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Wire will be used to build the frame (edges) of a square bottomed rectangular box. The frame consists of 8 horizontal edges and 4 vertical edges. 0. W W If the volume of the box is 50 cubic inches, express L, the total length of the wire that is needed to build the frame, as a function of w. 50 w² OL(w) = 8w+ OL(W) = 600 w² OL(w) = 4w + OL(W) = 200 w² OL(W) = 8W + 400 w² h 200 W²

Wire will be used to build the frame (edges) of a square bottomed rectangular box. The frame consists of 8 horizontal edges and 4 vertical edges. 0. W W If the volume of the box is 50 cubic inches, express L, the total length of the wire that is needed to build the frame, as a function of w. 50 w² OL(w) = 8w+ OL(W) = 600 w² OL(w) = 4w + OL(W) = 200 w² OL(W) = 8W + 400 w² h 200 W²

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter3: Polynomial Functions
Section3.5: Mathematical Modeling And Variation
Problem 6ECP: The resistance of a copper wire carrying an electrical current is directly proportional to its...
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Wire will be used to build the frame (edges) of a square bottomed rectangular box. The frame consists of 8 horizontal edges and 4 vertical edges. 0 W W If the volume of the box is 50 cubic inches, express L, the total length of the wire that is needed to build the frame, as a function of w. 50 w² OL(W) = 8w+ OL(W) = 600 w² OL(w) = 4w + OL(W) = 200 w² OL(W) = 8W + 400 w² h 200 W²
Transcribed Image Text:Wire will be used to build the frame (edges) of a square bottomed rectangular box. The frame consists of 8 horizontal edges and 4 vertical edges. 0 W W If the volume of the box is 50 cubic inches, express L, the total length of the wire that is needed to build the frame, as a function of w. 50 w² OL(W) = 8w+ OL(W) = 600 w² OL(w) = 4w + OL(W) = 200 w² OL(W) = 8W + 400 w² h 200 W²
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