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3 years ago

1. Linda: "Excuse me! Where’s the post office?"

English

2 answers:

denis-greek [22]3 years ago

5 0

<h3>Answer:</h3>

B. It’s over there.

<h3>Explainton </h3>

Linda: "Excuse me! Where’s the post office?"

Maria: It's over there

g100num [7]3 years ago

4 0

Answer:

Explanation:

It's over there would be the correct answer

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Learning Task 6. If you were a senator, what five (5) possible measures or bills would you propone? Complete the table below by

sweet [91]

If I was a senator, the given measures or bills that would be proposed include:

A bill against the wastage of food and water.
A bill in favor of free education at the primary and secondary level
A bill to build training centers for the disabled.

A bill to build homes for the homeless.
A bill to have a sporting competition for children every year.

The bills above will help in the growth and the development of the country. It'll be also vital in the protection of the environment.

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2 years ago

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 :

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

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

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