Question
3
The simplfied form of 154 is 3 1/2. Why is there a
coefficient of 3 outside the cube root? (1 point)
There is a coefficient of 3 outside the cube root because
9 divides into 54.
There is a coefficient of 3 outside the cube root because
54 is divisible by 3.
There is a coefficient of 3 outside the cube root because
three 3s divide into 54.
There is a coefficient of 3 outside the cube root because
two 3s divide into 54.
The simplfied form of 154 is 3 1/2. Why is there a
coefficient of 3 outside the cube root? (1 point)
There is a coefficient of 3 outside the cube root because
9 divides into 54.
There is a coefficient of 3 outside the cube root because
54 is divisible by 3.
There is a coefficient of 3 outside the cube root because
three 3s divide into 54.
There is a coefficient of 3 outside the cube root because
two 3s divide into 54.
Answers
Answered by
GPT-4o mini
There is a coefficient of 3 outside the cube root because three 3s divide into 54.
To explain further, the simplified form of the cube root of 54 can be calculated as follows:
- The prime factorization of 54 is \(2 \times 3^3\).
- Since \(3^3 = 27\) is a perfect cube, we can take out one \(3\) from the cube root.
- Hence, we can express \(\sqrt[3]{54}\) as \(\sqrt[3]{2 \times 3^3} = 3 \cdot \sqrt[3]{2}\).
Thus, the coefficient of 3 outside the cube root comes from the fact that you can take out \(3\) because three \(3s\) (which make \(3^3\)) divide into \(54\).
To explain further, the simplified form of the cube root of 54 can be calculated as follows:
- The prime factorization of 54 is \(2 \times 3^3\).
- Since \(3^3 = 27\) is a perfect cube, we can take out one \(3\) from the cube root.
- Hence, we can express \(\sqrt[3]{54}\) as \(\sqrt[3]{2 \times 3^3} = 3 \cdot \sqrt[3]{2}\).
Thus, the coefficient of 3 outside the cube root comes from the fact that you can take out \(3\) because three \(3s\) (which make \(3^3\)) divide into \(54\).