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Murrr4er [49]
3 years ago
14

A rectangular piece of metal is 15 in longer than it is wide. Squares with sides 3 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 1350 incubed​, what were the original dimensions of the piece of​ metal? What is the original​ width? nothing in
Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

Step-by-step explanation:

Let

x ----> the original length pf the piece of metal

y ----> the original width pf the piece of metal

we know that

x=y+15 ----> equation A

The volume of the box is equal to

V=LWH

we have

V=1,350\ in^3

L=(x-3-3)=(x-6)\ in\\W=(y-3-3)=(y-6)\ in

H=3\ in

substitute in the formula of volume

1,350=(x-6)(y-6)(3) ----> equation B

substitute equation A in equation B

1,350=(y+15-6)(y-6)(3)

solve for y

450=(y+9)(y-6)

y^2-6y+9y-54=450\\y^2+3y-504=0

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=21

Find the value of x

x=21+15=36

therefore

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

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