Please list some examples of rotation in real life, and point out the center and angle of rotation RT.

Please list some examples of rotation in real life, and point out the center and angle of rotation RT.

The wheel when the car starts: the center of rotation is the axis
Clock: the center of rotation is the center of the dial with three hands overlapping
The revolving door of the hotel: the revolving center is the column in the middle
In addition, there are many, such as windmills, electric fans and swing,
As for the angle of rotation, it depends on the actual situation

The coordinates of vertex C of parallelogram ABCD are (- 5,7), and the direction vectors of the line where AB and AD are located are (- 2,3) and (0,3) respectively
Finding the point direction equation of the line where BC and CD are located
Please be specific,

Let P (x, y) be a point on CD
Vector CP / / vector (- 2,3)
（x+5,y-7)//(-2,3)
(x+5)/(-2)=(y-7)/3
The point wise equation of CD is as follows:
3x+2y+1=0
Let a point m (x, y) on BC
Vector CP / / vector (0,3)
(x+5,y-7)//(0,3)
x+5=0
The pointwise equation of BC is as follows
x= - 5

As shown in the figure, in the equilateral triangle ABC, Bo and co divide ∠ ABC, ∠ ACB, OE ‖ AB, of ‖ AC equally, and try to explain be = EF = FC

It is proved that: ∵ ABC is equilateral triangle, ∵ ABC = ∠ ACB = 60 °, ∵ OE ‖ AB, of ‖ AC, ∵ OEF = ∠ ABC = 60 °, ∵ ofe = ∠ ACF = 60 °, ∵ OEF = ∠ ofe, ∵ EOF = 60 °, ∵ OEF is equilateral triangle, ∵ OE = of = EF, ∵ Bo, CO divide ∵ ABC, ∵ ACB, ∵ a equally

How to do question 3 on page 53 of mathematics compulsory course 2 of senior one

Tips:
The lines EH and FG intersect at K;
From the point K ∈ eh, EH is contained in the plane abd
We get: K ∈ plane abd
It can be proved that the point K ∈ plane BCD
And plane abd ∩ plane BCD = BD,
So, point K ∈ line BD
That is straight line eh, FG, BD three straight lines compared to a point

Take AB and AC of triangle ABC as sides, make equilateral triangle abd and triangle ace, connect CD and be to intersect at point O, and prove OA
Bisector doe

It is proved that Δ abd and Δ ace are equilateral triangles,
∴AD=AB、AC=AE,∠DAB=∠CAE=60°,
∴∠DAB+∠BAC=∠CAE+∠BAC,
That is, DAC = BAE,
∴ΔADC≌ΔABE(SAS),
∴∠ACO=∠AEO,
Through a as am ⊥ CD in M, an ⊥ be in N,
In RT Δ ACM and RT Δ AEN,
AC=AE,∠ACO=∠AEO,∠AMC=∠ANE=90°,
∴ΔACM≌ΔAEN(AAS),
∴AM=AN,
Ψ Ao bisection ∠ doe

Given that vector a = (1,1), vector b = (- 1,3), vector C = (k, 5), if vector a-2b and vector C are collinear, then K=

Vector a-2b = (3, - 5)
Collinear with vector C, so (3 / k) = - 5 / 5, so k = - 3

A. B, C, D are four points on the same sphere, where △ ABC is an equilateral triangle, ad ⊥ plane ABC, ad = 2Ab = 6, then the volume of the sphere is___ ．

A, B, C and D are expanded into triangular prism. The distance between the middle point of the center line between the upper and lower bottom surfaces and a is the radius of the ball. Ad = 2Ab = 6, OE = 3, △ ABC is an equilateral triangle, so AE = 23ab2 - (12ab) 2 = 3. Ao = 23. The volume of the ball is 323 π. So the answer is 323 π

It is known that vectors E1 and E2 are a set of bases of all vectors in plane a, (as follows)
And a = E1 + E2, B = 3e1-2e2, C = 2E1 + 3e2, if C = λ a + μ B, (λ, μ∈ R), try to find the value of λ, μ
I may have a wrong idea, which is different from the result of the answer. Make the idea clear

c=λa+μb=λ（e1+e2)+μ(3e1-2e2)=(λ+3μ)e1+(λ-2μ)e2=2e1+3e2
So λ = 13 / 5, μ = - 1 / 5

As shown in the figure, in RT △ ABC, ∠ C = 90 °, D is a point on the edge of AC, de ⊥ AB is in E, de: AE = 1:2. Find SINB, CoSb, tanb

∵∵∵ a = ∵ a, ∵ AED = ∵ ACB, ∵ ABC ∵ ade, ∵ BC: AC = de: AE = 1:2, let BC = x, then AC = 2x, then AB = BC2 + ac2 = 5x, ∵ SINB = acab = 255, CoSb = bcab = 55x, tanb = ACBC = 2

If log (x + 3)

First of all, we need to consider the domain of definition, x + 3 > 0, X-2 > 0, and get the range of X is x > 2, a > 0 and not equal to one
But we see that X-2 is obviously larger than x + 3, so the function must be a decreasing function, so the range of X is (2, + ∞), and the range of a is (0,1)