Advanced Placement (AP), 29.10.2020 17:00 Angel13613

Who Feels this to ? I know you think I got it all figured out

'Cause I walk around like my head's in the clouds

But I'm just a boy with his heart pouring out of his head

I wish that you could see the pain that I've seen

And all of the times I spent being not me

I hope you know that it's not always happy in my head

'Cause I don't know

The perfect road to go down

But I know

I'm trying my best

I'm trying my best to be okay

I'm trying my best but every day it's so hard

And I'm holding my breath

I'm holding my breath 'til I can say

All of the words I wanna say from my heart

If you really wanted I could let you inside

It's been so long and I've got nothing left to hide

Would you believe me if I told you that I've got flaws

Now it's time to let the curtains unfold

And tell all the stories that I didn't want told

I let it out so I unburden my soul I won't stop

'Cause I don't know

The perfect road to go down

But I know

I'm trying my best

I'm trying my best to be okay

I'm trying my best but every day it's so hard

And I'm holding my breath

I'm holding my breath 'til I can say

All of the words I wanna say from my heart

I'm trying my best

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Who Feels this to ?

I know you think I got it all figured out'Cause I walk around like my head's

Answers: 2

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Answer from: Quest

What’s the question exactly? i’d love to assist! would you mind rephrasing it? : )

Answer from: Quest

light-year

the distance that light travels in a vacuum in one year, equal to about 9.46 trillion km (5.88 trillion mi). light-years are used in measuring interstellar and intergalactic distances.

parsec

a unit of distance used in astronomy, equal to about 3.25 light years (3.08 × 1016 meters).

parallax

the apparent angular displacement of a celestial body due to its being observed from the surface instead of from the center of the earth (diurnal parallax or geocentric parallax) or due to its being observed from the earth instead of from the sun (annual parallax or heliocentric parallax)

absolute magnitude

the magnitude (brightness) of a celestial object as it would be seen at a standard distance of 10 parsecs.

apparent magnitude

the magnitude of a celestial object as it is actually measured from the earth.

redshift

an increase in the wavelength of radiation emitted by a celestial body as a consequence of the doppler effect.

the great thing about your imagination is that it doesn't have to make sense! if i were you i would be stuck in space, no communication with earth travelling around and around, you can only guess how long its been since your communication broke down. you land on an undiscovered planet, you do a scientific investigation examining the elements and conditions. you come across alien lifeforms.

best of luck!

Answer from: Quest

Answercommunity of population;

Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

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Who Feels this to ? I know you think I got it all figured out

'Cause I walk around like...

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