Advanced Placement (AP)
22.09.2021 07:52
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Which of the following is most responsible for the change in the emissions of volatile

Which of the following is most responsible for the change in the emissions of volatile organic compounds ( VOC ) from 1970 to 2020 as shown in the graph? A decrease in the number of vehicles fitted with catalytic converters and an increase in the number of hybrid vehicles beginning in 1970. A A decrease in the average driving distance in the United States from 1970 to 2020 that resulted from increased urbanization. B An increase in air emission standards that began in 1970 that regulated corporate average fuel economy ( CAFE ) standards. C An increase in VOC emissions from car exhausts that resulted from increased driving distances from 1970 to 2020.
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asims44
asims44
4,8(41 marks)

The most correct option is;

B. An increase in air emission standard which began in 1970 that regulated corporate average fuel economy standards

Explanation:

In 1966, due to the introduction of the control of pollution from motor vehicle emission standard which required air pumps with a higher efficiency and the 1970 requirements set to ensure automobile manufacturers attain the smog forming combustion control standards evidence has shown modern automobiles emit much less smog forming hydrocarbon and VOCs than they did in the 50s' or 60s'

The Corporate Average Fuel Economy (CAFE) standards came into effect in 1975 improving fuel efficiencies by up to 14%.

trisngle2565
trisngle2565
4,4(25 marks)

The equation of the tangent of g(x)^3 at x = 4 is y = 3 - x

Explanation:

The tangent of y = g(x) = -3·x + 11

Therefore, the slope of g(x) = 1/3

The value of y = -3*4 + 11 = -1

The equation of the line g(x) is given as follows;

y - 1 = 1/3*(x - 4)

y - 1 = 1/3x - 4/3

y = 1/3x - 4/3 + 1 = 1/3x - 1/3

g(x) = 1/3x - 1/3

g(x)^3 = (1/3x - 1/3)^3 = \dfrac{x^3 -3\cdot x^2 + 3 \cdot x - 1}{27}

The slope is therefore;

\dfrac{\mathrm{d} g(x)^{3}}{\mathrm{d} x} = \dfrac{27 \cdot (3 \cdot x^2 -6\cdot x +3 )}{729}

The slope of the tangent is the negative reciprocal of the slope of the line which gives;

Slope \ of \ tangent \ of \ g(x)^3= -\dfrac{729}{27 \cdot (3 \cdot x^2 -6\cdot x +3 )} = -\dfrac{9}{x^2 -2\cdot x + 1}

The value of the slope at x = 4 is  -\dfrac{9}{4^2 -2\cdot 4 + 1} = \dfrac{-9}{9}  = -1

Therefore, we have;

y at x = 4

y = \dfrac{4^3 -3\cdot 4^2 + 3 \cdot 4 - 1}{27} = \dfrac{27}{27} = 1

Therefore, the equation of the tangent is given as follows;

y - 1 =(-1) × (x - 4) = 4 - x

y = 4 - 1 - x = 3 - x

The equation of the tangent of g(x)^3 at x = 4 is y = 3 - x.

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