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High School Math Algebra High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Tocalculate: The numbers of vertices, edges and faces if every face has five edges and every edge is shared by two faces. Every face has five vertices and every vertex is shared by three faces. The sum of the number of vertices and faces is two more than the number of edges. The number of vertices are 20, edges are 30 and faces are12. Given information: It is given that every face has five edges and every edge is shared by two faces. Every face has five vertices and every vertex is shared by three faces. The sum of the number of vertices and faces is two more than the number of edges. Calculation: Consider the given information. Let x , y and z are the number of edges, faces and vertices respectively. Since every face has five edges so this can be written as x = 5 y . Also, every edge is shared by two faces so x = 5 2 y . Every face has five vertices and every vertex is shared by three faces, so 3 z = 5 y . z = 5 y 3 The sum of the number of vertices and faces is two more than the number of edges. z + y = x + 2 z + y − x = 2 Substitute the respective values into the above equation. 5 3 y + y − 5 2 y = 2 10 y + 6 y − 15 y = 12 y = 12 Put y = 12 into x = 5 2 y . x = 5 2 ( 12 ) x = 30 Put y = 12 into z = 5 y 3 . z = 5 ( 12 ) 3 z = 20 The number of vertices are 20, edges are 30 and faces are 12.

Tocalculate: The numbers of vertices, edges and faces if every face has five edges and every edge is shared by two faces. Every face has five vertices and every vertex is shared by three faces. The sum of the number of vertices and faces is two more than the number of edges. The number of vertices are 20, edges are 30 and faces are12. Given information: It is given that every face has five edges and every edge is shared by two faces. Every face has five vertices and every vertex is shared by three faces. The sum of the number of vertices and faces is two more than the number of edges. Calculation: Consider the given information. Let x , y and z are the number of edges, faces and vertices respectively. Since every face has five edges so this can be written as x = 5 y . Also, every edge is shared by two faces so x = 5 2 y . Every face has five vertices and every vertex is shared by three faces, so 3 z = 5 y . z = 5 y 3 The sum of the number of vertices and faces is two more than the number of edges. z + y = x + 2 z + y − x = 2 Substitute the respective values into the above equation. 5 3 y + y − 5 2 y = 2 10 y + 6 y − 15 y = 12 y = 12 Put y = 12 into x = 5 2 y . x = 5 2 ( 12 ) x = 30 Put y = 12 into z = 5 y 3 . z = 5 ( 12 ) 3 z = 20 The number of vertices are 20, edges are 30 and faces are 12.

Solution Summary: The author calculates the numbers of vertices, edges, and faces if every face has five edges and every vertex is shared by three faces.

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

15th Edition

ISBN: 9780133281163

Author: Prentice Hall

Publisher: Prentice Hall

See similar textbooks

Expert Solution & Answer

Chapter 3.5, Problem 45PPSE

Solution

Tocalculate:The numbers of vertices, edges and faces if every face has five edges and every edge is shared by two faces. Every face has five vertices and every vertex is shared by three faces. The sum of the number of vertices and faces is two more than the number of edges.

The number of vertices are 20, edges are 30 and faces are12.

Given information:

It is given that every face has five edges and every edge is shared by two faces. Every face has five vertices and every vertex is shared by three faces. The sum of the number of vertices and faces is two more than the number of edges.

Calculation:

Consider the given information.

Let x , y and z are the number of edges, faces and vertices respectively.

Since every face has five edges so this can be written as x=5y .

Also, every edge is shared by two faces so x=52y .

Every face has five vertices and every vertex is shared by three faces, so 3z=5y .

z=5y3

The sum of the number of vertices and faces is two more than the number of edges.

z+y=x+2z+y−x=2

Substitute the respective values into the above equation.

53y+y−52y=210y+6y−15y=12y=12

Put y=12 into x=52y .

x=52(12)x=30

Put y=12 into z=5y3 .

z=5(12)3z=20

The number of vertices are 20, edges are 30 and faces are 12.

Chapter 3.5, Problem 44PPSE

Chapter 3.6, Problem 1LC

Chapter 3 Solutions

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Ch. 3.1 - Prob. 1LC Ch. 3.1 - Prob. 2LC Ch. 3.1 - Prob. 3LC Ch. 3.1 - Prob. 4LC Ch. 3.1 - Prob. 5LC Ch. 3.1 - Prob. 6LC Ch. 3.1 - Prob. 7PPSE Ch. 3.1 - Prob. 8PPSE Ch. 3.1 - Prob. 9PPSE Ch. 3.1 - Prob. 10PPSE

Ch. 3.1 - Prob. 11PPSE Ch. 3.1 - Prob. 12PPSE Ch. 3.1 - Prob. 13PPSE Ch. 3.1 - Prob. 14PPSE Ch. 3.1 - Prob. 15PPSE Ch. 3.1 - Prob. 16PPSE Ch. 3.1 - Prob. 17PPSE Ch. 3.1 - Prob. 18PPSE Ch. 3.1 - Prob. 19PPSE Ch. 3.1 - Prob. 20PPSE Ch. 3.1 - Prob. 21PPSE Ch. 3.1 - Prob. 22PPSE Ch. 3.1 - Prob. 23PPSE Ch. 3.1 - Prob. 24PPSE Ch. 3.1 - Prob. 25PPSE Ch. 3.1 - Prob. 26PPSE Ch. 3.1 - Prob. 27PPSE Ch. 3.1 - Prob. 28PPSE Ch. 3.1 - Prob. 29PPSE Ch. 3.1 - Prob. 30PPSE Ch. 3.1 - Prob. 31PPSE Ch. 3.1 - Prob. 32PPSE Ch. 3.1 - Prob. 33PPSE Ch. 3.1 - Prob. 34PPSE Ch. 3.1 - Prob. 35PPSE Ch. 3.1 - Prob. 36PPSE Ch. 3.1 - Prob. 37PPSE Ch. 3.1 - Prob. 38PPSE Ch. 3.1 - Prob. 39PPSE Ch. 3.1 - Prob. 40PPSE Ch. 3.1 - Prob. 41PPSE Ch. 3.1 - Prob. 42PPSE Ch. 3.1 - Prob. 43PPSE Ch. 3.1 - Prob. 44PPSE Ch. 3.1 - Prob. 45PPSE Ch. 3.1 - Prob. 46PPSE Ch. 3.1 - Prob. 47PPSE Ch. 3.1 - Prob. 48PPSE Ch. 3.1 - Prob. 49PPSE Ch. 3.1 - Prob. 50PPSE Ch. 3.1 - Prob. 51PPSE Ch. 3.1 - Prob. 52PPSE Ch. 3.1 - Prob. 53PPSE Ch. 3.1 - Prob. 54PPSE Ch. 3.1 - 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