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
$d=6- \frac{50}{t^{2}+10 }$
$r = \frac{d}{2}= \frac{6- \frac{50}{t^{2}+10 }}{2}$
t = 5
⇒ $r = \frac{6- \frac{50}{t^{2}+10 }}{2} = \frac{6- \frac{50}{5^{2}+10 }}{2} =\frac{6- \frac{50}{25+10 }}{2} =\frac{6- \frac{50}{35}}{2}= \frac{ \frac{6}{1}- \frac{10}{7}}{2}=\frac{ \frac{6*7}{1*7}- \frac{10}{7}}{2}=\frac{ \frac{42}{7}- \frac{10}{7}}{2}$ $=\frac{ \frac{42-10}{7}}{2}=\frac{ \frac{32}{7}}{ \frac{2}{1} }= \frac{32}{7}* \frac{1}{2} = \frac{16}{7}$

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