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QveST [7]
3 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]3 years ago
6 0

Answer:

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

=7Y^2+4Y+1

Vlad1618 [11]3 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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3 years ago
The length of a rectangle is 7 inches more than its width. If the perimeter of the rectangle is 66 inches, find the dimensions.
Igoryamba

Answer:

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Step-by-step explanation:

Represent the quantities as follows:

P = 2W + 2L

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Thus, P = 66 inches = 4W + 14 inches, or

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Dividing both sides by 4, we get W = 13 inches.

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4 years ago
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4 years ago
The following linear system is to be solved using the elimination method. 5x + 7y=0 x+ 4y= -1 the first step would be to
OLga [1]

Answer:

D

Step-by-step explanation:

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7 0
3 years ago
Read 2 more answers
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