The two cities are 308km apart. The two trains left each other at the same time and met three hours later. It is known that car a is one third faster than car B,

The two cities are 308km apart. The two trains left each other at the same time and met three hours later. It is known that car a is one third faster than car B,

If the speed of car B is x, then the speed of car a is (1 + 1 / 3) X
3x+3(1+1/3)x=308
3X+4X=308
X=44
The two trains of a and B set out from station AB at the speed of 4:5, facing each other. B started 72 kilometers from station B, and then a started from station A. when the two trains met on the way,
Car B travels 45 kilometers more than car A. how many kilometers is the distance between station AB and station B?
Are the two trains a and B starting from station AB at the speed of 5:4?
The distance between ab stations is (72-45) △ 5-4) × (5 + 4) + 72 = 315 km
A is faster than B? The question is wrong
V A / v b = 4:5;
S B-S a = 5t-4t = 45;
t＝45
s＝5×45＋4×45＋72＝477
From the train speed ratio in the question, we can set a and B speed respectively: 4V and 5V
Then, after the same time, the walking distance can be set as 4S and 5S
Vehicle B travels 45km more than vehicle a, i.e. 5s-4s = s = 45
So 5S = 225, 4S = 180
Distance between two stations of AB: S = 4S + 5S + 72 = 180 + 225 + 72 = 477 (km)
(the topic is that there are some problems. It should be noted that car a starts to time after departure, and car B travels 45km more than car a)
From the train speed ratio in the question, we can set a and B speed respectively: 4V and 5V
Then, after the same time, the walking distance can be set as 4S and 5S
Vehicle B travels 45km more than vehicle a, i.e. 5s-4s = s = 45
So 5S = 225, 4S = 180
Distance between two stations of AB: S = 4S + 5S + 72 = 180 + 225 + 72 = 477 (km)
(the topic is that there are some problems. It should be noted that car a starts timing after departure, and car B travels 45km more than car a)
The two trains run from AB to ab at the same time. When they meet, car a runs 100 kilometers more than car B. the speed of car B is three fifths that of car A. how long is the railway?
Let's say that when the car goes by km, X meets
As the speed of car B is 3 / 5 of that of car a, car B has traveled 3x / 5km
formulation of equation
x-3x/5 = 100
The solution is x = 250 km
So the railway length is 250 + 3 * 250 / 5 = 400 km
Set the speed of car a as XKM / h and travel for t hours
Then xt-0.6xt = 100
xt=250km
Railway length: XT + 0.6xt = 1.6 * 250 = 400km
Let a line x, then B line 3 / 5x, railway long y, line t hours
X-3/5X=100
X+3/5X=Y
X=250
Y=400
Let's set the speed X of car a and walk for t hours when they meet
XT-3/5XT=2/5XT=100
So XT = 100 × 5 / 2 = 250
So the length of railway is XT + 3 / 5xt = 8 / 5xt = 400km
The total length of the railway is 400 meters
A. The distance between B and B is 560 km. Two cars of a and B are going from two places at the same time, and they meet after 7 hours. The speed ratio of a and B is 3:5, and the speed ratio of a and B is 3:5
The speed of car a is 560 △ 7 × 3 / 8 = 30 km / h;
The speed of car B is 560 △ 7 × 5 / 8 = 50 km / h
If you don't understand, you are welcome to ask,
Let the velocity of a be x, and the velocity of B be 5 / 3x
7 * x 7 * 5 / 3x = 560, so x = 30, so a's speed is 30, B's speed is 50
When k is taken as the value, the quadratic equation x2 + 4kx + (2k-1) 2 = 0 with respect to X has two real roots, and the roots of the equation are obtained (expressed by the algebraic formula containing K)
∵ a = 1, B = 4K, C = (2k-1) 2, ∵ △ = (4K) 2-4 (2k-1) 2 = 16k-4. When 16k-4 ≥ 0, that is, K ≥ 14, the equation has two real roots. In this case, the root of the equation is x = - 4K ± 16K − 42, that is, X1 = - 2K + 4K − 1, X2 = - 2k-4k − 1
The automobile transportation company has to send 800 tons of disaster relief materials to support the disaster area, using 12 vehicles to transport 25% of this batch of materials, according to this calculation, one by one
How many cars do you need to transport these materials at one time?
12/25%=48
48 times
48
There are 80 pear trees in the orchard. The number of peach trees is 3 / 5 of pear trees and 6 / 7 of apple trees. How many apple trees are there in this orchard?
Pear tree 80 peach tree is three fifths of pear tree, which means 80 * 3 / 5 = 48
For apple trees, x 6 / 7X * 48 = 56
Pear tree 80 peach tree 48 apple tree 56 total 184
The distance between a and B is 360 km. The passenger cars and freight cars start from the two places at the same time and travel in opposite directions. Their speed ratio is 5:4?
Because the speed ratio of a and B is 5:4, the distance is directly proportional to the speed, so the distance ratio is 5:4, 5 + 4 = 9, the distance of bus: 360 × 59 = 200 (km), the distance of truck: 360 × 49 = 160 (km). A: when the two cars met, they drove 200 km and 160 km respectively
A quadratic equation with one variable x ^ 2 - (2k + 1) + 4 (k-1 / 2) = 0 (a is not equal to 0)
A quadratic equation with one variable x ^ 2 - (2k + 1) + 4 (k-1 / 2) = 0 (a is not equal to 0)
(1) To prove, no matter what real number k goes to, the equation always has real root
(2) The solution to this problem is that ABC is the root of the triangle, B is the circumference of the triangle
X ^ 2 - (2k + 1) x + 4 (k-1 / 2) = 0 (a is not equal to 0) 1) because the discriminant of root = △ = B & # 178; - 4ac = [- (2k + 1] & # 178; - 4 × 4 (k-1 / 2) = (2k + 1) & # 178; - (16k-8) = 4K & # 178; + 4K + 1-16k + 8 = 4K & # 178; - 12K + 9 = (2k-3) & # 178; ≥ 0, so no matter what the reality of K is
Datong transportation company plans to use 20 vehicles to transport 36 tons of vegetables
The tonnage and profit of each vehicle are as follows:
Vegetable name a, B and C
Tons loaded per vehicle
The profit per ton of vegetables (RMB 10000) is 574
Requirements: ① the car must be fully loaded; ② each car only carries the same vegetable; ③ each vegetable must not be less than one car
How many options are there? Which one is the most profitable?
X, y and Z cars were set up to transport three kinds of vegetables
2x+y+1.5z=36
x+y+z=20
Profit B = 5x + 7Y + 4Z
Using the two equations in 1, we simplify 2 and get b = 4x + 32
Therefore, as long as we determine the value of X max, we can know the maximum value of B
From the two equations in 1
x=16-1/2z
So, the maximum value of X is 15
Substituting the integers with X less than 15 into the two equations in 1, we can find out that if y and Z are integers, we can use the scheme. If we have solved several schemes, we can find several schemes
That's all