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
Book Icon
Chapter 3.7, Problem 3P
To determine

To find: Number of nickels in the collection.

Expert Solution & Answer
Check Mark

Answer to Problem 3P

22 nickels are there in the collection.

Explanation of Solution

Given information:

Collection of 40 dimes and nickels cost $2.90.

Calculations:

Let the number of nickels be x .

Then

    NumberValueTotal
    Dimes40 − x 10400 − 10x
    Nickels x 5 5x

Now,

Adult tickets cost + Student tickets cost = Total tickets cost

Since

$2.90 = 290ȼ

Thus,

We have

  400−10x+5x=290

That becomes

  400−5x=290

Subtract 400 from both sides:

  −5x=−110

Divide the above Eq. by -5:

  x=−110−5=22

Thus,

The numbers nickels sold were 22.

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Chapter 3.7, Problem 2P
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Chapter 3.7, Problem 4P

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. 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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