In a lab experiment, 60 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 21 hours. How long would it be, to the nearest tenth of an hour, until there are 107 bacteria present?

17.5 hours

Step-by-step explanation:

107 = 60(2^n)

n(lg2) = lg(107/60)

n = 0.8345763908 × 21

n = 17.52610421 hours

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4