English, asked by sreekanyachakraborty, 4 days ago

how does mirabell introduce lady wishfort in The way of the world by william congreve?​

Answers

Answered by piyacutepie
12

Answer:

Lady Wishfort offers Mirabell her consent to the marriage if he can save her fortune and honour. Mirabell calls on Waitwell who brings a contract from the time before the marriage of the Fainalls in which Mrs. Fainall gives all her property to Mirabell.

Explanation:

Hope it may help you..

Answered by XBarryX
0

Answer:

\begin{gathered} \red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: TO \: \: FIND \:  \:  \:  \maltese }}}}} \\ \\ \huge \mathfrak{Derivative \: \: of } \\ \\ \bf \frac{x+cosx}{tanx} \end{gathered}✠   TOFIND   ✠✠   TOFIND   ✠Derivativeoftanxx+cosx

\color{Cyan}{\qquad \qquad \large \underline{ \pmb{{ \mathbb{ \maltese    \: REQUIRED \: \: INFO \:  \maltese }}}}}✠  REQUIREDINFO ✠✠  REQUIREDINFO ✠

\begin{gathered} \bigstar \: \: \: \mathfrak{ \underline{\huge Quotient \: \: Rule}} \\ \\ \bf \frac{d}{dx} \frac{u}{v} = \frac{v . \frac{du}{dx} - u \frac{dv}{dx} }{v {}^{2} } \\ \\ \bigstar \bf \: \: \: \frac{d}{dx} cosx = - sinx \\ \\ \bigstar \: \: \: \bf \frac{d}{dx} tanx = {sec}^{2} x\end{gathered}★QuotientRuledxdvu=v2v.dxdu−udxdv★dxdcosx=−sinx★dxdtanx=sec2x

\color{magenta}{ \large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: SOLUTION \:  \:  \:  \maltese }}}}}✠   SOLUTION   ✠✠   SOLUTION   ✠

\begin{gathered} \bf \frac{d}{dx} ( \frac{x + cosx}{tanx} ) \\ \\ = \sf \frac{tanx. \frac{d}{dx} (x + cosx) - (x + cosx) \frac{d}{dx} tanx}{ {(tanx)}^{2} } \\ ( \bf \because Quotient \: Rule ) \\ \\ = \sf\frac{tanx(1 - sinx) - (x + cosx) {sec}^{2} x}{ {tan}^{2} x} \end{gathered}dxd(tanxx+cosx)=(tanx)2tanx.dxd(x+cosx)−(x+cosx)dxdtanx(∵QuotientRule)=tan2xtanx(1−sinx)−(x+cosx)sec2x

\begin{gathered} \sf = \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. {sec}^{2} x}{ {tan}^{2}x } \\ \\ =\sf \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. \dfrac{1}{ {cos}^{2}x } }{ {tan}^{2}x } \\ \\ = \bf\frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: {sec}x}{ {tan}^{2}x } \end{gathered}=tan2xtanx−tanx.sinx−xsec2x−cosx.sec2x=tan2xtanx−tanx.sinx−xsec2x−cosx.cos2x1=tan2xtanx−tanx.sinx−xsec2x−secx

\large\red{ \mathfrak{ \text{W}hich \: \: is \: \text{T}he \: \: Required \: \: \text{ A}nswer }}WhichisTheRequired Answer

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