A. d=50t B. 50,100,150,200,250 C. 50 days
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
Options :
A. The initial number of bacteria is 7.
B. The initial of bacteria decreases at a rate of 93% each day.
C. The number of bacteria increases at a rate of 7% each day.
D. The number of bacteria at the end of one day is 360.
Answer:
C. The number of bacteria increases at a rate of 7% each day.
Step-by-step explanation:
Given the function :
f(x)=360(1.07)^x ; Number of bacteria in sample at the end of x days :
The function above represents an exponential growth function :
With the general form ; Ab^x
Where A = initial amount ;
b = growth rate
x = time
For the function :
A = initial amount of bacteria = 360
b = growth rate = (1 + r) = 1.07
If ; (1 + r) = 1.07 ; we can solve for r to obtain the daily growth rate ;
1 + r = 1.07
r = 1.07 - 1
r = 0.07
r as a percentage ;
0.07 * 100% = 7%
U would subtract 2 an 3 from 56 using a variable x
56-2-3=x
X=51