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 area covered by the fungus when t = 0? What area does it cover at the end of 5 weeks?

Answer 1

The area covered is 0 square inches at t = 0 and 5.2π square inches at t = 5.
The area covered is square inches at t = 0 and 4.2π square inches at t = 5.
The area covered is square inches at t = 0 and 5.2π square inches at t = 5.
The area covered is square inches at t = 0 and 10.4π square inches at t = 5.
The area covered is square inches at t = 0 and 5.2π square inches at t = 5.

Answer 2

are the answers

Answer 3

the blanks are: pi/4, pi/4, pi/6, and pi/2 (in the answers above)
...sorry copied and pasted

Answer 4

Area = pi (d/2)^2

Answer 5

depending on how yout equation is actually spose to look: id say its something like this:

Area = pi*[(6- 50/t^2+10)/2]^2

Answer 6

i am confused

Answer 7

all i did was take the function they gave you for the diameter and plugged it straight into the Area function of a circle

Answer 8

i got know idea what you equation is spose to look like, but thats on your end. plug in the t values and itll just sipt out an answer

Answer 9

i know i know but the answers they are giving me are confusing me

Answer 10

what do you get for d when t = 0?

Answer 11

1

Answer 12

yeah, 1 is good; now use that in the circles area formula

A = (d/2)^2 pi

A = 1/4 pi

Answer 13

3.14/4 = .78 something or another right?

Answer 14

or .25 pi

Answer 15

yes

Answer 16

what you get for t=5 in the diamter equation?

Answer 17

4.6 or something like that

Answer 18

i rounded

Answer 19

k, and 4.6/2 = 2.3

(2.3)^2 = 5.29 or so, so at t=5 the area = 5.29 pi or so

Answer 20

its either the middle or the last, since yout t=0 stuff got eaten

Answer 21

yeah that is right.

Answer 22

how did u get 1/4 for the first thing again?

Answer 23

nevermind

Answer 24

did you figure it out?