Ok so the radius is half the diameter, and the diameter is half the circle. And the area is the space inside a circle. And pie is 3.14.
So if your radius is 10 then you will multiply 10 x 10 to get 100. Then multiply 100 x pie or 3.14 which you will get, 314cm². So your answer is 314cm²
The formula for an area of a circle is radius x radius = ?
Then multiply pie, 3.14 by your new radius x radius, and you got your answer.
Your drawing was much more helpful and informative than your statements in words and symbols.
I see that you want to evaluate (3/2)^2 times (8/15)^2.
You could combine these two expressions into one, as follows:
3*8
( ----------- )^2
2*15
This, in turn, can be simplified by reduction:
( 4/5 )^2. Expanded, this result gives us 16/25.
Next problem
----------------------------
( 9/4 )^4 * ( 4/3 )^3
9^4 4^3
First, focus on ( ------- ) * ( ----------- )
4^4 3^3
Now reduce 4^3 / 4^4: The end result is 1/4.
Reduce 9^4 / 3^3. To do this, rewrite that 9 as (3^2), resulting in:
3^8 / 3^3. The end result is 3^5.
Putting this expression back together, we get 3^5 / 4 (answer)
the value of x is certainly equal to 3
<h3>Given</h3>
- room height is x feet
- room length is 3x feet
- room width is 3x feet
- a door 3 ft wide by 7 ft tall
<h3>Find</h3>
- The net area of the wall, excluding the door
<h3>Solution</h3>
The area of the wall, including the door, is the room perimeter multiplied by the height of the room. The room perimeter is the sum of the lengths of the four walls.
... gross wall area = (3x +3x +3x +3x)·x = 12x²
The area of the door is the product of its height and width.
... door area = (7 t)×(3 ft) = 21 ft²
Then the net wall area, exclusive of the door is ...
... net wall area = gross wall area - door area
... net wall area = 12x² -21 . . . . square feet
Answer:
One forth of 20 is 5, so there are 5 kindergarten and first grade classrooms.
5 x 2 is 10 so there are 10 second and third grade classrooms.
10 + 5 is 15.
20 - 15 is 5.
There are 5 forth and fifth grade classes.
Step-by-step explanation: