Statement x = (a = 5, B = + + a); after running, the values of X, a and B are? I want to know whether the value of a is 5 or 6?

Statement x = (a = 5, B = + + a); after running, the values of X, a and B are? I want to know whether the value of a is 5 or 6?

The results are all 6, and the operation results of the whole expression are as follows
0. First of all, remember that when a variable is given a new value, the old value is gone
1. A is first given the value of 5
2. + + is added first and then calculated, so a increases to 6 first, then 5 is gone
3. After a increases, B is given a after operation, which is 6
4. Because the value of a comma expression (a comma expression with multiple commas in parentheses is a comma expression, which is generally calculated last) always takes the last one in parentheses, so x takes the value of B, or 6
I don't believe you can write your own program
To expand, consider the following expression:
int x,a,b,c;
x = (a = 5, b=a++,c=++a,456,789);
//It must start with the brackets
//A equals five at first
//B = a + +, + + operation after, first calculate and then add, first give 5 of a to B, then add 1 to a, then B is 5 and a is 6
//C = + + A, + + operation first, add first and then calculate, because the last formula A is already 6, so add 1 to a to become 7, then give C, at this time a is 7, C is also 7
//Finally, we calculate the comma. X must take the next value, that is 789
//So finally, x = 789, a = 7, B = 5, C = 7
Because I don't know whether you are learning C or C + +, I won't write these two formulas and specific programming output methods

After executing the following statements, the value of a is (), and the value of B is () int, a = 5, B = 6, w = 1, x = 2, y = 3, z = 4; (a = w > x) & (b = Y > z);
The value of a is 0. Why is the value of B 6?

If a = w > x is executed first, WZ is not executed, B is still the initial value of 6

Let x, y and t be int type variables, then the statement is executed: x = y = 3; t = + + X | + + y; after that, the value of Y is_________ A) Indefinite value b) 4 C) 3 D) 1
Why?

X = y = 3; t = + + X | + + y; c) 3 because the first sentence y = 3, the second sentence t = + + X, then t = 4 is true, because the "| - or" operator as long as the front is true, the whole is true, so it will not execute the + + y statement, so y is still equal to 3. You can try the following code yourself: # - include void main() {int x, y; X = y = 3

If the statement "int x;" is set, the value of X after the statement "x = 7; X + = x - = x + X;" is_____ Thank you

Let's start with X - = x + X; at this time, the left x is reassigned
That is, x = 7 - (7 + 7) = - 7;
Then execute the expression after x + = that is, x = x + x = - 7 + (- 7) = - 4;

An acute angle trigonometric function multiple choice question in the third grade of junior high school
If a and B are acute angles, Sina = 0.725, CoSb = 0.697, then a + B = ()
A. Greater than 90 ° B. equal to 90 ° C. less than 90 ° D. may be greater than 90 ° and may be less than 90 °

∵sinA=0.725 cosB=0.697
sinA=cos（90°-A）=0.725
When B = 90 ° - A, CoSb = 0.725 ＞ 0.697
According to the cos α image, cos decreases monotonically on [0,90 °]
So b > 90 ° - A
So a + b > 90 degree

Help to solve the acute angle trigonometric function judgment problem
Judgment questions:
① sin（α + β）= sin α + sin β
②tan 2 α = 2 tan α
③cos A + cos B = cos （A + B）
I don't know what it means at all. Take a look. Thank you!

1.sin（α + β）= sinαcosβ+cosαsinβ
1 wrong
2.　tan2α=2tanα/[1-(tanα)^2]
2 wrong
3.cos（A+B）=cosAcosB-sinAsinB
3 mistakes

Solution of trigonometric function with acute angle
tan10°×tan20°×tan30°×tan40°×tan50°×tan60°×tan70°×tan80°=?
Say why

The original formula = tan10 & # 730; * tan20 & # 730; * tan30 & # 730; * tan40 & # 730; * cot40 & # 730; * cot30 & # 730; * cot10 & # 730; * cot20 & # 730; * cot10 & # 730;
=1

Knowledge points of the relationship between the edges and corners of right triangle

Trigonometric function formula of acute angle
Sine: sin α = - the opposite side of α / - the hypotenuse of α
Cosine: cos α = - the adjacent side of α / - the hypotenuse of α
Tangent: Tan α = - the opposite side of α / - the adjacent side of α
Cotangent: cot α = - the adjacent side of α / - the opposite side of α

How to generalize knowledge

1. Use positive and negative examples together;
2. Correct use of variants;
3. Scientific comparison;
4. Inspire students to summarize consciously

Please help to sum up some of the third grade chemical acid-base salt knowledge points~~

This is exactly what I'm going to teach you tomorrow. I've just sorted out the explanation of the general properties of acid, alkali and salt. The chapter "acid, alkali and salt" can be said to be the application and synthesis of the whole junior high school chemistry knowledge. On the basis of the basic concepts and theories of chemistry, through the explanation of the basic properties of acid, alkali and salt, it sums up the rules of learning inorganic chemistry