Answer:
All of the above
Step-by-step explanation:
We have to find the irrational numbers.
Irrational number: The number which can not be written as [tex]\frac{p}{q}[/tex] form, where p and q are integers and [tex]q\neq 0[/tex] then, the number is called irrational number.
a.[tex]\sqrt{14}[/tex]
[tex]\sqrt{2\times 7}[/tex]
[tex]\sqrt{2}\times \sqrt{7}[/tex]
We know that [tex]\sqrt{2}[/tex] and [tex]\sqrt{7}[/tex] are irrational numbers and product of two different irrational numbers is irrational number.
Hence, it is irrational number.
b.[tex]\sqrt{24}[/tex]
[tex]\sqrt{2\times 2\times 2 \times 3}[/tex]
[tex]2\sqrt{6}[/tex]
We know that [tex]\sqrt6[/tex] is a irrational number and 2 is rational number.The product of irrational number and rational number is irrational number.
Therefore, [tex]\sqrt{24}[/tex] is irrational number.
c.[tex]\sqrt{34}[/tex]
[tex]\sqrt{2}\times \sqrt{17}[/tex]
We know that [tex]\sqrt{2}[/tex] and [tex]\sqrt{17}[/tex] are irrational numbers and product of two different irrational numbers is irrational number.
Hence, it is irrational number.
d.[tex]\sqrt{44}[/tex]
[tex]\sqrt{2\times 2\times 11}[/tex]
[tex]2\sqrt{11}[/tex]
We know that [tex]\sqrt{11}[/tex] is a irrational number and 2 is rational number.The product of irrational number and rational number is irrational number.
Hence, it is irrational number.
e.[tex]\sqrt{54}[/tex]
[tex]\sqrt{2\times 3\times 3\times 3}[/tex]
[tex]3\sqrt{6}[/tex]
We know that [tex]\sqrt6[/tex] is a irrational number and 3 is rational number.The product of irrational number and rational number is irrational number.
Hence, it is irrational number.
