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
Plug 8 for y
8 = 2x + 4
Subtract 4
4 = 2x
Divide by 2
x = 2
Plug 16 for y
16 = 2x + 4
Subtract by 4
12 = 2x
Divide by 2
x = 6
Substitute 20 for y
20 = 2x + 4
16 = 2x
x = 8
Substitute 22 for y
22 = 2x + 4
18 = 2x
9 = x
So the values are 2, 6, 8, 9
Answer:
1-a 2-c hope im right
Step-by-step explanation:
Answer:
4c + 4c + 16c + 90 = 360
Add the like terms
24c + 90 = 360
Subtract 90 when taking it to the other side of the equal
24c = 270
Divide 270 by 24 to get ur answer.
c = 11.25
