Math Problem 13 2 Same as the question

The famous Euler's theorem: number of vertices + number of faces - number of edges = 2

Did you count wrong? Count again.

That is, x+z-y=2.

Let the number of vertices of the polyhedron x, the number of faces y, and the number of edges z. Cut out a face so that it becomes a flat shape (pull open) and find the sum of the inside angles of all faces

On the one hand, in the original diagram, the sum of the internal angles is found by using each face.

There are f faces, and the number of sides on each side is n1, n2, ,...,ny, the sum of the inner angles of each side is:

[(n1-2)·180 degrees + (n2-2)·180 degrees +.]+ny-2) ·180 degrees].

n1+n2+…+ny -2y) ·180 degrees.

2z-2y) ·180 degrees = (z-y) ·360 degrees (1).

On the other hand, in a pull-open graph, the vertices are used to find the sum of the inner angles.

Let a cut face be an n-sided shape, and the sum of its internal angles is (n-2)·180 angles, then among all x vertices, n vertices are on the edge, and x-n vertices are in the middle. The sum of the internal angles at the middle v-n vertices is (x-n)·360 degrees, and the sum of the internal angles at the n vertices on the edge is (n-2)·180 degrees.

So, the sum of the internal angles of the faces of the polyhedron:

(x-n)·360 degrees + (n-2)·180 degrees + (n-2)·180 degrees.

x-2)·360 degrees (2).

From (1) (2): (z-y) ·360 degrees = (x-2)·360 degrees.

So x+y-z=2.)

Far from it. Finally, the answer to the original question should be x+z-y=2

1) For small questions, just count the number of vertices, edges, and faces of each wooden block in Figure (b), (c), (d), and (e); When counting, pay attention: do not omit the part that cannot be seen directly in the figure, do not repeat, and can be counted by imagination and correctly filled in the table;

2) Find out the hidden quantitative relationship between the "number of vertices, edges, and faces" in each graph by observing, and this quantitative relationship can be expressed by formula

Answer: 1) Rule: x+z-2=y

Another question also gave me the best answer.

12 and 1/3 - 1 and 2/3 = 10 and 2/3

1.Solution: The sum of the number of students in the two schools is x.

If x 200

Available in: x=18 000 75=240

If 100 x 200, we get: x=18000 85=3600 17 does not fit the topic (rounded).

So the sum of the number of students who signed up for the tour in these two schools is equal to 240, which is more than 200.

2.Solution: There are y students who sign up for the tour in school A, and z people who sign up for the tour in school B.

When 100 y 200.

Available: y+z=240

85y+90z=20800

Solution: y=160

x = 80 when y 200.

Available: y+z=240

75y+90z=20800

Solution: y=160 3

z=560/3

It doesn't fit the topic (discarded).

Therefore, there are 160 students from school A and 80 students from school B.

Hope you will be able to adopt.

Because 200*85=17000<18000, there are more than 200 people.

The total number of students is n=18000 75=240 people, if there are x people in school A, then there are 240-x people in school B, 90*x+85*(240-x)=20800, x=80, 240-x=160, so there are 80 people in school A and 160 people in school B.

1.Let the basketball price be x, the volleyball price is y, 4x+3y=410=number one, 2x+5y=310=number two, number two multiplied by 2=4x+10y=number three=620, number three minus one, the remaining 7y=210, the solution is y is equal to thirty, and x is equal to eighty.

2.Let the maximum number of basketballs be 80 x 30 (30 x) = 1700, and the solution is x equal to 16

3.If you buy x for basketball and y for volleyball, then eighty x thirty y is equal to 520, x is equal to 2, 5, and y is equal to 12, 4 respectively, and there are two kinds.

Meet - 3 x 10 so the integer x is -1,0,1,2

-1 0 1 2 Because the root number 3 is about 2 to the third power of 8 and the third power of 3 is 27, so take 2 and not take the whole integer from -1 to 2.

13、a=2

14, root number 3 13

Root No. 3 Area 2 Root No. 5

, 3 13 under the root number

3、a=5，b=6,11

times the root number is 3 under the root number, and the root number is under 5

Question 13 2

Question 14 Root No. 3 13

Question 15 11

16 questions: 2 times root number 3 2 times root number 5

Solution: 1-1 (m+1)] m+1).

1 (m+1) Yinjing-1 (m+1) (m+1)m+1-1m

Dear, please [answer], yours is the power of the royal fight to rise in my answer to the old question of the town, thank you.