How Do You Feel About Becoming A Social Change Agent Leader And Advocate What Ar

How do you feel about becoming a social change agent, leader, and advocate? What areas of interest do you have that are related to bringing about a society committed to the mental health and welfare of its members? How might you inspire current and future students and colleagues so that they relate to and become engaged in doing something to bring about this vision? How do you plan to remain a competent human services professional?

Human services professionals are committed to understanding and addressing professional and societal issues. As you continue on your journey as a human services professional, it is your responsibility to constantly engage in professional development. This engagement might include continuing education courses, conferences, graduate degrees, certifications, and research, to name a few.

To prepare:

  • Identify at least three personal and/or professional commitments that you are willing to make as a result of your understanding of the need for social change, leadership, and advocacy.
  • Consider steps you might take in your professional development to become a more effective social change agent, leader, and advocate.
  • Consider how your new understanding might impact your future work as a human and social services professional.

With these thoughts in mind:

By Day 4

Post a brief description of at least three personal and/or professional commitments you selected. Explain how your understanding of social change, leadership, and advocacy has impacted your commitments and your willingness to make them. Explain the next steps you might take in your professional development to become a more effective social change agent, leader, and advocate. Then, explain how your commitment might impact your future work as a human and social services professional. Be specific, and provide examples to illustrate your points.

 
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How Do You Find The Explicit Formula And Calculate Term 20 For 3 9 27 81 243

The explicit formula for the progression is ##color(red)(t_n =3^n)## and ##color(red)(t_20 = “3 486 784 401”)##.

This looks like a geometric sequence, so we first find the common ratio ##r## by dividing a term by its preceding term.

Your progression is ##3, 9 , 27, 81, 243##.

##t_2/t_1 = 9/3= 3##

##t_3/t_2 = 27/9= 3##

##t_4/t_3 = 81/27= 3##

##t_5/t_4 = 243/81 = 3##

So ##r = 3##.

The ##n^”th”## term in a geometric progression is given by:

##t_n = ar^(n-1)## where ##a## is the first term and ##r## is the common difference

So, for your progression.

##t_n = ar^(n-1) =3(3)^(n-1) = 3^1 × 3^(n-1) = 3^(n-1+1)##

##t_n =3^n##

If ##n = 20##, then

##t_20 = 3^20 = “3 486 784 401″##

 
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How Do You Express Ln1 5 In Terms Of Ln2 And Ln3

##ln1.5 = ln(3) – ln 2##

##ln1.5 = ln3 – ln 2##

Remember that

##1.5 = 3/2##.

Then

##ln(1.5) = ln(3/2)##

Also, remember that

##ln(x/y) = lnx -lny##

So

##ln(1.5) = ln(3) – ln2##

 
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How Do You Find The Equation Of The Plane In Xyz Space Through The Point P 4 5 4

The answer is: ##5x+3y+4z-51=0##

Given a poiint ##P(x_p,y_p,z_p)## and a vector ##vecv(a,b,c)## perpendicular to the plane, the equation is:

##a(x-x_p)+b(y-y_p)+c(z-z_p)=0##

So:

##-5(x-4)-3(y-5)-4(z-4)=0rArr##

##-5x+20-3y+15-4z+16=0rArr##

##5x+3y+4z-51=0##

 
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How Do You Find The Equation Of A Parabola With Vertex At The Origin And Directr

##y^2=8x##

The standard form of the parabola is ##y^2=4ax##, giving a parabola with its axis parallel to the ##x##-axis, vertex at the origin, focus ##(a,0)## and directrix ##x=-a##. So in your case ##a=2##, giving ##y^2=4ax##.

Alternatively, you can from a definition of a parabola, which is the set of all points ##(x,y)## such that the distance from the point to the directrix ##x=-2## is the same as the distance to the focus (2,0).(The vertex is half-way between the focus and the directrix.)

##(x-(-2))^2=(x-a)^2+y^2####cancel(x^2)+4ax+cancel 4=cancel(x^2)-4ax+cancel 4+y^2####y^2=4ax+4ax=8ax##

 
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How Do You Find The Derivative Of Y Arcsin 2x 1

In these cases you need chain rule .

##dy/(dx)## =##d/(dz## x##dz/dx##===>

It’s better to take ##d/(dz## & ## dz/dx## separately .

##d/(dz##= ##d/(d(2x+1)##.##arcsin (2x+1)##.

##d/(dz##=##1/sqrt(1-(2x+1)^2##= ##1/(2sqrt(-x^2-x)## ——(1)

##dz/dx##= ##d/dx## .## (2x+1)## =##d/dx 2x## = ##2## ——(2)

Then multiply (1) &(2),

So finally you get ##dy/(dx)##= ##1/(2sqrt(-x^2-x)####2##

##dy/(dx)##=##1/(sqrt(-x^2-x)## ; where ##y## = ##arcsin(2x+1)##

 
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How Do You Find The Derivative Of Cotx

##dy/dx = -csc^2x##

##y = cotx##

##y = 1/tanx##

##y = 1/(sinx/cosx)##

##y = cosx/sinx##

Letting ##y= (g(x))/(h(x))##, we have that ##g(x) = cosx## and ##h(x) = sinx##.

##y’ = (g'(x) xx h(x) – g(x) xx h'(x))/(h(x))^2##

##y’ = (-sinx xx sinx – (cosx xx cosx))/(sinx)^2##

##y’ = (-sin^2x – cos^2x)/(sinx)^2##

##y’ = (-(sin^2x + cos^2x))/sin^2x##

##y’ = -1/sin^2x##

##y’ = -csc^2x##

Hopefully this helps!

 
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How Do You Find The Derivative Of Cos Pi X

The derviative is ##-pisin(pix)##

You must use the chain rule to calculate this derivative since there is a function (##pi*x##) inside of another function (the cosine function ##cos(pix)##).

First, take the derivative of the outside function (which is cosine in this case) without touching the inside and then multiply by the derivative of the inside function (##pix## for this problem).

The derivative of ##cos(x)## is ##-sin(x)## and the derivative of ##pi*x## is ##pi## (since ##pi## is a constant, albeit an irrational one) so the derivative for the entire thing is ##-sin (pi*x) * (pi)##.

Hope that helps!

 
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How Do You Find The Derivative Of Cos 2 2x

##y=cos^2(2x)####dy/dx=2cos(2x)*-sin(2x)*2####dy/dx=-4cos(2x)*sin(2x)####dy/dx=-2sin(4x)##

 
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How Do You Find The Critical Points Of A Rational Function

To find the critical points of a function, first ensure that the function is differentiable, and then take the derivative. Next, find all values of the function’s independent variable for which the derivative is equal to 0, along with those for which the derivative does not exist. These are our critical points.

The critical points of a function ##f(x)## are those where the following conditions apply:

A) The function exists.

B) The derivative of the function ##f'(x)## is either equal to 0 or does not exist.

As an example with a polynomial function, suppose I take the function ##f(x) = x^2 + 5x – 7## The derivative of this function, according to the power rule, is the function ##f'(x) = 2*x + 5##.

For our first type of critical point, those where the derivative is equal to zero, I simply set the derivative equal to 0. Doing this, I find that the only point where the derivative is 0 is at ##x = -2.5##, at which value ##f(x) = -13.25##.

For our second type of critical point, I look to see if there are any values of ##x## for which my derivative does not exist. I see there are none, so I am confident in stating that the only critical point on my function occurs at ##(-2.5, -13.25)##

For a slightly more tricky example, we will take the function ##f(x) = x^(2/3)## Differentiation yields ##f'(x) = (2/3)*x^(-1/3)## or ##f'(x) = 2/(3x^(1/3))##. In this example, there are no real numbers for which ##f'(x)=0##, but there is one where ##f'(x)## does not exist, namely at ##x=0##. The original function, however, does exist at this point, thus satisfying condition A from the summary. Therefore, this function possesses a critical point at ##(0, 0)##.

 
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