Xiao Ming, Xiao Hong and Xiao Fang did 72 math problems. Xiao Ming did 8 more than Xiao Hong, and Xiao Fang did 12 less than Xiao Ming and Xiao Hong What's wrong with each other?

Xiao Ming, Xiao Hong and Xiao Fang did 72 math problems. Xiao Ming did 8 more than Xiao Hong, and Xiao Fang did 12 less than Xiao Ming and Xiao Hong What's wrong with each other?

Let Xiaohong do X, Xiaoming do X + 8, Xiaofang do 72-x - (x + 8) = 64-2x
X+(X+8)-(64-2X)=12
4X=68
X=17
X+8=17+8=25
64-2X=64-17*2=30
Xiao Hong did 17 questions, Xiao Ming did 25 questions, and Xiao Fang did 30 questions

Xiao Ming has 18 yuan and Xiao Hong has 24 yuan. How many yuan does Xiao Ming give Xiao Hong? Xiao Hong's money is twice that of Xiao Ming

4 yuan

In warehouse A and warehouse B, the amount of chemical fertilizer stored in warehouse A is 60 tons more than that in warehouse A. if warehouse B transfers 60 tons to warehouse A, the amount of chemical fertilizer stored in warehouse B is 5 / 6 of that in warehouse a,
How many tons of chemical fertilizer were stored in the two warehouses

Warehouse A (60 + 60 + 60 × 5 / 6) / (1-5 / 6)
=170÷1/6
=1020 tons
Bin B 1020-60 = 960 tons

If the solution of equation X-5 / 2 = - X-1 / 5 is also the solution of equation 7x-a = 2, find the value of A

2 of X-5 = - 5 of X-1
Is that so? 2 / (X-5) = 5 / (- x-1)
5x-25=-2x-2
7x=23
x=23/7
Substituting x = 23 / 7 into 7x-a = 2
23-a=2
a=21

4*2.3+18.4*11.1+22.2*11.2

44.4*2.3+18.4*11.1+22.2*11.2
=44.4*2.3+4.6*(4*11.1)+(22.2*2)*5.6
=44.4*(2.3+4.6+5.6)
=44.4*12.5
=11.1*(4*12.5)
=11.1*50
=555

In the equal ratio sequence {an}, A3 = 3 / 2 and S3 = 9 / 2 are known

3/2=a3=q^2*a1
9/2=S3=a1+a2+a3=a1+q*a1+3/2
So 3Q ^ 2 = (9 / 2-3 / 2) * q ^ 2 = (a1 + Q * a1 + 3 / 2-3 / 2) * q ^ 2
=a1*q^2+a1*q^2*q
=3/2+(3/2)q
The solution
Q = 1 or (- 1 / 2)
If q = 1, then A1 = A3 / (Q ^ 2) = A3 = 3 / 2
If q = (- 1 / 2), then A1 = A3 / (Q ^ 2) = A3 = 6

How to simplify the calculation

3 / 4 × 5 / 9 + 3 / 4 ／ 9 / 4
=3 / 4 × 5 / 9 + 3 / 4 × 4 / 9
=3 / 4 × (5 / 9 + 4 / 9)
=3 / 4 × 1
=Three quarters

If the sequence {an} satisfies A1 = 1 and for any m, n has am + n = am + an + Mn, then 1 / A1 + 1 / A2 + 1 / A3 + +1/a2008=?
If the sequence {an} satisfies A1 = 1 and for any m, n has am + n (M + n is the subscript) = am + an + Mn (Mn is not the subscript), then 1 / A1 + 1 / A2 + 1 / A3 + +1/a2008=?
Give + points

Let m = n, n = 1, that is, a (n + 1) = an + A1 + N, that is, a (n + 1) = an + (n + 1)
So an = 1 + 2 +... + n = n (n + 1) / 2
1/a1+1/a2+...+1/a2008
=2(1-1/2+1/2-1/3+..1/2008-1/2009)
=2(1-1/2009)=4016/2009

From the four numbers of 0.3.5.7, choose three to form three digits, and the number is a multiple of 2
Dear good friends, if you understand, please come back as soon as possible

The multiples of 2 are: as long as it ends in 0: 350530370730570750
Multiple of 3: as long as the sum of three numbers is a multiple of 3, it can be: 357537735753375573
Multiple of 5: the ending is 0 or 5. The ending is OK: 350530305370730570750705,

If and only if the matrix is invertible, the more the answers, the better

Invertible matrix A of order n
A nonsingular
|A|≠0
A can be expressed as the product of elementary matrices
A is equivalent to the identity matrix of order n
r(A) = n
The column (row) vector group of a is linearly independent
The homogeneous linear equations AX = 0 have only zero solutions
There is a unique solution for the system of non homogeneous linear equations AX = B
Any n-dimensional vector can be represented linearly by a group of column (or row) vectors of A
The eigenvalues of a are not zero