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}*
nope. sqrt(256) = 16
A is all irrational.
and that's pi, not pie!
To determine which set contains 3 irrational numbers, we need to identify which numbers in each set are irrational.
An irrational number is any number that cannot be expressed as a fraction (or ratio) of two integers and is not a terminating or repeating decimal.
Let's analyze each set:
a. { √120, π, √3 }
The square root of 120 is irrational because 120 is not a perfect square. π (pi) is also irrational. The square root of 3 is also irrational because 3 is not a perfect square. Therefore, all three numbers in this set are irrational.
b. { -√256, 1/9, 1/12 }
The square root of 256 is 16, so it is a rational number. Both 1/9 and 1/12 are fractions, so they are also rational numbers. Therefore, none of the numbers in this set are irrational.
c. { 3.14, -47, 100 }
3.14 is an approximation of π, which is irrational. -47 and 100 are integers and therefore rational numbers. So, only one number in this set is irrational.
d. { π, √0.36, √121 }
We know that π is irrational. The square root of 0.36 is 0.6, which is rational. The square root of 121 is 11, so it is a rational number. Therefore, only one number in this set is irrational.
From the analysis above, the set that contains 3 irrational numbers is set a. { √120, π, √3 }.
To determine which set contains 3 irrational numbers, we need to identify the irrational numbers in each set:
a. { square root of 120 , pi , square root of 3 }
- The square root of 120 is irrational.
- Pi is an irrational number.
- The square root of 3 is irrational.
So, set a contains 3 irrational numbers.
b. { neg. square root of 256 , 1/9 , 1/12 }
- The negative square root of 256 is -16, which is a rational number.
- 1/9 is a rational number.
- 1/12 is a rational number.
So, set b does not contain 3 irrational numbers.
c. { 3.14 , -47 , 100 }
- 3.14 is a rational number.
- -47 is a rational number.
- 100 is a rational number.
So, set c does not contain 3 irrational numbers.
d. { pi , square root of 0.36 , square root of 121 }
- Pi is an irrational number.
- The square root of 0.36 is 0.6, which is a rational number.
- The square root of 121 is 11, which is a rational number.
So, set d does not contain 3 irrational numbers.
Therefore, the set that contains 3 irrational numbers is set a: { square root of 120 , pi , square root of 3 }.