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
