What size paper is 280mm × 420mm

What size paper is 280mm × 420mm

It's A3 paper

A formula for calculating the number of sheets

Normal paper: 787 × 1092mm opening size unit (mm) full opening 781 × 1086 opposite opening 530 × 760 three opening 362 × 781 four opening 390 × 543 three opening 362 × 390 eight opening 271 × 390 sixteen opening 195 × 271 note: finished product size = paper size trimming size large paper: 850 × 116

Half a sheet of A4 paper is several open, 210mm × 148mm
In addition, what's the size of 16?

Dadu 32
16 is about A4, the accuracy is 222 * 298mm

When Xiao Ming solved the equation 2x-1 / 3 = x + A / 3-1 to the denominator, THE-1 on the right side of the equation was not multiplied by 3, so the solution of the equation was x = 2, what was the value of a, and he correctly solved the equation
Solve by equation

From the meaning of the title
2x-1/3=x+a/3-1
6x-1 = x + A-1 when x = 2, then a = 10
2x-1/3=x+a/3-1
It should be: 6x-1 = x + A-3
The solution is: x = 6 / 5
It's a coincidence that we have this assignment=

When Xiao Ming solved the equation (2x-1) / 3 = (x + a) / 2-1 to remove the denominator, the - 1 on the right side of the equation was not multiplied by 6, so the solution of the equation was x = 4, what was the value of a, and he correctly solved the equation

When the denominator of equation (2x-1) / 3 = (x + a) / 2-1 is removed, the - 1 on the right side of the equation is not multiplied by 6, so 4x-2 = 3x + 3a-1, x = 3A + 1
3a+1=4,a=1,
Therefore, the original equation is (2x-1) / 3 = (x + 1) / 2-1,
4x-2=3x+3-6
x=-1

When Xiao Ming solved the equation (2x-1) / 3 = (x + a) / 2-1 to remove the denominator, the - 1 on the right side of the equation was not multiplied by 6, so he found that the solution of the equation was x = 2, what was the value of a

(2x-1)/3*6=(x+a)/2*6-1
2(2x-1)=3(x+a)-1
4x-2=3x+3a-1
x=3a+1
It is known that the solution of the equation is x = 2
Then 3A = 2 + 1
3a=1
a=1/3

The increasing interval of y = 3sin (2x - π 3) is___ ．

From (2x - π 3) ∈ [2K π - π 2, 2K π + π 2] (K ∈ z), the increasing interval of y = 3sin (2x - π 3) is [K π - π 12, K π + 512 π] K ∈ Z. so the answer is: [K π - π 12, K π + 512 π] K ∈ Z

Finding monotone increasing interval of function y = 3sin [(π / 3) - 2x]

-2kπ-π/2≤(π/3)-2x≤-2kπ+π/2
2kπ-π/2≤2x+π/3≤2kπ+π/2
2kπ-5π/6≤2x≤2kπ+π/6
kπ-5π/12≤x≤kπ+π/12

The decreasing interval of the function y = 3sin (2x + π / 6) (x belongs to [0, π])?
The decreasing interval of the function y = 3sin (2x + π / 6) (x belongs to [0, π])?

2kπ+π/2≤2x+π/6≤2kπ+3π/2
kπ+π/6≤x≤kπ+2π/3
K=0
π/6≤x≤2π/3
The decreasing interval [π / 6,2 π / 3] of the function y = 3sin (2x + π / 6) (x belongs to [0, π])

How to calculate 7.3 × 1.2-1.8

7.3x1.2-1.8
=0.6*(2*7.3-3)
=0.6*11.6
=8.76