Question

question b ,d , g solve the urgently

Answer 1

b.

$(27)^{\frac{2}{3}} \times (64)^{\frac{5}{6}} \times 2^{- 5} \div (\sqrt{81})^{\frac{3}{2}} \\ \\ = ((27)^{\frac{1}{3}})^2 \times ((64)^{\frac{1}{6}})^5 \times \frac{1}{2^5} \div ((9)^{\frac{1}{2}})^3 \\ \\ = (3)^2 \times (2)^5 \times \frac{1}{32} \div (3)^3 \\ \\ = 9 \times 32 \times \frac{1}{32} \div 27 \\ \\ = 288 \times \frac{1}{32} \div 27 \\ \\ = 9 \div 27 = \frac{9}{27} = \frac{1}{3}$

d.

$2^0 + 2^{- 1} + 2^{- 2} + \sqrt[5]{32} \\ \\ = 1 + \frac{1}{2} + \frac{1}{2^2} + 2 \\ \\ = \frac{4}{4} + \frac{2}{4} + \frac{1}{4} + \frac{8}{4} \\ \\ = \frac{4 + 2 + 1 + 8}{4} = \frac{15}{4} = 3\frac{3}{4}$

g.

$(2^3)^2 \div [2 \div (2 \div 2^{\frac{1}{3}})^{\frac{1}{4}}]^{\frac{1}{5}} \\ \\ = 2^6 \div [2 \div (2^{1 - \frac{1}{3}})^{\frac{1}{4}}]^{\frac{1}{5}} \\ \\ = 2^6 \div [2 \div (2^{\frac{2}{3}})^{\frac{1}{4}}]^{\frac{1}{5}} \\ \\ = 2^6 \div [2 \div 2^{\frac{1}{6}}]^{\frac{1}{5}} \\ \\ = 2^6 \div [2^{1 - \frac{1}{6}}]^{\frac{1}{5}} \\ \\ = 2^6 \div [2^{\frac{5}{6}}]^{\frac{1}{5}} \\ \\ = 2^6 \div 2^{\frac{1}{6}} = 2^{6 - \frac{1}{6}} = 2^{5\frac{5}{6}} = 2^{\frac{35}{6}} = \sqrt[6]{2^{35}}$

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