How to add ing to the grammar change rules of English present progressive tense

How to add ing to the grammar change rules of English present progressive tense

It's simple, remember:
1. Generally, it is added directly, not more
2. At the end of E, remove E and add ing:have-- having; leave -- leaving
3. For the stressed closed syllable ending with a consonant, double write the last consonant and add ing
Stop (there is only one syllable, usually regarded as the stressed syllable)
"Beginning" -- beginning
There are 10 declarative sentences, 10 general questions and 10 special questions in the present continuous tense
Give 100 in 10 minutes, 50 in half an hour, 25 in 45 minutes, and 10 in excess of the above time
All of the above are additional points
You are doing homework.You are washing hand.He Is studying.She is watching TV.She is playing computer games.He is drinking water.The boy is playing basketball.The girl is dancing.The singer ...
Special questions in the present continuous tense
Special question + be verb + subject + present participle form of verb + others?
Example:
What are you doing
When is he going to school
The answer to the special question in the present continuous tense should be based on the actual situation
The reform of special questions in the present continuous tense
Their structure is as follows:
Special question + bodyguard be [am, is, are] + host + doing +?
Secret: one borrow one exchange two change.
Borrow: put the borrowed special questions at the beginning of the sentence.
One change: the host and the bodyguard change positions.
Second change:
If the subject is in the first person, it should be changed into the second person. [of course, the host's bodyguards also changed accordingly]
2. A full stop at the end of a sentence becomes a question mark.
For example: expand
The reform of special questions in the present continuous tense
Their structure is as follows:
Special question + bodyguard be [am, is, are] + host + doing +?
Secret: one borrow one exchange two change.
Borrow: put the borrowed special questions at the beginning of the sentence.
One change: the host and the bodyguard change positions.
Second change:
If the subject is in the first person, it should be changed into the second person. [of course, the host's bodyguards also changed accordingly]
2. A full stop at the end of a sentence becomes a question mark.
for instance:
She is cleaning the room.→What is she doing?
I’m drinking some water.→What are you doing?
My mother is runing. →What is your mother doing?
The duck is swimming. → what is the duck doing
Given the function f (x) = x2 + ax + B, a = {x | f (x) = 2x} = {2}, try to find the value of a, B and f (x)
F (x) = 2x, that is, the equation x & sup2; + ax + B = 2XX & sup2; + (A-2) x + B = 0A is the set of solutions of the equation. Now a = {2} so the quadratic equation with one variable has two identical solutions x = 2, so the equation can be written as (X-2) (X-2) = 0x & sup2; - 4x + 4 = 0, so A-2 = - 4, B = 4A = - 2, B = 4f (x) = x & sup2; - 2x + 4
A={x|f(x)=2x}={2}
That is to say, the equation f (x) = 2x has only unique solution x = 2
x²+ax+b=2x
x²+(a-2)x+b=0
This equation has only one solution x = 2, which is obtained from Veda's theorem
2+2=-(a-2)
2×2=b
A = - 2, B = 4
f(x)=x²-2x+4
Given the set a = {x | X & # 178; + X + P + 3 = 0, X ∈ r}, if a is contained in negative real number, find the value range of real number P
A is contained in a negative real number, which means that the elements x in the set a are all negative. That is to say, for the univariate quadratic equation x & # 178; + X + P + 3 = 0, there are two negative solutions
We know that f (x) = LG (√ (x + 1) - x); (1) find its domain of definition; (2) prove that f (x) is an odd function; (3) prove that f (x) is a decreasing function
(√ (x + 1) - x > 0 holds for any x, so it is a real number field. F (- x) = LG (√ (x + 1) + x) f (x) = LG (√ (x + 1) - x) = LG [1 / (√ (x + 1) + x] = - LG (√ (x + 1) + x) f (- x) = - f (x) f (x) = LG [1 / (√ (x + 1) + x] = - LG (√ (x + 1) + x) lgx is an increasing function, and √ (x + 1) + X is an increasing function
Given the function f (x) = x2 + 2x + A, f (BX) = 9x2-6x + 2, where x belongs to R, a and B are constants, then the solution set of the equation f (AX + b) is?
Online, etc
Substitute BX into f (x) to get: F (BX) = (BX) ^ 2 + 2 (BX) + a
The result is: F (BX) = B ^ 2x ^ 2 + 2bx + a
Compared with F (BX), a = 2, B = - 3
Then f (AX + b) = f (2x-3) = 4x ^ 2-8x + 5
Let ^ 2 x = 0-4x-5
Solve the equation
If a set contains three elements 1, x, x2-x, then the value range of real number x is
The function f (x) = LG | x | is ()
A. Odd function, in the interval (0, + ∞) is a decreasing function B. odd function, in the interval (0, + ∞) is an increasing function C. even function, in the interval (- ∞, 0) is an increasing function D. even function, in the interval (- ∞, 0) is a decreasing function
∵ f (x) = LG | x | = lgx, X ＞ 0 − lgx, X ＜ 0, ∵ function is even function, image is symmetric about y axis, and it is decreasing function in interval (- ∞, 0), so choose: D
It is known that the two zeros of the function f (x) = x2 + ax + B are - 2 and 3 (1) to find the value of a + B. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2) to find the solution set of the inequality AF (- 2x) > 0
The two roots of solution (1) are - 2 and 3 (2 points) − 2 + 3 = − a − 2 × 3 = B  a = − 1b = − 6 (5 points) ｜ a + B = - 7 (6 points) (2) ∵ f (x) = x2-x-6 ------ (8 points) ∵ inequality AF (- 2x) > 0, that is - (4x2 + 2x-6) > 0 ⇔ 2x2 + x-3 < 0, -- - (10 points) the solution set is {x | 32 < x < 1}, -- (12 points)