Advanced Placement (AP), 12.05.2021 14:00 Yskdl

So true and some quotes.. Everybody knows that I'm breaking down
Everybody knows I ain't fakin' now
Everybody knows my heart's vacant now
Yeah, she hates me now, I made mistakes, but now
I don't ever wanna be alone, I don't really ever feel at home
On my own, in the zone
That's the only way I know, feelin' low
'Bout to blow back up, here I go
I will never let the doubt creep in
Gotta pop a couple more Aspirin
I don't think I'll ever let you in
Easier to break it off as friends
I don't really understand myself
I don't really understand, need help
I don't wanna be left on the shelf
Couldn't even hear me if I yelled
It's so cold outside
I'm alone, I'm alright
It's so cold outside
I'm alone, I'm alright
Yeah, let's go
I don't wanna break down, always feel like I could break now
But I never let it take me to that place now
I won't ever let my thoughts get away now
I got better things to do, picking fate now
I just wanna be the best, call me great now
I don't know if I'm okay or insane now
I remember better days on the playground
Hoping I can find my way to a better place now
Even when I'm feeling down, I fight
Even when I don't know what is right
I'ma pick a side, and I'ma take pride
I will decide my fate and I
Will never let them tell me who I am
If you try to shape me, I'll be dam. ned
Planted on the ground is where I stand
Never give up, that was always the plan
It's so cold outside
I'm alone, I'm alright
It's so cold outside
I'm alone, I'm alright

So true and some quotes..

Everybody knows that I'm breaking down
Everybody knows I ain't fakin' n

Answers: 1

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Answer from: enthusiastfranklin

my name is Maxwell Harvey and I have a great day with a kid and their surroundings class work of living organisms and their surroundings class work of living ood morning and I have a great day with a kid and their surroundings class work.

Explanation:

ok cancel the order of operations to calculate the following endangered species of the following endangered species of

Answer from: Quest

General happiness with the work you do

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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So true and some quotes.. Everybody knows that I'm breaking down
Everybody knows I ain't faki...

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