Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Question

Chapter 1.7, Problem 34E

To determine

To simplify : the expression 20(−45)÷(−9) .

Expert Solution & Answer

Answer to Problem 34E

The answer of the given expression is 100 .

Explanation of Solution

Given information :

Consider, the expression provided in the question,

20(−45)÷(−9)

Calculation :

To simplify the expression, multiply the first two terms of the expression and then divide the product of first two terms with the third term.

Theproductof two integers with differentsigns is neagative.

Therefore,

20(−45)÷(−9)=−900÷(−9)

Since, the quotient of two integers with same signs is positive. Therefore,

20(−45)÷(−9)=−900÷(−9)20(−45)÷(−9)=100

Thus, the answer of the expression is 100 .

Chapter 1.7, Problem 33E

Chapter 1.7, Problem 35E

Chapter 1 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 1.1 - Prob. 1C Ch. 1.1 - Prob. 2C Ch. 1.1 - Prob. 3C Ch. 1.1 - Prob. 4C Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E

Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - Prob. 57E Ch. 1.1 - Prob. 58E Ch. 1.1 - Prob. 59E Ch. 1.1 - Prob. 60E Ch. 1.1 - Prob. 61E Ch. 1.1 - Prob. 62E Ch. 1.1 - Prob. 63E Ch. 1.1 - Prob. 64E Ch. 1.1 - Prob. 65E Ch. 1.1 - Prob. 66E Ch. 1.1 - Prob. 67E Ch. 1.2 - Prob. 1C Ch. 1.2 - Prob. 2C Ch. 1.2 - Prob. 3C Ch. 1.2 - Prob. 4C Ch. 1.2 - Prob. 5C Ch. 1.2 - Prob. 6C Ch. 1.2 - Prob. 7C Ch. 1.2 - Prob. 8C Ch. 1.2 - Prob. 9C Ch. 1.2 - Prob. 10C Ch. 1.2 - Prob. 11C Ch. 1.2 - Prob. 1E Ch. 1.2 - Prob. 2E Ch. 1.2 - Prob. 3E Ch. 1.2 - Prob. 4E Ch. 1.2 - Prob. 5E Ch. 1.2 - Prob. 6E Ch. 1.2 - Prob. 7E Ch. 1.2 - Prob. 8E Ch. 1.2 - Prob. 9E Ch. 1.2 - Prob. 10E Ch. 1.2 - Prob. 11E Ch. 1.2 - Prob. 12E Ch. 1.2 - Prob. 13E Ch. 1.2 - Prob. 14E Ch. 1.2 - Prob. 15E Ch. 1.2 - Prob. 16E Ch. 1.2 - Prob. 17E Ch. 1.2 - Prob. 18E Ch. 1.2 - Prob. 19E Ch. 1.2 - Prob. 20E Ch. 1.2 - Prob. 21E Ch. 1.2 - Prob. 22E Ch. 1.2 - Prob. 23E Ch. 1.2 - Prob. 24E Ch. 1.2 - Prob. 25E Ch. 1.2 - Prob. 26E Ch. 1.2 - Prob. 27E Ch. 1.2 - Prob. 28E Ch. 1.2 - Prob. 29E Ch. 1.2 - 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Prob. 32E Ch. 1.7 - Prob. 33E Ch. 1.7 - Prob. 34E Ch. 1.7 - Prob. 35E Ch. 1.7 - Prob. 36E Ch. 1.7 - Prob. 37E Ch. 1.7 - Prob. 38E Ch. 1.7 - Prob. 39E Ch. 1.7 - Prob. 40E Ch. 1.7 - Prob. 41E Ch. 1.7 - Prob. 42E Ch. 1.7 - Prob. 43E Ch. 1.7 - Prob. 44E Ch. 1.7 - Prob. 45E Ch. 1.7 - Prob. 46E Ch. 1.7 - Prob. 47E Ch. 1.7 - Prob. 48E Ch. 1.7 - Prob. 49E Ch. 1.7 - Prob. 50E Ch. 1.7 - Prob. 51E Ch. 1.7 - Prob. 52E Ch. 1.8 - Prob. 1C Ch. 1.8 - Prob. 2C Ch. 1.8 - Prob. 3C Ch. 1.8 - Prob. 4C Ch. 1.8 - Prob. 5C Ch. 1.8 - Prob. 6C Ch. 1.8 - Prob. 7C Ch. 1.8 - Prob. 1E Ch. 1.8 - Prob. 2E Ch. 1.8 - Prob. 3E Ch. 1.8 - Prob. 4E Ch. 1.8 - Prob. 5E Ch. 1.8 - Prob. 6E Ch. 1.8 - Prob. 7E Ch. 1.8 - Prob. 8E Ch. 1.8 - Prob. 9E Ch. 1.8 - Prob. 10E Ch. 1.8 - Prob. 11E Ch. 1.8 - Prob. 12E Ch. 1.8 - Prob. 13E Ch. 1.8 - Prob. 14E Ch. 1.8 - Prob. 15E Ch. 1.8 - Prob. 16E Ch. 1.8 - Prob. 17E Ch. 1.8 - Prob. 18E Ch. 1.8 - Prob. 19E Ch. 1.8 - Prob. 20E Ch. 1.8 - Prob. 21E Ch. 1.8 - Prob. 22E Ch. 1.8 - Prob. 23E Ch. 1.8 - 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the expression 20 ( − 45 ) ÷ ( − 9 ) .

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To simplify : the expression 20(−45)÷(−9) .

Answer to Problem 34E

Explanation of Solution

Chapter 1 Solutions

Additional Math Textbook Solutions