-1,1,3,5,7... According to this rule, what is the nth number

-1,1,3,5,7... According to this rule, what is the nth number

It is not difficult to see from the meaning of the question that each item is increased by 2 from - 1. Therefore, the nth number in the formula can be reduced to 2n-3 by - 1 + 2 (n-1)

If we dig a 21 meter deep cylindrical pool to store the water, the volume is 2.84 times 10 to the fourth power, and the diameter of the bottom circle of the pool should be at least several meters
Thank you first!

Divide the volume by the depth of the pool, then you get the bottom area of the pool. Divide the bottom area by 3.1416, root the number, and multiply the number by 2, then you get the diameter

The perimeter of a circular pool is 31.4 meters. Now, after enlarging the diameter of the circular pool by two thirds, how many square meters more is the pool area than the original one

The original radius is 31.4/3.14/2 = 5 meters, and the area is 3.14 * 5 * 5 = 78.5 square meters
Expanded radius (10 + 10 * 2 / 3) / 2 = 8.34, area 3.14 * 8.33 * 8.33 = 218.04 square meters
218.04-78.5 more = 139.54 square meters
The area of the pool is 139.54 square meters more than before

A round pool around every 2 meters planted a poplar, a total of 40 planted around the pool is a few meters

80 meters

A cuboid pool is 5 meters long, 4 meters wide and 3.6 meters deep. Now there is 36 cubic decimeters of water in the pool. How many meters is the water depth?
Please! 20 points for the answer!

Water depth = 36 △ 5 △ 4 = 1.8m

There are two pools, large and small. There are 300 cubic meters of water in the large pool and 70 cubic meters in the small pool. Now, the same amount of water is injected into the two pools, so that the water volume of the large pool is three times that of the small pool. How many cubic meters of water should be injected into each pool?

According to the meaning of the question: the difference of their water volume is: 300-70 = 230 (cubic meters); when the water volume of the large pool is three times that of the small pool, the water volume of the small pool is: 230 ÷ (3-1) = 115 (cubic meters); the injected water is: 115-70 = 45 (cubic meters). A: each pool should be injected with 45 cubic meters of water

The circulating water volume of our high temperature furnace is about two cubic meters, and the pool is about 80 cubic meters!

Your cooling water tower doesn't need to be very large. According to the circulating water volume of 2 cubic meters per hour, plus 20% margin, 3 cubic meters per hour cooling water tower is OK, and the cooling pool is large enough

If we know the square of x plus x-3 = 0, then the cube of X + the square of 1991x + 1987X + 1990 =?

It is known that the square of x plus x-3 = 0, so the square of x = 3-x is substituted into the cube of X + the square of 1991x + 1987X + 1990 = x (3-x) + 1991 (3-x) + 1987X + 1990 = the square of 3x-x + 5973-1991x + 1987X + 1990 = 3x - (3-x) + 5973-1991x + 1987X + 1990 = 3x-3 + X + 7963-4x = 7960

The sum of the first n terms can be obtained by finding 1.3 of 3 and 178; 2.3 of 3 cubic parts

Let Sn = 1 / 3 + 2 / 3 ^ 2 + 3 / 3 ^ 3 +. + n / 3 ^ n,
Then 3Sn = 1 + 2 / 3 + 3 / 3 ^ 2 +. + n / 3 ^ (n-1),
Subtract 2Sn = 1 + 1 / 3 + 1 / 3 ^ 2 +. + 1 / 3 ^ (n-1) - N / 3 ^ n
=[1-(1/3)^n]/(1-1/3)-n/3^n
=3/2*[1-(1/3)^n]-n(1/3)^n
=3/2-(2n+3)/2*(1/3)^n
So Sn = 3 / 4 - (2n + 3) / 4 * (1 / 3) ^ n

The volume of an object is 200cm3. When half of its volume is immersed in water, the buoyancy it receives is______ N. What is the gravity of the drained water when it is completely submerged in water______ N．（g=10N/kg）

(1) When half of its volume is immersed in water, V row 1 = 12V = 12 × 200cm3 = 100cm3 = 1 × 10-4m3. According to Archimedes' principle, its buoyancy is f floating 1 = ρ liquid GV row 1 = 1.0 × 103kg / m3 × 10N / kg × 1 × 10-4m3 = 1n, (2) when it is completely immersed in water, V row 2 = v = 200cm3 = 2 × 10-4m3, its buoyancy is f floating 2 = ρ liquid GV row 2 = 1.0 × 103kg / m3 × 10N / kg × 2 × 10-4m3 = 2n According to the principle of Archimedes, f-floating = g row: g row 2 = f-floating 2 = 2n