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An 18 in. by 16 in. rectangular piece of cardboard is used to make an open-top box with a volume of 336 in. by cutting identical squares, of side length x inches, from each corner. A diagram is shown below. 3 18 in. 16 in.

An 18 in. by 16 in. rectangular piece of cardboard is used to make an open-top box with a volume of 336 in. by cutting identical squares, of side length x inches, from each corner. A diagram is shown below. 3 18 in. 16 in.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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An 18 in. by 16 in. rectangular piece of cardboard is used to make an open-top box with a volume of 336 in. by cutting identical squares, of side length x inches, from each corner. A diagram is shown below. 3 ? 18 in. 12. A polynomial equation that can be used to determine the dimensions of the box is A. x(18– 2x)(16 – 2x) + 336 = 0 B. x(18– 2x)(16 – 2x) – 336 = 0 C. 2r(18–x)(16 – x) +336 = 0 D. 2r(18–x)(16 – x) – 336 = 0 16 in.
Transcribed Image Text:An 18 in. by 16 in. rectangular piece of cardboard is used to make an open-top box with a volume of 336 in. by cutting identical squares, of side length x inches, from each corner. A diagram is shown below. 3 ? 18 in. 12. A polynomial equation that can be used to determine the dimensions of the box is A. x(18– 2x)(16 – 2x) + 336 = 0 B. x(18– 2x)(16 – 2x) – 336 = 0 C. 2r(18–x)(16 – x) +336 = 0 D. 2r(18–x)(16 – x) – 336 = 0 16 in.
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