Question

A certain fungus grows in a circular shape. Its diameter in inches after t weeks is given below 6 -(50/t^2+ 10). What is the square inches at the end of 5 weeks?

Answer 1

The answer is 5.2π in² = 16.3 in²

The area (A) of the circle with radius r is: A = r²π
It is known:
π = 3.14

`d=6- (50)/(t^(2)+10 )`

`r = (d)/(2)= (6- (50)/(t^(2)+10 ))/(2)`
t = 5
⇒
`r = (6- (50)/(t^(2)+10 ))/(2) = (6- (50)/(5^(2)+10 ))/(2) =(6- (50)/(25+10 ))/(2) =(6- (50)/(35))/(2)= ( (6)/(1)- (10)/(7))/(2)=( (6*7)/(1*7)- (10)/(7))/(2)=( (42)/(7)- (10)/(7))/(2)`
`=( (42-10)/(7))/(2)=( (32)/(7))/( (2)/(1) )= (32)/(7)* (1)/(2) = (16)/(7)`

A = r²π = (16/7)²π = (2.28)²π = 5.2π = 5.2 · 3.14 = 16.3 in²