Mathematics
TestU11403767816
49

Evaluate. fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5 5 to the power of negative 1 over 6 5 to the power of 3 over 2 9 92

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(2) Answers
MorganTyyler

The answer is 1/3 To calculate this will use several rules: [latex] \sqrt[n]{x^m } = x^{ \frac{m}{n} } [/latex] [latex]x^{-m}= \frac{1}{ x^{m} } [/latex] [latex] x^{a} * x^{b} =x ^{a+b} [/latex] [latex] \frac{ x^{a} }{ x^{b} } = x^{a-b} [/latex] The fourth root of 9 is [latex] \sqrt[4]{9}= \sqrt[4]{ 9^{1} }= 9^{ \frac{1}{4} } [/latex] Square root of 9 is [latex] \sqrt[2]{9}= \sqrt[2]{ 9^{1} } = 9^{ \frac{1}{2} } [/latex] The fourth root of 9 to the power of 5 is [latex] \sqrt[4]{9^{5} } = 9^{ \frac{5}{4} } [/latex] The fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5 is: [latex] \frac{9^{ \frac{1}{4} }*9^{ \frac{1}{2} }}{9^{ \frac{5}{4} }} = \frac{ 9^{ \frac{1}{4}+ \frac{1}{2}} }{9 \frac{5}{4} } =9^{ \frac{1}{4}+ \frac{1}{2}- \frac{5}{4} }=9^{ \frac{1}{4}+ \frac{1*2}{2*2}- \frac{5}{4} }=9^{ \frac{1}{4}+ \frac{2}{4}- \frac{5}{4} }=9^{ \frac{1+2-5}{4} }= 9^{ \frac{-2}{4} }[/latex] [latex]= 9^{- \frac{1}{2} }= \frac{1}{9^{ \frac{1}{2} } } =\frac{1}{ \sqrt[2]{9^{1} } } = \frac{1}{ \sqrt[2]{9} } = \frac{1}{3} [/latex]

Darryl750

(1) Fourth root of 9 multiplied by square root of 9 over the fourth root of 9 to the power of 5.= 9^(1/4) * 9^(1/2) / (9^(1/4))^5, cancel 9^(1/4)= 9^(1/2) / (9^(1/4))^4= 9^(1/2) / 9= 3 / 9= 1/3 (2) Five to the power of negative 1 over 6= 5 ^ (-1/6)= (1/5)^(1/6) (3) Five to the power of 3 over 2= 5 ^ (3/2)= sqrt(125) I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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