56. dig deeper! two rectangular, adjacent rooms share a
wall. one-foot-by-one-foot tiles cover the floor of each room.
describe how the greatest possible length of the adjoining
wall is related to the total number of tiles in each room. draw
a diagram that represents one possibility.
pressions and factors

Consider two adjacent rectangular rooms having Length=L, and, Breadth = B

Now Suppose the wall which is in between two rooms has a height or length =H.

Breadth of wall = B [ if the wall doesn't exceed the breadth of room]

Considering two rooms to be identical,

Area of each room= L × B square unit

Area of each tile = 1×1=1 square unit

Number of tiles required= L B ÷ 1= LB tiles( product of length and breadth of room is number of tiles required)

Suppose if,LB= N

B= N/L  (1)

Area of wall(W) = B×H= B H square unit

B =W/H (2)

Equating (1) and (2)

⇒N/L = W/ H

⇒H =w-z/N

⇒H =WL/LH

⇒H = W/B

 ⇒  H =area of wall / Breadth of room and wall

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

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step-by-step explanation: