4 A) 1/6
4 B) 1/45
4 C) 2/3
4 D) 3/5
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.
Is there any numbers to answer the question?
Answer:
-3(3)-6=-15
Step-by-step explanation:
Find the possible rational roots and use synthetic division to find the first zero.
I chose x=1 (which represents the factor "x-1")
1║2 -7 -13 63 -45
║ 2 -5 -18 45
2 -5 -18 45 0
(x-1) is a factor, (2x³ - 5x² - 18x + 45) is the other factor.
Use synthetic division on the decomposed polynomial to find the next zero.
I chose x = 3 (which represents the factor "x-3")
3║2 -5 -18 45
║ 6 3 -45
2 1 -15 0
Using synthetic division, we discovered that (x-1), (x-3), & (2x² + x -15) are factors. Take the new decomposed polynomial (2x² + x -15) and find the last two factors using any method.
Final Answer: (x-1)(x-3)(x+3)(2x-5)