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Dmitriy789 [7]
3 years ago
7

Simplify square root of 3 multiplied by the fifth root of 3.

Mathematics
1 answer:
almond37 [142]3 years ago
5 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}}}\qquad\qquad %  radical denominator
\cfrac{1}{\sqrt[{ m}]{a^{ n}}}= \cfrac{1}{a^{\frac{{ n}}{{ m}}}}\implies a^{-\frac{{ n}}{{ m}}} \\\\


\bf -----------------------------\\\\
thus\qquad \sqrt{3}\cdot \sqrt[5]{3}\implies \sqrt[2]{3^1}\cdot \sqrt[5]{3^1}\implies 3^{\frac{1}{2}}\cdot 3^{\frac{1}{5}}
\\\\\\
3^{\frac{1}{2}+\frac{1}{5}}\implies 3^{\frac{7}{10}}\implies \sqrt[10]{3^7}\implies \sqrt[10]{2187}
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Answer:

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Step-by-step explanation:

First, we must know the equations given:

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Following Crammer's Rule, we have the matrix form:

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Now we solve using the determinants:

x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2

y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

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And, with the second equation:

  • 3x + y = 5​
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