Answer:
here i finished!
hope it helps yw!
Step-by-step explanation:
The doubling period of a bacterial population is 15 minutes.
At time t = 90 minutes, the bacterial population was 50000.
Round your answers to at least 1 decimal place.
:
We can use the formula:
A = Ao*2^(t/d); where:
A = amt after t time
Ao = initial amt (t=0)
t = time period in question
d = doubling time of substance
In our problem
d = 15 min
t = 90 min
A = 50000
What was the initial population at time t = 0
Ao * 2^(90/15) = 50000
Ao * 2^6 = 50000
We know 2^6 = 64
64(Ao) = 50000
Ao = 50000/64
Ao = 781.25 is the initial population
:
Find the size of the bacterial population after 4 hours
Change 4 hr to 240 min
A = 781.25 * 2^(240/15
A = 781.25 * 2^16
A= 781.25 * 65536
A = 51,199,218.75 after 4 hrs
Answer:
GCF(15a; 28b²) = 1
Answer:
x = ± 10
Step-by-step explanation:
Given
x² - 100 = 0 ( add 100 to both sides )
x² = 100 ( take the square root of both sides )
x = ± 
← note plus or minus, hence
x = ± 10
-7=-9+n/8
Add 9 to both sides
2=n/8
Multiply 8 on both sides
Final Answer: 16=n
Answer:
The answer to 20 ÷ 20 is 1
