Math problems are difficult, math problems, not too difficult

Think about it along the following lines:

According to "10 cows can eat for 20 days", it can be calculated that 10 20 = 200 (head) cattle can be eaten in 1 day.

According to "15 cows can eat for 10 days", it can be calculated that 15 10 = 150 (head) cows can be eaten in 1 day. This is because the grass in the meadow grows less by 10 days (20-10 days) and the number of cattle varies by 50 (200-150). From this, it can be seen that the grass that grows every day can feed 5 cows (50 10) for 1 day.

How many cows can eat the original grass in the meadow (excluding the new grass) for 1 day?

10-5) 20=5 20=100 (head).

Or: (15-5) 10=10 10=100 (head) Now there are 25 cows, because the grass growing in the meadow is enough to support 5 cows. Just calculate how many days the remaining 20 cows will eat enough grass to find the result.

100 (25-5) = 100 20 = 5 (days).

Number of new grass per day: (10*20-15*10) (20-10)=5

Original grass amount: 10*20-5*20=100

Answer: 100 (25-5)=5

1.Linear programming is used for this problem. Set up x vehicles.

Y cars have been made. You can get 3 formulas.

40x+30y<=290...

10x+20y<=100...

x+y=8...

Just draw a linear programming diagram from .

Go to the diagram that meets the criteria.

Integer points. There are several options for a few points.

2.This question is based on the answer to the previous question and the plan, so I won't go into details.

I'm sure you should get the idea.

From the meaning of the title, get the system of inequality equations:

40x+30(8-x)≥290

10x+20(8-x)≥100

Solution 5 x 6 That is, there are two car rental schemes: the first is to rent 5 cars of type A and 3 cars of type B; The second is to rent 6 cars of type A and 2 cars of type B

2) The cost of the first car rental scheme is 5 2000 + 3 1800 = 15400 (yuan); The cost of the second car rental plan is 6 2000 + 2 1800 = 15600 (yuan).

Therefore, the first car rental option is more cost-effective

Set car A x car, car B y car, then.

40x+30y≥290

10x+20y≥100

x+y=8 solution: 2 y 3

So there are two schemes, car A 6, car B 2;

Car A 5, car B 3

Program fee: 2000 * 6 + 1800 * 2 = 15600 Program fee: 2000 * 5 + 1800 * 3 = 15400 So choose plan A5 B 3.

y=(x-40)*[500-20*(x-50)]

When the unit price is yuan, it is the most profitable.

According to the question, when Xiao Ming walked 50m, Xiao Lin walked 48 meters, and Xiao Long walked 45m, so the speed ratio of the three people was.

Xiao Ming: Xiao Lin: Xiao Long 50:48:45

1. When Xiao Ming reaches the end point, the distance traveled by Xiao Long is 100 50 45 90m100 90 10m

So, the little dragon is still 10 meters away from the finish line.

2. Xiao Ming's speed is 100 30 10 3 meters in seconds, so Xiaolong's speed is 10 3 50 45 3 meters in seconds.

1100÷2=50

2 Xiao Ming reached the end in 30 seconds, and Xiao Long walked 90 meters in 30 seconds.

So 90 30 = 3 (meter seconds).

The time it took for Xiao Ming to walk halfway through the whole process was x seconds.

Then Xiao Ming's speed is 50 x

Kobayashi speed is 48 x

The dragon speed is 45 x

The time it takes for Xiao Ming to reach the end point is 100 (50 x)=2x seconds, at this time, the distance traveled by the dragon is (45 x)*2x=90m, then the dragon is still 100-90=10m from the end point

It took 30 seconds for Xiao Ming to reach the end point, and 2x=30, then x=15, so the dragon's speed is 45 15=3 m seconds.

1. Do they go at the same time, if so, the time they spend is the same, v Ming = 50 t, v Lin = (50-2) t, v dragon = (50-5) t, so v Ming v dragon = 50 45, since the speed is the same, so s Ming s dragon = v Ming v dragon, so when s Ming = 100, 100 s dragon = 50 dragon = 90, 100-90 = 10

2. The above question knows the ratio of V Ming to V Dragon, and the time when Xiao Ming runs the whole distance is known, so V knows it, and V Dragon knows it.

Cross-sectional side length = 2 decimeters.

2 meters = 200 decimeters.

Surface area = 4 * 2 * 200 + 4 + 4 = 1606 square decimeters.