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kirza4 [7]
3 years ago
5

Which of the following sentences contains a dangling modifier?. . A. After reading the essay, Janine thought the point was uncle

ar.. B. After Janine read the essay, she was still unsure what the main point was.. C. After reading the essay, the point was still unclear.. D. After she read the essay, Janine thought the point was unclear.
English
1 answer:
Zanzabum3 years ago
8 0
<span>these are examples of dangling modifers:

I was late for the school bus again. Running for the bus, my book fell in the mud. (Was the book running for the school bus? It's the only nearby noun beside mud.)

Deciding to join the navy, the recruiter enthusiastically pumped Joe's hand. (Was the recruiter deciding to join the navy? The only other option is Joe's hand.)
 
Upon entering the doctor's office, a skeleton caught my attention. (Was the skeleton entering the doctor's office? The only other option is my attention.)

in the example you gave "the point" did not read the essay, the only other option is :unclear. So the answer is C.</span>
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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}

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\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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