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2. Find the real-number root. V1296 O 36 and 3 6 6 and -6

1 Answer

2 votes

we have


\sqrt[4]{1,296}

Remember that

1,296=(36)^2

and 36=6^2

so

1,296=(6^2)^2=6^4

substitute


\sqrt[4]{1,296}=\sqrt[4]{6^4}=6

answer is 6

Part 2

we have


\sqrt[3]{0.125b^3}

we have that

0.125=125/1,000=(5/10)^3

substitute


\sqrt[3]{0.125b^3}=\sqrt[3]{((5)/(10))^3b^3}=(5b)/(10)=0.5b

answer is 0.5b

User NaturalBornCamper
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