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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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determine the slope-intercept form of the equation of the line parallel to y=-4/3x + 11 that passes through the point (-6,2
Zielflug [23.3K]

is it...

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3 0
2 years ago
What value for x will make the equation true?
Rama09 [41]

Answer:

C . 4

Step-by-step explanation:

step 1 : -3 multiply with inside of the ( 7x-5 )

= -21x + 15

step 2 : then it will be 21x + 15 = 87 + 3x

step 3 : switch between the odds into like this

21x - 3x = 87 - 15

step 4 : subtract it all

18x = 72

step 5 : bring 18 to the other side

x = 72/18

step 6 : to find the ans of x is just calculate them

x = 72 ÷ 18

= 4

Or you can do like these ⬇️

-3 ( 7x - 15 ) = 87 + 3x

21x + 15 = 87 + 3x

21x - 3x = 87 - 15

18x = 72

x = 72/18 or 72÷18

x = 4

4 0
2 years ago
HELP! HELP!
alexgriva [62]

Answer:

The answer should be C. By combining like terms you can find the answer. 1.75a+2.75a = 4.5a

2.25b + 1.75b = 4b

2.25c + 1.25c = 3.5c

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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