Answer:
step 1
Find the average growth per year of the populations of rabbits farm A
Over the first 2 years
for t=0
numbers of rabbits=5 (is not exact)
for t=2
numbers of rabbits=40
average=[40-5]/2---------> 17.5
step 2
Find the average growth per year of the populations of rabbits farm B
Over the first 2 years
for t=0
numbers of rabbits=5 (is not exact)
for t=2
numbers of rabbits=30
average=[30-5]/2---------> 12.5
step 3
Compare the average growth per year of the populations of rabbits on both farms
farm A=17.5
farm B=12.5
the average rate of population growth of rabbits in farm A is greater than average rate of population growth of rabbits in farm B by about 5 rabbits per year.
therefore
the answer is the option C)
the average rate of population growth of rabbits in farm A is greater than average rate of population growth of rabbits in farm B by about 6 rabbits per year.
Step-by-step explanation:
The answer is B) The initial number of bacteria is 57, and the growth rate is 31% per hour.
The expression shown is an <em>exponential expression</em>. An exponential expression usually takes the form
, as this one does. A represents the <em>initial value</em>, in this case-the initial number of bacteria. B represents the <em>growth rate</em>. When it is an increasing growth rate, it will begin with 1, because it will have everything it had before, plus the percent increase, the decimal portion. In this case, a=57, and b=0.31.
Hope this makes sense!
Answer:
the answer to this question is the number 8
Answer:
Step-by-step explanation:
The difference quotient in general looks like
Sometimes the symbol "delta x"
is used instead of h.
For the exponential function, the difference quotient is

Answer: (x,y,z) = (-4,0,-2)
Step-by-step explanation: