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

(a) To write: The system of equations to represent the cost c for h hours. c = 25 h , c = 20 h + 10 Given information: It is given that one campus charges $25 per hour and other campus charges $20 per hour plus one-time registration charge of $10. Calculation : Let the playing hours be h . The cost for playing in first court is, c = 25 h . The cost for playing in second court is, c = 20 h + 10 . Hence the system of equation is c = 25 h c = 20 h + 10 (b) To find: Find the numbers of hours for which the cost is same using graphic calculator. The cost is $50 for 2 hours. Given information: It is given that one campus charges $25 per hour and other campus charges $20 per hour plus one-time registration charge of $10. c = 25 h c = 20 h + 10 Calculation : The graph for the system is given below. As it is shown in the graph that both the lines meet at the point 2 , 50 . Here x axis represents the hours and y axis represents the cost. Therefore the cost is $50 for 2 hours. (c) To find: If you want to practice for 10 hours which university you will choose. second university is cheaper to play. Given information: It is given that total playing hours are 10. c = 25 h ... ( 1 ) c = 20 h + 10.. ( 2 ) Calculation : c = 25 h ... ( 1 ) c = 20 h + 10.. ( 2 ) Equation first represents the cost to play in first university and equation (2) represents the cost to play in second university. Substitute h = 10 in both the equation the cost will be: First Equation: c = 25 h c = 25 ⋅ 10 = 250 Secon Equation: c = 20 h + 10 c = 20 ⋅ 10 + 10 = 200 + 10 = 210 The cost for playing in first university campus is $250 and the cost for playing in second university playground is $210. So choose second university.

(a) To write: The system of equations to represent the cost c for h hours. c = 25 h , c = 20 h + 10 Given information: It is given that one campus charges $25 per hour and other campus charges $20 per hour plus one-time registration charge of $10. Calculation : Let the playing hours be h . The cost for playing in first court is, c = 25 h . The cost for playing in second court is, c = 20 h + 10 . Hence the system of equation is c = 25 h c = 20 h + 10 (b) To find: Find the numbers of hours for which the cost is same using graphic calculator. The cost is $50 for 2 hours. Given information: It is given that one campus charges $25 per hour and other campus charges $20 per hour plus one-time registration charge of $10. c = 25 h c = 20 h + 10 Calculation : The graph for the system is given below. As it is shown in the graph that both the lines meet at the point 2 , 50 . Here x axis represents the hours and y axis represents the cost. Therefore the cost is $50 for 2 hours. (c) To find: If you want to practice for 10 hours which university you will choose. second university is cheaper to play. Given information: It is given that total playing hours are 10. c = 25 h ... ( 1 ) c = 20 h + 10.. ( 2 ) Calculation : c = 25 h ... ( 1 ) c = 20 h + 10.. ( 2 ) Equation first represents the cost to play in first university and equation (2) represents the cost to play in second university. Substitute h = 10 in both the equation the cost will be: First Equation: c = 25 h c = 25 ⋅ 10 = 250 Secon Equation: c = 20 h + 10 c = 20 ⋅ 10 + 10 = 200 + 10 = 210 The cost for playing in first university campus is $250 and the cost for playing in second university playground is $210. So choose second university.

Solution Summary: The author explains the system of equations to represent the cost for h hours.

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 3.1, Problem 40PPSE

Solution

(a)

To write: The system of equations to represent the cost c for h hours.

c=25h , c=20h+10

Given information:

It is given that one campus charges $25 per hour and other campus charges $20 per hour plus one-time registration charge of $10.

Calculation:

Let the playing hours be h .

The cost for playing in first court is, c=25h .

The cost for playing in second court is, c=20h+10 .

Hence the system of equation is

c=25h

c=20h+10

(b)

To find: Find the numbers of hours for which the cost is same using graphic calculator.

The cost is $50 for 2 hours.

Given information:

It is given that one campus charges $25 per hour and other campus charges $20 per hour plus one-time registration charge of $10.

c=25h

c=20h+10

Calculation:

The graph for the system is given below.

As it is shown in the graph that both the lines meet at the point 2,50 . Here x axis represents the hours and y axis represents the cost. Therefore the cost is $50 for 2 hours.

(c)

To find: If you want to practice for 10 hours which university you will choose.

second university is cheaper to play.

Given information:

It is given that total playing hours are 10.

c=25h...(1)

c=20h+10..(2)

Calculation:

c=25h...(1)

c=20h+10..(2)

Equation first represents the cost to play in first university and equation (2) represents the cost to play in second university.

Substitute h=10 in both the equation the cost will be:

First Equation: c=25hc=25⋅10=250Secon Equation: c=20h+10c=20⋅10+10=200+10=210

The cost for playing in first university campus is $250 and the cost for playing in second university playground is $210. So choose second university.

Chapter 3.1, Problem 39PPSE

Chapter 3.1, Problem 41PPSE

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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Prob. 20PPSE Ch. 3.5 - Prob. 21PPSE Ch. 3.5 - Prob. 22PPSE Ch. 3.5 - Prob. 23PPSE Ch. 3.5 - Prob. 24PPSE Ch. 3.5 - Prob. 25PPSE Ch. 3.5 - Prob. 26PPSE Ch. 3.5 - Prob. 27PPSE Ch. 3.5 - Prob. 28PPSE Ch. 3.5 - Prob. 29PPSE Ch. 3.5 - Prob. 30PPSE Ch. 3.5 - Prob. 31PPSE Ch. 3.5 - Prob. 32PPSE Ch. 3.5 - Prob. 33PPSE Ch. 3.5 - Prob. 34PPSE Ch. 3.5 - Prob. 35PPSE Ch. 3.5 - Prob. 36PPSE Ch. 3.5 - Prob. 37PPSE Ch. 3.5 - Prob. 38PPSE Ch. 3.5 - Prob. 39PPSE Ch. 3.5 - Prob. 40PPSE Ch. 3.5 - Prob. 41PPSE Ch. 3.5 - Prob. 42PPSE Ch. 3.5 - Prob. 43PPSE Ch. 3.5 - Prob. 44PPSE Ch. 3.5 - Prob. 45PPSE Ch. 3.6 - Prob. 1LC Ch. 3.6 - Prob. 2LC Ch. 3.6 - Prob. 3LC Ch. 3.6 - Prob. 4LC Ch. 3.6 - Prob. 5LC Ch. 3.6 - Prob. 6LC Ch. 3.6 - Prob. 7LC Ch. 3.6 - Prob. 8PPSE Ch. 3.6 - Prob. 9PPSE Ch. 3.6 - Prob. 10PPSE Ch. 3.6 - Prob. 11PPSE Ch. 3.6 - Prob. 12PPSE Ch. 3.6 - Prob. 13PPSE Ch. 3.6 - Prob. 14PPSE Ch. 3.6 - Prob. 15PPSE Ch. 3.6 - Prob. 16PPSE Ch. 3.6 - Prob. 17PPSE Ch. 3.6 - Prob. 18PPSE Ch. 3.6 - Prob. 19PPSE Ch. 3.6 - 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