The math problems in the first year of junior high school should be analyzed in detail. I ve seen it

Parallel relationships. Proof of the process –

Known: MCN= BCM=360- a- b- AMC, (CM is the bisector of BCD, and the sum of the inner angles of the quadrilateral is 360).

fnd=360-∠d-∠e-∠efn=360-∠a-∠b-∠efn=360-∠a-∠b-∠nfm,（∠a=∠d,∠b=∠e）

Old. MCN- FND=360- A- B- AMC-(360- A- B- NFM)= NFM- AMC= NFM-(180- FMC)= NFM+ FMC-180, (180 for the sum of angles and complements)=360- MCN- CNF-180 (360 for the sum of the inner angles of the quadrilateral)=180- MCN- CNF

So. 2∠mcn=180+∠fnd-∠cnf=180+∠fnd-(180-∠fnd)=2∠fnd

i.e. mcn= fnd

So, cm is parallel to fn.

The parallel lines that pass f to make mc intersect cd at point p have amc= mfp mcd= fpd

Because cm bisects bcd, bcm= mcd i.e. bcm= fpd

Because a= d, b= e

So efp+ fpd= amc+ bcm so efp= amc so efp= mfp so fp is afe and the bisector is known to coincide p with n so cm fn

The meaning of the title is that when the oca degree is moved, the two angles are equal, and the direction is not emphasized.

I am renting the reputation to set OCA for X, and launch other corners as shown below.

1: The temperature at noon in a certain place is 7, the temperature at 4 am is 9 lower than noon, and the temperature in the evening is 4 lower than noon. Try a positive or negative number to indicate that the place is 4 a.m. on this day 7-9 = -2;

Evening temperature 7-4=+3 ;

2: The storage temperature of the external standard of a certain brand of milk bottle is (4 + the following one-2), so it can be known in what temperature range is it suitable for storage?

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-9=-2（℃）

A: The temperature at 4 a.m. is -2 and +2=6 ( ) in the evening

A: It is suitable to store in the temperature range of 2 to 6 years.

Temperature at 6 p.m.: 8-3 = 5

Temperature at 4 a.m.: 5-7 = -2

1.4 a.m. 7-9=-2 ;

Evening temperature 7-4=+3 ;

-2 degrees Celsius, 3 degrees Celsius.

4 degrees Celsius to -2 degrees Celsius.

1. m+4=-1, get m=-3

2. X b+a is obtained from x-a b, x<(2b+a+1) 2 is obtained from 2x-a<2b+1, and the solution set is (2b+a+1) 2 x b+a, because the solution set is 3 x<5, so b+a=3, (2b+a+1) 2=5 is solved to a=-3, b=6, b a=-2

3、d4、y≤-8

a8、m≥-2

9、x≤12

That is, think of the positive half axis of the number axis as a rope, so that the number axis around the origin of the circle coincides with the position marked 0 on the circle.

1 coincides with the position marked with 1 on the circle.

2 coincides with the position marked 2 on the circle.

When the position of 3 is reached, the number axis has already circled the circle, so 3 and 0 coincide and go on and on, the position marked with 0 on the circle corresponds to the nth circle of multiples of 3, and the 0 on the circle corresponds to 3 (n-1) on the number line

1 on the circle corresponds to 3 (n-1) + 1 = 3n-2 on the number line, 2 on the circle corresponds to 3 (n-1) + 2 = 3n-1(1) 2 on the number line

2) 3n-2

In other words, imagine the positive half axis of the number axis as a rope, and let the number axis wrap around the circle.

The origin and the position on the circle marked with the nucleus 0 coincide.

1 coincides with the position marked with 1 on the circle.

2 coincides with the position marked 2 on the circle.

By the time we reach the position of 3, the number axis is already in a circle, so 3 and 0 coincide.

This goes on and on, and the positions marked with 0 on the circle correspond to multiples of 3.

In the nth circle, 0 on the circle corresponds to 3 (n-1) on the number axis, 1 on the circle corresponds to 3 (n-1) on the number axis + 1 = 3n-2, 2 on the circle corresponds to 3 (n-1) + 2 = 3n-13n-2 on the number axis

1.The 6th power is wrong because your answer is 3 significant digits.

2.1-4a+2a2 (the number of unknown numbers rises in the permutation, called the ascending power permutation).

3.The third question can only be solved in combination with the prerequisites. (Method: Absolute value of a non-negative number = itself, absolute value of a negative number = opposite number of itself).

The 6th power is right.

1-4a+2a²

The third question is that it cannot be simplified without preconditions

Trick: When they meet for the second time, A and B exercise Duan Feng's argument at the same time.

Let the distance of ab be x

According to the time = distance speed, when A meets for the second time, the distance traveled is one ab+(ab-34), then the time is (x+(x-34)) 70

In the second encounter, B travels a distance of ab+34 and the time is (x+34) 52

Column equation: x+(x-34)) 70

x+34)/52

The solution is x=122

A: ......

It's a matter of harmony. Drawing a line diagram is the best way to do this.

and is 22 articles;

Xiaohua has 2 more than me, and Xiaohong has 3 more than Xiaohua.

Little Red has 5 more than me.

So there are: I: (22-2-5) 3=5.

Xiaohua: 5+2=7.

Little Red: 22-5-7=10.

If Xiaohua fished x, then Xiaoming fished x-2, and Xiaohong fished x+3 x+x-2+x+3=22

3x+1=22

3x=21x=7.

Xiaohua fished 7, Xiao Ming fished 7-2=5, Xiaohong fished 7+3=10.