I come from China. Or I'm come from China?

I come from China. Or I'm come from China?

I come from China!
Because come is a verb, the present tense is used here. The verb am is not allowed to be added in front of it
If you remove come I am from China, that's right!

Do I come from China and I am from China mean the same

same.
COME FROM=BE FROM

I'm from China, or I'm come from China?

You have three choices
1.I am from China
2.I come from China
3.I am Chinese.

As shown in the figure, Mn is the diameter of ⊙ o, if ∠ e = 25 ° and ∠ PMQ = 35 °, then ∠ mqp = () A. 30° B. 35° C. 40° D. 50°

Connect Po and Qo
According to the theorem of circular angle, we get
∠POQ=2∠PMQ=70°，
OP = OQ,
Then ∠ OPQ = ∠ OQP = 55 °,
Then ∠ POM = ∠ e + ∠ ope = 80 °,
So ∠ PQM = 1
2∠POM=40°．
Therefore, C

As shown in the figure, Mn is the diameter of ⊙ O. if ∠ a = 10 ° and ∠ PMQ = 40 °, the regular polygon is an inscribed regular polygon with PM as its edge______ Polygon

Connect Qo, Po, ∵ Qo = Po,  OPQ = ∵ OQP,  PMQ = 40 °,  poq = 80 °, OPQ + ∠ OQP = 180 ° - 80 ° = 100 °,  OPQ = ∠ OQP = 50 °,  a + ∠ apo = ∠ POM = 10 ° + 50 ° = 60 °, ∵ Po = OM,  POM is an equilateral triangle,  PM = OP = OM,  taking PM as

P is a point out of the plane of the triangle ABC, m and N are the midpoint of AB and PC respectively. PA = BC = m, Pb = AC are known. It is proved that: (1) Mn is the common perpendicular line of AB and PC

Connecting an, BN, APC and BPC are congruent equilateral triangles, and N is the midpoint of PC
∴AN=BN
And ∵ m is the midpoint of AB, ∵ Mn ⊥ ab
Similarly, Mn ⊥ PC can be proved
And ∩ Mn ∩ AB = m, Mn ∩ PC = n
/ / Mn is the common perpendicular of AB and PC

As shown in the figure, the known point a is the trisection point on the semicircle, and B is At the midpoint of an, P is the moving point on the diameter Mn, and the radius of ⊙ o is 1? The minimum value of AP + BP is given

P is located at the intersection of a ′ B and Mn, and the value of AP + BP is the smallest; as a symmetric point of a about Mn, according to the symmetry of the circle, then a ′ must be on the circle, connect BA ′ to intersect Mn with P, connect PA, then pa + Pb is the smallest, at this time PA + Pb = PA ′ + Pb = a ′ B, connect OA, OA ', ob, ∵ an = 13mn,  AON = ﹤ a ′ on = 60

There is a point P in the plane where △ ABC is located, so that the vector PA + Pb + PC = ab. find the position of point P

Vector PA + vector Pb + vector PC = vector ab
ν vector PA + vector Pb + vector PC = vector AP + vector PB
ν vector PA + vector PC = vector AP
ν vector PC = vector 2AP
As shown in the figure, P is the trisection point of line AC (close to a)
--------------
A P C

It is known that: P is any point in the triangle ABC. If PA is connected, find the relationship between ab + BC + AC and PA + Pb + PC

In △ abd, there are: ab + BD > ad = PA + PD; in △ PCD, CD + PD > PC; by adding the two formulas, we can get: ab + BD + CD + PD > PA + PD + PC, where BD + CD = BC, we can get: ab + BC > PA + PC; similarly, we can get: BC + AC > PA + Pb; AC + AB > Pb + PC; 3

The bottom surface ABC of the triangular pyramid p-abc is a right triangle AB = BC, PA = 2Ab, PA vertical plane ABC. Calculate the degrees of angles between BC vertical Pb, Pb and plane PAC

First question:
∵ PA ⊥ plane ABC, ∵ BC ⊥ PA
∵ △ ABC is a right triangle, and ab = BC, ᙽ BC ⊥ ab
From BC ⊥ PA, BC ⊥ AB, ab ∩ PA = a, BC ⊥ plane PAB, ∩ Pb
Second question:
The angle between Pb and PAC is arcsin (√ 10 / 10)