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3 years ago

Describe how globalization changes traditional cultures.

Advanced Placement (AP)

1 answer:

ss7ja [257]3 years ago

5 0

Answer:

During globalization, traditional culture is marked by the common consumption of cultures that have been diffused by the Internet, popular culture media, and international travel.

Explanation:

When diffused by the internet and media, they can be shaped and changed by others diffusing the culture. They are no longer traditional in a sense but have change to more international.

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All have binocular vision with fields of view that significantly overlap, resulting in true three dimensional (3-D) depth perception or stereoscopic vision. ... Depth perception is an invaluable tool for animals that need to move quickly. It allows them to judge the distances to important objects in their environment.

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4 years ago

2 ways information can be stored as short-term memory.

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There's only one way

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The duration of short term memory seems to be between 15 and 30 seconds. Items can be kept in short term memory by repeating them verbally (acoustic encoding), a process known as rehearsal.

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3 years ago

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3 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!

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:

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

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):

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

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

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.

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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2 years ago

What determines a person's face shape?

Ierofanga [76]

Their hairline and how long and wide their face is

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3 years ago

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