Tasks for movement in opposite directions. How to solve traffic problems? Methodology for solving motion problems
Math lesson in 4th grade.
Lesson topic:
"Solving problems on movement in opposite directions."
Lesson Objectives:
Learn to solve problems for movement in opposite directions;
To teach how to write inverse problems for movement in opposite directions;
Improve computing skills;
Develop attention, memory and logical thinking;
Develop skills to work in small groups;
foster a responsible attitude to educational work.
Equipment:
textbook "Mathematics Grade 4" (edited by M.I. Moro), interactive whiteboard, presentation "Movement in opposite directions", cards with values and cards for working in pairs, table "Movement".
During the classes:
1. Organizational moment.
- Good afternoon guys! I am glad to welcome you to the lesson of the queen of sciences - MATHEMATICS. I wish that the lesson brings you the joy of communicating with each other and that everyone leaves the lesson with a significant amount of knowledge. Now smile and wish each other successful work.
2. Oral account.
a) Game "Find the extra":
You need to choose the values that are used
in motion tasks.
kg, km, t, s, km/h, cm, day, m, c, h, min, m/min, km/s, m/s, dm
(on the card board).
on km, s, km/h, m, h, min, m/min, km/s, m/s
b) – What 3 groups can these units of measurement be divided into?
p/o Units of speed, time and distance.
What problems do we use these values for?
p / o For solving problems on the movement.
Are you able to solve such problems?
Now let's check.
c) Movement tasks:
slide 2
“A snail crawls at a speed of 5 m/h. How far will it travel in 4 hours?
slide 3
“A turtle will crawl 40 m in 10 minutes. How fast is a turtle crawling?”
slide 4
“A camel moves through the desert at a speed of 9 km/h. How long will it take him to cover 54 km?
slide 5
A hare runs 72 km in 3 hours. How fast is the rabbit running?
slide 6
“A pigeon flies at a speed of 50 km/h. How far can a dove fly in 6 hours?
Slide 7
The eagle flies at a speed of 30 m/s.
How long will it take him to fly 270 m?
p / o - 20 m; 4 m/min; 6 h; 24 km/h; 300 km; 9s.
3. Message of the topic and objectives of the lesson:
Today we continue to work with motion tasks
and get acquainted with the new type of tasks "Movement
in opposite directions."
4. Explanation of the new material.
Open your textbooks on page 27, find #135 and read the first problem.
Slide 8
“Two pedestrians left the village at the same time and went in opposite directions. The average speed of one pedestrian is 5 km/h, the other is 4 km/h. How far apart will the pedestrians be after 3 hours?
5 km/h 4 km/h
km
- What is known? What to find? How do we find distance?
p / o Velocity and time are known. You need to find the distance. To find the distance, you need to multiply the speed by the time.
- To find the distance, what do we find in the 1st action?
p / o Deletion speed.
- We write down the solution.
Slide 9
9 ∙ 3 = 27 (km) – distance
Answer: distance - 27 kilometers.
- Read the second problem.
Slide 10
“Two pedestrians left the village at the same time in opposite directions. The average speed of one pedestrian is 5 km/h, the other is 4 km/h. In how many hours will the distance between them be 27 km?
5 km/h 4 km/h
27 km
- What is known? What to find? How do we find time?
p / o Known speed and distance. Need to find the time. To find the time, you need to divide the distance by the speed.
- To find the time, what is the 1st action?
p / o Deletion speed.
We write down the solution.
slide 11
p / o 5 + 4 \u003d 9 (km / h) - removal speed
27:9 = 3 (h)
Answer: time is 3 hours.
- Read the third problem.
slide 12
“Two pedestrians left the village at the same time in opposite directions. After 3 hours, the distance between them was 27 km. The first pedestrian walked at an average speed of 5 km/h. How fast was the second pedestrian?
5 km/h? km/h
27 km
What is known? What to find? How do we find speed?
p / o Known distance, one of the speeds and time. Find the second speed. To find the unknown speed, you need to subtract the known speed from the total speed.
- To find an unknown speed, what is the 1st action?
p / o Deletion speed.
- We write down the solution.
slide 13
p / o 27: 3 \u003d 9 (km / h) - removal speed
9 - 5 = 4 (km/h)
Answer: speed is 4 kilometers per hour.
- Are these tasks similar?
p / o These are tasks for moving in the opposite direction.
- How are these tasks different?
p / o If the distance is unknown in problem No. 1, then in problem No. 2 it is given. But what is known in problem number 1 will become unknown in problem
№ 2.
- What are these tasks called?
p / o Reverse.
Slide 14
5. Physical education minute.
Hands to the sides - in flight (hands to the sides)
Sending a plane
Right wing forward (right turn)
Left wing forward (left turn)
One, two, three, four (jumping in place)
Our plane took off.
6.Primary fixing of the material.
Read problem #143 on page 28.
“Two skiers left the village at the same time and went in opposite directions. One of them walked at an average speed of 12 km / h, and the other - 10 km / h. In how many hours will the distance between them be 44 km? How far will each skier cover during this time?
What is known about the problem?
p / o Direction, speed and total distance.
What do you need to know?
p / o Travel time and distance that each skier will cover.
Let's make a drawing for this task.
12 km/h 10 km/h
Km? km
44 km? h
If the distance and time these skiers have in common. What do you need to know first?
p / o Overall speed.
Think about what this speed will be called if we are talking about the speed of approach in oncoming traffic?
p / o Deletion speed.
Right. We find the speed of removal, that is, how many kilometers the skiers will move away from each other in 1 hour.
Knowing the distance and speed, how to know the time?
p / o It is necessary to divide the distance by the speed of removal.
Knowing the time and speed of each skier, we can find out the distance each skier has traveled. How to do it?
p / o You need to multiply the speed by the time.
Write down the solution to this problem.
p / o 1) 12 + 10 \u003d 22 (km / h) - removal speed
2) 44: 22 = 2 (h) - time
3) 12 ˑ 2 = 24 (km) - 1 skier
4) 10 ˑ 2 = 20 (km) - 2 skier
Answer: after 2 hours, 24 km and 20 km.
7. Work on the material covered.
a) Work in pairs:
Which row solves examples faster?
Chain account:
1 desk - 480: 6 =
2nd desk - 80: 20 =
3rd party - 4 x 50 =
4 desks - 200 x 4 =
5 desk - 800: 20 =
p / o 80, 4, 200, 800, 40.
b) Work according to the textbook: No. 138 (independent work).
1 option - 1 line
10000 – 2178 ∙ 6: 4 + 267 =10000 – 13068: 4 + 267 = 10000 – 3267 +267 = 6733 + 267 = 7000
240 ∙ 3 + 4540: 20 = 720 + 227 = 947
Option 2 - line 2
487 ∙ 8 + 45270: 3: 10 = 3896 + 15090: 10 = 3896 + 1509 = 5405
560: 7 + (3820 – 850) = 80 + 2970 = 3050
c) The task of ingenuity (orally), a conversation about traffic rules (additional task).
“Two students came out of the school and went in different directions. The first went at a speed of 2 m/min, and the second - 3 m/min. In how many minutes will the distance between them be 10 meters?
p / o Solution: 1) 2 + 3 \u003d 5 (m / min) - removal speed
2) 10:5=2(min)
Answer: after 2 minutes the distance between them will be 10 meters.
When the children walked home from school, they had to follow the rules of the road.
What do you advise them?
(Children's answers.)
8. The result of the lesson:
What new did you learn in the lesson? What have you learned?
p / o Learned to solve problems for movement in opposite directions.
How fast do objects move when moving in opposite directions?
p/o Objects move at the speed of removal.
Self-esteem.
Do you think you have learned well the material of today's lesson? If yes, then we get up, and if not, we raise our right hand.
In the next lessons, we will continue to work on tasks for movement.
(Rating.)
Homework:page 27 no. 136.
- Thank you for the lesson. The lesson is over.
Individual card work
1 option. VALUES:
1. Convert to meters 45 km 40m = __________ m
2. How many meters are in 1/2 of a kilometer? ______m
3. Emphasize: which is more than 190 minutes or 3 hours?
Option 2. VALUES:
1. Convert to meters 35 km 600m = _________ m
2. How many meters are there in 1/4 of a kilometer? _______m
3. Emphasize: which is more than 130 minutes or 2 hours?
1 row
Chain account:
1 desk - 480: 6 =
2nd desk - 80: 20 =
3rd party - 4 x 50 =
4 desks - 200 x 4 =
5 desk - 800: 20 =
2 row
Chain account:
1 desk - 480: 6 =
2nd desk - 80: 20 =
3rd party - 4 x 50 =
4 desks - 200 x 4 =
5 desk - 800: 20 =
3 row
Chain account:
1 desk - 480: 6 =
2nd desk - 80: 20 =
3rd party - 4 x 50 =
4 desks - 200 x 4 =
5 desk - 800: 20 =
kg km t s km/h cm day m c h min m/min km/s m/s dm slide 2
A snail crawls at a speed of 5 m/h. How far will she cover in 4 hours? 5 ∙ 4 = 20 (m)
A turtle crawls 40 m in 10 minutes. How fast does a turtle crawl? 40: 10 = 4 (m/min)
A camel moves through the desert at a speed of 9 km/h. How long will it take him to cover 54 km? 54: 9 = 6 (h)
A hare runs 72 km in 3 hours. How fast is the rabbit running? 72: 3 = 24 (km/h)
A dove flies at a speed of 50 km/h. How far can a dove fly in 6 hours? 50 ∙ 6 = 300 (km)
An eagle flies at a speed of 30 m/s. How long will it take him to fly 270 m? 270: 30 = 9 (s)
MOVING IN OPPOSITE DIRECTIONS? How far apart will the pedestrians be after 3 hours? 5 km/h 4 km/h
MOVEMENT IN OPPOSITE DIRECTIONS 1) 5 + 4 \u003d 9 (km / h) - REMOVAL SPEED 2) 9 x 3 \u003d 27 (km) Answer: 27 kilometers.
MOVEMENT IN OPPOSITE DIRECTIONS 27 km What was the speed of the second pedestrian? 5 km/h?
MOVEMENT IN OPPOSITE DIRECTIONS 1) 27: 3 = 9 (km / h) - REMOVAL SPEED 2) 9 - 5 = 4 (km / h) Answer: 4 kilometers per hour.
MOVEMENT IN OPPOSITE DIRECTIONS 27 km In how many hours will the distance between them be 27 km? 5 km/h 4 km/h
MOVEMENT IN OPPOSITE DIRECTIONS 1) 5 + 4 = 9 (km / h) - REMOVAL SPEED 2) 27: 9 = 3 (h) Answer: after 3 hours.
You are already familiar with the quantities "speed", "time", "distance" and know how these quantities are related to each other. We have already solved problems in which objects moved in the same direction or towards each other. Now consider tasks when objects move in opposite directions. And let's get acquainted with the concept of "deletion speed".
Two pedestrians left the village at the same time and went in opposite directions. The average speed of one pedestrian is 5 km/h, the other is 4 km/h. How far apart will pedestrians be after 3 hours (Fig. 1)?
Rice. 1. Illustration for problem 1
To find the distance that two pedestrians will be in three hours, you need to find out how much distance each will walk during this time. To find how far a pedestrian has traveled, you need to know his average speed and his travel time. We know that the pedestrians left the village at the same time and were on the road for three hours, which means that each of the pedestrians was on the road for three hours. We know the average speed of the first pedestrian - 5 km/h and we know his travel time - 3 hours. We can find the distance traveled by the first pedestrian. Multiply his speed by his travel time.
We know the average speed of the second pedestrian - 4 km/h and we know his travel time - 3 hours. Multiply his speed by his travel time to get the distance he traveled:
Now we know the distance that each of the pedestrians has traveled, and we can find the distance between the crossings.
In the first hour, one pedestrian will move 5 km away from the village, in the same hour the second pedestrian will move 4 km away from the village. We can find the speed of removal of pedestrians from each other.
We know that in every hour the pedestrians moved away from each other by 9 km. We can find out how far they will move away from each other in three hours.
Multiplying the speed of removal by the time, we found the distance between pedestrians.
Answer: In 3 hours the pedestrians will be 27 km apart from each other.
Two pedestrians left the village at the same time in opposite directions. The average speed of one pedestrian is 5 km/h, the other is 4 km/h. After how many hours will the distance between them be 27 km (Fig. 2)?
Rice. 2. Illustration for problem 2
To find the time of movement of pedestrians, you need to know the distance and speed of pedestrians. We know that for every hour one pedestrian moves away from the village by 5 km, and another pedestrian moves away from the village by 4 km. We can find their removal rate.
We know the speed of removal and we know the whole distance - 27 km. We can find the time after which the pedestrians will move away from each other by 27 km, for this we need to divide the distance by the speed.
Answer: in three hours the distance between the crossings will be 27 km.
Two pedestrians left the village at the same time in opposite directions. After 3 hours the distance between them was 27 km. The first pedestrian walked at a speed of 5 km/h. What was the speed of the second pedestrian (Fig. 3)?
Rice. 3. Illustration for problem 3
To find out the speed of the second pedestrian, you need to know the distance he traveled and his travel time. To find out how far the second pedestrian walked, you need to know how far the first pedestrian walked and the total distance. We know the total distance. To find the distance traveled by the first pedestrian, you need to know his speed and his travel time. The average speed of the first pedestrian is 5 km/h, his travel time is 3 hours. If the average speed is multiplied by the travel time, we get the distance that the pedestrian has traveled:
We know the total distance and we know the distance that the first pedestrian walked. We can now find out how far the second pedestrian traveled.
Now we know the distance that the second pedestrian walked and the time spent on the way. We can find its speed.
Answer: the speed of the second pedestrian is 4 km/h.
We learned to solve problems for movement in opposite directions and got acquainted with the concept of “removal speed”.
Homework
Bibliography
- Mathematics: textbook. for the 4th class. general education institutions with Russian. lang. learning. At 2 p.m. Part 1 / T.M. Chebotarevskaya, V.L. Drozd, A.A. joiner; per. with white lang. L.A. Bondareva. - 3rd ed., revised. - Minsk: Nar. asveta, 2008. - 134 p.: ill.
- Mathematics. Textbook for 4 cells. early school At 2 o'clock / M.I. Moro, M.A. Bantova. - M.: Education, 2010.
- Mathematics: textbook. for the 4th class. general education institutions with Russian. lang. learning. At 2 p.m. Part 2 / T.M. Chebotarevskaya, V.L. Drozd, A.A. joiner; per. with white lang. L.A. Bondareva. - 3rd ed., revised. - Minsk: Nar. asveta, 2008. - 135 p.: ill.
- Mathematics. 4th grade. Textbook at 2 hours Bashmakov M.I., Nefedova M.G. - 2009. - 128 p., 144 p.
- Internet portal Slideshare.net ().
- Internet portal For6cl.uznateshe.ru ().
- Internet portal Poa2308poa.blogspot.com ().
§ 1 Movement in opposite directions
In this lesson, we will get acquainted with the problems of movement in opposite directions.
When solving any movement problem, we are faced with such concepts as "speed", "time" and "distance".
Velocity is the distance an object travels per unit of time. The speed is measured in km / h, m / s, etc. Denoted by the Latin letter ʋ.
Time is the time it takes an object to cover a certain distance. Time is measured in seconds, minutes, hours, etc. Denoted by the Latin letter t.
Distance is the distance traveled by an object in a certain amount of time. The distance is measured in kilometers, meters, decimeters, etc. Designated with the Latin letter S.
In motion problems, these concepts are interrelated. So, to find the speed, you need to divide the distance by the time: ʋ = S: t. To find the time, you need to divide the distance by the speed: t = S: ʋ. And to find the distance, the speed is multiplied by the time: S = ʋ · t.
When solving problems for movement in opposite directions, another concept is used - "removal speed".
Removal rate is the distance that objects are removed per unit of time. Denoted ʋud..
To find the speed of removal, knowing the speed of objects, you need to find the sum of these speeds: ʋud. = ʋ1 + ʋ2. To find the removal rate, knowing the time and distance, it is necessary to divide the distance by the time: ʋsp. = S: t.
§ 2 Problem solving
Consider the relationship between the concepts of "speed", "time" and "distance" when solving problems for movement in opposite directions.
PROBLEM 1. A truck and a car left the bus station for different directions. In the same time, a truck traveled 70 km, and a car - 140 km. How fast was the car moving if the speed of the truck was 35 km/h?
Depict the movement of freight and passenger car on the diagram.
The speed of the truck will be denoted by the letter ʋ1 = 35 km/h. The speed of a passenger car is denoted by the letter ʋ2 = ? km/h The travel time is denoted by the letter t. Distance traveled truck- letter S1 = 70 km. The distance traveled by the passenger car is S2 = 140 km.
Let's take a look at the first option.
Since, in order to find an unknown speed, it is necessary to know the distance traveled by a passenger car, and it is known and equal to 140 km, and to know the time of movement, which is not indicated in the conditions of the problem, it is necessary to find this time. From the condition of the problem, we know the distance that the truck S1 = 70 km has traveled and the speed of the truck ʋ1 = 35 km/h. Using this data, we can find the time. t = S1: ʋ1 = 70: 35 = 2 hours. Knowing the time and distance traveled by the car, we can find out the speed of the car, since ʋ2 = S2: t = 140: 2 = 70 km/h. We found that the speed of the car is 70 km/h.
Let's consider the second option.
Since, in order to find the unknown speed, it is necessary to know the speed of the truck, from the conditions of the problem it is known, and the speed of removal, which is not specified by the conditions of the problem, then it is necessary to find the speed of removal. To find the speed of cars moving away, you can divide the distance that both cars have traveled by the time. ʋud. = S: t . The distance traveled by both cars is equal to the sum of the distances S1 and S2. S = S1 + S2 = 70 + 140 = 210 km. Time can be found by dividing the distance a truck has traveled by its speed. t = S1: ʋ1 = 70: 35 = 2 hours. So, ʋud. = S: t = 210: 2 = 105 km/h. Now, knowing the speed of removal, we can find the speed of the car. ʋ2 = ʋsbl. - ʋ1 = 105 - 35 = 70 km/h. We found that the speed of the car is 70 km/h.
PROBLEM 2. Two people left the village at the same time in different directions. One was moving at a speed of 6 km/h, the speed of the other was 5 km/h. How many hours will it take for the distance between them to become 33 km?
Let's depict the movement of people on the diagram.
The speed of the first person will be denoted by the letter ʋ1 = 5 km/h. The speed of the second person will be denoted by the letter ʋ2 = 6 km/h. The distance they traveled is denoted by the letter S = 33 km. Time - letter t = ? hours.
To answer the question of the problem, it is necessary to know the distance and the speed of removal, since t = S: ʋud .. Since we know the distance from the condition of the problem, we need to find the speed of removal. ʋud. = ʋ1 + ʋ2 = 5 + 6 = 11 km/h. Now knowing the removal rate, we can find the unknown time. t \u003d S: ʋud \u003d 33: 11 \u003d 3 hours. We get that it took 3 hours for the distance between people to become 33 km.
PROBLEM 3. Two trains simultaneously started moving in opposite directions from different stations, the distance between which is 25 km. One was moving at a speed of 160 km/h. How far apart will the trains be after 4 hours if the speed of the other train is 130 km/h?
Let's show the movement of trains on the diagram.
The speed of the first train is denoted by the letter ʋ1 = 130 km/h. Let's denote the speed of the second train as ʋ2 = 160 km/h. The distance between the stations is denoted by the letter Sm = 25 km. Time - letter t = 4 hours. And the desired distance - the letter S =? km.
To answer the question of the problem, it is necessary to know the distance between stations, the distance traveled by the first train, and the distance traveled by the second train, since S = Sm + S1 + S2. The distance between the stations is known from the condition of the problem, but the distances S1 and S2 are not, but they can be found using other data of the problem. However, the desired distance can be found in a more rational way, namely by adding the distance between stations and the total distance traveled by both trains, since S = Sm + Sob .. Since the distance between stations is known from the condition of the problem, it is necessary to find the total distance. To do this, you need to multiply the time by the removal rate. Sb \u003d t ʋsp. And the speed of removal is equal to the sum of the speeds of the trains. ʋud. = ʋ1 + ʋ2 = 160 + 130 = 290 km/h. Now we can find the total distance Sb = t · ʋsp. = 4 · 290 = 1160 km. Knowing the total distance, we can find the desired distance. S \u003d Sm + Sb \u003d 25 + 1160 \u003d 1185 km. We got that after 4 hours the distance between trains will be 1185 km.
§ 3 Brief summary on the topic of the lesson
When solving problems for movement in opposite directions, it should be remembered that the following conditions are met in problems of this type:
1) objects start their movement simultaneously in opposite directions, which means they are on the way for the same amount of time; time is denoted by the Latin letter t = S: ʋud;
2) distance S is the sum of all distances specified by the conditions of the problem;
S \u003d S1 + S2 + Smi S \u003d ʋsp. t;
3) objects are removed at a certain speed - the speed of removal, denoted by the Latin letter ʋud. = S: t or ʋud = ʋ1 + ʋ2, respectively
ʋ1 = S1: t and ʋ2 = S2: t.
List of used literature:
- Peterson L.G. Mathematics. 4th grade. Part 2. / L.G. Peterson. – M.: Yuventa, 2014. – 96 p.: ill.
- Mathematics. 4th grade. Methodical recommendations for the textbook of mathematics "Learning to learn" for grade 4 / L.G. Peterson. – M.: Yuventa, 2014. – 280 p.: ill.
- Zak S.M. All tasks for the mathematics textbook for grade 4 L.G. Peterson and a set of independent and control works. GEF. – M.: UNVES, 2014.
- CD-ROM. Mathematics. 4th grade. Lesson scenarios for the textbook for part 2 Peterson L.G. – M.: Yuvent, 2013.
Used images:
>> Lesson 27. Moving in opposite directions
1. From points A and B, the distance between which is 6 km, 2 pedestrians left simultaneously in opposite directions. The speed of the first pedestrian is 3 km/h and the speed of the second pedestrian is 5 km/h. How does the distance between them change in 1 hour? What will it be in 1 hour, 2 hours, 3 hours, 4 hours? Will there be a meeting? Finish the drawing and complete the table. Write down the formula for the dependence of the distance between pedestrians d on the time of movement t.
2. Solve the problem in two ways. Explain which one is better and why?
From two cities located at a distance of 65 km from each other, two cars left simultaneously in opposite directions. One of them was walking at a speed of 80 km / h, and the other - 110 km / h. How far apart will the cars be 3 hours after leaving?
3. 2 boats set sail simultaneously from one pier in opposite directions. After 3 hours, the distance between them became 168 km. Find the speed of the second boat if it is known that the speed of the first boat is 25 km/h.
4. Compose mutually inverse problems according to the schemes and solve them:
5. Come up with a problem for movement in opposite directions, in which you need to find:
a) the speed of one of the moving objects;
b) the initial distance between them; c) travel time.
6. From two cities, 1680 km apart, 2 trains left at the same time towards each other. The first train travels all this distance in 21 hours, and the second train in 28 hours. After how many hours will the trains meet?
7. Choose the expressions corresponding to this task and put a "+" sign next to it. Cross out the rest of the expressions.
8. Solve the equations:
a) (a 16 - 720): 30 \u003d 400 - 392;
b) (95 - 380: b) + 35 = 16 + 94.
9. Variables x and y are related: y \u003d (x - 2) x + x 3.
| X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| at |
What do you notice? Try to express the relationship between the variables x and y with a simpler formula.
10. a) Decipher the statement of the famous American scientist and entrepreneur Thomas Edison, the author of over 1000 inventions!
b) Write sequentially the remainders of the division of these numbers in empty cells - and you will find out the years of life of Thomas Edison:
1) 76: 15 4) 322: 35 7) 19 203: 96
2) 176: 24 5) 470: 67 8) 74 429: 92
3) 148: 16 6) 609: 75
11. The icebreaker made its way through the ice for 3 days. On the first day he swam the whole way, on the second day he swam the rest of the way, and on the third day he swam the remaining 90 km. What distance did the icebreaker sail in 3 days of travel? How many kilometers did he swim on the first and second day?
12. Make a program of action and calculate:
a) (600: 30 - 7) 5 - (24 - 4 4) (32: 16) + 60: 4 10;
b) 500 - (28 5 + 25 4 - 120: 2): 6 - (28: 14 + 420: 140) 30.
thirteen*. Old task.
One man was asked how much money he had. He replied: “My brother is three times richer than me, my father is three times richer than my brother, my grandfather is three times richer than my father, and we all have exactly 1000 rubles. So find out how much I have of money".
fourteen*. Game "Find the unknown picture".
Peterson Ludmila Georgievna. Mathematics. 4th grade. Part 2. - M.: Yuventa Publishing House, 2005, - 64 p.: ill.
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