Answer:
a) 
b) The population estimated in 2008 is of 24651.
Step-by-step explanation:
The population growth model may be given by the following exponential function:
In which P(t) is the population after t years, 
is the initial population and r is the growth rate.
(A.) Find a function that models the population t years after 2000 (t=0 for 2000)
We have that:
. So
(B.) Use the function from part A to estimate the fox population in the year 2008.
2008 is 8 years after 2000. So:
The population estimated in 2008 is of 24651.
First, let's identify the like terms.
2 and 17
2n and 3n
Next, you would need to combine them. Ex: Add 2n to both sides, and subtract 17 from both sides.
2 - 2n = 3n + 17
2 = 5n + 17
-15 = 5n
Now, all you would need to do is isolate the n. To do this, you would divide both sides by 5.
-15 = 5n
-3 = n
n = -3
The solution would be -3.
I hope this helps!
Answer:
3:15, 5:25, 7:35, 8:40, 10:50
Step-by-step explanation:
When I first tried this I thought it was for every one there was fifteen. I messed up then looked, realized, and changed it. For every 3 there is 15 so 15 ÷ 3 = 5 so when I saw the 5 I multiplied it by five to get 25 then continued throughout. Your first step when doing these problems again is to divide (but make sure you divide the right number) then once you find out that sweet number, multiply is by the time to get the photos, and divide by five on the photos to get the time.
<em>Hope you got this right!</em>
The available balance online may not reflect all debit transactions or all checks that were written. Some recent debit transactions and check written may not have been posted to his account yet. The available balance would need to subtract any outstanding debit transactions and any written checks that have not cleared his account. This would yield the correct available balance.
The graph starts flat but then curves steeply upwards. You can tell this by the sudden jumps in y-coordinates, illustrating that it keeps going further up faster and faster as the x-coordinates progress steadily.
