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aksik [14]
3 years ago
7

Research Mount Everest and explain what makes it such a difficult and dangerous mountain to climb. Be sure to cite the websites

you research.
English
2 answers:
maksim [4K]3 years ago
8 0
Some reasons why mount everest is so difficult to climb are that there is not much oxygen up there and many other people visit the mountain. Another reason is ice can fall down the mountain and hit a climber.

(look up 360 expeditions for your website)
AveGali [126]3 years ago
3 0

Answer:

1.) Avalanches ( Once your trapped your dead)

2.) Ice Cracking and Ice Water Breaks ( If you fall in water in those type of weather conditions you could die from hypothermia.)

3.)  Altitude sickness ( This is the opposite of you diving deep into the ocean, it is something similar to that. )

4.) Lack of oxygen

5.) Falling rocks

Explanation:

Hope this helps : )

This is prior knowledge : )

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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

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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

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\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}

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\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

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