Why 1 to 2 is not right Why 1 is right and 2 is wrong

You're not familiar with the if else statement.

#include

- What about statements that satisfy the truth condition? What you're trying to say here is that if the if() condition is met, don't execute anything, just run the thing in else. And else includes both of them, so the output must be i is both a leap year and not a leap year, so your if() statement is equal to not being written.

else, either use two, or don't use one, and the following two can meet your desired results:

statement_1;

elsestatement_2;

2. if()

statement_1;}

elsestatement_2;}

Also, is there a difference between "a leap year" and "not a normal year"!!

Neither is true.

#include

#include

using namespace std;

int main()

int i;

cout<<"Enter the year:";

cin>>i;

coutreturn 0;

#include

using namespace std;

void main()

int i;

cout<<"Enter the year:";

cin>>i;

Don't want this.

It should be the second else, which has no corresponding if

cin>>"Enter the year:";

What's wrong with this 1?

The first method uses one of the two important limits, the second method you forcibly divide x into two variations, you can't do that, it's easy to think that this is right, pay attention.

The denominator transformation of the second question is wrong.

1+1/x)^(x^x)≠[1+1/x)^x]^x=(x+1/x)^(x^2)

Ask your teacher about this.

Your teacher will help you out.

2 is right, add left and subtract right, the coefficient before x is 1 2, so 2 3 before multiplying a 1 2, moving 2 3 units to the right is minus 1 3.

1 is doing the right thing and the result is wrong.

The relationship between the coordinates of any point (x0, y0) on the hyperbola is: y0 = 2 x0, that is, x0 * y0 = 2, and the area of the rectangle = length * width, and the length is the x coordinate x0, and the width (height) is y0, so the area of the rectangle is fixed, both are 2.

I don't know about the first one, but the second one is because the recessed quadrilateral is different.

The naming rules for alkanes are five words: long, many, near, simple, and small.

Long: Contains the longest carbon chain as the main chain.

Many: When the number of carbons in the main chain is the longest and equal, the main chain with the largest number of substituents should be selected.

Near: When both the first and second strands are satisfied, the substituents closest to the end of the chain should be selected to start numbering.

Jan: When the substituents are all closest and equal to the end of the chain, choose the end of the simple substituents to start numbering.

Small: After the first four are satisfied, the algebra of the substituent group number and the small one should be numbered for the main chain.

Finally, there is naming.

The first one is numbered from the left end, which does not satisfy the principle of "near", so it is wrong, and the second one is correct.

a=1,b=1

a+b=2a, and b are not greater than 1, so they are not true.

k=3 is hyperbola, not elliptic, because at this time a b, if it is an ellipse, it should be a b

k=2 5, isn't that round, so it's not right.