Math, asked by Sukmit77, 9 months ago

–p^6+ 2p^4 + 4p^3 + 2p^2 by
$\sqrt{2p {}^{2} }$

Answers

Answered by aditya123456791

Answer:

so owno

$5..x6.9. {62. < 6.66.626.51 < 36 {6.6.324 = + > 1 > js{ \: owooqe \sec( \sec( \binom{ \sec( \alpha \cot(e\pi) ) }{?} ) ) }^{?} \times \frac{?}{?} }^{2}$

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–p^6+ 2p^4 + 4p^3 + 2p^2 by ​

Answers

–p^6+ 2p^4 + 4p^3 + 2p^2 by
$\sqrt{2p {}^{2} }$