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Reptile [31]
2 years ago
13

Simplify completely.

" \sqrt{68} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Keith_Richards [23]2 years ago
3 0
\bf \sqrt{68}\qquad 
\begin{cases}
68=2\cdot 2\cdot 17\\
\qquad 2^2\cdot 17
\end{cases}\qquad \sqrt{2^2\cdot 17}\implies 2\sqrt{17}
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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the solution ~

Let's find the measure of hypotenuse first, by using Pythagoras theorem ;

\qquad \sf  \dashrightarrow \: h {}^{2}  =  {8}^{2}  +  {6}^{2}

\qquad \sf  \dashrightarrow \: h {}^{2}  =  {36}^{}  +  {64}^{}

\qquad \sf  \dashrightarrow \: h {}^{2}  = 100

\qquad \sf  \dashrightarrow \: h {}^{}  =  \sqrt{100}

\qquad \sf  \dashrightarrow \: h {}^{}  =  {10}

Now, let's find the asked values ~

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{opposite \: side}{hypotenuse}

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{6}{10}

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{3}{5}   \: or \: 0.6 \: units

For Cos y :

\qquad \sf  \dashrightarrow \:  \cos(y) =  \dfrac{adjcant \: side}{hypotenuse}

\qquad \sf  \dashrightarrow \:  \cos(y)   = \dfrac{6}{10}

\qquad \sf  \dashrightarrow \:  \cos(y)   = \dfrac{3}{5}  \: or \: 0.6 \: units

As we can see that both sin x and Cos y have equal values, therefore The required relationships is equality.

I.e Sin x = Cos y

Hope it helps ~

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