A detailed explanation of English present continuous tense

A detailed explanation of English present continuous tense

The present continuous tense consists of: subject + be + verb ing (present participle) form first person + am + doing + sth second person + are + doing + sth third person + is + doing +
Do you want to restore the verb
Do not restore
Let me give you an example: "he is doing his home work" is a general progressive tense, which is changed into a general question as follows: "is he doing his home work?"
Note: keep the participle form of "doing"
Hope to help you.
Instead of reducing verb (not be verb), put be verb in advance.
Verb present continuous
Thirty examples
1.i am listening to the music2.he is watching TV.3.My mother is cooking dinner.4.My sister is talking on the phone5.John is plaing video games.6.My little brother is sleeping now.7.I am washing my sho...
I'm eating.
I'm running.
I'm flying.
He's jumping
She's laughing
They're swimming
It's singing
I'm talking
The teacher is teaching
That woman is buying in the super
I'm eating.
I'm running.
I'm flying.
He's jumping
She's laughing
They're swimming
It's singing
I'm talking
The teacher is teaching
That woman is buying in the supermaket
Jim is climbing
Tom is going
Lily is amorisming.
Kim is learning
Jeff is listening to a music
Richard is dancing
Bush is crying
present progressive:
Be (AM / is / are) + ing (present participle) forms the predicate. Be is an auxiliary verb, which helps to form tense, voice and mood, negative sentence and interrogative sentence be / have / had
Sentence structure:
1) Affirmative sentence: subject + be + verb ing +
2) Negative sentence: subject + be + not + verb ing +
3) Be + subject + verb ing +?
4) Special questions
present progressive:
Be (AM / is / are) + ing (present participle) forms the predicate. Be is an auxiliary verb, which helps to form tense, voice and mood, negative sentence and interrogative sentence be / have / had
Sentence structure:
1) Affirmative sentence: subject + be + verb ing +
2) Negative sentence: subject + be + not + verb ing +
3) Be + subject + verb ing +?
4) What + be + subject + doing +?
The change rules of the verb ing present participle are as follows
1) Verb + ing: do doing teach teaching;
2) For monosyllabic words ending with a vowel and a consonant, double write the consonant and then + ing: put putting;
1) If it ends with the letter E, remove E and then + ing: making taking taking.
The meaning of the present continuous tense
1) It means the action that is going on at the moment. "Now" "at the / this moment" is an adverbial of time
The students are having an English class.
The teacher is teaching them English.
2) It means the action in progress at the present stage, but the moment of speaking is not necessarily in progress. "Recently" "these days" are adverbials of time
I'm studying English these days.
3) The present continuous tense of some directional verbs can indicate the future :
I'm coming. I'm going / leaving.
He is leaving for Shanghai tomorrow.
4) Now, always, forever, continuously and other adverbs are used together to express the speaker's appreciation or complaint
Be always doing He is always talking at class.
5) Verbs expressing state, feeling, emotion or mental activity can not be used in the progressive tense
Believe, doubt, hear, know, understand, long, think, consider, look, see, show, mind, have, sound, taste, require, posses, care, mind, like, hate, desire Put it away
Let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ a
1) If 2 ∈ a, then there are two other elements in a; (2) set a cannot be a single element set; (3) set a has at least three different elements
1. If a = 1 / 2, then a = 1 / 2, a = 1 / 2, a = 1 / 2, a = 1 / 2, a = 1 / 2
1）∵2∈A ∴1\（1-2）=-1∈A ∴1\（1-（-1））=1\2∈A 1\（1-1\2）=2∈A
The other two elements in a are 1-2, - 1
2) If set a is a single element set, then a = 1 / (1-A) has no real solution, so set a cannot be a single element set
3) A ≠ 1 / (1-A) 1 \ (1 - (1 \ (1-A))) = (A-1) \ \ a tested (A-1) \ \ a ≠ 1 / (1-A...)
1）∵2∈A ∴1\（1-2）=-1∈A ∴1\（1-（-1））=1\2∈A 1\（1-1\2）=2∈A
The other two elements in a are 1-2, - 1
2) If set a is a single element set, then a = 1 / (1-A) has no real solution, so set a cannot be a single element set
3) A ≠ 1 / (1-A) 1 \ (1 - (1 \ (1-A))) = (A-1) \ \ a tested (A-1) \ \ a ≠ 1 / (1-A) ≠ a
There are at least three different elements in set a. Put it away
In high school mathematics, let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ a
1) (2) if there are at least two elements in the set a ∈; (3) if there are at least two other elements in the set a ∈, then there is no possibility that a is more than a.
It is proved that let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ a,
(1) If 2 ∈ a, then 1 / (1-2) = 1 ∈ a,
In the same way
If - 1 ∈ a, then 1 / (1 + 1) = 1 / 2... Expansion
In high school mathematics, let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ a
1) If 2 ∈ a, then there are two other elements in a; (2) set a cannot be a single element set; (3) set a has at least three different elements.
It is proved that let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ a,
(1) If 2 ∈ a, then 1 / (1-2) = 1 ∈ a,
In the same way
If - 1 ∈ a, then 1 / (1 + 1) = 1 / 2 ∈ a,
So there are two other elements in a, - 1,1 / 2;
(2) Set a cannot be a single element set;
It is known from (1) that there are two other elements in a, - 1,1 / 2;
So (2) set a cannot be a single element set;
(3) To the contrary:
Suppose there are only three different elements in set a: 2, 1 / 2, - 1
Let a be a set of real numbers and satisfy the following conditions: if a ∈ a, a ≠ 1, then 1 / 1-A ∈ a,
have to
If 3 ∈ a, then 1 / (1-3) = 1 / 2 ∈ a,
This and
Suppose there are only three different elements in set a: 2, 1 / 2, - 1
so
There are at least three different elements in set a. Put it away
Let f (x) be a function defined on R and satisfy f (0) = 1. For any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1), 1. Find the expression of F (x) 2
Let f (x) be a function defined on R and satisfy f (0) = 1. For any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1), 1. Find the expression of F (x). 2. If G (x) = f (x) - (4a + 3) + A ^ 2, X ∈ [0,1] (a is the letter coefficient), find the range of the minimum value H (a) of G (x)
In this paper, we have the f (y) = y (f (y) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x) max = g (1) = A & # 178; - 4A
Let a > 0, f (x) = ex / A + A / ex be the value of a for even function 1 on R
Because f (x) is an even function on R
So f (- x) = e ^ - X / A + A / e ^ - X
=1/ae^x +ae^x
=f(x)
E ^ X / A + A / e ^ x = 1 / AE ^ x + AE ^ x
Well organized
1/a (e^x+1/e^x)=a(e^x+1/e^x)
1/a=a
a^2=1
A1 = 1, A2 = - 1 (rounding off)
So a = 1
Even function, f (- 1) = f (1), f (- 1) = 1 / AE + AE, f (1) = E / A + A / E
So: 1 / AE + AE = E / A + A / E, that is: ae-a / E = E / A-1 / AE
That is: a (E-1 / E) = (1 / a) (E-1 / E)
(a-1/a)(e-1/e)=0,
So: a = 1 / A, a = ± 1,
Because a > 0, a = 1;
Because f (x) is an even function on R
So f (- x) = e ^ - X / A + A / e ^ - X
=1/ae^x +ae^x
=f(x)
E ^ X / A + A / e ^ x = 1 / AE ^ x + AE ^ x
Well organized
1/a (e^x+1/e^x)=a(e^x+1/e^x)
1/a=a
a^2=1
A1 = 1, A2 = - 1 (rounding off)
So a = 1
Don't copy mine!!
Given the set a = {x | x square - (P + 2) x + 1 = 0}, B = positive real number, and a intersection B = empty set, the value range of real number P is obtained
There is no positive real number root in equation a
Is it possible to have a root discriminant
P
Let f (x) be a function defined on R and satisfy f (0) = 1, and for any real number x, y, f (X-Y) = f (x) - Y (2x-y + 1), find the analytic expression of F (x)
Let y = x be:
f (x -x) = f(x) - x(2x -x +1)
f(0) = f(x) - x² -x
Because f (0) = 1
So 1 = f (x) - X & # 178; - X
So f (x) = x & # 178; + X + 1
Let a > 0, f (x) = ex / A + A / ex be the value of a obtained from function (1) on R. it is proved that f (x) is an increasing function on [0, + ∝]
Let a > 0, f (x) = ex / A + A / ex be a function on R
(1) Finding the value of a
(2) It is proved that f (x) is an increasing function on [0, + ∝]
If f (x) = f (- x) e ^ X / A + A / e ^ x = e ^ (- x) / A + A / e ^ (- x) e ^ X / A + A / e ^ x = 1 / [a * e ^ x] + A * e ^ x E ^ x (1 / a - a) + (a - 1 / a) / e ^ x = 0 (e ^ X - 1 / e ^ x) (1 / a - a) = 0e ^ X - 1 / e ^ x, it is not equal to 0
So f (- x) = e ^ - X / A + A / e ^ - X
=1/ae^x +ae^x
=f(x)
E ^ X / A + A / e ^ x = 1 / AE ^ x + AE ^ x
Well organized
1/a (e^x+1/e^x)=a(e^x+1/e^x)
1/a=a
a^2=1
A1 = 1, A2 = - 1 (rounding off)
So a = 1
Given the set a = = {X / x square + (P + 2) x + 1 = 0}, if a ∩ positive real number = ф, find the value range of real number P. note that ф is an empty set
The equation has no positive root
The discriminant is less than 0
p²+4p+4-4