A mathematical problem about the linear inequality of one variable? When a supermarket purchases a batch of fruits, the quality loss is 5% in the transportation process, assuming that the supermarket does not include other costs (1) If the supermarket takes 5% of the purchase price as the selling price, please use the calculation to explain whether the supermarket is losing money (2) If the supermarket wants to make at least 20% profit, what is the minimum price increase of this fruit? (the result is accurate to 0.1%) It's a simple inequality

（1）（1-5%）×（1+5%）=1-0.0025

Mathematics problems of grade two in junior high school
In the past two years, the export-oriented economy of a certain place has developed rapidly. Some famous companies have settled down in the New District, and the demand for all kinds of talents is increasing. Now a company recruits personnel for the society. The information is as follows:
Information 1: recruitment object: 150 mechanical manufacturing and planning and design personnel
Information 2: Salary: 600 yuan / month for mechanical personnel and 1000 yuan / month for planning and design personnel
The company recruitment machinery manufacturing and planning and design personnel were x, y people
(1) The algebraic expression containing x is used to express y;
(2) If the company pays the monthly salary of the recruited personnel is p yuan, it is necessary to make the recruitment of planning and design personnel not less than 2 times of the mechanical manufacturing personnel, and find the value range of P

⑴y=150-x.
(2) according to the meaning of the question, y ≥ 2x
The solution is: X ≤ 50
And ∵ x ≥ 0150-x ≥ 0
∴0≤x≤50
∴p=600x+1000(150－x)
=-400x+150000
Method 1: x = (150000-p) / 400
The solution is 130000 ≤ P ≤ 150000
(method 2) P decreases with the increase of X, and 0 ≤ x ≤ 50,
That is, 130000 ≤ P ≤ 150000

Solving inequality X & # 178; - LXL - (1 / 2) x > 0

Less than - (1 / 2) or more than 3 / 2

The process of solving inequality 7x-x & # 178; ≥ 0

7x-x^2≥0
x^2-7X

Let the solution set of inequality ax & # 178; + 7x + 6 > 0 be (1 / 4,1 / 3) to find the value of a and B

Ax & # 178; + 7x + b > 01, suppose a > 0, there are: X & # 178; + 2 (7 / a) x + (7 / a) & # 178; - (7 / a) & # 178; + B / a > 0 (x + 7 / a) & # 178; > (49 AB) / A & # 178; in this case, even if the inequality has a solution, it is different from the given solution. 2, suppose a < 0, there are: X & # 178; + 2 (7 / a) x + (7 / a) & # 178; - (7 / a) & # 178

Find the inequality X & # 178; - x + 2 > 0

x^2--x+2>0
x^2--x+1/4+7/4>0
(x--1/2)^2+7/4>0
Because (X -- 1 / 2) ^ 2 > = 0
So (X -- 1 / 2) ^ 2 + 7 / 4 > 0 holds
So the solution of inequality x ^ 2 -- X + 2 > 0 is: all real numbers

Let f (x) = asin (π x + m) + bcos (π x + k), if f (2009) = 1, then f (2010)=

f(x)=Asin(πx+m)+Bcos(πx+k)
f(2009)=Asin(π+m)+Bcos(π+k)=-Asinm-Bcosk=1
f(2010)=Asin(m)+Bcos(k)=-1
Please advise!

We know the function f (x) = asin (π x + α) + bcos (π x + b), where a, B, α and B are all non negative integers. We also know that f (2009) = 1,
Then f (2010)=
To explain in detail, it's better to have an explanation

f(2009)=asin(2009π+α)+bcos(2009π+B)=asin(π+α)+bcos(π+B)=-asinα-bcosB=1
f(2010)=asin(2010π+α)+bcos(2010π+B)=asin(α)+bcos(B)=-1

Given the function f (x) = asin (π x + a) + bcos (π x + β) + 1, and f (2012) = 2012, find the value of (2013)

The solution f (2012) = asin (2012 π + α) + bcos (2012 π + β) + 1
=asinα+bcosβ+1
So asin α + bcos β = 2011
f(2013)=asin(2013π+α)+bcos(2013π+β)+1
=asin(π+α)+bcos(π+β)+1
=-（asinα+bcosβ ）+1
=-2011+1
=-2010

Given f (x) = asin (π x + a) + bcos (π X - β), where a, B, α and β are all non-zero real numbers. If f (2012) = 1, what is f (2013) equal to

f(2012)=asin(π2012+a)+bcos(π2012-β)
=asina+bcosβ=1
f(2013)= asin(π2013+a)+bcos(π2013-β)
=-(asina+bcosβ)=-1