cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n) I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3. Is this the correct way to write the original problem another way? I finally got (432m^4n^2)^2/3. Is this correct?
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Originally Posted by sharkman cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n) I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3. Is this the correct way to write the original problem another way? I finally got (432m^4n^2)^2/3. Is this correct? Nearly there. Recall that In your question you have . Using the fact I gave first, this means than the answer is... Hint: Your multiplication of the and terms was not wrong, it was the index - - that you got wrong.
Originally Posted by craig Nearly there. Recall that In your question you have . Using the fact I gave first, this means than the answer is... Hint: Your multiplication of the and terms was not wrong, it was the index - - that you got wrong. The answer should be cubert{432m^4n^2}, right?
Correct
Originally Posted by craig Correct Great! Thanks!
Originally Posted by sharkman The answer should be cubert{432m^4n^2}, right? Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. so . Perhaps more importantly, so . The best way to write your answer is .
Originally Posted by HallsofIvy Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. so . Perhaps more importantly, so . The best way to write your answer is . I understand. You decided to break the problem down a bit more.
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