The two passenger and freight cars leave each other at the same time from a and B, and meet at a distance of 3km from the midpoint. The speed of the bus is 9 / 10 of that of the truck. How many kilometers are there between a and B?

The two passenger and freight cars leave each other at the same time from a and B, and meet at a distance of 3km from the midpoint. The speed of the bus is 9 / 10 of that of the truck. How many kilometers are there between a and B?

3 times 2 divided by (1-9 / 10) = 60 (km)
60 times 9 / 10 = 54 (km)
60 + 54 = 114 (km)

Given that ABCD is a real number and C is greater than D, then a is greater than B. what is the condition that a-c is greater than B-D? Please write down the required formula and calculation process, al
Necessary and sufficient conditions? Necessary but not sufficient? Neither sufficient nor necessary conditions? Sufficient but not necessary conditions?

c>d
-cb
Obviously, a-c > B-D is not necessarily true
For example, a = 2, B = 1, C = 1, d = - 1
Not enough
a-c>b-d
c>d
Then - d > - C
Add
a-c-d>b-c-d
a>b
So it's necessary
Therefore, it is necessary but not sufficient

Nine fourths of the tons of coal are transported from the canteen, one third in the first week and three fifths in the second week

The question is how many tons are left?
9 / 4 * (1-1 / 3) * (1-3 / 5) = 3 / 5 (ton)
Absolutely

A mathematical problem of mean inequality in senior one
a. B, X and y are all positive real numbers. A / y + B / y = 1 (A and B are fixed values). Find the minimum value of X + y
The first floor is totally wrong. I need strict proof.
Besides, there's no answer in the options.

X + y is equal to the formula (XY) 1 under the double change sign
Let a / x = Sina ^ 2, B / y = cosa ^ 2, a be the acute angle
We get x = A / Sina ^ 2, y = B / cosa ^ 2
We can get a * B / Sina * cosa when x + y is greater than or equal to 2 times of the change sign
Sina * cosa = 1 / 2sin2a, the maximum is 1 / 2, so the minimum of 1 / Sina * cosa is 2
So the minimum of X + y is 4 times of a * B

The two trains, passenger train and freight train, depart from a and B at the same time, and meet five hours later. The distance between the two places is 770km. It is known that the speed of passenger train is 1.2 times that of freight train. What is the speed of the two trains?

Strange Ask a question about the inequality of grade one in senior high school
On the image of the function y = 1 / x, find the coordinates of the point whose minimum value is 1 / x + 1 / y
Why is the minimum value 2 when the point is (1,1)?
I think it should be - 2 when (- 1, - 1)!
Why is there no minimum in the third quadrant

It should be in the first quadrant
If it is the third quadrant, there is no minimum
x>0,y>0
y=1/x
1/y=x
So 1 / x + 1 / y = 1 / x + x > = 2 √ (1 / X * x) = 2
So the minimum is 2
Here x = 1 / X
x=1,y=1
Namely (1,1)

The speed ratio of passenger and freight cars is 4:3. After meeting, the speed of passenger cars decreases by 20%, and that of freight cars decreases by 20%
In this way, when the bus arrives at B, the truck is 25 kilometers away from A. how many kilometers are there between a and B?

The speed ratio after meeting is 4 (1-20%): 3 (1 + 1 / 3) = 3.2:4 = 4:5. When the bus meets, it takes 4 / (3 + 4) = 4 / 7, and the truck takes 3 / 7. According to a certain time, the distance ratio is equal to the speed ratio. The equation is that there are x kilometers between a and B. 3x / 7: (4x / 7-25) = 4:5

If (3,5) (4,7) (- 1, b) are collinear, then the value of B is?

b=-3
I do it with functions. I don't know, right

There are two baskets of apples. The weight of basket B is 3 / 5 of that of basket A. if you take 5 kg from basket a and put it into basket B, then basket B is 7 / 9 of that of basket a?

The original weight of the armour basket was x kg
(x -5)*7/9 = 3/5 x +5
X = 50 - armour basket
50 * 3 / 5 = 30kg - basket B

What's the date of 231 days from today

10.1