Arectangular 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 2706 incubed, what were the original dimensions of the piece of metal?

length=53 in

width= 23 in

Step-by-step explanation:

First, we know that each corner has now a square of 6 in long, so the heigth of the box will be of this size (as you can see in the picture attached). Then, the length and the width of the box would be the same as the piece, but with 12 in less. Therefore the equations to solve the problem are:

We factorize the 6 and we get:

Finally we solve for x, and we get the initial dimensions of the piece of metal:

a negative exponent is translated as follows:

so, we have

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.