An arithmetic problem, Divide a circle into several small sectors, and then form an approximate rectangle. The length of this rectangle is 9.42 decimeters, and the perimeter is 24.54 decimeters. The perimeter of this circle is ()? The area is?

The length of a rectangle is half the circumference of a circle, and the width is the radius of the circle
24.54 △ 2 - 9.42 = 2.85 (decimeter) the width of the rectangle, that is, the radius of the circle
The circumference of the circle is: 2.85 × 2 × 3.14 = 17.898 (square decimeter)
The area of the circle is: 2.85 × 2.85 × 3.14 = 25.50465 (cubic decimeter)

Find the answer to an arithmetic problem,
（121+122+...+170...41+42+...+98
What's the number? Using a simple method, 170 + is added to 41. Sorry, I forgot the plus sign.

From 121 to 170: (121 + 170) * 50 / 2
From 41 to 98, I don't care
That is (first item + last item) * number / 2
The formula of continuous number addition, haven't you learned yet?

Who can help me think of a 1 to 9 addition, subtraction, multiplication and division problem

(9-6-3)/(2+7-8)+4*5=20

Dear God, the third grade of primary school arithmetic problems for advice ah~
How much divided by 7 is more than 4
How much is divided by 8 and how much is left over by 7
How much divided by 6 is equal to how much more than 3
How much divided by 9 is equal to how much more than 3
They have the same divisor

The minimum number is 39
39 divided by 7 equals 5 and 4
39 divided by eight is four and seven
39 divided by 6 is 6 and 3
39 divided by 9 is four and three

1×2+2×3+3×4+… +99×100．

1×2+2×3+3×4+… +99×100，=1×（1+1）+2×（2+1）+3×（3+1）+… +98×（98+1）+99×（99+1），=12+1+22+2+32+3+… +982+98+992+99，=（12+22+32+… +982+992）+（1+2+3+… +98+99），=99×（99+1）×（2×99+1）÷6+（1+99）×99÷2，=328350+4950，=333300．

99 kilogram is equal to how many tons

1 ton (T) = 1000 kg (kg) 0.099

How about 1-4 + 9-16 + 25.. + 99 * 99?

I didn't expect a good formula. Let's give a result first. 166650-161700 = 4950

It is known that OC is the angular bisector of ∠ AOB, OD is the angular bisector of ∠ AOD, and OE is the angular bisector of ∠ BOC. If ∠ DOE = 30 °, the degree of ∠ AOB is calculated

Od is the angular bisector of ∠ AOD? Od is the angular bisector of ∠ AOC! 60 degrees, all are angular bisectors. Just draw a picture, ∠ AOD = ∠ cod, ∠ COE = ∠ EOB, so ∠ AOD + ∠ EOB = ∠ cod + ∠ COE = 30 degrees

As shown in Figure 1, it is known that ∠ AOB = 80 °, OC is the bisector of ∠ AOB, OD and OE bisector ∠ BOC and ∠ COA 1 respectively, and calculate the degree of ∠ doe
2. When OC rotates around point O, OD and OE are still bisectors of ∠ BOC and ∠ COA. Is the size of ∠ DOE the same? If so, please write down the whole situation!

(1) ∵ OC is the bisector of ∠ AOB ∵ AOC = ∠ BOD = ∠ AOB = × 80 ° = 40 °, ∵ OD, OE bisector ∵ BOC, ∵ doc = ∠ BOC = × 40 ° = 20 °∠ EOC = ∠ AOC = × 40 ° = 20 °, ∵ DOE = ∠ doc = ∠ EOC = 20 ° + 20 ° = 40 °; (2) when OC rotates ∵ OD, OE

It is known that in the interior of AOB, OD is the bisector of AOC, OE is the bisector of BOC, and the degree of DOE is calculated

∵ od bisection ∠ AOC
∴∠COD＝∠AOC/2
∵ OE bisection ∠ BOC
∴∠COE＝∠BOC/2
∴∠DOE＝∠COD+∠COE＝（∠AOC+∠BOC）/2＝∠AOB/2＝70/2＝35°
The math group answered your question,

An arithmetic problem, Divide a circle into several small sectors, and then form an approximate rectangle. The length of this rectangle is 9.42 decimeters, and the perimeter is 24.54 decimeters. The perimeter of this circle is ()? The area is?