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baherus [9]
3 years ago
10

Hellpppp plz quickly

Mathematics
2 answers:
hammer [34]3 years ago
8 0
8 < \sqrt[3]{a} < 9
512 < a < 729

So, a can be any number between (512, 729).
∴ answer is (A) 679
OleMash [197]3 years ago
4 0
8 < ∛a < 9
We need to solve for a.
Let's get the 1/3th root of the whole equation:
\sqrt[1/3]{8} \ \textless \   \sqrt[1/3]{cubic-root-of-x} \ \textless \   \sqrt[1/3]{9}


\sqrt[1/3]{cubic-root-of-x} = x
\sqrt[1/3]{8} = 512
\sqrt[1/3]{9} = 712

So
\sqrt[1/3]{8} \ \textless \   \sqrt[1/3]{cubic-root-of-x} \ \textless \   \sqrt[1/3]{9}

512 < x < 729

Now from the values we got, 679 is between 512 and 729.

So  8 < ∛679 < 9 (∛679 ≈ 8.78) (Answer A)


Hope this helps! :D

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