A math problem to explain the process and why

That is, every 5 160 = hours, a passenger car meets a van.

Kilometers means that between each encounter, the bus travels a kilometer.

Between the three encounters, the bus has to travel a kilometer.

Therefore, when the passenger car meets the first truck, the distance from the previous station cannot exceed one kilometer.

This ensures that the buses meet three wagons between the two stops.

When the bus departed, the trucks on the way went separately.

60 km, 55 km, 50 km, 45 km, 40 km, 35 km, 30 km and one just departed.

At this time, the passenger cars are separated from them.

Kilometers (followed by pushed but non-existent kilometers).

Passenger cars are needed to meet them separately.

160...95 160 hours.

During these periods, the passenger cars traveled separately.

160...9500 160 km.

Calculate them and get.

The corresponding location of the 11 stations is:

Proofread each of the 18 encounters to see if they met the requirements.

You can get an encounter that meets the requirements.

The eighth encounter, near the fifth station, kilometer.

The seventeenth encounter, near the ninth station, kilometer away.

Because there was no nineteenth encounter.

So, the answer is that between the fifth and sixth stations, there are three encounters.

As can be seen from the question, the exchange ratio is 3:5:8

Solution: Let's exchange Mickey Mouse with the robot cat for 8x and Monkey King for 8y.

Therefore, the equation 8x+8y=88, then x+y=11 solution: let Sun Wukong exchange 3x with the robot cat and 3z with Mickey Mouse.

So we can get the formula 3x+3z=39, then x+z=13 can also get y+z=18

So we get the equation {x+y=11

x+z=13

y+z=18

x=3, y=8, z=10

So Mickey Mouse exchanged 24 with Monkey King and 64 with the robot cat.

Analysis: Suppose Monkey King takes out x Xiantao and swaps cookies with the robot cat; Then Sun Wukong took out (39-x) Peach and exchanged Mickey Mouse for bubble gum.

The robot cat pulls out a y-cookie and swaps bubble gum with Mickey Mouse; Then Mickey Mouse takes out 8 5Y bubble gum and swaps cookies with the robot cat.

According to the question, two equations about x and y can be listed as follows.

5/3x+y=90

8/3(39-x)+8/5y=88

to solve the values of x and y.

Peach: Cake: Bubblegum = 3:5:8

Let an intermediate equivalent A = 3 Peach = 5 Cookies = 8 Bubble Gum then Sun Wukong has 13 A, the robot cat has 18 A, and Mickey Mouse has 11 A to exchange items X, Y, Z.

x+y=13, y+z=18, z+x=11, get, x=3, y=10, z=8

That is, Monkey King has 8 A's and the robot cat has 3 A's, so Sun Wukong has 64 bubblegum and the robot cat has 24 bubblegum ---

This topic is wrapped to death, and children will not be allowed to do this kind of problem in the future.

Standard 1 hour = alarm 59 minutes and 30 seconds, alarm clock 1 hour = watch 1 hour and 30 seconds, standard 24 hours = watch x time.

Substitution: The alarm clock is 24 * 30 = 720 seconds slower, that is, the alarm clock has gone for 23 hours and 48 minutes, and the watch has gone for more seconds than the alarm clock = 11 minutes and 54 seconds.

So the watch went for 23 hours, 59 minutes and 54 seconds. That is, the watch time is 11:59:54.

Set the alarm clock line to share the watch line x minutes, and the alarm clock line 60 minutes watch line minutes, so x

Solve x 59 again.

As a result, the hourly watch is 60 59 minutes and seconds slower than the standard time, and the 24-hour watch is second slower than the standard time.

At 12 o'clock tomorrow afternoon, Mr. Wang's watch will show the time at 11:59:54.

When the watch is at a certain time and a certain minute, after a period of time, if the hour hand just goes to the position of the original minute hand, and the minute hand just goes to the position of the original hour hand, that is, the position of the two needles is interchanged, because the minute hand and the hour hand are not distinguishable, so the time of the time can not be correctly judged when the two hands are interchanged.

The position of the two needles is interchanged, when the hour hand and the minute hand go a total of 60 squares, because the hour hand goes 1 square, and the minute hand goes 12 squares, so the time interval between the two needle positions is 60 12 (12 + 1) = 55 and 5 13 (minutes), which can appear at more than 12 o'clock to more than 1 o'clock, more than 1 o'clock to more than 2 o'clock, and more than 2 o'clock to more than 3 o'clock ......From 10 o'clock to 11 o'clock at night, there are 11 times.

The same can calculate the two needle position interchange hour hand, when the minute hand is in the same grid, the number of times that the two needles can be exchanged is respectively, so the number of times that the correct time can not be distinguished is (11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) 2 = 132 times.

Note: The title only asks us to count the number of times we can't tell the time, so there is no need to calculate the specific time.

Think of it this way: because when the hour hand and minute hand are reversed, we get the time wrong, so that means we can't get it wrong except when the hour hand coincides with the minute hand. That is, at 1 o'clock in the morning, in addition to the coincidence, we will get it wrong 11 times, at 2 o'clock, in addition to the one that has been counted at 1 o'clock and the coincidence, there will be 10 times to get wrong, and so on, in the morning there will be 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 (times), and then 2, that is, it is a day.

This is a question in junior high school?。。 Judging by the angle of counting ... Anyway, after a small format 3 degrees...

There is a ratio of degrees between the minute hand and the hour hand... The minute hand is 12 degrees and the hour hand is 1 degree. Just make a column equation...

First use one needle as a minute to make an equation, then another needle as a minute hand to make a first equation... It would be impossible to judge that both equations are in agreement... It should be regarded as chasing the problem...

It can be discussed one by one. A needle is between 12 and 1. Another one is casual...

And so on... It's not clear to say that ...

According to the title: the minute hand goes one circle, and the hour hand goes 1 8 times, that is, the minute hand goes 360 degrees, and the hour hand goes 360 8 = 45 degrees. Therefore, the minute hand has traveled 360-45 degrees more than the hour hand in 80 minutes = 315 degrees, that is, the minute hand has traveled more than the hour hand per minute:

315 80 degrees. When the minute hand coincides, when it coincides again, the minute hand and the hour hand are perpendicular twice, so the minute hand and the hour hand are at right angles for the third time, that is, when the hour and minute hands coincide again, they are at right angles. When the minute hand coincides again, the minute hand moves 360 degrees more than the hour hand, and when it is at right angles for the third time, the minute hand moves 360 + 90 = 450 degrees more than the hour hand.

So, the time required is 450 (315 80) = 800 7 minutes.

As can be seen from the title, the angular velocity of the minute hand is 360 80 = degrees of minutes, and the angular velocity of the hour hand is 1 8 degrees, that is, degrees of seconds.

When the minute hand coincides with the hour hand, the minute hand is at right angles to the hour hand at an angle of 90 degrees and 270 ,..degrees, respectivelyk*180-90，..

Therefore, at a right angle for the third time, the angle at which the minute hand is ahead of the hour hand is: 3 * 180 90 450 degrees.

Depending on the speed of the minute and hour hands, the required time can be obtained:

450 (min.)

The best way to do this is to enumerate 1 4 9 cumulative 3 two digits 4 -9 12+3=15 three digits 10 -31 66+15=81 four digits 32 -99 272 + 81 = 353 and 612-353=259=5*51+

Let the number of parts in this batch be x

The ratio of time taken by A and B alone is 4:5. Rule.

A's velocity is 5 4 of B's.

At the time of completion, A machined 5 9 parts.

then (5 9-3 7) x = 56

Solution: The number of parts in this batch is x=441.

The time required is 4:5, which means that their speed is 5:4. The speed of A is 5x, and the speed of B is 4x.

According to the title, when completing the task, A processes 3/76 more than B, so the time ratio of processing a batch of parts A and B alone is 4:5, so A processes 4 9 56 (4 9-3 7) 56 (1 63) 3528 The speed here refers to the number of parts completed per unit time.

Good luck with your studies!

Hope it helps.

Thank you!

Set both High Oak Bush and Capacity to 1

Let the crack leakage velocity be x, the position is known from the bottom of the pool y, the water inlet speed of the stove inlet pipe is 1 50, and the water outlet speed of the outlet pipe is pure as 1 60

After considering the cracks, make a list of equations.

50y+(1-y)/(1/50-y)=801-y)/(1/60+x)+60y=

The deformation is: x=1 50-(1-y) (80-50y)x=(1-y) (

Eliminate x to give 11 300 = (1-y) (80-50y) + (1-y) (

After the division is passed, it becomes a one-dimensional quadratic equation, and the solution is y=2 5

The answer is that the speed of B is 7

Let the first meeting between C and B be t hours. The distance traveled by A is 9 t, B is 125-63 t, C is 63 t, and B has a speed of (125 t-63).

Then after the first encounter with A, A walked 63t*2*9 (9t+63t), which is 7 4 times that of 9t, and B also walked 7 4 times that of (125-63t). At this time, the distance between A and B is 125-(7 4)*125-54t), and so on.

125-7/4*（125-54t)-7/4*（125/t-63+9)*[125-7/4*（125-54t)]*t/125=45

t=25 14, and B's velocity is 7

When A is met for the third time, the distance between A and B is 45-45 (63+7)*(7 4)*(7+9)=27

When A and B are 20 kilometers apart, it is actually when C and B collide and return.

27/70*(63*2)-(9+63)*(27-20)/(7+9)= km