The donation of three classes is one third of that of class 2 and class 3, 60% of that of class 1 and class 3, and 2400 yuan for class 3

The donation of three classes is one third of that of class 2 and class 3, 60% of that of class 1 and class 3, and 2400 yuan for class 3

If class 1 is regarded as 1, then the sum of class 23 is 1 / 3 = 3
Therefore, 1 class accounts for 1 / 4 of the total
In the same way
Class 2 is 1, class 13 is 1 △ 60% = 5 / 3
Therefore, class 2 accounts for 1 (1 + 5 / 3) = 3 / 8
2400÷(1-1/4-3/8)=19200
A: the total is 19200 yuan

How to calculate (+ 1 / 5) - (+ 1 / 4) + (- 3 / 5) - (- 6 and 3 / 4)
I'm in a hurry,

Hello
(+ 1 / 5) - (+ 1 / 4) + (- 3 / 5) - (- 6 and 3 / 4)
=0.2-0.25-0.6+6.75
=-0.25+6.15
=5.9
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1/2+（1/3+2/3）+（1/4+2/4+3/4）+…… +（1/60+2/60+…… 59 / 60)

1/2+（1/3+2/3）+（1/4+2/4+3/4）+…… +（1/60+2/60+…… 59/60）
=1/2+2/2+3/2+…… +59/2
=1/2×（1+59）×59/2
=15×59
=885

The parabola passes through the point (0,5 / 3), (10,0). The function has the maximum value. 3. Find the analytical formula of the parabola
I did it myself. The equation listed below is am ^ 2 + 3 = 5 / 3; a (10 + m) ^ 2 + 3 = 0
I can't work it out

Parabola y = ax ^ 2 + BX + C
Through the point (0,5 / 3), (10,0), the function has a maximum value of 3
c=5/3
100a+10b+5/3=0
60a+6b+1=0.(1)
The function has a maximum of 3, a

Let m and n be positive integers, and M is not equal to 2. The distance between the image of quadratic function y = x ^ 2 + (3-MT) x-3mt and the two focal points of x-axis is D1. The distance between the image of quadratic function y = - x ^ 2 + (2t-n) x + 2nt and the two intersection points of x-axis is D2. If D1 is greater than or equal to D2, the value of M and N is obtained
I don't know how to do the following inequality about t

y=x^2+(3-mt)x-3mt
x1+x2=mt-3,x1x2=-3mt
So D1 ^ 2 = (x1-x2) ^ 2 = (x1 + x2) ^ 2-4x1x2 = m ^ 2T ^ 2 + 6mt + 9 = (MT + 3) ^ 2
y=-x^2+(2t-n)x+2nt
x1+x2=2t-n,x1x2=-2nt
d2^2=(x1-x2)^2=(x1+x2)^2-4x1x2=4t^2+4nt+n=(2t+n)^2
d1>=d2>0
So D1 ^ 2-d2 ^ 2 > = 0
So (MT + 3) ^ 2 - (2t + n) ^ 2 > = 0
(m ^ 2-4) T ^ 2 + (6m-4n) t + 9-N ^ 2 > = 0
So the discriminant is less than or equal to 0
So (6m-4n) ^ 2-4 (m ^ 2-4) (9-N ^ 2)

One variable one time problem (encounter problem)
A bikes from a to B, B bikes from B to a. a travels 2 kilometers more per hour than B. they start at 8 a.m. at the same time. They are 36 kilometers apart at 10 a.m. and 36 kilometers apart at 12 noon. Find the distance between a and B

Suppose that the velocity of B is XKM / h and that of a is (x + 2) km / h
It's obvious that there is a difference of 36km between the two times
It means that they met halfway
From 10:00 to 12:00, they walked 36 + 36 = 72km
So 72 = 2x + 2 (x + 2)
Calculated x = 17km / h
So the total distance is 2x + 2 (x + 2) + 36 = 108km

In the plane rectangular coordinate system, there are ()
A. 1
B. 2
C. 3
D. 4
reason?

This topic is to find out the intersection points between the circle with radius 10 and the coordinate axis
The equation of a circle is (X-6) square + (y + 8) square = 10 square, when x = 0, y = 0 or - 16, when y = 0, x = 0 or 12, so the answer is (0,0) (0, - 16) (12,0)

1.ab(c²+d²)﹢cd(a²+b²) 2.(ax+by)²+(bx-ay)²

1.ab(c²+d²)﹢cd(a²+b²) =abc²+abd²+a²cd+b²cd=bc(ac+bd)+ad(ac+bd)=(ac+bd)(bc+ad)2.(ax+by)²+(bx-ay)² =a²x²+b²y²+2abxy+b²x²+...

How many mu of grain cultivated land are there in China?

According to Xinhua News Agency on February 24, 2011, the agriculture and Rural Committee of the National People's Congress revealed that at present, China's cultivated land area is about 1.826 billion mu, 123 million mu less than the 1.949 billion mu in 1997

Physics, to detailed process and formula
An object with a weight of 10N moves in a straight line at a constant speed under the action of horizontal tension, and it moves 15m in 3 seconds. If the friction resistance between the object and the ground is 2n, the solution is given
1. Power of pulling force
2. Under the same conditions, if the object moves at a constant speed for 10s, how much work does the pulling force do?

1. Because of the uniform linear motion, the pulling force F = f = 2n, so the power P = w / T = f · s / T = 10W
2. The velocity v = s / T = 5m / s, so when t '= 10s, the distance s' = V · t' = 50m, the work w = f · s' = 100J