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pav-90 [236]
2 years ago
14

ign="absmiddle" class="latex-formula">i asked my teacher and she said this "Hi there. separate the radicals - numerator and denominator first. Then, you'll need to make "8" into 2^3. You'll want to create a group of 4 for both the 2 and the y in the denominator, because your index is "4". So, you'll need one more "2" and "y^3". Then you multiply top and bottom by the 4th root of 2y^3 and simplify from there." can you show me what to do
Mathematics
1 answer:
Furkat [3]2 years ago
3 0

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

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The area of the composite polygon is:

\boxed{A_{T}=376 \ m^2}

<h2>Explanation:</h2><h2 />

Hello! remember you have to write complete questions in order to get good and exact answers. Here you haven't provided any figure, so I'll choose the figure below in order to illustrate this problem. A composite polygon is a polygon that can be divided into two or more basic shapes. So here we have a composite polygon formed by a triangle and a rectangle. So:

A_{total}=A_{T} \\ \\ A_{triangle}=A_{tr} \\ \\ A_{rectangulo}=A_{r} \\ \\ \\ A_{tr}=\frac{b\times h}{2} \\ \\ b:Base \ of \ the \ triangle \\ \\ h:height \ of \ the \ triangle \\ \\ \\ A_{r}=B\times H \\ \\ B:base \ of \ the \ rectangle \\ \\ H:height \ of \ the \ rectangle

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A_{T}=\frac{16\times 11}{2} + 16\times 18 \\ \\ A_{T}=88+288 \\ \\ \boxed{A_{T}=376 \ m^2}

<h2>Learn more:</h2>

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