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2 years ago

Read the quote from "My Heart Is Bursting."

English

2 answers:

Alekssandra [29.7K]2 years ago

7 0

Answer:pride

Explanation:

Because in day so people can see

stealth61 [152]2 years ago

7 0

Answer:

A). Pride.

Explanation:

Tone is described as the author's approach or attitude regarding a specific subject matter reflected through the language and word-choice adopted by the author. It allows him/her to provide the readers with a perspective to look at and feeling to feel while reading at the text.

In the given quote, the author's attitude('tone') could best be narrated as conveying 'pride' which is clearly reflected through the style and word-choice adopted by him('fighting, would have done it by day'). These phrases clearly mirror that the speaker is filled with 'self-esteem and honor' which implies that he is proud of his achievement which he made('by day...not in the dark'). Therefore, option A is the correct answer.

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★ Formula Applied :

$\begin{gathered}\sf \bullet\ \; cos^2x-sin^2x=cos2x\\\\\to\ \sf \pink{sin^2x-cos^2x=-cos2x}\end{gathered}$

$\begin{gathered}\bullet\ \; \sf sin2x=2.sinx.cosx\\\\\to \sf \blue{sinx.cosx=\dfrac{sin2x}{2}}\end{gathered}$

$\displaystyle \bullet\ \; \sf \int \dfrac{dx}{x}$

$\bullet\ \; \sf \ln (ab)=\ln a+\ln$

★ Explanation :

$\begin{gathered}\displaystyle \sf \int \dfrac{sin^2x-cos^2x}{sinx.cosx}dx\\\\\to \sf \int \dfrac{-cos2x}{\frac{sin2x}{2}}dx\\\\\to \sf \int \dfrac{-2cos2x}{sin2x}dx\end{gathered}$

Lets use substitution method ,

Let , u = sin2x

⇒ du = 2.cos2x.dx

$\begin{gathered}\to \displaystyle \sf \int \dfrac{-du}{u}\\\\\to \sf -\int \dfrac{du}{u}\\\\\to \sf -ln|u|\\\\\end{gathered}$

$\to \sf \red{-ln|sin2x|+c}$

$\to \sf - \ln |2sinx.cosx|+c$

$\to \sf - \ln |sinx.cosx|+(\ln 2+c)$

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$\displaystyle \to \sf \int \left( \dfrac{sin^2x}{sinx.cosx}-\dfrac{cos^2x}{sinx.cosx}\right)dx$

$\begin{gathered}\displaystyle \to \sf \int \left( \dfrac{sinx}{cosx} -\dfrac{cosx}{sinx} \right)dx\\\\\to\ \sf \int (tanx-cotx)dx\\\\\to \sf \int tanx.dx-\int cotx.dx\\\\\to \sf ln|secx|-ln|sinx|+c\end{gathered}$

$\to \sf ln\left| \dfrac{secx}{sinx}\right|+c$

$\begin{gathered}\to \sf ln\left| \dfrac{1}{sinx.cosx} \right|+c\\\\\end{gathered}$ ]

$\leadsto \sf \pink{-ln|sinx.cosx|+c}\ \; \bigstar$

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