Advanced Placement (AP), 04.04.2020 19:49 JAXKBOII213

1Explain how global trade networks have shifted agriculture to extensive commercial operations. 2Explain how global cities are attracting more international migrants then smaller places. 3Compare the impact of global trade networks on the labor market.

Answers: 2

Show answers

Answers

Answer from: jhenderson2024

1) global trade network has shifted agriculture to extensive commercial operations due to the fact that the trading across borders encourages increase in demand for an agricultural product by ensuring an increased consumers size. This has permitted extensive development and commercialization of agriculture as agricultural goods now find their way into countries they originally were not marketed in prompting methods of mass and timely production of this goods.

2) global cities are attracting more international migrants then smaller places basically and most importantly because of their industrial and socioeconomic development. These developments encourages industrial productivity which can support a growing labor force. Migrant hoping for better quality of life are drawn to these global cities hoping to increase their earning power. Other factors such as security, developed health care system and infrastructure all contribute to migrant influx into global cities.

3) the impact of global trade networks on the labor market is without question one of the most important benefits of global trading. Global trading allows increased productivity of industries. Increase in productivity means the industries can support a more robust and fulfilled labor workforce, therefore, countries that are focused on global trading will have a more fulfilled labor market.

Answer from: Quest

The did multiple things that the economy

Answer from: Quest

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)

Another question on Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 02:00

How does this passage demonstrate the use of propaganda? select two options.

Answers: 3

Answer

Advanced Placement (AP), 23.06.2019 14:00

Which of the following is characteristic of an ethnic religion? a.) the religion has spread far beyond its hearth. b.) the religion is not focused on a particular region. c.) the religion is centered on the message of a prophet. d.) the religion attempts to appeal to people everywhere. e.) the religion does not actively try to convert others

Answers: 2

Answer

Advanced Placement (AP), 24.06.2019 02:00

Drivers must always yield to emergency vehicles traveling

Answers: 1

Answer

Advanced Placement (AP), 25.06.2019 03:30

Why is the road most slick when it first begins to rain? a. oils the road has absorbed begin to float. b. the water has not had time to run off. c. a higher concentration of dirt makes the road more slick. d. none of the above

Answers: 1

Answer

You know the right answer?

1Explain how global trade networks have shifted agriculture to extensive commercial operations. 2Exp...

Questions