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Rus_ich [418]
3 years ago
5

Urban sprawl tends to delineate age and poverty lines. What additional health risks does this pose to certain populations?

Advanced Placement (AP)
2 answers:
Monica [59]3 years ago
4 0

Answer:

Urban sprawl is the progression of urbanized areas on the outskirts of cities faster than population growth. The main attractions of sub-urban areas compared to city centers are for those who reside there: advantageous cost of land and real estate, the possibility of living in a detached house and a living environment closer to natural environments. It is the improvement of transport conditions and in particular access to the automobile that has enabled this residential urban sprawl. The ease of getting to the city center ever faster and farther and farther thanks to improvements in transport services allows the continuous enlargement of agglomerations.  

This type of housing requires for each resident a greater use of transport, in particular of the automobile, than in the city center. On the other hand, individual housing, which corresponds to the aspirations of many households in sub-urban areas, proportionately emits much more greenhouse gases than collective housing in urban centers.

Urban sprawl thus has serious consequences on the environment and global warming, particularly on local ecosystems.

Lesechka [4]3 years ago
3 0
Many negative consequences for residents and the environment, such as higher water and air pollution, increased traffic fatalities and jams, loss of agricultural capacity, increased car dependency
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leave.  honestly not enough time for that.  leave and dont pay that physician, hes not doing his job right

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2 years ago
Effective and efficient maintenance of records requires records management support personnel within a command to perform which o
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Explanation:

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2 years ago
1. The rate at which people enter a movie theater on a given day is modeled by the function S defined by S(t) = 80 -12 cos 6 The
Arlecino [84]

Hi there!

a.

To find the total amount of people that have ENTERED by t = 20, we must take the integral of the appropriate function.

\text{Amount that entered} = \int\limits^{20}_{10} {S(t)} \, dt \\\\ = \int\limits^{20}_{10} {80 - 12cos(\frac{t}{5})} \, dt

Evaluate using a calculator:

= 899.97 \approx \boxed{900\text{ people}}

b.

To solve, we can find the total amount of people that have entered of the interval and subtract the total amount of people that have left from this value.

In other terms:
\text{Amount of people} = \int\limits^{20}_{10} {S(t)} \, dt - \int\limits^{20}_{10} {R(t)} \, dt

We can evaluate using a calculator (math-9 on T1-84):


\text{\# of people} = \int\limits^{20}_{10} {80-12cos(\frac{t}{5})} \, dt - \int\limits^{20}_{10} {12e^{\frac{t}{10}}+20} \, dt

= 899.97 - 760.49 = 139.47 \approx \boxed{139 \text{ people}}

c.

If:
P(t) = \int\limits^t_{10} {S(t) - R(t)} \, dt

Then:

\frac{dP}{dt}  = P'(t)= \frac{d}{dt}\int\limits^t_{10} {S(t) - R(t)} \, dt  = S(t) - R(t)

Evaluate at t = 20:


S(20) = 80 - 12cos(\frac{20}{5}) = 87.844\\\\R(20) = 12e^{\frac{20}{10}} + 20 = 108.669

S(20) - R(20) = 87.844 - 108.669 = -20.823

This means that at t = 20, there is a <u>NET DECREASE</u> of people at the movie theater of around 20.823 (21) people per hour.

d.

To find the maximum, we must use the first-derivative test.

Set S(t) - R(t) equal to 0:

80 - 12cos(\frac{t}{5}) - 12e^{\frac{t}{10}} - 20 = 0\\\\60 - 12(cos(\frac{t}{5}) + e^{\frac{t}{10}})= 0

Graph the function with a graphing calculator and set the function equal to y = 0:

According to the graph, the graph of the first derivative changes from POSITIVE to NEGATIVE at t ≈ 17.78 hours, so there is a MAXIMUM at this value.

<u>Thus, at t = 17.78 hours, the amount of people at the movie theater is a MAXIMUM.</u>

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