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svetlana [45]
3 years ago
9

Which is equivalent to √60n^11/256n^4 after it has been simplified completely? Assume n =0

Mathematics
1 answer:
motikmotik3 years ago
5 0

Answer:

\large\boxed{\dfrac{n^3}{8}\sqrt{15n}}

Step-by-step explanation:

\sqrt{\dfrac{60n^{11}}{256n^4}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\=\sqrt{\dfrac{60}{256}n^{11-4}}=\dfrac{\sqrt{60n^7}}{\sqrt{256}}=\dfrac{\sqrt{4n^{6+1}\cdot15}}{16}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{\sqrt{4n^6\cdot15n}}{16}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt{4n^{3\cdot2}}\cdot\sqrt{15n}}{16}\qquad\text{use}\ (a^n)^m=a^{nm}

=\dfrac{\sqrt{4(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{\sqrt{4}\cdot\sqrt{(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{2\!\!\!\!\diagup^1\sqrt{15n}}{16\!\!\!\!\!\diagup_8}=\dfrac{n^3\sqrt{15n}}{8}

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