If x-y2- y-z2- z-p2 where p -236- 948- then -x is

p=2^36× (3^2)^48=2^36× (3)^96 As z=p^2⇒ z=(2^36)^2 × (3^96)^2=2^72× 3^192 similarly, y=z^2=2^72× 3^192=2^144× 3^384 and x=y^2⇒√(x)=y=2^144× ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Powers and Exponents

If x=y2, ...

Question

If $x = y^{2}$ , $y = z^{2}$ , $z = p^{2}$ where $p = 2^{36} \times 9^{48}$ , then $\sqrt{x}$ is_____

Open in App

Solution

Suggest Corrections

Similar questions

x^{2} + y^{2} + z^{2} = 5, x, y, z \in R,

(x - y)^{2} + (y - z)^{2} + (z - x)^{2}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = ?

\frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y}

\frac{a (b - c)}{y^{2} - z^{2}} = \frac{b (c - a)}{z^{2} - x^{2}} = \frac{c (a - b)}{x^{2} - y^{2}}

x^{3} + y^{3} + z^{3} = 6 x y z

x + y + z \neq 0

x y z = x + y + z

(x - y)^{2} + (y - z)^{2} + (z - x)^{2}

u = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} =

Join BYJU'S Learning Program