Math Problem !! Who would??? Help me!!

When recruiting 50 A workers, the wages paid are minimal.

Because, if the recruitment of a type of workers is x, and the B type is (150-x) people, the inequality can be listed by the title: (150-x) is greater than or equal to 2x, and the solution is that x should be less than or equal to 50,.

Let the wage payable be y yuan, so according to the topic, the function can be listed as y 600x+1000 (150-x), and the simplified yield is y=-400x+150000, so when x is larger, the y value is smaller, so when x takes the maximum value of 50, the wage y paid to the worker is the smallest, which is 130000.

Solution: If there are X workers of type A, then type B is (150-x) people, and the monthly salary paid is Y yuan, which is derived from the title:

150-x）>=2x………1）

y=600x+1000（150-x）

From (1): x<=50

So when x = 50, the minimum salary paid is 130,000

Recruit X number of workers in type A work.

150-x>=2x

x<=50

Monthly wages paid

p=600x+(150-x)*1000

150000-400x

Because p decreases as x increases.

p min = 150000 - 400 x max.

x=50 ~

Let a be x b be y and the wage be z

z=600x+1000y

x+y=150

y>=2x

z=600x+1000(150-x)=150000-400xx<=50

So when x = 50, z is minimum 130000

If you're right!! The type of work B accounts for at least 2 3, that is, 100 people, and the remaining 50 people may be A or B, if you want to have a low salary, you need to be A, then it is 50 people!!

s=600a+1000b，，a+b=150，，s=150000-400a，b》2a

Therefore, when S is 50, take the minimum 130000

1. Idea analysis: If there are n terms in the series, then there are a1+a2+a3=34, an+an-1+an-2=146, and the two formulas are added together.

a1+an)+(a2+an-1)+(a3+an-2)=180, and a1+an=a2+an-1=a3+an-2, so a1+an=60and sn=n(a1+an) 2, so n*60 2=390, n=13

Analysis: Because am-1-am2+am+1=0, am2=am-1+am+1=2amBecause an≠0,

So am=2�

and s2m-1=(2m-1)(a1+a2m-1) 2=(2m-1)*2am 2=2(2m-1)=38,

So m=10

1. Solution: a1 + a2 + a3 = 3 a2 = 34

a2=34/3an+a(n-1)+a(n-2)=3a(n-1)=146

a(n-1)=146/3

sn=(a1+an)*n/2=[a2+a(n-1)]*n/2=30n=390

n=132, solution:

an=a1+(n-1)d

am-1+am+1=2am

and am-1+am+1-am =0

am²-2am=0

am=0 or am=2

When am=0:

am=a1+(m-1)d=0

S2M-1=(2A1+(2M-2)D)(2M-1) 2=[(A1+(M-1)D](2M-1)=0, which does not conform to S2M-1=38,

am=0 (rounded).

m=2s2m-1=2(2m-1)=38

m=10

What is the problem?

Set up a warehouse for goods as a unit of 1

A moves 1 10 an hour, B moves 1 12 an hour, C moves 1 15 an hour

1.Assuming that the mass of the original alloy copper and zinc is 2k and 5k respectively, then the mass of copper and zinc of the new alloy after adding 12 grams of copper is 2k k respectively

So (2k+12): 5k=2:3 solution k=9

So the mass of the new alloy is 2k+12+5k=75 (grams).

2.There are 30 days in September, and no one should have cleaned for 10 days, but now B has cleaned for 18 days, that is, he has cleaned for 8 days, and C has cleaned for 12 days, that is, he has cleaned for 2 days, that is to say, A's work B has done 8 10, and C has done 2 10, so according to the number of days of cleaning, B should get 8 10 * 60 = 48 yuan.

3.Let this number be x, then it depends on the title.

x-1/2x）-1/4（x-1/2x）-1/5[（x-1/2x）-1/4（x-1/2x）]=1

The solution yields x=10 3

4.If the number of stamps is 5k and 6k respectively, then after Xiao Ming gives 10 stamps to Xiaohong, the number of stamps of Xiao Ming and Xiaohong is 5k-10 and 6k+10 respectively according to the title.

5k-10): 6k+10=4:5 solution k=90

So they have a total of 5k+6k=11k=990 stamps.

1.Let copper and zinc be x and y, respectively

x/y=2/5

x+12)/y=2/3

The solution is x=18 y=45

The mass of the new alloy is 45 + 18 + 12 = 75

2.There are 30 days in September, so it should be 60*(18 30)=36 yuan3Let the original number be a

1 so a = 20 7

4.If Xiao Ming has x sheets, then Xiao Hong has (5 6) x sheets.

x-10) [(5 6)x+10]=4 5 solution x=54 (5 6)x=45

So Xiao Ming has 54 and Xiao Hong has 45.

1.Let the original copper 2x grams, zinc 5x grams (according to the ratio of copper to zinc is 2:5) (2x+12): 5x=2:3

Solution: x=9

That is, the original copper is 18 grams and zinc is 45 grams.

Added 12 grams of copper.

Then the mass of the new alloy is 18 + 45 + 12 = 75g2In September, A is unable to clean due to work needs (equivalent to "A cannot participate in cleaning for 30 days"), but in fact A only needs to clean for 10 days.

B cleaned for 18 days, C cleaned for 12 days (i.e., B cleaned for 8 days, C cleaned for 2 days) B should get [8 (8+2)] = 80% of the money, which is 80% * 60 = 48 yuan 3Let the original number be x

Then x*(1 2)【a number minus its 1 2】*(3 4)【minus the remaining 1 4】*(4 5)【minus 1 5】=1

x=10/3

4.There are 5x stamps for Xiao Ming and 6x for Xiao Hong, for a total of 11x stamps [the ratio of Xiao Ming and Xiao Hong's stamps is 5:6].

5x-10):(6x+10)=4:5

25x-50=24x+40

x=9011x=990

There are 990 stamps.

1 The square of the third side can be: 3 +5 = 34;

or 5 -3 =16;

The possible values of 2 x are 2 +4 = 20, or 4 -2 = 12, and the possible values of x are 2;

3 Choose a because the Pythagorean theorem is not satisfied.

It's very simple.,It's all about asking for several possibilities.,But I don't have time to help you calculate at work.。。。 =。=

Is the third side an integer? Yes, the side length of 34567 can be squared 9, 16, 25, 36, 49