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2 years ago

True or False: Concurrent enrollment course grades will appear on high school transcripts but not college transcripts.

Advanced Placement (AP)

2 answers:

Greeley [361]2 years ago

8 0

False i think... but idek honestly

quester [9]2 years ago

6 0

I think false because most things appear I college transcripts rather than the high school ones

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Panuto: Piliin ang titik ng tamang sagot at isulat sa patlang.

rusak2 [61]

Answer:

1.A.sanglaan

2.A.monopolyo

3.B.advertisement

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2 years ago

Do students that are planning to apply to an early college high school need to pass an SAT/ACT to get in?

Delicious77 [7]

Not necessarily, it all depends on the schools that require it

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2 years ago

Read 2 more answers

What trend are we seeing with glaciers over the last century

Andreas93 [3]

There are more glaciers melting around the world due to global warming.

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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!

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

(Hi please help)Text of the Gettysburg Address (November 1863):

Romashka [77]

Answer:

Explanation:

The Gettysburg Address was delivered by Lincoln on November 19, 1863. Text of Abraham Lincoln's Gettysburg Address: Fourscore and seven years ago our fathers brought forth on this continent a new nation, conceived in liberty and dedicated to the proposition that all men are created equal.

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2 years ago