In △ ABC, ∠ a = 2 ∠ B, the residual angle of ∠ B is 5 times of that of ∠ a, and the relationship between ∠ A and ∠ B is______ .

In △ ABC, ∠ a = 2 ∠ B, the residual angle of ∠ B is 5 times of that of ∠ a, and the relationship between ∠ A and ∠ B is______ .

The residual angle of ∵ B is 5 times that of ∵ a,
∴90°-∠B=5（90°-∠A），
5 ∠ a - ∠ B = 360 ° is obtained,
∵∠A=2∠B，
∴10∠B-∠B=360°，
The solution is ∠ B = 40 °,
Then ∠ a = 80 °,
∴∠A+∠B=120°．
So the answer is: ∠ a + ∠ B = 120 °

In the triangle ABC, c * CoAA = B, find the angle C

From cosine theorem and known C (B 2 + C 2 - a 2) / (2BC) = B. by simplifying, we get a + B 2 = C 2, and from Pythagorean theorem we get that the angle c is a right angle

1. Find such a three digit number, which is equal to the sum of the factorials of each digit. That is, ABC = a! + B! + C!

#include #include int fun(int n){\x05int num = 1;\x05while(n > 0)\x05{\x05\x05num *= n;\x05\x05n--;\x05}\x05return num;}int main(int argc,char* argv[]){\x05int i,j,k;\x05int n;\x05for(n=100;n

If the side length of equilateral triangle ABC is 1, vector AB = a, vector BC = B, vector CA = C, then what is a * B + b * C + C * a equal to? The answer is - 3 / 2

a.b=|a|.|b|.cos120°=1*1*（-0.5）=-0.5
b.c=|b|.|c|.cos120°=1*1*（-0.5）=-0.5
c.a=|c|.|a|.cos120°=1*1*（-0.5）=-0.
So A.B + B.C + C.A = - 0.5-0.5-0.5 = - 1.5
Notice that the angle between the two vectors is 120 degrees

If 1 / A + 1 / b = 1 / 6, 1 / B + 1 / C = 1 / 9, and 1 / A and 1 / C = 15, then what is the ABC of AB + BC + Ca

1/a+1/b=1/6
1/b+1/c=1/9
1/a+1/c=1/15
Sum of three forms
2（1/a+1/b+1/c）=31/90
1/a+1/b+1/c=31/180
(bc+ac+ab)/abc=31/180
abc/(ab+bc+ca)=180/31

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，
BC + Ca: BC + ca
abc＝6，
That is, 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，
BC + Ca: BC + ca
abc＝6，
That is, ABC
ab+bc+ca=1
6．

Difference between derivative rule and differential rule of compound function

1. The derivation method of compound function and implicit function are the same. They are all chain derivation methods. We distinguish them carefully in Chinese and English

The approximate formula (1 + x) ^ a ≈ 1 + ax of differential is known. When the absolute value of X is small enough, the approximate value of (8012) ^ (1 / 3) is obtained

Original formula = (8000 + 12) ^ 1 / 3
=8000^1/3+1/3*8000^(-2/3)*12
=20+12/(3*400)
=20.01

If the function y = f (x) is the inverse of the function y = 3 ^ x, then the value of F (1 / 2) is

Log logarithm of 1 / 2 base 3