Question

Rewrite the irrational square root 48−−√ as the product of an integer and another irrational square root.(1 point)
Responses

43–√
4 start root 3 end root

212−−√
2 start root 12 end root

412−−√
4 start root 12 end root

224−−√

Answer 1

Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point)
Responses

−26–√3
negative 2 root index 3 start root 6 end root

412−−√3
4 root index 3 start root 12 end root

43–√3
4 root index 3 start root 3 end root

26–√3

Answer 2

What is the irrational cube root −24−−−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

26–√3
2 root index 3 start root 6 end root

−24–√3
negative 2 root index 3 start root 4 end root

23–√3
2 root index 3 start root 3 end root

−23–√3

Answer 3

What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

−25–√3
negative 2 root index 3 start root 5 end root

210−−√3
2 root index 3 start root 10 end root

25–√3
2 root index 3 start root 5 end root

85–√3

Answer 4

To rewrite the irrational square root \(\sqrt{48}\) as the product of an integer and another irrational square root, we first simplify \(\sqrt{48}\).

1. Factor \(48\) into its prime factors:
\[
48 = 16 \times 3
\]

2. Take the square root:
\[
\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}
\]

Thus, the correct answer is:
\[
\sqrt{48} = 4\sqrt{3}
\]

So the response that matches is:
**4 start root 3 end root**.

Answer 5

To rewrite the irrational cube root \(\sqrt[3]{48}\) as the product of an integer and another irrational cube root, we start by factoring \(48\):

1. Factor \(48\) as follows:
\[
48 = 16 \times 3
\]
where \(16 = 2^4\) can be split further:
\[
48 = 2^4 \times 3
\]

2. Since we want to take the cube root, we can express this as:
\[
\sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \times \sqrt[3]{3}
\]

3. Now let's simplify \(\sqrt[3]{16}\):
- \(16 = 2^4\), so:
\[
\sqrt[3]{16} = \sqrt[3]{2^4} = 2^{4/3} = 2^{1 + 1/3} = 2 \times \sqrt[3]{2}
\]

4. Therefore, we can rewrite it as:
\[
\sqrt[3]{48} = 2 \times \sqrt[3]{2} \times \sqrt[3]{3} = 2 \times \sqrt[3]{6}
\]

Putting this all together, we find:
\[
\sqrt[3]{48} = 2 \sqrt[3]{6}
\]

Thus, the correct response is:
**negative 2 root index 3 start root 6 end root**.

Answer 6

To rewrite the irrational cube root \(-\sqrt[3]{24}\) as the product of an integer and another irrational cube root, we start by factoring \(24\):

1. Factor \(24\):
\[
24 = 8 \times 3
\]
where \(8 = 2^3\).

2. Now we rewrite the cube root:
\[
-\sqrt[3]{24} = -\sqrt[3]{8 \times 3} = -\sqrt[3]{8} \times \sqrt[3]{3}
\]

3. Since \(\sqrt[3]{8} = 2\), we have:
\[
-\sqrt[3]{24} = -2 \sqrt[3]{3}
\]

Thus, the expression \(-\sqrt[3]{24}\) can be rewritten as:
\[
-2 \sqrt[3]{3}
\]

Therefore, the correct response is:
**−23–√3**.

Answer 7

To rewrite the irrational cube root \(\sqrt[3]{40}\) as the product of an integer and another irrational cube root, we start by factoring \(40\):

1. Factor \(40\):
\[
40 = 8 \times 5
\]
where \(8 = 2^3\).

2. Now we rewrite the cube root:
\[
\sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \times \sqrt[3]{5}
\]

3. Since \(\sqrt[3]{8} = 2\), we have:
\[
\sqrt[3]{40} = 2 \times \sqrt[3]{5}
\]

Thus, the expression \(\sqrt[3]{40}\) can be rewritten as:
\[
2 \sqrt[3]{5}
\]

Therefore, the correct response is:
**2 root index 3 start root 5 end root**.