On Sunday, the class leader Wang Xiaoming organized 50 students to go to Nanhu Park. There are two kinds of boats available for hire in the park. Each big boat can seat 5 people at most, and the rent is 30 yuan for each boat, and the rent is 20 yuan for each boat. Please use what you have learned to design a scheme so that the boat does not leave vacant seats and pays the least rent. Remember: use the solution of one variable linear inequality equation~

On Sunday, the class leader Wang Xiaoming organized 50 students to go to Nanhu Park. There are two kinds of boats available for hire in the park. Each big boat can seat 5 people at most, and the rent is 30 yuan for each boat, and the rent is 20 yuan for each boat. Please use what you have learned to design a scheme so that the boat does not leave vacant seats and pays the least rent. Remember: use the solution of one variable linear inequality equation~

If x-seat large ship is set, then (50-x) small boat
Because 20 * 50 / 3 > 30 * 50 / 5
So 20 * 50 / 3 > = 30 * x / 5 + 20 * (50-x) > = 30 * 50 / 5
88 > = x > = 50
Since there are only 50 people, x = 50
That is to say, 50 people are in a big boat

What is the minimum value of the sum of the inner angles of an n-polygon which is at least 120 ° greater than the sum of its outer angles?

（n-2）•180°-360°≥120°，
The solution is n ≥ 14
3，
So the minimum value of n is 5
So the answer is: 5

Mathematics in the second year of junior high school 1. In October 2008, 200 people in Shenzhen were obliged to take part in the activity of "picking up the Wutong Mountain". Three students, including Xiaoming, were scheduled to clear up 44 hectares of mountain forest within 7 hours, and began to finish 8 hectares in 1 hours. Now they finish the task 2 hours ahead of schedule, and ask how many hourly hourly hours they will finish at least every hour in the next few hours. Many I will add points!

After that, the average completion time is x ha per hour
（7-1-2）x≥44-8
x≥9
A: after that, we will finish 9 hectares per hour on average

One is the application of inequality A batch of parts processing tasks are handed over to Party A and Party B. in order to arrange conveniently, the working days of Party A and Party B must be full days. It is known that it takes 12 days for Party A to complete the task alone, and 14 days for Party B to complete the task alone (1) According to the meaning of the title, we list the bivariate linear equations about X and y (2) Find the natural number solutions of the listed equations (3) How to arrange the working time of Party A and Party B to minimize the total time limit for completing the task?

(1) According to the meaning of the title, X / 12 + Y / 14 = 1 (2) y = 14-7x / 6, x = 0, y = 14; X = 6, y = 7; X = 12, y = 0 (3) a and B work at the same time for 6 days, and B works for another day?

Someone brought 100 yuan to the store to buy some drinks, and spent 60 yuan. Later, she bought 4 kilograms of pears (3 yuan per kilogram) and 5000 grams of apples. After paying, she still had a balance. If he bought 6 kilograms of pears and 5000 grams of apples, the money he brought was not enough. What's the price range of apples? Solution of column inequality

5x+3*4+60<100
5x+3*6+60>100
therefore
5.6>x>4.4

Application problems of inequality group (relatively simple) The admission fee of a park is 10 yuan per person, and 20 people (including 20 people) can get a 20% discount on group tickets. When is it cheaper to buy group tickets for 20 people than individual tickets when the number of people is less than 20?

There are x persons
10x>10*0.8*20
x>16

If a tourist group stays in a hotel in the scenic area, if there are 4 passengers in each room, 20 people can't make arrangements; if there are 6 passengers in each room, one is not empty and dissatisfied. How many rooms does this hotel have? Set up a group of inequalities, the solution will not bother you

There are x rooms in this hotel
4x+20=Y
Zero

The following inequalities are proved by using the monotonicity of functions 1. E × > 1 + X, X is not equal to 0 2.Lnx

Let f (x) = e ^ x-x-1
Then f '(x) = e ^ X-1
From F '(x) = 0, x = 0,
F (0) = 0 is the minimum value, which is also the minimum value
Therefore, when x is not 0, there is f (x) > F (0) = 0
That is, e ^ x > 1 + X
Two
Let f (x) = x-lnx
f'(x)=1-1/x=(x-1)/x
X = 1 is the minimum and also the minimum
f(1)=1
So there is f (x) > 1 > 0, that is, x > LNX
Let g (x) = e ^ X-X
Then G '(x) = e ^ X-1. When x > 0, there is g' (x) > 0
That is, G (x) increases monotonically
g(0)=1
So there is g (x) > 1 > 0, that is, e ^ x > X
Therefore, when x > 0, there is LNX

Let f (x) = a part x-x 1 (a is a constant) 1. Solve the inequality FX > 0 2. When a = 2, judge and prove the monotonicity of function FX The known function f (x) = x of a to 1 of X (a is a constant) 1. Solve the inequality FX > 0 2. When a = 2, judge and prove the monotonicity of function FX, and find the value range of function on [1,2]

When a = 2, f (x) = x / 2-1 / x, let any x ₁ x ∪ (- ∞, 0) ∪ (0, + ∞), and X ₁ ₁∪ (- ∞, 0) ∪ (0, + ∞), and X ₁∑ ₁∪ (0, + ∞), and X ₁₁﹤ f (x ₁) - f (x Ψ) = x ₁ / 2-1 / XS

Using monotonicity of function to prove inequality, when x > O, 1 + (1 / 2) x > √ 1 + X

f(x) = 1+(1/2)x - √(1+x),x> 0
f'(x) = 1/2 - 1/2 / (1+x)^(3/2)
When x > 0, f '(x) > 1 / 2 - 1 / 2 / (1 + 0) ^ (3 / 2) = 0
So f increases with x > 0, so f (x) > F (0) = 0, i.e. 1 + (1 / 2) x > √ (1 + x)