A great + many + noun singular or plural + verb singular or plural?

A great + many + noun singular or plural + verb singular or plural?

When many a + singular noun is used as subject, the predicate verb is singular; when a great / good many + plural noun is used as subject, the verb is plural
I haven't seen great many
Here is a difference
A good / great many of + countable plural
A great deal of + uncountable nouns
A great many of and a great many of
A great many is followed by a modifier
And a great many of No
Is that right
A great number / deal of and a great number / deal
Is it the same
1、 After a great many, the noun can have modifiers, such as: a great many students a great many young students 2. A great many of is followed by: 1. Pronoun: a great many of them2
There are only a great number of and a great deal of, but no a great number / deal
Many a, amount of and so on words singular and plural situation!
The amount of
A large amount of
A small amount of
The number of
A large number of
A small number DF
Many is used in the countable noun eg: many people many sheep many students
A (large / big) amount of is used for uncountable nouns, and its countable noun form is a (large / big) number of
a large amount of eg:There is a large amount of milk in the bottle
A lot of = lots of
Mang (countable)
Many (uncountable)
Mang more (more)
Given that the function f (x) = LG (2-ax) is a decreasing function on [0,1], then the value range of a is
y=lg(2-ax)
Let u = 2-ax, y = LGU
If a > 0
Then x A kind of U A kind of Y A kind of
If a < 0
Then x A kind of U A kind of Y A kind of
So a > 0 is in line with the meaning of the question
When a > 0
To make y meaningful, then
2-ax＞0
a＜2/x
If x ∈ [0,1], obviously 2 / X decreases, then a < 2
The value range of a is a ∈ (0,2)
How to find the left and right limit of function
X → 0 - means that x tends to 0 from the left side of 0, so x & lt; 0, if x → 0 +, X tends to 0 from the right side of 0, x0. Similarly, X → 1 - means that x tends to 1 from the left side of 1, so x & lt; 1, if x → 1 +, X tends to 1 from the right side of 1, x1. For example: Lim [x → 1 -] f (x) note that X & lt; 1
=lim[x→1-] (x-1)=0
Lim [x → 1 +] f (x) is x1
=lim[x→1+] (2-x)=1
The left and right limits are not equal, so the function is a jumping point at x = 1
Both X-1 and 2-x are elementary functions. When calculating the limit of this elementary function, as long as the function value can be calculated directly, it can be calculated directly instead of the value
It is known that set a consists of three elements 1, X and x ^ 2, and set B consists of three elements 1, 2 and X. if set a is equal to set B, the value of X is obtained
Don't copy from the Internet, please write more standard. I just learned_ ∩)o...
x=√2
If f (x) = LG (1-2 / x + a) is an odd function, then the range of x where f (x) is less than 0 is?
F (x) = LG [2 / (1-x) + a] is an odd function,
∴f(x)+f(-x)=lg{[2/(1-x)+a][2/(1+x)+a]}=0,
∴[2/(1-x)+a][2/(1+x)+a]=1,
∴(4+4a)/（1-x^2)+a^2-1=0,
The definition of [x] / (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (A-1) + (a,
∴a=-1.
The definition field is - 1
Finding the left and right limit of a function
The left and right limits when x tends to 1
When x → 1 +, X / (x-1) → + ∞, the limit of denominator is + ∞, so the right limit is 0
When x → 1 - when x is still greater than 0, X / (x-1) → - ∞, the limit of denominator is 0-1 = - 1, so the left limit is - 1
The left limit is - 1 and the right limit is 0
If x ∈ a, then 6-x ∈ a; 2. If there are three elements in a, use enumeration method to represent set a
Why is the answer in pairs? Why?
One hundred and fifty-three
Two hundred and thirty-four
Two sets
If 1 is in a, then 6 -- 1 = 5 is also in a
The same reason can be proved
The set satisfying the first condition is {1.2.3.4.5}
If there are only three elements, {1.3.5} {2.3.4}
There are three elements:
{1,3,5}
{2,3,4}
In addition:
There is one element: {3}
There are two elements {1,5}, {2,4}
There are four elements {1,2,4,5}
There are five elements {1,2,3,4,5}
Note that the definition field of function f (x) = √ [(x-1) / (x + 1)] is a, and the definition field of function g (x) = LG (x-a-1) (2a-x) is B
If B contains a, find the value range of real number a
F (x) is defined as [- Infinity - 1], [1, + infinity]
When G (x) domain is a > = 1 (a + 1,2a);
If B contains a
2a>=1
a>1/2
And a > = 1
So a > = 1
A
A = (- inf, - 1) and [1, inf]
B = (a-1,2a) or (2a, A-1)
If B contains a? A should have no solution.
The region of a is x > = 1, or x0, that is, a + 1