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ikadub [295]
3 years ago
11

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Mathematics
1 answer:
Drupady [299]3 years ago
5 0

Answer:

The equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

Let us solve the expression step by step to get the equivalent

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

as

\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}      ∵ \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=x^{2\cdot \frac{1}{7}}

=x^{\frac{2}{7}}

also

\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}         ∵  \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=y^{3\cdot \frac{1}{5}}

=y^{\frac{3}{5}}

so the expression becomes

\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)            ∵ \:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}

Thus, the equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

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