Find the value of the following expression co | vedclass

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

Question

(B) Given the expression: $\frac{\sqrt{10} - \sqrt{5}}{2}$We can rewrite $\sqrt{10}$ as $\sqrt{2} 	imes \sqrt{5}$.So,the expression becomes: $\frac{\

Vedclass · Accepted Answer

$0.463$

Vedclass · Answer

(B) Given the expression: $\frac{\sqrt{10} - \sqrt{5}}{2}$We can rewrite $\sqrt{10}$ as $\sqrt{2} 	imes \sqrt{5}$.So,the expression becomes: $\frac{\sqrt{2} 	imes \sqrt{5} - \sqrt{5}}{2}$Factor out $\sqrt{5}$ from the numerator: $\frac{\sqrt{5}(\sqrt{2} - 1)}{2}$Substitute the given values $\sqrt{5} = 2.236$ and $\sqrt{2} = 1.414$:$= \frac{2.236(1.414 - 1)}{2}$$= \frac{2.236(0.414)}{2}$$= 1.118 	imes 0.414$$= 0.462852$Rounding to three decimal places,we get $0.463$.

Find the value of the following expression correct to three decimal places,taking $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$:
$\frac{\sqrt{10} - \sqrt{5}}{2}$

Similar Questions

Express $0.5\overline{7}$ in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.

Prove that $(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})^{\frac{1}{2}} = (1^{3}+2^{3}+3^{3}+4^{3})^{\frac{1}{2}} + (5^{3})^{\frac{1}{3}}$.

For each question,select the proper option from four options given,to make the statement true: $\sqrt{5^{2}+12^{2}}$ is a / an $\ldots \ldots$ number.

Which of the following is irrational?

Represent $\sqrt{6}$ on the number line.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the value of the following expression correct to three decimal places,taking $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$:$\frac{\sqrt{10} - \sqrt{5}}{2}$