Respuesta :

mirai123

Answer:

[tex]v = \sqrt{\dfrac{a -cb}{c\sqrt{2}} } = \dfrac{\sqrt{ a -cb}}{\sqrt{c}\sqrt[4]{2} }[/tex]

Step-by-step explanation:

[tex]c = \dfrac{a}{b+\sqrt{2v} }[/tex]

The equation for [tex]v[/tex] will be

[tex]c = \dfrac{a}{b+\sqrt{2v} } \implies c(b+\sqrt{2v}) = a \implies cb + c\sqrt{2v} = a[/tex]

Once [tex]\sqrt{2v} = \sqrt{2} \sqrt{v}[/tex]

[tex]cb + c\sqrt{2}\sqrt{v} = a \implies c\sqrt{2}\sqrt{v} = a -cb \implies \sqrt{v} =\dfrac{a -cb}{c\sqrt{2}}[/tex]

Square both sides

[tex]\sqrt{v} =\dfrac{a -cb}{c\sqrt{2}} \implies v = \sqrt{\dfrac{a -cb}{c\sqrt{2}} } = \dfrac{\sqrt{ a -cb}}{\sqrt{c}\sqrt[4]{2} }[/tex]

Trancescient

Step-by-step explanation:

Multiply both sides of the equation by [tex]b + \sqrt{2\nu}[/tex] to get

[tex]cb + c\sqrt{2\nu} = a[/tex]

Put the cb term to the right hand side and then divide by c to get

[tex]\sqrt{2\nu} = \dfrac{a - cb}{c}[/tex]

Taking the square of the equation, we get

[tex]2\nu = \left(\dfrac{a - cb}{c}\right)^2[/tex]

Finally, dividing the equation by 2 and we get an equation for [tex]\nu.[/tex]

[tex]\nu = \dfrac{1}{2}\left(\dfrac{a - cb}{c}\right)^2[/tex]

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