On a highway, there is a warehouse every 100 km, with a total of five warehouses. No.1 warehouse has 10t goods, No.2 warehouse has 20t goods, No.34 warehouse has no goods, and No.5 warehouse has 40t goods. Now if you want to store all the goods in one warehouse, if you need 0.5 freight per ton of goods to transport 5 km, how much freight will it cost at least?

On a highway, there is a warehouse every 100 km, with a total of five warehouses. No.1 warehouse has 10t goods, No.2 warehouse has 20t goods, No.34 warehouse has no goods, and No.5 warehouse has 40t goods. Now if you want to store all the goods in one warehouse, if you need 0.5 freight per ton of goods to transport 5 km, how much freight will it cost at least?

The minimum cost is 1000 yuan. There's a problem with the freight. Hi, from the 1st to the 5th, 10 times 400 divided by 5 times 0.5 = 400, from the 2nd to the 5th, 20 times 300 divided by 5 times 0.5 = 600, add up to 1000

1. Uncle Zhang bought 6.5kg of beans in the vegetable market, paid 20 yuan and recovered 1.15 yuan. How much yuan per kilogram of beans?
2. A piece of rectangular glass is 3.5 meters long and 2.7 meters wide. It is known that the price per square meter is 7.2 yuan. How much should I get back if I pay 70 yuan for this piece of glass?
3. The distance between a and B is 450 km. A freight car and a passenger car run from each other. The freight car runs 65 km per hour and the passenger car 55 km per hour. How many hours will the two cars meet? (using equation solution)

1. Beans = (20-1.15) △ 6.5 = 2.9 yuan
2. Area = 3.5 × 2.7 = 9.45m2
Return = 70-9.45 × 7.2 = 1.96 yuan
3. Let the two cars meet in X hours
X*(65+55)=450
120X=450
X=450÷120
X=3.75
The two cars met in 3.75 hours

（2+√10）^10（2-√5）^11
THIS

Is the title wrong
（2+√10）^10•（√2-√5）^11
＝（2+√10）^10•（√2-√5）^10 •（2-√5）
＝[（2+√10）•（√2-√5）] ^10•（2-√5）
=36（2-√5）
＝72－36√5