There are two motor vehicles, A and B, on a two way four lane road

Solution: (1) Method 1: Y is a primary function of x from the graph, let y=kx+b (1 point).

The image passes through the dots (0,300), (2,120), 2 points).

Solution, (3 points).

y=-90x+300．

That is, the expression y y about x is y=-90x+300 (4 points).

Method 2: As you can see from the diagram, when x=0, y=300;When x=2, y=120

So, the highway is 300 kilometers long

The 2-hour journey of car A is 300-120=180 (km).

The speed of car A is 180 2 = 90 (km/h) (3 minutes).

The expression for y about x is y=300-90x (y=-90x+300) (4 points).

2) s=-150x+300 (5 points).

3) In s=-150x+300 when s=0, x=2

That is, the two cars A and B meet after 2 hours (6 minutes).

In y=-90x+300, when y=0, x= So, the time it takes for car B to reach the end point after meeting is -2=2 (hours).

The speed of car B and car A after meeting a = (300-2 60) 2 = 90 (kmh).

a=90 (kmh) (7 points).

Solution: (1) Y is a primary function of x from the graph, let y=kx+b pass through the points (0,300), (2,120), and the solution is y=-90x+300

2）s=-150x+300．

3) In s=-150x+300 when s=0, x=2 that is, the two cars of A and B meet after 2 hours.

In y=-90x+300, when y=0, the time taken by car B to reach the end point after x=10 3 meets is 10 3+2 3-2=2 (hours) a=(300-2 60) 2=90 (kilometers hour).

If the distance length is x (km), the length of ordinary highway (x-60) is 2km, and the length of highway is x 2+30km

then [(x-60) 2] 60+(x 2+30) 100=3

Q: How many hours does the highway take, but what about the ordinary road?

Solution: The highway was opened for x hours, and the ordinary highway was opened for (3-x) hours.

100x-60（3-x）=60

x = A: ......No plagiarism.

Let's assume that the speed of car B is x kilometers per hour.

450/(x+15)=400/x

x=120x+15=135

The speed of car B is 120 km/h.

The speed of car A is 135 km/h.

(450-400) 15 10 3 (hours).

450 10 3 135 km h.

400 10 3 120 km h.

A walks 450 meters.

B walks 400 meters.

The distance ratio is 9:8

For the same time, the velocity ratio is also 9:8

A speed: 15 * 9 = 135km

B speed: 135-15 = 120km

I also think it's a, d is right, 7, choose d, 1, no.

cd three cars may be driving in the same direction, all to the west: A is the slowest B is the fastest, 1, I think it is cd because the trees take the car as the standard, to the east, then car C must go west, but to give all the examples, d is absolutely correct, the question C may be two words, so it is not accurate, if it is a single choice, so choose d, 1, if there are three cars A, B and C at the same time on an east-west street lead sun, A, B and C three cars at the same time on an east-west street on the street, The person in car A sees car C moving westward relative to car A, the person in car B sees that car A and car C are moving east relative to car B, and the person in car C sees the roadside trees moving westward, so the correct statement is ( ).

Car A must drive east, and car B must drive west.

c. Vehicle C may travel west, and d. Three vehicles may travel in the same direction.

The answer to this question I analyzed is A, D, may I ask it?

What are the requirements? AB Distance?

Let ab travel a distance of x

1 3) x 60 + (2 3) x 100 = Simplification: x 180 + x 150=

Multiply both sides by 900: 5x+6x=1980

11x=1980

x=80 So, AB is 180 km.

The time for driving on a normal highway is x, and the length of an ordinary highway is y.

The length of the ordinary highway travel time The driving speed, ie

y=60x ..

Length of the highway = travel time Travel speed = (

The length of the highway (two-thirds of the total length) = the length of the ordinary highway (one-third of the total length) 2, there is:

2y=100( .

A binary linear equation is formed by and

y=60x ..

2y=100( .

Solution: x = 1 hour.

y = 60 km.

Then the length of the highway = 2y = 120 km.

The driving time on ordinary roads is 1 hour and the length is 60 km;

The driving time of the highway is hourly and the length is 120 km.

Let the total distance be x, then the time to travel on the ordinary road is x 3 divided by 60 The time to travel on the highway.

Time multiplied by velocity = distance so (

x=180m

Question: What is the total distance traveled? Solution: Let the total distance be xkm x (3*60)+(2x) (3*100)= x=180km by the time relation

A is between B and C, 60 km from B and 90 km from C.

The coordinates of the point are (,0).

Let the abscissa of the m point be a, and the velocity of car B is b, and the equation system is obtained: a b=90, (2-a) b=60Find a=.

The abscissa of point m is the time when car B arrives at place a.

2.The speed of car A can be set to c. Again, 1 c = 60, find c = 60 (km h).

From problem 1, it can be seen that when car A arrives at place C, the distance from place A is 90 kilometers, so it can be seen that the final ordinate of the function image of car A is 90, and the speed of car A is 60 kilometers per hour, so the abscissa of the function image of car A is x=90 60+1=. That is, the coordinates of the final point of the image of the car function are (90, when 0 x<1, y1=-60x+60;

When x=1, y1=0;

It's not embarrassing to count a little potato cherry blossom when 1 is counted, and I'm also writing about it.

<> socks are only used to use trapezoidal announcements to close t=6 rounding.