A. The waterway between port B and port a is 208 km long. A ship sails from port a to port B and arrives 8 hours downstream; it returns from port B to port a and arrives 13 hours upstream. The speed of the ship in still water and the velocity of water flow are calculated

A. The waterway between port B and port a is 208 km long. A ship sails from port a to port B and arrives 8 hours downstream; it returns from port B to port a and arrives 13 hours upstream. The speed of the ship in still water and the velocity of water flow are calculated

Downstream speed: 208 △ 8 = 26 (km); upstream speed: 208 △ 13 = 16 (km); still water speed: (26 + 16) △ 2, = 42 △ 2, = 21 (km); water speed: 21-16 = 5 (km). Answer: the speed of the ship in still water is 21 km, and the water speed is 5 km

It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o

It is proved that: the bisector of ∵ ABC and ∵ ACB intersects at point O, ∵ OBC = 12 ∵ ABC, ∵ OCB = 12 ∵ ACB, ∵ OBC + ∵ OCB = 12 (∵ ABC + ∵ ACB). In △ OBC, ∵ BOC = 180 ° - (∵ OBC + ? OCB) = 180 ° - 12 (? ABC + ? ACB) = 180 ° - 12 (180 ° - a) = 90 ° + 12 ? a

I want 30 Olympiad math problems in the seventh grade volume 2 of PEP!
Reading pictures should have pictures! You'd better have answers or ideas

1. It is known that the equation 2A (x-1) = (5-a) x + 3B has innumerable solutions, then a=_____ ,b=_____ Answer: 2A (x-1) = (5-a) x + 3b 2ax-2a = 5x ax + 3b 3ax-5x = 2A + 3B x (3a-5) = 2A + 3B the equation 2A (x-1) = (5-a) x + 3B about X has innumerable solutions, so no matter what value x takes, it always holds, so

During the "Rhododendron Festival" in Huanggang, 70 employees of a company organized a group to visit and appreciate the scenic spots. The regulations of the scenic spots are as follows: ① tickets are 60 yuan per person, no discount; ② you can take the scenic spot sightseeing bus when you go up the mountain. The sightseeing bus has four seats and eleven seats, each of which is 60 yuan, and eleven seats is 10 yuan per person. The employees of the company are just full of each car, and the total cost is not more than 5000 yuan How many cars are there for each car?

Suppose four seater cars rent x cars and eleven seater cars rent y cars, then: 4x + 11y = 7070 × 60 + 60x + 11y × 10 ≤ 5000, transform 4x + 11y = 70 into 4x = 70-11y, substitute 70 × 60 + 60x + 11y × 10 ≤ 5000, then: 70 × 60 + 15 (70-11y) + 11y × 10 ≤ 5000, the solution is y ≥ 5011, then ∵ x = 70 − 11y4 ≥ 0, ∥ y ≤ 7011, so y = 5, 6. When y = 5, x = 154. When y = 6, x = 1 Rent one four seater and six eleven seater