(1) A kind of disinfectant is made up of 1:5 ratio of disinfectant and water bank. How many kilograms of water are put into 10 kilograms of disinfectant? (2) The 314cm thin iron wire is just wound 10 times on a round iron bar. What is the radius of the cross section of the round iron bar?

(1) A kind of disinfectant is made up of 1:5 ratio of disinfectant and water bank. How many kilograms of water are put into 10 kilograms of disinfectant? (2) The 314cm thin iron wire is just wound 10 times on a round iron bar. What is the radius of the cross section of the round iron bar?

10kg*(5/6)=25/3kg
314cm/10/2π=5cm

Who can help me solve these math problems
1. How many square decimeters is the surface area of a square box with four decimeters of edge length?
2. There is a cube water tank, each side of which is 5 decimeters long. If the cube water tank is filled with water, and then it is poured into a cuboid water tank which is 8 decimeters long and 25 decimeters wide, how deep is the water?
3. The length, width and height of a cuboid are three continuous natural numbers. Its volume is 120 cubic centimeters. What is its surface area in square centimeters?

4*4*5=80
Volume of water: 5 * 5 * 5 = 125
Water depth is the volume of water used divided by the bottom area of the pool: 125 / (8 * 25) = calculated by yourself
120=2*2*2*3*5
So if three natural numbers are multiplied, they can only be: 4, 5, 6
So the surface area is 2 * (4 * 5 + 5 * 6 + 4 * 6) = calculated by yourself

A rectangular orchard is 450 meters long and 350 meters wide. Each fruit tree covers an area of 15 square meters. How many tons of fruit can be obtained from this orchard according to 72 kg of fruit per fruit tree?

I'm glad to be able to answer your question here. The correct answer to this question should be: orchard area: 450 × 350 = 157500m ᦇ 178; fruit trees: 157500 ﹥ 15 = 10500; total fruit yield: 10500 × 72 = 756000 kg & nbsp; 756000 kg = 756 tons. If you have any questions, please ask me

In the triangle ABC, the angle ABC is 90 ° and the diamond ACEF is made outward with AC as one side. D is the center of the diamond ACEF and connects BD. when BD bisects the angle ABC, what is the special quadrilateral of ACEF? If AB = 3, BD = 4, root 3, find the length of BC. If there is no picture, please draw the picture according to the diagram for 60 minutes

∵ BD bisection ∠ ABC
Three wires in one, ∠ ADB = 45 °= ∠ BDC
ABCD is a rectangle, so ACEF is a square
BD=AC
∴BC=√39

A mathematical problem about geometry,
As shown in the figure, the triangle ABC is an isosceles triangle. Given ad = ed, find the degree of CED. The answer is CED = 80 degrees. How can we get 80 degrees,
Sorry, there's no point
Triangle ABC is an isosceles triangle, be is the bisector of ∠ ABC, ABC is 40 degrees (so ∠ abd and ∠ CBD are both 20 degrees), ACB is 40 degrees, ad = ed, find the degree of CED
CED=80
I would like to ask how 80 degrees can be obtained. Any method will do.

Intercept BF = Ba on BC and connect DF
The following angles are known: abd = CBD = 20 ° and ACB = 40 °
According to SAS, △ abd ≌ △ FBD can be proved
Therefore, BDF = BDA = 60 °, BFD = a = 100 ° and ad = DF
Therefore, CDF = 60 ° and CFD = 80 °
However, CDE = BDA = 60 degree
Therefore, CDF = CDE
So according to SAS, △ CDF ≌ △ CDE can be proved
Therefore, e = CFD = 80 degree
Jiangsu Wu Yunchao answers for reference!

Urgent, a mathematical problem of space geometry
In the dead pyramid p-abcd, the bottom ABCD is a square with side length a, the side pad ⊥ the bottom ABCD and PA = PD = √ 2 / 2ad, if e and F are the midpoint of PC and BD respectively
(1) Prove that the EF ‖ face pad (2) proves: the face PDC ⊥ face pad (3) finds the tangent of the dihedral angle b-pd-c

1. Connect AC, obviously f bisects AC in △ PAC, EF is the median line ﹣ EF ﹣ PA ﹣ EF ﹣ plane pad. 2. Side pad ⊥ bottom ABCD, and CD ⊥ intersection line ad ﹣ CD ⊥ side pad is CD ∈ plane PDC ﹣ plane PDC ⊥ plane pad. 3

As shown in the figure, in the spatial quadrilateral ABCD, ab = CD, ab ⊥ CD, e and F are the midpoint of BC and ad respectively, then the angle formed by EF and ab is______ ．

Take the midpoint m of AC and connect EM and FM. ∵ e is the midpoint of BC, ∵ EM ∥ AB and EM = 12ab; similarly: FM ∥ CD and FM = 12CD, ∵ FEM is the angle formed by AB and EF, and ∵ ab ⊥ CD, ab = CD, ∵ FM = em, FM ⊥ em, ∵ EFM is isosceles right triangle, ∵ FEM = 45 ° so the answer is 45 °

Known: as shown in the figure, CB ⊥ AB, CE bisection ⊥ BCD, de bisection ⊥ CDA, ⊥ 1 + ⊥ 2 = 90 ° verification: Da ⊥ ab

∵ CE bisection ∠ BCD, de bisection ∠ CDA, ∵ 1 = 12 ‰ ADC, ∵ 2 = 12 ‰ BCD, ∵ 1 + ∵ 2 = 12 ‰ ADC + 12 ‰ BCD = 12 (∵ ADC + ∵ BCD) = 90 °, ∵ ADC + ∵ BCD = 180 °, ∵ ad ‖ BC, ∵ a + ∵ B = 180 °,