Step-by-step explanation:
3(p2q + pq2 - pq) - 6pq + 3pq2 =
3[-20 × 2 + (-20) × 2 - (-20)] - 6 × (-20) + 3 × 2 × (-20) =
3[-40 + (-40) + 20] - 6 × (-20) + 3 × 2 × (-20) =
3(-40 - 40 + 20) - 6 × (-20) + 3 × 2 × (-20) =
3(-80 + 20) - 6 × (-20) + 3 × 2 × (-20) =
3 × (-60) - 6 × (-20) + 3 × 2 × (-20) =
-180 + 120 + 6 × (-20) =
-180 + 120 + (-120) =
-180 + 120 - 120 =
-60 - 120 = -180
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
