A T-shirt is priced at 200 yuan and can sell 500 pieces per month. According to the survey, 50 more T-shirts can be sold for every 3 yuan price reduction. There are 1000-1200 T-shirts at present. If you want to sell them out in a month, how much should you sell?

Let's reduce the price by X Yuan
The solution of (200-x) (500 + 50x) ≥ 1000 (200-x) is x ≥ 10
（200-x）（500+50x）≤1200（200-x） x≤14
A. the price is 10-14

If there are only six positive integer solutions to the inequality 3x-a ≤ 0 about X, then a should satisfy______ ．

The solution set of inequality 3x-a ≤ 0 is: X ≤ A3, its positive integer solution is 1, 2, 3, 4, 5, 6, then 6 ≤ A3 < 7, the solution is: 18 ≤ a < 21. So the answer is: 18 ≤ a < 21

A mathematical problem of linear inequality of one variable
There is a highway with a, B and C in turn. The speed of a is 40km / h from a to C, and that of B is 10km / h from B to C. If two people set out at the same time, a arrives 6 minutes earlier than B. If two people set out in t hours, the value range of T is calculated
I forgot to say that AB is 8 km and BC is 4 km, so please answer again. I forgot to add the conditions.

The distance length of AB and BC is expressed by AB and BC respectively. After t hours, BC = 10 * t, AC = 40 * (T-6), and because of AC

There are linear inequalities of one variable in mathematics
The known equations X-Y = 2K
The sum of solution X and y of {is negative, and the range of value of K is obtained
x+3y=1-5y .

Solving equations
{ x-y=2k
X + 3Y = 1-5y, we get {x = (1 + 8K) / 5
y=(1-2k)/10
So x + y = (1 + 8K) / 5 + (1-2k) / 10 = (3 + 14K) / 10
From the meaning of "the sum of X and Y is negative", we can get: (3 + 14K) / 10 < 0
If we solve this inequality, we get k ＜ - 3 / 14
A: the range of K is k ＜ - 3 / 14

On the inequality system x-a > 2 and b-2x greater than 0, the solution set is - 1 less than x less than 1.. then what is the power of (a + b) 2009
If the solution set of the inequality system X is greater than M-1 and X is greater than m + 2 is x greater than - 1, then M is equal to. The sooner the better. There is no chance after this day

1∵x-A>2,B-2X>0
one

Inequality of degree one variable
1. Given the square of | X-2 | + (2x-3y-a) = 0, find the value range of A. (the square is only the square of (2x-3y-a), not the absolute value.)
2. In the system of equations 2x + y = 1-m
3X+4Y=2
If the unknowns X and y satisfy x + Y > 3, find the value range of M
3. It is known that y = 3x-2, when what is the value of X, - 3 ≤ Y0 and y
Please take a good look. Here are four questions. If you answer well, you will get extra points,

Given that the square of | X-2 | + (2x-3y-a) is 0, find the value range of A. (that square is only the square of (2x-3y-a), not including the absolute value) because the absolute value and the square number are greater than or equal to 0, so: X-2 = 0; 2x-3y-a = 0x = 23y + a = 4Y = (4-A) / 3. Is there a value range of Y in the title? For example: Y >

Xiao Ming's father bought a taxi (including the right to operate) with 500000 yuan. After it was put into operation, the total income of each year was 185000 yuan, while the total expenditure of various expenses was 60000 yuan. (1) how many years did the taxi start to make profits? (profit refers to the difference between the total income minus the total expenses of car purchase and various expenses is positive), (2) if the term of taxi operation right is 10 years and the old car can recover 5000 yuan when it expires, what is the average profit of the car in these 10 years?

(1) Suppose the car starts to make a profit after X years of operation, and the profit is y 10000 yuan, then y = (18.5-6) x-50, that is, y = 12.5x-50 ∵ y ∵ 0 ∵ 12.5x-50 ∵ 0 ∵ x ∵ 4 ∵ the car starts to make a profit after 4 years. (2) when x = 10, y = 12.5 × 10-50 = 75, (75 + 0.5) △ 10 = 75500 yuan ∵ the average profit of the car in this 10 years is 75500 yuan

On the mathematical problems of linear inequality of one variable
X/3-1/2>X X/5>3+(X-2/2) -2X+1

X/3-1/2>X
2x/3

A mathematical problem of linear inequality of one variable
The equation (K + 2) X-2 = 1-k (4-x)

KX+2X-2=1-4K+KX
KX-KX+2x=1+2-4k
2x=3-4k
x=1.5-2k
∵X＜0
∴1.5-2K＜0
∴2K＞1.5
K＞0.75
When k is greater than 0.75, X is negative

A mathematical problem about scheme design and one variable inequality
A car rental company needs to buy 10 cars and minibuses. Among them, at least 3 cars need to be purchased. Each car is 70000 yuan, and each minibus is 40000 yuan. The company can invest no more than 550000 yuan
(1) There are several car purchase plans that meet the requirements? Please explain the reasons
(2) If the monthly rent of each car is 110 yuan, assuming that the newly purchased 10 cars can be rented every month, and the monthly rent of these 10 cars is not less than 1500 yuan, which purchase scheme should be chosen?

(1) If x cars are to be purchased, then (10-x) vans are to be purchased. According to the meaning of the question, 7x + 4 (10-x) ≤ 55 and the solution is x ≤ 5. Because x ≥ 3, then x = 3, 4 or 5. So there are three schemes to purchase cars: scheme 1: 3 cars, 7 vans; scheme 2: 4 cars, 6 vans; scheme 3: 5 cars

A T-shirt is priced at 200 yuan and can sell 500 pieces per month. According to the survey, 50 more T-shirts can be sold for every 3 yuan price reduction. There are 1000-1200 T-shirts at present. If you want to sell them out in a month, how much should you sell?