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3 years ago

The availability of a wide variety of goods and services is a characterisitic of a

Advanced Placement (AP)

1 answer:

blsea [12.9K]3 years ago

5 0

The answer is command economy.

Explanation:

Command economy is the combination of both communistic and capitalistic economy.

A command economy is a system that determines what are the goods to be produced and how much should be produced and price of the goods offered for sale. They also determine investment and income.

They mostly suffer from problems with poor incentives for planners, managers and workers in state- owned enterprises.

Thus the availability of a wide variety of goods and services is a characteristic of a command economy.

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Solve the following differential equation with initial conditions: y''=e^-2t+10e^4t ; y(0)=1, y'(0)=0

skad [1K]

Answer:

Option A. $y = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Explanation:

This is a second order DE, so we'll need to integrate twice, applying initial conditions as we go. At a couple points, we'll need to apply u-substitution.

Round 1:

To solve the differential equation, write it as differentials, move the differential, and integrate both sides:

$y''=e^{-2t}+10e^{4t}$

$\frac{dy'}{dt}=e^{-2t}+10e^{4t}$

$dy'=[e^{-2t}+10e^{4t}]dt$

$\int dy'=\int [e^{-2t}+10e^{4t}]dt$

Applying various properties of integration:

$\int dy'=\int e^{-2t} dt + \int 10e^{4t}dt\\\int dy'=\int e^{-2t} dt + 10\int e^{4t}dt$

Prepare for integration by u-substitution

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt$ , letting $u_1=-2t$ and $u_2=4t$

Find dt in terms of $u_1 \text{ and } u_2$

$u_1=-2t\\du_1=-2dt\\-\frac{1}{2}du_1=dt$ $u_2=4t\\du_2=4dt\\\frac{1}{4}du_2=dt$

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt\\\int dy'=\int e^{u_1} (-\frac{1}{2} du_1) + 10\int e^{u_2} (\frac{1}{4} du_2)\\\int dy'=-\frac{1}{2} \int e^{u_1} (du_1) + 10 *\frac{1}{4} \int e^{u_2} (du_2)$

Using the Exponential rule (don't forget your constant of integration):

$y'=-\frac{1}{2} e^{u_1} + 10 *\frac{1}{4}e^{u_2} +C_1$

Back substituting for $u_1 \text{ and } u_2$ :

$y'=-\frac{1}{2} e^{(-2t)} + 10 *\frac{1}{4}e^{(4t)} +C_1\\y'=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\$

Finding the constant of integration

Given initial condition $y'(0)=0$

$y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\0=y'(0)=-\frac{1}{2} e^{-2(0)} + \frac{5}{2}e^{4(0)} +C_1\\0=-\frac{1}{2} (1) + \frac{5}{2}(1) +C_1\\-2=C_1\\$

The first derivative with the initial condition applied: $y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\$

Round 2:

Integrate again:

$y' =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\\frac{dy}{dt} =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\dy =[-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int [-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int -\frac{1}{2} e^{-2t} dt + \int \frac{5}{2}e^{4t} dt - \int 2 dt\\\int dy = -\frac{1}{2} \int e^{-2t} dt + \frac{5}{2} \int e^{4t} dt - 2 \int dt\\$

$y = -\frac{1}{2} * -\frac{1}{2} e^{-2t} + \frac{5}{2} * \frac{1}{4} e^{4t} - 2 t + C_2\\y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + C_2$

Finding the constant of integration :

Given initial condition $y(0)=1$

$1=y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + C_2\\1 = \frac{1}{4} (1) + \frac{5}{8} (1) - (0) + C_2\\1 = \frac{7}{8} + C_2\\\frac{1}{8}=C_2$

So, $y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Checking the solution

$y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

This matches our initial conditions here $y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + \frac{1}{8} = 1$

Going back to the function, differentiate:

$y' = [\frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}]'\\y' = [\frac{1}{4} e^{-2t}]' + [\frac{5}{8} e^{4t}]' - [2 t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} [e^{-2t}]' + \frac{5}{8} [e^{4t}]' - 2 [t]' + [\frac{1}{8}]'$

Apply Exponential rule and chain rule, then power rule

$y' = \frac{1}{4} e^{-2t}[-2t]' + \frac{5}{8} e^{4t}[4t]' - 2 [t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} e^{-2t}(-2) + \frac{5}{8} e^{4t}(4) - 2 (1) + (0)\\y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2$

This matches our first order step and the initial conditions there.

$y'(0) = -\frac{1}{2} e^{-2(0)} + \frac{5}{2} e^{4(0)} - 2=0$

Going back to the function y', differentiate:

$y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2\\y'' = [-\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2]'\\y'' = [-\frac{1}{2} e^{-2t}]' + [\frac{5}{2} e^{4t}]' - [2]'\\y'' = -\frac{1}{2} [e^{-2t}]' + \frac{5}{2} [e^{4t}]' - [2]'$

Applying the Exponential rule and chain rule, then power rule

$y'' = -\frac{1}{2} e^{-2t}[-2t]' + \frac{5}{2} e^{4t}[4t]' - [2]'\\y'' = -\frac{1}{2} e^{-2t}(-2) + \frac{5}{2} e^{4t}(4) - (0)\\y'' = e^{-2t} + 10 e^{4t}$

So our proposed solution is a solution to the differential equation, and satisfies the initial conditions given.

7 0

2 years ago

Both South Asia and Western Europe have high population densities. How does the distribution of these two populations differ?

Aleksandr-060686 [28]

South Asia's population mostly lives in the countryside as farmers. Western Europe's population mostly lives in large cities.

Explanation:

Based on the information from the World Bank, about 70% of people living in South Asia have about the third quarter of the population living on agriculture and stays in rural areas. There is about 36% urban ratio in South Asia, with 303 people per square kilometers. On the other hand, the urban ratio in Western Europe is about 79.8%, with a population density of 181 people per square kilometers. Hence, in this case, the right answer is Option E. South Asia's population mostly lives in the countryside as farmers. Western Europe's population mostly lives in large cities.

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3 years ago

Around 20% of all yearly deaths in the United States can be attributed to chronic __________ use.

Zielflug [23.3K]

I think the answer is D. marijuana

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4 years ago

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Step2247 [10]

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