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
First distribute the negative through the parenthesis on the left and the 2 on the right.
-5 -15y +1 = 14y -32 - y
Combine like terms on the right and left
-4 -15y = 13y -32
Now move the variables to one side and the constants to the other.
Subtract 13y from both sides
-4 -28y = -32
Add 4 to both sides
-28y = -28
Divide both sides by -28
y = 1
Answer:
x = 4
Step-by-step explanation:
Answer: 6/14 and 42/99
Step-by-step explanation: