(1 / 3) at both ends of a straight road 12 kilometers away, Party A and Party B set out at the same time and walk opposite each other. The speed of Party A is 5 km / h, and that of Party B is 3 km / h (1 / 3) seeking: at both ends of a straight road 12 kilometers away, Party A and Party B set out at the same time and walk opposite each other. The speed of Party A is 5 km / h, and that of Party B is 3 km / h. There is a dog running at the speed of 6 km / h

(1 / 3) at both ends of a straight road 12 kilometers away, Party A and Party B set out at the same time and walk opposite each other. The speed of Party A is 5 km / h, and that of Party B is 3 km / h (1 / 3) seeking: at both ends of a straight road 12 kilometers away, Party A and Party B set out at the same time and walk opposite each other. The speed of Party A is 5 km / h, and that of Party B is 3 km / h. There is a dog running at the speed of 6 km / h


The speed of a 9 kilometer dog is known. To find the distance, you only need to know the movement time. The movement time of a dog is the time from the beginning to the meeting. The meeting time of a dog is the sum of distance divided by the sum of speed



In reading, Congcong misread the decimal point and read it as 6400.2?


The original number is 640.02



The two trains of Party A and Party B depart from ab at the same time and meet at 98km away from a for the first time. After meeting, the two trains continue to move forward and return to their destination
Return immediately, the second meeting is 48 kilometers away from A. how many kilometers is the distance between a and B?


Hello
(98 × 3 + 48) △ 2 = 171 (km)



A positive number has two square roots


Opposite numbers



A and B trains leave 1050 km apart at the same time, and a train travels 80 km per hour,
Two trains a and B leave from 1050 km apart at the same time. Car a travels 80 km per hour. After 2:8 hours, the two trains are three fifths of the distance. How many km per hour does car B travel?


This topic, I will give two solutions: one is the normal equation solution: let B car travel x kilometers per hour, then (80 + x) × 2.8 = 1050 × (1-3 / 5) 2.8x = 196, x = 70, so B car travel 70 kilometers per hour



Given that a > b > C, the average of a, B and C is 1.8, then B must be greater than 1.8______ (judge right or wrong)


According to the meaning of average, it can be seen that: the quantity representing the trend of a group of data sets, which is an indicator reflecting the trend of data sets; for example: 2.7, 1.4, 1.3, the average of the three numbers is (2.7 + 1.4 + 1.3) △ 3 = 1.8; for another example: 2.2, 2.1, 1.1, the average of the three numbers is (2.2 + 2.1 + 1.1) △ 3 = 1.8; for another example: 2.4, 1.8, 1.2, the average of the three numbers is (2.4 + 1.8 + 1.8) 2) It must be smaller than the largest one and larger than the smallest one, but the size of the data in the middle can't be determined



A. B. the highway between the two places is 50 kilometers long. One person rides a bicycle and the other rides a motorcycle. The motorcycle and bicycle leave one hour and 30 minutes late,
When you arrive an hour earlier, you know that the speed of a motorcycle is 2.5 times that of a bicycle. You can find the speed of a motorcycle and a bicycle


If the bicycle speed is x, then
50/x=50/2.5x+1.5+1
X = 12 km / h
The motorcycle speed is: 12 * 2.5 = 30 km / h



11 plus the square of one natural number equals the square of another


I'm glad to answer your question
The square of 11 + [5] = [6]
11+25=36



Car a and car B travel from two places 475km apart. They meet in five hours. Car a's speed is 80% of car B's. what's the speed of car a and car B?


If B's speed per hour is set to x km, then a's speed is
4 / 5x according to the above description, 5x + 4 / 5 × 5x = 475
The solution is x = 52
Our teacher taught!



Let a, B, C, D ∈ (0, 1). Compare the size of ABCD with a + B + C + D-3, and give your proof


Let's compare the size of ABC and a + B + C-2. ∵ 0 < AB < 1, ∵ ABC = (AB) C > AB + C-1 > A + B-1 + C-1 = a + B + C-2, and further, ABCD = (ABC) d > ABC + D-1 > A + B + C + D-3