The advantage of doing housework is urgent! Don't worry about how to do housework! About 100 words is OK

The advantage of doing housework is urgent! Don't worry about how to do housework! About 100 words is OK

I did a survey recently about doing housework among my friends to see people's attitude about it.The result is that most of them are tired of housework .I feel so amazing,because I think doing housework is very interesting,it makes youth more independent , it means that we are not relay on our parents any more.And it also help us to get on well with our parents,we'll have more topics with them.Then we'll find communicate with our parents isn't hard.

How to help parents do housework

Kid, is it for homework?
If you want other people's works to be your role, give up school
If you want to read, look for a dictionary and grammar book. I believe you can finish it

Solving inequality (X & sup2; + 4x-5) (X & sup2; - 2x + 2) > 0

(x^2+4x-5)(x^2-2x+2)＞0
Because: x ^ 2-2x + 2 = (x-1) ^ 2 + 1 > 0
Therefore, the original inequality is equivalent to: x ^ 2 + 4x-5 > 0
That is: (x + 5) (x-1) > 0
x1

How to calculate arctan Pie / 2 = 90 degree with calculator

First, there must be a scientific calculator
It's easy to communicate, but it's also easy to be a star
Then, in scientific calculation, press shift, and the tan key shows tan-1
Then, (PAI / 2)
Finally=

It is known that the root of the equation AX & # 178; + BX + C = 0 (a ≠ 0) of X is correct as follows:
If one root is zero, C = 0;
② If one root is 1, a + B + C = 0;
③ If one root is - 1, A-B + C = 0;
④ There is only one real root
A. 1 B.2 C.3 D.4

When x = 0
0+0+c=0
c=0
x=1
Then a + B + C = 0
x=-1
Then A-B + C = 0
So 123
Choose C

If the sum of binomial coefficients of the n-th expansion of (a + b) is 64, then n =?

∵2^n=64=2^6,∴n=6.

Calculation: (1 / 2006-1) (1 / 2005-1) (1 / 2004) (1/3-1)(1/2-1).
GM, help me

Original formula = - (1-1 / 2006) (1-1 / 2005)... (1-1 / 3) (1-1 / 2)
=-(2005/2006)(2004/2005)...(2/3)(1/2)
After continuous reduction
Original = - 1 / 2006

As shown in the figure, in △ ABC, the linear equation of the height on the BC side is x-2y + 1 = 0, and the linear equation of the bisector of ∠ A is y = 0. If the coordinates of point B are (1,2), the coordinates of point a and point C are obtained

Point a is the intersection of two lines y = 0 and x-2y + 1 = 0, the coordinates of point a are (- 1, 0).. KAB = 2 − 01 − (− 1) = 1. The equation of bisector line of ∵ - A is y = 0, ∵ - KAC = - 1. The equation of line AC is y = - X-1. BC is perpendicular to x-2y + 1 = 0, ∵ KBC = - 2. The equation of line BC is Y-2 = - 2 (x-1). From y = - X-1, y = - 2x + 4, the coordinates of point a and point C are obtained They are (- 1,0) and (5, - 6) respectively

Approximate solution of equation lgx + x = 3 by dichotomy
Accurate to 0.1

Let f (x) = lgx + X
f(2)=2.3013
∴2

How to select the free unknowns of homogeneous linear equations in 200 linear algebra? According to what principle? Please give an example

The general selection method of free unknowns is as follows
Firstly, the coefficient matrix is transformed into row simplified ladder matrix by elementary row transformation
The column of the first nonzero element in the nonzero row corresponds to the constrained unknown quantity
Other unknowns are free unknowns
From the above selection method, we can see that:
The column of constrained unknowns is a maximal independent group of the column vector group of A
The column in which the free unknowns are located can be uniquely linear expressed by this maximally independent group
In this way, the constrained unknowns can be uniquely solved for any group of free unknowns
It is clearer to express the equations in vector form
For example, α 1,..., α R is a maximal independent group of α 1,..., α n
Then x R + 1,..., x n are free unknowns
The equation is written as
x1α1+...+xrαr = -xr+1αr+1+...-xnαn
For any group of values of XR + 1,..., xn,
The linear combination - XR + 1 α R + 1 +... - xn α n can be uniquely linear expressed by α 1,..., α R
The constraint unknowns x1,..., XR can be uniquely determined
Example: homogeneous linear equations
x1-x2+x3-x4=0
x1-x2-x3+x4=0
x1-x2-2x3+2x4=0
Analysis: coefficient matrix A=
1 -1 1 -1
1 -1 -1 1
1 -1 -2 2
r2-r1,r3-r1
1 -1 1 -1
0 0 -2 2
0 0 -3 3
r2*(-1/2),r3+3r2,r1-r2
1 -1 0 0
0 0 1 -1
0 0 0 0
According to the general selection method, X1 and X3 are constrained unknowns, X2 and X4 are free unknowns
The same solution equations are
x1=x2
x3=x4
Taking any group of numbers x2 and x4, X1 and X3 can be uniquely solved
So, can we take X1 and X4 as free unknowns?
According to the principles mentioned above, it is OK
Because columns 2 and 3 are also maximal independent groups
Yes, please accept^_^