A junior high school math problem junior high school math problem a junior high school math proble

Because A and B go in the same direction, A is at point A and B is at point C, which is equivalent to a catch-up problem (different from the general straight line chasing and overtaking, the route of this problem is triangular), B is 12 meters in front of A, and A is chasing B. A and B should be in the same straight line, that is, the distance between A and B must be less than or equal to 6 meters. So A catches up with at least 6 meters.

The speed of A is meters per minute, B is 5 meters per minute, and A has more meters per minute than B, 6 minutes), which means that after at least 4 minutes, A and B may be in the same straight line. A traveled a total of meters after 4 minutes and stopped at the edge of BC 2 meters away from point B. B travels 5*4=20 meters in 4 minutes, and stops at AC 2 meters away from point C, so A and B are not in a straight line.

After another minute, A stops at the AC edge at point C and B stops at the AB edge 1 meter away from point A, still not in a straight line. Observe and calculate that B will always be on the edge of ab for the next minute, and A will be on the edge of ab after finishing the remaining meters on the edge of ab, and A and B will be on the same line for the first time. That is, the next minute A and B are on the same straight line.

Add up all the walking time, and walk for 5 and 13/7 minutes, and A and B are in the same straight line for the first time. That's the answer to this question. The steps to solve the problem and the explanation are written together, and the landlord will sort it out again.

According to the diagram you draw, you know that ab, bc is the waist.

If AB and BC are waists, then 0 is known: A per minute, B 5 m per minute

0 then: if the third side is 6m, it can be counted as 6 minutes, and if the third side is 12m, it can be counted as 4 minutes.

Do the rest for yourself, and hopefully it will help you.

The topic is more detailed, it's an equilateral triangle. If the angle of a c b is uncertain, the length of ab is 12<*<0, so that there are so many answers that you will be dizzy.

This....The landlord is not serious, in class, the absolute value is equal to 0, only before and after are zero will be equal to 0, so a is equal to 2, b is equal to 1

Solution: Because the sum of the two absolute values is 0, |a-2|AND |b-1|It's all 0, not the other way around.

Yields: a-2=0 b-1=0

a=2 b=1

The ratio of the absolute value of a number is non-negative.

a-2|0 and |b-1|≥0

Again|a-2|+|b-1|=0, then |a-2|=0，|b-1|=0

a=2,b=1

Forehead. The landlord didn't attend the class? It's too easy. The absolute value of 0 is 0....Others will put?

Let the number of people with a score of 8 rings be x.

6 1+7 3+8x+9 2) (1+3+x+2)=solution x=4

A: The number of people with a score of 8 rings is 4.

The amount of work A completes per day is 1 10, and the amount of work completed by B per day is 2 1 10 = 1 5

The amount of work completed by two people in m days is: m(1 10 + 1 5) = 3m 10 The remaining amount of work is: 1-3m 10 = (10-3m) 10 The time required for B is:

10-3m) 10 (1 5) = (10-3m) 2 days.

It will take x days.

A 1 day to complete 1 10

B completes 1 10 2 = 1 5 in 1 day

then m 10 + m 5 + x 5 = 1

Multiply both sides by 5m 2 + m + x = 5

x=5-(3m/2)

Efficiency of A: 1 10

Efficiency of B: 1 5

1 (1 10 1 5)m 1 5 5 1 5m The number of days required is 5 1 5m

1. The law is constantly one negative and one positive.

II. -1 2-5

Cong Bi 4-(4, 5, -1

VI. -4 3

7. It may be 0, 2, -2 because a and b don't know plus or minus.

8. Da Zheng Feng-1

9. 10. The first surplus: 1000*10%.

Second loss: 1000*(1+10%)*90%(1-90%)

1: The law is: --get positive, -+got-, keep an eye on ++ get +.

3: I hand let the An-123 in the order of the imitation potatoes and order

Because angle one = angle two, ab parallel ef (inner. parallel).

Because de is parallel to bc, angle one = angle b, angle three = angle c (two. equal)