A cuboid pool, 80 meters long and 40 meters wide, can hold up to 4000 cubic meters of water. How many meters deep is the pool? How much cement is applied to the four walls and the bottom square meter?

A cuboid pool, 80 meters long and 40 meters wide, can hold up to 4000 cubic meters of water. How many meters deep is the pool? How much cement is applied to the four walls and the bottom square meter?

4000 / (80 * 40) = 1.25 (m)
(80 * 1.25 + 40 * 1.25) * 2 + 80 * 40 = 3500 (M2)

12+56+1112+1920+2930+… +97019702+98999900．

12+56+1112+1920+2930+… +97019702+98999900=（1-12）+（1-16）+（1-112）+… +（1-19900）=1×99-（12+16+112+… +19900），=99-（11×2+12×3+13×4… +199×100），=99-（1-12+12-13+13−14+… +199-1100），=99-（1-...

The area of this triangle is () square decimeter
The area of this triangle is () square decimeters

3.5 times 3.5 divided by 2
=12.25/2
=6.125 (square decimeter)

The quotient of 45 divided by 9 / 7 is 15 times more than that of a number. Find the number and use the solution of the equation

45/(7/9)=2x+15
45*9/7-15=2x
x=150/7

A cuboid tin oil tank with a circumference of 28 decimeters on the bottom and a height of 1 meter. Paint the outside of the oil tank. What is the paint area
If the weight of oil per liter is 0.68 kg, how many kg of oil can the tank hold at most?

A cuboid tin oil tank, the bottom is a cube with a circumference of 28 decimeters and a height of 1 meter. Paint the outside of the oil tank. What is the area of the paint?
1 meter = 10 decimeters
Side length of bottom surface 28 △ 4 = 7 decimeters
28×10+7×7×2
=280+98
=378 square decimeters
If each liter of oil weighs 0.68 kg, how many kg of oil can the tank hold at most?
0.68×（7×7×10）
=0.68×490
=333.2kg

How to calculate 4.8 * 0.23 + 0.48 * 7.7 simply

4.8*0.23+0.48*7.7
=4.8*0.23+4.8*0.77
=4.8*（0.23+0.77）
=4.8

In RT △ ABC, the bisector of acute angle a intersects with the bisector of the adjacent complementary angle of acute angle B at point D, then ∠ ADB=______ Degree

If the size of acute angle ∠ A is x, then the adjacent complementary angle of acute angle ∠ ABC is 90 ° + X; then ∠ ADB = 180 ° - (x2 + 90 ° - x + 45 ° + x2) = 45 °

Prove inequality with vector: √ (A1 ^ 2 + A2 ^ 2 + a3 ^ 2) * √ (B1 ^ 2 + B2 ^ 2 + B3 ^ 2) ≥ A1 * B1 + A2 * B2 + a3 * B3|

It is known that s △ doc = 15 square centimeter, Bo = 23bd

Let H be the height of the trapezoid, which is also the height of △ DBC, because ob = 23bd, BD = Bo + OD, so Bo = 2od, and because in △ AOD and △ DBC, ad ‖ BC, Bo = 2od, so ad = 12bc, because △ doc is the same height as △ BOC, Bo = 2od, s △ doc = 15 square centimeter, so s △ BOC = 2, and △ doc = 2 × 15 = 30

Solving the equation lgx ^ 2-lgx = 0

This is lgx ^ 2 = 2lgx
lgx^2-lgx=0
2lgx-lgx=0
lgx=0
x=1