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kakasveta [241]
3 years ago
13

Simplify √63x^15y^9/√7xy^11 Please don’t guess I need to get this right!!

Mathematics
2 answers:
belka [17]3 years ago
8 0

Answer:

\large\boxed{\dfrac{\sqrt{63x^{15}y^9}}{\sqrt{7xy^{11}}}=\dfrac{3\cdot x^7}{y}=\dfrac{3x^7}{y}}

Step-by-step explanation:

\dfrac{\sqrt{63x^{15}y^9}}{\sqrt{7xy^{11}}}\\\\\text{use}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\=\sqrt{\dfrac{63\!\!\!\!\!\diagup^9x^{15}y^9}{7\!\!\!\!\diagup_1xy^{11}}}=\sqrt{\dfrac{9x^{15}y^9}{xy^{11}}}\\\\\text{Cancel x and y respectively}\\\\=\sqrt{\dfrac{9x^{14}}{y^2}}\\\\\text{use}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\ \text{and}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt9\cdot\sqrt{x^{11}}}{\sqrt{y^2}}\\\\\text{use}\ (a^n)^m=a^{nm}

=\dfrac{\sqrt{3^2}\cdot\sqrt{x^{7\cdot2}}}{\sqrt{y^2}}=\dfrac{\sqrt{3^2}\cdot\sqrt{(x^7)^2}}{\sqrt{y^2}}\\\\\text{use}\ \sqrt{a^2}=a\\\\=\dfrac{3\cdot x^7}{y}=\dfrac{3x^7}{y}

podryga [215]3 years ago
4 0

Answer:

3x^7 / y

Step-by-step explanation:

√63x^15y^9/√7xy^11

= √ [(63/7) x^(15-1) y^(9-11)

= √9x^14y^-2

= √9x^14 / y^2

= 3x^7 / y

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