Advertisements
Advertisements
प्रश्न
Find the value to three places of decimals of the following. It is given that
`sqrt2 = 1.414`, `sqrt3 = 1.732`, `sqrt5 = 2.236` and `sqrt10 = 3.162`
`2/sqrt3`
उत्तर
We know that rationalization factor of the denominator is `sqrt3`. We will multiply numerator and denominator of the given expression `2/sqrt3` by `sqrt3` to get
`2/sqrt3 xx sqrt3/sqrt3 = (2 xx sqrt3)/(sqrt3 xx sqrt3)`
`= (2sqrt3)/3`
`= (2 xx 1.732)/3`
`= 3.4641/3`
= 1.1547
The value of expression 1.1547 can be round off to three decimal places as 1.155
APPEARS IN
संबंधित प्रश्न
Rationalise the denominator of the following:
`1/sqrt7`
Rationalise the denominator of the following
`(3sqrt2)/sqrt5`
In the following determine rational numbers a and b:
`(5 + 3sqrt3)/(7 + 4sqrt3) = a + bsqrt3`
Find the value of `6/(sqrt5 - sqrt3)` it being given that `sqrt3 = 1.732` and `sqrt5 = 2.236`
Simplify `(3sqrt2 - 2sqrt3)/(3sqrt2 + 2sqrt3) + sqrt12/(sqrt3 - sqrt2)`
Simplify: \[\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3} - \sqrt{2}}\]
Simplify \[\sqrt{3 + 2\sqrt{2}}\].
If \[\frac{\sqrt{3 - 1}}{\sqrt{3} + 1}\] =\[a - b\sqrt{3}\] then
The number obtained on rationalising the denominator of `1/(sqrt(7) - 2)` is ______.
Rationalise the denominator in the following and hence evaluate by taking `sqrt(2) = 1.414, sqrt(3) = 1.732` and `sqrt(5) = 2.236`, upto three places of decimal.
`sqrt(2)/(2 + sqrt(2)`