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The speed ratio of the car to the truck is 4:7. The car and the truck leave from two places and meet 15 kilometers away from the midpoint. How many kilometers did the truck travel?
Thirty-five
The speed ratio of cars to trucks is 4:7, and the distance ratio of cars to trucks is also 4:7. If the total distance is divided into 11 points, the car takes four, the truck takes seven, and the truck takes three more.
It is 15 km away from the midpoint, which means that the freight car has traveled 2 × 15 = 30 km more. So each piece is 30 / 3 = 10 kilometers.
7 × 10 = 70 km.
Suppose the total length is 11, then the midpoint is 5.5, and the bus only goes 4. 5 from the midpoint. So 1.5 is 15 kilometers
Seven is 70 kilometers
Formula:
7+15/((7+4)/2-4)=70
When meeting, the distance of the car is s, then the distance of the truck is (7 / 4) s
The distance between the two places is s + (7 / 4) s = (11 / 4) s
The distance from the midpoint is (11 / 8) s
(7 / 4) s - (11 / 8) s = 15, that is, (3 / 8) s = 15, s = 40
So when we met, the car drove 40 kilometers and the truck 70 kilometers
70km
According to the fact that the two vehicles are facing each other, and the distance between the meeting place and the midpoint is 15 km, we can know that the distance difference between the truck and the vehicle is 30 km, which is proportional to (7-4): 7 = 30: X, and the solution is x = 70
Suppose the speed of a car is 4x, then the speed of a truck is 7x.
According to the conditions, the speed of freight cars is faster
And because they met at the midpoint of 15 kilometers, the truck traveled 30 kilometers more than the car.
7x-4x = 30, the solution is x = 10
That is, the speed of a car is 40 km / h, and that of a truck is 70 km / h
And because they had the same time, they drove for y hours
70Y-40Y=30
The solution is y = 1 hour
Therefore, the truck travel distance S = 70 * 1
Suppose the speed of a car is 4x, then the speed of a truck is 7x.
According to the conditions, the speed of freight cars is faster
And because they met at the midpoint of 15 kilometers, the truck traveled 30 kilometers more than the car.
7x-4x = 30, the solution is x = 10
That is, the speed of a car is 40 km / h, and that of a truck is 70 km / h
And because they had the same time, they drove for y hours
70Y-40Y=30
The solution is y = 1 hour
Therefore, the driving distance of freight cars is s = 70 * 1 = 70 km
A————————
The most complete answer is put away
A. B is 490 km away from two places, and a and B depart from a and B respectively. A travels 40 km per hour, B travels 40 km per hour
How many kilometers did car a and car B meet each other?
Hello!
Let car B pass for X hours
（40+60）x=490-40
100x=450
x=4.5
A walked 40x (4.5 + 1) = 220km
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How many large and small trucks of 6 tons and 4 tons are used to transport 58 tons of fruits to the city?
There are five options
1: Cart: 1, trolley: 13;
2: Cart: 3, trolley: 10;
3: Cart: 5, trolley: 7;
4: Cart: 7, trolley: 4;
5: Cart: 9, trolley: 1
14 x = 2. 14 + X, how to solve, write the process clearly
3.14x=2.14+x
3.14x-x=2.14
2.14x=2.14
X=1
3.14x=2.14+x
3.14x-x=2.14
2.14x=2.14
X=1
The distance between a and B is s kilometers. Someone plans to arrive in an hour. Now he has to arrive 2 hours ahead of time and walk more every hour ()
S / (A-2) (km)
Let a ＞ 0, B ＞ 0, if the root sign 3 is the equal proportion of 3 ^ A and 3 ^ B, then the minimum value of 1 / A + 2 / B is
The root sign 3 is the proportional median of 3 ^ A and 3 ^ B
The power of 3 = 3 is (a + b), that is, a + B = 1
1 / A + 2 / b = (a + b) (1 / A + 2 / b) = 1 + 2 + 2A / B + B / a ≥ 3 + 2 radical 2
3^b=3*3^a b=a+1
1 / A + 2 / b = 1 / A + 2 / (a + 1) when 1 / a = 2 / (a + 1), the sum is the minimum, a = 1, and the minimum is 2
3=3^a * 3^b
a+b=1
1/a+2/b=(a+b)/a +(2a+2b)/b
=1+b/a+ 2a/b +2
≥ 3+2√2
If and only if B / a = 2A / B, the equal sign holds
Minimum 3 + 2 √ 2
Because the root sign 3 is the middle term of 3 ^ A and 3 ^ B, (√ 3) &# 178; = 3 ^ A * 3 ^ B, so 3 ^ (a + b) = 3
That is, a + B = 1, we can use the basic inequality to solve it according to a > 0, b > 0
1/a+2/b=(a+b)/a+2(a+b)/b=1+b/a+2a/b+2≥2√(b/a*2a/b)+3=2√2+3
The minimum value of 1 / A + 2 / B is 2 √ 2 + 3
Because the root sign 3 is the equal proportion of 3 ^ A and 3 ^ B, so 3 ^ A * 3 ^ B = 3, then a + B = 1, and a > 0, b > 0, then 1 / A + 2 / b = (a + b) / A + 2 * (a + b) / b=
1 + B / A + 2 * (1 + A / b) = 3 + A / B + 2 * B / a > = 3 + 2 times root 2
Because a ＞ 0, B ＞ 0, the root sign 3 is the equal proportion of 3 ^ A and 3 ^ B, so a times b equals the square of 3, that is 9
So a = 9 / B, so 1 / A + 2 / b = B / 9 + 2 / B, and then use the mean inequality, so B / 9 + 2 / b > = 2 times the root 2 and divide by 3 is the minimum
A batch of goods will be transported by a number of 8-ton trucks. If each truck is only loaded with 5 tons, then 10 tons of goods will be left. If each truck is full of 8 tons, then the last truck is not empty and dissatisfied. How many trucks are there?
If there are x cars, we get 8 (x − 1) + 8 ＞ 5x + 108 (x − 1) ＜ 5x + 10, and the solution is 103 ＜ x ＜ 6
3.14x (10 * 2) square X3 * 3
3.14x (10 * 2) square X3 * 3
=3.14x20²x9
=3.14x400x9
=11304
A. The distance between B and a is s km. Xiaoliang plans to travel from B to a in t hours. If he arrives at a 2 hours ahead of time, Xiaoliang should travel () km more
s/(t-2)-s/t
Take "s" as 1000 meters, "t" as 4 hours, and then calculate. It's 500
S km
Let a > 0, b > 0, if (radical 3) is the middle term of equal ratio of 3 ˇ A and 3 ˇ, then the minimum value of 1 / A + 1 / b——
So (√ 3) ^ 2 = 3 ^ A * 3 ^ B = 3 ^ (a + b),
3 = 3 ^ (a + b), so a + B = 1. And a > 0, b > 0, so a + b > = 2 √ AB, that is, ab = 4
So 1 / A + 1 / b = (a + b) / AB = 1 / AB > = 4. So the minimum value of 1 / A + 1 / B is 4
Right upstairs!
Because (root sign 3) is the median of 3 ˇ A and 3 ˇ
So 3 ˇ a * 3 ˇ = 3 shows that a + B = 1
1 / A + 1 / b = (a + b) / AB = 1 / AB the minimum value of 1 / A + 1 / B is required, that is, the maximum value of a * B is 0.25
So the minimum value of 1 / A + 1 / B is 4
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