5.2 x 10-2 and 5.2 x 10-1
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!
Answer:
2(z+3)
Step-by-step explanation:
$2,500 • 7.5% (or 0.075) = $187.50 (for one year).
For two years, the interested added onto the original price is $375 because $187.50 • 2 (years) = $375.
$2,500 + $375 = $2,875
The total of the cost of the surgery and the interest is $2,875.
Hope this helps! ❤️
Answer:
One of it culd be -3 or 3
Step-by-step explanation:
