Advanced Placement (AP), 11.06.2020 03:57 avavackity

when a standard user attempts to install software on her computer, what prevents her from installing software that might change the windows registry?

Answers: 3

Show answers

Answers

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

Answer from: Quest

The criminological theory that alan used to address the problem is routine activity theory. in this situation, the community was not supported by the political party. as a result, alan was forced to resort to crime because of the status that his family is currently in.

Answer from: Quest

Horizontal asymptote at $y=-k,\ k> 0$ tells us that the original graph has been translated down.since $0\ \textless \ a\ \textless \ b$ , $a, b$ are positive. also we can say that the point $(1,a)$ has a smaller $y$ -coordinate than $(3, b)$ .this means that the transformed exponential function is increasing as it goes to the right. as it goes from $x=1$ to $x=3$ , the value of the exponential function increases. it goes from $a$ to something bigger, $b$ . therefore, as $x \to \infty,\ y \to \infty$ . the left hand would be the side that approaches the horizontal asymptote.so the answer is d

Other questions on the subject: Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 01:00, cupcake20019peehui

Select each of the paced steps below. define the problem. conduct an internet search for information. list the alternatives. select the criteria. evaluate the alternatives. make a decision. evaluate the decision.

Answers: 3

continue

Advanced Placement (AP), 23.06.2019 19:30, gafran

Want fre points & brainliest? answer this drivers ed question correctly and i got you : ) where do you not need to look before you change lanes or pass another vehicle? a. out your window toward your target lane b. in your rearview mirror at cars behind you c. over your shoulder, at your blind spot d. inside the vehicle you plan on passing

Answers: 2

continue

Advanced Placement (AP), 23.06.2019 20:00, manueelaa

With my drivers ed ! another name for dash lights is a. vehicle problem indicators b. vehicle warning indicators c. vehicle status indicators d. vehicle caution indicators

Answers: 2

continue

Advanced Placement (AP), 23.06.2019 20:00, kluke7170

*i will mark you brainliest plus 10 free points * which of the following is not a basic defensive driving skill? a. a solid grasp on the instruments and controls at your disposal. b. the pre-drive procedures and adjustments to your ride. c. scanning for hazards and adequate space management. d. the post-drive procedures and adjustments during your ride. e. all of the above

Answers: 1

continue

You know the right answer?

when a standard user attempts to install software on her computer, what prevents her from installing...