[tex]-\frac{3}{11}, \sqrt{36} \ and \ 3 \sqrt{200-4}[/tex] are rational numbers.
Solution:
Rational number:
Rational number is of the form [tex]\frac{p}{q} , \ q \neq0[/tex] where p and q are integers.
Option A:
[tex]$-\frac{3}{11}[/tex]
It is of the form [tex]\frac{p}{q} , \ q \neq0[/tex]. Therefore, it is a rational number.
Option B:
[tex]\sqrt{36}=\sqrt{6^2}[/tex]
square and square root get canceled.
[tex]\sqrt{36}=6[/tex]
Any integer can be written as that integer by 1.
[tex]$\sqrt{36}=\frac{6}{1}[/tex]
It is of the form [tex]\frac{p}{q} , \ q \neq0[/tex]. Therefore, it is a rational number.
Option C:
[tex]$\frac{-2 \pi}{5}[/tex]
we know that π is an irrational number.
So that [tex]\frac{-2 \pi}{5}[/tex] is not rational number.
Option D:
[tex]3 \sqrt{200-4}=3 \sqrt{196}[/tex]
[tex]3 \sqrt{200-4}=3 \sqrt{14^2}[/tex]
square and square root get canceled.
[tex]3 \sqrt{200-4}=3 \times 14 =42[/tex]
Any integer can be written as that integer by 1.
[tex]$3 \sqrt{200-4}=\frac{42}{1}[/tex]
It is of the form [tex]\frac{p}{q} , \ q \neq0[/tex]. Therefore, it is a rational number.
Hence [tex]-\frac{3}{11}, \sqrt{36} \ and \ 3 \sqrt{200-4}[/tex] are rational numbers.
