Question

Hellpppp plz quickly

Answer 1

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

Answer 2

`8 < \sqrt[3]{a} < 9`

`512 < a < 729`

So, a can be any number between (512, 729).
∴ answer is (A) 679