contestada


Question 1 (Essay Worth 10 points)
(07.02 MC)

The lengths of three sides of a quadrilateral are shown below:

Side 1: 4y + 2y2 − 3

Side 2: −4 + 2y2 + 2y

Side 3: 4y2 − 3 + 2y

The perimeter of the quadrilateral is 22y3 + 10y2 + 10y − 17.

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)



Question 2 (Essay Worth 10 points)
(07.01, 07.06 MC)

The side of a square measures (2x − 5) units.

Part A: What is the expression that represents the area of the square? Show your work to receive full credit. (4 points)

Part B: What are the degree and classification of the expression obtained in Part A? (3 points)

Part C: How does Part A demonstrate the closure property for polynomials? (3 points)



Question 3 (Essay Worth 10 points)
(07.09 HC)

A container of oil has spilled on a concrete floor. The oil flow can be expressed with the function n(t) = 7t, where t represents time in minutes and n represents how far the oil is spreading.

The flowing oil is creating a circular pattern on the concrete. The area of the pattern can be expressed as A(n) = πn2.

Part A: Find the area of the circle of spilled oil as a function of time, or A[n(t)]. Show your work. (6 points)

Part B: How large is the area of spilled oil after 8 minutes? You may use 3.14 to approximate π in this problem. (4 points)

Respuesta :

Ashraf82

Answer:

Question 1

Part A: The total length of sides 1, 2, and 3 is (8y² + 8y - 10)

Part B: The length of the fourth side is 22y³ + 2y² + 2y - 7

Part C: Yes the answers for Part A and Part B show that the polynomials are closed under addition and subtraction

Question 2

Part A: The expression of the area of the square is 4x² - 20x + 25

Part B: The degree and classification of the expression obtained in part A

are second degree and trinomial

Part C: The polynomials are closed under multiplication

Question 3

Part A: The function of the area of the circle of spilled oil is 49 πt²

Part B: The area of the spilled oil after 8 minutes is 9847.04 units²

Step-by-step explanation:

* Lets explain how to solve the problems

# Question 1

∵ The length of the three sides of a quadrilateral are

- Side 1: 4y + 2y² - 3

- Side 2: -4 + 2y² + 2y

- Side 3: 4y² - 3 + 2y

- The perimeter of the quadrilateral is 22y³ + 10y² + 10y − 17

* Part A:

- To find the total length of sides 1, 2, and 3 of the quadrilateral

  add them

∴ s1 + s2 + s3 = (4y + 2y² - 3) + (-4 + 2y² + 2y) + (4y² - 3 + 2y)

- Collect the like terms

∴ S1 + S2 + S3 = (2y² + 2y² + 4y²) + (4y + 2y + 2y) + (-3 + -4 + -3)

∴ S1 + S2 + S3 = 8y² + 8y + (-10) = 8y² + 8y - 10

* The total length of sides 1, 2, and 3 is (8y² + 8y - 10)

* Part B:

∵ The perimeter of the quadrilateral is the sum of its 4 sides

∴ The length of its fourth side is the difference between its

   perimeter and the sum of the other 3 sides

∵ The perimeter of the quadrilateral is 22y³ + 10y² + 10y − 17

∵ The sum of the three sides is (8y² + 8y - 10)

∴ The length of the 4th side = (22y³ + 10y² + 10y − 17) - (8y² + 8y - 10)

- Remember that (-)(+) = (-) and (-)(-) = (+)

∴ S4 = 22y³ + 10y² + 10y - 17 - 8y² - 8y + 10

- Collect the like terms

∴ S4 = (22y³) + (10y² - 8y²) + (10y - 8y) + (-17 + 10)

∴ S4 = 22y³ + 2y² + 2y + (-7) = 22y³ + 2y² + 2y - 7

* The length of the fourth side is 22y³ + 2y² + 2y - 7

* Part C:

- Polynomials will be closed under an operation if the operation

 produces another polynomial

∵ In part A there are 3 polynomials add to each other and the answer

  is also polynomial

∴ The polynomials are closed under addition

∵ In part B there are 2 polynomial one subtracted from the other and

  the answer is also polynomial

∴ The polynomials are closed under subtraction

* Yes  the answers for Part A and Part B show that the polynomials

  are closed under addition and subtraction

# Question 2

∵ The side of a square measure (2x - 5) units

* Part A:

∵ The are of the square = S × S, where S is the length of its side

∵ S = 2x - 5

∴ The area of the square = (2x - 5) × (2x - 5)

- Multiply the two brackets using the foil method

∵ (2x - 5)(2x - 5) = (2x)(2x) + (2x)(-5) + (-5)(2x) + (-5)(-5)

∴ (2x - 5)(2x - 5) = 4x² + (-10x) + (-10x) + 25

- Add the like terms

∴ (2x - 5)(2x - 5) = 4x² + (-20x) + 25 = 4x² - 20x + 25

∴ The area of the square = 4x² - 20x + 25

* The expression of the area of the square is 4x² - 20x + 25

* Part B:

∵ The greatest power in the expression obtained in Part A is 2

∴ Its degree is second

∵ The expression obtained in part A has three terms

∴ The expression obtained in Part A is trinomial

* The degree and classification of the expression obtained in Part A

  are second degree and trinomial

* Part C:

- Polynomials will be closed under an operation if the operation

 produces another polynomial

∵ (2x - 5) is polynomial

∵ (4x² - 20x + 25) is polynomial

∴ The product of two polynomials give a polynomial

∴ The polynomials are closed under multiplication

# Question 3

∵ n(t) = 7t, where t represents time in minutes and n represents how

  far the oil is spreading

∵ The area of the pattern can be expressed as A(n) = πn²

* Part A:

- To find the area of the circle of spilled oil as a function of time, then

  find the composite function A[n(t)]

- That means replace n in A(n) by the function n(t)

∵ n(t) = 7t

∴ A[n(t)] = A(7t)

∵ A(n) = πn²

- Replace n by 7t

∴ A(7t) = π (7t)² = 49 πt²

∴ A[n(t)] = 49 πt²

* The function of the area of the circle of spilled oil is 49 πt²

* Part B:

∵ The area of the circle of spilled oil in t minutes = 49 πt²

- To find the area of the circle of spilled oil after 8 minutes substitute

  t by 8

∴ Area of the spilled oil after 8 minutes = 49 π (8)²

∵ π = 3.14

∴ Area of the spilled oil after 8 minutes = 49(3.14)(64) = 9847.04

* The area of the spilled oil after 8 minutes is 9847.04 units²

Otras preguntas