0. Irrational
,
1. Irrational
,
2. Rational
,
3. Irrational
1) Considering that Rational Numbers are the ones that can be written as ratios like:
![\begin{gathered} (a)/(b) \\ 1,(3)/(2),5,\text{ }\sqrt[]{4} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/smevlhua5ch1eof2js5flfsokloxyue9ld.png)
2) Then we can examine each row and each expression as well. So we can begin with:
![3\sqrt[]{9}\cdot\sqrt[]{2}=3\cdot3\cdot\sqrt[]{2}=9\sqrt[]{2}=12.72792\ldots](https://img.qammunity.org/2023/formulas/mathematics/college/ijnp7p3860mopbo1opypkp08out51aidbw.png)
Since the square root of 2 yields an irrational number, 9 times that yields an irrational one as well.
Irrational
![\frac{\sqrt[]{9}}{\pi}=(3)/(\pi)=0.9452\ldots](https://img.qammunity.org/2023/formulas/mathematics/college/gc2iiro9s3re39m7y1wxxzr8obxmofuwce.png)
Note that in the second row we have an irrational number as well, a decimal and not a repeating number.
Irrational
![\frac{\pi\sqrt[]{24}}{\sqrt[]{6\pi^2}}=\frac{\pi\cdot2\sqrt[]{2}\cdot\sqrt[]{3}}{\pi\sqrt[]{2}\cdot\sqrt[]{3}}=2](https://img.qammunity.org/2023/formulas/mathematics/college/96m4h7xjs1ao1crmv8x6zk85h7mol2twqg.png)
Since 2 is a rational number the same as 2/1. This is a Rational one.
And Finally:
![-\sqrt[]{3}+2=0.2674](https://img.qammunity.org/2023/formulas/mathematics/college/h3qfs0prdv80rspvln8mqpjsdiahnveo7c.png)
Notice that the square root of 3 is already irrational so the sum of that with 2 yields another irrational number.
Irrational
3) Hence, the answer is:
0. Irrational
,
1. Irrational
,
2. Rational
,
3. Irrational