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  • one year ago

help me plz

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  1. anonymous
    • one year ago
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  2. Owlcoffee
    • one year ago
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    Well, let's just work it like this: \[B ^{-3}=\frac{ 1 }{ 27 }\] This is an excercise that focuses more in the part of analysis rather than applying exponential properties, so, let's apply the property of the reciprocate so we can turn it into this: \[\frac{ 1 }{ B^3 }=\frac{ 1 }{ 27 }\] Now, what we want to, since both sides are almost equal, except for we have a B^3 in the denominator of the left side and a "27" in the denominator of the right side. Since this is an equality, we can for sue say that the fractions are equal, but since it has a "1" on both numerators we can assure that there is no "k" wich is a constant of proportionality on any of them. Moving on, we can treat this as a proportion, so we will do a cross product, ending up with: \[27=B^3 \] And applying square with base 3 on both sides: \[B=\sqrt[3]{27}\] You can do this in your calculator, and the result will be "3": \[B=3\]

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