If sqrt{left(1+frac{27}{169}right)}=1+frac{x} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13}$First,simplify the expression inside the square root:$1+\frac{27}{169} =

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the equation: $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13}$First,simplify the expression inside the square root:$1+\frac{27}{169} = \frac{169+27}{169} = \frac{196}{169}$Now,take the square root of the fraction:$\sqrt{\frac{196}{169}} = \frac{14}{13}$Substitute this back into the original equation:$\frac{14}{13} = 1+\frac{x}{13}$Subtract $1$ from both sides:$\frac{14}{13} - 1 = \frac{x}{13}$$\frac{14-13}{13} = \frac{x}{13}$$\frac{1}{13} = \frac{x}{13}$Therefore,$x = 1$.

If $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13},$ then $x$ equals

Similar Questions

If $\sqrt{\frac{x}{169}} = \frac{54}{39}$,then $x$ is equal to

The largest number of three digits which is a perfect square is

$A$ General wishing to draw up his $5180$ men in the form of a solid square found that he had $4$ men less. If he could get four more men and form the solid square,the number of men in the front row is

The square root of $\frac{(0.75)^{3}}{1-0.75} + [0.75 + 0.75^{2} + 1]$ is

$\sqrt{33124} \times \sqrt{2601} - (83)^{2} = (?)^{2} + (37)^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module