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VMariaS [17]
3 years ago
7

Show that y2+ x-3 =0 is an implicit solution to dy/dx =-1/ (2y) on the interval (- [infinity] , 3)

Mathematics
1 answer:
Sedaia [141]3 years ago
4 0

Answer:

Implicit function

\frac{d y}{d x} = \frac{-1}{2y}

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given  y² + x - 3=0 ...(i)

Differentiating equation (i) with respective to 'x', we get

                   2 y \frac{d y}{d x} +1 = 0

           ⇒     2 y \frac{d y}{d x} = -1

           ⇒     \frac{d y}{d x} = \frac{-1}{2y}

                 \frac{d y}{d x} = \frac{-1}{2 (3)}

                 \frac{d y}{d x} = \frac{-1}{6}

 

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Answer:

The answers of the questions are given below :

  • a) = 4096
  • b) = 1.25
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  • 4) = r⁴s³
  • 5) = a⁸/b¹²

Step-by-step explanation:

\large{\tt{\underline{\underline{\red{QUESTION}}}}}

\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\  3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \:  \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}

\begin{gathered}\end{gathered}

\large{\tt{\underline{\underline{\red{SOLUTION}}}}}

Question. 1

>> 4⁶

\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\  \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024  \times 4} \\ \qquad{= 4096} \end{gathered}

  • Hence, the answer is 4096.

\begin{gathered}\end{gathered}

Question. 2

>> (2⁶/5³)^-⅓

\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\  \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\  \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\  \\ \qquad\implies{\bigg( 1 \times  \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\  \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\  \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)}  \\  \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\  \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}

  • Hence, the answer is 1.25.

\begin{gathered}\end{gathered}

Question. 3

>> (m^2/3)•(m^4/3)

\begin{gathered} \qquad{=  \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{2}{3} +  \frac{4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{2 + 4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{6}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{2}\Big)}\end{gathered}

  • Hence, the answer is m².

\begin{gathered}\end{gathered}

Question. 4

>> (r¹² s⁹)^⅓

\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)}  \\  \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)}  \\  \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}

  • Hence, the answer is r⁴s³.

\begin{gathered}\end{gathered}

Question. 5

>> (a⁴/b⁶)^2

\begin{gathered} \qquad{ =  \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\  \qquad{ =  \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ =  \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}

  • Hence, the answer is a⁸/b¹².

\underline{\rule{220pt}{3pt}}

4 0
2 years ago
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