Advertisements
Advertisements
Question
The square root of 64 divided by the cube root of 64 is
Options
64
2
\[\frac{1}{2}\]
642/3
Solution
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
RELATED QUESTIONS
If a = 3 and b = -2, find the values of :
(a + b)ab
Prove that:
`(x^a/x^b)^cxx(x^b/x^c)^axx(x^c/x^a)^b=1`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
Find the value of x in the following:
`5^(2x+3)=1`
Simplify:
`root(lm)(x^l/x^m)xxroot(mn)(x^m/x^n)xxroot(nl)(x^n/x^l)`
If (23)2 = 4x, then 3x =
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]