A and B two cars from AB two places, a 56 kilometers per hour, B 48 kilometers per hour, two cars from the midpoint 32 meters meet

A and B two cars from AB two places, a 56 kilometers per hour, B 48 kilometers per hour, two cars from the midpoint 32 meters meet

AB distance?
Jiaduo walked 32 * 2 = 64 km
Encounter time 64 / (56-48) = 64 / 8 = 8 hours
Distance (56 + 48) * 8 = 104 * 8 = 832km
The distance difference between a and B is 32 × 2 = 64
Encounter time = 64 ÷ (56-48) = 8
AB distance between two places = (56 + 48) × 8 = 832 (km)
A and B start from ab at the same time. Car a runs 48 meters per hour and car B 40 meters per hour. The two cars meet at 32 meters from the midpoint. How far is the distance between the two places
48:40= X/2+32:X/2-32
24X-1536=20X+1280
4X=2816
X=704
Set time t
48t=40t+32+32
T = 8 hours
S = 48T + 40t = 704M
（32*2）/（48-40）*（48+40）
=704
A and B depart from ab at the same time. A travels 40 kilometers per hour, B 60 kilometers per hour. 4.5 hours after meeting, a arrives at B. how many kilometers is the distance between AB and a?
Don't use equations, just formulas,
After meeting, car a passed by = 4.5 × 40 = 180 km
Encounter time = 180 △ 60 = 3 hours
Distance = 40 × (4.5 + 3) = 300 km
4.5*40/0.6=300
In the "urban and rural cleaning project", an environmental sanitation team rents a number of vehicles with a carrying capacity of 8 tons to transport a batch of construction waste. If each vehicle only contains 4 tons, then 20 tons of construction waste is left; if each vehicle is full of 8 tons, then the last vehicle is not empty. How many vehicles does the environmental sanitation team rent?
If there are x cars, there are (4x + 20) tons of goods. From the meaning of the title, we get 0 ＜ (4x + 20) - 8 (x-1) ＜ 8, and the solution is 5 ＜ x ＜ 7. ∵ x is a positive integer, ∵ x = 6
A grain depot first transports a quarter of the stored grain, and then transports 126 tons of it. At this time, the grain in the depot is 80% more than the original. How many tons of grain was stored in the depot?
The original grain storage capacity is set at x tons
（1-1/4）X+126=（1+80%）X
X = 120 tons
Can anyone do the math problem of grade 6 in primary school? A and B start from a and B at the same time and go opposite each other. They will meet each other 5 hours later. If a car leaves one hour earlier, they will be able to do it
A and B start from a and B at the same time, and run in opposite directions, and meet 5 hours later. If a car starts 1 hour earlier, it will meet a car at the midpoint by 13 kilometers. If a car starts 1 hour earlier, it will meet B car after passing the midpoint by 37 kilometers, and find out the speed difference between a car and B car
10 km / h
(1) The first time: (V A + v b) * 5 = whole course
(2) The second time: [(full / 2) - 13] / V B-1 = [(full / 2) + 13] / V A
(3) The third time: [(full / 2) + 37] / V A-1 = [(full / 2) - 37] / v b
Speed difference between vehicle a and vehicle B = V A-V B = 10 km / h
Method 2
(1) A and B start from a and B at the same time and run in opposite directions. They meet in 5 hours;
According to (1): if the total distance becomes twice the original, the encounter time becomes 5 * 2 = 10 hours
(2) If car B starts one hour ahead of schedule, it will meet car a at the midpoint of 13 kilometers;
According to (2)
A car movement time: T1, movement distance: half of the total distance + 13 km
B car movement time: 1 + T1, movement distance: half of the total distance - 13 km
(3) If car a starts one hour ahead of schedule, it will meet car B after passing the midpoint of 37 km;
According to (3)
A car movement time: 1 + T2, movement distance: half of the total distance + 37 km
Second car movement time: T2, movement distance: half of the total distance - 37 km
According to (2) and (3), we can know that:
(2) And (3) the total distance is twice that of a and B,
The movement time of car a and car B is: 1 + T1 + T2 = 2 * 5 = 10 hours
The distance of a car movement: (half total distance + 13 km) + (half total distance + 37 km) = total distance + 50 km
Distance of car B: (half total distance - 13km) + (half total distance - 37km) = total distance - 50km
The time for both is the same: 10 hours
The difference of movement distance between a and B is: (total distance + 50 km) - (total distance - 50 km) = 100 km
Then the speed difference between vehicle a and vehicle B is: 100 km / 10 h = 10 km / h
Remember to give me a score. That's right. There are some problems in the total solution
10 km / h
Process:
(1) The first time: (V A + v b) * 5 = whole course
(2) The second time: [(full / 2) - 13] / V B-1 = [(full / 2) + 13] / V A
(3) The third time: [(full / 2) + 37] / V A-1 = [(full / 2) - 37] / v b
The answer is: the speed difference between car a and car B = V A-V B = 10 km / h
Finished! ... unfold
Remember to give me a score. That's right. There are some problems in the total solution
10 km / h
Process:
(1) The first time: (V A + v b) * 5 = whole course
(2) The second time: [(full / 2) - 13] / V B-1 = [(full / 2) + 13] / V A
(3) The third time: [(full / 2) + 37] / V A-1 = [(full / 2) - 37] / v b
The answer is: the speed difference between car a and car B = V A-V B = 10 km / h
Finished! Put it away
Given the equation (K & # 178; - 1) x & # 178; + (K + 1) x + (k-7) y = K + 2, when k takes what number, the equation is one variable linear equation
（k²-1）x²+(k+1)x+(k-7)y=k+2
Y=[(k²-1）x²+(k+1)x-(K+2)]/(7-K)
If the equation is a linear equation of one variable, then the x ^ 2 coefficient is 0 and the X coefficient is not equal to 0
K & # 178; - 1 = 0, K + 1 is not equal to 0
K = 1, the equation is one variable linear equation
In the "urban and rural cleaning project", an environmental sanitation team rents a number of vehicles with a carrying capacity of 8 tons to transport a batch of construction waste. If each vehicle only contains 4 tons, then 20 tons of construction waste is left; if each vehicle is full of 8 tons, then the last vehicle is not empty. How many vehicles does the environmental sanitation team rent?
If there are x cars, there are (4x + 20) tons of goods. From the meaning of the title, we get 0 ＜ (4x + 20) - 8 (x-1) ＜ 8, and the solution is 5 ＜ x ＜ 7. ∵ x is a positive integer, ∵ x = 6
63 + 120 = 183, 80-19 = 61183 ÷ 61 = 3______ .
63 + 120 = 183, 80-19 = 61183 ÷ 61 = 3, which is rewritten as the comprehensive formula is: (63 + 120) ÷ (80-19), so the answer is: (63 + 120) ÷ (80-19)
A and B set out from a and B at the same time, facing each other. The first time they met head-on was 100 meters away from B. after meeting, a's speed increased to twice of the original. A turned around immediately after arriving at B, and when they caught up with B, B still had 50 meters to get to a
As shown in the figure, point C is where Party A and Party B meet face to face for the first time, and point D is where Party A overtakes Party B after turning around
When B goes from B to D,
A goes from a to C, after speeding up, from C to B, and then turns around to d
BD × 2 + 50m (AD)
Let's assume that a starts to walk at twice the original speed from point a,
There will be more distance from a to C (the green line in the picture) at the original speed,
That is: 50m + CD, so it is equal to:
BD × 2 + 50m + 50m + CD
= BD × 2 + 100m + CD (as can be seen from the figure, CD + 100m = BD)
＝BD×3
From this we can see that the speed of a is increased to 2 times of the original speed, and then it is 3 times of that of B
Then the original speed of a is that of B: 3 △ 2 = 1.5 times
B is the first time from a to B (C = 100 meters)
In this way, the distance between AB and ab:
100×（1.5＋1）
＝100×2.5
= 250 (m)
A-50-C-D-100-B
AC spacing 50, DB spacing 100; D is the first meeting point, C is the second meeting point;
[2*(50+CD)]:100=(100*2+CD):CD
1+CD/50=200/CD+1
CD^2=200*50=100^2,
CD=100
AB = 50 + 100 + 100 = 250 question: understand the formula, thank you very much! Can we use reasoning that pupils can accept?