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
40 + 2x - 30 + x + 20 = 180
3x + 30 = 180
3x = 150
x = 50
m<B = 2x - 30
m<B = 2(50) - 30
m<B = 100 - 30
m<B = 70
Answer:
<em>Solve for b. by simplifying both sides of the inequality, then isolating the variable.</em>
Inequality Form:
Interval Notation:
(
−
∞
,
)
Hope this helps :)
<em>-ilovejiminssi♡</em>
Where is the grid are u gonna take a pic?
Answer:
Step-by-step explanation:
Combine Like Terms:
Therefore, 
will be your final answer.
