The population of an idealized colony of bacteria grows exponentially, so that the population doubles every half-hour. The experiment begins at 6:00pm. If at 6:10pm the population is measured at 20 bacteria, how many will there be at 8:00pm?

at 8:00 pm there will be 254 bacteria

Step-by-step explanation:

the equation governing the population of the colony N in function of time t is

N(t)=N₀*2^(t/30)  , where t is in minutes from 6:00 pm , and N₀ is the initial population

thus for t₁=6:10 pm = 10 min from 6:00 pm and t₂=8:00 pm=120 min from 6:00 pm , we have

N₁=N₀*2^(t₁/30)

N₂=N₀*2^(t₂/30)

dividing both equations

N₂/N₁=2^(t₂/30-t₁/30)

N₂ = N₁*2^[(t₂-t₁)/30]

replacing values

N₂ = N₁*2^[(t₂-t₁)/30] = 20 bacteria * 2^[(120 min-10 min)/30] =253.98 ≈ 254 bacteria

then at 8:00 pm there will be 254 bacteria

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.