Answer:
0.2682 or 26.82%
Step-by-step explanation:
The monthly interest rate (r) that your friend is willing to pay is:
The following expression converts a monthly interest rate (r) into an annual interest rate (i):
![r=\sqrt[12]{1+i}-1](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B12%5D%7B1%2Bi%7D-1)
For r = 0.02:
![0.02=\sqrt[12]{1+i}-1 \\1.02^{12}-1=i\\i=0.2682](https://tex.z-dn.net/?f=0.02%3D%5Csqrt%5B12%5D%7B1%2Bi%7D-1%20%5C%5C1.02%5E%7B12%7D-1%3Di%5C%5Ci%3D0.2682)
Your friend is willing to pay an annual rate of 0.2682 or 26.82%.
Multiply you first equation by 2 to get 4x + 10y = 22
Now the x terms can be eliminated and cancelled out using elimination.
4x + 10y = 22
4x + 3y = 1
To eliminate you need to "subtract", so you need to multiply one of the equations by -1.
4x + 10y = 22
-4x -3y = -1
------------------
0 + 7y = 21
y = 3
Now plug 3 into either one of the equations to get x.
Well i did this in my notes and i found this
-7-13k would be the answer
First we need to find k ( rate of growth)
The formula is
A=p e^kt
A future bacteria 4800
P current bacteria 4000
E constant
K rate of growth?
T time 5 hours
Plug in the formula
4800=4000 e^5k
Solve for k
4800/4000=e^5k
Take the log for both sides
Log (4800/4000)=5k×log (e)
5k=log (4800/4000)÷log (e)
K=(log(4,800÷4,000)÷log(e))÷5
k=0.03646
Now use the formula again to find how bacteria will be present after 15 Hours
A=p e^kt
A ?
P 4000
K 0.03646
E constant
T 15 hours
Plug in the formula
A=4,000×e^(0.03646×15)
A=6,911.55 round your answer to get 6912 bacteria will be present after 15 Hours
Hope it helps!
