In the U.S., the command-and-control environmental laws of the early 1970s, together

The correct answer is C. are given considerable credit for cleaner air and water in recent decades.

Explanation:

The command-and-control environmental laws are a set of policies first proposed in the early 1970s that protected the environment by limiting the pollution levels. Also, the government demanded certain changes in production methods or the use of technologies to reduce pollution.

Moreover, these regulations are considered to be the main factor that contributed to the reduction in air and water pollution because since the laws were approved air and water pollution had decreased in the country. Also, it is believed these laws protected ecosystems and natural resources, which contributes to the conservation of nature. Thus, these laws "are given considerable credit for cleaner air and water in recent decades".

The times are t = 9, t = 10, t = 11 and t = 12

Step-by-step explanation:

For the train to be more than 16 km South and since south is taken as negative,

d(t) > -16

t³ - 9t² + 6t > -16

t³ - 9t² + 6t + 16 > 0

Since -1 is a factor of 16, inserting t = -1 into the d(t), we have

d(-1) = (-1)³ - 9(-1)² + 6(-1)+ 16 = -1 - 9 - 6 + 16 = -16 + 16 = 0. By the factor theorem, t + 1 is a factor of d(t)

So, d(t)/(t + 1) = (t³ - 9t² + 6t + 16)/(t +1) = t² - 10t + 16

Factorizing  t² - 10t + 16, we have

t² - 2t - 8t + 16

= t(t - 2) - 8(t - 2)

= (t -2)(t - 8)

So t - 2 and t - 8 are factors of d(t)

So (t + 1)(t -2)(t - 8) > 0

when t < -1, example t = -2 ,(t + 1)(t -2)(t - 8) = (-2 + 1)(-2 -2)(-2 - 8) = (-1)(-4)(-10) = -40 < 0

when -1 < t < 2, example t = 0 ,(t + 1)(t -2)(t - 8) = (0 + 1)(0 -2)(0 - 8) = (1)(-2)(-8) = 16 > 0

when 2 < t < 8, example t = 3 ,(t + 1)(t -2)(t - 8) = (3 + 1)(3 -2)(3 - 8) = (4)(1)(-5) = -20 < 0

when t > 8, example t = 9,(t + 1)(t -2)(t - 8) = (9 + 1)(9 -2)(9 - 8) = (10)(7)(1) = 70 > 0

Since t cannot be negative, d(t) is positive in the interval 0 < t < 2 and t > 8

Since t ∈ (0, 12]

In the interval 0 < t < 2 the only value possible for t is t = 1

d(1) = t³ - 9t² + 6t = (1)³ - 9(1)² + 6(1) = 1 - 9 + 6 = -2

Since d(1) < -16 this is invalid

In the interval t > 8 , the only possible values of t are t = 9, t = 10.t = 11 and t = 12.

So,

d(9) = 9³ - 9(9)² + 6(9) = 0 + 54 = 54 km

d(10) = 10³ - 9(10)² + 6(10) = 1000 - 900 + 60 = 100 +60 = 160 km

d(11) = 11³ - 9(11)² + 6(11) = 1331 - 1089 + 66 = 242 + 66 = 308 km

d(12) = 12³ - 9(12)² + 6(12) = 1728 - 1296 + 72 = 432 + 72 = 504 km