Elementary school 6th grade mathematics is eager for answers, and it wants a process.

No matter what shape the cylinder is cut into or what shape it becomes, the area may change, but the volume will never change.

In fact, the volume of the rectangle is the volume of the cylinder, and all the problems are solved by only requiring the height.

You probably just don't know the high:

80 2 (8 2) = 40 4 = 10 centimeters.

So the volume of the cylinder: cubic centimeters.

There are more left and right sides of the box, the length is equal to the radius, and the width is equal to the height.

Radius: 8 2 = 4

Left area: 80 2=40

High: 40 4=10

The volume of the box is equal to the volume of the cylinder.

Cubic centimetre. I hope mine is helpful to you, if you don't understand anything about this question, you can ask, if you are satisfied, remember! Thank you!! [Mathematics Group].

80 2 = 40 (square centimeters).

8 2 = 4 (cm).

40 4 = 10 (cm).

8* centimeter) centimeter) cubic centimeter).

80 2 (8 2) means: the two sides of the cuboid, that is, 80 square centimeters more than the original, the increase of 80 square meters 2 is equal to the area of one side of the cuboid, the length of the side of the cuboid is the height of the cylinder, the width of the side of the cuboid is the radius of the cylinder, the radius is 8 2, 80 2 (8 2) is the height of the cylinder. The area of the rectangle (80 2) width (8 2) = height (10) is the meaning of this equation.

The height of the cylinder: 80 2 (8 2) = 40 4 = 10 (cm).

The volume of the box = the volume of the original cylinder: cubic centimeters).

15/28

1/2 of a number multiplied by its reciprocal is equal to one.

2/38/4

You can't multiply them all.

Hello: Total to complete: 400 (1 10%) 440 (pcs).

The second half of the month: 440 400 40% 280 (pcs).

400*(1+10%-40%)=280

A workshop that plans to produce 400 parts in June is regarded as 1, and 40% is completed in the first half of the month: 1-40% is now to make the actual output exceed the target by 10%, +40%; Multiply by 400

The plan is 10% over, which means that 400 (1+10%) = 440 pieces need to be produced, and 400 * (1+10%) = 160 pieces have been completed in the first half of the month, so 440-160 = 280 pieces need to be completed in the second half of the month.

From ab:bc=1:2, ab=6cm, bc=12cm.

And you need the cone to be high AB=6cm, and the base radius is BC=12cm

So v=1 3*12*12* *6=288 (cm2).

Circumference of the maximum circle =

Maximum circle radius =

Maximum circle area =

The diameter ratio of a circle is 1:2:3

So the circumference ratio is: 2 :3 = 1:2:3

The circumference of the circle is the required wire length.

Then let the circumferences of the three circles be x, 2x, and 3x

x+2x+3x=

6x=x=so the circumference of the circle1.

Circle 2 circumference.

Circle 3 circumference.

Visible circle 3 is the largest.

Then 2 r=

r=r=r=3 area s = r ==

So the maximum circle area.

The diameters of the three gardens are A, 2A, and 3A

then the solution is a = 2 decimeters.

The maximum circle diameter is 3a = 6 dm.

Radius = 3 decimeters.

Area = square decimeter.

c=pi*d Let the three diameters be d, 2d, 3d, and the sum of the circumferences of the three circles =pi*(d+2d+3d)=i.e., pi*6d=

So d = decimeter) so 3d = 6

Maximum circle area = pi * (6 2) * (6 2) = square decimeters.

10 solution: set the number of female workers x

The number of male workers is 156-12-x = 144-x

x-1/11 x = (144-x)*2

10x out of 11 = 288-2x

x = 288 * 11/32

x=99156-99=57 people.

There are 99 female workers and 57 male workers.

11 solution: set the sixth grade x yuan.

x+420-x)*1/7 = 60

x=（420-x+60）*2

x=960x-2x

3x=960

x=320320-60=260 yuan.

-12 = 144 (names) Removing 12 male workers who went out equates to only 144 workers.

1-1 11=10 11 The rest of the female workers.

10 11 2=5 11 Remaining male workers.

1+5 11 =16 11 All the female workers and the remaining male workers.

144 16 11 = 99 (names) Number of female workers.

156 -99=57 (people) The original number of male workers.

1 7 = 420 (yuan) The sum of donations made by grades 5 and 6.

420 -60 2) 3=100 (RMB) Grade 5 donation.

420-100 -60=260 (yuan) The sixth grade donated more than the fourth grade.

Let a person walk on the street at a speed v1 and a car at a speed v2

15*（v2-v1）=10*（v1+v2）v2=5v1

Bus No. 2 every 15*(v2-v1) v2=15*(5v1-v1) 5v1=12 minutes.

12 minutes Let x minutes the speed of the car be a and the speed of the person is b ax (a-b)=15 ax (a+b)=10 to get a = 5b ax (4 5a) = 15

x=12

Suppose a car will be sent every x minutes, and if the person does not leave, the car will catch up with the person every x minutes, or a car will come onto every x minutes. And now it is 15 minutes to drive a car from behind, which means that the car chased (15-x) minutes for the distance traveled by the person, and a car came onward in 10 minutes, indicating that the distance traveled by the person was less than (x-10) minutes, so the equation (15-x) 15 (x-10) 10 solution of x=12 can be listed