Advanced Placement (AP), 10.04.2021 01:50, cadez

“[Case Study 1]: Improving Access to Safe Water- Public and Private Alliance in Guinea ‘In the late 1980s less than 15% of the population in Guinea had access to safe water. By 1996, that share had increased almost fourfold to 55%. In a little more than a decade, Guinea had brought one of the least developed water supply services in Sub-Saharan Africa to the point at which it could provide safe water to more than half the population. These significant achievements are the result of a public-private alliance that originated when Guinea transferred responsibility over the water supply agency to an autonomous company that is partly government owned.’

[Case Study 2]: Successful Air Quality Management in Chile

‘Recently, the government has been quite successful in combating the problem of air pollution in the capital of Santiago based on the Framework Environmental Law, which provides the basis for a gradual improvement in environmental quality, while avoiding conflict and increasing collaboration between industry, government, and pressure groups. Private and public transport fleets were revamped, fuels improved, urban sprawl curbed, and sustainable mechanisms for controlling airborne emissions from industry established. As a result of these efforts, Chile today has good air quality management capabilities.’

[Case Study 3]: Alternatives to Pesticide Use

‘Integrated pest management and biological control have proved to be successful alternatives to pesticides. Crop rotation, intercropping, and the introduction of natural predators inhibit the proliferation of weeds and pests and introduce biological controls that rely on nature’s own checks and balances. This system of integrated pest management has reduced pesticide use in Brazil by 80%, by 90% in China’s Jiangsu Province, and by 30-50% in the province of Orissa in India.’”

Case studies included in an official report of the United Nations on human development, 1998

The agricultural methods implemented in the third case study could be directly applied to reduce the threat from all of the following environmental or economic challenges that arose during the twentieth century EXCEPT

- global economic inequalities

- species loss and declines in biodiversity

- nuclear proliferation

- industrial pollution and greenhouse gas emissions

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answered: Guest

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answered: Guest

Answer effort at inappropriate collaborative activity in which the work submitted to the school does not represent the work of the enrolled student

answered: Guest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

answered: Guest

This statement is true i just took the test