A math problem binary inequalities .

Because 26 10>30 7, it saves money, not waste.

Second question: Let's say the group is x people.

List unary inequalities:

10x>30 7 (because a group ticket is only cost-effective if you pay more for a single ticket than if you buy a group ticket).

Solution: x>21.

And because the number of people is an integer, the smallest integer of 21 is 22

A: When there are at least 22 people, buying more tickets will cost less.

Buy 30 tickets: 30*7=210 yuan.

Buy 26 tickets: 26 * 10 = 260 yuan.

So it's not a waste.

When buying 21 tickets: 21*10=210 yuan.

When there are more than 22 people, buying more tickets will cost less.

The number of people is set to x.

10 * x = 30 * 7 = 210x = 21

That is, when the number of people is 21, the cost of the two ways to buy tickets is equal.

When the number of people is greater than 21, it costs less, i.e. buying more tickets costs less.

When the number of people is less than 21, it costs less to buy a single ticket, that is, to buy more tickets and spend more.

So it's not a waste to buy 30 tickets at this time.

If you buy 30 tickets, it costs 210 yuan.

Buy 26 sheets, spend 260

When there are at least 22 people, buy more tickets and pay less.

So don't waste.

No waste. If you buy 30 tickets, it costs 210 yuan.

Buy 26 sheets, spend 260

When there are at least 22 people, buy more tickets and pay less.

No waste, 26 * 10 = 260 yuan.

7*30=210 yuan.

If you buy more, you will have less money.

Forehead. Not a waste.

First of all: 10*26=260 (yuan).

7*30=210 yuan.

Proof of no waste.

And then 21 people were the same.

So, when there are 22 people (inclusive), buy more tickets and spend less.

No waste. 26x10=260 (yuan).

30x 7=210 (yuan).

So don't waste.

210 10 = 21 (person).

It's the same when 21 people.

So spend less when there are at least 22 people.

1>26x10=260 yuan 30x7=210260>210 So don't waste it.

2> Setup requires x people.

26x210x<26x

x<105/3x

x>=22

No, if you buy 26 tickets, it will cost 260 yuan.

If you buy 30 tickets, you will need 210

210 is less than 260

When there are more than 21 people, buy more tickets and spend less.

This problem can be done by drawing a number line, which is equivalent to x to -2009, -2008, -2007 ......Distances of 2007, 2008, 2009. And when -2009 x 2009, f(x)=2(1+2+3+4+5+......2009) is a certain value, and when the absolute value of x is greater than 2009, f(x) is also greater than 2(1+2+3+4+5+......2009）

So, at -2009 a*2-3a+2 2009,-2009 a-1 2009, f(a*2-3a+2)=f(a-1), there are infinite solutions.

a*2-3a+2 2009 or a*2-3a+2 -2009, a-1 -2009 or 2009 a-1, only a*2-3a+2=a-1 or a*2-3a+2+a-1=0, and there are countless solutions (this step only needs to consider a*2-3a+2=a-1 or a*2-3a+2 to -2009 or 2009 is equal to the distance from a-1 to 2009 or -2009).

So, to sum up, if there is a solution, choose D

1. Let x children and y apples be, and the equation is:

5x+12=y

8x y gives 8x 5x+12

x＞4y＜32

2. Solution: The two formulas are added to obtain x=-m-1

Subtract the two equations to yield: y=(3 2)m-2

From x is positive to get -m-1 and 0 to get m -1

From y is a negative number, we get (3 2) m-2 0 to get m 4 3, so we get m.

Value range. m＜-1

3. (1) y=25x+15(10-x), simplify.

Get y=150+10x

2) Known 180 Y 200, x 4.

180≤150+10x≦200

Solution: Refers to Ashiga 3 x 5

This gives x=4 or x=5

When x=4, y=190

When x=5, y=200

Therefore I choose. Large buses.

When there are 4 cars, the fare is the least.

Hello Old Hub:

In fact, it is based on the invested funds, to judge which is big, and now the correct answer is as follows:

If the principal of the investment is x, then use the first method, that is, the sales at the beginning of the month, so the profit g(x)=x*10%+(x+x*10%)*20%=, with the second method, that is, the sales at the end of the month, the profit is f(x)=x*40%-800= where the domain of this question is x 0

Therefore, when g(x)>f(x) is solved, 0 x<10000 is solved, and the first method, that is, the beginning of the month, is used, and the profit is more.

When g(x)=f(x), i.e. x=10000, both ways will work, and the profit will be the same.

When g(x)10000, use the second method, i.e., sell at the end of the month, and then make a profit.

I did it myself, I don't know if it's right.

Solution: Set the cost to x yuan.

x(1+10%)+x(1+10%)(1+20%)+x(1+10%)(1+20%)(1+40%)≥800

x≥187..4

That is, when the cost is greater than or equal to yuan, the profit can be greater than or equal to 800 yuan.

For example:

x²-3x+2<0

x-2)(x-1)<0

x-2>0 and x-1<0....

or x-2<0 and x-1>0....

Solution, get:

solution, gets: 1 In summary, the set of solutions of the original inequality is (1,2).

If the number of people is x, the number of students is x-1

Travel agency fee = 240 + 120 * (x-1) = 240 + 120x-120 = 120 + 120x

B travel agency fee = 240 * 60% * x = 144x order 120x + 120 = 144x to get x = 5

So when the number of people is 5, it is okay to participate in both travel agencies; When the number of people is greater than 5, it is more cost-effective to participate in travel agency A; When the number of people is less than 5, it is more economical to participate in travel agency B.

1.By 2x<4 x<2 because their solution set is the same, so (a-1)x-1 of 2>x a Since there are 5 integer solutions, so a = —3

Pure hand-hitting: Thank you.