answer:First, let's calculate the total cost of the maths books. June bought four maths books at 20 each, so the cost is: 4 maths books * 20/book = 80 Next, let's calculate the cost of the science books. She bought six more science books than maths books, so she bought 4 + 6 = 10 science books. At 10 each, the cost is: 10 science books * 10/book = 100 Now, let's calculate the cost of the art books. She bought twice as many art books as maths books, so she bought 2 * 4 = 8 art books. At 20 each, the cost is: 8 art books * 20/book = 160 Now, let's add up the cost of all the books except the music books: Cost of maths books + Cost of science books + Cost of art books = 80 + 100 + 160 = 340 June started with 500, so the remaining amount after buying the maths, science, and art books is: 500 - 340 = 160 Therefore, June spent boxed{160} on music books.

answer:Use the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) to find the roots of 5x^2 - 8x - 7 = 0. - Calculating the discriminant: [ b^2 - 4ac = (-8)^2 - 4 cdot 5 cdot (-7) = 64 + 140 = 204 ] - The roots are: [ x = frac{8 pm sqrt{204}}{10} ] - The positive difference between the roots: [ frac{sqrt{204}}{5} ] At this point, ensure 204 simplifies to a form m without any prime squared: [ 204 = 2^2 cdot 3 cdot 17 ] Thus, ( m = 51 ) (since 51 = 3 cdot 17) and ( n = 5 ). So, the required sum of m + n is: [ m + n = 51 + 5 = boxed{56} ]

answer:Factor out the largest power of 7 that divides both terms, which is 7^4: [ 7^7 - 7^4 = 7^4 cdot 7^3 - 7^4 cdot 1 = 7^4(7^3 - 1) ] Calculate 7^3 - 1 = 343 - 1 = 342, which factors as 2 cdot 3^2 cdot 19. Thus, the prime factorization of 342 is 2, 3, 19. Considering the factorization with 7^4, the overall distinct prime factors are 2, 3, 7, 19. The sum of these distinct prime factors is: [ 2 + 3 + 7 + 19 = boxed{31} ]

answer:1. **Assume the contrary**: All air routes connect cities of different republics. Let's denote the cities with at least 70 outgoing air routes as "millionaire cities". 2. **Define the quantity of millionaire cities**: According to the problem, there are at least 70 millionaire cities, each connected to at least 70 other cities by air routes. 3. **Analyzing a single republic**: If a republic contains a millionaire city, then it connects to at least 70 cities. Since all air routes connect cities of different republics, and the republic contains at most 30 cities (100 total cities minus the 70 cities being connected to), the other connections must reach into the other republics. 4. **Distribution in three republics**: There are three republics, and the total possible number of cities (30 per republic) solely based on connections outside the republics is constrained to these numbers as: - Republic 1: 30 cities maximum - Republic 2: 30 cities maximum - Republic 3: 30 cities maximum 5. **Full city count including millionaire cities**: If millionaire cities need to fulfill the connection requirements, each additional republic should also not exceed 30 cities. 6. **Total city inconsistency**: Summing up the maximum possible number of cities in each republic: [ 30 + 30 + 30 = 90 , text{cities total in the three republics} ] But we know that the total number of cities in the country is 100. Thus, the assumption runs into an inconsistency. We do not have enough cities left (only 90 calculated) if some connections are solely to satisfy connecting only different republics. 7. **Conclusion**: Therefore, the assumption that all air routes connect cities of different republics is false. This logically implies: [ boxed{text{There must be at least one air route that connects two cities within the same republic.}} ]