Room is ´x´m long, ´y´m broad and ´z´m high. What is the area of 4-walls and ceiling of the room?

2xz + 2yz + xy    See Below

Step-by-step explanation:

In this room, you would have one pair of walls that have the same area and a second pair of walls that would have the same area.

The first pair would be the long wall (x) times the height (z).

The second pair would be the broad wall (y) times the height (z)

So you really have just two equations:

x times z = area of one wall

x times z = area of the opposite wall

y times z = area of the other wall

y times z = area of the opposite wall

Total wall area = (x times z) + (x times z) + (y times z) + (y times z)   OR

Total wall area = 2(x times z) + 2(y times z) OR

Total wall area = 2xz + 2yz

The ceiling would be the long wall (x) times the broad wall (y)

Ceiling area = x times y  OR   xy

Total area of all 4 walls and ceiling:

2xz + 2yz + xy

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation: