There are 98 students in the two classes of grade six. If six students are transferred from class one to class two, the ratio of class one to class two is 3:4. How many students are there in class one? Thank you

Hello, Mr. building, Weiwei, our league members set up class 1 for you with X people, class 2 is 98-x. according to the following, the number of class 1 and class 2 is 3:4. It can be listed as: (X-6) / 98 - (X-6) = 3 / 4, x = 48, then the original class has 48 people! If my answer can help you, forget to adopt it, thank you!
Hope to adopt

There are 98 students in class one and two in grade six and 152 in class two and three and four in grade six. How many students are there in class two?

The number of two classes is X
98 + 152 = 250, but the number of class 2 is increased once more
250-x = 4x x = 50 people
A: the number of class 2 is 50
I hope it will help you

A simple problem,
At present, dadandan has a deposit of 564 yuan and is still earning 1.3 yuan per hour
At present, the deposit of Da Pangpang is 69 yuan, earning 1.9 yuan per hour
When can Da Pangpang be the same as Da Dandan? How many hours does it take?

It takes x hours
564+1.3x=69+1.9x
0.6x=405
x=825
(564-69) / (1.9-1.3) = 825 (hours)
If it takes x hours, then: 69 + 1.9x = 564 + 1.3x, then: x = 825 hours, then the deposit is 1636.5
（564-69）/（1.9-1.3）=825

A positive number, if its decimal part, integral part and positive real number are in equal proportion sequence, then the positive number is equal to

Let the integer part of the number be a and the decimal part be B, then the number is (a + b)
A * a = b * (a + b)
b*b+ab-a*a=0
B = ((- 1 + radical 5) / 2) a (negative radical removed)
Solving inequality 0

If the common ratio Q of an infinitely proportional sequence satisfies | Q|

a(m)=ka(m+1)/(1-q)=kqa(m)/(1-q)
k=(1-q)/q
=1/q -1
-1

It is known that the common ratio of {an} is Q, and 0 ＜ Q ＜ 1 / 2. (2) if q = 1, and for any positive integer k, AK - (AK + 1 + AK + 2) still exists
It is known that the common ratio of {an} is Q, and 0 ＜ Q ＜ 1 / 2
(2) If q = 1, and for any positive integer k, AK - (AK + 1 + AK + 2) is still a term of the sequence;
① Find the common ratio Q;
② If BN = - ㏒ an + 1 (√ 2 + 1), Sn = B1 + B2 bn,Tn=s1+s2+… +Sn, s2011 for t2011

(1) suppose there are positive random numbers I, K, m such that AI + AI + M = 2ai + K
AI ＞ 0, an is an equal ratio sequence,
∴1+q^m=2q^k
0＜q＜0.5
And 1 + Q ^ m ＞ 1 ＞ 2q ＞ 2q ^ K
The hypothesis does not hold, and there is no arithmetic sequence of three terms in an
(2) Suppose AK - (AK + 1 + AK + 2) = AK + m, M is a positive integer
Then 1-q-q ^ 2 = q ^ m
When m ≥ 2, there is 1-q-q ^ 2 = q ^ m ≤ Q ^ 2
Q ≥ 0.5 or Q ≤ - 1
And 0 < Q < 0.5
Therefore, M = 1
That is 1-q-q ^ 2 = Q
q=-1±√2
0＜q＜0.5
∴q=√2-1
a1=1,∴an=（√2-1）^（n-1）
BN = - Logan + 1 (√ 2 + 1) I don't know whether an + 1 is the base or √ 2 + 1 is the base
The base is √ 2 + 1, BN = n
Sn=n(n+1)/2,S2011=2023066
Tn=(1+2+…… +n)/2+（1^2+2^2+…… +n^2）/2=n(n+1)/4+n(n+1)(2n+1)/12
T2011=1357477286
An + 1 as the base, BN = 1 / N, the result is not easy to calculate, ha ha, I'll study it myself

If the length of three sides of a triangle is known to be an equal ratio sequence, find the value range of the common ratio Q

By using the relationship between the difference sum of the lengths of three sides, we obtain a system of inequalities about the common ratio Q
1/Q+1>Q,1-1/Q

Decomposition factor: - 12x's cube Y's square + 18x's Square Y's cube-6xy

Original = - 6xy (2x & # 178; y-3xy & # 178; + 1)

The square of 18x - [the square of 6xy - (the square of XY - the square of 12x, the cube of Y)]

18X^2-[6XY^2-(XY^2-12X^2Y^3)]
=18X^2-6XY^2+XY^2+12X^2Y^3
=18X^2-5XY^2+12X^2Y^3
=X(18X-5Y^2+12XY^3)

Let X and y satisfy the constraint conditions 3x-y-6 ≤ 0, X-Y + 2 ≥ 0, X ≥ 0, y ≥ 0. If the maximum paper of the objective function z = ax + by (a > 0, b > 0) is 12, then the minimum value of 2 / A + 3 / B is? (why can we directly bring the point (4,6) into the paper and then directly 4A + 6B = 12?)

The conditions in the title show that the maximum value must be taken at the intersection

There are 98 students in the two classes of grade six. If six students are transferred from class one to class two, the ratio of class one to class two is 3:4. How many students are there in class one? Thank you