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Tems11 [23]
3 years ago
6

What is one difference between AP classes and dual enrollment?

Advanced Placement (AP)
2 answers:
disa [49]3 years ago
7 0

Answer:AP Classes require a certain test score for college credit, while dual enrollment does not.

Explanation:

Marina CMI [18]3 years ago
5 0

Answer:

C. AP is a test only, while dual enrollment is a course

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Rasek [7]

Haha I’m currently working on this right now but this is what I have so far hopefully this helps a lil :)

1. I agree with the statement “sports, just like everything else is political” because throughout these past few years it has been nothing but that. For instance, take the 2021 presidency for example. There were many polls stationed around many stadiums through out the United States as well as voting ads appearing on sports channels.

2. A “paradoxical yet symbiotic relationship” was that NFL and Major League Baseball are trying not to be political but most people think that what they are doing is just that, political.

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5. My view on it is that it’s a good way for players to speak up and bring awareness to the political matters at hand.

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2 years ago
Two landforms are described in the table shown below.
Gekata [30.6K]
The correct answer is A
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3 years ago
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James purchased 3 pounds of steak for $21. Which equation can be used to determine the cost, y, to purchase x pounds of steak?
ra1l [238]
3x=21y I think is how you would do this
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Which set of yearly incomes should be normally distributed? A. Population of entire U.S. B. population of professional baseball
baherus [9]
The answer is none of the above, i dont know how but i was just asked the same question and got it wrong so yeah lol

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Particle x moves along the positive x-axis so that its position at time t ~ 0 is given by x(t) = 5t 3 - 9t 2 + 7. at what time t
Arada [10]

Answer: 1.2 seconds

Explanation:

$$At the farthest point the velocity of particle will be zero. \ Differentiating the given equation of position and equating it with zero:$$\begin{aligned}x(t) &=5 t^{3}-9 t^{2}+7 \\x^{\prime}(t) &=15 t^{2}-18 t \\0 &=15 t^{2}-18 t \\t(15 t-18) &=0 \\t &=0 \\t &=\frac{18}{15}=1.2 \mathrm{~s}\end{aligned}$$

$$Thus, at $t=1.2 \mathrm{~s}$ particle will be at the farthest point. \\\\\underline{Justification:}\\The particle moves towards the left from $t \in[0,1.2)$. After that particle keeps moving in  the positive x-direction.

5 0
2 years ago
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