Answer :
To determine whether the number 169 is a perfect square, a perfect cube, both, or neither, we need to do the following:
1. Check if 169 is a perfect square:
- A perfect square is a number that can be expressed as the product of an integer with itself.
- In other words, if [tex]\( n^2 = 169 \)[/tex] for some integer [tex]\( n \)[/tex], then 169 is a perfect square.
- Let's find the integer whose square is equal to 169. Taking the square root of 169, we get:
[tex]\[
\sqrt{169} = 13
\][/tex]
- Since [tex]\( 13 \times 13 = 169 \)[/tex], 169 is indeed a perfect square.
2. Check if 169 is a perfect cube:
- A perfect cube is a number that can be expressed as the product of an integer with itself three times.
- In other words, if [tex]\( n^3 = 169 \)[/tex] for some integer [tex]\( n \)[/tex], then 169 is a perfect cube.
- Let's find the integer whose cube might be equal to 169 by taking the cube root:
[tex]\[
\sqrt[3]{169}
\][/tex]
- Estimating the cube root of 169, let's consider [tex]\( n \)[/tex] values:
[tex]\[
5^3 = 125 \quad \text{and} \quad 6^3 = 216
\][/tex]
- Since 169 is between 125 and 216, and there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 169 \)[/tex], 169 is not a perfect cube.
So, based on these observations, the correct classification is:
C. The number is a perfect square because [tex]\( \square \times \square = 169 \)[/tex].
1. Check if 169 is a perfect square:
- A perfect square is a number that can be expressed as the product of an integer with itself.
- In other words, if [tex]\( n^2 = 169 \)[/tex] for some integer [tex]\( n \)[/tex], then 169 is a perfect square.
- Let's find the integer whose square is equal to 169. Taking the square root of 169, we get:
[tex]\[
\sqrt{169} = 13
\][/tex]
- Since [tex]\( 13 \times 13 = 169 \)[/tex], 169 is indeed a perfect square.
2. Check if 169 is a perfect cube:
- A perfect cube is a number that can be expressed as the product of an integer with itself three times.
- In other words, if [tex]\( n^3 = 169 \)[/tex] for some integer [tex]\( n \)[/tex], then 169 is a perfect cube.
- Let's find the integer whose cube might be equal to 169 by taking the cube root:
[tex]\[
\sqrt[3]{169}
\][/tex]
- Estimating the cube root of 169, let's consider [tex]\( n \)[/tex] values:
[tex]\[
5^3 = 125 \quad \text{and} \quad 6^3 = 216
\][/tex]
- Since 169 is between 125 and 216, and there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 169 \)[/tex], 169 is not a perfect cube.
So, based on these observations, the correct classification is:
C. The number is a perfect square because [tex]\( \square \times \square = 169 \)[/tex].
