Given a column number A1, A2, A3 ，an，… Among them, A1 = 0, A2 = 2A1 + 1, A3 = 2A2 + 1 ，an+1=2an+1，… . then the individual digits of a2004-a2003 are () A. 2B. 4C. 6D. 8

Given a column number A1, A2, A3 ，an，… Among them, A1 = 0, A2 = 2A1 + 1, A3 = 2A2 + 1 ，an+1=2an+1，… . then the individual digits of a2004-a2003 are () A. 2B. 4C. 6D. 8

∵a1，a2，a3，… ，an，… If the number of digits in a2004 is 7 and the number of digits in A2003 is 3, then the number of digits in a2004-a2003 is 7-3 = 4

The area of a classroom is 45 square meters, with a side length of 0.5 meters of square tiles, the total number of such tiles?

45 ÷ (0.5 × 0.5) = 180 pieces

What are the common traditional festivals of 56 nationalities?

The Tajik people do not fast, so the Eid al Fitr is not as lively as the other two religious festivals. The most solemn religious festival of the Tajik people is the baloti Festival, also known as the Lantern Festival, which is held two months before Ramadan every year

Calculate 1 / sin10 ° - (radical 3) / sin80 ° without looking up the table=
. and what --
It would be better if you could give me the skills to solve the problem

This is a basic problem. Notice that 10 + 80 = 90? OK, convert sin10 ° to cos80 ° and then divide the molecules into (sin80 ° - (radical 3) * cos80 °). At this time, we propose the coefficient 2, and the molecules become the sine and cosine corresponding to 2 * (1 / 2 * sin80 ° - (radical 3) / 2 * cos80 °) 1 / 2 and (radical 3) / 2, OK

Two isosceles right triangles of equal area can form a square______ ．

Two isosceles right triangles of equal area are exactly the same and can form a square

Application of sine theorem and cosine theorem
In the triangle ABC, ad is the middle line on the side of BC, AC = 2Ab = 2ad = 4, then BD=
In the triangle ABC, ad is the middle line on the side of BC, AC = 2Ab = 2ad = 4, then BD =?
If we don't use the middle line theorem, can we do it? Consider the section "sine theorem"_ 〒

Using cosine theorem
Let BD = x, then BC = 2x
4 square = 2 square + (2x) square - 2 * 2 * 2x * cos angle B
2 square = 2 square + x square - 2 * 2 * x * cos angle B
Cos angle B = (4 + 4x-16) / 8x = (4 + x-4) / 4x
4x-12 = 2x
2X = 12
X = 6
X = root 6

If the solution set of inequality ax + 1 〈 0 is (1 / 3, + ∞), then a=

When x > 1 / 3 and ax + 10, x0 is not an inequality about X
So - 1 / a = 1 / 3 = > A = - 3

What's the area of the remaining part after subtracting the largest square from the original paperboard with a diameter of 8 cm

Diameter = 8, radius = 4
Diagonal of the largest square = 8, side length = 8 / √ 2 = 4 √ 2
The area of the remaining part is π × 4 & sup2; - 4 √ 2 × 4 √ 2
=50.24-32
=18.24 (cm2)

In △ ABC, ab = AC ∠ a = 36 °, two methods are designed to divide △ ABC into three isosceles triangles

Make the middle line of AB intersect AB with D and connect CD to three different isosceles triangles
：△ABC.△ACD.△BCD

Prove that 6 π is a period (process) of function f (x) = 2Sin (x / 3 - π / 6) by definition

f(x+6π)=2sin[(x+6π)／3-π／6]=2sin(x／3+2π-π／6)=2sin(x／3-π／6）=f（x）
So 6 π is a period of function f (x) = 2Sin (x / 3 - π / 6)