The point of - 3 on the number axis is marked as a, the point of 2 is marked as B, and the point of a is marked as a______ Edge movement______ The distance between a and B is______ ． | Math enthusiast | Mathematical art

The point of - 3 on the number axis is marked as a, the point of 2 is marked as B, and the point of a is marked as a Edge movement The distance between a and B is__ ．

The point representing - 3 on the number axis is marked as a, and the point representing 2 is marked as B. move point a to the right for 5 units of length to get point B, then the distance between two points a and B is 5, so the answer is: right, 5, 5

Point A & nbsp; is the moving point of - 2 on the number axis. When point A & nbsp; moves 4 units along the number axis to B, the real number represented by point B is ()
A. 1b. - 6C. 2 or - 6D

When point a moves to the right: so point B is 2; when point a moves to the left: so point B is - 6

If a, B, C ∈ R +, and ab + BC + Ca = 1, it is proved that a + B + C ≤ 1 / 3ABC
If the left side of the inequality is once and the right side is - 3 times, which can be adjusted by ab + BC + Ca = 1, then it can be
A + B + C ≤ (AB + BC + Ca) ^ 2 / 3ABC, but then, I can't melt it
Brother, don't waste your time,
How did a + B + C = 1 come from?

a+b+c≤(ab+bc+ca)^2/3abc
Expand: A ^ 2BC + AB ^ 2C + ABC ^ 2 less than or equal to a ^ 2B ^ 2 + B ^ 2C ^ 2 + C ^ 2A ^ 2
A ^ 2B (B-C) + C ^ 2A (a-b) greater than or equal to B ^ 2C (C-A)
Because this is a rotational symmetry, let a be greater than or equal to B, C be greater than or equal to 0
Reduce the left side of the original formula by that formula and enlarge the right side
2C ^ 3 (B-C) greater than or equal to C ^ 3 (B-C)
In this case, equality, if not reduced and expanded, then equality is greater than or equal to

How to prove a + B + c-3abc = (a + B + C) (a + B + c-ab-bc-ca)

A ^ 3 + B ^ 3 + C ^ 3-3abc = [(a + b) ^ 3-3a ^ 2b-3ab ^ 2] + C ^ 3-3abc = [(a + b) ^ 3 + C ^ 3] - (3a ^ 2B + 3AB ^ 2 + 3ABC) = (a + B + C) [(a + b) ^ 2 - (a + b) C + C ^ 2] - 3AB (a + B + C) = (a + B + C) (a ^ 2 + B ^ 2 + 2Ab AC BC + C ^ 2) - 3AB (a + B + C) = (a + B + C) (a ^ 2 + B ^ 2 + C ^ 2-AB AC BC)

In △ ABC, the lengths of the opposite sides of angles a, B and C are a = 3, B = 5 and C = 6 respectively, then the value of vector ab · vector BC + vector BC · vector Ca + vector Ca · vector AB is

Given the line segments a, B, C, find the angle ABC so that BC = a, CA = B, ab = C
Method:
(1) Make () = ()
(2) On the same side of BC, take () as (), take () as a half stroke (), and then take (), with (), two arcs intersecting at point ();
(3) Connecting (), (), angle ABC is the triangle

(1) (BC) = (a)
(2) On the same side of BC, draw (ARC) with (b) as the center and (c) as the semi meridian, then draw (ARC) with (c) as the center and (b) as the semi meridian, and the two arcs intersect at point (a);
(3) Connecting (BA), (CA), the triangle ABC is the triangle

In the RT triangle ABC, the angle c = 90 °, ab = C, BC = a, CA = B, if a: B = 3:4, C = 10, find the length of a and B

A:
a:b=3:4
b=4a/3
According to Pythagorean theorem:
a^2+b^2=c^2
So:
a^2+(4a/3)^2=10^2
a^2+16a^2/9=100
25(a^2)=900
a^2=36
A = 6 (a = - 6 does not conform to rounding)
So: B = 4A / 3 = 8
To sum up, a = 6, B = 8

In △ ABC, ab = C, CA = B, BC = a, when (c * b): (b * a): (a * c) = 1:2:3, find the ratio of three sides of △ ABC
All letters represent vectors

bccosA:abcosC :accosB=1:2:3
From the cosine theorem bccosa = (b ^ 2 + C ^ 2-A ^ 2) / 2
In the same way, replace the rest
In this case, we only need to solve the equation
We get a ^ 2: B ^ 2: C ^ 2 = 5:3:4

Known: a = x / 5 + 20, B = x / 5 + 19. C = x / 5 + 21, find the value of a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ca

It is known that A-B = 1, B-C = - 2, C-A = 1
a^2+b^2+c^2-ab-bc-ca=a(a-b)+b(b-c)+c(c-a)
=a-2b+c
=3

As shown in the figure, it is known that △ ABC (1) draws △ a'b'c ', so that AB / a'B' = BC / b'c '= Ca / c'a' = 2 (2) compares the size of ∠ A and ∠ a '
1. As shown in the figure, △ ABC is known
(1) Draw △ a'b'c 'so that AB / a'B' = BC / b'c '= Ca / c'a' = 2
(2) By comparing the size of ∠ A and ∠ a ', can we judge whether △ ABC is similar to △ a'b'c' and why?
(3) Let AB / a'B '= BC / b'c' = Ca / c'a '= k, change the value of K and try again
2. Judge whether △ ABC is similar to △ a'b'c 'according to the following conditions
(1)∠A=100°,AB=5cm,AC=7.5cm,∠A'=100°,A'B'=8cm,A'C'=12cm(

1. Draw your own picture
∠A=∠A'
The reason why △ ABC is similar to △ a'b'c 'is that if the corresponding sides of two triangles are proportional, then the two triangles are similar
AB / a'B '= BC / b'c' = Ca / c'a '= k, the same conclusion can be obtained
2、∵AB/A'B'=5/8
CA/C'A'=7.5/12=5/8
∴ AB/A'B'=AC/A'C'
∠A=∠A'
∴△ABC∽△A'B'C

The point of - 3 on the number axis is marked as a, the point of 2 is marked as B, and the point of a is marked as a______ Edge movement______ The distance between a and B is______ ．