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Harrizon [31]
3 years ago
9

Can someone help me out with number 2?

Mathematics
1 answer:
Norma-Jean [14]3 years ago
5 0
\sin255^o=\sin(300^o-45^o)=\sin300^o\cos45^o-\cos300^o\sin45^o= \\\\=- \frac{ \sqrt{3} }{2}\cdot  \frac{ \sqrt{2} }{2}- \frac{1}{2} \cdot  \frac{ \sqrt{2} }{2}=-      \frac{ \sqrt{6} }{4} -  \frac{ \sqrt{2} }{4}=  -\frac{  \sqrt{6}+\sqrt{2} }{4}

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3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B49%7D%20" id="TexFormula1" title=" \sqrt{49} " alt=" \sqrt{49} " align="absmiddle
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Root of 49

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