Aculture started with 2000 bacteria after 5 hours it grew to 2600

P(t) = P₀ e^(kt) 

Where P₀ is the initial population, 
P(t) is the population after "t" time. 
t is your rate (can be hours, days, years, etc. in this case, hours) 
k is the growth constant for this particular problem. 

So using the information given, solve for k: 

P₀ = 2000 
P(4) = 2600 

P(t) = P₀ e^(kt) 
2600 = 2000e^(k * 4) 
1.3 = e^(4k) 

Natural log of both sides: 

ln(1.3) = 4k 
k = ln(1.3) / 4 

Now that we have a value for "k", use that, the same P₀, then solve for P(17): 

P(t) = P₀ e^(kt) 
P(17) = 2000 e^(17ln(1.3) / 4) 

Using a calculator to get ln(1.3) then to simplify from there, we get: 

P(17) ≈ 2000 e^(17 * 0.262364 / 4) 
P(17) ≈ 2000 e^(4.460188 / 4) 
P(17) ≈ 2000 e^(1.115047) 
P(17) ≈ 2000 * 3.0497 
P(17) ≈ 6099.4 

Rounded to the nearest unit: 

P(17) ≈ 6099 bacteria  hope i could help =)))

40 & 50 in the right corner are the complementary angles as when placed in a sum (added together) they make 90°

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