A math problem won t, ask for help

The least common multiple of 30 and 40 is 120, so it is 120 days later

Because May, June, July, and August are 31 days, 30 days, 31 days, and 31 days, respectively.

So there are 3 days left until August 31, that is, the next time they get a haircut at the same time is August 28, 2008

July 31st.

Since the first time they cut their hair was on the same day, they had to cut their hair at the same time next time, which means that the time passed between them is the same and because Xiao Wei has a haircut every 30 days, and Xiao Ming has a haircut every 40 days! So just ask for the least common multiple of 30 and 40 is 120, which means that they get a haircut at the same time after 2 months in between! This way we can also know which day after this haircut will be the same as the haircut!

That is, there are 240,480 in the common multiple....

The least common multiple of 30 and 40 is 120, which is the 120th day from May 1st, because May and July are 31 days, and June is 30 days, so it is August 28, 2008.

The least common multiple of 30,40 is 120

That is, 120 days later.

5. There are 31 days in July and 30 days in June.

August 28th.

The least common multiple of 30 and 40 is 120

120 - 30 of May (1 day cannot be counted) = 90

90 days of June - 30 = 60

60-31 of July=29

So it's August 29th.

That's right, after 120 days, there are 30 days in May, 30 days in June, 31 days in July, and 31 days in August, which should be two days before the end of August, that is, August 29th!

After 120 days, there are 31 days in the month, and there are 30 days in June. So it's 120-31-31-30=28

Also 28th. So it's August 29, 2008. ~

Find the least common multiple of 30 and 40.

It is equal to 1205 months with 31 days, 30 days in June, 31 days in July, and 31 days in August.

Then subtract 31 from 120 minus 30 minus 31 equals 28

It's August 28th.

The least common multiple of 30,40 was 120 days later, i.e. on August 28, 2008.

The least common divisor of 30 and 40 is 120 days later, on August 29.

@错了我, don't talk about it, forget it, let's talk about it, it's August 29!I meant that.

Radius = meters.

Area = 2 * 2 * square meters.

tana, tanb are the two real roots of the quadratic equation x +mx+m+1 tana+tanb=-m

tana*tanb=m+1

tanc=tan[(180°-(a+b)]=-tan(a+b)

(tana+tanb)/(1-tana·tanb)=-(-m)/(1-m-1)

1 and 0< C<180°

c=135°（3π/4）

Therefore, a and b are acute angles. and less than 45 degrees, that is, 0=0m2-4m+4>=8

m-2)^2>=8

m>=2+2 root number 2, m = <2-2 root number 2

To sum up, -1 hope it helps you

I'm sorry, brother, but I saw that there was no process for the answer.

Written incorrectly, bc=10cm, 2The quadrilateral ACDE is a parallelogram and not an ACDF with an area of 48

3.The circumference is equal to 44

Friend, in the right angle ABC, ab=6cm, AC=8cm, then BC is equal to 10, is there a problem with the problem?

(1) 1 group + 2 groups = 50 * 2 = 100

2) 2 groups + 3 groups = 43 * 2 = 86

3) 1 group + 3 groups = 45 * 2 = 90

Add the formulas.

2 (1 group + 2 groups + 3 groups) = 276

So 1 group + 2 groups + 3 groups = 138

So the average size of the 3 groups is 138 3 = 46

Isn't that just two groups?

If the average size of the two groups is 43, then let x, y, zx+y=50 2, y+z=43 2, x+z=90, respectively

Solving the system of equations is as follows: the average is (52+48+38) 3=46 people.

The title should have been written incorrectly. Since there are 3 groups, 4 in the third sentence should be 3

That's how it goes.

The group has a total of 50 2 = 100 people.

There are 43 2 = 86 people in the group.

There are 45 2 = 90 people in the group.

Then the 3 groups have a total of (100 + 86 + 90) 2 = 138 people, and the average number of people in the 3 groups is 138 3 = 46 people.

1,2 groups total: 50 2 = 100 people.

2,3 groups total: 43 2 = 86 people.

1, 3 groups in total: 45 2 = 90 people.

The three groups are total: (100 + 86 + 90) 2 = 138 people, and the average of the three groups is 138 3 = 46 people.

I don't know if you want this, or if you want the number of people in each group?

Group 1: 138-86 = 52 people.

Group 2: 138-90 = 48 people.

Group 3: 138-100 = 38 people.

Momo Elementary School has 3 co-curricular groups, with an average of 50 students in the two groups, an average of 43 students in the two groups, and an average of 45 students in the two groups.

50 + 43 + 45) 3 = 46 (person).