Mathematics, 27.05.2020 17:59, poohmoney
Which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater than or equal to 0
Answers: 2
![\frac{2y}{x^{2}}\sqrt[3]{2y}](/tpl/images/0667/1588/f72a2.png)
![\frac{\sqrt[3]{32x^{3}y^{6}}}{\sqrt[3]{2x^{9}y^{2}}}\\=\sqrt[3]{\frac{32x^{3}y^{6}}{2x^{9}y^{2}}}\\=\sqrt[3]{\frac{16y^{\left(6-2\right)}}{x^{\left(9-3\right)}}}\\=\sqrt[3]{\frac{16y^{4}}{x^{6}}}\\=\sqrt[3]{\frac{2^{4}y^{4}}{x^{6}}}\\=\sqrt[3]{\frac{2^{3}y^{3}\cdot2y}{\left(x^{2}\right)^{3}}}\\=\frac{2y}{x^{2}}\sqrt[3]{2y}](/tpl/images/0667/1588/bab6d.png)
![\frac{\sqrt[3]{16y^4}}{x^2}](/tpl/images/0667/1588/1168c.png)
![\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }](/tpl/images/0667/1588/9030a.png)
![\sqrt[n]{a} = a^{\frac{1}{n}}](/tpl/images/0667/1588/554fb.png)
for 32
; So,
![a^{\frac{m}{n} = \sqrt[n]{a^m}](/tpl/images/0667/1588/2b6f7.png)
; So,![\frac{\sqrt[3]{(2y)}^{4}}{x^2}](/tpl/images/0667/1588/5c2cf.png)
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Which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or...
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