MATH SOLVE

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# Question 1 (Essay Worth 10 points)(07.02 MC)The lengths of three sides of a quadrilateral are shown below:Side 1: 4y + 2y2 − 3Side 2: −4 + 2y2 + 2y Side 3: 4y2 − 3 + 2yThe 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)

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Answer:Question 1Part 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 - 7Part C: Yes the answers for Part A and Part B show that the polynomials are closed under addition and subtractionQuestion 2Part A: The expression of the area of the square is 4x² - 20x + 25Part B: The degree and classification of the expression obtained in part Aare second degree and trinomialPart C: The polynomials are closed under multiplicationQuestion 3Part A: The function of the area of the circle of spilled oil is 49 πt²Part B: The area of the spilled oil after 8 minutes is 9847.04 units²Step-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