Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Question

Chapter 2.6, Problem 39E

To determine

An expression in terms of x for the amount of wheat (in tou ) that should be replaced by the horse’s owner; by the cow’s owner.

Expert Solution

Answer to Problem 39E

The amount of wheat that should be replaced by the horse’s owner is 2x tou .

The amount of wheat thatshould be replaced by the cow’s owner is 4x tou .

Explanation of Solution

Given:

A goat, a horse and a cow eats a farmer’s stalk of wheat.

The horse eats twice as many stalks as the goat.

The cow eats twice as many stalks as the horse.

The farmer demands 5 tou of wheat from the owners of the animals.

Consider x amount of wheat (in tou ) is to be replaced by the goat’s owner.

Calculation:

As per the given problem

Consider x amount of wheat (in tou ) is to be replaced by the goat’s owner.

The horse eats twice as many stalks as the goat. Therefore, in terms of x the horse’s owner has to replace 2x tou of wheat.

The cow eats twice as many stalks as the horse.Therefore, in terms of x the cow’s owner has to replace 2⋅2x=4x tou of wheat.

Hence,

The horse’s owner has to replace 2x tou and the cow’s owner has to replace 4x tou of wheat.

Conclusion:

The amount of wheat that should be replaced by the horse’s owner is 2x tou .

The amount of wheat that should be replaced by the cow’s owner is 4x tou .

To determine

The value of x ; the amount of wheat (in tou ) to be replaced by each animal’s owner.

Expert Solution

Answer to Problem 39E

The value of x is 0.71428 tou .

Amount of wheat to be replaced by the goat’s owner = 0.7 tou.

Amount of wheat to be replaced by the horse’s owner = 1.4 tou .

Amount of wheat to be replaced by the cow’s owner = 2.9 tou .

Explanation of Solution

Given:

The goat’s owner has to replace x touof wheat The horse’s owner has to replace 2x tou of wheat.

The cow’s owner has to replace 4x tou of wheat.

The total amount of wheat demanded by the farmer is 5 tou .

Concept used:

The sum of the amounts of wheat to be replaced by each animal’s owner is equal to the amount of wheat demanded by the farmer.

Calculation:

As per the given problem

Amount of wheat to be replaced by the goat’s owner = x tou .

Amount of wheat to be replaced by the horse’s owner = 2x tou .

Amount of wheat to be replaced by the cow’s owner = 4x tou .

The total amount of wheat demanded by the farmer = 5 tou .

The sum of the amounts of wheat to be replaced by each animal’s owner is equal to the amount of wheat demanded by the farmer.

Writing an equation using the above information,

5=x+2x+4x

Adding the like terms on the right hand side of the equation:

5=7x

Dividing each side by 7 ,

57=7x7

Therefore,

x=57=0.71428

Hence,

The value of x is 0.71428 tou .

Amount of wheat to be replaced by the goat’s owner = 0.71428 tou .

Amount of wheat to be replaced by the horse’s owner = 2x = 2⋅0.71428=1.42856 tou .

Amount of wheat to be replaced by the cow’s owner = 4x = 4⋅0.71428=2.857 tou .

Rounding off the answers to the nearest tenth of a tou,

Amount of wheat to be replaced by the goat’s owner = 0.7 tou.

Amount of wheat to be replaced by the horse’s owner = 1.4 tou .

Amount of wheat to be replaced by the cow’s owner = 2.9 tou .

Conclusion:

The value of x is 0.71428 tou .

Amount of wheat to be replaced by the goat’s owner = 0.7 tou.

Amount of wheat to be replaced by the horse’s owner = 1.4 tou .

Amount of wheat to be replaced by the cow’s owner = 2.9 tou .

Chapter 2.6, Problem 38E

Chapter 2.6, Problem 40E

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 - 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