Why does Yep feel like an outsider?

Yep feels like an outsider for lots of reasons. For one thing, as a Chinese American he does not feel completely accepted in society. No matter where he goes or what he does, he never seems to fit in anywhere. Yep was raised in a predominantly black neighborhood; at the same time, he is too American to fit easily into Chinatown and too Chinese to fit in anywhere else. There is also the inevitable generation gap between himself and his hardworking immigrant parents. Yep's parents spend every hour that God sends running a grocery store, which leaves little time for them to devote to their son. Because Yep cannot speak Chinese, he is unable to connect with his grandmother, Marie Lee, who leads a traditional Chinese lifestyle. It is no wonder that Yep feels like such an outsider.

What was Samson Occom's point of view in From a Short Narrative of My Life?

Samson Occom uses the first person point of view in his story.
The first person point of view lends immediacy to our reading experience. In the narrative, Occom speaks earnestly about his faith in God and his strong desire to educate the Native children entrusted to his care. He also relates with strong emotion his dismay at the treatment he received from the Church.

Now you See what difference they made between me and other missionaries; they gave me 180 Pounds for 12 years Service, which they gave for one years Services in another Mission, — In my Service (I speak like a fool, but I am Constrained) I was my own Interpreter. I both a School master and Minister to the Indians, yea I was their Ear, Eye & Hand, as Well as Mouth. . . what can be the Reason that they used me after this manner?

In the above quote, Occom tells us that he "speaks like a fool." He does not claim to be wiser than the Church leaders, and his humility is apparent. However, his distress is clear; he questions how the religious authorities could have treated a fellow Christian with such shameful disregard. The grave disparity in pay causes Occom great pain. 
He voices his anguish at being blamed for being "extravagant." His narrative is filled with earnest, powerful sentiments:

I Can’t Conceive how these gentlemen would have me Live. I am ready to (forgive) their Ignorance, and I would wish they had Changed Circumstances with me but one month, that they may know, by experience what my Case really was; but I am now fully convinced, that it was not Ignorance.

Through the first person point of view, Occom is able to focus our attention on the injustices he endured as an Indian missionary.
https://literarydevices.net/point-of-view/

Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 22

Given: f(x)=2x^3-6x,[0, 3]
Find the critical values for x by setting the derivative equal to zero and solving for the x value(s).
f'(x)=6x^2-6=0
6(x^2-1)=0
6(x+1)(x-1)=0
x=-1, x=1
The critical values for x are x=1 and x=-1. Plug in the critical value(s) and the endpoints of the interval into f(x). Because x=-1 is not in the interval [0, 3], it is not necessary to plug in the x=-1
f(x)=2x^3-6x
f(0)=2(0)^3-6(0)=0
f(1)=2(1)^3-6(1)=-4
f(3)=2(3)^2-6(3)=36
Examine the f(x) values to determine the absolute extrema.
The absolute minimum value is the point (1, -4).
The absolute maximum value is the point (3, 36).

Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 36

Determine the integral $\displaystyle \int \frac{\sin \Phi}{\cos^3 \Phi} d \Phi$


$
\begin{equation}
\begin{aligned}

\int \frac{\sin \Phi}{\cos^3 \Phi} d \Phi =& \int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi

\end{aligned}
\end{equation}
$


Let $u = \cos \Phi$, then $du = - \sin \Phi d \Phi$, so $\sin \Phi d \Phi = -du$. Thus


$
\begin{equation}
\begin{aligned}

\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \int \frac{1}{u^3} \cdot -du
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& - \int \frac{1}{u^3} du
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& - \int u^{-3} du
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{-u^{-3 + 1}}{-3 + 1} + c
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{-u^{-2}}{-2} + c
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{u^{-2}}{2} + c
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2u^2} + c
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2 (\cos \Phi)^2} + c
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2 \cos^2 \Phi} + c
\\
\\
\int \frac{1}{\cos^3 \Phi} \sin \Phi d \Phi =& \frac{1}{2} \sec^2 \Phi + c


\end{aligned}
\end{equation}
$


then


$
\begin{equation}
\begin{aligned}

\int \sec^3 x dx =& \int udv
\\
\\
\int \sec^3 x dx =& uv - \int v du
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \tan x \cdot \sec x \tan x dx
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \sec x \tan^2 x dx
\qquad \text{Apply Trigonometric Identity } \sec^2 x = \tan^2 x + 1
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \sec x (\sec^2 x - 1) dx
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int (\sec^3 x - \sec x) dx
\\
\\
\int \sec^3 x dx =& \sec x \tan x - \int \sec^2 x dx + \int \sec x dx
\qquad \text{Combine like terms}


\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

\int \sec^3 x dx + \int \sec^2 x dx =& \sec x \tan x + \int \sec x dx
\\
\\
2 \int \sec^3 x dx =& \sec x \tan x + \int \sec x dx
\\
\\
2 \int \sec^3 x dx =& \sec x \tan x + \ln (\sec x + \tan x) + c
\\
\\
\int \sec^3 x dx =& \frac{\sec x \tan x + \ln (\sec x + \tan x)}{2} + c


\end{aligned}
\end{equation}
$


@ 2nd term

$\int \sec x dx = \ln (\sec x + \tan x) + c$

Combine the results of the integration term by term


$
\begin{equation}
\begin{aligned}

\int \tan^2 x \sec x dx =& \frac{\sec x \tan x + \ln(\sec x + \tan x)}{2} - \ln (\sec x + \tan x) + c
\\
\\
\int \tan^2 x \sec x dx =& \frac{\sec x \tan x + \ln(\sec x + \tan x) - 2 \ln (\sec x + \tan x)}{2} + c
\\
\\
\int \tan^2 x \sec x dx =& \frac{\sec x \tan x - \ln (\sec x + \tan x)}{2} + c
\\
\\
\text{ or} &
\\
\\
\int \tan^2 x \sec x dx =& \frac{1}{2} (\sec x \tan x - \ln (\sec x + \tan x)) + c


\end{aligned}
\end{equation}
$

Why do Connie and her mother clash?

Connie's mother has a tendency to compare Connie to her older sister, June, and that comparison seems to manifest with Connie being considered inferior.  In the story's first paragraph, Connie's mother asks why Connie cannot keep her room clean like June does, and why she uses stinky hair spray even though June does not.
Connie's vanity is also a sore spot between mother and daughter.  Connie is fond of looking at herself in the mirror, and her mother criticizes her for it, asking "you think you're so pretty?" Connie believes that her mother is jealous of her beauty because she was once beautiful but "now her looks were gone."
On the fateful day of her aunt's barbecue, Connie opts to stay home and wash her hair to let it dry in the sun all day.  When she rolls her eyes at her mother in response to the invitation to go to the barbecue, her mother sharply responds, "stay home alone, then."
Generally speaking, Connie is a typically rebellious teenager who is testing boundaries, and her mother is a parent struggling with the transition. 

In Warriors Don't Cry by Melba Pattillo Beals, what were some problems the characters faced? What were the solutions to the problems?

The main problem that the nine black students face is the resistance of the white students to their presences at the high school. Melba Pattillo Beals names some of her enemies: a boy named Andy and a girl named Sammy Jo Parker, who takes part in an interview with The New York Times in which she says that she suspects that the black students are being paid by the NAACP to attend Central High.
The students contend with the general problem of the white segregationists not understanding that they are not conspiracists but simply fellow teenagers who want to attend a quality school. However, the most pervasive problem is the daily violence to which the black students are subjected. Melba describes instances in which her life is threatened, including an attempt by a group of white girls to set her on fire while she is using a bathroom stall. Worse, many of the white teachers and administrators are unsympathetic to the fears expressed by Melba and the other students and even accuse them of imagining the violent harassment. Thus, the students have to contend with these problems on their own.
To help her maintain her mental strength, Melba depends greatly on her mother and grandmother. She also finds solace in her faith in God and reads about the non-violent civil disobedience of Mahatma Gandhi.

Some specific problems Melba faces throughout the book all trace to the prevailing racist attitude in Little Rock.
Melba is unable to enter the school on her first scheduled day. On her first day at school with the other students, she has to be escorted secretly from the building to avoid an angry mob outside; she even overhears an adult in the principal’s office suggest that one of the students he offered up to appease the mob. After that, Melba has a personal escort Danny, a member of the 101st Airborne Division. Danny saves Melba twice from potentially life-threatening situations. The first of these is a stick of dynamite that is hurled down the stairs at Melba. The second is when Melba gets an unidentified acidic chemical splashed in her eyes. These severe instances of bullying are extreme examples of how the racist attitudes of the city impacted Melba’s daily life.
She also receives little support from the adult teachers. On top of this, Melba has to deal with threats of violence at her home.

In Warriors Don't Cry, the main problem that Melba and the other members of the Little Rock Nine faced was integrating Central High School in 1957-1958. Some of the white students at the school harassed the Little Rock Nine, while others even assaulted them or threw objects at them. In the face of danger, pressure, and harassment, the Little Rock Nine had to get through the school year by employing different solutions. Melba Pattillo, the author of the book, survived by imagining that she was a warrior doing battle in the school. She steeled herself for violence and attacks each day; she could not respond with violence, but she did respond with constant vigilance and mental toughness. The other students handled the violence and harassment in different ways; for example, Minnijean Brown dropped a tray loaded with chili on some students and then was expelled from school for calling some girls "white trash." The other students lasted out the year, and then the school closed rather than integrate.

College Algebra, Chapter 4, 4.4, Section 4.4, Problem 78

Determine all rational zeros of the polynomial $P(x) = 2x^4 + 15x^3 + 31x^2 + 20x + 4$ and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Rule of Signs, the quadratic formula or other factoring techniques.


So, $\displaystyle \frac{-1}{2}$ is a zero and $\displaystyle P(x) = \left( x + \frac{1}{2} \right) \left( 2x^3 + 14x^2 + 24x + 8 \right)$. We continue by factoring the quotient, the possible rational zeros of $P$ are $\displaystyle \pm \frac{1}{2}, \pm 1, \pm 2, \pm 4, \pm 8$ we check again the negative candidates first, beginning with the smallest.
Using Synthetic Division,


So $-2$ is a zero and $\displaystyle P(x) = \left(x + \frac{1}{2} \right) (x+2) (2x^2 + 10x + 4)$. We now factor the quotient using quadratic formula.

$
\begin{equation}
\begin{aligned}
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
x &= \frac{-10 \pm \sqrt{(10)^2 - 4(2)(4)}}{2(2)}\\
\\
x &= \frac{-5\pm\sqrt{17}}{2}
\end{aligned}
\end{equation}
$

Therefore, $\displaystyle P(x) = \left( x + \frac{1}{2} \right) (x + 2)\left( x + \frac{5+\sqrt{17}}{2}\right) \left( x + \frac{5-\sqrt{17}}{2}\right)$