A square table is made of a table top and four legs. It is known that one cubic meter of wood can be used to make 50 tables or 300 legs. Now five cubic meters of wood, what should we do Cutting material, can make the table top and table leg just match?

A square table is made of a table top and four legs. It is known that one cubic meter of wood can be used to make 50 tables or 300 legs. Now five cubic meters of wood, what should we do Cutting material, can make the table top and table leg just match?

If x cubic meter wood is used as table top, 5-x cubic meter wood is used as table legs
A total of 50x tabletops and 300 (5-x) legs can be made;
The formula of countable proportion: 50x: 300 (5-x) = 1:4,
The solution is: x = 3, 5-x = 2,
That is: 3 cubic meters of wood to do desktop, with 2 cubic meters of wood to do legs

In RT △ ABC, the opposite sides of a, B and C are a, B and C respectively. If a = 3b, the values of SINB, CoSb and tanb are obtained

Obviously, if a > b, then there is a big side to a big angle, and angle B must be smaller than angle a, then angle B is not a right angle
When angle a = 90
Then SINB = B / a = 1 / 3, CoSb = 2, radical 2 / 3, tanb = radical 2 / 4
When C = 90
So tanb = 1 / 3, SINB = root 10 / 10
CoSb = 3 radical 10 / 10

log_ a（2-√3）

0=log_ a(1)=log_ b(1)
That is, log_ B [2 - (radical 3)]

In acute angle △ ABC, AC = 1, ab = C, ∠ a = 60 ° and △ ABC circumcircle radius R ≤ 1, then the value range of C is ()
A. 12＜c＜2B. 0＜c≤12C. c＞2D. c=2

According to the cosine theorem, BC = AB2 − 2 · ab · AC · cosa + ac2 = C2 − C + 1. According to the meaning of the topic, it is obtained that | C − 1 ＜ C2 − C + 1C2 + (C2 − C + 1) ＞ 1212 + (C2 − C + 1) ＞ c2c2 − C + 1sin60 ° = 2R ≤ 2 {12 ＜ C ≤ 2. When C = 2, △ ABC is a right triangle, so 12 ＜ C ＜ 2

123+123+123+12233-123456=?

-110854

In the triangle ABC, ab = AC, ad is perpendicular to BC at D, point P is on BC, PE is perpendicular to BC, the extension line of Ba is at e, AC is at F, the proof is: 2ad = PE + PF
I'm sorry. I sent it on my cell phone. No pictures

Make GC ⊥ BC cross be extension line to g, make eh ⊥ GC to H
Then ∠ GEH = ∠ B = ∠ FCD
EH=CP
∴Rt△GEH≌Rt△FCP
∴HG=PF
∴PE+PF=CH+HG=CG
The vertical line at the bottom of an isosceles triangle is the center line
The ad is the median of △ BCG
∴2AD=CG=PE+PF

Using pseudo code to express the algorithm process of solving inequality ax > b (a is not equal to 0)
RT.

A = input; input a
B = input; input B
If a > 0 then output [x > b / a]; if a > 0 then output x > b / A
if a

If a quadrilateral ABCD satisfies AB * BC = CD * Da, | ab | = | CD |, then the shape of the quadrilateral is a parallelogram. How to prove? (AB BC CD Da is a vector)
AB BC CD Da is a vector
AB*BC=|AB|*|BC|*cos
AB + BC + CD + Da = 0 (AB BC CD Da is vector)
And | ab | = | CD|
Then | BC | = | da|
If it is trapezoidal, then | BC | = | Da | does not hold

AB*BC=|AB|*|BC|*cos
CD*DA=|CD|*|DA|*cos
Because | ab | = | CD|
Then | BC | * cos = | Da | * cos
AB + BC + CD + Da = 0 (AB BC CD Da is vector)
And | ab | = | CD|
Then | BC | = | da|
cos=cos
If two opposite sides are equal, the figure is a parallelogram

1. There is a 12 decimeter long iron wire, which is surrounded by different rectangles (length and width are the whole decimeter number) or rectangles. Can you think of several different encirclement methods? Please write down their length and width

1,5; 2; 4; 3; 3 (a square is a special rectangle)

Among the following physical quantities, the vector is ()
A. Force B. distance C. acceleration D. time

Vector is a physical quantity with both size and direction. Force and acceleration are vectors. Distance and time are scalars with only size and no direction, not vectors. So AC is correct and BD is wrong