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Anarel [89]
4 years ago
8

Use the properties of radicals to simplify the expression​

Mathematics
1 answer:
Ludmilka [50]4 years ago
4 0

Given:

The expression is

\dfrac{\sqrt[3]{9}}{\sqrt[3]{4}}

To find:

The simplified form of given expression.

Solution:

We have,

\dfrac{\sqrt[3]{9}}{\sqrt[3]{4}}

It can be written as

\dfrac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\dfrac{9}{4}}                    \left[\because \dfrac{\sqrt[a]{x}}{\sqrt[a]{y}}=\sqrt[a]{\dfrac{x}{y}}\right]

Therefore, the simplified form of given expression is \sqrt[3]{\dfrac{9}{4}}.

Note: We can further simplify this expression but be need use exponential properties.  

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4 years ago
the intensity, I, of light varies inversely as the square of the distance, D2 , from the light source. If I is 150 units when D
podryga [215]

\bf \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{\underline{I} varies inversely with }\underline{D^2}}{I = \cfrac{k}{D^2}} \qquad \qquad \textit{we also know that } \begin{cases} I=150\\ D=6 \end{cases}

\bf 150=\cfrac{k}{6^2}\implies 150=\cfrac{k}{36}\implies 5400=k~\hfill \boxed{I=\cfrac{5400}{D^2}} \\\\\\ \textit{when I = 24, what is \underline{D}?}\qquad \qquad 24=\cfrac{5400}{D^2}\implies D^2=\cfrac{5400}{24} \\\\\\ D^2=225\implies D=\sqrt{225}\implies D=15

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