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April 18, 2014

April 18, 2014

Posted by **Audrey** on Monday, January 21, 2013 at 8:06pm.

- College Algebra -
**Reiny**, Monday, January 21, 2013 at 8:47pm"The load L a horizontal beam can safely support varies jointly as the width w and the square of the depth d and inversely as the length l"

---- >Load = k(wd^2/L)

given :

L = 12, w = 6inches or 1/2 , d = 8 inches or 2/3, load = 600

600 = k(1/2)(4/9) /12

(1/54)k = 600

k = 32400

Load = 32400(wd^2/L)

for the case L = 14 , w = 11/24 , d = 1/3

Load = 32400(11/24)(1/9)/14 = 117.857 or 118 to the nearest pound

check my arithmetic, easy to make a mistake on this one.

- College Algebra -
**Riggins**, Wednesday, January 23, 2013 at 11:23amShow me the steps in doing this problem

30ax-15ay+42x-21y

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