as per question statement
we need to set up an equation first
they have given that it is relative to 1200 that means it starts to increase from 1200 at t=0 initially 1200 bacteria were present
we need to find population at t=6
we need to plug t=6 in p(t).
1638.38 bacteria were present at that time t=6
The highest value of the domain of the function called its<u> Maximum value</u>.
<u>Explanation:</u>
The maximum value of a function is the place where a function reaches its highest point, or vertex, on a graph. There is no point above the maximum value of the function. Thus the highest point on the graph is known as the maximum value of the domain of the function.
The maximum value is one of the extreme values of the domain of the function. The other extreme value is known as the minimum value. It is on one side of the graph and the maximum value is on the other side of the graph.
When
, the original equation tells you that you have
You also know that when
, you have
. Substituting everything you know into the differentiated equation, you get
So
is changing at a rate of
.
Answer:
amount need to invest in a month = $
40.84
Step-by-step explanation:
given data
Future value A = $500
interest rate r = 4% = 0.04
Time period t = 6 month = 0.5 year
solution
we get here future value formula that is
A = .....................1
here n is 12 in a year and r is are and t is time period
put here value and we get
$500 =
solve it we get
P = $490.11
so amount need to invest in a year= $490.11
and amount need to invest in a month =
amount need to invest in a month = $
40.84