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Sagot :
Answer:
[tex]\frac{1}{x^{12} }[/tex]
Step-by-step explanation:
Given [tex](\frac{x}{x^{-5}})^{-2}[/tex]
The Negative Exponent Rule states that, [tex]a^{-n} = \frac{1}{a^{n}}[/tex]
Basically, since you're raising the fraction inside the parenthesis to a negative exponent, the whole fraction becomes a denominator of 1: [tex]\frac{1}{(\frac{x}{x^{-5}})^{2}}[/tex]
Then, you'll have to work on the denominator:
[tex]\frac{1}{(\frac{x}{x^{-5}})^{2}} = \frac{1}{\frac{x^{2} }{(x^{-5})^{2}}}[/tex]
Next, you'll have to work on the divisor of x² (in the denominator), [tex](x^{-5})^{2}[/tex] by using the Power-to-Power Rule of Exponents: [tex](a^{m})^{n} = a^{mn}[/tex], which results in:
[tex](x^{-5})^{2} = x^{-10}[/tex]
Since it is a negative exponent, you'll have to apply the Negative Exponent Rule once again to the denominator.
[tex]\frac{1}{(\frac{x}{x^{-5}})^{2}} = \frac{1}{\frac{x^{2} }{(x^{-5})^{2}}} = \frac{1}{\frac{x^{2} }{x^{-10}}} = \frac{1}{\frac{x^{2} }{\frac{1}{x^{10}}}}[/tex]
At this point, you could apply the Fraction rule onto the denominator: [tex]\frac{a}{\frac{b}{c}} = \frac{a*c}{b}[/tex]
So the denominator now becomes:
[tex]\frac{1}{\frac{x^{2} }{\frac{1}{x^{10}}}} = \frac{1}{x^{2}*x^{10} }[/tex]
Finally, you could apply the Product Rule of Exponents, [tex]a^{m}a^{n} = a^{m + n }[/tex] onto the denominator:
[tex]\frac{1}{x^{2}*x^{10} } = \frac{1}{x^{2+10}} = \frac{1}{x^{12} }[/tex]
Therefore, the correct answer is: [tex]\frac{1}{x^{12} }[/tex]
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