How many minutes does it take for a clock to coincide with the hour hand from four o clock? Equation

Here's the equation:

20 (1-1 12) = 20 * 12 11 = 21 and 9 11 points.

At 4 o'clock, the minute hand and the hour hand are 20 squares apart, the minute hand goes 1 square in a minute, and the hour hand goes 1 12 squares in a minute, according to: chasing time = chasing distance speed difference can be substituted.

A: From four o'clock sharp, after another 21 and 9 11 minutes, the minute and hour hands coincide for the first time.

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At 4 o'clock, the minute hand and the hour hand are 20 squares apart, the minute hand goes 1 small grid in a minute, and the hour hand goes 1 large grid in an hour, that is, a 5 small square, according to: chasing time = chasing distance speed difference can be substituted.

20 (1-5 60) = 20x12 11 = 21 and 9 11 minutes.

So from four o'clock sharp, it will take another 21 and 9 11 minutes for the minute and hour hands to coincide for the first time.

After that, the minute hand coincides with the hour hand, i.e. a difference of 60 stops.

60 (1-5 60) = 60x12 11 = 65 and 5 11 minutes.

After the first coincidence.

After another 65 and 5 11 minutes coincided for the second time, the first.

Third, the fourth time, and so on.

12 grids, 360 degrees, 30 degrees each, i.e. 6 uses radians to calculate. The angular velocity of the minute hand is 2 per hour and the hour hand is 6 per hour. The coincidence means that the difference between the angles of the two is an integer multiple of 2, and the column equation is:

4 6+ t 6=2 t+2k so that k = 1,2,3,- can be solved to coincide moments.

Hello, you can set the two hands to meet after x minutes.

4+x/60)*5=x

The solution is x=240 11Hope.

We know that the minute hand turns 360 °, the hour hand turns 30 °, that is, the minute hand turns 1 °, the hour hand turns 1 12 °, that is to say, the minute hand turns 1 °, you can catch up with the hour hand 11 12 °, at four o'clock, the hour hand is ahead of the minute hand 120 °, then the minute hand turns 120 (11 12) = 1440 11 °, just catch up with the hour hand, at this time the two coincide, the angle is 1440 11 =

Set 4 points vertically and ax minutes, and the minute hand and the hour hand meet.

1-1/12)x=20

x=20*12/11

x=240/11

So the minute hand meets the hour hand after 240 11 minutes.

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6 [Brother 120.]

1440 11 degrees.

The minute hand of the clock starts at 4 o'clock sharp, and when it turns 1440 11 degrees, the minute hand coincides with the hour hand.

The minute hand rotates 6 degrees per minute, and the angle at which the hour hand turns per minute is 1 12 of the minute hand, so the hour hand rotates in degrees per minute.

Turning this problem into a catch-up problem, the distance is 120 degrees, so t=120 minutes is derived from time = distance speed.

So after 240 11 minutes, that is, 21 minutes and 9 11 minutes, the hour hand and the minute hand coincide for the first time.

Because the minute hand travels once per hour, that is, 60 divisions, and the hour hand only travels 5 divisions, the minute hand speed can be set to 60 divisions per hour, and the hour hand speed is 5 divisions per hour. This is a catch-up question, according to the meaning of the question, the requirement is how long does it take for the minute hand to catch up with 20 divisions (the distance between the hour and minute hand at 4 o'clock) Time = distance Speed difference = 20 (60-5) = 4 11 hours = 60 * 4 11 minutes = 21 and 9 11 minutes. "

The minute hand of the clock coincides with the hour hand when it rotates clockwise from 4 o'clock sharp.

Solution: It can be regarded as a chase problem, the hour hand rotates every minute, the minute hand rotates 6 degrees per minute, at 4 o'clock, in the clockwise direction, the hour hand is in front, the minute hand is behind, the degree of the angle between the two needles is 120°, when the two needles coincide, the angle of the minute hand rotation is equal to the angle of the rotation of the hour hand plus the angle of the difference between the original two hands.

If the two hands coincide at 4 points x minute, there will be 6x

The solution is x = i.e. the two needles coincide at 4 o'clock.

At this point, the number of degrees the minute hand turns clockwise is: 6 = degrees.

Solution: The minute hand rotates 360° in one hour and 6° per minute;

The hour hand rotates 360 12 = 30 ° per hour, and 30 60 = per minute;

The hour hand is at 4 o'clock, and the hour hand is 30*4=120° ahead of the score hand;

Let x minutes elapse and the clock coincide.

Then there is: 6x=120+

The solution is x=240 11

Degree of rotation = 6 * x = 6 * 240 11 = 1440 11 = A: Thus the minute hand is in the hour, and the minute hand and the hour hand coincide.

Set the minute hand speed to 60 divisions, the hour hand speed to 5 divisions, the hour to 4 o'clock, and the hour hand to be 20 divisions before the minute hand.

The first coincidence.

4:21:21 p.m.

Let the hour and minute hands meet for the first time after x minutes.

The hour hand travels 6 degrees per minute, and the minute hand travels 6 degrees per minute.

Solution x=80 3

We know that the minute hand turns 360° and the hour hand turns 30°, that is, the minute hand turns 1°, and the hour hand turns 1 12°, that is to say, the minute hand turns 1°, and it can catch up with the hour hand 11 12°, and at four o'clock, the hour hand is 120° ahead of the minute hand, then 120 (minutes.

Think of it as a pursuit-type problem.

Solution: Let the first coincide after x minutes.

6x = + 120

x = 11 240

This is the best way for me to understand!!

Set 4 o'clock x minutes, the hour and minute hands coincide for the first time.

So 120

So x 240 11

The distance problem on the clock face.

Minute hand, rotation per minute: 360 60 = 6 degrees.

Hour hand, rotation per minute: 360 12 60 = degrees.

At 4 o'clock, the minute hand is behind the hour hand: 4 12 360 = 120 degrees The first time the two hands coincide, required: 120 (minutes.

The hour hand moves 30 degrees per hour, every minute; The minute hand travels 6 degrees per minute. The difference in speed between the two stitches is in degrees.

At four o'clock, the minute hand is 4 * 30 = 120 degrees behind the hour hand, so the minute hand of the clock passes from the position of four o'clock.

It takes 120 minutes for the minute hand to coincide with the hour hand.

30×4÷（

21 and 9 11 points.

The minute hand of the clock starts at four o'clock sharp, and it takes 21 and 8 11 minutes for the minute hand to coincide with the hour hand.

Every hour the hour hand goes 30 degrees Hour hand speed v1 = 30 degrees Hour Minute speed v2 = 360 degrees Hour At 4 o'clock The hour hand minute angle difference is 120 degrees then let t time coincide after the two Minute hand overturns - 120 = Hour hand rotation degrees The equation is 360 degrees Hour * t - 120 degrees = 30 degrees Hour * t solution gives t = 4 11 hours.

The clock is divided into 60 parts, each at an angle of 6°, the minute hand at 6° per minute, and the hour hand at 4 o'clock per minute, at an angle of 120°

Coincided after 240 11 minutes.

This question can be done as a chase problem. Solution: Suppose that the hour and minute hands coincide for the first time after x minutes.

From the title: 360 60*x=4*30+30 60xx=240 11

Answer: After 240 11 minutes, the hour and minute hands coincide for the first time.

At 4 o'clock, the hour hand is at the position of 4, the minute hand is at the position of 12, the distance between the hour hand and the minute hand is 120 °, the speed of the hour hand is 6° per minute, and the minute hand is 6° per minute.

6 t - t =120

t = 240/11

The idea of the second floor is right, but the result is not right, it should be.

Let the minute hand go x degrees, and the minute hand coincides with the hour hand.

Because the minute hand goes 360 degrees, and the hour hand goes 30 degrees.

So the minute hand goes 1 degree, and the hour hand goes 1 12 degrees.

Solve the equation: x=360+x(1 12).

The solution gives x = 392 degrees and 44 minutes, that is, degrees.

At 4 o'clock.

The minute and hour hands differ by 20 stops.

The minute hand travels 1 block per minute, and the hour hand travels 5 60 = 1 12 blocks per minute, and the surface is 60 divisions.

See the process of chasing.

Speed difference = 1-1 12 = 11 12 square points.

Distance difference = 20 blocks.

Time required = 20 (11 12) = 240 11 minutes 21 minutes 49 seconds.

After x minutes have passed, the minute hand coincides with the hour hand.

6x=4*30+30\60*x

x=240\11

60 minutes! An hour is not like that!

How many degrees does the minute hand of the clock pass from the position of four o'clock before the minute hand coincides with the hour hand?