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.

7 0

3 years ago

During the summer Andrew worked for 5 days each week he worked 8 hours each day

Tema [17]

He made $40 those 5 days.

8•5=40

7 0

2 years ago

What’s the common multiples for 12 and 30

jasenka [17]

Hi!

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 30: 1, 2, 3, 5, 6, 10, 30

Explanation: <u><em>This question is super super easy. This question should be the least common multiples. First find the prime factorization of 12. It gave us 2*2*3=12. Then the prime factorization of 30. It gave us 2*3*5=30. You can also multiply each factors the greater number of times and it occurs to in steps 1 or step 2 and it gave us LCM=2*2*3*5=60. The least common multiple factors is 60 is the right answer. Hope this helps! And thank you for posting your question at here on Brainly. And have a great day. -Charlie</em></u>

6 0

3 years ago

Read 2 more answers

Samantha and her children went into a movie theater and will buy bags of popcorn and candies. Each bag of popcorn costs $6 and e

Juliette [100K]

Multiply the price of popcorn by the number of bags so 6b.

Multiply the price of candy by the number bought, so 3.25c

Add those together to get total :

6b + 3.25c

Now the most she can spend is 50 so set the equation to less than it equal to what she can spend:

6b + 3.25c <= 50

3 0

3 years ago