[help] there is a point P in the triangle ABC, which satisfies "vector PA + vector Pb + vector PC = 0". Can you explain that P is the center of gravity of the triangle ABC?

[help] there is a point P in the triangle ABC, which satisfies "vector PA + vector Pb + vector PC = 0". Can you explain that P is the center of gravity of the triangle ABC?

That is vector PA + vector Pb = - vector PC
So the sum vector of PA and Pb is opposite to that of PC
Let the vector be PD, intersect AB with O
Then p, D, C and o are collinear
By the parallelogram rule of vector composition, PD and ab are equally divided
So Ao is a middle line of the triangle ABC
In the same way, the other two lines also pass P
So p is the center of gravity of the triangle ABC
Er
Yes~
First, ask LZ to draw a picture, and take the midpoint of BC as D
So vector PA + vector Pb = 2 vector PD
Because
PA+PB+PC=0
There's the vector PC = - 2, the vector PD
It shows that PCD is collinear and PC / PD = 2:1
Doesn't that mean point P is the center of gravity
In the binomial expansion of (x ^ 2-x + 2) ^ 10, what is the coefficient of item x ^ 3?
The answers on the first three floors are all wrong. The answer is 9600
Please don't trust the answers in books, especially in exercise books
Because I don't know the education background of the building owner, I will introduce the basic binomial expansion formula first, and then solve your problem
For (a + b) ^ n, the expansion is the summation of C (n, K) * a ^ (n-k) * B ^ k, where k is from 0 to N, and C (n, K) = n! / [(n-k)! * k!] is a common combination formula
The introduction of basic knowledge is over. Next, solve the problem
x^2 - x + 2 = (x^2+2) - x
[(x^2+2)-x]^10 =
C(10,0)*(x^2+2)^10*(-x)^0 + C(10,1)*(x^2+2)^9*(-x)^1 + C(10,2)*(x^2+2)^8*(-x)^2 + C(10,3)*(x^2+2)^7*(-x)^3…… + C(10,9)*(x^2+2)^1*(-x)^9 + C(10,10)*(x^2+2)^10*(-x)^10
Please look at the expansion above. For those with (- x) ^ 4, (- x) ^ 5 (- x) ^ 10, it will not contain x ^ 3 after further expansion
Any power of (x ^ 2 + 2) is always even power of X after expansion. Multiplication with even power of (- x) will not produce x ^ 3
To sum up, only C (10,1) (x ^ 2 + 2) ^ 9 * (- x) ^ 1 and C (10,3) * (x ^ 2 + 2) ^ 7 * (- x) ^ 3 can produce x ^ 3 terms
For (x ^ 2 + 2) ^ 9, the term x ^ 2 in the expansion is C (9,8) * (x ^ 2) * (2 ^ 8),
After multiplying with C (10,1) * (- x) ^ 1, the x ^ 3 term of the whole formula is produced. The coefficient is
-C(10,1)*C(9,8)*2^8 = -23040
For (x ^ 2 + 2) ^ 7, the x ^ 0 term in the expansion is C (7,7) * [(x ^ 2) ^ 0] * (2 ^ 7),
It is multiplied by C (10,3) * (- x) ^ 3 to produce the x ^ 3 term of this formula. Its coefficient is
-C(10,3)*C(7,7)*2^7 = -15360
Add two coefficients - 23040 - 15360 = - 38400
This is the coefficient of the last x ^ 3 term
In the plane rectangular coordinate system, we know that O is the origin of the coordinate, points a (3,0), B (0,4), take point a as the center of rotation, and △ ABO rotates clockwise to get △ ACD. Note that the rotation angle is α, and ∠ ABO is β
If the rotation satisfies ∠α = 60 °, the coordinate of point C can be obtained
C (3 + (3 √ 3 + 4) / 10, (4 √ 3-3) / 10) ∠ ABO = β, then ∠ Bao = 90 ° - β, and Tan β = AO / Bo = 3 / 4, so sin β = 3 / 5, cos β = 4 / 5ab = 5 after clockwise rotation, AC = AB = 5, take the right point of a on X axis as e, then ∠ CAE = 180 ° - (α + (90 ° - β)) = β + 30 ° so C (XC, YC) coordinate is
(1) ∵ point a (3,0), B (0,4), OA = 3, OB = 4,
In RT △ AOB, ab = oa2 + ob2 = 5 is obtained by Pythagorean theorem,
Da = OA = 3
As shown in Figure 1, DM ⊥ X axis is made at point m through point D,
Then MD ‖ ob,
There are ADAB = Amao = dmbo,
The results show that am = ADAB &; AO = 35 × 3 = 95,
∴OM= 65，
Using Cramer's rule to solve the equations X1 + x2 + X3 + X4 = 5 x1-2x2-x3 + 4x4 = - 2 2x1-3x2-x3-5x4 = - 2 3x1 + x2 + 2x3 + 11x4 = 0
X1+X2+X3+X4=5 (1)X1-2X2-X3+4X4=-2 (2)2X1-3X2-X3-5X4=-2 (3)3X1+X2+2X3+11X4=0 (4)(1)+(2) (1)+(3) (1)x2-(4)2X1-X2+5X4=3 (5)3X1-2X2-X4=3 (6)-X1+X2-9X4=10 (7)(5)x2-(6) (5)+(7)X1+11X4=6 (8)X1-4X4=13 (9)(8)-...
If AB = 6, BC = 4, then AE =?
First, we use cosine theorem: C square = a square + b square - 2abcos θ to calculate the angle of triangle, and then we have
Be square = AE square + AB square - 2ae * AB * cosa sum
Be squared = CE squared + BC squared - 2 * be * ce * COSC and
AE + CE = 6 can get the value of AE
In the expansion of [x / 2-x ^ (- 1 / 3)] ^ n, only the binomial coefficient of the fifth term is the largest
Finding the constant term in the expansion
According to the Yang Hui triangle, the largest binomial coefficient of the nth row is n / 2 + 1, so n = 8
The constant term in the expansion is easy to do. The term of (x / 2) ^ 2 is C28 / (2 ^ 2). The third term is the constant term in the expansion
In Cartesian coordinate system, given points a (4,0), B (0,3), if a right triangle is congruent with RT △ ABO, and they have a common edge, please write out the unknown vertex coordinates of the triangle (do not need to write the calculation process). (hint: consider the three cases of Ao, Bo, AB being the common edge respectively)
As shown in the figure, if AB is the common edge, there are three answers (72259625), (4,3), (2825, - 2125); if Bo is the common edge, there are two answers (- 4,3) and (- 4,0); if Ao is the common edge, there are two answers (0, - 3) and (4, - 3)
Use Clem's rule to solve the following equations: 2x1 + x2-5x3 + X4 = 8 x1-3x2-6x4 = 9 2x2-x3 + 2x4 = - 5 X1 + 4x2-7x3 + 6x4 = 0
Do you have to use Cramer's law? It's a bit boring. I think the theoretical significance of Cramer's law is greater than the practical significance. To solve the equation will only increase the amount of calculation
Triangle ABC is obtuse angle triangle, a = 3, B = 4, C is unknown, C is obtuse angle, then the value range of C is? The answer is 5 < C < 7, why is 5 < C < 7?
A + B = 3 + 4 = 7; and the third side is less than the sum of the two sides. Because it is an obtuse angle, it is a right angle when it is equal to 5,
cosC=(9+16-c^2)/245
C
Find the sum of the coefficients of the expansion of binomial "(x + 2) to the 15th degree"
To find the sum of the coefficients of the expansion of binomial "(x + 2) to the 15th degree", we must give a detailed answer
The coefficients of the expansion = A0 + a1x ^ 1 + a2x ^ 2 +... + a15x ^ 15
The sum of the coefficients = A0 + A1 + A2 +.. + A15
When x = 1, 15 times of "(x + 2) = 3 ^ 15
The sum of the coefficients = A0 + A1 + A2 +.. + A15 = 3 ^ 15
It's very simple to sprinkle with Yangwei 3 jiao