answer:Solution: Cutting the sphere with a plane results in a cross-sectional area of pi, which means the radius of the small circle is 1. Given the distance from the center of the sphere to this cross-section is 1, the radius of the sphere is r= sqrt{1+1} = sqrt{2}. Therefore, the volume of the sphere is: frac{4}{3}pi r^3 = frac{8sqrt{2}}{3}pi. Hence, the answer is: boxed{frac{8sqrt{2}}{3}pi}. By finding the radius of the small circle and using the distance of 1 cm from the center of the sphere to the cross-section, along with the radius of the small circle, the radius of the sphere can be determined through the Pythagorean theorem, which then allows for the calculation of the volume of the sphere. This question tests the relationship between the radius of the small circle, the distance from the center of the sphere to the cross-section, and the radius of the sphere, examining computational skills. It is a basic question.

answer:1. **Introduction and Restatement of the Problem**: We aim to show that among the students in the group, there are at least two students who know the same number of people. Here, "knowing" is a reciprocal relationship which means if student A knows student B, then student B also knows student A. 2. **Using the Pigeonhole Principle**: The problem can be approached using the Pigeonhole Principle. According to the Pigeonhole Principle, if ( n ) items are put into ( m ) containers, with ( n > m ), then at least one container must contain more than one item. 3. **Identify the Pigeons and Pigeonholes**: - **Pigeons**: Each student. - **Pigeonholes**: The number of people each student knows. 4. **Range of Possible Values**: Each student can know a minimum of 0 and a maximum of ( n-1 ) students (excluding themselves). Therefore, there are ( n ) possible numbers of acquaintances ranging from 0 to ( n-1 ). 5. **Eliminate Impossible Scenarios**: It is important to note that it is impossible for there to be a scenario where one student knows 0 people and at the same time another student knows ( n-1 ) people. This is because if one student knows everyone, then no student can know no one. Thus, only ( n - 1 ) values are realistically possible for the number of acquaintances each student can have. 6. **Apply the Pigeonhole Principle**: Given there are ( n ) students but only ( n-1 ) possible numbers of people each student can know, by the Pigeonhole Principle, at least two students must know the same number of other students. # Conclusion: Hence, it is proven that among the students, there exist at least two students who know the same number of people. [ boxed{} ]

answer:To solve the problem, we first need a common denominator. Noticing that 87 is a multiple of 29, we use 87 as the common denominator: [frac{8}{29} = frac{8 cdot 3}{29 cdot 3} = frac{24}{87}] Now, subtract the fractions: [frac{24}{87} - frac{5}{87} = frac{24 - 5}{87} = frac{19}{87}] Next, we simplify frac{19}{87}. Since 19 is a prime number and does not divide 87, the fraction is already in its simplest form: [boxed{frac{19}{87}}]

answer:Jungkook collected 6 divided by 3, which equals 2. So, Yoongi collected 4, Jungkook collected 2, and Yuna collected 5. Therefore, Yuna collected the biggest number, which is boxed{5} .