19539 bacteria will be present after 18 hours
<u>Solution:</u>
Initial value of bacteria N = 6000
Value after 4 hours
= 7800
<em><u>The standard exponential equation is given as:</u></em>
![N=N_{o} E^{-k t}](https://tex.z-dn.net/?f=N%3DN_%7Bo%7D%20E%5E%7B-k%20t%7D)
where
N is amount after time t
No is the initial amount
k is the constant rate of growth
t is time
Plugging in the values in formula we get,
![7800 = 6000E^{-4k}](https://tex.z-dn.net/?f=7800%20%3D%206000E%5E%7B-4k%7D)
Solving for "k" we get,
![\frac{13}{10}=E^{-4k}](https://tex.z-dn.net/?f=%5Cfrac%7B13%7D%7B10%7D%3DE%5E%7B-4k%7D)
Taking "ln" on both sides, we get
![ln\frac{13}{10} = -4k](https://tex.z-dn.net/?f=ln%5Cfrac%7B13%7D%7B10%7D%20%3D%20-4k)
![\frac{ln\frac{13}{10}}{4} = -k](https://tex.z-dn.net/?f=%5Cfrac%7Bln%5Cfrac%7B13%7D%7B10%7D%7D%7B4%7D%20%3D%20-k)
On solving for ln, we get k = -0.0656
The equation becomes, ![N = 6000E^{0.0656t}](https://tex.z-dn.net/?f=N%20%3D%206000E%5E%7B0.0656t%7D)
Now put "t" = 18,
![N = 6000E^{0.0656 \times 18}\\\\N = 6000 \times 3.256 = 19539](https://tex.z-dn.net/?f=N%20%3D%206000E%5E%7B0.0656%20%5Ctimes%2018%7D%5C%5C%5C%5CN%20%3D%206000%20%5Ctimes%203.256%20%3D%2019539)
Hence the bacteria present after 18 hours is 19539
Answer:
17.22 mpg................................
Answer:
X=8
Step-by-step explanation:
24×10=30X
240=30X
Divide by 30
X=8
Answer:
Option A is correct, i.e. 3 mph.
Step-by-step explanation:
Tinh can row at a rate of 6 mph in still water.
Suppose the speed of the river current is X mph.
Then Upstream speed = (6-X) mph.
And Downstream speed = (6+X) mph.
It takes her 2 hours to row upstream from a dock to a park. She then rows back to the dock, and it only takes 40 minutes.
It means Distance upstream = Distance downstream.
Speed upstream x Time upstream = Speed downstream x Time downstream.
(6-x) * 2 = (6+x) * 40/60
12 - 2x = 4 + 2x/3
12 - 4 = 2x/3 + 2x
8 = 2x/3 + 6x/3 = 8x/3
x = 3 mph.
Hence, option A is correct, i.e. 3 mph.
C. is the correct answer because (x+3)^2 is x^2+6x+9