It being given that 3-1-732-5-2-236-6-2-449 and 10-3-162- find to three places of decimal- the value of each of the following-i 16-5ii 65-3iii 143-35iv 3-53-5v 1-232-3vi 5-25-2

(i) nbsp;16+5=16+5 #215;6 56 5=6 562 52 =6 56 5=6 5=2.449 2.236 =0.213 (ii) nbsp; 65+3=65+3 #215;5 35 3=65 352 32 =65 35 3=65 32=35 3 =3 #215;2.236 ...

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Byju's Answer

Standard VI

Mathematics

Addition and Subtraction of Decimals

It being give...

Question

It being given that $\sqrt{5} = 2.236, \sqrt{6} = 2.449$ , find to three places of decimal, the value of of the following.
$\frac{1}{\sqrt{6} + \sqrt{5}}$

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Solution

$\frac{1}{\sqrt{6} + \sqrt{5}} = \frac{1}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{\sqrt{6} - \sqrt{5}}{6 - 5}$
$= 2.449 - 2.236$
$= 0.213$

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Similar questions

Q. It being given that

\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236 and \sqrt{10} = 3.162

, find the value of three places of decimals, of each of the following.
(i)

\frac{2}{\sqrt{5}}

(ii)

\frac{2 - \sqrt{3}}{\sqrt{3}}

(iii)

\frac{\sqrt{10} - \sqrt{5}}{\sqrt{2}}

Q. Find the values of each of the following correct to three places of decimals, it being given that

\sqrt{2} = 1.4142,

\sqrt{3} = 1.732

\sqrt{5} = 2.2360,

\sqrt{6} = 2.4495

and

\sqrt{10} = 3.162

,

(i)

\frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}}

(ii)

\frac{1 + \sqrt{2}}{3 - 2 \sqrt{2}}

Q. Find the value to three places of decimals of each of the following. It is given that

\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236

and

\sqrt{10} = 3.162

\frac{\sqrt{5} + 1}{\sqrt{2}}

Q. Find the value to three places of decimals of each of the following. It is given that

\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236

and

\sqrt{10} = 3.162

.

(i)

\frac{2}{\sqrt{3}}

(ii)

\frac{3}{\sqrt{10}}

(iii)

\frac{\sqrt{5} + 1}{\sqrt{2}}

(iv)

\frac{\sqrt{10} + \sqrt{15}}{\sqrt{2}}

(v)

\frac{2 + \sqrt{3}}{3}

(vi)

\frac{\sqrt{2} - 1}{\sqrt{5}}

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It being given that √5=2.236,√6=2.449, find to three places of decimal, the value of of the following. 1√6+√5

1√6+√5=1√6+√5×√6−√5√6−√5=√6−√56−5=2.449−2.236=0.213

It being given that $\sqrt{5} = 2.236, \sqrt{6} = 2.449$ , find to three places of decimal, the value of of the following.
$\frac{1}{\sqrt{6} + \sqrt{5}}$

$\frac{1}{\sqrt{6} + \sqrt{5}} = \frac{1}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{\sqrt{6} - \sqrt{5}}{6 - 5}$
$= 2.449 - 2.236$
$= 0.213$