If the hundreds and tens of a three digit number are ABC respectively, and (a + B + C) can be divided by 9 positively, this number must be divisible by 9. Why

If the hundreds and tens of a three digit number are ABC respectively, and (a + B + C) can be divided by 9 positively, this number must be divisible by 9. Why

If these three numbers are a, B, C, then the value of the three digits is 100A + 10B + C = 99A + 9b + (a + B + C), where 99A, 9b and (a + B + C) can be divisible by 9, so the three digits must be divisible by 9

A three digit, a digit in a digit, B in a 10 digit digit, and C in a 100 digit digit, are the three digit ABC, right? Why don't you make it clear

This number should be 100C + 10B + a

If the side length of the equilateral triangle ABC is 1, vector AB = a, vector BC = B, vector CA = C, then a * B + b * C + C * a is equal to

-3 / 2 because the three inner angles of the equal proportion triangle are all 60 vectors, the anticlockwise rotation of ab 180-60 = 120 degrees can be transformed into vector Ca (pay attention to the consistency of vector direction) clockwise rotation of 120 degrees can be converted into vector BC. Let a = (cosx, SiNx), then B = (COS (x-120), sin (x-120)) C = (COS (x + 120), sin (x + 120)) a ·

In the equilateral triangle ABC with side length 1, let BC vector be a vector, CA vector be B vector, and ab vector be C vector, then A.B + B.C + C.A =? Why is the included angle 120

a·b+b·c+c·a=BC·CA+CA·AB+AC·BC
=|BC|*|CA|*cos(π-C)+|CA|*|AB|*cos(π-A)+|AC|*|BC|*cos(π-B)
=cos(2π/3)+cos(2π/3)+cos(2π/3)=-1/2-1/2-1/2=-3/2
Why the included angle is 2 π / 3 depends mainly on the position of the starting point of the vector

Let AB = C, BC = a, CA = B, then AB + BC + Ca is equal to? A.0 B.-3/2 C.3 D.-3

The angle between them is 120 ° and cos120 ° is - 1 / 2, and the side length is √ 2
√ 2 × √ 2 × (- 1 / 2) × 3 = - 3, select D

In the equilateral triangle ABC with side length 1, let vector AB = vector C, vector BC = vector a, vector CA = vector B, then vector a * vector B + vector b * vector C+ Vector c * vector a =?

a.b+b.c + c.a
= BC.CA + CA.AB + AB.BC
=|BC||CA|cos120° + |CA||AB|cos120° + |AB||BC|cos120°
= -3/2

It is known that a, B and C are the three sides of the triangle ABC, and a ^ 2 + B ^ 2 + C ^ 2 is equal to ab + BC + ca. try to judge the shape of the triangle

A^2+B^2+C^2=AB+BC+CA
2A^2+2B^2+C^2=2AB+2BC+2CA
2A^2+2B^2+C^2-2AB-2BC-2CA=0
(A-B)^2+(B-C)^2+(C-A)^2=0
(A-B)^2>=0,(B-C)^2>=0,(C-A)^2>=0
So, A-B = B-C = C-A = 0
A=B=C
The triangle is equilateral

Given that ab of a + B = 1 / 3, B + C of BC = 1 / 4, C + a of Ca = 1 / 5, find AB + BC + AC of ABC

ab/(a+b)=1/3
Take the reciprocal
(a+b)/ab=3
a/ab+b/ab=3
1/b+1/a=3
In the same way
1/b+1/b=4
1/a+1/c=5
Add up
2(1/a+1/b+1/c)=12
1/a+1/b+1/c=6
All points
(ab+bc+ca)/abc=6
Take the reciprocal
abc/(ab+bc+ca)=1/6

It is known that a, B and C are real numbers, and ab a+b＝1 3，bc b+c＝1 4，ca c+a＝1 5. Find ABC The value of AB + BC + Ca

Take the reciprocal of the known three fractions to get a + B
ab＝3，b+c
bc＝4，c+a
ca＝5，
That is 1
A+1
b＝3，1
B+1
c＝4，1
C+1
a＝5，
Add the three formulas to get 1
A+1
B+1
c＝6，
Results: ab + BC + ca
abc＝6，
That is, ABC
ab+bc+ca=1
6．

1. A / | a | + B / | B | + C / | C | = 1, find the value of | ABC | (BC / | ab | AC / | BC | × AB / | Ca | 2. Simple calculation: 1 / 36 ^ (1 / 4 + 1 / 12-7 / 18-1 / 36) + (1 / 2 + 1 / 12-7 / 18-1 / 36) △ 1 / 36

1. From the conditional formula, we can know that a, B and C have one negative number and two positive numbers,
|abc|/abc÷(bc/|ab|×ac/|bc|×ab/|ca|)=|abc|/abc=-1
2、1/36÷（1/4+1/12-7/18-1/36）+(1/2+1/12-7/18-1/36)÷1/36
=1/36÷（9/36+3/36-14/36-1/36）+（18/36+3/36-14/36-1/36)÷1/36
=1/36÷（-3/36）+（6/36)÷1/36
=3