A and B two cars start from ab at the same time, facing each other. The first meeting place is 100 kilometers away from a A and B start from a and B at the same time, and run in opposite directions. The first meeting place is 100 km away from A. after meeting, the two vehicles continue to drive at the same speed. After arriving at B and a respectively, they immediately return along the original road. This is another meeting at 60 km away from B. find the distance between a and B

A and B two cars start from ab at the same time, facing each other. The first meeting place is 100 kilometers away from a A and B start from a and B at the same time, and run in opposite directions. The first meeting place is 100 km away from A. after meeting, the two vehicles continue to drive at the same speed. After arriving at B and a respectively, they immediately return along the original road. This is another meeting at 60 km away from B. find the distance between a and B

Let X be the distance
In the first meeting, a drives 100 and B drives X-100; in the second meeting, a drives x + 60 and B drives 2x-60
100:(x-100)=(x+60):(2x-60)
100(2x-60)=(x-100)(x+60)
200x-6000=x^2-40x-6000
240x=x^2
x=240

What is the result of rational number classification
emergency

Positive, negative, 0

Train a and B run from a and B with a distance of 1000km. It is known that train a runs 120km per hour and train B runs 80km per hour,
After 30 minutes of B's departure, train a will also leave. How many hours after a's departure, will train a meet train B? (solved by linear equation with one variable)

Set: meet in X hours
（1000-80×1/2）=（120+80）X
X = 4.8 hours
After 4.8 hours, car a meets car B

32.50 29.66 31.64 30.00 31.01 30.76 30.24 32.87 31.05 mean and variance

The average of 32.50 and 29.66.31.05 was 31.08
Then variance S & # 178; = s ^ 2 = 1 / n [(x1-m) ^ 2 + (x2-m) ^ 2 +. + (xn-m) ^ 2]
=1/9[(32.50-31.08)^2+(29.66-31.08)^2+.+(31.05-31.08)^2]
=1.06

A car runs 280 kilometers in 2 hours. At this speed, it has been driving for 8 hours from a to B. how long is the highway between a and B
Answer in two ways

1
Length 280 △ 2x8 = 1120 km
two
The design length is x km
2:8=280：x
x=280x8÷2
x=1120
If you don't understand this question, you can ask,

How many perfect squares are there in the numbers 1-1999?
For example, 4 = 2, 9 = 3, 25 = 5, 4,9,25 are the completed squares

^For square sign 1 ^ = 12 ^ = 43 ^ = 94 ^ = 165 ^ = 367 ^ = 498 ^ = 649 ^ = 8110 ^ = 10011 ^ = 12112 ^ = 14413 ^ = 16914 ^ = 19615 ^ = 22516 ^ = 25617 ^ = 28918 ^ = 38419 ^ = 36120 ^ = 40021 ^ = 44122 ^ = 48423 ^ = 52924 ^ = 57625 ^ = 62526 ^ = 67627 ^ = 72928 ^ = 78429 ^ = 84130 ^ = 900

A and B leave from a and B at the same time, and meet in 2 hours. After meeting, the two vehicles continue to move on, and when a arrives at B
When car a arrives at B, car B is 60 kilometers away from A. It is known that the speed ratio of the two cars is 3,

60 △ [3-2] = 60 km, 60 × 3 = 180 km, 180 △ 2 = 90 km, 90 △ [3 + 2] × 3 = 54 km, 90-54 = 36 km
A: 54 kilometers for car a and 36 kilometers for car B

Let E / f be the midpoint of the edges AD and BC of the spatial quadrilateral ABCD, then what is the size of EF and [1 / 2 (AB + CD)]?
Why pinch?
I'd like to know. I'm stuck in this problem

Let the midpoint of AC be g, and connect eg and FG, then eg and FG are the median lines of △ ACD and △ ABC respectively, so eg = CD / 2, FG = AB / 2, and because eg, FG and EF are the three sides of △ EFG, then eg + FG > ef (the sum of the two sides is greater than the third side), that is, CD / 2 + AB / 2 > EF

There are two piles of coal, a and B. it turns out that a is five eighths of B. If 22 tons of coal are transported from B to a, then seven ninths of a and B. how much is a and B originally

In the past, B is 1 △ (1 + 5 / 8) = 8 / 13 of the total amount, and then B is 1 △ (1 + 7 / 9) = 9 / 16 of the total amount. There are 22 △ (8 / 13-9 / 16) = 416t in two piles of coal. In the past, B has 416 x 8 / 13 = 256t, in the past, a has 416-256 = 160t. A: in the past, a has 160t, B has 256t

What is the check function? How to use the check function to solve the problem that the mean inequality can not solve?

1. Concept: the general form of check function is f (x) = x & nbsp; + & nbsp; a & # / X & nbsp; & nbsp; (A & gt; 0)
2. Parity and monotonicity: it is easy to get that the check function is odd
The monotonicity of check function can be obtained by derivation or by definition. It has four monotone intervals
It is an increasing function on (- ∞, - A] and [a, + ∞), and a decreasing function on [- A, 0) and (0, a]
3. Image: because it is an odd function, the image is symmetrical about the origin, and then according to the monotonicity, the image of the function can be obtained
(2) the image of the check function has two vertices, which are symmetrical about the origin, a (a, 2a) and B (- A, - 2A)
(3) the image of the check function has two asymptotes, which are Y-axis and straight line y = X. the image of the check function is sandwiched between asymptotes and has two symmetrical "hooks"
4. Examples of solving problems that can not be solved directly by means inequality:
This is the smallest value of the function, which is the smallest value of the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest value of the function, which is the smallest of the function, which is the case (x-178; (178; (178; + 4) (x-178; (178; + 4) in the case (X & \\\\\\\\\\\\\\\\\\\\\\nbsp; & nbsp;
(x²+5)/√(x²+4)=(x²+4+1)/√(x²+4)
=√(x²+4)+1/√(x²+4)
≥2√(x²+4)•1/√(x²+4)]=2
So the minimum value of F (x) is 2
② Error cause analysis: since the minimum value of √ (X & # 178; + 4) is 2, it can not be equal to 1 / √ (X & # 178; + 4), and the above inequality cannot be "=". It is certainly not feasible to use the formula directly
③ Application of check function
Let t = √ (X & # 178; + 4), t ≥ 2, then T & # 178; = x & # 178; + 4,
g(t)=f(x)=(x²+5)/√(x²+4)=(t²+1)/t= t+1/t ,t≥2
Since f (x) = g (T) = t + 1 / T & nbsp; is an increasing function on [2, + ∞), note: in fact, an increasing interval is [1, + ∞)
Therefore, when t = 2, there is a minimum value of 5 / 2