Answer:
I can't tell you that ANSWER because I need to see the answers they gave you to circle or something
is the volume of the sample when the water content is 10%.
<u>Explanation:</u>
Given Data:

First has a natural water content of 25% =
= 0.25
Shrinkage limit, 

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
![V \propto[1+e]](https://tex.z-dn.net/?f=V%20%5Cpropto%5B1%2Be%5D)
------> eq 1

The above equation is at
,

Applying the given values, we get

Shrinkage limit is lowest water content

Applying the given values, we get

Applying the found values in eq 1, we get


Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...
Explanation:
Is the same as the example,
If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...
Then
(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)
Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...
The way to write this is
Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)
(photo)