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2 years ago

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

Mathematics

2 answers:

belka [17]2 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]2 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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alexgriva [62]

Both integers have negative values

Explanation

Let

x and y represents the number

so, the options are

$\begin{gathered} +\text{ + +} \\ +\text{ + -} \\ -\text{ + +} \\ -\text{ +-} \end{gathered}$

a) Both integers have negative values Both integers

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c)One integer has a positive value, and one integer has a negative value

$\begin{gathered} -x+y\text{ or x-y} \\ the\text{ sign of the result depends on the greates absolute value, it means the answer will have the same of the bigger number( absolute value)} \end{gathered}$

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