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A car goes from place a to place B at the speed of 60 km / h. When the car runs for 4.5 hours, the road is slippery in the rain, and the speed is reduced by 20 km / h Arrive at place B 3 / 4 hours later. Find the distance between Party A and Party B
(1) If the original estimated time is x hours and the distance between the two places is y kilometers, then: 60x = y; if 60 * 4.5 + (60-20) * (x-4.5 + 3 / 4) = y, then: x = 6, y = 360 (2) if the original estimated time is x hours, then: 60x = 60 * 4.5 + (60-20) * (x-4.5 + 3 / 4) then: x = 6, 60x = 360 (3) if the distance between the two places is y kilometers, then
On a map with a scale of 0 4080 120 km, the distance between AB and ab is 3.5 cm. What is the actual distance between AB and ab
Scale = 1:40 * 100000 = 1:40000000
Actual distance = 3.5 × 40 × 100000 = 140 km
When a is 1 / 4 away from B, B is 80 km away from A. at this time, a travels 40 km less than B, and the whole journey is calculated
A is 40 km less than B, that is, 80 + 40 = 120 km from B
Whole journey: 120 △ 1 / 4 = 480 (km)
The two warehouses of a and B need to transfer wheat to ab. a can transfer 80 tons, B can transfer 40 tons. A needs 50 tons of wheat, B needs 70 tons. The freight is as follows
A B (yuan / ton)
A-10-40
B -- 20 -- 30
1. Suppose that warehouse A is transported to place a for X tons, and find the functional relationship between the total freight y and X
2) Which is the least total freight? Which is the most? And find out the most and least freight
------Land a -- land B - (yuan / ton)
A-10-40---
B-20-30---
Let: a transfer x tons to a. the total freight is y yuan. Then transfer (80-x) tons to B, B (70-80 + x) tons to B, B (50-x) tons to A. y = 10x + 40 (80-x) + 30 (70-80 + x) + 20 (50-x) y = 10x + 3200-40x + 2100-2400 + 30x + 1000-20xy = - 20x + 3900, because the coefficient of X in the analytic expression of this function is negative
The distance between a and B is 200 kilometers. Car a runs 40 kilometers per hour from place a to place B. after car a runs for 1.5 hours, car B runs to place a every hour
How long does car B meet car a after driving 30 km? Solution of linear equation with one variable
Let car B run for X hours and meet car a
30x+40(x+1.5)=200
70x=140
x=2
Car B runs for 2 hours and meets car a
The distance between a and B is 200 kilometers. A and B drive from a to B at the same time. A travels 40 kilometers per hour and B 55 kilometers per hour
The two cars are 45 kilometers apart after a few hours
3 hours, 15km difference per hour, 45km difference in three hours
The distance between a and B is 200 kilometers. A car travels x kilometers per hour from a to B. how many kilometers is it from B after a hour?
Distance to B
=200 - distance traveled
=(200 ax) km
The distance between a and B is 100 kilometers. Car a runs from a to B for 20 kilometers per hour. When car a runs for 1.5 hours, car B runs from B to a for 15 kilometers per hour. How long does it take for car B to meet car a? The solution of the equation,
Let car B meet car a after X hours,
20×1.5 +(20 + 15)x = 100
The solution is: x = 2
A: car B met car a after 2 hours
A car drove from a to B at a speed of 40 km / h. After 3 hours, it was forced to reduce 10 km / h due to rain
As a result, we arrived 45 minutes later than expected
Use the equation of one variable to solve the problem, and explain the idea of solving the problem
Set the driving time after rain as t
Equivalency: the distance between two modes is equal
The countable equation is as follows
40*3+(40-10)t=40*[3+t-(45/60)]
T = 3 can be obtained
Distance between the two places = 40 * 3 + (40-10) t = 210
A car drove from place a to place B at a speed of 40 kilometers per hour. After 3 hours, due to rain, the average speed was reduced by 10 kilometers,
As a result, we arrived in the second place 45 minutes later than expected, so we can find the distance between the first and the second
Let the distance after acceleration be x, then the equation x / 40 + 0.75 hours = x / 30, and the solution is x = 90, so the total distance is 40 * 3 + 90 = 210 km
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