The school plans to organize teachers and excellent students to visit Dahongshan in spring. There are 22 teachers, two travel agencies and two schools

The school plans to organize teachers and excellent students to visit Dahongshan in spring. There are 22 teachers, two travel agencies and two schools

There are X students
0.8x=(x+22)*0.75
0.8x=0.75x+16.5
0.8x-0.75x=16.5
x=330
A: there are 330 students on spring outings
There are X students
0.8x=(x+22)*0.75
x=330
A: there are 330 students on spring outings
Let {an} be an equal ratio sequence, A1 * A9 = 64, A3 + A7 = 20, find an
Because A1 * A9 = A3 * A7
So, A3 + A7 = 20, A3 * A7 = 64
The solution is A3 = 4, a7 = 16 or A3 = 16, a7 = 4
Then, help me with the next solution
a3×a7=a1×a9=64
And A3 + A7 = 20, a7 = 20-a3
a3×(20-a3)=64
a3²-20a3+64=0
(a3-4)(a3-16)=0
A3 = 4 or A3 = 16
(1) When A3 = 4, a7 = 16
q^4=a7/a3=4,q=±√2,a1=a3/q²=2
(2) When A3 = 16, a7 = 4
q^4=a7/a3=1/4,q=±√2/2,a1=a3/q²=32
So the sequence {an} is not unique, there are four cases
①a1=2,q=√2,an=2×(√2)^(n-1)
②a1=2,q=-√2,an=2×(-√2)^(n-1)
③a1=32,q=√2/2,an=32×(√2/2)^(n-1)
④a1=32,q=-√2/2,an=32×(-√2/2)^(n-1)
In fact, it's very simple. I'm worried about writing too much, so I'll tell you how to do it yourself/
You now have A3 = 4, a7 = 16 or A3 = 16, a7 = 4
Then we can calculate the ratio of an and Q in two cases
Don't you find out!
-x/2-5x/2-x=-3-4
Solve the equation, did not learn to understand
-x/2-5x/2-x=-3-4
(-1/2-5/2-1)x=-7
-4x=-7
x=7/4
-4x=-7
x=7/4
The school is going to organize teachers and excellent students to go on a spring outing in Dahongshan. There are 22 teachers. There are two travel agencies, a and B, with the same price, but different preferential ways. Travel agency a says that teachers are free and students are charged 20% off. Travel agency B says that teachers and students are charged 75% off. After accounting, the school leaders think that a and B are free, How many students take part in the spring outing?
Set the student as s, then
S*0.8=（22+S)*0.75 S=330
There are 330 students
In the arithmetic sequence {an}, a1 + A4 + A7 = 39, A3 + A6 + A9 = 27, then the sum of the first nine terms of the sequence {an} is equal to?
a1+a4+a7=39①,a3+a6+a9=27②
Two formulas are added
a1+a3+a4+a6+a7+a9=66
∵ {an} is an arithmetic sequence
∴a1+a9=a3+a7=a4+a6
∴3(a1+a9)=66
∴a1+a9=22
∴S9=(a1+a9)*9/2=99
This is a new arithmetic sequence. Because a1 + A4 + A7 = 39, A3 + A6 + A9 = 27
So we can get A2 + A5 + A8 = 33. So S9 = 39 + 33 + 27 = 99
How to solve 5x-2 (20-x) = 51
5x-2(20-x)=51
5x-40+2x=51
5x+2x=51+40
7x=91
x=13
Two teachers are going to organize the students of grade seven (1) class to have a spring outing in the bamboo sea of Yixing. The quotations of travel agency a and B are the same, and they both say that they can get a discount. Travel agency a charges 70% off for teachers and students, while travel agency B charges free for teachers. Which travel agency should they choose?
Use one variable inequality!
Suppose there are X students, the cost is y yuan, and the admission fee is a yuan
A: y a = (2 + x) * a * 0.7
B: y B = x * a
Let y a = y B, that is, (2 + x) * a * 0.7 = x * a, then x = 14 / 3 = 4 and 2 / 3
When XY B
When x > = 5, y a
In the arithmetic sequence [an}, a1 + A4 + A7 = 39, A3 + A6 + A9 = 27?
a1+a4+a7=3a4=3a1+9d=39
a1+3d=13 (1)
a3+a6+a9=3a6=3a1+15d=27
a1+5d=9 (2)
(2)-(1)
d=-2
a1=19
S9=9a1+(8*9/2)d=99
The process of 14.5x + 4.5X = 51
(14.5+4.5)x=51
x=51/(14.5+4.5)
x=51/19
14.5x+4.5x=19x=51 x=51/19
May I ask you a math problem
If the volume V of a tetrahedron is expressed as the function f (x) of X, then the increasing interval of F (x) is? And the decreasing interval of F (x) is?
Let AB = x in a tetrahedron ABCD, the length of the other edges be 1, take the midpoint e of AB, connect CE and De, then
AB⊥CE,AB⊥DE,
⊥ ab ⊥ plane CDE
CE=DE=√[1-(x/2)^],
The height h on the edge CD of △ CDE = √ [CE ^ - (CD / 2) ^] = √ [3 / 4-x ^ / 4],
∴V=f(x)=(1/3)S△CDE*AB
=(1/6)x√（3/4-x^/4)
=(1/12)√[x^(3-x^)]
=(1/12)√[9/4-(x^-3/2)^],
The increasing interval of F (x) is (0, √ 6 / 2), and the decreasing interval is (√ 6 / 2, √ 3)