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How can you write the following using a radical sign and no negative exponents? 1. 3y^(2/5) 2. (3y)^(2/5) 3. a^(4/7) * b^(-4/7) 4. a^(1/10) * b^(-1/5)
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\[3\sqrt[5]{y^2}\] is the first one
second is \[\sqrt[5]{3y^2}\]
explanations would be great if possible
1.)\[\ 3 \sqrt [5] {y^2}\]2.)\[\sqrt [5] {9y^2}\]3.)\[\sqrt [7] {a^4} * \sqrt [7] {\frac{1}{b^4}}\]4.)\[\sqrt [10] {a} * \sqrt [5] {\frac{1}{b}}\]
Here's an explanation for the first... \[3y^{2/5}=3*y^{2/5}\]Since the following is true...\[y^{1/5}= \sqrt [5] {y}\]The first statement is the same as...\[3*y^{1/5}*y^{1/5}=3*\sqrt[5]{y}*\sqrt[5]{y}=3\sqrt[5]{y^2}\]Does that make sense?
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yes! thank you:P
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