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MAVERICK [17]
2 years ago
10

20 POINTS. In The First Men in the Moon, Mr. Bedford and Mr. Cavor risk their lives to discover if travel to the moon is possibl

e. Think about a time you took a risk by trying to accomplish something you didn’t know you could do. What were your fears? What motivated you to try? Write about your experience in your journal. ACTUAL ANSWERS ONLY.
English
1 answer:
Snezhnost [94]2 years ago
4 0

Answer:

I once tried doing a back flip, and it did not turn out how I wanted. one of my fears was that when I tried to do it my knees would hit my chin. My uncles told me it was a good idea so I tried it and it went surprisingly well.

Explanation:

I don´t know if this is what you meant but i just wanted to share that.

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I will make you brainlist if you help me the 2 question by explain them i swear i will make you brainlist
Xelga [282]

★ 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)

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

★ Alternate Method :

\displaystyle \sf \int \dfrac{sin^2x-cos^2x}{sinx.cosx}

\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

4 0
3 years ago
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