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kondor19780726 [428]
3 years ago
10

a rectangular piece of metal is 10 in. longer than it is wide. squares with sides 2 in. long are cut from the four corners and t

he flaps are folded upward to form an open box. if the volume of the box is 832 in. cubed, what were the original dimensions of the piece of metal
Mathematics
1 answer:
lions [1.4K]3 years ago
7 0
Volume of the box = length  * width * height

length of original peice = x + 10 , width = x

length of box  (x + 10) - 4  = x + 6
width of box = x - 4
height of box =  2

so we have 

2 * (x -4) * ( x + 6) = 832

2x^2 + 4x - 48 = 832

2x^2 + 4x - 880 = 0
x^2  + 2x - 440 = 0

(x + 22)(x - 20) = 0

x = 20

so width of original piece of metal = 20 ins , length = 30 inches.
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