chain, expand the chain by bringing (N) into consideration. - This way, by iterating, we can ensure (N) fits into the chain without breaking the current chain's integrity. 6. **Proving the Full Chain:** - If our chain initially contains 5 nodes (with (5) students), consider the remaining student (say (E)). - If (E) is friends with the first or last student of the chain, then (E) is connected, forming a chain of 6. - Otherwise, (E) must be friends with students in the middle, allowing us to extend the chain in both directions. 7. **Cycle and Contradiction:** - Any further extension of the chain would mean all nodes are interconnected. Given that each node has at least 3 friends, they must interconnect to avoid any node being isolated beyond having 3 friends. - Thus, trying to disconnect any will lead back across the chain or forming a minimal cycle, allowing the book to be passed through every student. Conclusion: By above construction and chain adjustments, we ensure all students will be part of a single chain where the book can be passed to each one. Hence, the teacher can always give the new book to any one student, ensuring it will be read by all 6 students. [ boxed{text{All students will read the book}} ]

answer:- Calculate the total number of ways to choose 2 quitters from 16 contestants: binom{16}{2} = 120 - Calculate the number of ways for two quitters to be from the same tribe: - For any chosen tribe, since there are 4 members: binom{4}{2} = 6 - There are 4 tribes, so ways for both quitters to be from the same tribe across all tribes: 6 times 4 = 24 - Calculate the probability: text{Probability} = frac{text{Ways for both to be from the same tribe}}{text{Total ways to pick any 2 quitters}} = frac{24}{120} = boxed{frac{1}{5}}

answer:To compare the given real numbers, we first express them in a uniform format: - frac{sqrt{2}}{2} can be rewritten as frac{1}{sqrt{2}}. - frac{sqrt{3}}{3} can be rewritten as frac{1}{sqrt{3}}. - frac{sqrt{5}}{5} can be rewritten as frac{1}{sqrt{5}}. Now, we compare these fractions along with the number 1 based on their denominators: - Since sqrt{2} < sqrt{3} < sqrt{5}, it follows that frac{1}{sqrt{2}} > frac{1}{sqrt{3}} > frac{1}{sqrt{5}} because as the denominator increases, the value of the fraction decreases. Thus, we have the following order: [1 > frac{1}{sqrt{2}} > frac{1}{sqrt{3}} > frac{1}{sqrt{5}}] Which translates back to the original numbers as: [1 > frac{sqrt{2}}{2} > frac{sqrt{3}}{3} > frac{sqrt{5}}{5}] Therefore, among the given real numbers, the smallest real number is frac{sqrt{5}}{5}. Hence, the correct answer is boxed{D}.

answer:To find the domain of (k(x)), we need to find the values of (x) that make the denominators zero, as these are the points where (k(x)) is undefined. 1. Solving (x+9 = 0): [ x = -9 ] 2. Solving (x^2+9 = 0) for real (x): [ x^2 = -9 quad text{(no real solution)} ] 3. Solving (x^3+9 = 0): [ x^3 = -9 implies x = sqrt[3]{-9} = -sqrt[3]{9} ] Thus, the function (k(x)) is undefined at (x = -9) and (x = -sqrt[3]{9}). Therefore, the domain of (k(x)) is all real numbers except (x = -9) and (x = -sqrt[3]{9}). We express this as a union of intervals: [ boxed{(-infty, -9) cup (-9, -sqrt[3]{9}) cup (-sqrt[3]{9}, infty)} ]