What a difficult math problem, I admire him for who made it! 5 points!!

Solution: Let the original driving speed be v, the time taken is t, and the distance between A and B is so v*t=s (1).

According to the title: increase the speed of the car by 25% of the original, 24 minutes earlier.

That is, (2) drive 80 kilometers at the original speed first, and the time is 80 V

If you increase the speed to 4 (3*V), you can arrive 10 minutes earlier.

That is, (4*v) 3*(t-10-80 v)+80=s (3) is obtained from (2).

v=(Substitute (1) for v=(.)

v=s/120 (4)

(1) and (4) are substituted by (3) (4v*t) 3-(40v) 3-320 3+80=s.

4s/3-s/9-80/3=s

2s/9=80/3

s=120A: A and B are 120 kilometers apart.

Let the distance between the two places s, the original velocity v, and the time t

s=v·t=

The solution yields t=120 (points).

s=80+ (4 3)v·(110-80 v)=v·t to obtain v=1 (km).

s=v·t=120x1=120 (km).

Let the speed of the vehicle be x, and it will take y minutes to get to place B at this speed.

x*y=(1+25%)x*(y-24)

y=5/4(y-24)

y=120120x-80)/x=(120x-80)/(1+1/3)x+10

4*(120x-80) 4x=3 *(120x-80) 4x+10x=1*120=120

Just make a list of equations and you're good to go.

Set: The original speed is x and the total distance is y, and the time is converted to minutes.

5/4x×（y/x-24）=y

y-80) (4x 3)+80 x=y x-10 do the math yourself!

Suppose s=vt

Then for the first time: vt=v(t-2 5)(1+1 4) gives t= 2

Second: vt-80 = v(t-1 6) (1 + 1 3) to vt-2v 3 = 80

Multiply t both sides by t, which is 2s-2s 3=160

s = 120 hours at an hourly rate).

When A and B met for the first time, they walked a full journey, and when A and B met for the second time, they walked a total of three full journeys

Therefore, when A and B meet for the second time, A walks three times the distance A travels when they meet for the first time

Therefore, when A and B meet for the second time, A walks a distance of 36*3=108 (kilometers), and the meeting point of the second meeting of A and B is 68 kilometers away from Dongzhen.

Method 1: Arithmetic solution.

108+68) 2=88 (km).

Method 2: Set the distance between the east and west villages x kilometers.

2x=36*3+68

2x=108+68

2x=176

x=88A:

The distance between the east and west villages is 88 kilometers.

You must be right, give it to me!

It took 5 minutes for the boat to catch up with the map from U-turn, indicating that when the boat caught up with the map, the boat and the map were in 5 minutes, the boat traveled more than the map, and the boat still water speed was 5

This distance is the distance from the time the map falls into the water to the time the boat turns around, and at this time the boat is going against the current, the map is drifting, and their speed is still water speed, so the person finds that the map falls into the water after (b) 5 minutes.

The figure is stationary relative to the water, because it only floats with the water, and the speed of the boat relative to the water is constant, whether it is downstream or countercurrent, so it took 5 minutes to catch up, and then it was also 5 minutes later when it was discovered.

When the current is countercurrent. v boat ground = v boat water - v water land.

When going downstream. v boat ground = v boat water - v water land.

where v ship to ground denotes the velocity of the ship with respect to the ground.

Column equations are on the line.

If you don't understand, you can ask.

The first time 950*2 (40+150)=10 points 40*10=400, meet at a distance of 550 meters from place b.

After that, they meet head-on every 10 minutes, the second time is 150 meters away from B, the third time is 250 meters away from B, and the fourth time is 650 meters away from B, so the second time is closest to B, and the closest is 150 meters.

It took 19 3 minutes for B to turn back, and it took 95 4 minutes for A to get the location of A's first meeting, and it took 95 4 minutes for A to come back and meet again.

Count the places of several encounters in this way, and get the final result.

The hero's own attributes + equipment attributes are counted as a whole.

That is, the sum of the orange and green numbers is counted as a whole. And not separately calculated.

The same formula is used for the calculation of offense and defense.

Based on the data given, the formula fitted is:

y= e^(

where x is the sum of the hero's own attributes and equipment attributes, which can be offensive or defensive.

y is the percentage of the bonus. e =

y= multiplied by ( multiplied by x ) to the power of ().

1.Because f(x) defines the domain as [0,1].

Since the function f(x 2+1) matches the correspondence f, x 2+1 is also in the defined domain.

So 0<=x 2+1<=1

1<=x^2<=0

Because x 2>=0

So x=0

Define the domain as. 2.(x^3+x^2-2x)/(x^4-1)=x(x^2+x-2)/(x^2+1)(x^2-1)=x(x+2)(x-1)/(x^2+1)(x+1)(x-1)=x(x+2)/(x^2+1)(x+1)

This step is about x-1, where x is not equal to 1).

Because (x 3+x 2-2x) x 4-1 0 then x(x+2) (x 2+1)(x+1) 0 then x(x+2)(x 2+1)(x+1) 0 and x is not equal to 1, and x 2+1 must be greater than 0

So x(x+2)(x+1) 0, at this point you can draw a number line to help understand that x -2 or -1 is summed up and x is not equal to 1

1。Since f(x) is the function that defines the domain on [1,2), then 2a-1>1-3a 2a-1>2 1-3a 1, the solution of the inequality gives the range a>3 2

How about 50 points for me? Actually, I won't!

There are a total of 8 people, each person shakes hands up to 6 times, and the minimum handshake is 0 times Except for Mr. A, the number of other handshakes is not the same, so there are 1 from 0 to 6 timesIf Mrs. A shakes hands 0 times.

That is to say, Mrs. A has not shaken hands with everyone, so the remaining 6 people, each of whom has a maximum of 5 handshakes, contradictory.

So Mrs. A's handshake is not 0 times, and one of the other three couples must have shook hands 0 times 2Let's assume that B shakes hands 0 times.

As we said earlier, each person shakes hands a maximum of 6 times, and now B shakes hands 0 times means that the rest of the people have not shaken hands with B, except for Mrs. B, the rest of them shake hands a maximum of 5 times.

So Mrs. B shook hands 6 times, which means that Mrs. B shook hands with everyone else3Now there are 1-5 times left.

If Mrs. A shakes hands 1 time.

In other words, only Mrs. B shook hands with Mrs. A.

So for couples C and D, each of them did not shake hands with Mrs. A and B, nor did they shake hands with their spouses, and each person shook hands at most 4 times, and did not shake hands 5 times, so Mrs. A's handshake cannot be 1 time, then one of the C, D couples must shake hands 1 time.

4.Let's assume that C shakes hands 1 time.

Then C only shook hands with Mrs. B.

For Mr. and Mrs. D, they can only shake hands with Mr. and Mrs. A, Mrs. B, and Mrs. C up to 4 times each, so the one who shakes hands 5 times is Mrs. C 5What has now been established is:

b: 0 times. Mrs. B: 6 times, shake hands with Mr. and Mrs. A, C and D once.

C: 1 time, shake hands with Mrs. B.

Mrs. C: 5 times, shake hands with Mr. and Mrs. A, Mr. D, and Mrs. B once, and there are 2--4 times left

If Mrs. A shakes hands 2 times.

In other words, Mrs. A, Mrs. B, and Mrs. C all shook hands once.

For Mr. and Mrs. D, Mrs. A, B, and C did not shake hands, and each of them shook hands a maximum of 3 times, and did not shake hands 4 times, which was a contradiction.

So Mrs. A's handshake is not 2 times.

Then one of the D couples must have shook hands twice.

6.You may wish to set up a d handshake 2 times.

Then D only shakes hands with Mrs. B, Mrs. C.

What has now been established is:

b: 0 times. Mrs. B: 6 times, shake hands with Mr. and Mrs. A, C and D once.

C: 1 time, shake hands with Mrs. B.

Mrs. C: 5 times, shake hands with Mr. and Mrs. A, D, Mrs. B, and Mrs. B: 2 times, shake hands with Mrs. B and Mrs. C once.

There are still the number of Mrs. A and Mrs. D that have not been determined.

The number of times is still 3,4

Mrs. A and A, B, C, D did not shake hands, and the number of times could only be 3 times, so Mrs. A shook hands 3 times, and shook hands with Mrs. B, Mrs. C, Mrs. D 4 times, and shook hands with Mrs. A, Mrs. B, Mrs. C and A respectively, Mrs. B, Mrs. C, Mrs. D, also 3 times In summary, Mr. A shook hands with his wife 3 times.

Anyone can go up to 6 times, a total of 8 people, so no one is different from the impossible, no solution.

I probably did the math, it's a very complicated probability problem!

Both shook hands 3 times, the first couple shook hands 0 and 6, the second couple 1 and 5, the third couple 2 and 4, and Mr. A and his wife both shook hands 3 times.

The first couple shakes hands 6 times (assuming the husband) shakes hands with all outsiders, and the wife does not shake hands.

The second couple shakes hands 1 time (assuming the husband) no longer shakes hands with the others (the husband has already shaken hands with the first couple), and the wife shakes hands with everyone else, for a total of 5 times (the remaining four people still have the husband of the first couple).

The third husband shakes hands twice (the husband of the first couple and the wife of the second couple have already shaken hands), and the wife shakes hands with all the rest, that is, the husband and wife of the a, a total of 4 times (the husband and wife of the first couple and the wife of the second couple).

A husband and wife shook hands three times, with the husband of the first couple, the wife of the second couple, and the wife of the third couple.

The handshake order of the first three couples in this question can be messed up, but the relationship between 0,6 and 1,5 and 2,4 can not be messed up, and the number of handshakes between husband and wife A will not change, so that the number of handshakes that others and A say is 0,1,2,3,4,5,6, which is different.

If there is no solution, according to the conditions, each person shakes hands up to 6 times, and the answer is 6543210, there is 0, that is, someone does not shake hands, then the maximum number of handshakes per person becomes 5 times, then the person who said 6 times does not exist