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Artist 52 [7]
3 years ago
7

What is the perimeter of a polygon with the vertices at (-2,1) (-2,4) (2,7) (6,4) (6,1)?

Mathematics
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

24

Step-by-step explanation:

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The answer i got is -9/13
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Solve the equation for x, where x is a real number (5 points): <br> -5x^2 + 7x + 41 = 32
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Subtract 32 to both sides to the equation becomes -5x^2 + 7x + 9 = 0.

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                   x = [ -b ± √(b^2 - 4ac) ] / (2a)
                   x = [ -7 ± √(7^2 - 4(-5)(9)) ] / ( 2(-5) )
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6 0
3 years ago
Solve the system by substitution<br> -4.5x-2y=-12.5<br> 3.25x-y=-0.75
gogolik [260]

Answer:

The answer to your question is: x = 1, y = 4

Step-by-step explanation:

I'll change decimals to fractions

4.5 = 9/2

12.5 = 25/2

3.25 = 13/4

0.75 = 3/4

                             -9/2 x - 2 y = - 25/2         (I)

                             13/4 x  -   y  = - 3/4            (ll)

Process

               from (II)     y = 13/4x  + 3/4

Sunstitution             -9/2x -2(13/4 x + 3/4) = - 25/2

                               -9/2 x - 13/2 x - 3/2 = -25/2

Multiply by 2         -9 x - 13 x - 3 = -25

                             -22 x = -25  + 3

                            -22 x = -22

                             x = 22/22

                            x = 1

                             y = 13/4 (1) + 3/4

                             y = 16/4

                             y = 4                                          

7 0
2 years ago
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