If sqrt{1+frac{x}{9}}=frac{13}{3},then the va | vedclass

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(B) Given the equation: $\sqrt{1+\frac{x}{9}}=\frac{13}{3}$Squaring both sides of the equation,we get:$1+\frac{x}{9} = \left(\frac{13}{3}\right)^2$$1+

Vedclass · Accepted Answer

$160$

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(B) Given the equation: $\sqrt{1+\frac{x}{9}}=\frac{13}{3}$Squaring both sides of the equation,we get:$1+\frac{x}{9} = \left(\frac{13}{3}\right)^2$$1+\frac{x}{9} = \frac{169}{9}$Subtract $1$ from both sides:$\frac{x}{9} = \frac{169}{9} - 1$$\frac{x}{9} = \frac{169-9}{9}$$\frac{x}{9} = \frac{160}{9}$Multiplying both sides by $9$,we get:$x = 160$

If $\sqrt{1+\frac{x}{9}}=\frac{13}{3}$,then the value of $x$ is

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