Mathematics problems for the winter vacation homework of the second year of junior high school, and

1) When the point P coincides with the point O, ABCD is a square, so the triangle COD is an equilateral right triangle.

of is the height on the side cd, which is obtained by the combination of three lines: of is the middle line. ∴cf=df.

Connect the PD, then the triangle PAB triangle Pad (SAS). pd=pb, ∠adp=∠abp.

pf‖ad ∴∠adp=∠fpd ∴∠abp=∠fpd. (1)

And 0bp+ opb=90°, ope+ opb=90°

obp=∠ope.

obp+∠abp=45°,pe‖bc,∴∠ope+∠fpe=45, ∴abp=∠fpe.（2）

From (1) and (2) fpd= fpe ∴fpd≌⊿fpe（sas）

df=ef and set the edge length to m, de=a-ce, ef=

cf= In an isosceles right triangle PCF, cf=fpFrom the Pythagorean theorem, cf +fp =pc i.e. 2cf =pc∴pc=√2cf

pc = 2 ( and ac=pa+pc = 2a

pc=(pa+pc) finishing, get.

pc=pa+√2ce.

You just need to type in the math answers in the first book of the eighth grade, and a lot of them will come out, and you can pick the one you are satisfied with!

The young man does not work hard, and the boss is sad.

1.(1) BCE congruent AFE

Proof: AD BC

f=∠bce，∠b=eaf

and e is the midpoint of the imitation wide ab.

AE=BE, BCE, Quanqinliang, et al. AFE

Solution: (2) ab=6

E is the midpoint of AB.

ae=be=3

AB BC again

In the RT Bridge BCE, it is obtained according to the Pythagorean theorem.

ec = 16 + 9 = 5 under the root number

and BCE congruent AFE

ec=ef=5ef=5

Let the l1 function be y=kx and replace a(4,3) with people.

3=4kk=3/4

y=3/4x

The x-axis of point A is point C

AC X-axis.

oc=4，ac=3

In RT AOC, it is obtained according to the Pythagorean theorem.

oc=5 and oa=ob

Point b is (0,-5).

Let the l2 function be y=kx+b, and substitute a(4,3) and b(0,-5 into it. 3=4k+b①

5=b Substituting .

3=4k-5

k=2y=2x-5

Hopefully oh, I'm typing slowly, haha.

(1) It can be analyzed from different angles such as:

The average deviation rate of student A is 16%, and the average deviation rate of student B is 11%.

The deviation rate of student A is 7%, and the deviation of student B is 16%;

The minimum deviation rate of student A is 13%, and the minimum deviation rate of student B is 4%.

The maximum deviation rate of students A and B is 20%;

Student A's ability to estimate the number of words did not improve significantly, while student B's ability to estimate the number of words improved significantly

2) Can be analyzed from different angles such as:

From the average deviation rate**:

The average deviation rate of student A is 16%, and the estimated word count is in the range of 84 116;

The average deviation rate of student B is 11%, and the estimated word count is in the range of 89 111;

From the median deviation rate**:

The median deviation rate of student A is 15%, and the estimated word count is in the range of 85 115;

The median deviation rate of student B is 10%, and the estimated word count is in the range of 90 110;

Change from the deviation rate**:

There is no obvious trend characteristic of the deviation rate of student A, and there are many methods, such as the average value of the maximum and minimum values of the deviation rate is , and the estimated word count is in the range of 84 116 or 83 117

The deviation rate of student B is 0% to 4%, and the estimated word count is in the range of 96 104 or others

Alas. The second year of junior high school is so difficult, huh? I'm just getting ready.

It's hard, I'm only in the sixth grade, don't count on it

What is the deviation rate of the second year of junior high school? Why am I not impressed?

1 Bucket and object weigh 250kg

2. The underwater volume of the iron bucket is 250kg (1g cm) = 250000cm

3 barrels have a radius of r

4 r*r*3,14*25=250000

5 r=

Because the beam is bent into a rubber boring, the triangle AFD folds along DF, and point A falls on point E. Therefore, the full delta AFD is equal to the triangle DFE

That is, ad=de af=fe angle bad=angle bed and because the angle a=90 degrees, the angle d=90 degrees, so the adef is a square.