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AVprozaik [17]
4 years ago
15

Rewrite in simplest radical form 1/x^-3/6

Mathematics
1 answer:
VladimirAG [237]4 years ago
4 0
\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad

\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}
\\\quad \\
% rational negative exponent
a^{-\frac{{ n}}{{ m}}} =
 \cfrac{1}{a^{\frac{{ n}}{{ m}}}} \implies \cfrac{1}{\sqrt[{ m}]{a^{ n}}}
\\ \quad \\ \quad \\

%  radical denominator
\cfrac{1}{\sqrt[{ m}]{a^{ n}}}= \cfrac{1}{a^{\frac{{ n}}{{ m}}}}\implies a^{-\frac{{ n}}{{ m}}}\qquad thus\\
----------------------------\\

\\ \quad \\

\bf 
\\ \quad \\

\cfrac{1}{x^{-\frac{3}{6}}}\implies \cfrac{1}{x^{-\frac{1}{2}}}\implies \cfrac{1}{\frac{1}{x\frac{1}{2}}}\implies \cfrac{1}{1}\cdot \cfrac{x^{\frac{1}{2}}}{1}
\\ \quad \\
x^{\frac{1}{2}}\implies \sqrt{x}
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Anna [14]

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Which equation can pair with x-y=-2 to create a consistent and dependent system?
MrRissso [65]

See the explanation

<h2>Explanation:</h2>

A system that has one or infinitely many solutions is called <em>consistent. </em>If an equation in a system tells us no new information then the equations of the system are <em>dependent. </em>In other words, to find an equation that creates a consistent and dependent system with the given equation we have to get the same line:

The given line is:

x-y=-2

If we multiply both sides of the equation by a constant we will have the same line when plotting, therefore let's multiply by 3:

3(x-y)=3(-2) \\ \\ 3x-3y=-6

So a system of two linear equation that is consistent and dependent is:

\left \{ {{x-y=-2} \atop {3x-3y=-6}} \right.

<h2>Learn more:</h2>

Graph of lines: brainly.com/question/14434483#

#LearnWithBrainly

3 0
4 years ago
1. Select all of the fraction models that are equivalent. <br><br><br><br><br><br><br><br> d.
guajiro [1.7K]

Answer:

A,D,C

Step-by-step explanation:

3 0
3 years ago
1. Pia printed two maps of a walking trail. The length of the trail on the first map is 8 cm. The length of the trail on the sec
Alex

Answer:

Step-by-step explanation:

b) 1- scale factor from the first map to the second map:

\frac{8}{6}  = 1.33

   2- landmark on the first map is a triangle with side lengths of 3 mm, 4 mm, and 5 mm.

Side lengths of the landmark on the second map

Divide the length by scale factor:

side lengths of 3 mm:  \frac{3}{1.33} = 2.25 mm

side lengths of 4 mm:  \frac{4}{1.33} = 3.007 mm

side lengths of 5 mm:  \frac{5}{1.33} = 3.75 mm

5 0
3 years ago
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