$2 \sqrt{3}+\sqrt{3}$ is equal to
$2 \sqrt{3}+\sqrt{3}$ is equal to
- A
$2 \sqrt{6}$
- B
$6$
- C
$3 \sqrt{3}$
- D
$4 \sqrt{6}$
Similar Questions
If $a=\frac{\sqrt{5}}{8}$ and $\frac{8}{a}=b \sqrt{5},$ then find $b$.
If $a=\frac{\sqrt{5}}{8}$ and $\frac{8}{a}=b \sqrt{5},$ then find $b$.
Represent geometrically numbers on the number line:
$\sqrt{2.3}$
Represent geometrically numbers on the number line:
$\sqrt{2.3}$
Express $2 . \overline{137}$ in the form $\frac{p}{q} ;$ where $p$ and $q$ are integers and $q \neq 0$
Express $2 . \overline{137}$ in the form $\frac{p}{q} ;$ where $p$ and $q$ are integers and $q \neq 0$
Which of the following is not equal to $\left[\left(\frac{5}{6}\right)^{\frac{1}{5}}\right]^{-\frac{1}{6}} ?$
Which of the following is not equal to $\left[\left(\frac{5}{6}\right)^{\frac{1}{5}}\right]^{-\frac{1}{6}} ?$
Simplify:
$\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}$
Simplify:
$\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}$