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Q: If you used 500 W of power for 24 h you have used?
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Approximate the length and width of a rectangle whose perimeter is 96 feet and whose area is 500 square feet accurate to three decimal places?

Perimeter, P = 96 ft.Area, A = 500 ft^2.A = LW500= LWL = 500/WP = 2L + 2W96 = 2L + 2W substitute 500/W for L;96 = 2(500/W) + 2W divide by 2 to both sides;48 = 500/W + W multiply by W to both sides;48W = 500 + W^20 = 500 + W^2 - 48W or,W^2 - 48W + 500 = 0W = [48 +,- square root of [48^2 - (4)(1)(500)]/2W = [48 +,- square root of (2304 -2000)]/2W = (48 +,- square root of 304)/2W = (48 + 17.436)/2 or W = (48 - 17.436)/2W = 65.436/2 or W = 30.564/2W = 32.718 ft or W = 15. 282 ftL = 500/W = 500/32.718 or L = 500/15.282L = 15.282 ft or L = 32.718 ft