Answer:
A. 7,348
Step-by-step explanation:
P = le^kt
intitial population = 500
time = 4 hrs
end population = 3,000
So we have all these variables and we need to solve for what the end population will be if we change the time to 6 hours. First, we need to find the rate of the growth(k) so we can plug it back in. The given formula shows a exponencial growth formula. (A = Pe^rt) A is end amount, P is start amount, e is a constant that you can probably find on your graphing calculator, r is the rate, and t is time.
A = Pe^rt
3,000 = 500e^r4
now we can solve for r
divide both sides by 500
6 = e^r4
now because the variable is in the exponent, we have to use a log
ln(6) = 4r
we can plug the log into a calculator to get
1.79 = 4r
divide both sides by 4
r = .448
now lets plug it back in
A = 500e^(.448)(6 hrs)
A = 7351.12
This is closest to answer A. 7,348
9514 1404 393
Answer:
1. asymptotes: x=-6, x=6, y=0; zero: x=0
2. 20 blue fish
Step-by-step explanation:
1. The vertical asymptotes are where the denominator is zero. It factors as (x-6)(x+6), so will have zeros at x=-6 and x=6.
The horizontal asymptote is at the limit when x goes to infinity. Here, the ratio of highest-degree terms is 6/x, so the value goes to zero as x gets large.
The zero is where the numerator is zero, at x=0.
In summary:
__
2. 60% are red, so 40% are blue. The difference in percentages is 60%-40% = 20%. The percentage of blue fish (40%) is twice the difference (=2×20%), so the number of blue fish is twice the difference in numbers.
blue fish = 2×10 = 20 fish
<em><u>Check</u></em>
There are then 20+10=30 red fish, so 20+30=50 fish total. Reds are 30/50 = 60% of the total.
<u><em>Additional comment</em></u>
I find it often works well to work with ratios, which I can often do in my head. Other folks like to see equations. Here, we could write equations like ...
r = 60%(r + b) . . . . reds are 60% of the total
r - b = 10 . . . . . . . . 10 more reds than blues
We want the value of b, so we can eliminate r by substituting for it:
r = 10 + b . . . . from the second equation above
(10 +b) = 0.60((10 +b) + b) . . . . substitute for r
10 +b = 6 + 1.2b . . . simplify
4 = 0.2b . . . . subtract b+6
20 = b . . . . . . multiply by 5 (there are 20 blue fish)
