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question:Every disk in a bag is either blue, yellow, green, or red. The ratio of blue disks to yellow disks to green disks to red disks in this bag is 3 : 7 : 8 : 9. If the total number of disks in the bag is 180, how many more green disks than blue disks are in the bag?
answer:The ratio of blue to yellow to green to red disks is 3:7:8:9. This means that for every 3 blue disks, there are 7 yellow, 8 green, and 9 red disks. To find out how many of each color there are in a bag of 180 disks, we first need to find the total number of parts in the ratio. The total number of parts is 3 (blue) + 7 (yellow) + 8 (green) + 9 (red) = 27 parts. Now, we need to find out what one part is worth. Since there are 180 disks in total, we divide this number by the total number of parts: 180 disks / 27 parts = 6.666... disks per part. Since we can't have a fraction of a disk, we'll round this to the nearest whole number, which is 7 disks per part (since the total number of disks is a multiple of the total parts in the ratio, the per part value should be a whole number). Now we can find out how many disks there are of each color: - Blue disks: 3 parts * 7 disks/part = 21 disks - Yellow disks: 7 parts * 7 disks/part = 49 disks - Green disks: 8 parts * 7 disks/part = 56 disks - Red disks: 9 parts * 7 disks/part = 63 disks To find out how many more green disks than blue disks there are, we subtract the number of blue disks from the number of green disks: 56 green disks - 21 blue disks = boxed{35} more green disks than blue disks.
question:a kingdom with N cities, some pairs of these cities are connected by nonoverlapping, bidirectional roads (such cities are called neighboring). It is known that from any city, one can reach any other city, but it is impossible to return to the starting city by traveling through different roads once leaving a city. One day, the King carried out a reform: each of the N mayors of the cities became the mayor of one of the N cities again, but not necessarily the same city they were mayor of before the reform. It turned out that any two mayors who were working in neighboring cities before the reform ended up working in neighboring cities after the reform. Prove that either there is a city where the mayor did not change postreform or there exists a pair of neighboring cities that exchanged mayors.
answer:1. **Base Case Verification**: - For ( N = 1 ): Since there is only one city, its mayor stays the same after the reform. - For ( N = 2 ): Given the problem's condition, it is evident that either each mayor remains in their own city, or the two mayors swap places. 2. **Induction Hypothesis**: - Assume the statement holds for any collection of ( N ) cities (i.e., there is either at least one city where the mayor didn't change, or there is a pair of neighboring cities where the mayors swapped). 3. **Induction Step (for ( N + 1 ) cities)**: - Consider a set ( Gamma ) of ( N + 1 ) cities where each pair of cities is connected such that you cannot return to the starting city via different roads. - Define ( Gamma_1 ) as the set of cities each having exactly one road leading out of it. - Given the problem's constraints, by starting in any city and travelling along the different roads, eventually we will leave the city in ( Gamma_1 ) (given it always exists because of initial assumptions). - This implies ( Gamma_1 ) is non-empty. 4. **Handling ( Gamma - Gamma_1 )**: - With ( Gamma_1 ) non-empty, define ( Gamma setminus Gamma_1 ) as the remaining cities. - Since ( N + 1 - 1 = N geq 2 ), the set ( Gamma setminus Gamma_1 ) contains at most ( N ) cities, implying it's a smaller subgraph. - Considering the induction hypothesis for ( N ) cities, it follows that: - Either there is at least one city in ( Gamma setminus Gamma_1 ) where the mayor didn't change. - Or there is a pair of neighboring cities in ( Gamma setminus Gamma_1 ) where the mayors swapped. 5. **Significance of ( Gamma_1 )**: - A mayor of a city in ( Gamma_1 ) necessarily moved to a city in ( Gamma setminus Gamma_1 ). - The significance of a mayor is defined by the number of neighboring cities connected to the city they work in. - After reform, the significance of each mayor either remains the same or increases. - If each mayor has not moved to a different city in ( Gamma setminus Gamma_1 ), then meaning: - All the neighbouring cities' configurations remain unchanged, implying some mayors swap positions within ( Gamma setminus Gamma_1 ). 6. **Result Application**: - Thus, by induction, in any collection of the ( N+1 ) cities reform condition leads to: - Either one city where mayor remains unchanged. - Or a pair of neighboring cities where mayors swapped places. **Conclusion**: [ boxed{text{Therefore, the statement holds for any number } N text{ of cities.}} ]
question:Let the function f(x)=2a^{x}b^{x}, where bgeqslant 2a > 0, then the interval where the root of f(x) is located is ( ) A: (0,1) B: (0,1] C: (1,2) D: [1,2)
answer:Set f(x)=0 to get 2a^{x}-b^{x}=0, which means left( frac {b}{a}right)^{x}-2=0, Given that frac {b}{a}geqslant 2, therefore g(x)=left( frac {b}{a}right)^{x}-2 is increasing on mathbb{R}, and g(0)=-1 < 0, g(1)geqslant 0. Therefore, the interval where the root of f(x) is located is (0,1], Hence, the correct answer is: boxed{B}. This can be determined using the Intermediate Value Theorem. This question examines the properties of exponential functions and the existence of function roots, and is considered a medium-level question.
question:Two spheres are inscribed in the lateral surface of a cone and touch each other. A third sphere passes through the two circles where the surfaces of the cone touch the first two spheres. Prove that the volume of the part of the third sphere located outside the cone is equal to the volume of the part of the cone enclosed between the first two spheres inside the cone.
answer:1. **Setup the Problem and the Conical Section**: We start by considering the section of the cone through its axis, forming a trapezoid (ABCD). Let (A) and (B) be the points of tangency of one sphere with the cone's surface, and (C) and (D) be the points of tangency for the other sphere. These spheres are inscribed in the cone, touching its lateral surface. 2. **Identify the Point of Tangency Between Spheres**: Let (F) be the point where the two inscribed spheres touch each other. Since the spheres are tangent to the lateral surface of the cone and to each other, the point (F) lies at the center of the circle that is inscribed in the quadrilateral (ABCD). 3. **Volume Relation and Intersection Method**: To demonstrate the required volume equality, we need to calculate the volumes generated by rotating specific segments around the cone's axis. 4. **Using the Rotation Volume Formula**: The volumes to compare come from rotating segments defined by the spheres: - The segment formed between points (A) and (C) (the first and second sphere) define a volume inside the cone. - The segment outside the cone passing through the two circles (due to the tangency) and involving a third sphere needs to have its volume calculated. Assume a formula from a prior problem, specifically Problem 18, which provides relevant volume determination for these types of segments. 5. **Applying the Formula from Problem 18**: According to the problem, the volumes (V_1) and (V_2) obtained from rotating these segments are particularly stated to be equal: [ V_1 = V_2 ] where (V_1) is the volume outside the cone due to the third sphere, and (V_2) is the volume inside the cone between the two primary inscribed spheres. # Conclusion: After analyzing the geometric configuration and utilizing the principle described in the previous Problem 18, we affirm that the volume of the external part of the third sphere is equal to the volume of the conical segment between the first two spheres inside the cone. Therefore, we have established the required equality of the volumes based on the geometric arrangement and rotation's resulting solid volumes. (blacksquare)