Two cars leave from a and B at the same time. Five hours later Mathematics problems in grade six of primary school Two cars leave from a and B at the same time, and meet at the midpoint of 30 kilometers after five hours. How many kilometers does the fast train travel per hour and the slow train travel per hour

Two cars leave from a and B at the same time. Five hours later Mathematics problems in grade six of primary school Two cars leave from a and B at the same time, and meet at the midpoint of 30 kilometers after five hours. How many kilometers does the fast train travel per hour and the slow train travel per hour

30×2÷5
=60÷5
=12 km
60-12 = 48 km / h
A: the local train runs 48 kilometers per hour
Equation:
Set the local train to run x kilometers per hour
（60-x）×5=30×2
300-5x=60
5x=300-60
5x=240
x=48
A: the local train runs 48 kilometers per hour
That's the right answer. No one else has an equation,

Solving equation 2-3 of 2X-4 = negative 6 of X-7 process

Equation 2-3 of 2X-4 = minus 6 of X-7, both sides multiply by 6 at the same time, we get the following equation:
12-2 (2X-4) = - (X-7), expand, get: 12-4x + 8 = - x + 7, transfer, merge: 13 = 3x, so: x = 13 / 3

1. Party A and Party B are 6km apart. Now they are facing each other at the speed of 2.5km/h and 3.5km/h respectively. At the same time, the dog with belt A is running at the speed of 8km / h
1. Party A and Party B are 6km apart. Now they are facing each other at the speed of 2.5km/h and 3.5km/h respectively. At the same time, the dog with a runs to Party B at the speed of 8km / h. After meeting Party B, they quickly return to Party A and run to Party B again And so on, until Party A and Party B meet, ask: the distance of the dog running 2. Measure the depth of the well with a rope, fold the rope three times to measure, the remaining rope outside the well is four feet, fold the rope four times to measure, the remaining rope outside the well is one foot. Find the depth of the well and the length of the rope 3. A vendor sells two pieces of clothes in the business, each of which is sold for 135 yuan. According to the cost calculation, one makes a profit of 25%, and the other loses 25% It takes 20 seconds for a train to enter a 300 m tunnel to pass through. A fixed light at the top of the tunnel shines on the train for 10 seconds to find the length of the train

No matter how the dog runs, as long as you know the speed and time of the dog, you can find out the distance the dog runs
6 / (2.5 + 3.5) = 1 (hour), 8 * 1 = 8 (kilometer)
2, (depth of well + 4) * 3 = (depth of well + 1) * 4 = length of rope,
So the depth of the well is 8 feet and the length of the rope is 36 feet
3. The cost of a coat with 25% profit is 135 / (1 + 25%) = 108 yuan,
The cost of a coat with a loss of 25% is 135 / (1-25%) = 180 yuan,
The total cost of the two coats is 288 yuan, while the total sales price is 270 yuan
4, (train length + 300) / 20 = train length / 10 = train speed,
So the length of the train is 300 meters
In this problem, we can also directly use the ratio: 20:10 = 2,
(train length + tunnel length): train length = 2,
So the length of the train is 300 meters

Use 1, 3, 5 and decimal point to form a system. The largest two decimal places are (), and the smallest two decimal places are ()

Use 1, 3, 5 and decimal point to form a system. The maximum two decimal places are (5.31) and the minimum two decimal places are (1.35)

A and B start from ab at the same time and travel kilometers away from each other. A travels 50 kilometers per hour, B 60 kilometers per hour. When B arrives at a, a returns
It takes two hours to get to B. how many kilometers is the distance between AB and B?

50 × 2 ÷ (60-50) = 10 (hours)
60 × 10 = 600 (km)
Idea: when car B arrives at place a, car a has 50 × 2 = 100 (km) left
This is because car B is 10 kilometers faster than car a every hour. After several hours, car B travels 100 kilometers more than car A. in this way, the travel time of car B will be known, and the distance will be known when the time is found

Using the convergence criterion of monotone bounded sequence to prove the existence of sequence limit
x(1)>0,x(n+1)=1/2*(x(n)+a/x(n)),n=1,2,...,a>0.
Where n of X (n) is the subscript

The conclusion is: xn ≥ √ a
x(n+1)－xn＝1/2×[a/xn－xn]＝1/2×(√a＋xn)(√a－xn)/xn≤0
So, xn decreases monotonically
So, xn is monotonically bounded and the limit exists

The first train runs 110 kilometers per hour, the second train runs 90 kilometers per hour, and the two trains are leaving
When did the two cars meet at 20 kilometers?

According to the meaning of the title:
100X-90X=40
20X=40
x=2
A: the two cars will meet in two hours

Given that the average of a, B, C and C is 6 and the variance is 5, calculate the average and variance of data 2A + 3, 2b + 3 and 2C + 3

1. A, B and C are sequences with variance of 5, and we get: B = a + 5, C = a + 10 2. The average of a, B, C is 6, and we get: a + B + C = 6 × 3 = 183. We get the following results: a + A + 5 + A + 10 = 3A + 15 = 18, and we get a = 1, B = 6, and C = 114. The average of 2A + 3 = 5, 2b + 3 = 15, 2C + 3 = 252A + 3, 2b + 3, and 2C + 3 is (5 + 15 + 25)

A. B is 50km away. One and a half hours after a starts from a to B by bike, B goes from a to B by motorcycle. The speed is 2.5 times that of A. B arrives one hour earlier than a
Solving by fractional equation
H means hour

Let the velocity of a be X,
[50-0.5X]/X-50/2.5X=1
X = 20 km / h
Speed A: 20 km / h
Speed B: 50 km / h

For example, the cube of any natural number greater than 1 can be written as the square difference of two natural numbers

When m is even, let m = 2K, K ∈ n
m^3=(2k)^3=k^2*8k=k^2*{(k+2)^2-(k-2)^2}={k(k+2)}^2-{k(k-2)}^2
When m is odd, let m = 2K + 1, K ∈ n
m^3=(2k+1)^3=(2k+1)^2*(2k+1)=(2k+1)^2*{(k+1)^2-k^2}={(2k+1)(k+1)}^2-{k(2k+1)}^2