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.
Well, if it is increasing at a fixed rate of 2 mm/s, then the volume would still be increasing at a rate of 2 mm/s when the diameter is 80 mm, if I understand your question correctly.
Answer:
hope it helps
Step-by-step explanation:
Step-by-step explanation: Your answer is in the photo
0 because 0 times anything is 0
