Respuesta :
Answer:it’s b
Step-by-step explanation:
Just took quiz on edge 2020
The product of [tex]\frac{3k}{k+1}[/tex] and [tex]\frac{(k)^{-1} }{(3k)^{3} }[/tex] is [tex]\frac{1}{9(k^{4} +k^{3}) }[/tex] .
What are the exponent rules?
Product of powers rule - Add powers together when multiplying like bases i.e.
[tex]x^{m}.x^{n} =x^{m+n}[/tex]
Quotient of powers rule - Subtract powers when dividing like bases i.e.
[tex]\frac{x^{m} }{x^{n} } =x^{m-n}[/tex]
Power of powers rule - Multiply powers together when raising a power by another i.e.
[tex](x^{m} )^{n} =x^{mn}[/tex]
According to the given question.
We have an expression.
[tex]\frac{3k}{k+1} \times\frac{(k)^{-1} }{(3k)^{3} }[/tex]
The product of [tex]\frac{3k}{k+1}[/tex] and [tex]\frac{(k)^{-1} }{(3k)^{3} }[/tex] is given by
[tex]\frac{3k}{k+1} \times\frac{(k)^{-1} }{(3k)^{3} }[/tex]
[tex]=\frac{3k^{1-1} }{(k+1)(3)^{3} k^{3} }[/tex] (product power rule)
[tex]=\frac{3k^{0} }{(k+1)27k^{3} }[/tex]
[tex]=\frac{3(1)}{(k+1)27k^{3} }[/tex] (because [tex]x^{0} =1)[/tex]
[tex]=\frac{1}{9(k+1)k^{3} }[/tex] (cancelling out 27 by 3)
[tex]=\frac{1}{9(k^{4} +k^{3}) }[/tex]
Hence, the product of [tex]\frac{3k}{k+1}[/tex] and [tex]\frac{(k)^{-1} }{(3k)^{3} }[/tex] is [tex]\frac{1}{9(k^{4} +k^{3}) }[/tex] .
Find out more information about exponent rule here:
https://brainly.com/question/14513824
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