Advanced Placement (AP), 14.12.2021 01:10 jacky852

A 40 N box is pushed 5 m up a ramp. The angle of the ramp is 37° above the horizontal. The applied force is parallel to the ramp and opposed by a frictional force of 10.0 N. Find the net increase in potential energy by the crate.

Answers: 1

Show answers

Answers

Answer from: Quest

If you don't know the answer, then don't respond.

Answer from: Quest

To solve this integral, what you can do is split it up as follows: (1) ∫40(x2−4)dx=∫40(x2)dx−∫40(4)dx. to integrate 4 as an indefinite integral, a good rule of thumb is by an equation below for polynomials; . (2) ∫xa=xa+1a+1+c , where a is any number. c denotes some constant that is crucial when doing

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

stop doing work have fun

explanation:

Another question on Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 06:30

An important skill for the beginning home inspector is a. familiarity with reading electric meters. b. good telephone interviewing technique. c. experience in issuing building permits. d. ability to interpret building codes.

Answers: 3

Answer

Advanced Placement (AP), 23.06.2019 12:30

Jacob’s four-month-old infant is crying continuously. while jacob’s wife suggests that he pick up the baby, jacob’s mother thinks that picking up the baby each time she cries is bad for the baby. why is jacob’s wife right? a. babies sleep better when carried. b. babies learn to cry whenever they want someone to hold them. c. babies may choke if crying while lying on their back. d. babies develop a sense of trust that someone is there to provide care.

Answers: 1

Answer

Advanced Placement (AP), 23.06.2019 13:10

You want free brainliest + free points? answer this drivers ed question and you shall receive : )) gas, food, and lodging are in rural areas. a. easily accessible b. always separated c. found everywhere d. less accessible

Answers: 2

Answer

Advanced Placement (AP), 25.06.2019 11:50

Some studies have found that what percentage of students have cheated in some form? a. 10% b. 25% c. 45% d. 75%

Answers: 2

Answer

You know the right answer?

A 40 N box is pushed 5 m up a ramp. The angle of the ramp is 37° above the horizontal. The applied f...

Questions