(REV)00 Edition

ISBN: 9780395977224

Author: Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher: McDougal Littell

Concept explainers

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

Chapter 3.5, Problem 10P

To determine

To find: the mass of each steel ball

Expert Solution & Answer

Answer to Problem 10P

The mass of each steel ball is 25 grams

Explanation of Solution

Given information: “Dionne has six steel balls of equal mass”, “If she puts five of them in one pan of a beam balance and one ball and a 100 g mass in the other pan, the pans balance each other”

Concept Involved:

Assign variable for the unknown that we need to find.

Use the keywords in the statement to rewrite the statement to meaningful math equation.

“is” means =

“more than” means +

“three times” means 3×

Solve the equation by getting the variable in one side of the equation and undoing whatever is done to it by performing reverse operation.

Calculation: Assign variable for the unknown that we need to find

Let x represents the mass of each steel ball

Write an equation based on the given sentence which tells about mass in each pan of a balance. “Five balls with each ball having mass of x grams in left pan & one ball with mass of x gram and 100 in right pan in the balance”

5x=x+100

In the process of getting x in one side of the equation subtract x on both sides

5x−x=100+x−x

Combine like terms on both sides

4x=100

Divide 4 on both sides of the equation

4x4=1004

Simplify fraction on both sides of the equation

x=25

Conclusion:

The mass of each steel ball is 25 grams

Chapter 3.5, Problem 9P

Chapter 3.5, Problem 11P

Chapter 3 Solutions

Algebra: Structure And Method, Book 1

Ch. 3.1 - Prob. 1OE Ch. 3.1 - Prob. 2OE Ch. 3.1 - Prob. 3OE Ch. 3.1 - Prob. 4OE Ch. 3.1 - Prob. 5OE Ch. 3.1 - Prob. 6OE Ch. 3.1 - Prob. 7OE Ch. 3.1 - Prob. 8OE Ch. 3.1 - Prob. 9OE Ch. 3.1 - Prob. 10OE

Ch. 3.1 - Prob. 11OE Ch. 3.1 - Prob. 12OE Ch. 3.1 - Prob. 13OE Ch. 3.1 - Prob. 14OE Ch. 3.1 - Prob. 15OE Ch. 3.1 - Prob. 16OE Ch. 3.1 - Prob. 17OE Ch. 3.1 - Prob. 18OE Ch. 3.1 - Prob. 19OE Ch. 3.1 - Prob. 20OE Ch. 3.1 - Prob. 21OE Ch. 3.1 - Prob. 1WE Ch. 3.1 - Prob. 2WE Ch. 3.1 - Prob. 3WE Ch. 3.1 - Prob. 4WE Ch. 3.1 - Prob. 5WE Ch. 3.1 - Prob. 6WE Ch. 3.1 - Prob. 7WE Ch. 3.1 - Prob. 8WE Ch. 3.1 - Prob. 9WE Ch. 3.1 - Prob. 10WE Ch. 3.1 - Prob. 11WE Ch. 3.1 - Prob. 12WE Ch. 3.1 - Prob. 13WE Ch. 3.1 - Prob. 14WE Ch. 3.1 - Prob. 15WE Ch. 3.1 - Prob. 16WE Ch. 3.1 - Prob. 17WE Ch. 3.1 - Prob. 18WE Ch. 3.1 - Prob. 19WE Ch. 3.1 - Prob. 20WE Ch. 3.1 - Prob. 21WE Ch. 3.1 - Prob. 22WE Ch. 3.1 - Prob. 23WE Ch. 3.1 - Prob. 24WE Ch. 3.1 - Prob. 25WE Ch. 3.1 - Prob. 26WE Ch. 3.1 - Prob. 27WE Ch. 3.1 - Prob. 28WE Ch. 3.1 - Prob. 29WE Ch. 3.1 - Prob. 30WE Ch. 3.1 - Prob. 31WE Ch. 3.1 - Prob. 32WE Ch. 3.1 - Prob. 33WE Ch. 3.1 - Prob. 34WE Ch. 3.1 - Prob. 35WE Ch. 3.1 - Prob. 36WE Ch. 3.1 - Prob. 37WE Ch. 3.1 - 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Prob. 6MRE Ch. 3.3 - Prob. 7MRE Ch. 3.3 - Prob. 8MRE Ch. 3.3 - Prob. 9MRE Ch. 3.3 - Prob. 10MRE Ch. 3.3 - Prob. 11MRE Ch. 3.3 - Prob. 12MRE Ch. 3.3 - Prob. 1E Ch. 3.3 - Prob. 2E Ch. 3.3 - Prob. 3E Ch. 3.4 - Prob. 1P Ch. 3.4 - Prob. 2P Ch. 3.4 - Prob. 3P Ch. 3.4 - Prob. 4P Ch. 3.4 - Prob. 5P Ch. 3.4 - Prob. 6P Ch. 3.4 - Prob. 7P Ch. 3.4 - Prob. 8P Ch. 3.4 - Prob. 9P Ch. 3.4 - Prob. 10P Ch. 3.4 - Prob. 11P Ch. 3.4 - Prob. 12P Ch. 3.4 - Prob. 13P Ch. 3.4 - Prob. 14P Ch. 3.4 - Prob. 15P Ch. 3.4 - Prob. 16P Ch. 3.4 - Prob. 17P Ch. 3.4 - Prob. 18P Ch. 3.4 - Prob. 19P Ch. 3.4 - Prob. 20P Ch. 3.4 - Prob. 21P Ch. 3.4 - Prob. 22P Ch. 3.4 - Prob. 23P Ch. 3.4 - Prob. 24P Ch. 3.4 - Prob. 25P Ch. 3.4 - Prob. 26P Ch. 3.4 - Prob. 27P Ch. 3.4 - Prob. 28P Ch. 3.4 - Prob. 29P Ch. 3.4 - Prob. 30P Ch. 3.4 - Prob. 31P Ch. 3.4 - Prob. 32P Ch. 3.4 - Prob. 33P Ch. 3.4 - Prob. 1MRE Ch. 3.4 - Prob. 2MRE Ch. 3.4 - Prob. 3MRE Ch. 3.4 - Prob. 4MRE Ch. 3.4 - Prob. 5MRE Ch. 3.4 - Prob. 6MRE Ch. 3.4 - Prob. 7MRE Ch. 3.4 - Prob. 8MRE Ch. 3.4 - Prob. 9MRE Ch. 3.4 - Prob. 10MRE Ch. 3.4 - Prob. 11MRE Ch. 3.4 - Prob. 12MRE Ch. 3.5 - Prob. 1OE Ch. 3.5 - Prob. 2OE Ch. 3.5 - Prob. 3OE Ch. 3.5 - Prob. 4OE Ch. 3.5 - Prob. 5OE Ch. 3.5 - Prob. 6OE Ch. 3.5 - Prob. 7OE Ch. 3.5 - Prob. 8OE Ch. 3.5 - Prob. 9OE Ch. 3.5 - Prob. 10OE Ch. 3.5 - Prob. 11OE Ch. 3.5 - Prob. 12OE Ch. 3.5 - Prob. 1WE Ch. 3.5 - Prob. 2WE Ch. 3.5 - Prob. 3WE Ch. 3.5 - Prob. 4WE Ch. 3.5 - Prob. 5WE Ch. 3.5 - Prob. 6WE Ch. 3.5 - Prob. 7WE Ch. 3.5 - Prob. 8WE Ch. 3.5 - Prob. 9WE Ch. 3.5 - Prob. 10WE Ch. 3.5 - Prob. 11WE Ch. 3.5 - Prob. 12WE Ch. 3.5 - Prob. 13WE Ch. 3.5 - Prob. 14WE Ch. 3.5 - Prob. 15WE Ch. 3.5 - Prob. 16WE Ch. 3.5 - Prob. 17WE Ch. 3.5 - Prob. 18WE Ch. 3.5 - Prob. 19WE Ch. 3.5 - Prob. 20WE Ch. 3.5 - Prob. 21WE Ch. 3.5 - Prob. 22WE Ch. 3.5 - Prob. 23WE Ch. 3.5 - Prob. 24WE Ch. 3.5 - Prob. 25WE Ch. 3.5 - Prob. 26WE Ch. 3.5 - Prob. 27WE Ch. 3.5 - Prob. 28WE Ch. 3.5 - Prob. 29WE Ch. 3.5 - Prob. 30WE Ch. 3.5 - Prob. 31WE Ch. 3.5 - Prob. 32WE Ch. 3.5 - 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