Immerse an iron ball in a cube container with an edge length of 3 decimeters. The height of the water surface rises from 6 cm to 8 cm. What is the volume of the iron ball Decimeter? (why?)

Immerse an iron ball in a cube container with an edge length of 3 decimeters. The height of the water surface rises from 6 cm to 8 cm. What is the volume of the iron ball Decimeter? (why?)

Volume = 3 × 3 × (0.8-0.6) = 9 × 0.2 = 1.8 cubic decimeter
6cm = 0.6dm
8 cm = 0.8 decimeter
Because the volume of the iron ball is the volume of water at the rising height of the water surface

A square iron block with 8 cm edge length was immersed in a cuboid container with 2 decimeters long and 1.6 decimeters wide and containing water
How many centimeters will the water rise

Cube volume = 8 × 8 × 8 = 512 CC
Cuboid bottom area = 2 × 1.6 = 3.2 square decimeter = 320 square centimeter
Because cube volume = volume of water rising,
So the water surface will rise: 512 △ 320 = 1.6cm

The sum of the edge length of a cuboid is 48dm. It is known that the length and width of a cuboid are 8dm and 3DM respectively

The volume of cuboid: 8 × 3 × 1 = 24 (cubic decimeter); the surface area of cuboid: (8 × 3 + 8 × 1 + 3 × 1) × 2, = (24 + 8 + 3) × 2, = 35 × 2, = 70 (square decimeter); a: the volume of cuboid is 24 cubic decimeter, the surface area is 70 square decimeter

Compared with the first stage, ()
A. As if back to the original starting point
B. The form is exactly the same
C. The higher stage repeats some characteristics of the lower stage
D. They are essentially the same

C

That man is our teacher

Question:
who is the man over there?
That man is our teacher.

Ad is the middle line of triangle ABC, e is the point on ad, and AE: ed = 1:3, the extension line of be intersects AC at F, and the value of AF / FC is calculated

Do DG ‖ BF, intersect AC at point G
∵ D is the midpoint of BC
The midpoint of FC is g
That is GF = GC
∵AE/ED=1/3
∴AF/FG=1/3
∴AF/FC=1/6

English translation
My father likes sports show

Does your father like sports show?
I want to change it into your

Section 2, known triangle abd similar triangle ace, prove triangle abd similar triangle ade

[correction: verification ⊿ ABC ⊿ ade] prove that: ⊿ abd ⊿ ace ⊿ AB / AC = ad / AE, that is AB / ad = AC / AE & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; bad = ∠ CAE ⊿ bad = ∠ BAC + ∠ CAD & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; CAE = ∠ DAE + ∠ CAD ⊿ BAC =} DAE ⊿ ABC ⊿ ade [corresponding edge is proportional to

The greatest common divisor of a and B is 8, the least common multiple is 560, and one of them is 80
Please give the answer in half an hour,
There should be more than one answer to the specific process and formula!

Suppose another number is n;
The common divisor is 8, 10 after 80 convention, and N / 8 after n convention;
Common multiple = 8 * 10 * n / 8 = 560;
So n = 56

The triangle ABC is inscribed on the circle O, AB is the diameter of the circle O, BC = 6, AC = 8 (1) the values of cosa and Sina; (2) the values of sin angle BOC and Tan angle BOC; and;

Because AB is the diameter of circle O, BC = 6, AC = 8, so ∠ ACB = 90 ° so AB = (6 ^ 2 + 8 ^ 2) ^ 0.5 = 10, so cosa = AC / AB = 8 / 10 = 4 / 5sina = BC / AB = 6 / 10 = 3 / 52, because the center angle BOC = 2 times the circumference angle a, so sin ∠ BOC = sin (2 ∠ a) = 2Sin ∠ a · cos ∠ a = 2 * (3 / 5) * (4 / 5) = 24 /