Unit 1 English composition, PEP edition, Volume 2, grade 6

Unit 1 English composition, PEP edition, Volume 2, grade 6

Hello,I am XXX.I have a friend.(He/She) is shorter than me.I am 2cm taller than (him/her).I am 1 year older than (him/her).(He/She) is younger than me.My hair is longer than(his/hers).But (he/she) is ...

Unit one English test paper for grade six

1、 Write the following words (24 points) waiting for the police to stop where work hobbies

Unit 1 Composition of PEP
On the basis of oral communication, to write an exercise with the title of unforgettable "the first time", we should write clearly the experience of "the first time", and also write the enlightenment after the experience of "the first time"

2-3 of 2X-4 = 12 of X-7 to solve the equation

2X-4 of 2-3 = X-7 of 12 times 12
24-4（2x-4）=x-7
24-8x+16=2x-7
40-8x=2x-7
8x+2x=40+7
10x=47
x=4.7

How to know which row and which column to eliminate?
How to judge?

In theory, it is arbitrary. In fact, the simplest one is generally chosen
For example, a non-zero column with the most rows of 0 and 1 can be used to calculate less subdeterminants that multiply 0 or 1

Know x2 = Y3, find the value of 7X2 − y2x2 − 2XY + 3y2?

Let x = 2K, y = 3k, the original formula = 7 · (2k) 2 − (3K) 2 (2k) 2 − 2 · 2K · 3K + 3 · (3K) 2 = 19k219k2 = 1

The absolute value of X-5 and the absolute value of 6-y are opposite to each other. Find the value of X + Y / X

The absolute values of X-5 and 6-y are opposite to each other
Namely
|x-5|+|6-y|=0
x-5=0,6-y=0
x=5,y=6
therefore
X + X of Y
=5 + 5 / 6
=5 out of 11

9y ^ 2-3y + 1 / 4 = 0 factorization!

9y^2-3y+1/4
=（3y-1/2）^2
=0
So 3Y = 1 / 2, y = 1 / 6

What is the 2014 power of (1 – √ 4)

1

Let the column vectors of matrix B be linearly independent, Ba = C. It is proved that the column vectors of matrix C are linearly independent if and only if the column vectors of matrix A are linearly independent

In this paper, we prove that CX = 0 and Ax = 0 have the same solution. On the one hand, it is obvious that the solution of AX = 0 is the solution of Cx = Bax = 0. On the other hand, let X1 be the solution of Cx = 0, then CX1 = 0. So (BA) X1 = 0, so B (ax1) = 0. Because column B is full rank, so there is ax1 = 0. That is, X1 is the solution of AX = 0. So there is R (c) = R (a). Because column A and column C have the same number, so the