solve equations, Party A and Party B set out from a and B at the same time and walked towards each other. Three hours later, they met on the way. It is known that the distance between a and B is 24km, and the speed ratio of Party A and Party B is 2:3. How many km do they travel per hour?

Suppose: if a line is x km / h, B line is 2x / 3 km / h
3 * (x + 2x / 3) = 24
The solution is: x = 24 / 5 km / h
A

As for the meeting, a reward will be offered before 22:30 on December 7
Party A and Party B ride bicycles from AB to ab. Party A starts 15 minutes earlier than Party B. the speed ratio of Party A and Party B is 2:3. When they meet, Party A walks 6000 meters less than Party B. It is known that Party B walks for 1 hour and 30 minutes. Find out the speed of Party A and Party B and the distance between the two places

1 hour 30 minutes = 1.5 hours
15 points = 1 / 4
Suppose that the velocity of a is x km / h and that of B is 3 / 2 x m / h
3/2x ×1.5-（1.5+1/4）x=6
2.25x-1.75x=6
0.5x=6
x=12
Speed of B: 12 × 3 / 2 = 18 km / h
Distance: 18 × 1.5 × 2-6 = 48 km

There are 80 students in the second grade of a school, one fourth more than the third grade. The sum of the students in the second and third grade accounts for 30% of the total number of the school
How many students are there in the school?

There are 80 students in the second grade, a quarter more than the third grade,
The third grade has 80 / (1 + 1 / 4) = 64
There are 144 students in grade 2 and grade 3
The sum of the students in the second and third grades accounts for 30% of the total number of students in the school
144 / (30%) = 480
[beauty of mathematics] I'm glad to answer for you. If you don't understand, please ask! If you are satisfied, please accept. Thank you_ ∩)O~

7 out of 13 times 7 out of 24 [34 times 12]

7 out of 13 times 7 out of 24 [34 times 12]
=7/13*7/24*12/34
=7/13*7/24*6/17
=49/884

Sequence an is less than or equal to BN, less than or equal to CN, BN converges, the limit of CN an is 0, it is proved that an and CN converge
Sequence an is less than or equal to BN, less than or equal to CN, BN converges, the limit of CN an is 0, how to prove that an and CN converge

By
　　　　an

X + 15 / 2 + 5 / 2 = 4 / 3

(x+15)/2+5/2=4/3
Multiply both sides of the equation by 6 to get the following result:
3(x+15)+15=8
3x+45+15=8
3x+60=8
3x=8-60
3x=-52
x=-52/3
If x + 15 is not the whole, then:
x+20/2=4/3
x+10=4/3
x=4/3-10
x=-26/3

Fill in 2.4.3.6.5. (10) () ()

Analysis: 2.4.3.6.5. (10) () ()
If we observe it carefully, we will find the rule: the sum of the first two numbers separated by a term is equal to the value of the third term, that is, the first term 2 + the third term 3 = the fifth term 5, the second term 4 + the fourth term 6 = the sixth term 10, and so on. The brackets after 10 are 5 + 3 = 8, then 10 + 6 = 16, and the last bracket is 8 + 5 = 13

If the sequence {an} satisfies A1 = 32, an + 1 = an2-an + 1 (n ∈ n *), then M = 1A1 + 1A2 + +The integral part of 1a2010 is ()
A. 0B. 1C. 2D. 3

It is known from the problem that an + 1-1 = an (an-1), х 1An + 1 − 1 = 1An − 1-1an, х 1An − 1-1an + 1 − 1 = 1An. By accumulation, M = 1A1 + 1A2 + +1a2010 = 1A1 − 1-1a2011 − 1 = 2-1a2011 − 1

1 / (a ^ 2 + x ^ 2) ^ 3 / 2 indefinite integral
We can't be in a hurry

This problem can be divided into two situations
(1) When a = 0,
The original formula = ∫ DX / X & sup3;
=-1 / (2x & sup2;) + C (C is an integral constant)
(2) When a ≠ 0,
Let x = atant
Then DX = ASEC & sup2; TDT
sint=xcost/a
=(x/a)(1/sect)
=(x/a)[1/√(1+tan²t)]
=(x/a)[a/√(a²+x²)]
=x/√(a²+x²)
The original formula = ∫ ASEC & sup2; TDT / (A & sup3; sec & sup3; t)
=1/a²∫dt/sect
=1/a²∫costdt
=Sint / A & sup2; + C (C is the integral constant)
=X / [A & sup2; √ (A & sup2; + X & sup2;)] + C (C is an integral constant)

The first line 1, the second line 23 4, the third line 56 7 8 9, the fourth line 10 11 12 13 14 15 16, the fifth line. How many lines should 2005 be in

The fourth number in the 11th row is 1 + 2 + 3 + 4 + +10 + 4 = 592. The number m in line n is n * (n-1) / 2 + m3. The first number in line n is n * (n-1) / 2 + 1, and the last number is n * (n-1) / 2 + N, so the sum is (n * (n-1) / 2 + 1 + n * (n-1) / 2 + n) * n = n * n * n + N4

solve equations, Party A and Party B set out from a and B at the same time and walked towards each other. Three hours later, they met on the way. It is known that the distance between a and B is 24km, and the speed ratio of Party A and Party B is 2:3. How many km do they travel per hour?