A - A bar graph would be most appropriate
Answer:
88
Step-by-step explanation:
Answer:
B. 
Step-by-step explanation:
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Answers:
Part A: 12y² + 10y – 21
Part B: 4y³ + 6y² + 6y – 5
Part C: See below.
Explanations:
Part A:
For this part, you add Sides 1, 2 and 3 together by combining like terms:
Side 1 = 3y² + 2y – 6
Side 2 = 4y² + 3y – 7
Side 3 = 5y² + 5y – 8
3y² + 2y – 6 + 4y² + 3y – 7 + 5y² + 5y – 8
Combine like terms:
3y² + 4y² + 5y² + 2y + 3y + 5y – 6 – 7 – 8
12y² + 10y – 21
Part B:
You have the total perimeter and the sum of three of the sides, so you just need that fourth side value, which we can call d.
P = 4y³ + 18y² + 16y – 26
Sides 1, 2 & 3 = 12y² + 10y – 21
Create an algebraic expression:
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
Solve for d:
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
– 12y² – 12y²
10y – 21 + d = 4y³ + 6y² + 16y – 26
– 10y – 10y
– 21 + d = 4y³ + 6y² + 6y – 26
+ 21 + 21
d = 4y³ + 6y² + 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 y², y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms.
