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Blababa [14]
3 years ago
6

Sqrt(15)/5 *sqrt(20)

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
5 0
So if you mean: \frac{ \sqrt{15} }{5} times \sqrt{20} then
simplify
rememeber
x^{ \frac{m}{n} }= \sqrt[n]{ x^{m} }
and
[x^{n}]^{m}= x^{nm}
and
\sqrt{x}  \sqrt{y}= \sqrt{xy}

multiply

\frac{ \sqrt{15} }{5} times \frac{ \sqrt{20} }{1}=\frac{ \sqrt{300}}{5}

simplify √300
√300=√3 times √100
we know that √100=10 so
√300=√3 times 10
therefor
\frac{ \sqrt{300} }{5} = <u></u>\frac{10\sqrt{3}}{5}= \frac{5}{5} times  \frac{2\sqrt{3}}{1} = 1 times \frac{2\sqrt{3}}{1}=2√3










erma4kov [3.2K]3 years ago
5 0
So,

\frac{ \sqrt{15}}{5} *  \sqrt{20}

Simplify the square root of 20.
\sqrt{20} =  \sqrt{4} *   \sqrt{5} = 2 *  \sqrt{5} = 2 \sqrt{5}

\frac{ \sqrt{15}}{5} *   \frac{ 2\sqrt{5} }{1}

Multiply.
\frac{ 2\sqrt{75}}{5}

Simplify.
\frac{ 2\sqrt{25}*  \sqrt{3} }{5}
\frac{ 2 * 5*  \sqrt{3} }{5}
\frac{10\sqrt{3} }{5}
\frac{2\sqrt{3} }{1}
2\sqrt{3}
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Then, distribute(-1)

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Step-by-step explanation:

We have the angle in standard post has a sine ratio of

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Using Pythagoras Theorem, the adjacent side length can be found using:

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