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 3354 incubed, what were the original dimensions of the piece of metal?

Answer = 55,25 inches

Solution -

let's take x as length and y as width of the metal piece. As per the question x is 30 more than y,

⇒ x = y + 30

Then four square pieces of side 6 are cut from each corner,

so the new length and width are

x-12 , y-12

Then the volume of the new box created will be

(x-12)(y-12)6

in the question the volume of the given figure is given to be 3354

so (x-12)(y-12)6 = 3354

putting the value of x in the the above equation

⇒ (y+30 - 12)(x-12) = 3354/6 = 559

⇒ (y+18)(y-12) = 559

⇒ y² + 6y - 775 = 0

⇒ y² + 31y - 25y -775 = 0

⇒ (y+31)(y-25) = 0

⇒ y = -31, 25

as length can not be -ve , so y = 25

then x = 25+30 = 55

Hence the dimensions of the metal piece are 55, 25 inches

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step-by-step explanation:

the answer for this question is : 360°

step-by-step explanation:

we have been given a triangle   we need to tell which one of these is a triangle below to map onto itself.

360° rotation is the complete rotation which eventually leads to the same figure which is given.

the correct will be, that is 360° that will cause the triangle below to map onto itself.