A type of cell triples every hour. If there are 5 cells initially, which function can be used to

find the number of cells, c, at the end of t hours?

Oc (t) = 5 (1)

O c (t) = 5 (3)

Oc (t) = 5 + () 's

Oc (t) = 5 + (3)

T (t) = 5 (3) (t)

Step-by-step explanation:

There are 5 cells initially.

And it triple every hour.

So

T0 = 5

T (t) = T0 (3t)

Each cell has it's multipling factor because it's Will be multipling by 3 every hour

T (t) = 5 (3) (t)