a. Answer 6 years Explanation The enrollment will be equal in Spanish and French classes at the point where the graph of both will intersect. From the given graph it can be observed that they intersect at the point ( 6 , 300 ) . Conclusion: Thus, after 6 years there will be equal enrollment in Spanish and French classes. b. To determine To verify: the answer in part (a) using the equation 355 − 9 x = 229 + 12 x . Answer 6 years Explanation The equation that can be used to determine the time after which there will be equal enrollment is: 355 − 9 x = 229 + 12 x 355 − 229 = 12 x + 9 x 126 = 21 x 126 21 = x 6 = x Conclusion: Hence, the result in part (a) is verified.

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN: 9781680331141

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 1.3, Problem 38E

To determine

the number of years after which there will be equal enrollment in Spanish and French classes.

Expert Solution

Answer to Problem 38E

6 years

Explanation of Solution

The enrollment will be equal in Spanish and French classes at the point where the graph of both will intersect.

From the given graph it can be observed that they intersect at the point (6,300) .

Conclusion:

Thus, after 6 years there will be equal enrollment in Spanish and French classes.

To determine

To verify: the answer in part (a) using the equation 355−9x=229+12x .

Expert Solution

Answer to Problem 38E

6 years

Explanation of Solution

The equation that can be used to determine the time after which there will be equal enrollment is:

355−9x=229+12x355−229=12x+9x126=21x

12621=x6=x

Conclusion:

Hence, the result in part (a) is verified.

Chapter 1.3, Problem 37E

Chapter 1.3, Problem 39E

Chapter 1 Solutions

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - Prob. 57E Ch. 1.1 - Prob. 58E Ch. 1.1 - Prob. 59E Ch. 1.1 - Prob. 60E Ch. 1.1 - 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