Arectangular piece of metal is 25 in longer than it is wide. squares with sides 5 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 930 incubed/ in^3, what were the original dimensions of the piece of metal?

The original length was 41 inches and the original width was 16 inches

Step-by-step explanation:

Let

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

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

we know that

When squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box

The dimensions of the box are

The volume of the box is equal to

so

simplify

-----> equation A

Remember that

The piece of metal is 25 in longer than it is wide

so

----> equation B

substitute equation B in equation A

solve for y

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=16

see the attached figure

Find the value of x

therefore

The original length was 41 inches and the original width was 16 inches

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )