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QveST [7]
2 years ago
9

Find the perimeter of the irregular hexagon with side lengths y, 2y + 11, -3y - 9, y^2, 5y - 1, and бу^2.​

Mathematics
2 answers:
Lady bird [3.3K]2 years ago
6 0

Answer:

PERIMETER=Y+2Y+11+-3Y-9+Y^2+5Y-1+6Y^2

=7Y^2+4Y+1

Vlad1618 [11]2 years ago
5 0
Y+2y+11+(-3y-9)+y²+(5y-1)+6y²
=y+2y+11-3y-9+y²+5y-1+6y²
=5y+1+7y²

Hope my answer helped u :)
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B^2-5b+3 for b=-3 plz help​
nika2105 [10]

Hey there!

b^2 - 5b + 3

= -3^2 - 5(-3) + 3

= -9 - (-15) + 3

= -9 + 15 + 3

= 6 + 3

= 9

Answer: 9

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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