1) area of a cicular pool
area of a circle=πr^2 d=2.5 then r=2.5/2
A=3.14 *(2.5/2)^2
A=4.906m^2 A=4.9m^2

2) arae of a coin
<span>(if i read it right d=15mm?)
</span>A=πr^2 d=15 then r=15/2
A=3.14*(15/2)^2 =176.6mm^2

Rest all are same problems u try to solve it

4 0

3 years ago

!!!!HELP NEEDED!!!!

Svetllana [295]

It is true that the product of two consecutive even integers are always one less than the square of their average.

<u>Step-by-step explanation</u>:

Let the two consecutive odd integers be 1 and 3.

The product of 1 and 3 is (1 $\times$ 3)=3
The average of 1 and 3 is (1+3)/2 =4/2 = 2
The square of their average is (2)² = 4

∴ The product 3 is one less than the square of their average 4.

Let the two consecutive even integers be 2 and 4.

The product of 2 and 4 is (2 $\times$ 4)=8
The average of 2 and 4 is (2+4)/2 =6/2 = 3
The square of their average is (3)² = 9

∴ The product 8 is one less than the square of their average 9.

Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.

4 0

3 years ago

Look at the image. thanks :)

gregori [183]

Answer:

B. (8, 4)

Step-by-step explanation:

Hope this helps!

3 0

2 years ago

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