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
9514 1404 393
Answer:
Step-by-step explanation:
A recursive formula consists of two parts:
- initialization (rule for the first term(s))
- rule for the next term
When we look at the differences between terms in the sequence 3, -4, -11, ..., we find that they are constant at -7. That is each term can be found from the previous one by subtracting 7. This is our recursive rule. The first term is obviously 3, so the recursive formula is ...
a[1] = 3
a[n] = a[n-1] -7
Answer:
The volume of the tumor experimented a decrease of 54.34 percent.
Step-by-step explanation:
Let suppose that tumor has an spherical geometry, whose volume (
) is calculated by:
Where 
is the radius of the tumor.
The percentage decrease in the volume of the tumor (
) is expressed by:
Where:
- Absolute decrease in the volume of the tumor.
- Initial volume of the tumor.
The absolute decrease in the volume of the tumor is:
The percentage decrease is finally simplified:
![\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%](https://tex.z-dn.net/?f=%5C%25V%20%3D%20%5Cleft%5B1-%5Cleft%28%5Cfrac%7BR_%7Bf%7D%7D%7BR_%7Bo%7D%7D%5Cright%29%5E%7B3%7D%20%5Cright%5D%5Ctimes%20100%5C%2C%5C%25)
Given that 
and 
, the percentage decrease in the volume of tumor is:
![\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%](https://tex.z-dn.net/?f=%5C%25V%20%3D%20%5Cleft%5B1-%5Cleft%28%5Cfrac%7B0.77%5Ccdot%20R%7D%7BR%7D%5Cright%29%5E%7B3%7D%20%5Cright%5D%5Ctimes%20100%5C%2C%5C%25)
The volume of the tumor experimented a decrease of 54.34 percent.
