Question

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

2. What is the irrational square root √28 simplified?

3. Rewrite the irrational cube root ^3√48 as a product of an integer and another irrational cube root.

4. What is the irrational cube root ^3√−24 rewritten as a product of an integer and another irrational cube root?

5. What is the irrational cube root ^3√40 rewritten as a product of an integer and another irrational cube root?

Answer 1

1. To rewrite the irrational square root \( \sqrt{48} \) as a product of an integer and another irrational square root, we can factor 48:
\[
\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}.
\]

2. To simplify \( \sqrt{28} \):
\[
\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}.
\]

3. To rewrite the irrational cube root \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we factor 48:
\[
\sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \times \sqrt[3]{6} = 2\sqrt[3]{6}.
\]

4. To rewrite the irrational cube root \( \sqrt[3]{-24} \):
\[
\sqrt[3]{-24} = \sqrt[3]{-8 \times 3} = \sqrt[3]{-8} \times \sqrt[3]{3} = -2\sqrt[3]{3}.
\]

5. To rewrite the irrational cube root \( \sqrt[3]{40} \):
\[
\sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \times \sqrt[3]{5} = 2\sqrt[3]{5}.
\]

Answer 2

as a list

Answer 3

Sure! Here’s the information presented as a list:

1. \( \sqrt{48} = 4\sqrt{3} \)

2. \( \sqrt{28} = 2\sqrt{7} \)

3. \( \sqrt[3]{48} = 2\sqrt[3]{6} \)

4. \( \sqrt[3]{-24} = -2\sqrt[3]{3} \)

5. \( \sqrt[3]{40} = 2\sqrt[3]{5} \)