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2 years ago

Any cu.te g irl wan.na b.e my fr.iend

Advanced Placement (AP)

2 answers:

aev [14]2 years ago

5 0

Answer:

thanks for extraaaaaa points

misty here

ozzi2 years ago

3 0

Thanks for the points.

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Explain how poverty plays a role in U.S. politics. Should play a stronger or weaker role in politics?

aivan3 [116]

Poverty plays a controversial role in U.S as some politicians would increase or cut taxes on benefits/aids for others. I know it’s not that much but hope this helps

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3 years ago

What are some similarities between rostow’s model and wallerstein’s model?

ahrayia [7]

Answer:-Theory proposes the world system as a set of mechanisms which distributes resources from the periphery to the core. -Wallerstein stated that the core is the more developed, industrial part of the world, and the periphery is the raw materials-exporting, poor part of the world.

Explanation:

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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>

8 0

1 year ago

Please help, I don’t understand

kolbaska11 [484]

Answer: the hawk will eat the snake

Explanation:

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3 years ago

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Which of the following shows the factors that contribute to land degradation in the order in which they most often occur?

melamori03 [73]

Answer: C. II, IV, I, III

Explanation:

Just completed the Apex quiz.

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2 years ago

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