State whether the following statement is true:
There is a number $x$ such that $x^{2}$ is irrational but $x^{4}$ is rational. Justify your answer by an example.
True,
Let us take $x=\sqrt[4]{2}$
Now, $\quad x^{2}=(\sqrt[4]{2})^{2}=\sqrt{2},$ an irrational number.
$x^{4}=(\sqrt[4]{2})^{4}=2,$ a rational number.
So, we have a number $x$ such that $x^{2}$ is irrational but $x^{4}$ is rational.
Visualise the representation of $2.6 \overline{4}$ on the number line up to $5$ decimal places, that is up to $2.64444$
Rationalise the denominator in each of 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.
$\frac{4}{\sqrt{3}}$
Express $0.7 \overline{39}$ in the form $\frac{P}{q} ;$ where $p$ and $q$ are integers and $q \neq 0$
Simplify
$\frac{11^{\frac{1}{3}}}{11^{\frac{1}{5}}}$
The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to