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konstantin123 [22]
3 years ago
7

Simplify. Write in radical form. (x^3y^-2/xy)^-1/5

Mathematics
2 answers:
kifflom [539]3 years ago
8 0
Hope this helped! Much luck!

Leokris [45]3 years ago
5 0

Answer:

The radical form of the expression (\frac{x^3y^{-2}}{xy})^{\frac{-1}{5}} is \sqrt[5]{\dfrac{y^3}{x^2}}

Step-by-step explanation:

 Given : (\frac{x^3y^{-2}}{xy})^{\frac{-1}{5}}

We have to simplify the given expression and write in radical form.

RADICAL FORM is the simplest form of expression that do not involve any negative exponent and power is less than n, where n is the nth root of that expression.

Consider the given expression  (\frac{x^3y^{-2}}{xy})^{\frac{-1}{5}}

Cancel out the common factor x, we get,

(\frac{x^2y^{-2}}{y})^{\frac{-1}{5}}

Using laws of exponents, a^{-m}=\frac{1}{a^m} , we have,

(\frac{x^2}{y\cdot y^2})^{\frac{-1}{5}}

Using laws of exponents, x^m \cdot x^n=x^{m+n} , we have,

(\frac{x^2}{y^3})^{\frac{-1}{5}}

Again using laws of exponents, a^{-m}=\frac{1}{a^m} , we have,

(\frac{y^3}{x^2})^{\frac{1}{5}}

Also, written as  \sqrt[5]{\dfrac{y^3}{x^2}}

Thus, the radical form of the expression (\frac{x^3y^{-2}}{xy})^{\frac{-1}{5}} is \sqrt[5]{\dfrac{y^3}{x^2}}

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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
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Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

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So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

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\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

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\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

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So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
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