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To rewrite the irrational square root \( \sqrt{48} \) as a product of an integer and another irrational square root, we can factor 48: \[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}. \]
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To simplify \( \sqrt{28} \): \[ \sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}. \]
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To rewrite the irrational cube root \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we factor 48: \[ \sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \times \sqrt[3]{6} = 2\sqrt[3]{6}. \]
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To rewrite the irrational cube root \( \sqrt[3]{-24} \): \[ \sqrt[3]{-24} = \sqrt[3]{-8 \times 3} = \sqrt[3]{-8} \times \sqrt[3]{3} = -2\sqrt[3]{3}. \]
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To rewrite the irrational cube root \( \sqrt[3]{40} \): \[ \sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \times \sqrt[3]{5} = 2\sqrt[3]{5}. \]
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?
3 answers
as a list
Sure! Here’s the information presented as a list:
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\( \sqrt{48} = 4\sqrt{3} \)
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\( \sqrt{28} = 2\sqrt{7} \)
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\( \sqrt[3]{48} = 2\sqrt[3]{6} \)
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\( \sqrt[3]{-24} = -2\sqrt[3]{3} \)
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\( \sqrt[3]{40} = 2\sqrt[3]{5} \)