KayvionG414810 KayvionG414810
  • 25-10-2022
  • Mathematics
contestada

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

Respuesta :

LidiaL478567 LidiaL478567
  • 25-10-2022

we have

[tex]\sqrt[4]{1,296}[/tex]

Remember that

1,296=(36)^2

and 36=6^2

so

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

substitute

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

answer is 6

Part 2

we have

[tex]\sqrt[3]{0.125b^3}[/tex]

we have that

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

substitute

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

answer is 0.5b

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