A stream traverses two lakes flowing downstream, and carrying fresh water as it enters the upper lake. The upper lake contains 4∗109 gallons of water, and the lower lake contains 2 ∗109 gallons of water. The flow rate of the stream is the same at all points and is 4 ∗ 106 gallons per day. A factory situated at the upper lake releases a pollutant at a rate of 200 lbs. per day. Let Q1(t) and Q2(t) be the amount, in pounds, of pollutant in the upper and lower lakes, respectively, where time t is measured in days. Assuming that each lake is well mixed, Q1 and Q2 obey the system(a) Q'1 = 4 ∗ 106 (200 − Q1), Q'2 = 4 ∗ 106 (Q1 − Q2).(b) Q'1 = 200 − Q1/1000, Q'2 = Q1/1000 − Q2/500.(c) Q'1 = 200 − 4 ∗ 106Q1, Q'2 = 4 ∗ 106 (Q1 − Q2).
(d) Q'1 = 200t − 4 ∗ 106Q1, Q'2 = 4 ∗ 106 (Q1 − Q2).

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answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

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