Given that sqrt{13} = 3.605 and sqrt{130} = 1 | vedclass

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

Question

(D) We need to evaluate $\sqrt{1.3} + \sqrt{1300} + \sqrt{0.013}$.First,express each term in terms of $\sqrt{13}$ or $\sqrt{130}$:$\sqrt{1.3} = \sqrt{

Vedclass · Accepted Answer

$37.304$

Vedclass · Answer

(D) We need to evaluate $\sqrt{1.3} + \sqrt{1300} + \sqrt{0.013}$.First,express each term in terms of $\sqrt{13}$ or $\sqrt{130}$:$\sqrt{1.3} = \sqrt{\frac{130}{100}} = \frac{\sqrt{130}}{10} = 0.1 	imes 11.4 = 1.14$$\sqrt{1300} = \sqrt{100 	imes 13} = 10 	imes \sqrt{13} = 10 	imes 3.605 = 36.05$$\sqrt{0.013} = \sqrt{\frac{130}{10000}} = \frac{\sqrt{130}}{100} = 0.01 	imes 11.4 = 0.114$Now,add these values together:$1.14 + 36.05 + 0.114 = 37.304$Thus,the value is $37.304$.

Given that $\sqrt{13} = 3.605$ and $\sqrt{130} = 11.40$,find the value of $\sqrt{1.3} + \sqrt{1300} + \sqrt{0.013}$.

Similar Questions

$\sqrt{\frac{96.8}{0.578}} = ?$

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

The digit in the unit's place in the square root of $15876$ is

The areas of two square fields are $420.25 \; m^2$ and $441 \; m^2,$ respectively. The ratio of their sides is

If $\sqrt{1+\frac{x}{961}}=\frac{32}{31},$ then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module