The correct question is: evaluate
![\sqrt[3]{- \frac{8}{125} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B-%20%5Cfrac%7B8%7D%7B125%7D%20%7D%20)
The
answer is: option B. -2 / 5Explanation:![\sqrt[3]{- \frac{8}{125} }=- \frac{ \sqrt[3]{8} }{ \sqrt[3]{125} } =- \frac{ \sqrt[3]{2^3} }{ \sqrt[3]{5^3} } =- \frac{2}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-%20%5Cfrac%7B8%7D%7B125%7D%20%7D%3D-%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B8%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B125%7D%20%7D%20%3D-%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B2%5E3%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B5%5E3%7D%20%7D%20%3D-%20%5Cfrac%7B2%7D%7B5%7D%20)
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You can do 5X=10.If you don't know 5X means 5 times X. I hope this helps.
Answer: x = (1,9]
Step-by-step explanation: If x is all real numbers greater than 1 (simply: x is greater but not equal to 1) then it is 1 < x. And if x is less than or equal to 9 then its x ≤ 9. To translate 1 < x ≤ 9 into interval notation, you need to keep in mind that parentheses are for these symbols: <>. And brackets are for these symbols: ≤≥. With that in mind, your answer is (1,9]. Also, keep in mind that interval notation goes from lowest value number to highest, so you wouldn’t write this notation as [9,1).
