The lengths of three sides of a quadrilateral are shown below:
side 1: 3y2 + 2y − 6
side 2: 3y − 7 + 4y2
side 3: −8 + 5y2 + 4y
the perimeter of the quadrilateral is 4y3 + 18y2 + 16y − 26.
part a: what is the total length of sides 1, 2, and 3 of the quadrilateral? (4 points)
part b: what is the length of the fourth side of the quadrilateral? (4 points)
part c: do the answers for part a and part b show that the polynomials are closed under addition and subtraction? justify your answer. (2 points)

Step-by-step explanation:

Part A:

To find the total length of sides 1, 2, and 3 of the quadrilateral add all three sides together.

Side 1: 3y2 + 2y − 6  

Side 2: 3y − 7 + 4y2  

Side 3: −8 + 5y2 + 4y

3y^2+2y-6+3y-7+4y^2-8+5y^2+4y

Combine the like terms:

12y^2+9y-21

Part B:

To find the length of the fourth side of the quadrilateral we have the total perimeter and the sum of three of the sides, so you just need that fourth side value. Let the fourth side be d.

P= 4y3 + 18y2 + 16y − 26.

Sides = 12y^2+9y-21

12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26

Solve for d

Separate d from rest of the terms.

d = -12y^2-10y+21+4y^3+18y^2+16y-26

d = 4y^3+6y^2+6y-5

Part C:

If closed means that the degree that these polynomials are at stay that way, then yes, this is true in these cases because you will notice that each side had a y^2, y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm