$$\begin{gathered} Simplify \\ {\left[\frac{256x^{16}}{81y^4}\right]}^{-1/4} \end{gathered}$$

Dear Student,

Answer (i):

Simplifying:

$$\begin{gathered} ={\left[\frac{256 x^{16}}{81 y^4}\right]}^{-\frac{1}{4}} \\ ={\left[\frac{4^4 x^{16}}{3^4 y^4}\right]}^{-\frac{1}{4}} \left\{as 256=4^4 and 81=3^4\right\} \\ =\frac{{\left(4\right)}^{4\times -\frac{1}{4}} {\left(x\right)}^{16\times -\frac{1}{4}}}{{\left(3\right)}^{4\times -\frac{1}{4}} {\left(y\right)}^{4\times -\frac{1}{4}}} \left\{as {\left(x^a\right)}^b=x^{a\times b}\right\} \\ =\frac{{\left(4\right)}^{-1} {\left(x\right)}^{-4}}{{\left(3\right)}^{-1} {\left(y\right)}^{-1}} \\ =\frac{3y}{4x^4} \left\{as x^{-1}=\frac{1}{x}\right\} \end{gathered}$$


Answer (ii):

The figure can we redrawn as:



In a circle, the angle at the centre of the circle is twice the angle subtended by the same arc on the circumference. This is called the Angle at the Centre theorem.

Hence, $\angle AOC=2\angle ABC=2\times 50=100$
Now, as $△AOC$ is an isosceles triangle as AO = OC which are the radius of the circle.
And in an isosceles triangle, angle opposite to equal sides are always equal,
Hence,  $\angle OAC=\angle OCA=x$

As the sum of interior angles of triangle is 180, thus in $△AOC$, 
$$\begin{gathered} \angle OAC+\angle OCA+\angle AOC=180 \\ or, x+x+100=180 \\ or, 2x=80 \\ or, x=40 \end{gathered}$$


Answer (iii):

Let's say;

$$\begin{gathered} S=0.1233333.... \\ or, 10S=1.233333.... \left\{Mutlyiplying both the sides by 10\right\} \\ or, 10S-S=1.23333.....-0.123333.... \\ or, 9S=1.110000.... \\ or, 9S=1.11 \\ or, S=\frac{1.11}{9} \\ or, S=\frac{111}{900} \end{gathered}$$

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