Xiaoqiang and Xiaoming each have several books. It is known that Xiaoqiang's books account for 60% of their total books. When Xiaoqiang lent Xiaoming 20 books, the ratio of Xiaoqiang's books to Xiaoming's books is 2:3. How many books do they have in total?

Xiaoqiang and Xiaoming each have several books. It is known that Xiaoqiang's books account for 60% of their total books. When Xiaoqiang lent Xiaoming 20 books, the ratio of Xiaoqiang's books to Xiaoming's books is 2:3. How many books do they have in total?

20 (60% - 22 + 3) = 20 (35-25), = 20 (15), = 100 (Books)

Fast forward! There are 114 books in the book corner of class 63, 80% of which are literature and art books
There are 114 books in the book corner of class 63, of which the literature and art books account for 80% of other books?

Literature and art books = 114 * 80% = 91

It is difficult to find the differential of y = x / √ x ^ 2 + 1

y'=[√(x^2+1)-x/√(x^2+1)]/(x^2+1)=[x^2+1-x]/[(x^2+1)√(x^2+1)]
Differential dy = [x ^ 2 + 1-x] / [(x ^ 2 + 1) √ (x ^ 2 + 1)] DX

Finding the n-th derivative of y = Xe ^ x

y=xe^x
y'=e^x+xe^x
y''=e^x+y'=2e^x+xe^x
y'''=e^x+y''=3e^x+xe^x
.
Y (n) = ne ^ x + Xe ^ X / /: Y (n) the Nth derivative of Y

Subtracting the numerator and denominator of______ After that, it can be divided into 14

Let X be subtracted from both the numerator and the denominator of 1 325, and then it becomes 14. From this, we can get the equation: & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp

The derivative of differential equation y = Y / x + X / y

Homogeneous equation, let Y / x = u, then y = Xu, y '= u + Xu'
If the original equation is changed to U + Xu '= u + 1 / u, then x * Du / DX = 1 / u
The separation of variables is: UDU = DX / X
The integral of both sides is: 1 / 2U ^ 2 = ln | x | + ln | C1|
Then u ^ 2 = 2ln|c1 * x|
The solution of the equation is: e ^ (u ^ 2) = CX ^ 2, where C = C1 ^ 2

Calculation: 1 + m-2n divided by N-M △ M & # 178; - N & # 178; 1 + m-2n divided by (m-2n) &# 178;

=1+(m+n)/(m-2n)=(2m-n)/(m-2n)

Fill in the numbers 2 3 4 5 6 and write the multiplication formula. (1) to make the product maximum, [] [] [] × [] [] (2) to make the product minimum, [] [] [] × [] []

532*64 356*24

In the rectangular coordinate system, the polar coordinate equation of curve C is ρ = 2cos θ - 4sin θ. Write the rectangular coordinate equation of curve C______ ．

ρ = 2cos θ - 4sin θ, i.e. ρ 2 = 2 ρ cos θ - 4 ρ sin θ, is transformed into rectangular coordinate equation as x2 + y2 = 2x-4y, i.e. x2 + y2-2x + 4Y = 0, so the answer is x2 + y2-2x + 4Y = 0

(1 / 2) f (a) = {sin (- a) cos (PIE + a) cos (Pie / 2-A)} / {cos (pie-a) sin (PIE + a) Tan (PIE + a)}
(1 / 2) f (a) = {sin (- a) cos (PAI + a) cos (PAI / 2-A)} / {cos (pai-a) sin (2 Pai + a) Tan (PAI + a)}, f (a) = {sin (- a) cos (PAI + a) cos (PAI / 2-A)} / {cos (pai-a) sin (2 Pai + a) Tan (PAI + a)}（

1.f(a)=(-sina*(-cosa)*sina)/(-cosa*sina*sina/cosa)
=(sina*cosa*sina)/(sina*sina)
=cosa
2. Because a is in the second quadrant and Sina = 3 / 5
So f (a) = cosa = - √ (1-sin ^ 2a) = - √ (1-9 / 24) = - 4 / 5