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anonymous

  • one year ago

Evaluate. ^3 sqrt 343 + 3/4 ^3 sqrt -8

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  1. anonymous
    • one year ago
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    "sqrt" is square root btw

  2. anonymous
    • one year ago
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    Does it look like this \[\bf \sqrt[3]{343}~+~\frac{3}{4} \sqrt[3]{-8}\] ?

  3. anonymous
    • one year ago
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    yes exactly

  4. anonymous
    • one year ago
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    Well here is the exponent rule you can use \[\sf \large \sqrt{b^{n}}~=~b^{\frac{1}{n}}\]

  5. anonymous
    • one year ago
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    how would i plug that in?

  6. anonymous
    • one year ago
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    Oh whoops I made a mistake it should be \[\sf \large \sqrt[n]{b}~=~b^{\frac{1}{n}}\]

  7. anonymous
    • one year ago
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    what do i do with the 3/4 just add it at the end?

  8. SolomonZelman
    • one year ago
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    \(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[3]{-8\color{white}{\large|}}\) you know that (-8)=(-2)³ \(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[3]{(-2)^3\color{white}{\large|} }\) you can cancel the powers in the second term and this will give you a -2. Note: You can do this- to cancel the powers - only when they are odd, because if they are even you would get an absolute value of -2 and might make a mistake (since with even powers that is equivalent to 2, not -2). \(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[\cancel3]{(-2)^{\cancel 3}\color{white}{\large|} }\) \(\sqrt[3]{343}+\dfrac{3}{4}(-2) \)

  9. SolomonZelman
    • one year ago
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    then, 343 = 7³ so you can re-write the 343, and then cancel the powers just the same way I have done it with the second term of your expression.

  10. anonymous
    • one year ago
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    i got 5 3/4 but, a. -5 1/2 b. 5 1/2 c. 8 1/2 @SolomonZelman

  11. SolomonZelman
    • one year ago
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    what did you get for the \(\sqrt[3]{343}\) part, what was this part equivalent to?

  12. anonymous
    • one year ago
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    7^3 @SolomonZelman

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