When an indefinite pronoun is the subject, should the predicate be in the singular or plural form?

When an indefinite pronoun is the subject, should the predicate be in the singular or plural form?

Explanation: 1. The common indefinite pronouns are: some (something, somebody, someone, somewhere), any (anything, anybody, anyone, anywhere), no (nothing, nobody, no one), every (everything, everybody, every
singular.
It depends on what the indefinite pronoun stands for. For example, some stands for countable nouns, which are plural, while uncountable nouns are singular
But every each is singular
The singular number is not the plural number
People as the subject, the predicate verb with the singular or plural?
People is a collective noun, and the predicate of collective noun is plural
English: () or () as the subject, the predicate verb in the singular form, () or () as the subject, the predicate verb in the plural form
When (infinitive) or (gerund) is the subject, the predicate verb is singular;
For example:
1.Buying clothes is often a time-consuming job because those clothes that a person likes are rarely the cones that fit him or her.
2.To understand the situation completely requires more thought than has been given thus far.
When (collective noun) or (National word and article) is the subject, the predicate verb is in the plural
1.The Chinese people are brave and hardworking The cattle are grazing in the sunshine.
2.The Japanese were once very aggressive.
Please ask in time,
Ten years and ten meters are singular or plural
If compound nouns of money, time, price or measure are used as subjects, these nouns are usually regarded as a whole, and the predicate is usually singular
The maximum value of the function y = 6x / 1 + X & # 178; is
Y = 6x / (1 + X & # 178;) function definition field is r, y '= [6 (1 + X & # 178;) - 12x & # 178;] / (1 + X & # 178;) = - 6 (x + 1) (x-1) / (1 + X & # 178;) &# 178; y' = 0, then X1 = - 1, X2 = 1 changes with X, y ', y changes as follows: X (- ∞, - 1) - 1 (- 1,1) 1 (1, + ∞) y' - 0 + 0 - y minus minimum increase
The function f (x) = 6x / (1 + X & # 178;), F & # 180; (x) = [6 (1 + X & # 178;) - 6x * 2x] / (1 + X & # 178;) &# 178; = 6 (1-x & # 178;) / (1 + X & # 178;) &# 178;, f (x) = [6 (1 + X & # 178;) / (1 + X & # 178;);,
Let F & # 180; (x) = 0, then the stationary point x = ± 1,
When x ∈ (- ∞, - 1) ∪ (1, + ∞), F & # 180; (x) < 0, f (x) decreases monotonically;
When x ∈ (- 1,1), F & # 180;... Is expanded
The function f (x) = 6x / (1 + X & # 178;), F & # 180; (x) = [6 (1 + X & # 178;) - 6x * 2x] / (1 + X & # 178;) &# 178; = 6 (1-x & # 178;) / (1 + X & # 178;) &# 178;, f (x) = [6 (1 + X & # 178;) / (1 + X & # 178;);,
Let F & # 180; (x) = 0, then the stationary point x = ± 1,
When x ∈ (- ∞, - 1) ∪ (1, + ∞), F & # 180; (x) < 0, f (x) decreases monotonically;
When x ∈ (- 1,1), F & # 180; (x) > 0, f (x) increases monotonically;
So x = 1 is the maximum point, and the maximum is f (1) = 3
I wish you progress in your study and hope you will adopt it.
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Matlab factorial sum how to write? Novice ask
Factorial is a factorial function
sum=0;
for i=1:n
sum=sum+factorial(n);
End
Sum
Given that the function y is equal to the third power of negative x plus the square of six x plus the maximum value of M 13, what is the value of M?
y=-x^3 6x^2 m
＝＞y'=-3x^2 12x
There is a maximum at x = 4
＝＞f(4)=13
＝＞-19
M=-19
-19
What is the sum of 1 to 20 factorials
#include "stdio.h"
void main()
{
int i;
double s=0,x=1;
for(i=1;i
Judge the parity of the following functions: (1) f (x) = 2x ^ 4 + 3x ^ 2 (2) f (x) = x ^ 2-2x (3) f (x) = (x ^ 2 + 1) / X (4) f (x) = x ^ 2 + 1
To detailed process! Thank you! Urgent!
Let x = - X
First, f (- x) = f (x) = 2x ^ 4 3x ^ 2, even function
Similarly, f (- x) = x ∧ 2 2x and - f (x) do not add up to 0 or F (- x), which is not odd or even
Odd function
Even function
Calculate and output the factorial of 10. The factorial of 10 = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 (using Java)
public class Test{   public static void main(String[] args) {   System.out.println (add(10)); } public static int &nb...