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.
when you add two negatives its just like regular number, -3+-5=-8 same as 3+5=8
Answer:
find the classmark of each interval
- forexample
- (140+150)/2
- (150+160)/2
do the same up to (190+200)/2
then
draw a graph by using frequently (number of weeks) on y-axis against classmark on x-axis
Answer:
1/3125 or, 5^-5
Step-by-step explanation:
5^-1 = blank * 5^4
1/5/625= blank or, 5^-1/5^4=blank
blank=1/3125 5^-5=blank
You have sqrt(8), sqrt(18), and sqrt(2).
You need to simplify the radicals.
sqrt(2) is already simplified.
For both sqrt(8) and sqrt(18), you need to factor out the greatest perfect square.
8 = 4 * 2
You can take the square root of 4 and put it outside the root.
18 = 9 * 2
You can take the square root of 9 and put it outside the root.
