help me plz

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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