(a) Answer The dimension is 12 in. by 6 in Explanation Given information: A tin box by cutting x inch by x inch pieces of tin off the corners of a rectangle and folding up each side Calculation: The length of original piece of tin is 12 inch and width is 6 inch. Thus, the dimension is 12 in. by 6 in. (b) To determine To write: a function that represents the volume of the box and to find the degree. Answer The degree of function is 3. Explanation Given information: A tin box by cutting x inch by x inch pieces of tin off the corners of a rectangle and folding up each side Calculation: The volume is product of length, width and height, so V = l w h = ( 12 − 2 x ) ( 6 − 2 x ) x = x ( 12 − 2 x ) ( 6 − 2 x ) Therefore, the volume of box is given by V ( x ) = x ( 12 − 2 x ) ( 6 − 2 x ) . The degree of function is 3.

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Question

Chapter 4.2, Problem 56E

(a)

To determine

To find: the dimensions of the original piece of tin.

(a)

Expert Solution

Answer to Problem 56E

The dimension is 12 in. by 6 in

Explanation of Solution

Given information:

A tin box by cutting x inch by x inch pieces of tin off the corners of a rectangle and folding up each side

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 4.2, Problem 56E , additional homework tip 1

Calculation:

The length of original piece of tin is 12 inch and width is 6 inch.

Thus, the dimension is 12 in. by 6 in.

(b)

To determine

To write: a function that represents the volume of the box and to find the degree.

(b)

Expert Solution

Answer to Problem 56E

The degree of function is 3.

Explanation of Solution

Given information:

A tin box by cutting x inch by x inch pieces of tin off the corners of a rectangle and folding up each side

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 4.2, Problem 56E , additional homework tip 2

Calculation:

The volume is product of length, width and height, so

V=lwh=(12−2x)(6−2x)x=x(12−2x)(6−2x)

Therefore, the volume of box is given by V(x)=x(12−2x)(6−2x) . The degree of function is 3.

Chapter 4.2, Problem 55E

Chapter 4.2, Problem 57E

Chapter 4 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.2 - Prob. 1E Ch. 4.2 - Prob. 2E Ch. 4.2 - Prob. 3E Ch. 4.2 - Prob. 4E Ch. 4.2 - 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