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0.10 = 6.0/v^2 + 0.00050v^2 - 0.033

what is v=?

pls detailed instruction on how to. :)

Guest May 8, 2014

Best Answer 

 #1
avatar+93356 
+8

$$v\ne0$$

$$0.1=\frac{6}{v^2}+0.0005v^2-0.033\\\\
0.133=\frac{6}{v^2}+0.0005v^2\\\\
0.133v^2=6+0.0005v^4\\\\$$

substitute x for v2

$$0.0005x^2-0.133x+6=0\\$$

 

$${\mathtt{0.000\: \!5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.133}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{133}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5\,689}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{5\,689}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{133}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{57.574\: \!540\: \!107\: \!467\: \!690\: \!7}}\\
{\mathtt{x}} = {\mathtt{208.425\: \!459\: \!892\: \!532\: \!309\: \!3}}\\
\end{array} \right\}$$

 

 $$v^2=57.574540\:\:\rightarrow v=\pm7.58779\:\: \mbox{approx}$$   or

$$v^2=208.42545989\:\:\rightarrow\:\:v=\pm14.4369\:\:\mbox{approx}$$

I think that is all okay.

Melody  May 8, 2014
 #1
avatar+93356 
+8
Best Answer

$$v\ne0$$

$$0.1=\frac{6}{v^2}+0.0005v^2-0.033\\\\
0.133=\frac{6}{v^2}+0.0005v^2\\\\
0.133v^2=6+0.0005v^4\\\\$$

substitute x for v2

$$0.0005x^2-0.133x+6=0\\$$

 

$${\mathtt{0.000\: \!5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{0.133}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{133}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5\,689}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{5\,689}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{133}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{57.574\: \!540\: \!107\: \!467\: \!690\: \!7}}\\
{\mathtt{x}} = {\mathtt{208.425\: \!459\: \!892\: \!532\: \!309\: \!3}}\\
\end{array} \right\}$$

 

 $$v^2=57.574540\:\:\rightarrow v=\pm7.58779\:\: \mbox{approx}$$   or

$$v^2=208.42545989\:\:\rightarrow\:\:v=\pm14.4369\:\:\mbox{approx}$$

I think that is all okay.

Melody  May 8, 2014

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