Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Statements and Logic

Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

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Question

Chapter 2.6, Problem 52E

To determine

To calculate: The value of the given expression.

Expert Solution & Answer

Answer to Problem 52E

The value of 5(−14)=−70_ .

Explanation of Solution

Given information:

The following expression:

5(−14)

Calculation: To calculate the value of the given expression, multiplication is performed. For multiplication of integers, the absolute values of the numbers are multiplied followed by the multiplication of the signs of the numbers. Since 5×14 gives 70 , and (+)×(−)=(−) , the expression can be written as:

5×(−14)⇒−70 .

Chapter 2.6, Problem 51E

Chapter 2.6, Problem 53E

Chapter 2 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 2.1 - Prob. 1PSQ Ch. 2.1 - Prob. 2PSQ Ch. 2.1 - Prob. 3PSQ Ch. 2.1 - Prob. 4PSQ Ch. 2.1 - Prob. 5PSQ Ch. 2.1 - Prob. 6PSQ Ch. 2.1 - Prob. 7PSQ Ch. 2.1 - Prob. 8PSQ Ch. 2.1 - Prob. 9PSQ Ch. 2.1 - Prob. 10PSQ

Ch. 2.1 - Prob. 11PSQ Ch. 2.1 - Prob. 1C Ch. 2.1 - Prob. 2C Ch. 2.1 - Prob. 3C Ch. 2.1 - Prob. 4C Ch. 2.1 - Prob. 5C Ch. 2.1 - Prob. 6C Ch. 2.1 - Prob. 7C Ch. 2.1 - Prob. 8C Ch. 2.1 - Prob. 9C Ch. 2.1 - Prob. 10C Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - Prob. 51E Ch. 2.1 - Prob. 52E Ch. 2.1 - Prob. 53E Ch. 2.1 - Prob. 54E Ch. 2.1 - Prob. 55E Ch. 2.1 - Prob. 56E Ch. 2.1 - Prob. 57E Ch. 2.1 - Prob. 58E Ch. 2.1 - Prob. 59E Ch. 2.1 - Prob. 60E Ch. 2.1 - Prob. 61E Ch. 2.1 - Prob. 62E Ch. 2.1 - Prob. 63E Ch. 2.1 - Prob. 64E Ch. 2.1 - Prob. 65E Ch. 2.2 - Prob. 1C Ch. 2.2 - Prob. 2C Ch. 2.2 - Prob. 3C Ch. 2.2 - Prob. 4C Ch. 2.2 - Prob. 5C Ch. 2.2 - Prob. 6C Ch. 2.2 - Prob. 7C Ch. 2.2 - Prob. 8C Ch. 2.2 - Prob. 9C Ch. 2.2 - Prob. 10C Ch. 2.2 - Prob. 11C Ch. 2.2 - Prob. 12C Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - 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