The donation of class 6 is one third of the total donation of class 6. The donation of class 6 is three fifths of the total donation of class 6 The donation of class 6 is one third of the total donation of class 6. The donation of class 6 is three fifths of the total donation of class 6. It is known that class 6 has donated 2400 yuan in total. How much is the total donation of grade 6?

The donation of class 6 is one third of the total donation of class 6. The donation of class 6 is three fifths of the total donation of class 6 The donation of class 6 is one third of the total donation of class 6. The donation of class 6 is three fifths of the total donation of class 6. It is known that class 6 has donated 2400 yuan in total. How much is the total donation of grade 6?

Donation of class 6: 1 / (1 + 3) = 1 / 4
Donation of class 6: 3 / (5 + 3) = 3 / 8
Donation of class 6: 1-1 / 4-3 / 8 = 3 / 8
Total donation for Grade 6: 2400 △ 3 / 8 = 6400 (yuan)

An earthquake that shocked the whole world has brought a heavy blow to the people in a disaster area. According to statistics, the number of donations from class 61 is one-third of the total donations from the other two classes; the number of donations from class 62 is 60% of the total donations from the other two classes. It is known that 63 classes have donated 2400 yuan in total. In this donation activity, how much is the total donation from grade 6

The donation amount of class 61 is one of the total donations of grade 6 (one class accounts for one, and the other two classes account for three)
1÷（1+3）=1/4
The number of donations from class 62 is one percent of the total number of donations from grade 6
6÷（6+10）=6/16=3/8
The donation of class 6-3 is the total donation of grade 6
1-1/4-3/8=8/8-2/8-3/8=3/8
The total number of donations for Grade 6 is:
2400 △ 3 / 8 = 6400 (yuan)
A: the sixth grade donated 6400 yuan in total

Two two digit numbers, their greatest common factor 9, least common multiple 360, find two numbers, urgent

The basic principle of application is: the greatest common factor multiplied by the least common multiple equals to the product of two numbers
Let the two numbers be x and Y respectively. According to the theorem xy = 9 * 360 = 3240, decompose 3240 into factors
3240 = 2 * 2 * 2 * 3 * 3 * 3 * 5, because the greatest common factor is 9, that is to say, the two numbers contain 3 * 3, and if the three 2 separate the two sides, then there will be more common factor 2, which is inconsistent with the problem. So there are two cases, that is, x = 72, y = 45; or x = 360, y = 9 (XY is interchangeable)

If the line L1 and line L2 intersect at a point P, and the slope of L1 is 1K, the slope of L2 is 2K, and the lines L1, L2 and X-axis form an isosceles triangle, then all possible values of positive real number k are___ ．

Let the inclination angles of line L1 and line L2 be α, β, because k > 0, so α, β are acute angles. Because the lines L1, L2 and X-axis form an isosceles triangle, there are the following two cases: (1) when α = 2 β, Tan α = Tan 2 β, 1K = 4k1-4k2, because k > 0, the solution is k = 24; (2) when β = 2 α, Tan β = Tan 2 α, 2K = 2k1-1k2, because k > 0, the solution is k = 2 ．

Change this is a sheet into a plural sentence

These are sheep.

If the vertex of the parabola y = 3x2 + ax + 4 is on the negative semiaxis of the x-axis, then a=______ ．

According to the meaning of the question: - A2 × 3 = - A6 < 0, then a > 0, 4 × 3 × 4 − a24 × 3 = 0, the solution is: a = ± 43, a = 43, so the answer is: 43

Divide a natural number by 300.262.205 to get the same remainder. How many natural numbers are there?

300-262=38
262-205=57
This number must be the common divisor of 38 and 57
The possible natural numbers are 1, 19, only two

In the triangle ABC, it is known that | ab | = 4 root sign 2, and the three interior angles a, B, C satisfy 2sina + sinc = 2sinb. Establish an appropriate coordinate system and find the trajectory equation of vertex C
Equation I'm looking for a trajectory equation, not a trajectory

Taking the midpoint of AB as the coordinate origin and the line of AB as the x-axis, the rectangular coordinate system is established
Because 2sina + sinc = 2sinb, from the sine theorem we get that:
2A + C = 2B, and C = | ab | = 4, radical 2
So B-A = (1 / 2) C = 2, radical 2
That is | Ca | - | CB | = 2
The locus of C is a hyperbola with focus on X axis (excluding vertex)
Let the hyperbolic equation be x2 / a2-y2 / B2 = 1
Then 2A = | Ca | - | CB | = 2 radical 2, 2C = | ab | = 4 radical 2
a2=2,c2=8,b2=c2-a2=6
So the trajectory equation of point C is x2 / 2-y2 / 6 = 1 (y is not equal to 0)

Can integral multiplication and division and factorization teach me the method?
I've got a big head when I look at those letter formulas! I'll confuse those formulas
Find a way find a way

I've forgotten the specific methods, but I remind you that there are thousands of methods, which vary from person to person. You don't understand other people's methods, and you can't use them. The methods are all summed up in failure and experience. I think it's meaningless to ask others for methods. I suggest you forget all about this part, and then sort out the knowledge step by step from the basic reading

The quadratic function f (x) satisfies f (- 2) = 0 and 3x + 5

Let f (x) = AX2 + BX + C (a ≠ 0) have f (- 1) = 2 and f (- 2) = 0, there are 4a-2b + C = 0, A-B + C = 2, that is, B = 2 + 3a, C = 4 + 2A according to the topic, according to the topic, all real number x should satisfy 3x + 5, according to the topic, all real number x must satisfy 3x + 3x + 5, according to the topic, according to the topic, all real number x, all real number x must satisfy all real number x, according to the topic, all real number x must satisfy 3x + 3x + 3x + 5 (3x (x) ≤ 178; + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7x + 7-7-f (x + 7x + 7x + 7x + 7-F (x) \\\thatis to say, △ = (7-B) &; -4 (2-A) (7-c) ≤ 0 is always true. After sorting out, a = 1, B = 5, C = 6 is substituted into formula 2, which satisfies the meaning of the question