Advertisements
Advertisements
प्रश्न
The square root of 64 divided by the cube root of 64 is
विकल्प
64
2
\[\frac{1}{2}\]
642/3
उत्तर
We have to find the value of `(2sqrt64)/(3sqrt64)`
So,
`(2sqrt64)/(3sqrt64) = (2(sqrt2 xx 2 xx 2 xx 2 xx 2 xx 2) )/(2(sqrt2 xx 2 xx 2 xx 2 xx 2 xx 2) )`
`= 2^(6xx 1/2)`
`= 2^(6xx 1/3)`
`= 2^(6xx 1/2)/2^(6xx 1/3)`
`(2sqrt64)/(3sqrt64) = 2^3/2^2`
`=2^(3-2)`
`=2^1`
= 2
The value of `(2sqrt64)/(3sqrt64)` is 2.
Hence the correct choice is b.
APPEARS IN
संबंधित प्रश्न
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
Simplify:
`(16^(-1/5))^(5/2)`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
Find the value of x in the following:
`5^(2x+3)=1`
Which one of the following is not equal to \[\left( \sqrt[3]{8} \right)^{- 1/2} ?\]
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
If \[x + \sqrt{15} = 4,\] then \[x + \frac{1}{x}\] =
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
