Suppose the larger pump alone can empty the tank in L hours, and the smaller pump can finish the job in S hours, then each hour the large pump empties 1/L portion of the tank, and the small pump empties 1/S per hour
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
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%
ANSWER:
15
32
31
10
23
6.5
35
4.8
63
66
200
72
Solution:
Say you have a triangle that is 2.5 by 1
The Number that comes first is the base and there’s only one bass and there’s two sides in those size are 1 so 1×2 is 2+2.5=4.5 so all you’re doing is adding up all the sides together on any shape
