(1+2+3+4+····+1999+2000+1999+···+2+1)÷2000

(1+2+3+4+····+1999+2000+1999+···+2+1)÷2000

(1+2+3+4+····+1999+2000+1999+···+2+1)÷2000
=[2*(1+2+3+4+····+1999)+2000]/2000
=[(1+1999)*1999+2000]/2000
=(2000*2000)/2000
=2000

(1-1/2)+(1/2-1/3)+(1/3-1/4).+(1/1999-1/2000)

Decompose (1-1 / 2) + (1 / 2-1 / 3) + (1 / 3-1 / 4) + (1 / 1999-1 / 2000) into 1-2 / 1 + 2 / 1-3 / 1 + 3 / 1. + 1 / 1999-1 / 2000. Because 1-2 / 1 is subtracted and 2 / 1 is added, so + 2 / 1 and - 2 / 1 offset each other, and - 3 / 1 + 3 / 1 also offset each other. + 1 / 19991 and - / 1999 offset each other
(1-1/2)+(1/2-1/3)+(1/3-1/4).+(1/1999-1/2000)
=1-2/1+2/1-3/1+3/1.+1/1999-1/2000
=1-（2/1-2/1）-（3/1-3/1）-.（1/1999-1/1999）-1/2000
=1-1/2000
=1999/2000

1 * 2:1 + 2 * 3:1 + 3 * 4:1. + 1999 * 2000:1

Solution: 1 * 2 of 1 + 2 * 3 of 1 + 3 * 4 of +... + 1999 * 2000 of 1
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/1998-1/1999)+(1/1999-1/2000)
=1-1/2000
=1999/2000

As shown in the figure, it is a pattern made of pieces. The first pattern needs 7 pieces, the second pattern needs 19 pieces, and the third pattern needs 37 pieces. If you go down in this way, the tenth pattern needs 7 pieces______ A chess piece

In the first picture, the total number of pieces is 6 + 1 = 7; in the second picture, the total number of pieces is 6 × (1 + 2) + 1 = 19; in the third picture, the total number of pieces is 6 × (1 + 2 + 3) + 1 = 37 In the nth picture, the total number of chess pieces is 6 × (1 + 2 + 3 +...) n) So when n = 10, the total number of pieces is 6 × (1 + 2 + 3 +...) +10) + 1, = 6 × 55 + 1, = 330 + 1, = 331 pieces. A: it takes 331 pieces to place the 10th pattern

A geometric problem about quadratic function in Junior Three
As shown in the figure, in the semicircle with diameter AB, draw a triangle area so that one side of the triangle is ab, the vertex C is on the semicircle circle, and the lengths of the other two sides are 6 and 8 respectively. Now we need to build a rectangular tank defn connected to △ ABC, where de is on ab. if DN = x, find the functional relationship between the area y and X of tank defn

Let CG ⊥ AB be the diameter of gab, C on the circumference of semicircle ⊥ C = 90 ', that is, AC ⊥ BCAC = 6, BC = 8ab = √ (AB ^ 2 + BC ^ 2) = 10cg = AC * BC / AB = 4.8dn ∥ CG ∥ EF, NF ∥ ABDN:GC=AD :AG=EF:GC=BE:BG=x/4.8(AD+BE):(AG+BG)=(AB-DE):AB=DN/GC(10-DE):10=x/4.8DE=10-10x/4....

A 30cm long wire is divided into two parts, each part is bent into an equilateral triangle, and the minimum value of the sum of their area is obtained (solved by quadratic function)

Let the length of the triangle be x, and 30-x, then the sides of the triangle are x / 3, 10-x / 3, S1 = √ 3 / 4 * x / 3, S2 = √ 3 / 4 * (10-x / 3), then the sum of the areas is y = S1 + S2 = √ 3 / 4 * (x / 3 + 10-x / 3) = 5 √ 3 / 2, so their areas are fixed, 5 √ 3 / 2

A + A + A + B + B + B = 32, a + B + B + B = 24, what is ab equal to

From a + B + B + B = 24, a = 24-3b is obtained. Substituting a + A + A + B + B + B + B = 32, 3 * (24-3b) + 4B = 32 and B = 8 are obtained
Substituting B = 8 into a = 24-3b gives a = 0
AB=0*8=0

If e is inside or outside of the ladder ABCD, the rest of it will not change. Is the conclusion that EC = ed true
It's very urgent. I hope someone can solve it immediately,

This problem is very simple!
When e is outside the trapezoid ABCD, because EA = EB, the angle EAB = angle EBA, because the trapezoid ABCD and ad = BC, the angle bad = angle ABC, so the angle ead = angle EBC, and because EA = EB, ad = BC, the triangle ead is equal to the triangle EBC, so Ed = EC
When e is inside the trapezoid ABCD, because EA = EB, the angle EAB = angle EBA, and because the trapezoid ABCD and ad = BC, the angle bad = angle ABC, so the angle ead = angle EBC, and because EA = EB, ad = BC, so the triangle ead is equal to the triangle EBC, so Ed = EC

280 hectares = square kilometers? 2.4 hectares = square meters? 320 square decimeters = square meters = square centimeters?

1 hectare = 15 mu = 10000 square meters = 0.01 square kilometers
280 ha = 2.8 km2
2.4 ha = 24000 M2
320 square decimeters = 3.2 square meters = 32000 square centimeters

The school originally has a long hair shaped site with an area of 1500 square meters. Now, combined with the renovation of the environment, one side of the site has been increased by 5 meters, and the other side has been reduced
The school used to have a long hair shaped site with an area of 1500 square meters. Now, in combination with the environment improvement, the site has been increased by 5 meters on one side and reduced by 5 meters on the other side. As a result, the site area has been increased by 10%, and the length and width of the current site have been calculated

Let the original two sides be x and Y respectively, then
（x+5）(y-5)=1500(1+10%)
x*y=1500
The solution is: x = 25, y = 60,
So now the length is 60-5 = 55, the width is 25 + 5 = 30
I'm working on the task, please help! Thank you very much!