Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)
Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)
2nd Edition
ISBN: 9780076602919
Author: McGraw-Hill Education
Publisher: McGraw-Hill Education
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Chapter 3.2, Problem 3GP
To determine

To find:

Zero, domain, and range of the given equation y=1.5x45 and to write its mean in the context of given situation.

Expert Solution & Answer
Check Mark

Answer to Problem 3GP

Zero of the equation is x=30 and it means thatminimum more than 30 candies should be sold surely to get the profit accordingly.

Also, domain of this relation is {30,31,32,33} and range is {0,1.5,3,4.5,6,7.5,..} .

Explanation of Solution

Given information:

A`s class is selling candy to raise money for a class trip. They paid $45 for the candy, and they are selling each candy bar for $1.50. The function, y=1.5x45 representsstudents profit ywhen they sell x candy bars.

Conceptused:

Zero of an equation is obtained by putting it equal to 0 and then solving it for independent variable x.

Also set of all values of x is its domain and set of all values of y is its range.

Calculation:

Plug in y = 0 in and on solving it for x,

  0=1.5x451.5x=45 x=451.5=30

So, only zero of the equation is x=30.

As x are the number of candies sold and 30 are minimum candies, to start getting profit. So, x may be greater or equal to 30. So, domain of this function will be an infinite set {30,31,32,33} and by selling a single candy, it will be a profit of $1.5, so its range will be the set of all multiples of 1.5 as {0,1.5,3,4.5,6,7.5,..} .

Conclusion:

Value x=30 means that minimum more than 30 candies should be sold surely to get the profit accordingly. Also, domain of this relation is {30,31,32,33} and range is {0,1.5,3,4.5,6,7.5,..} .

Chapter 3 Solutions

Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)

Ch. 3.1 - Prob. 2CYUCh. 3.1 - Prob. 3CYUCh. 3.1 - Prob. 4CYUCh. 3.1 - Prob. 5CYUCh. 3.1 - Prob. 6CYUCh. 3.1 - Prob. 7CYUCh. 3.1 - Prob. 8CYUCh. 3.1 - Prob. 9CYUCh. 3.1 - Prob. 10CYUCh. 3.1 - Prob. 11CYUCh. 3.1 - Prob. 12CYUCh. 3.1 - Prob. 13PPSCh. 3.1 - Prob. 14PPSCh. 3.1 - Prob. 15PPSCh. 3.1 - Prob. 16PPSCh. 3.1 - Prob. 17PPSCh. 3.1 - Prob. 18PPSCh. 3.1 - Prob. 19PPSCh. 3.1 - Prob. 20PPSCh. 3.1 - Prob. 21PPSCh. 3.1 - Prob. 22PPSCh. 3.1 - Prob. 23PPSCh. 3.1 - Prob. 24PPSCh. 3.1 - Prob. 25PPSCh. 3.1 - Prob. 26PPSCh. 3.1 - Prob. 27PPSCh. 3.1 - Prob. 28PPSCh. 3.1 - Prob. 29PPSCh. 3.1 - Prob. 30PPSCh. 3.1 - Prob. 31PPSCh. 3.1 - Prob. 32PPSCh. 3.1 - Prob. 33PPSCh. 3.1 - Prob. 34PPSCh. 3.1 - Prob. 35PPSCh. 3.1 - Prob. 36PPSCh. 3.1 - Prob. 37PPSCh. 3.1 - Prob. 38PPSCh. 3.1 - Prob. 39PPSCh. 3.1 - Prob. 40PPSCh. 3.1 - Prob. 41PPSCh. 3.1 - Prob. 42PPSCh. 3.1 - Prob. 43PPSCh. 3.1 - Prob. 44PPSCh. 3.1 - Prob. 45PPSCh. 3.1 - Prob. 46PPSCh. 3.1 - Prob. 47PPSCh. 3.1 - Prob. 48PPSCh. 3.1 - Prob. 49PPSCh. 3.1 - 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