algebra

A rectangle has a width of 8 meters. The length is twice as long as the width. What is the length of the diagonal? Show your work

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1. W = 8
L = 16

x^2 = w^2 + L^2
x^2 = 8^2 + 16^2 = (2^3)^2 + (2^4)^2
x^2 = 2^6 + 2^8
x^2 = 2^6 + 4*2^6
x^2 = 5 * 2^6
x^2 = sqrt 5 * 2^3
x = 8 sqrt 5

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posted by Damon
2. The diagonal is c.

a^2 + b^2 = c^2

8^2 + 16^2 = c^2

64 + 256 = c^2

320 = c^2

17.889 = c

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