Answer:
a) 1.96cm³
b) 54.32%
Explanation:
From the question, we are told that:
The shape of a cancerous tumor is roughly spherical and has volume
V= 4/3 πr^3
a) When first observed, the tumor has radius 0.73 cm, and 45 days later, the radius is 0.95 cm. By how much does the volume of the tumor increase during this period?
Volume of the tumour at a radius of 0.73cm
= V= 4/3 πr^3
V = 4/3 × π × 0.73³
V = 1.6295105991cm³
Approximately = 1.63cm³
After 45 days, radius of the tumour increased to 0.95cm
Volume of the tumour at radius 0.95cm
V= 4/3 πr^3
V = 4/3 × π × 0.95³
V = 3.5913640018cm³
V = 3.59cm³
The percent increase in the volume of the cancerous tumor
= Volume of the tumor after 45days - Volume of the tumour before 45 days
= 3.59 - 1.63
= 1.96cm³
Therefore, the volume of the tumor increases by 1.96cm³
b) After being treated with chemotherapy, the radius of the tumor decreases by 23%. What is the corresponding percentage decrease in the volume of the tumor?
From the question above, we are told that after chemotherapy, the radius of the tumour decreased by 23%
This means the radius of 0.95 decreases by 23%
Therefore,
23% of 0.95cm
= 23% × 0.95 cm = 0.2185cm
Therefore, the radius of the tumour after chemotherapy = 0.95cm - 0.2185cm
= 0.7315cm
The current volume of the tumour is calculated as
= V= 4/3 πr^3
V = 4/3 × π × 0.7315³
V = 1.6395761818cm³
V = approximately 1.64cm³
Volume of the tumour at 0.95cm³ is 3.59cm³
The decrease = 3.59 - 1.64 = 1.95
Percentage decrease = 1.95/3.59 × 100
= 54.317548747 %
Approximately = 54.32%