Can anyone help me solve a junior high school math problem?

The shortest time for all 100 people to arrive at a certain place is that people keep walking, cars keep driving, and finally 100 people arrive at the same time.

If the time of the first batch of people is t, the distance between the first batch and the second batch is 5[t+(55-5)t (55+5)].

Then the difference between the second batch of cars and the third batch of cars is 5[t+(55-5)t (55+5)].

Then the distance between the third batch of cars and the fourth batch of cars is 5[t+(55-5)t (55+5)].

The time of the fourth batch of buses is also t

3*5[t+(55-5)t/(55+5)]+55t=33

t=2 5, total time: 2 5 + [33-55*(2 5)] 5=13 5

Obviously, getting 25 people to their destination and back to pick them up is not the fastest.

The fastest is: 25 people in the car go to the halfway point A, put down the 25 people and return, meet 75 people at the point B, carry 25 people, go to C, and so on, and finally 100 people arrive at the same time.

You can think of it this way:

First use this car to send 25 people, the remaining 75 people at the same time on foot, the car arrives at the destination immediately after returning, after meeting the 75 people on the way, then use the car to send 25 people, the remaining 50 people are still walking, the car arrives at the destination immediately after returning, after meeting the 50 people on the way, then use the car to send 25 people, the remaining 25 people are still walking, the car arrives at the destination immediately back, after meeting the last 25 people on the way, use the car to send them to the destination, so that the arrangement takes the shortest time.

Let cm=x, then cn=14-x, and amc and bnc are both right-angled triangles, which are according to the title.

6^2+x^2=8^2+(14-x)^2

Solution x = 8 km.

Answer: Station C should be built at a distance of 8 kilometers from point m.

Do you understand?

Hope it helps.

o(∩_o~

Let's look at the law of 2 to the nth power: 2, 4, 8, 6, 2, 4, 8, 6....

That is to say, every 4 numbers is cycled, 64 4 = 8, just divided, it is the 4th number, and the end is 6

Or let's solve it another way:

This formula is 1x3x5x7x9 ......There is a 5 in it, and multiplying 5 by any odd number will result in 5, so the final structure of this equation is 5+1=6 and the single digit is 6

The law of the nth power single digit of 2 is 2,4,8,6,2....A cycle of every 4.

Then the single digit of 2 64 is 6

This formula is 1x3x5x9 ......There is a 5 in it, and multiplying 5 by any odd number by a single digit results in 5, so the last single digit result of this equation is 5+1=6 and the single digit is 6

To do this kind of questions to be good at finding the rules, generally speaking, the special situation is the situation that is easy to test, similar to the situation that 5 and even numbers exist at the same time, and there is a situation where all the digits are 1, so experience it well, and pay attention to it in ordinary study.

Yes, the law of the nth power is a bit of a cycle, then all plus 1 is a cycle, and they are all odd numbers. N odd numbers are multiplied or odd numbers, and there are 5 times multiplied. 5 times the odd number of units is 5, (2-1) (2+1) (2 2+1) (2 3+1) (2 4+1) ...

2 32+1) is 5. Adding 1 is 6.

(a-1)'2 is a positive number or 0, i.e. a non-negative number, |b+3|It's also a positive number or 0, and it's also a non-negative number, so how can two positive numbers be opposites, so they can only be 0, because only the opposite number of 0 is 0. So.

a-1=0 b+3=0 then a=1 b=3

So a 3 + a 2b-2a = 1-3-2 = -4

Because (a-1) 2 is greater than or equal to 0, |b+3|Greater than or equal to 0, (a-1) 2 and |b+3|are inverse numbers to each other.

So (a-1) 2 is equal to 0, ,|b+3|is equal to 0, so a is 1 and b is -3

So a 3 + a 2b - 2a is equal to 0

Because (a-1) 2 is the same as |b+3|Numbers are opposite to each other, so there is (a-1) 2+|b+3|=0

So a-1=0 b+3=0 (This step should have been learned, because (a-1) 2>=0 ,|.)b+3|>=0, and the sum of the two is equal to 0, so both are equal to 0).

then a=1 b=-3

A 3+(A 2)B-2A=1+1*(-3)-2*1=-4A 3+A (2B)-2A=1+1+1-2=0 I don't know what the original question is, please take a look at it yourself, AB should be able to solve it.

Because the square and absolute values are both greater than or equal to 0.

So a-1 = 0 and b+3 = 0

So a=1 b=-3

So a 3 + a 2b-2a = 1-3-2 = -4

There's the answer to this one

This ** is good.