Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00 Edition
ISBN: 9780395977224
Author: Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher: McDougal Littell
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Concept explainers

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
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Chapter 3.7, Problem 17P
To determine

To calculate the money that both has together.

Expert Solution & Answer
Check Mark

Answer to Problem 17P

$3.80

Explanation of Solution

Given:

Nadine has seven more nickels than Delano has dimes.

If Delano gives four of his dimes, Delano will have the same amount of money as Nadine.

Calculation:

Let us assume that Delano has x dimes.

Cost of a dime is $0.10 and the cost of a nickel is $0.05

Given that Nadine has seven more nickels than Delano has dimes.

Thus, Nadine has x + 7 nickels.

Given that Delano gives four dimes, then they both have the same amount of money.

  (x−4)(0.10)=(x+7)(0.05)+4(0.10)0.10x−0.40=0.05x+0.35+0.400.10x−0.05x−0.40=0.05x+0.75−0.05x0.05x−0.40=0.750.05x+0.40−0.40=0.75+0.400.05x=1.150.05x0.05=1.150.05x=23

Therefore, the total amount both have is,

  Total Amount=0.10(23)+0.05(30)=2.30+1.50=3.80

Conclusion:

Therefore, the total amount that both have is $3.80

Previous
Chapter 3.7, Problem 16P
Next
Chapter 3.7, Problem 18P

Chapter 3 Solutions

Algebra: Structure And Method, Book 1

Ch. 3.1 - Prob. 1OECh. 3.1 - Prob. 2OECh. 3.1 - Prob. 3OECh. 3.1 - Prob. 4OECh. 3.1 - Prob. 5OECh. 3.1 - Prob. 6OECh. 3.1 - Prob. 7OECh. 3.1 - Prob. 8OECh. 3.1 - Prob. 9OECh. 3.1 - Prob. 10OE
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Prob. 2WECh. 3.5 - Prob. 3WECh. 3.5 - Prob. 4WECh. 3.5 - Prob. 5WECh. 3.5 - Prob. 6WECh. 3.5 - Prob. 7WECh. 3.5 - Prob. 8WECh. 3.5 - Prob. 9WECh. 3.5 - Prob. 10WECh. 3.5 - Prob. 11WECh. 3.5 - Prob. 12WECh. 3.5 - Prob. 13WECh. 3.5 - Prob. 14WECh. 3.5 - Prob. 15WECh. 3.5 - Prob. 16WECh. 3.5 - Prob. 17WECh. 3.5 - Prob. 18WECh. 3.5 - Prob. 19WECh. 3.5 - Prob. 20WECh. 3.5 - Prob. 21WECh. 3.5 - Prob. 22WECh. 3.5 - Prob. 23WECh. 3.5 - Prob. 24WECh. 3.5 - Prob. 25WECh. 3.5 - Prob. 26WECh. 3.5 - Prob. 27WECh. 3.5 - Prob. 28WECh. 3.5 - Prob. 29WECh. 3.5 - Prob. 30WECh. 3.5 - Prob. 31WECh. 3.5 - Prob. 32WECh. 3.5 - Prob. 33WECh. 3.5 - Prob. 34WECh. 3.5 - Prob. 35WECh. 3.5 - Prob. 36WECh. 3.5 - Prob. 37WECh. 3.5 - Prob. 38WECh. 3.5 - Prob. 39WECh. 3.5 - Prob. 40WECh. 3.5 - Prob. 41WECh. 3.5 - Prob. 42WECh. 3.5 - Prob. 43WECh. 3.5 - Prob. 44WECh. 3.5 - Prob. 45WECh. 3.5 - Prob. 46WECh. 3.5 - Prob. 47WECh. 3.5 - Prob. 48WECh. 3.5 - Prob. 1PCh. 3.5 - Prob. 2PCh. 3.5 - Prob. 3PCh. 3.5 - Prob. 4PCh. 3.5 - 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Prob. 2OECh. 3.8 - Prob. 3OECh. 3.8 - Prob. 1WECh. 3.8 - Prob. 2WECh. 3.8 - Prob. 3WECh. 3.8 - Prob. 4WECh. 3.8 - Prob. 5WECh. 3.8 - Prob. 6WECh. 3.8 - Prob. 7WECh. 3.8 - Prob. 8WECh. 3.8 - Prob. 9WECh. 3.8 - Prob. 10WECh. 3.8 - Prob. 11WECh. 3.8 - Prob. 1MRECh. 3.8 - Prob. 2MRECh. 3.8 - Prob. 3MRECh. 3.8 - Prob. 4MRECh. 3.8 - Prob. 5MRECh. 3.8 - Prob. 6MRECh. 3.8 - Prob. 7MRECh. 3.8 - Prob. 8MRECh. 3.8 - Prob. 9MRECh. 3.8 - Prob. 10MRECh. 3.8 - Prob. 11MRECh. 3.8 - Prob. 12MRECh. 3.8 - Prob. 1STCh. 3.8 - Prob. 2STCh. 3.8 - Prob. 3STCh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - 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Use of ALGEBRA in REAL LIFE; Author: Fast and Easy Maths !;https://www.youtube.com/watch?v=9_PbWFpvkDc;License: Standard YouTube License, CC-BY
Compound Interest Formula Explained, Investment, Monthly & Continuously, Word Problems, Algebra; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=P182Abv3fOk;License: Standard YouTube License, CC-BY
Applications of Algebra (Digit, Age, Work, Clock, Mixture and Rate Problems); Author: EngineerProf PH;https://www.youtube.com/watch?v=Y8aJ_wYCS2g;License: Standard YouTube License, CC-BY