Which of the following sets contains 3 irrational numbers ?

a. { square root of 120 , pie , square root of 3 }
b. {neg. square root of 256 , 1/9 , 1/12}
c. {3.14 , -47 , 100}
d. {pie , square root of 0.36 , square root of 121

It would be A

To determine which set contains 3 irrational numbers, we need to understand what irrational numbers are.

Irrational numbers are real numbers that cannot be expressed as a fraction (or ratio) of two integers. They cannot be written as terminating or repeating decimals.

Now, let's analyze each set:

a. { square root of 120 , pie , square root of 3 }
The square root of 120 is irrational since it cannot be simplified into a fraction. Pi (or pie) is also irrational. However, the square root of 3 is also irrational. Therefore, there are 3 irrational numbers in this set.

b. {neg. square root of 256 , 1/9 , 1/12}
The negative square root of 256 is a rational number since it can be simplified to -16, which is a fraction. Both 1/9 and 1/12 are rational numbers since they can be written as fractions. Therefore, there are no irrational numbers in this set.

c. {3.14 , -47 , 100}
3.14 is the decimal form of pi. Therefore, it is irrational. -47 and 100 are integers, which are rational numbers. Therefore, there is only 1 irrational number in this set.

d. {pie , square root of 0.36 , square root of 121}
Pi (or pie) is irrational. The square root of 0.36 can be simplified to 0.6, which is a rational number. Similarly, the square root of 121 can be simplified to 11, which is also a rational number. Therefore, there is only 1 irrational number in this set.

Based on the analysis, the set that contains 3 irrational numbers is:
a. { square root of 120 , pie , square root of 3 }