Math amateur masters, please enter, math masters.

If 12 people play in pairs, the game will have c(12,2)=66 games, but now the number of games is still 66-47=19 games, this number is caused by the rule of the same school, that is, if everyone from the same school plays, the number of games between them is 19So the following discussion is only for hypothetical co-school competitions.

If a school sends 1 person, the school will have 0 games, and if it is n(n>1) people, the school game will be c(n,2).

In c(n,2) (n>1), less than 19 I listed.

c(2,2)=1, with an average of 1 game per person.

c(3,2)=3, with an average of 2 sessions per person.

c(4,2)=6, with an average of 3 games per person.

c(5,2)=10, with an average of 4 games per person.

c(6,2)=15, with an average of 5 games per person.

What is the maximum number of people sent from all schools for discussion, first prove that the maximum number of people cannot be less than 5, because the total number of competitions in the same school is 19, 2 people participate in a game, a total of 12 people, the average number of games per person participates in 19 * 2 12 = 19 6>3, if the number of people sent by all schools does not exceed 4, the average number of participants in the same school competition will be less than 3, get a contradiction. So the maximum number of people can only be 5 or 6

Divide 12 into the sum of several numbers, and then count the number of games from most to less, and filter them one by one.

When the maximum number of people in the party is 6 people. (c(6,2)+c(6,2)>c(6,2)+c(5,2)>c(6,2)+c(4,2)>19, so 6,5,4 is not considered).

Corresponds to c(6,2)+c(3,2)+c(3,2)=21

The requirements are not met.

Corresponds to c(6,2)+c(3,2)+c(2,2)+0=19

6,3,2,1 is the combination sought.

Corresponds to c(6,2)+c(3,2)+0+0+0=18

The requirements are not met.

The next count will only decrease in the number of entries per person, which means that the total number of races will be less than 19, so that's it.

When the maximum team is 5 players. (c(5,2)+c(5,2)>19, so 5 is not considered).

Corresponds to c(5,2) + c(4,2) + c(3,2) = 19

5, 4, 3 are the desired combinations.

Corresponds to c(5,2)+c(4,2)+c(2,2)+0=17

The requirements are not met.

Same as above, you can stop calculating.

In summary, there are two groups of solutions that meet the conditions:

4 schools, the number of teams is: 6, 3, 2, 1

3 schools, the number of teams is: 5, 4, 3

There are M schools participating, and the number of players from each school is A1, A2, etcam

then a1+a2+.am=12

Since every player will be playing against everyone else but their own school, there will be so many matches for the first school as A1* (12-A1).

The same goes for other schools, so the total number of games will be.

a1*(12-a1)+a2*(12-a2)+.am*(12-am))/2=47

Note that this must be divided by 2 because all the matches are repeated 2 times when added (between player 1 of school X and player 1 of school Y, 1 time is counted in the game of school X and 1 time in the game of school Y).

Simplifying the above formula, i.e., 12(a1+a2+..am)-(a1^2+a2^2+..am^2)=94

i.e. 12*12-(a1 2+a2 2+..am^2)=94

To sum up, you will get two formulas:

a1+a2+..am=12

a1^2+a2^2+..am^2=50

and a1, a2 ,..am are all greater than 0, so m<12, and obviously m>1

Because 7*7=49, a1, a2 ,..AM cannot be greater than or equal to 7

You might as well set a1<=a2<=....=am

Numbers less than 49 squared are only 1, 4, 9, 16, 25, 36

Analysis of 50:

1) If am=36, then.

50 = 36 + 14 = 36 + 9 + 4 + 1 (4 schools) (the number of players at this time is 6 + 3 + 2 + 1 = 12, which meets the question).

36 + 4 + 1 + 4 + 4 + 1 (6 schools) (at this time, the number of players is 6 + 2 + 1 + 2 + 1 = 13, which does not meet the question).

36 + 4 + 4 + 1 + 1 + 1 + 1 + 1 + 1 (9 schools) (at this time, the number of players is 6 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 15, which does not meet the topic).

36+4+1+1+1+1+1+1+1+1+1+1+1 (12 schools) (at this time, the number of players is 6+2+1+1+1+1+1+1+1+1+1+1=18, which does not meet the topic).

2) If am=25, then.

50 = 25 + 25 (2 schools) (the number of players at this time is 5 + 5 = 10, which does not meet the question).

25 + 16 + 9 (3 schools) (the number of players at this time is 5 + 4 + 3 = 12, which meets the question).

25 + 16 + 4 + 4 + 1 (5 schools) (at this time, the number of players is 5 + 4 + 2 + 2 + 1 = 14, which is not satisfactory).

.The next few cases are not satisfied).

3) If am=16, then.

50 = 16 + 16 + 16 + 1 + 1 (the number of players is 4 + 4 + 4 + 1 = 13, which does not meet the question).

.In the case thereafter, the number of players exceeds 12).

In the same way, we can know that there is no solution that matches the meaning of the problem in the case of am=9,4,1.

Thus, the possible solution is:

4 schools, the number of players is 6, 3, 2, 1

3 schools, the number of players is 5, 4, 3 respectively

f(x)=asin2x+cos2x=(a 2+1) sin(2x+ )sin =1 (a 2+1) cos = a (a 2+1).

The above reasoning can be derived from a compulsory 4 after-class exercise.

The minimum positive period of f(x) is 2 2= then 1 4 periods is 4 12+ 4= 3 (minus is OK).

i.e. f( 3) = 0 3 2a-1 2 = 0 3a = 1 a = 3 3

Observation diagram is available.

13-8=5, 11-6=5, 10-5=5, so it can be deduced.

15-？=5, so ? =15-5=10 , i.e. ? fill in 10

The meaning of the title is not very clear.,If you look for a rule.,It should be 10.,The two numbers of each box are different by 5.

I'm sorry, but you can't see it clearly, and you can draw it better. You are imagining yourself, imagining from the difference between the number and the quotient, and carefully looking for the relationship.

fill in 10;

Because: 13-8=11-6=10-5=15-? The difference is 5;

Thanks for adopting!

Draw a straight line with a ruler and measure the length of segment A with a compass.

The metal edge is used to draw the length of 2a (ab, bc) with a compass on the straight line that has been drawn, and then the end of the 2a long line segment (point c) is used as the vertex, and the length of the line segment b (cd) is drawn in the reverse direction with a compass, at this time.

The length of 2a-b is the length of ad.

Draw a circle with a radius. Then the diameter is 2a, and then draw a circle, the center of which is b, and the radius of the previous circle is b. Connect the center of the circle and extend it to intersect the circle you drew earlier. and the circle of the later painting intersects with the point b. Then the length of the AB line segment is 2A-B

Mark the two ends of the line segment A and B, set the point of the compass at the point B, and the other end at the point A, measure the length of A, then draw an arc outside A, extend the arc of AB at point C with a ruler, and obtain the line segment 2A; In the same way, the line segment B is measured, and then, the compass tip is set at point C, and the arc is crossed by BC at point D, then the line segment AD is equal to 2A-C

(1) Draw the 2A+B line segment with a ruler, and note that the intersection point should be drawn in advance at the junction of the 2A and B line segments (if the A and B line segments have been drawn, use the ruler to extend the A line segment to 2 times);

2) Use a compass to measure the length of the B line segment;

3) Place the compass at the intersection of the 2A and B line segments, and draw a cross line to one end of the 2A line segment (if the A and B line segments have been drawn, press (1) bracket operation, and then cut off the B line segment length on 2A);

4) Use a compass to measure the distance from the intersection line to the starting end of the 2A line segment, that is, the C line segment (if the A and B line segments have been drawn, measure them on the extended A line segment);

5) Take the compass of the C line segment and draw an arc below the 2A+B line segment (if the A and B line segments have been drawn, then below the A and B line segments);

6) Draw a horizontal line from the center of the circle to the arc with a ruler to draw the C line segment.

Drifting Mimi, hello :

Because there is already a section of a, then the compass can copy the length, you use the two points of the compass to fall on the two ends of a, and then draw a circle, use a ruler to make a diameter, then this diameter is 2a, at the same time, use both end points to fall on both ends of b, truncate back, one end point falls on one end of the diameter, one end falls on the diameter, then the rest is 2a-b

Draw a circle with a radius.

As its diameter ab, draw a circle with one end a as the center of the circle, and b as the radius to draw the circle ab at c

then ac is c=2a-b

For example, when dismantling and respecting the oak, B has more lines than A: 28 2 = 56 kilometers.

Per hour, B has more lines than A: 63-55 = 8 km.

Time taken by A and B to meet: 56 8 = 7 hours.

The distance between the two places: (55+63) 7=826 thousand liangling meters.

28 2 = 56 km.

63-55 = 8 km.

56 8 = 7 hours.

55+63) 7=826 km.

It must be 28 kilometers from the midpoint.

As can be seen from the question, B is 8 kilometers faster than A per hour, and the two cars meet at the place where the potato is dismantled 28 kilometers away from the midpoint, so B has walked 28 2 = 56 kilometers more than A, so A and B have walked a total of 56 8 = hours.

Therefore, the distance between the two places is 55 +63 = 826 km.

Because they meet at a distance of 28 km from the midpoint, car B has exceeded the midpoint by 28 km and car A is 28 km away from the midpoint, so car B travels 56 km longer than car A, so the driving time is 56 (63-55) = 7 hours.

The distance between the two places is AB: (55+63) 7=826 kilometers.

The distance between the north and the south: (28+28) (63-55)x(63+55)=826 (km).

4×3×2×1×4-3×2×1×3=78

This question should be dealt with by the indirect method.

This is a question of permutations and combinations, first of all, the condition "five people are divided into four classes, at least one person in each class" can be known, there are always two people in a class, you can first divide five people into four groups, these four groups may have 5 4 2 kinds of 10 kinds. If there is no condition of "A is not in the same class", then the division of these four groups of people into four classes is 4 3 2 1 24 kinds, that is, there are a total of 10 24 kinds and 240 kinds.

However, with the condition of "A is not in the first class", it is necessary to subtract all the cases of A in the first class and divide it into A in the first class, and Class A is divided into two students, which can be divided into 4 3 2 types and 24 types; A class is only divided into student A, then the other four students are grouped, divided into three groups with 4 3 2 kinds 6 kinds, these three groups are divided into.

There are 3 2 1 6 kinds of divisions in the second, third and fourth classes, so there are 6 6 kinds and 36 kinds of divisions in the first class of A. The two cases add up to 24, 36, 60.

I don't know if there is any omission in understanding, I hope it helps.

Five people are divided into four classes, and at least one person in each class is divided into: c(5,2) 3 2 1 4=10 24=240

Method 1: Because there are 4 classes, the probability of A being in a class = 1 4 , 240 4 = 60 so there are 240-60 = 180 ways to not be in a class.

Method 2: A in the division of a class has.

Except for A, the remaining 4 students are divided into 1 class 4 * 3 * 2 * 1 = 24 The remaining 4 students are less than 1 class, that is, 4 students in three classes c(4,2)) 3 2 1 = 6 6 = 36

The division of A in class is 26 + 24 = 60

Therefore, the division of A not in the same class is: 240-60 = 180 kinds.

Regardless of the situation of A, if five people are divided into four classes and each class has at least one person, then two of the five people must be in one class, so two of the five people can be selected and then arranged to get 240 kinds; If A is in a class, as long as the remaining four people are divided into four classes, it can be divided into two situations, one is A and one of the remaining four people is in a class, there are 24 kinds, and the other is A is one person in a class, then the remaining four people should be divided into the remaining three classes, and two of the four people should be selected and then arranged into 36 kinds. Finally, use 240-24-36 = 180