please anybody can help me to solve this question btw im from malaysia but the questions have the eng subtitle so you can read t
he eng subtitle, no worries! please help me to answer the questions 
1 answer:
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Note: The height of the room must be 3 m instead of 3 cm because 3 cm is too small and it cannot be the height of a room.
Given:
Perimeter of the floor of a room = 18 metre
Height of the room = 3 metre
To find:
The area of 4 walls of the room.
Solution:
We know that, the area of 4 walls of the room is the curved surface area of the cuboid room.
The curved surface area of the cuboid is
Where, h is height, l is length and b is breadth.
Perimeter of the rectangular base is 2(l+b). So,
Putting the given values, we get
Therefore, the area of 4 walls of the room is 54 sq. metres.
Answer:
In the given equation -14(3a+6)=12(6-4a)+12 the value of a is 28
Step-by-step explanation:
Given equation is -14(3a+6)=12(6-4a)+12
To simplify the given equation:
-14(3a+6)=12(6-4a)+12
Taking all terms to one side
-14(3a+6)-12(6-4a)-12=0
-[14(3a+6)+12(6-4a)+12]=0
Now dividing by negative sign on the above equation we get
14(3a+6)+12(6-4a)+12=0 (using distributive property)
42a+84+72-48a+12=0 ( adding the like terms )
-6a+168=0
-6a=-168
6a=168
Therefore a=28
Therefore in the given equation -14(3a+6)=12(6-4a)+12 the value of a is 28