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PUHSD – Common Core Integrated Mathematics

Learn about the shift to Common Core Integrated Mathematics, its benefits, and the transition plan for courses. Understand the depth, focus on real-world problems, and new assessment models. Get ready for the evolving math curriculum.

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PUHSD – Common Core Integrated Mathematics

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  1. PUHSD – Common Core Integrated Mathematics

  2. Why change to Integrated Math? • Common Core materials are more supportive of an Integrated Approach. • There is no algebra gap that occurs in the 2nd course (like in Geometry). • There is a greater focus on data and real world problems.

  3. Pathways • Integrated Math 1 begins in 2014-2015 school year. Algebra 1 will no longer be offered. • Incoming Freshmen will be enrolled into courses based on placement test (similar to last year). Most students will start in IM-1. • Integrated Math 2 begins in 2015-2016 school year. Geometry will no longer be offered. • A diagnostic test will replace the placement test for the 2015-2016 school year.

  4. 2014 - 2015 8th Grade Placement Test IM-1 CC Geom CC Alg 2 Current 9th, 10th, 11th Math A IM-1 Alg 1 CC Geom CC Alg 2

  5. Integrated Math 1 /Math A Algebra1 Geometry Algebra2 AP Stats Trig/Pre-Calc AP Calc AB/BC College Readiness Math Curriculum Path – 2014-2015 Placement Test

  6. Schedule Choice – Class of 2018

  7. 2015 - 2016 8th Grade Diagnostic IM-1 IM-2 IM-1 IM-2 9th, 10th, 11th (from 2014-15) IM-2 IM-A IM-1 IM-2 IM-1 CC Alg 2

  8. IM-1 Diagnostic Test College Readiness Math IM-2 IM-3 AP Stats Trig/Pre-Calc AP Calc AB/BC Curriculum Path IM-1/IM-A

  9. Schedule Choice – Class of 2019

  10. 2016 - 2017 8th Grade Diagnostic IM-1 IM-2 IM-1 IM-2 9th, 10th, 11th (from 2015-16) IM-2 IM-A IM-1 IM-2 IM-3 IM-3 IM-1 IM-2 IM-3 Pre Calc Other Adv

  11. Realities of Common Core • Common core requires a much deeper understanding of mathematics that students use. • The problems are “real world” and not just a collection of algorithms. • 60% of the assessment is “Performance Based”

  12. 4 Claims • Claim 1: Concepts and Procedures • Students can explain and apply mathematical concepts and interpret and carry out mathematical procedures with precision and fluency. • Claim 2: Problem Solving • Students can solve a range of complex, well-posed problems in pure and applied mathematics, making productive use of knowledge and problem-solving strategies. • Claim 3: Communicating Reason • Students can clearly and precisely construct viable arguments to support their own reasoning and to critique the reasoning of others. • Claim 4: Modeling and Data Analysis • Students can analyze complex, real-world scenarios and can construct and use mathematical models to interpret and solve problems.

  13. 6 in 6 in What Changes? Typical Problem from current State Geometry assessments: A right circular cone has a height of 6 inches and a diameter of 6 inches. What is the volume of the cone? (A) (B)(C) (D)

  14. d h A Common Core Type Question An ice cream parlor sells ice cream cones which have three flavors of ice cream. The bottom of the cone is filled with vanilla ice cream so that half of the volume of the cone is vanilla. The top part of the cone is filled with strawberry ice cream. Finally to top the cone off a hemisphere (half of a sphere) of chocolate ice cream is placed on top to the cone. The cone has a height (h) of 6 inches and a diameter (d) of 6 inches. (a) What is the volume of all the ice cream used for the cone? (b) Explain why the height of the vanilla in the cone is not3 inches. T (c) The company that owns the ice cream parlor makes several versions of the ice cream treat which has similar dimensions (the diameter and the height of the cone are equal). To simplify ordering they want to find a formula that will find the volume of the ice cream (both the cone and the hemisphere) based on the total height (T) of the ice cream treat. Find a formula that will calculate the volume based on the value of T.

  15. Answers to Problems • Star Test Type – Answer B • Common Core Problem • The volume = 36 in3 (18 for the cone and 18 for the hemisphere) • The cone is much narrower on the bottom than the top. If we fill the vanilla to only 3 inches, the volume would be which is not 9 in3. • First start with doing the problem in terms of the radius. Since the height of the cone is the same as the diameter, the volume of the cone is The volume of the hemisphere is The volume in terms of r for the figure isThe height of the full figure T = 3r. ThereforeSubstituting in for r we get

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