Advanced Placement (AP), 31.03.2021 17:00 3kp

W T F, hannahjaxon whats your problem man like chill nobody cares you are like the person who snitches on people in class but on branily like it don't matter just let us be

Answers: 1

Show answers

Answers

Answer from: bravoy

lol

Explanation:

Answer from: DragonWarrior203

2 = 2 = 2 = 2 = 2= 2=2=2=2=2=2=2==2=2=2=2=2=2=2=2=2==2=2=2=2=2=2=2=2=2=2=2=2=2==2=2=2=2=2=2==2=2=2=2=2=2=2=2=2=2==2=2=2=2=2=2=2=2=2= im bored

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)

Answer from: Quest

The marshall plan was designed to europe rebuild after wwii

Another question on Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 03:30

Read the statements and determine the correct answer for each blank. microorganisms that cause diseases are known as . they thrive best between the temperatures of °f °f. because are usually around 70°f, it the growth of spoilage-causing bacteria. a)

Answers: 2

Answer

Advanced Placement (AP), 23.06.2019 04:31

Based on your observations of porto alegre and houston, what would you conclude about the longest and shortest days for the two cities on opposite sides of the equator

Answers: 2

Answer

Advanced Placement (AP), 23.06.2019 09:00

You are a livestock epidemiologist working to get pedv under control. you decide to check the place where the current outbreak began. where do you go? you are a livestock epidemiologist working to get pedv under control. you decide to check the place where the current outbreak began. where do you go? ohio north carolina south carolina iowa

Answers: 2

Answer

Advanced Placement (AP), 23.06.2019 15:00

What is the function of the organelle labeled a in the diagram?

Answers: 3

Answer

You know the right answer?

W T F, hannahjaxon whats your problem man like chill nobody cares you are like the person who snitch...

Questions