For a math problem, please give a detailed process, and if you are satisfied, you can add a reward

If C B x meets after focus, then C A (x+2) minutes later.

The total distance traveled by C and A when they meet is equal to the total distance traveled by C and B when they meet, and it is also equal to the distance between the east and west towns).

x = 36, i.e. C and B meet after 36 minutes, and the total distance is meters.

Set the distance to be x km.

A-C encounter time = x (60+75).

B-C encounter time = x (

x/135=2x/285+2

57x=54x+15390

3x=15390

x=5130

The distance between the east and west towns is 5130 meters.

Calculate the speed of A, B, C. A: 1 m s.

Second:. Third:. It took X seconds for C and B to meet.

The two towns are y meters apart. Then: y=(.

At this point, A and B are at a distance: (i.e., the distance between A and C). 2 minutes = 120 seconds.

According to the title, it can be obtained: (. Solve x and bring in the first equation to find y.

Pure hand-played. Hope.

When C and B meet, they take a t-point;

Then there is: 75t+

t=36 then the whole process is 75*36+

Question 1: Let the upper level of the book be x and the lower level be y book.

According to the meaning of the title, y=80, when the book is taken out, the upper layer is left with (2 3)x, and the lower layer is left with (3 4)y, according to the title: 3 (2 3) x = (3 4) y

Substituting y gives us x=30That is, the upper floor originally had 30 books.

Question 2: B is wrong.

A said that he played four sets, indicating that he had played with four other people, and Ding also said that he had played four sets, indicating that he had also played with other people. In this case, B played at least two sets, one against A and one against Ding.

Authoritative answer, forget, thank you!

1. After the lower layer is taken away, there is 80 * (1-1 4) = 60, and there are 20 left in the upper layer at this time, so the original upper layer has 20 (1-1 3) = 30

2. B is wrong. Elimination method, A and D have no stake, both are 4 sets, because only one person is wrong, so A and D are not wrong, if both are wrong, then B is wrong, B should be three sets.

(1-1 4) = 60 copies.

60 3 = 20 copies.

20 (1-1 3) = 30 copies.

The upper floor originally had 270 books.

2. If A and D both played 4 games, then, both of them played a game with others, then B could not have a big game, at least two sets, and E played three games, in addition to playing a game with A Ding, he had to find someone to play a game, C was two sets bigger, then he could only play with B, then in fact B should have played three games.

After reading the questions, the sorting conditions can be found.

The main principle is that the line segment between two points is the shortest, and the perpendicular line segment from the point to the straight line is the shortest.

The main idea is to equate BM+MN to a straight line segment.

Solution: Find a little e on the AC edge so that ae=an and connect EMBecause ad is the angular bisector of the bac.

So me=mn (equivalent to moving point n to point e to solve).

Finding the minimum value of BM+MN is the minimum value of BM+ME.

Because m and n are the moving points on ad and ab respectively, the perpendicular line of ac is crossed by b and ac is crossed at f point (emphasis added), bf is the minimum value of bm + me, that is, the minimum value of bm + mn.

and ab=4 (the second under the root number) bac=45° bfa=90°

So bf=ab=4

That is, the minimum value of BM+MN is 4.

If m is taken as the fixed point first, then mn is the smallest mn ab

Set me ac

ad is the bisector of bac.

me=mnIn bme, if bm+me is minimal, then b, m, and e are in the same straight line be ac

and bac=45° ab=4 2

be=4 i.e. the minimum value of bm+mn is 4

In the acute triangle ABC, ab=4 2, BAC=45°, the bisector of BAC intersects BC at D, m, and n are the moving points on AD and AB, respectively, and the minimum value of BM+MN is obtained.

Solution: abc is an acute triangle, a=45°, 45°b n (the sum of the two sides of the triangle is greater than the third side), > nb (the hypotenuse is greater than the right angle)=4, so it is proved.