Respuesta :
Answer:
StartFraction 25 Over 9 EndFraction
Step-by-step explanation:
[tex] \bigg( \huge{\frac{ {3}^{2}a ^{ - 2} }{3 {a}^{ - 1} }} \bigg)^{k} \\ \\ = \bigg( \huge{\frac{ 9a }{3 {a}^{ 2} }} \bigg)^{k} \\ \\ = \bigg( \huge{\frac{ 3 }{ {a}}} \bigg)^{k} \\ \\ = \bigg( \huge{\frac{ 3 }{ {5}}} \bigg)^{ - 2} \\ \\ = \bigg( \huge{\frac{ 5 }{ {3}}} \bigg)^{ 2} \\ \\ = \huge{\frac{ 25 }{ {9}}} [/tex]
The value of the expression [tex](\frac{3^{2}a^{-2} }{3a^{-1} } )^{k}[/tex] when a = 5 and k = -2 is [tex]\frac{25}{9}[/tex].
What are the exponent rules?
Product of powers rule -Add powers together when multiplying like bases. That is
[tex]x^{m} .x^{n} = x^{m+n}[/tex]
Quotient of powers rule- Subtract powers when dividing like bases.
[tex]\frac{x^{m} }{x^{n} } =x^{m-n}[/tex]
Power of powers rule - Multiply powers together when raising a power by another exponent.
[tex](x^{m} )^{n} = x^{mn}[/tex]
What is substitution?
Substitution means putting numbers in place of letters to calculate the value of an expression .
According to the given question.
We have an expression.
[tex](\frac{3^{2}a^{-2} }{3a^{-1} } )^{k}[/tex]
The above expression can be written as
[tex](\frac{3^{2}a^{-2} }{3a^{-1} } )^{k}[/tex]
[tex]= (\frac{9a^{-2} }{3a^{-1} } )^{k}[/tex]
[tex]= (\frac{3a^{-2} }{a^{-1} } )^{k}[/tex] (cancelling 9 by 3)
[tex]=(3a^{-2+1} )^{k}[/tex] (quotient of power rule)
[tex]=(3a^{-1} )^{k}[/tex]
Substitute a = 5 and k = -2 in the above expression.
[tex]\implies (3(5)^{-1} )^{-2}[/tex]
[tex]\implies (3)^{-2} 5^{-1(-2)}[/tex] (power of powers rule)
[tex]\implies \frac{(5)^{2} }{(3)^{2} }[/tex]
[tex]\implies \frac{25 }{9}[/tex]
Hence, the value of the expression [tex](\frac{3^{2}a^{-2} }{3a^{-1} } )^{k}[/tex] when a = 5 and k = -2 is [tex]\frac{25}{9}[/tex].
Find out more information about substitution and exponent rules here:
https://brainly.com/question/26629824
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