A few math questions about inequality equations, which are in urgent need of answers!!

1. Solution: From the meaning of the question, 2(x+70) 350 and 70x 7560 can be solved, and the solution is 105 x 108

The length of the football field for international matches is between 100m.

This stadium cannot be used for international football matches.

2. Solution: Let the speed of the detonator run as v.

1 Solution v 300

After lighting the fuse, the detonator must travel at least 300 meters per second to reach the safe area.

From the question, it can be set to rent x 45-seater passenger cars just full, then 45x=60(x-1)-30, and x=6, the total number of people is 45x=270. And know that the rent of 45-seat passenger car is 250 yuan, and the rent of 60-seat passenger car is 300 yuan, then set up x 45-seat car y 60-seat car is the most economical, and the inequality can be established, 45x+60y is greater than or equal to 270, 250x+300y is less than or equal to 1500, then the final solution is x=2, y=3, and the minimum rent is equal to 1400

Set a separate rental of 45-seater passenger cars for x cars, then.

45x=60(x-1)-30, and the solution is x=6

Now it is necessary to rent two kinds of buses at the same time, you can set up a rental 45-seat passenger car Y and a 60-seat passenger car 5-Y rental car.

Therefore, 45y+60(5-y)>=45*6, and y<=2 If you rent 2 45-seater passenger cars, you need to rent: 250*2+300*3=1400 (yuan).

If you rent a 45-seater bus, you need to rent: 250 * 1 + 300 * 4 = 1450 (yuan).

Therefore, 45-seater buses and 60-seater buses should be rented 2 and 3 respectively, so that the rent is minimal.

Let's say a car is required to rent a 45-seater passenger car separately.

60*（a-1）-30=45a

a=6Total number of people = 45*6=270 people.

In order to save rent, it is assumed that X and Y buses are rented separately.

x+y=6-1

45x+60y≥270

Obtained by formula 2: 45*(x+y)+15y 270 --3 formula 1 is substituted for formula 3 to obtain: y 3

When y=3, x=5-3=2

Calculation: 45 * 2 + 60 * 3 = 270 meet the conditions, the cost is 250 * 2 + 300 * 3 = 1400 yuan.

When y=4, x=5-1=4

Calculation: 45 * 1 + 60 * 4 = 285 meet the conditions, the cost is 250 * 1 + 300 * 4 = 1450 yuan.

When y=5, x=0

Check: 60 * 5 = 300 Qualified, the cost is 300 * 5 = 1500 yuan Answer: In order to save rent and ensure that each student has a seat, it is necessary to rent 2 45-seat buses and 3 60-seat buses to make the rent the least.

According to the title, if you rent a 45-seater car separately, the total number of people is 45a, and there are 30 seats left in a 60-seater car rented A-1.

then (45a+30) 60=a-1

If a=6, there are 45*6=270 people.

The total number of rental cars is one less than the 45-seater car, that is, 5 cars, if the 5 cars are 60 seats, then there are 60 5 = 300, a total of 300-270 = 30 more seats than the actual number of seats, and each 60-seater car is 60-45 = 15 seats more than each 45-seater car, because the reduced number of seats is due to the 45-seater car as 60 seats, so the number of 45-seater cars is 30 15 = 2

Then 60-seater cars are 5-2=3 cars.

The number of cars that require 45 seats is x

45x=60（x-1）-30

x=6 then the number of students is 270

Set the need for 45-seater passenger car for x60-seater passenger car for Y

45x+60y=270

x+y=5 gives x=2y=3

So 2 for 45-seater buses and 3 for 60-seater buses.

48 people go 8 times per person, and a total of 48 * 8 = 384 (person-times) need to go. Set each time to go to x people, that is, buy x cards, then the number of times to take is 384 x (times), let the total expenditure be y, then there is y=40*384 x+240x, if and only if there is a minimum value when 40*384 x=240x, the solution is x=8, that is, buy 8 cards, the minimum value of y is (40*384 x)*(240x)=48*40, that is, each person can pay at least 40 yuan.

If you want to buy x cards, the number of swims is 48*8=384, so you have to go 384 x times, so the total expenditure = 40*384 x+240x, so you can use the basic inequality, you can calculate x=8, and then everything else can be calculated.

Solve systems of equations. 2x+3y=3m-1---

4x-5y=m-9---

Get: x=(9m-16) 11

y=(5m+7)/11

x<0 and y>0

9m-16) 11<0 solution: m<16 9

5m+7) 11>0 solution: m>-7 source leak 5 So: -7 5m is the integer of the local fissure, which can be: -1, 0, 1

Solution: There are x people in the class, and the opening fee is a total of Y yuan.

12(x-1）-y=84(1)

10（x-1)40(3)

y=12x-12-84=12x-96(4) brings (4) into (2).

10（x-1)<12x-96

10x-10<12x-96

2x>86

x>43

Bring (4) into (3).

11（x-1）-（12x-96)>40

11x-11-12x+96>40

x+85>40

x<45

43x is a positive integer.

x=44,y=12x-96=12x44-96=432Answer: There are 44 students in the class, and the opening fee is 432 yuan.

12x-84=11x-40 gives x=44

Therefore, there are 44 + 1 = 45 people, and the opening fee is 444 yuan.

Solution: Let the tourist walk out x km, x (6 + 1) + x (6-1) 3 x 7 + x 5 3

5x 35 + 7x 35 3 12x 35 3x 35*3 12 = what you want.

It can be simplified to:

Let the downstream time be x and the countercurrent time be y

6+1）*x=（6-1）*y

Because x+y<3, find (6+1)*x

Simplification to 7x=5y

Because x<3-y

So 5y<7(3-y).

y<7/4

6+1)*x=(6-1)*y<35 4=So you can go out of a kilometer at most.

1 solution: Set the original production of x vehicles per day.

then 15(x+6)>20x

x<18 Because x is an integer, the maximum integer value of x = 17 A: Originally, a maximum of 17 units were produced per day.

2.There are x rooms on the ground floor, and there are (x+5) rooms on the second floor, which is 4x<18 from the title

5x>18

4(x+5)>18

Solving inequalities gives 9 2>x>18 5 so x=4 (between) 3The actual charge of travel agency A is w=500*2+500*the actual charge of travel agency B is n=500*

When m>n is 350x+1000>400x+800x<4 (people), choose B travel agency.

When x > 4 people, choose a travel agency.

When x = 4 people, both are fine.

4.If the discount of x is calculated, then 2190x+<=2190(1+x<=, so at least 8% discount is costable.

5.(1) Let the profit after x years be 72x>120+40x x>15 4 x is an integer, so x=4

So make a profit after 4 years.

2) (72*15-120-40*15+20) 15=76 3 (10,000 yuan).

6.Set the invested funds at the beginning of October ** profit m = (1+

Profit at the end of the month n=

When m>n, i.e., when x<20000, the profit at the beginning of the month is more, and when the profit is more at the end of the month when x<20000.

Same when x=20000.