A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 2400 incubed, what were the original dimensions of the piece of metal?

22 inch by 52 inch

Step-by-step explanation:

If the Width of the metal=x

The length is 30inch longer than its width= x+30

If Squares with sides 6 in long are cut from the four corners and the flaps are folded.

The length of the pre-folded box =x+30-12=(x+18)inch

The width of the pre-folded box = (x-6-6)inch = (x-12) inch

The height of the flap= 6 inch

Volume of the box =2400 inch cubed

Volume of a Cuboid=LWH

Therefore:

6(x-12)(x+18)=2400

6x²+36x-1296=2400

6x²+36x-1296-2400=0

6x²+36x-3696=0

6(x-22)(28+x)=0

x-22=0 or x+28=0

x=22 inch (Since length cannot be negative)

The original dimensions of the metal sheet are 22 inch by 52 inch

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value