Saw the 1.2 meter long cuboid wood into three sections, the surface area increased by 48 square decimeters, the original volume of the wood is______ ．

Saw the 1.2 meter long cuboid wood into three sections, the surface area increased by 48 square decimeters, the original volume of the wood is______ ．

2 m = 12 decimeters, 48 △ 4 × 12, = 12 × 12, = 144 (cubic decimeters). Answer: the original volume of wood is 144 cubic decimeters. So the answer is: 144 cubic decimeters

A square table is composed of one table top and four legs. If one cubic meter of wood can be used to make 50 tables or 300 legs of a square table, and there are 5 cubic meters of wood, how many cubic meters of wood can be used to make the table top and how many cubic meters of wood can be used to make the table legs, and the table top and legs can be made into a square table? How many square tables can be made?

According to the meaning of the question, x + y = 550x × 4 = 300y, the solution is x = 3Y = 2, ∵ 1 cubic meter of wood can make 50 table tops of square table, a square table has a table, ∵ 50x = 50 × 3 = 150. ∵ 150 square tables can be made. Answer: 3 cubic meters of table top, 2 cubic meters of table legs, 150 square tables

A square table is made of a table top and four legs. It is known that one cubic meter of wood can be used as a table top
Please use the equation of one variable to solve the problem
1. A square table is made of a table top and four legs. It is known that one cubic meter of wood can be used as 50 or 300 legs of the table top. How to cut the existing 5 cubic meters of wood to make the table top and legs match exactly?
Thank you for the process

Let x cube wood be used as table top. According to the meaning of the question, the equation is 4 × 50x = 300 (5-x)
The solution is: x = 3 (cubic) desktop
Table leg: 5-3 = (2 cubic meters)

If AB = 4, ad = 3, Aa1 = 5, ∠ bad = 90, ∠ baa1 = ∠ daa1 = 60 in the parallelepiped abcd-a1b1c1d1, then | AC1|=______ ．

Connecting AC, ∵ AB = 4, ad = 3, ∠ bad = 90 °∵ AC = 5, according to cos ∠ a'ab = cos ∠ a'ac · cos ∠ cab, i.e. 12 = cos ∠ a'ac · 22 ∵ a'ac = 45 °, then ∠ c'ca = 135 ° and AC = 5, AA ′ = 5, AC ′ = 85 is obtained according to cosine theorem, so the answer is: 85

As shown in the figure, Bo is the middle line on the hypotenuse of R △ ABC, extending Bo to point D, so that do = Bo, connecting AD and CD

Can you send out a picture and tell me the whole topic

Senior one mathematics compulsory 2 exercise 3.3a group 6, the process is not very clear, hope to explain more detail!

According to the formula of distance between two points, there is | ab | = √ 5 | PQ | = √ 5 | Mn | = 5
Because | ab | + | PQ | < | Mn |, the segments AB, PQ and Mn cannot be enclosed as triangles

Taking AB and AC of triangle ABC as sides, make equilateral triangle abd and ACE out of triangle, connect CD and be, intersect at O. prove OA bisector angle doe

It is proved that am ⊥ be is in M and an ⊥ CD is in n when passing point a
∵ equilateral △ abd, equilateral △ ace
∴AB＝AD,AC＝AE,∠BAD＝∠CAE＝60
∵∠BAE＝∠CAE+∠BAC,∠DAC＝∠BAD+∠BAC
∴∠BAE＝∠DAC
∴△ABE≌△ADC （SAS）
∴BE＝CD,S△ABE＝S△ADC
∵AM⊥BE,AN⊥CD
∴S△ABE＝BE×AM/2,S△ADC＝CD×AN/2
∴BE×AM/2＝CD×AN/2
∴AM＝AN
Ψ OA bisection ∠ doe

We know that vector a = (3,4), vector b (8,6), vector C = (2, K), where k is a constant
If the angles of vector a and B are equal to vector C respectively, find the value of K

The angles of vector a and B are equal to vector C respectively,
Then the angular bisectors of the angles C and a, B are collinear
The unit vector of vector a is a '= (3 / 5,4 / 5)
The unit vector of vector B is B '= (4 / 5,3 / 5)
So the unit vector of the angular bisector is a '+ B' = (7 / 5,7 / 5)
Let C = m (7 / 5,7 / 5)
(2,k)=m(7/5,7/5)
m=2/(7/5)=10/7
k=7/5 m=7/5*10/7=2

The length of the bottom side of the regular triangular prism abc-a1b1c1 is 2, the length of the side side edge is root 3, D is the point on the edge of BC, ADC1 = 90 ° and the size of dihedral angle c1-ad-c is calculated

Because the regular triangular prism abc-a1b1c1, so CC1 ⊥ plane ABC,
And because ad is on plane ABC, ad ⊥ CC1,
In addition, ADC1 = 90 ° means ad ⊥ DC1,
It can be seen from the above that the ad ⊥ plane is Dcc1;
So ad ⊥ CD, that is, D is the midpoint of BC;
There is also ad ⊥ DC1, so ∠ cdc1 is dihedral angle c1-ad-c,
In the triangle cdc1, ∠ Dcc1 = 90 °, CD = 0.5 * BC = 1, CC1 = root 3,
We get Tan ∠ cdc1 = root 3, so ∠ cdc1 = 60 degree

The module of plane vector AB is 3, the module of vector BC is 4, and the module of vector CA is 5. Find vector ab · vector BC + vector BC · vector Ca + vector Ca · vector ab

Vector AB + vector BC + vector CA = zero vector
(vector AB + vector BC + vector CA) ² = 0
Vector AB & # 178; + vector BC & # 178; + vector Ca & # 178; + 2 (vector ab · vector BC + vector BC · vector Ca + vector Ca · vector AB) = 0
9 + 16 + 25 + 2 (vector ab · vector BC + vector BC · vector Ca + vector Ca · vector AB) = 0
Vector ab · vector BC + vector BC · vector Ca + vector Ca · vector AB = - 25
Another way: the module of vector AB is 3, the module of vector BC is 4, and the module of vector CA is 5
|Vector ab | &# 178; + | vector BC | &# 178; = | vector AC | &# 178;,
In triangle ABC, angle B is a right angle
Vector Ba vertical vector BC
Vector ab · vector BC + vector BC · vector Ca + vector Ca · vector ab
=Vector BC · vector Ca + vector Ca · vector ab
=Vector ca. (vector BC + vector AB)
=Vector Ca · vector AC
=-|Vector Ca |, # 178;
=-25