Where was Naomi‘s mother throughout the novel Obasan?

Obasan is a novel by Joy Kogawa that takes an intensely personal look at the lives of Japanese immigrants during and after the period of time when they were forcibly kept in internment camps in Canada (and the United States) during World War II.
Naomi Nakane is a Japanese Canadian woman who lived with her family in Vancouver, Canada when she was a young child. She and her family, as Japanese immigrants, were forced to leave their homes and live in an internment camp during WWII, and even for some time after the war was over. Naomi and her brother were raised by their Obasan (Japanese for "Aunt") and Uncle, because their father was separated from them for a long time and forced to go to a work camp, where he eventually died from tuberculosis. Naomi's mother went to Japan in 1941 to take care of a sick relative before her family was forced into internment camps and never came back.
As an adult, Naomi searches through her own past. Her family eventually decides to tell her the truth of why her mother never returned from Japan. Naomi's mother, Grandma Kato, and some extended family were in Nagasaki the day the city was bombed by the US Air Force in 1945. Naomi's mother and Grandma Kato survived the nuclear bombing, but her mother was horribly wounded and lost part of her face. In Grandma Kato's letter, she is described as "utterly disfigured." After the bombing, Naomi's mother spent a while recovering in a hospital where she "was expected to die, but she survived." Aunt Emily tells Naomi that "a missionary found [her] name on a plaque of the dead" on a Canadian maple tree. They do not know exactly when she died. Naomi's mother specifically told her relatives that she didn't want Naomi and her brother to know about what happened to her, and so she remained a mystery to her children up until this point.

Can you please analyze the opening lines of To the Lighthouse from "to her son these words . . ." to ". . . him guide his scissors neatly round the refrigerator."

It is interesting that this novel's very first line is actually a piece of dialogue from Mrs. Ramsay, the book's first named character:

"Yes, of course, if it's fine tomorrow," said Mrs Ramsay. "But you'll have to be up with the lark, she added."

The passage you have identified for analysis, then, begins with the phrase, "To her son . . ." At this juncture, we see a reversal of the trend commonly observed in English literature, where a woman is defined by her relationship to a man: James Ramsay, named later in this passage, is referred to only as "he" several times, while defined only by his relationship to Mrs. Ramsay, his mother. The positioning here leads the reader to question, from the beginning of the novel, what kind of dynamic exists between mother and son and if there is a reason the writer chooses to define James Ramsay in this fashion.
Indeed, the attributes described in the passage that follows seem to suggest that James Ramsay has an internal life that is opposed to the "image of stark and uncompromising severity" which his appearance presents to the world; James Ramsay is not the sort of boy his pragmatic father might prefer him to be but a dreamer, in whom the prospect of a longed-for "expedition" can bring forth an "extraordinary joy." This, combined with the fact that his mother's word is, to him, law—the expedition "bound to take place" given her statement—gives some indication of where James's loyalties lie within the household and where he, as a six-year-old child, observes the authority to be.
Whatever the "wonder" is to which James is looking forward, the narrator's tone suggests that, while it has not really been "years and years," it is something of great import to the child. Moreover, James's longings are not classified as being childish; rather, it is almost as if he is precocious in belonging "even at the age of six to that great clan" for whom "any turn in the wheel or sensation has the power to crystallise and transfix the moment upon which its gloom and radiance rests."
As if to illustrate the "transfix[ing]" of this moment, then, Woolf describes the scene in the kitchen in precise detail: "James Ramsay, sitting on the floor cutting out pictures from the illustrated catalogue of the Army and Navy stores . . ." The details of the room are mundane, and yet they are imbued with "heavenly bliss" for James. Woolf enumerates the elements of the scene: "the wheelbarrow, the lawnmower, the sound of poplar trees, leaves whitening before rain, rooks cawing, brooms knocking, dresses rustling." The cataloguing is so methodical as to seem, indeed, extremely precocious for a child of six, and yet the sense conveyed is of the scene being transfixed in Ramsay's mind, lending the reader an understanding of how it must be to belong to "that great clan." The child seems more than a child; the word "joy" appears again, not happiness, but something beyond it, approaching the sublime. James's "private code" and "secret language" enable him to understand the meaning of the seemingly insignificant detail in a way which is too cryptic for others to decode; the "human frailty" at which James seems to frown is that of those outside the group or "clan" who cannot understand.
Mr. Ramsay's bathetic comment in the following line—"But . . . it won't be fine"—somewhat punctures the grandiose effect of the opening scene, but already the nature of the young James Ramsay, as understood by his mother, has been vividly conveyed to the reader.

Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 50

Use the guidelines of curve sketching to sketch the curve $\displaystyle xy = x^2 + x + 1$ then find the equation of slant asymptote.

A. Domain. We can rewrite $f(x)$ as $\displaystyle y = \frac{x^2 + x + 1}{x}$. Therefore, the domain of the function is $(- \infty, 0) \bigcup (0, \infty)$

B. Intercepts. Solving for $y$ intercept, when $x = 0$,

$\displaystyle y = \frac{0^2 + 0 + 1}{0}$, $y$ intercept does not exist

Solving for $x$ intercept, when $y = 0$,


$
\begin{equation}
\begin{aligned}

0 =& \frac{x^2 + x + 1}{x}
\\
\\
0 =& x^2 + x + 1

\end{aligned}
\end{equation}
$


$x$ intercept does not exist

C. Symmetry. The function is not symmetric to either $y$ axis and origin by using symmetry test.

D. Asymptote. For vertical asymptote, we have $x = 0$

For horizontal asymptote, since $\displaystyle \lim_{x \to \pm \infty} f(x) = \pm \infty$, the function has no horizontal asymptote.

For slant asymptote, by using long division,







We can rewrite $f(x)$ as $\displaystyle y = x + 1 + \frac{1}{x}$, so...

$\displaystyle \lim_{x \to \pm \infty} f(x) - (x + 1) = \frac{1}{x} = 0$

Therefore, the equation of slant asymptote is $y = x + 1$.

E. Intervals of increase or decrease,

If $\displaystyle f(x) = \frac{x^2 + x + 1}{x}$, then by using Quotient Rule,


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{x (2x + 1) - (x^2 + x + 1)(1)}{x^2} = \frac{2x^2 - \cancel{x} - x^2 - \cancel{x} - 1}{x^2} = \frac{x^2 - 1}{x^2}
\\
\\
f'(x) =& 1 - \frac{1}{x^2}

\end{aligned}
\end{equation}
$


When $f'(x) = 0$


$
\begin{equation}
\begin{aligned}

0 =& 1 - \frac{1}{x^2}
\\
\\
\frac{1}{x^2} =& 1

\end{aligned}
\end{equation}
$


The critical numbers are

$x = 1$ and $x = -1$

Hence, the intervals of increase and decrease are..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
x < -1 & + & \text{increasing on } (- \infty, -1) \\
-1 < x < 1 & - & \text{decreasing on $(-1, 1)$ except at $x = 0$} \\
2 < x < 6 & - & \text{decreasing on (2, 6)} \\
x > 1 & + & \text{increasing on } (6, \infty) \\
\hline

\end{array}
$


F. Local Maximum and Minimum Values

Since $f'(x)$ changes from positive to negative at $x = -1, f(-1) = -1$ is a local maximum. On the other hand, since $f'(x)$ changes from negative to positive at $x = 1, f(1) = 3$ is a local minimum.

G. Concavity and inflection point

If $f'(x) = \displaystyle 1 - \frac{1}{(x^2)}$, then


$
\begin{equation}
\begin{aligned}

f''(x) =& \frac{2}{x^3}

\end{aligned}
\end{equation}
$


when $f''(x) = 0,$

$\displaystyle 0 = \frac{2}{x^3}$

$f''(x)= 0$ does not exist, therefore the function has no inflection point.

Hence, the concavity is..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
x < 0 & - & \text{Downward} \\
x > 0 & + & \text{Upward}\\
\hline
\end{array}
$

H. Sketch the graph.

lim_(x->-oo)tanhx Find the limit

A limit is the value that the function approach as x approaches "a".
In the given problem, the x-gt- oo indicates that independent variable x approaches large negative numbers for  given function: f(x)=tanh(x) .
 The function f(x)= tanh(x) is the hyperbolic tangent function. Its domain is all real number that can be expressed with the interval notation (-oo,oo) . It is a symmetric odd function. It also has an inflection point that can be found at x=0. There are no local extrema that can found in the continuous function of hyperbolic tangent.
 The "attached image" is the graph of f(x)=tanh(x) .
By graphical inspection, as graph continues to left of y-axis or x approaches -oo , it approaches y = -1 .
Therefore, the limit will be:
lim_(x-gt-oo) [tanh(x)] = -1 .

College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 22

State whether the matrices $\displaystyle A = \left[ \begin{array}{cc}
\sqrt{25} & 1 \\
0 & 2^{-1}
\end{array} \right]$ and $B = \left[ \begin{array}{cc}
5 & e^0 \\
\log 1 & \displaystyle \frac{1}{2}
\end{array} \right]$ are equal.

Matrices $A$ and $B$ are equal, because when both matrices are simplified they will have the same result.

In matrix $A$

$\displaystyle A = \left[ \begin{array}{cc}
\sqrt{25} & 1 \\
0 & 2^{-1}
\end{array} \right] = \left[ \begin{array}{cc}
5 & 1 \\
0 & \displaystyle \frac{1}{2}
\end{array} \right] $

And in matrix $B$

$\displaystyle B = \left[ \begin{array}{cc}
5 & e^0 \\
\log 1 & \displaystyle \frac{1}{2}
\end{array} \right] = \left[ \begin{array}{cc}
5 & 1 \\
0 & \displaystyle \frac{1}{2}
\end{array} \right]$

Which were the alliances that were formed in World War I in order?

Most of the alliances of World War I had been formed for years before the war even began.  Germany and Austria-Hungary formed the Dual Alliance in 1879, with the agreement that they would remain a united force if Russia was at war against them.  Soon Italy joined the Dual Alliance and it became the Triple Alliance.  Italy also made an agreement with France if Germany were to attack them.  Britain, Russia, and France formed the Triple Entente.  
After the war began, Italy decided that Germany had violated the terms of their allegiance.  Italy was able to remain neutral for the first year of the war because of this.  Eventually, Italy joined the Triple Entente because of the promise of land in 1915.  Land to the north of Italy belonged to Austria-Hungary, even though it was primarily populated by Italians.
Serbia and Montenegro became allied with Russia during the early parts of the war.  In 1917, the United States joined the war on the side of the Triple Entente.  Siam, Greece, and Liberia did the same in 1917.  Japan had also joined this alliance in 1914, shortly after the start of the war.  Russia withdrew from the alliance before the war ended.

Intermediate Algebra, Chapter 2, 2.1, Section 2.1, Problem 20

Solve the equation $2(3-2x) = x - 4$, and check your solution. If applicable, tell whether the equation is an identity or contradiction.


$
\begin{equation}
\begin{aligned}

2(3-2x) =& x - 4
&& \text{Given equation}
\\
6-4x =& x - 4
&& \text{Distributive property}
\\
-4x - x =& -4-6
&& \text{Subtract $(x+6)$ from each side}
\\
-5x =& -10
&& \text{Combine like terms}
\\
\frac{-5x}{-5} =& \frac{-10}{-5}
&& \text{Divide both sides by $-5$}
\\
x =& 2
&&

\end{aligned}
\end{equation}
$


Checking:


$
\begin{equation}
\begin{aligned}

2[3-2(2)] =& 2-4
&& \text{Substitute } x = 2
\\
2(3-4) =& 2-4
&& \text{Multiply}
\\
2(-1) =& 2-4
&& \text{Subtract inside the parentheses}
\\
-2 =& -2
&& \text{True}

\end{aligned}
\end{equation}
$