Advanced Placement (AP), 16.10.2020 17:01, markayla61

When a human services professional engages in therapeutic and personal roles with a client, it is considered a(n) relationship.
dual
healing
ethical
O symbiotic

Answers: 2

Show answers

Answers

answered: Guest

you are probably not ugly you might be clingy or annoying. it depends on what gender you are. i can flirt well but can you? ya need to reach out to others and be kind to them and dont be like sex sex sex okay.

answered: Guest

Manual pipetting creates a musculoskeletal load on the neck, shoulders, and upper limbs and the type of injuries that this can lead to is called repetitive stress injuries.

answered: Guest

Answer(32) percent

answered: Guest

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)

Other questions on the subject: Advanced Placement (AP)

Advanced Placement (AP), 23.06.2019 19:30, lwaites18

On a time limit! willing to give free 25 points plus brainliest for this drivers ed question if answered correctly : ) systematic searching is a skill that takes to master. a. no time at all b. repetitive practice c. innate talent d. extra gadgets

Answers: 2

continue

Advanced Placement (AP), 24.06.2019 20:30, yorbal6109

Which presentation option is also known as a screen of a presentation?

Answers: 1

continue

Advanced Placement (AP), 26.06.2019 04:00, devenybates

Indicates a bal of .08 or higher. a. compulsive behavior b. weakened vision c. reduced control of vehicle speed d. loss of judgment

Answers: 1

continue

Advanced Placement (AP), 27.06.2019 08:00, jigaboo420

Maria is late for school sometimes because she is not a morning person. if maria is at school from 8: 00 a. m. to 3: 00 p. m. every day and volunteers at an animal shelter on the weekends, when would be the best time for her to study? a. before school b. right after school c. during the weekend d. before she goes to bed

Answers: 2

continue

Do you know the correct answer?

When a human services professional engages in therapeutic and personal roles with a client, it is c...