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Suppose you are building an aquariam of volume 2880 in^3 . The aquarium will be 12 in high and the base of the rectangle with the length 4 in. More than twice the width fin the dimensions of base

User Emerino
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2 Answers

4 votes

Answer: 10 inches

Explanation:

User Vyassa Baratham
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8.3k points
6 votes

Answer:

Explanation:

The volume of the aquarium is 2880 in^3.

Height of the aquarium is 4 inches

The base of the aquarium is rectangular in shape.

Let the length be L

Let the width be W

The length of the rectangle is 4 inches more than twice the width. It means that

L = 2W + 4

The volume of the aquarium is expressed as L × W × H

Therefore

L × W × H = 2880

Since H = 4,

4LW = 2880

LW = 2880/4 = 720

LW = 720 - - - - - - - -1

Substituting L = 2W + 4 into equation 1, it becomes

W(2W + 4) = 720

2W^2 + 4W - 720 = 0

W^2 + 2W - 360 = 0

W^2 + 20W - 18W- 360 = 0

W(W +20) - 18(W + 20) = 0

W-18 = 0 or W + 20 = 0

W = 18 or W = -20

Since the width cannot be negative, it is 18 inches

L = 720/W = 720/18

L = 40 inches

Length is 40 inches

Width is 18 inches

User Remiii
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