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3 years ago

Someone read this and just tell me in short words about what is says.

English

1 answer:

jok3333 [9.3K]3 years ago

8 0

Answer:

Explanation:

A person with aphasia may have trouble understanding, speaking, reading, or writing. ... Or you might say a word that does not make much sense, like "radio" for "ball." Switch ... For example, it may be hard to tell time, count money, or add and subtract. ... Ask me to draw, write, or point when I am having trouble talking.

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Read the passage. Paraphrase the central idea.

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Which of the following choices BEST identifies the above project plan step? A. project initiation B. project closure C. project

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4 years ago

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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