Advanced Placement (AP), 16.04.2021 05:20 Lollipop1287

Throughout the book and in these final chapters, Krakauer points out several instances where Chris McCandless was in the vicinity of where he was trying to go or was closer to help than he knew. Why might Krakauer dwell on tracing just how close Chris was to being saved or coming out alive?

Answers: 1

Show answers

Answers

Krakauer is concluding, from what he knows of McCandless, that he is a sane individual, if not brilliant. However, Krakauer had an experience very similar and can relate to McCandless and his hunger for adventure. He does not think McCandless was batty but as an explorer The narrative works almost as a countdown, starting on July 8, 1992 with McCandless’s return to the bus after finding the Teklanika too flooded, pauses the narrative for a discussion of his physical decline, then enumerating each remaining day at a time. His quotations from McCandless’s diary adds to the poignancy and drama of these last days, as does the inclusion of the detail that McCandless wanted to leave the wild. His diary records that he is now trapped, despite his desire to return, lending the chapter a fatal tone at the same time that Krakauer builds narrative urgency.

Explanation:

Answer from: Quest

Hellothe rates at which criminals are rehabilitated into society after their sentencinghope this

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)

Another question on Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 00:00

Which of the following credit scores will most likely get you the least expensive interest rate on a car loan

Answers: 1

Answer

Advanced Placement (AP), 23.06.2019 13:30

Free brainliest + free points if answered ! vehicles are expected to be moving at around on highways. a. 30-50 mph b. 40-60 mph c. 50-70 mph d. 60-80 mph

Answers: 2

Answer

Advanced Placement (AP), 24.06.2019 04:30

Anyone knows answer for code.org unit 3: intro to programming, lesson 7: apis and function parameters, bubble 11. !

Answers: 2

Answer

Advanced Placement (AP), 24.06.2019 16:00

The biome pictured above is found in areas such as kenya and south africa. the land is grass-covered with very few trees and gazelles and other grazers are common here. why does this biome have a low population of trees? a) large grazing animals destroy the trees. b) it is a deforested rainforest with poor soil. c) it has high temperatures and low precipitation. d) it has a long rainy season followed by flooding.

Answers: 1

Answer

You know the right answer?

Throughout the book and in these final chapters, Krakauer points out several instances where Chris M...

Questions