Is the predicate verb in the subject clause singular or plural

Is the predicate verb in the subject clause singular or plural

As long as the early sentence is like this, such as: less they are win.we won't be happy.
unless we be happy,.
Is the predicate verb after none singular or plural?
None of the predicate verbs after none of are singular
If it is followed by the plural form of countable nouns, if it means everyone's state, the predicate verb is plural, if it means everyone's state, the predicate verb is singular. ---------Bo Bing grammar book
According to the singular and plural of the names after them
The singular none of is determined by the following noun
When none is used as a pronoun, it can be followed by the singular and plural
None has arrived, none can refer to person or thing, and of can be used together. When it refers to uncountable noun, it can be followed by singular form; when it refers to countable noun, it can be followed by singular or plural form
All right. It depends on personal habits.
It's usually singular.
If none, it depends on how you use it.
When none (of) is the subject
For uncountable nouns, the predicate is singular; for plural nouns, the predicate is singular (formal style) or plural (informal style)
Is the predicate after many kinds of clothes plural or singular?
complex
Is the predicate verb after many a boy singular or plural
Many a: one by one, many by many
Many a boy learns to swim before he can read
Many children can swim before they can read
complex
How can there be many a boy?
singular
Many boys, many a time, many a days
many a
The singular of countable nouns and the singular of predicate
Many a countable noun singular predicate singular
Find the domain of definition, first derivative and second derivative of function y = 2x & # 179; - 3x & # 178; and discuss them in table,
Y = 2x ^ 3-3x ^ 2 the domain is all real numbers
y'=6x^2-6x;
y''=12x-6.
For positive integer n, it is proved that f (n) = 32n + 2-8n-9 is a multiple of 64
Find the definition field F (x) = √ (2x + 1) / (x-1) f (x) = √ 2x & # 178; + 3x of the following functions
The first domain is (- ∞, - 1 / 2] ∪ (1, + ∞)
The second domain is (- ∞, - 3 / 2] ∪ [0, + ∞)
Prove that (a + b) ^ n is greater than or equal to a ^ n + B ^ n, where n is greater than 1, but may not be an integer, so binomial theorem cannot be used
There should be conditions for a ≥ 0 and B ≥ 0
Because n > 1, let n = m + 1;
(a+b)^n
=(a+b)^(1+m)
=(a+b)*(a+b)^m
=a*(a+b)^m+b*(a+b)^m (a+b>a,a+b>b)
≥a*a^m+b*b^m
=a^n+b^n
If the function f (x) is a decreasing function on [- 2,2], and f (X & # 178; - 3x) < f (4), the value range of X is obtained
The math teacher gave us a question in class and asked us to do our homework. This kind of question hasn't been done. I'll have two math classes tomorrow morning. Help~
There is a problem with this topic
Because you said that f (x) is a decreasing function on [- 2,2], you didn't explain the domain of the function or whether it is a periodic function
The question arises: is f (X & # 178; - 3x) < f (4) in the domain of definition?
So there is something wrong with the title
It can not be solved simply by - 2 ≤ X & # 178; - 3x ≤ 2 and X & # 178; - 3x > 4
-Is 1 multiplied by - n to the nth power of (- 1) multiplied by the factorial of N
That's right
(-1)(-2)(-3)(-4).(-n)
=[-1*1][-1*2][-1*3][-1*4].[-1*n]
=(-1)^n ×(1*2*3*4*.*n)
=(-1)^n ×n!