Responses
64–√
6 start root 4 end root
24
24
46–√
4 start root 6 end root
166–√
9 answers
To find the square root of 96, we can simplify it first.
We can factor 96 into its prime factors: \[ 96 = 16 \times 6 = 4^2 \times 6. \]
Now, we can take the square root: \[ \sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}. \]
So, the square root of 96 is: \[ 4\sqrt{6}. \]
Among the provided responses, the correct answer is: 4 start root 6 end root.
To find the cubic root of 96, we can simplify it, although cubic roots can be a bit more straightforward depending on the factors.
The prime factorization of 96 is: \[ 96 = 2^5 \times 3^1. \]
To find the cubic root, we can apply the property of indices: \[ \sqrt[3]{96} = \sqrt[3]{2^5 \times 3^1} = \sqrt[3]{2^5} \times \sqrt[3]{3^1} = 2^{5/3} \times 3^{1/3}. \]
Now, let’s simplify \(2^{5/3}\): \[ 2^{5/3} = 2^{1 + 2/3} = 2^1 \times 2^{2/3} = 2 \times \sqrt[3]{4}. \]
Therefore, we can express the cubic root of 96 as: \[ \sqrt[3]{96} = 2 \times \sqrt[3]{4} \times \sqrt[3]{3} = 2 \times \sqrt[3]{12}. \]
Thus, the cubic root of 96 is: \[ 2\sqrt[3]{12}. \]
Among the provided responses, the correct answer is: 2 root index 3 start root 12 end root.
To find the cubic root of 189, we first perform its prime factorization:
\[ 189 = 3^3 \times 7 \]
Now we can find the cubic root:
\[ \sqrt[3]{189} = \sqrt[3]{3^3 \times 7} = \sqrt[3]{3^3} \times \sqrt[3]{7} = 3 \times \sqrt[3]{7} \]
Thus, the cubic root of 189 is:
\[ 3 \sqrt[3]{7} \]
Among the provided responses, the correct answer is:
3 start root 7 end root.
To find the square root of 250, we can simplify it first.
The prime factorization of 250 is: \[ 250 = 25 \times 10 = 5^2 \times 10. \]
Now, we can take the square root: \[ \sqrt{250} = \sqrt{25 \times 10} = \sqrt{25} \times \sqrt{10} = 5 \sqrt{10}. \]
So, the square root of 250 is: \[ 5\sqrt{10}. \]
Among the provided responses, the correct answer is: 5 start root 10 end root.
To find the square root of 128, let's start with simplifying it.
Factor 128: \[ 128 = 64 \times 2 = 8^2 \times 2 = 4^2 \times 8 = 2^7 \]
Take the square root: \[ \sqrt{128} = \sqrt{64 \times 2} = \sqrt{64} \times \sqrt{2} = 8\sqrt{2} \]
So, the square root of 128 is: \[ 8\sqrt{2} \]
Among the provided responses, the correct answer is: 8 start root 2 end root.
