The following statement is correct: a point on the number axis a can only be a rational number B. a number can only be represented by a point on the number axis C. the number of two units from the origin on the number axis is a rational number

The following statement is correct: a point on the number axis a can only be a rational number B. a number can only be represented by a point on the number axis C. the number of two units from the origin on the number axis is a rational number

A wrong, it can also be an irrational number. Make a square, use a compass to take the length of the diagonal, and put it on the number axis to be √ 2, which is an irrational number. There are many other examples
A number corresponds to a number axis and can only be represented by one point
The number of two units of C from the origin on the number axis is positive 2 and negative 2
A. there can be two points on the number axis to represent rational numbers. B. all rational numbers can be represented by points on the number axis
In the following statements, the correct one is:
A. There can be two points on the number axis to represent rational number 2 b. all rational numbers can be represented by points on the number axis
There is no point representing 0 on the c axis, and no point representing irrational number on the d axis
B
Each point on the axis ()
A. Irrational B. rational C. real D. integer
According to the fact that the real number corresponds to the point on the number axis one by one, C
Factorial calculation
Write a program, for a given n (n ≤ 100), calculate and output the factorial K! (k = 1,2 Because the required integer may greatly exceed the number of general integers, the program uses one-dimensional array to store long integers, and each element of the long integer array only stores one digit of the long integer
N=a[m]×10m-1+a[m-1]×10m-2+ … +a[2]×101+a[1]×100
And use a [0] to store the number of digits m of the long integer n, that is, a [0] = M. according to the above convention, each element of the array stores one digit of the factorial K! Of K, and stores it in the second element and the third element of the array from the low order to the high order For example, = 120, the storage form in the array is:
3 0 2 1 ……
The first element 3 indicates that the long integer is a 3-digit number, followed by 0, 2 and 1 from low to high, which is expressed as integer 120
3,
void fact(int n,int jc[])
{ int i,j,s;
Int sum [300] = {0}; / / temporarily stores the calculation results
sum[0]=1;
for(i=1;i
The maximum value of function f (x) = 2x cubic - 3x quadratic + A is 6. What is the value of a?
f'(x)=6x^2-6x=0
6x(x-1)=0
x=0,1
f''(x)=12x-6
f''(0)=-60
When f '' (x)
On the calculation of factorial
Find 1 / 2! + 2 / 3! + 3 / 4! + +n/(n+1)!
Which great Xia can tell me how to do it? Thank you here
Turn n / (n + 1)! Into 1 / N! - 1 / (n + 1)!; the whole formula can be changed into 1 / 1! - 1 / 2! + 1 / 2! - 1 / 3! 1 / N! -- 1 / (n + 1)! = 1-1 / (n + 1)!; use limit to get it
When x=______ When y = 2X-4 and y = 3x-3 have the same function value? The value of this function is______ .
∵ function y = 2X-4 and y = 3x-3 have the same function value, ∵ y = 2x − 4Y = 3x − 3, the solution is: x = − 1y = − 6, so when x = - 1, function y = 2X-4 and y = 3x-3 have the same function value, this function value is - 6. So the answer is: - 1, - 6
Factorial estimation
How to estimate factorial by inequality
Lnn! / lnn is close to some elementary function
Using stirling formula
When n is large,
Ln n! Is approximately equal to n ln n - N + (1 / 2) ln (2 pi n)
therefore
Ln n! / ln n is approximately n + (1 / 2) - N / ln n + (1 / 2) ln (2pi) / ln n
For example, when n = 10, the difference between the two methods is no more than 10 ^ (- 2)
Find the maximum and minimum of function y = 2x3-3x2-12x + 5 on [0, 3]
Let ∵ f ′ (x) = 6x2-6x-12, let ∵ f ′ (x) = 6x2-6x-12 = 0, obtain x = - 1 or x = 2, and the list is as follows: x0 (0,2) 2 (2,3) 3F ′ (x) - 0 + F (x) 5 decreasing minimum - 15 increasing - 4, so the decreasing interval of function y on [0,3] is [0,2], increasing interval is [2,3], so the
C language design: define a factorial function, and then calculate 1! - 2! + 3! - 4! + 5! - 6! + 7! - 8! + 9
#include "stdio.h"
Double & nbsp; fun (int & nbsp; n) / / use double to improve precision and count digits
{
    double sum=1.0;
    int i;
    for(i=1;i<=n;i++)
           sum*=i;
     return sum;
}
 
int main()
{
       int i,mark=1;
       double sum=0,item=0;
       for(i=1;i<=9;i++)
       {
            item=mark*fun(i);
            sum+=item;
            mark=-mark;
       }
       printf("1!-2!+3!-4!+5!-6!+7!-8!+9!=%.0lf\n",sum);
 
}