Advanced Placement (AP), 30.09.2019 14:50 vava1804

Do you believe that cutting or reducing the amount of pe in schools in favor of core academic subjects is a good idea? why or why not?

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This questions seems to be a question where you are required to mention more than one point. The question has two options of course, I'll tell you my personal answer to the question in broken up points.

I disagree with this statement because I believe that Physical Education (PE) is a very important subject for each individual.
It has been proven that any sort of physical education improves the activity of your brain, if an individual's level of brain activity is higher, it means he learns more from other education and performs better on his tests.
Physical Education is also a practical lesson that relieves stress off students from all the other core subjects that do not that involve as much practical activity.
Physical Education makes you healthier, the healthier you are the better your life style and (technically) life time will be.
Physical Education is also now proven to wake up students, this means students are less lazy and goes back to one of my previous points mentioning a students activity and performance in a lesson or a test.

I hope this helped, any other questions let me know. :3

Answer from: Quest

the answer

1. copyright sole right to publish or sell a form of artistic work

hope it

Answer from: Quest

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

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Do you believe that cutting or reducing the amount of pe in schools in favor of core academic subjec...

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