a. Answer The area of the wall is 98 7 16 square feet. Explanation Given information : The length of the wall is 8 3 4 feet and the height of the wall is 11 1 4 feet. Calculation : Since, the length and height of the wall is provided. Therefore, Area of the wall = Lenght of the wall ⋅ Height of the wall Substitute the provided values in above equation and solve: Area of the wall = 8 3 4 ⋅ 11 1 4 (Substitute) Area of the wall = 35 4 ⋅ 45 4 (Write number as improper fraction) Area of the wall = 35 ⋅ 45 4 ⋅ 4 (Use rule for multiplying fractions) Area of the wall = 1575 16 (Multiply) Area of the wall = 98 7 16 (Write fraction as mixed fraction) Thus, The area of the wall is 98 7 16 square feet. b. To determine To explain : whether there is enough paint available if two coats of paints is applied. Answer The paint is enough to apply two coats. Explanation Given information : The can of paint will cover 200 square feet with one coat. To apply two coats of paint the area covered will be two times the area of the wall. Therefore, Area of the wall covered in two coats = 2 ⋅ Area of the wall Area of the wall covered in two coats = 2 ⋅ 1575 16 Area of the wall covered in two coats = 196 7 8 sqaure feet Since, the paint in the can will cover 200 square feet which is greater than 196 7 8 sqaure feet . Thus, the paint is enough to apply two coats of paints. The paint is enough to apply two coats.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Question

Chapter 5.4, Problem 21E

To determine

To find : the area of the wall.

Expert Solution

Answer to Problem 21E

The area of the wall is 98716 square feet.

Explanation of Solution

Given information :

The length of the wall is 834 feet and the height of the wall is 1114 feet.

Calculation :

Since, the length and height of the wall is provided. Therefore,

Area of the wall=Lenght of the wall⋅Height of the wall

Substitute the provided values in above equation and solve:

Area of the wall=834⋅1114 (Substitute)Area of the wall=354⋅454 (Write number as improper fraction)Area of the wall=35⋅454⋅4 (Use rule for multiplying fractions)Area of the wall=157516 (Multiply)Area of the wall=98716 (Write fraction as mixed fraction)

Thus,

The area of the wall is 98716 square feet.

To determine

To explain : whether there is enough paint available if two coats of paints is applied.

Expert Solution

Answer to Problem 21E

The paint is enough to apply two coats.

Explanation of Solution

Given information :

The can of paint will cover 200 square feet with one coat.

To apply two coats of paint the area covered will be two times the area of the wall.

Therefore,

Area of the wall covered in two coats=2⋅Area of the wallArea of the wall covered in two coats=2⋅157516Area of the wall covered in two coats=19678 sqaure feet

Since, the paint in the can will cover 200 square feet which is greater than 19678 sqaure feet . Thus, the paint is enough to apply two coats of paints.

The paint is enough to apply two coats.

Chapter 5.4, Problem 20E

Chapter 5.4, Problem 22E

Chapter 5 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 5.1 - Prob. 1C Ch. 5.1 - Prob. 2C Ch. 5.1 - Prob. 3C Ch. 5.1 - Prob. 4C Ch. 5.1 - Prob. 5C Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E

Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - Prob. 49E Ch. 5.1 - Prob. 50E Ch. 5.1 - Prob. 51E Ch. 5.1 - Prob. 52E Ch. 5.1 - Prob. 53E Ch. 5.1 - Prob. 54E Ch. 5.1 - Prob. 55E Ch. 5.1 - 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Prob. 25E Ch. 5.7 - Prob. 26E Ch. 5.7 - Prob. 27E Ch. 5.7 - Prob. 28E Ch. 5.7 - Prob. 29E Ch. 5.7 - Prob. 30E Ch. 5.7 - Prob. 31E Ch. 5.7 - Prob. 32E Ch. 5.7 - Prob. 33E Ch. 5.7 - Prob. 34E Ch. 5.7 - Prob. 35E Ch. 5.7 - Prob. 36E Ch. 5.7 - Prob. 37E Ch. 5.7 - Prob. 38E Ch. 5.7 - Prob. 39E Ch. 5.7 - Prob. 40E Ch. 5.7 - Prob. 41E Ch. 5.7 - Prob. 42E Ch. 5.7 - Prob. 43E Ch. 5.7 - Prob. 44E Ch. 5.7 - Prob. 45E Ch. 5.7 - Prob. 46E Ch. 5.7 - Prob. 47E Ch. 5.7 - Prob. 48E Ch. 5.7 - Prob. 49E Ch. 5.7 - Prob. 50E Ch. 5.7 - Prob. 51E Ch. 5.7 - Prob. 52E Ch. 5.7 - Prob. 53E Ch. 5.7 - Prob. 54E Ch. 5.7 - Prob. 55E Ch. 5.7 - Prob. 56E Ch. 5.7 - Prob. 57E Ch. 5 - Prob. 1PSQ Ch. 5 - Prob. 2PSQ Ch. 5 - Prob. 3PSQ Ch. 5 - Prob. 4PSQ Ch. 5 - Prob. 5PSQ Ch. 5 - Prob. 6PSQ Ch. 5 - Prob. 7PSQ Ch. 5 - Prob. 8PSQ Ch. 5 - Prob. 9PSQ Ch. 5 - Prob. 10PSQ Ch. 5 - Prob. 11PSQ Ch. 5 - Prob. 12PSQ Ch. 5 - Prob. 13PSQ Ch. 5 - Prob. 14PSQ Ch. 5 - Prob. 15PSQ Ch. 5 - Prob. 16PSQ Ch. 5 - Prob. 17PSQ Ch. 5 - Prob. 1MCQ Ch. 5 - 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